NPTEL Course Developer for Fluid Mechanics DYMAMICS OF FLUID FLOW

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1 Module 04; Lecture DYMAMICS OF FLUID FLOW Energy Equation (Conservation of Energy) In words, the conservation of energy can be stated as, Time rate of increase in stored energy of the system = Net time rate of energy addition by heat transfer into the system + Net time rate of energy addition by work transfer into the system. In symbolic form, the statement is given by, or, D e ρ dv = ( Q in Q out ) + ( W W sys in out Dt ) () sys sys D e. ρ. dv = ( Q net, in + W net, in Dt ) () sys sys where e is the total stored energy per unit mass of the system and is related to internal energy per unit mass ( u V ), kinetic energy per unit mass unit mass (gz ), i.e. and potential energy per The net rate of heat transfer into the system ( net, in ) ) V e= u+ + gz (3) Q and the net rate of work transfer into the system ( Q net, in is considered to be + ve quantity and the outward flow of heat and work is taken as - ve. Expanding the left hand side, the Eq. () can be written for a control volume (CV) as, t e. ρ. dv + eρ V. n da= Q + W p V. n da (4) ( ˆ),, ( ˆ net in net in ) CV CS CS The last term in the RHS of Eq. (4) is the flow work associated with the control surface. Now, using Eq. (3) in Eq. (4), V p V u+ + gz ρdv + u+ + + gz ρ( V. nˆ) da= Q,, ( V. ˆ net in + Wnet in p n) da t ρ CV CS CS (5)

2 Module 04; Lecture Discussion: The first term in the LHS of Eq. (5) represents the time rate of change of the total stored energy ( e) of the contents of CV. This term is zero for the steady flow i.e. V u+ + gz ρ dv t CV = 0 (6) p V The integrand u+ + + gz ρ ( V. ˆ) CS ρ n da fluid crosses the control surface i.e. ( V. nˆ 0) can be non-zero only when the. If the properties like internal energy, pressure, kinetic and potential energies are uniformly distributed over the cross-sectional area involved, then the equation becomes simple and can be written as, p V u gz ρ ( V. nˆ ) da u p V gz m u p V = gz m ρ ρ ρ CS flow out flowin (7) If there is only one stream entering and leaving the CV, then, ( V. ˆ) p V p V p V u+ + + gz ρ n da= u+ + + gz m out u+ + + gz ρ ρ ρ CS out in (8) The simplified expression for one-dimensional steady flow energy equation is then written as, p p Vout V in m ( u u ) g( z z ) = Q + W out in out in net, in net in ρ ρ out in m in, (9) This equation is valid for both compressible and incompressible flows. Here, we introduce one more fluid property called enthalpy and is defined by, p h = u+ (0) ρ In terms of enthalpy, Eq. (9) is written as, Vout V in m ( hout hin ) + + g( zout zin ) = Q net, in + W net, in ()

3 Module 04; Lecture BASIC EQUATIONS (DIFFERENTIAL FORM) Conservation of Mass (Continuity Equation) Consider an infinitesimal control volume in the shape of a rectangular parallelepiped (Fig. ) fixed in x-y-z coordinate system for a general flow. Let u, v, and w be the velocities of the fluid in directions x, y, and z respectively and ρ be the density of the fluid. Now consider the flow through the surfaces and, which are parallel to y-z plane. The efflux rate per unit area through section is -ρ u and varies continuously in the x direction. The net efflux rate through the section can be given by ρ ( ρ ) The net efflux rate through these surfaces is given by ( ) similar computations for other pairs of sides and adding the results, Net efflux rate = ( ρ ) ( ρ ) ( ρ ) u v w + + dx dy dz x y z Net rate of decrease in mass inside the control volume = Equating the above two expressions, ( ρu) ( ρv) ( ρw) ρ + + = x y z t u+ u x dx. ρu x dx dy dz. Performing ρ dx dy dz t () y ρu () () ρu + / x(ρu)dx x z Fig. : Infinitesimal fixed control volume. 3

4 Module 04; Lecture For steady flow, Eq. () becomes, ( ρu) ( ρv) ( ρw) + + x y z = 0 (3) For incompressible flow, the fluid density remains constant. Hence, Eq. (3) reduces to, u v w + + = 0 x y z or,.v = 0 where ˆ ˆ ˆ = i + j + k and V = uiˆ+ vj ˆ+ wkˆ x y z Euler s Equation of Motion According to Newton s law, the net force F acting on a fluid element of mass m and acceleration a is given by, (4) F = ma. (5) In general, following forces may be present in for a fluid in motion. Gravity force, F g Pressure force, Fp Viscous force, Fv Force due to turbulence, Ft Force due to compressibility, Fc The net force becomes, F = Fg + Fp + Fv + Ft + Fc (6) The Euler s equation of motion represents a special case of a flow field in which the viscous forces and force due to turbulence is absent. The linear momentum of a fluid element of mass dm, moving at velocity V is given as ( dm V ). The fundamental statement of Newton s law for an inertial reference is given in terms of linear momentum as, The above equation may be written as, df D = ( dm V ) (7) Dt 4

5 Module 04; Lecture D df ( dm V ) dx dy dz u V v V V V = = ρ + + w + (8) Dt x x x t The Euler s equation is restricted to the case of no shear stress with only gravity as a body force. The negative z direction corresponds to the direction of gravity. So, ( ρ ) ˆ ( )( ρ ) Gravity force = g dx dy dz k = g z dx dy dz The surface force on the fluid element is only due to pressure p (Fig. ). Hence, where p Pressure force ˆ p ˆ p i j kˆ = + + dx dy dz = p dx dy dz x y z ( ) ( )( ) iˆ, ˆj and k ˆ are the unit vectors in x, y and z directions respectively. p +( p/ z)dz p y p p +( p/ y)dy p x p +( p/ y)dy z Fig. : Pressure variation in the x-y-z directions. Incorporating the gravity and pressure forces in Eq. (8), V V V V DV p g z = u + v + w + = (9) ρ x x x t Dt The above equation can be expanded in rectangular coordinates as, p u u u u x direction, + Bx = u + v + w + (0a) ρ x x y z t p v v v v y direction, + By = u + v + w + ρ y x y z t p w w w w z direction, + Bz = u + v + w + ρ z x y z t In streamline coordinate system, Eq. (9) is expressed as, (0b) (0c) 5

6 Module 04; Lecture p V V Along the directionof streamline(), s + Bs = V + ρ s s t (a) p V Vn Along the direction perpenticular to' s ', + Bn = + (b) ρ n R t where B stands for gravity force in the respective direction. Eqs. (0a-c and a-b) are known as Euler s equation of motion. Integration of Steady-State Euler Equation: Bernoulli s Equation Consider a general steady-state flow of an inviscid, incompressible, irrotational fluid. The Euler s equation of motion can be expressed in streamline coordinate system as, V p g z = V () ρ s where s is the coordinate along a stream line. Now, take the dot product of each term given for streamline coordinates. Thus, V p.ds g z.ds = V.ds ρ s The term p.ds becomes dp, the differential change in pressure along a streamline; z.ds becomes dz, the differential change in elevation along the streamline. The righthand side of the equation simplifies to VdV., where dv is the velocity change along a streamline. Hence, Taking g dp V gdz= VdV = d ρ as constant and integrating along a streamline, the above equation can be written as, p dp V gz constant ρ + + = (3) 0 This equation is often called the compressible form of Bernoulli s equation. If ρ is expressed as a function of p only, i.e. ρ ρ( p) =, the first expression is integrable. Flows having this characteristic are called barotropic flows. On the other hand, if the flow is incompressible, i.e. ρ = constant, the Eq. (3) becomes, 6

7 Module 04; Lecture p V z constant ρ g + g + = (4) Multiplying Eq. (4) by ( g ) and replacing ( ρ g ) by γ (i.e. specific weight or weight density of the fluid), p V z constant γ + g + = (5) Discussions The terms in Eq. (5) are units of length and are frequently designated as pressure, velocity and elevation heads respectively. Between any two points along a streamline, Eqs. (4) and (5) can be written as, p V p V + + z = + + z (6) ρg g ρg g p V p V + + z = + + z (7) γ g γ g For most of the fluid flow, there are some energy losses (expressed in terms of heads, h L ), which should be taken into account for considering Bernoulli s equation between two state points of a streamline. One of the important features of Bernoulli s equation is to define two grade lines of a flow. One of the grade lines is the Bernoulli s constant in Eq. (5) and named as energy grade line (EGL). The other grade line is the hydraulic grade line (HGL) which shows the height corresponding to elevation and pressure head i.e. EGL minus Velocity Head. Comparison of Bernoulli and Steady Flow Energy Equation Bernoulli s equation is based on certain assumptions;. Steady uniform flow a common assumption applicable for many kinds of practical flows.. Incompressible flow is acceptable for both the equations (i.e. Euler s and Bernoulli s equation) if the Mach number is less than

8 Module 04; Lecture 3. Frictionless flow very restrictive because friction can be introduced from the solid walls on to the flow. 4. Flow along a streamline i.e. different streamlines will have different Bernoulli s constants (Eq. 5). 5. No shaft work no pumps or turbines are involved between two sections. 6. No heat transfer is involved between two sections. Example- Air enters in a compressor with negligible velocity with a flow rate of 0kg/s and is discharged through a pipe of cross-sectional area of 0.m with same flow rate. The pressure and temperature at the inlet to compressor is atmospheric where as the corresponding values in the discharge section is 3.5 bar and 40 0 C. If the compressor takes 600hp power input, determine the rate of heat rejection. Solution: The control volume showing the inlet and exit sections are given in the Ex. Fig.. The steady flow energy equation for the compressor can be written as given by Eq. () V V i.e. m ( h h) + + g( z z) = Q + W Since, air behaves as an ideal gas with constant specific heat, so h h = c T T. Also, air enters with negligible velocity i.e. the above equation becomes, ( ) V Q = mc p ( T T) + m W ( ) p V V ~0 and also, z z = 0. Hence Control volume () p = bar T = 98 K V = 0 () p = 3.5 bar T = 33 K A = 0.m m = 0 kg/s Compressor power input = 600hp Ex. Fig. : Control Volume for compressor. 8

9 Module 04; Lecture Using continuity equation and equation of state for air, we can write, m m R. T m = ρ AV = ρ AV i.e. V =. ρ =. A A p For air, R = 0.87 kj kg.k, c p =.005kJ kg.k and atmospheric conditions at inlet to compressor can be taken as, Example- so V p = bar ; T = 0 5 C ( ) == = 5.7 m s ( ) ( 5.7) ( ) Q = = 9.95kW 000 A tank of volume 0.m 3 is connected to a high-pressure air reservoir at MPa through a valve. Initial pressure in the tank is 00kPa (absolute). The line connecting the reservoir and the tank is sufficiently large so that the temperature may be assumed to be uniform at 5 0 C. When the valve is opened, the tank temperature rises at the rate of C/s. Determine the instantaneous flow rate of air into the tank by neglecting the heat transfer. Solution: The CV is chosen with reference to the following schematic diagram. Control volume Air reservoir p = MPa T = 98 K Tank V = 0.m3 p = 00 kpa T = 98 K Ex. Fig. : Control Volume for the tank. Following assumptions are made; Since, no heat transfer and shafts work are involved, W = Q = 0. Velocities in the line and tank are small and can be neglected. 9

10 Module 04; Lecture Since, there is no elevation difference, the potential energy changes can be neglected. Flow is uniform at the inlet of the tank m = ρ V A Properties are uniform in the tank. Air can be taken as ideal gas with state equation, p = ρ. R. T and du = cvdt With these assumptions, Eq.(5) reduces to, u. ρ. dv + ( u+ pv)( ρva) = 0 t CV Since the properties are uniform, so t can be replaced by d, and then dt From continuity equation, dm dt d [ um. ] ( u + pv ) m = 0 dt dm du or, u. + M. um. pvm. = 0 dt dt = m Hence the simplified expression becomes, m = = kg s ( ) ρ ( ) M. cv dt dt. V. cv dt dt m = = pv. RT. 3 ptank 00 0 ρ = = =.35 kg m RT

11 Module 04; Lecture EXERCISES. The front of a jet engine acts as a diffuser receiving air at 900km/h, -5 0 C, 50kPa bringing it to 80m/s before entering the compressor. If the flow area is reduced to 80% of the inlet area, find the temperature and pressure in the compressor inlet.. Air at a temperature of 5 0 C passes through a heat exchanger at a velocity of 30 m/s where its temperature is raised to C. It then enters a turbine with the same velocity of 30 m/s and expands until the temperature falls to C. On leaving the turbine, the air is taken at a velocity of 60 m/s to a nozzle where it expands until the temperature falls to C. If the air flow rate is kg/s, calculate (a) the rate of heat transfer to the air in the heat exchanger, (b) the power output from the turbine assuming no heat loss, and (c) the velocity at the exit of the nozzle, assuming no heat loss. Take the enthalpy of air as pressure equal to.005 KJ/kg.K and T is the temperature. h= c T, where is the specific heat at constant p 3. Air flows steadily through a turbine that produces a power output of 750hp. The diameter of inlet and exit port is 5cm. The static pressure and temperature at the inlet section are MPa and 50 0 C respectively. The corresponding values in the exit section are 0.3MPa and 0 C and respectively. If the inlet velocity is 30cm/s, determine, (i) exit velocity; (ii) rate of heat transfer. 4. A balloon is to be partially filled at atmospheric pressure with helium from a wellinsulated storage tank that has a volume of 6.m 3. At the time of filling the balloon, atmospheric pressure is 95kPa. The helium in the tank is initially at gage pressure of 500kPa and 0 0 C before the valve on the tank is opened. Assuming the conditions to be uniform throughout the tank at any instant and heat transfer is negligible; determine the mass of the balloon that has been put into the balloon when a pressure gage in the storage tank reads 300kPa. For helium under these conditions, h=.667u = 5.93T and R=.007 kj kg.k with h and u in kj kg and T in K. 5. The steam supply to an engine comprises two streams, which mix before entering the engine. One stream is supplied at the rate of 0.0 kg/s with an enthalpy of 95 kj/kg and a velocity of 0 m/s. The other stream is supplied at the rate of 0. kg/s with an enthalpy of 569 kj/kg and a velocity of 0 m/s. At the exit from the engine, the fluid leaves as two streams, one of water at the rate 0.00 kg/s with enthalpy of 40 kj/kg c p

12 Module 04; Lecture and the other stream; the fluid velocities at the exit are negligible. The engine develops a shaft power of 5 kw and the heat transfer is negligible. Evaluate the enthalpy of the second stream. 6. A nozzle is a device for increasing the velocity of a steadily flowing stream. At the inlet to a certain nozzle, the enthalpy of the fluid passing is 3000 kj/kg and the velocity is 60 m/s. At the discharge end, the enthalpy is 76 kj/kg. The nozzle is horizontal and there is negligible heat loss from it. (i) Find the velocity at the exit of the nozzle; (ii) If the inlet area is 0. m and specific volume at the inlet is 0.87 m 3 /kg, find the mass flow rate; (iii) If the specific volume at the nozzle exit is m 3 /kg, find the exit area of the nozzle. 7. A turbine operates under steady flow conditions, receiving stream at the following state: pressure. MPa, temperature 88 0 C, enthalpy 785 kj/kg, velocity 33.3 m/s and elevation 3 m. The steam leaves the turbine at the following state: pressure 0 kpa, enthalpy 5 kj/kg, velocity 00 m/s and elevation 0 m. Heat is lost to the surroundings at the rate of 0.9 kj/s. If the rate of steam flow through the turbine is 0.4 kg/s, what is the power output in kw? 8. Steam enters to a turbine with a velocity 40m/s and enthalpy of 3560 kj/kg and leaves as liquid-vapour mixture with a velocity of 75m/s and enthalpy of 478 kj/kg. If the flow through the turbine is adiabatic and elevation changes are negligible, determine the specific work output for the turbine.

13 Module 04; Lecture 3 DYMAMICS OF FLUID FLOW ENERGY OR HEAD LOSS OF FLOWING FLUID The change in velocity of the liquid in a flow (either in magnitude or direction) induces large-scale turbulence due to formation of eddies. So, a portion of energy possessed by the flowing liquid is ultimately dissipated as heat and is considered to be the loss of energy. Some of the losses of energy caused by the change in velocity are, Due to sudden enlargement Due to sudden contraction At the entrance to a pipe from large vessel At the exit from a pipe Due to an obstruction in the flow passage Due to gradual contraction or enlargement Due to bends In various pipe fittings The above losses of energy are termed as minor losses because the magnitude of these losses is quite small compared to the loss due to friction in long pipes (which are distinguished as major losses ). The minor losses are confined to a very short length of the passage of the flowing liquid. The analytical expressions representing the loss of energy for above cases are discussed below. Loss of Energy due to Sudden Enlargement Consider a pipe of cross-sectional area A and carrying a liquid of specific weight γ. It is connected to another pipe of larger cross-sectional area A ( A A ) >. Since there is a sudden change in the cross-sectional area of flow passage, the liquid emerging from smaller pipe is unable to follow the abrupt change in boundary (Fig. ). Consequently, the flow separates from the boundary, forming turbulent eddies that results in the loss of energy to be ultimately dissipated as heat. If p, V and p, respectively, are the pressures and velocities of flow of liquid in the narrower and wider pipe, then by continuity equation, the discharge is, Q= AV = AV V

14 Module 04; Lecture 3 () () V V p A p A Fig. : Flow through sudden enlargement in a pipe. Now, the force acting on the liquid in the control volume in the direction of flow is, { + ( ) } = ( ) pa p A A pa p p A By Newton s second law, the rate of change of momentum is equal force acting on the liquid, i.e. or, γ Q g ( p p ) A = ( V V ) Now, if h L p p Q V = ( V V) = ( V V) () γ ga. g is the head loss between two sections and due to sudden enlargement, then applying Bernoulli s equation For a horizontal pipe, z = z p V p V + + z = + + z +h γ g γ g, so that L Using Eq. () in (), we get h L p p V V = + γ γ g g () h ( V V ) L = (3) g Using continuity equation, Eq. (3) may be alternatively expressed as,

15 Module 04; Lecture 3 h L V A V A = = (4) g A g A Eqs. (3) and (4) is the expression for head loss due to sudden enlargement and was first obtained by J.C. Borda ( ) and L. Carnot (738-83). Sometimes, it is known as Borda-Carnot equation. Loss of Energy due to Sudden Contraction Consider a pipe carrying certain liquid of specific weight γ whose cross-sectional area at a certain section reduces abruptly from to A as shown in Fig.. Although, a sudden A contraction is reverse of sudden enlargement geometrically, the Bernoulli s equation cannot be applied because the streamlines between section and are curved. So, the liquid is accelerated due to which the pressure at the annular face varies in an unknown manner, which cannot be determined easily. However, no major loss of energy occurs in the region between the upstream (section ) and the accelerating flow in the converging portion. As the liquid flows from the wider pipe to narrower pipe, a vena-contracta is formed, after which the stream of liquid widens again to fill up completely the narrower pipe. In between the vena-contracta and the wall of the pipe, lot of eddies are formed that accounts for considerable dissipation of energy. In this region, the flow pattern is almost similar to that of sudden enlargement. So, the head loss can be expressed as, h L ( V V ) c = (5) g Vena Contracta () V p p A V c p c Ac V A Fig. : Flow through sudden contraction in a pipe. 3

16 Module 04; Lecture 3 By continuity equation, AV c. c = AV. ; where A c and Vc are cross-sectional area and velocity at vena-contracta. So, Eq. (5) can be written as, where C c A h L V A V V = = = k. (6) g Ac g Cc g c = is the coefficient of contraction. The value of Cc or k is not constant but A depends on the ratio A D or. In general, the loss of head due to sudden contraction A D is taken to be equal to 0.5 V g i.e. the value of k is adopted as 0.5. Loss of Energy at the Entrance to a Pipe Energy loss at the entrance to the pipe is also called as inlet loss. It occurs, when the liquid enters to the pipe from a large vessel (or tank). The flow pattern is similar to that of sudden contraction. In general, for a sharp-cornered entrance, the loss of head at the entrance is taken equal to 0.5 Vg, where V is the mean velocity of flow of liquid in the pipe. Loss of Energy at the Exit from a Pipe The outlet end of a pipe carrying liquid may be either left free or connected to a large reservoir. The liquid leaving the pipe possesses some kinetic energy corresponding to the velocity of the flow in the pipe which is ultimately dissipated either in the form of free jet or turbulence in the reservoir depending on the outlet condition in the pipe. The loss may be determined by using Eq. (6) with the conditions for which A. So, the loss of head at the exit of the pipe is equal to liquid in the pipe. V, where V is the mean velocity of flow of g 4

17 Module 04; Lecture 3 Loss of Energy due to Obstruction in Flow Passage The loss of energy due to flow obstruction in a pipe occurs due to the sudden reduction in the cross-sectional area followed by an abrupt enlargement of the stream beyond the obstruction (Fig. 3). Consider a pipe flow (cross-sectional area of the pipe is A ) in which an obstruction is placed with maximum cross-sectional area reduced to ( a. As the flow passage is A a), a vena-contracta is formed beyond which the flow becomes uniform after certain distance from vena-contracta. If V c and V be the velocities at vena-contracta and at some section (where the flow is uniform), then the loss of head due to obstruction can be deduced from Eq. (6) i.e. h L V A = g Cc [ A a] (7) Fig. 3: Flow through a pipe with obstruction. Loss of Energy due to Gradual Contraction or Enlargement The energy loss due to sudden contraction of enlargement can be minimized substantially by gradual decrease or increase in area of cross-section of the flow because the dissipation of energy due to turbulent eddies will be eliminated. In such case, the head loss is expressed as, h L ( V V ) = k (8) g 5

18 Module 04; Lecture 3 where and V are the mean velocities at the inlet and outlet; k is the loss coefficient V whose value depends on the angle of convergence/divergence and cross-sectional areas at the upstream and downstream respectively. Loss of Energy in Bends The bends are provided in a flow passage to change the direction of the flow. This causes the loss of energy due to flow separation from the boundary and subsequent formation of turbulent eddies. In general, the head loss may be expressed as, V hl = k (9) g where V is the mean velocity of flow and k is the loss coefficient. The value of k depends on the total angle of bend, radius of curvature of the pipe axis and pipe diameter. Loss of Energy in Various Pipe Fittings All pipe fittings such as valves, couplings etc. cause loss of energy and is also represented by Eq. (9). However, the value of k depends on the type of pipe-fittings. Example- A pipe having diameters 0cm and 0cm at two sections A and B, carries water that flows at a rate 40Lts/s. Section A is 5m above datum and section B is m above datum. If the pressure at section A is 4 bar, find the pressure at section. Solution: Schematically, the pipe is shown in the following figure. It has two section; A and B. Given that: Diameters, D = 0 cm = 0.m ; D = 0 cm = 0.m A B 6

19 Module 04; Lecture 3 5 P A = 4 bar = 4 0 N/m Z = 5m Z = m A 3 Q = 40 lits/s = m s By continuity equation, Q 0.04 VA = = =.7 m/s A π ( 0.) A 4 Q 0.04 VB = = = m/s A π ( 0.) B 4 Now, PA VA PB VB Z A Z ρg + + = ρg + + Example- B 5 P B = = P B = 3. bar A drainage pump has inlet as a vertical pipe with a tapered section, which is filled with water as shown in the following figure. Diameters at ends of the tapered inlet are.m and.06m respectively. The pipe is running full of water. The free water surface is.8 m above the center at the inlet and center at upper end is m above the top at the free surface. The pressure at the upper end of the pipe is 8cm of Hg and the head loss between two sections is /0 of the velocity head at top section. Find the discharge of water in the pipe. P B B Solution: At section- 76 P = 76 cm. of Hg = 3.6 = 0.336m of water ; D =.m 00 At section- 7

20 Module 04; Lecture 3 8 P = 8 cm. of Hg = 3.6 = 3.808m of water ; D = 0.6m 00 Head loss, ( V h = 0 ) g By continuity equation, AV = AV (.) V π ( 0.6) π = 4 4 V = 4V V Bernoulli s equation gives, P V P V + + Z = + + Z + h ρg g ρg g V V g + = + g + + V g 0 V V = g 0g V V g 0g = V 44V.78 g 5g = 5 V 88 V =.78 0g 83V =.78 0g V = 83 V =.43m/s Discharge, ( ) 3 Q= AV = π..43 =.6m /s = 60 lits/s 4 Example-3 Water flows downwards through a vertical tapered pipe at a rate lits/min. The diameter of the pipe at the upper end is 55m, which gradually reduces to 8mm at the lower end through a length of m. The pressure difference between the ends is 38kPa. Assuming 8

21 Module 04; Lecture 3 that the head loss varies as the square of the velocity, determine the quantity of water passing through the pipe when there is no pressure difference. Solution: The flow of water through the pipe is shown in the following figure. If V is the velocity of flow at any section, the head loss for small length ' dx' of the pipe, then head loss for the section 'dx' is given by, dh = kv dx Integrating, hf f = kv dx x + Diameter at any section of the pipe, d = 0.08, Q Q So, velocity, V = = π d π x + ( ) 4 So head loss can be expressed as, 4 kq 6kQ dx 6kQ hf =. dx= ( ) ( ) = 4 0 ( ) ( ) 3 π x π x π ( 0.08) kQ hf = 4 ( π ) ( 0.08) ( 4 ) Apply Bernoulli s equation between sections &, ( P P V V = + ( Z Z) + h ρg g Q = = m /s ; Z = 0 ; Z = m 60 9

22 Module 04; Lecture 3 V 3 Q = = = 5.4m/s A π ( 0.08) 4 3 ( ) k = + () π 4 ( 0.08 ) () = k k = 0.08 When P P = 0, then Q π ( ) π = 0.08 V = ( 0.055) V 4 4 V = 3.86V ( ) V V g g = V V = V = V =.45m/s 3 Q V = = =.4m/s A π ( 0.055) 4 π ( ) 3 Q = AV = = m /min = 3.45Lts/s 4 EXERCISES. Water is flowing through a pipe having inlet and exit diameter of 5cm and cm respectively. If the velocity of water at inlet section is 5m/s, find the velocity head at exit and also rate of discharge.. Water flows at a rate 50Lts/s through a taper pipe of 80m length having diameters of 50 cm at upper end and 0 cm at the lower end. The pipe has a slope in 30. Find the pressure at the lower end, if the pressure at higher level is bar. 3. Brine is to be drained from the bottom of a large tank through a pipe. The drawn pipe ends at a point 0m below the surface of the brine in the tank. Considering a streamline starting at the surface of the brine in the tank and passing through the center of the drain line to the point of discharge, calculate the velocity of flow along the streamline at the point of discharge from the pipe. 0

23 Module 04; Lecture 3 4. An open circuit wind tunnel draws from atmosphere through a contoured nozzle. In the test section, where the flow is straight and nearly uniform, a static pressure tap is drilled into tunnel wall. A manometer connected to the tap shows that static pressure within the tunnel is 45mm. of water below atmospheric. Assume that air is incompressible and at 5 0 C, pressure of 00kPa (absolute). Calculate the velocity of water in the wind tunnel section. Take density of water as 999kg/m 3, and characteristic gas constant as 87J/kg K. 5. For a smooth inclined pipe of uniform diameter of 80mm, a pressure of 80kPa was observed at the inlet section. The velocity and pressure at the exit section was found to be.5 m/s and 30kPa respectively. If the inlet and exit section of the pipe are at an elevation at 8m and 4m respectively, find the head loss between two sections and direction of flow. 6. Oil with specific gravity of 0.8 flows through a tapered pipeline whose diameter changes from 30 mm at section to 60 mm at section. The pressures at section and are observed to be bar and 0.65 bar respectively. The discharge through the pipeline is 300 lits/s and the difference in elevation between two sections is 46m. Determine (i) loss at head and direction of flow. 7. A conical tube of length 3m is fixed vertically with its smaller end upwards. The flow enters at a velocity of 4m/s at the smaller end and leaves at the lower end at a velocity of m/s. The pressure head at the smaller end is m at the liquid. The head loss in the tube is 0.95 ( V V ) g where and V are velocities at the smaller end and lower end V respectively. Determine the pressure head at lower end. 8. In a vertical pipe conveying water, the pressure gages are located at A and B, where the diameters are cm and 6cm respectively. The point B is.5m below A. When the flow rate down the pipe is 0lits/s, the pressure at B is 0kPa. Assuming that the losses in the pipe between A and B as ( kv A ) g, where V A is the velocity at A, find the value of k. If the gages are replaced by tubes filled with water and connected to U-tube mercury manometer, calculate the reading in the manometer. Specific gravity of mercury can be taken as 3.6.

24 Module 04; Lecture 4 DYMAMICS OF FLUID FLOW EXPERIMENTAL TEST RIG - LOSSES IN PIPING SYSTEM The pressure loss estimation in the piping system is demonstrated by an experimental test rig installed in the Fluid Mechanics Laboratory in the Department of Mechanical Engineering at IIT Guwahati. This apparatus was procured from TecQuipment, England. It enables the pressure loss measurements in piping systems that consists of small diameter pipes, pipe bends, sudden enlargements/contractions etc. Description of Apparatus The schematic diagram of the apparatus is shown in Fig. and the photograph of the set up is shown in Fig.. The apparatus consists of two separate hydraulic circuits with a number of pipe systems; one painted in dark blue and other painted in light blue. Both the circuits are supplied water with same hydraulic bench. The components in each of the circuits are given in Table. Table : Component of test apparatus.. Gate valve. Standard Elbow Dark blue circuit Mitre bend 4. Straight pipe 5. Globe valve 6. Sudden expansion 7. Sudden contraction Light blue circuit 8. 50mm 90 0 radius bend 9. 00mm 90 0 radius bend 0. 50mm 90 0 radius bend

25 Module 04; Lecture 4 Fig. : Schematic arrangement of the apparatus.

26 Module 04; Lecture 4 Fig. : Photograph of test rig for measuring pressure losses in piping system. Referring to Fig., the nomenclature for various components of the apparatus arrangements are, A: Straight pipe with 3.7mm bore B: 90 0 sharp bend C: Proprietary 90 0 elbow D: Gate valve E: Sudden enlargement 3. 7mm to 6.4mm F: Sudden contraction 6.4mm to 3.7mm G: Smooth 90 0 bend with 50mm radius H: Smooth 90 0 bend with 00mm radius J: Smooth 90 0 bend with 50mm radius K: Globe valve L: Straight pipe with 6.4mm bore 3

27 Module 04; Lecture 4 The pressure change across each of the components is measured by a pair of piezometer tubes whereas pressure change across the valves are measured by U-tube mercury manometer. In order to identify the manometer tubes for various components, they are numbered as given in Table. Table : Identification of manometer tubes. Monometer tube number Unit - Elbow bend 3-4 Straight pipe 5-6 Mitre bend 7-8 Expansion 9-0 Contraction - 50mm bend mm bend mm bend Objectives of the experiments Straight pipe losses (a) Head losses as function of volume flow rate (b) To calculate friction factor and Reynolds number for straight pipe portion (dark blue circuit) and their functional relationship Sudden expansion (c) Compare the measured head rise across the sudden expansion with the rise calculated based on the assumption of no head loss and with head loss. Sudden expansion (d) Compare the measured head fall across the sudden contraction with the fall calculated based on the assumption of no head loss and with head loss. Losses due to friction and pipe bends (e) To measure loss coefficients due to bends and friction and then plot it as a function of non-dimensional diameter. Losses due to valves 4

28 Module 04; Lecture 4 (f) To establish the functional relationship between loss coefficient and volume flow rate for globe valve and gate valve. Theoretical background The continuity and Bernoulli s equation for an incompressible fluid flow at two sections and in a pipe flow is given by, where, Q= AV = AV Q= = p V p V z+ + = z h ρg g ρg g 3 Volumetricflow rate(m /s) p Static pressure(n/m ) V = = L( ) Mean velocity(m/s) g Acceleration due to gravity (9.8m/s ) A= ρ = 3 Cross sectional area (m ) Density(kg/m ) z = Height above datum (m) h = Head loss (m) L( ) The head losses for various components are as follows; Head loss in straight pipes For a constant diameter ( d ) pipe of length () l, the head loss is given by, where f h L 4 f lv = gd is the roughness of the internal surface of the pipe. Head loss due to sudden expansion It is given by the expression, Head loss due to sudden contraction h L = ( V V ) When there is a sudden contraction in area of the flow, the head loss is expressed as, h L g KV = g where K is a dimensionless coefficient that depends on area ratio as shown in Table 3. 5

29 Module 04; Lecture 4 Head loss due to pipe bends Table 3: Loss coefficient for sudden contraction. ( A A ) K This loss is given by the expression, h B = KV B g where K B is a dimensionless coefficient that depends on the ratio of bend radius to pipe radius and the angle of bend. Head loss due to valves The pressure loss due to opening and closing of valve is given by the expression, h L KV = g where K is the loss coefficient that depends on the type of the valve and degree of openings given by Table 4. Table 4: Typical values of loss coefficients for gate and globe valves. Globe valve, fully open 0 Gate valve, fully open 0. Gate valve, half open 5.6 Experimental procedure Connect the hydraulic bench supply to the inlet of the apparatus and direct the outlet hose into the hydraulic bench-weighing tank. Close the globe valve, open the gate valve and admit water to dark blue circuit by starting the pump and opening the outlet valve on hydraulic bench. Allow water to flow for -3 minutes. 6

30 Module 04; Lecture 4 Open the gate valve by adjusting the screws on the U-tube to fill both limbs of the tube with water. Ensure that no air remains in the tube. Close the gate valve, open the globe valve and repeat the above procedure for light blue circuit. The apparatus is now ready for pressure loss measurement to be made on components in either circuit. Now, open the water control valve fully on the hydraulic bench. With the globe valve fully closed, adjust the gate valve to obtain maximum flow through dark blue circuit. Record the readings in the piezometer tubes and U-tube. Collect sufficient quantity of water in the weighing tank to ensure that weighing takes place over a minimum period of minute. Repeat the above procedure for 0 different flow rates covering entire flow ranges. It can be done by adjusting the gate valve. Prepare the data sheet in the specified format as given in Table 5 for dark blue and light blue circuit. Table 5: Format for experimental data sheet preparation. Test number Time to collect certain Piezometer tube Manometer readings mass of water (s) readings (cm of (cm of Hg) water) Presentation of experimental data Basic data Internal diameter of the pipe: 3.7mm Internal diameter of the pipe (between sudden expansion and contraction): 3.7mm Pipe material: Copper tube Distance between pressure tappings for straight pipe and bends: 0.94m Bend radii: 90 0 elbow = 0; 90 0 proprietary elbow =.7mm; 90 0 smooth bend = 50mm, 00mm, 50mm Kinematic viscosity of water at room temperature (5 0 C) = m /s Graphical representation 7

31 Module 04; Lecture 4 For each test, various parameters such as flow rate, head loss at various sections, loss coefficients etc. are calculated and are plotted in the graphical form. The trends of the graphical plots are given in Figs Fig. 3: Head loss as a function of volume flow rate for straight pipe. 8

32 Module 04; Lecture 4 Fig. 4: Variation of friction factor with Reynolds number for straight pipe. Fig. 5: Head rise across a sudden enlargement in a pipe. 9

33 Module 04; Lecture 4 Fig. 6: Head loss due to sudden contraction in a pipe. 0

34 Module 04; Lecture 4 Fig. 7: Loss coefficient due to pipe bends as a function of non-dimensional pipe radius. Fig. 8: Loss coefficient for gate and globe valve as function of flow rate.

35 Module 04; Lecture 4 Conclusion and general remarks An attempt has been made in this experimental apparatus to study pressure losses in various components in a piping system consisting of straight pipes, sudden expansion and contraction, bends and valves. The general trends and magnitudes obtained during experiments will be the indication of the pressure loss from various components in the pipe system. However, in practical situations, such combinations are not adopted over a short span length. The normally accepted design criterion for pressure tappings is 30 to 50 pipe diameters away from obstruction. Also, sufficient pipe length has been left between each component in the circuit to eliminate any adverse influence of the neighboring components. Thus, any discrepancies between experimental and theoretical results may be attributed due to three main factors; Relatively small physical scale of the pipe work Relatively small pressure differences Low Reynolds numbers

36 Module 04; Lecture 5 DYMAMICS OF FLUID FLOW PRACTICAL APPLICATIONS OF BERNOULLI S EQUATION Bernoulli s equation finds wide applications in all types of problems of incompressible flow where there is involvement of energy considerations. The other equation, which is commonly used in the solution of the problems of fluid flow, is the continuity equation. In this section, the applications of Bernoulli s equation and continuity equation will be discussed for the following measuring devices. Venturi meter Nozzle Orifice meter Pitot tube Venturimeter It is an instrument, which is used to measure the rate of discharge in a pipeline and is often fixed permanently at different sections of the pipeline to measure the discharge. The principle of venturi meter was demonstrated by Italian physicist G.B. Venturi (746-8) in 797, but it was first applied by C. Herschel (84-930) in 887 to develop the device for measuring the discharge or rate of flow of fluid through pipes. The basic principle is that by reducing the cross-sectional area of the flow passage, a pressure difference is created and the measurement of pressure difference enables the determination of the discharge through pipes. Construction It consists of three parts as shown in Fig.. () An inlet section followed by a convergent cone () A cylindrical throat (3) A gradually divergent cone

37 Module 04; Lecture 5 Fig. : Venturi meter. The inlet section of the venturi meter is of same diameter as that of the pipe followed by a convergent cone. The convergent cone is a short pipe that tapers from the original size to that of throat of venturi. It has an included angle of 0 ± 0 and approximate length of.7 ( D d ) parallel to the axis, where D is the diameter of the inlet section and 3 d is the diameter of the throat d = to D 3 4. The throat of the venturi meter is a short parallel-sided tube having its crosssectional area smaller than that of the pipe. The length of the throat is approximately d. The diverging cone is a gradually diverging pipe with its cross-sectional area increasing from that of throat to the original size of the pipe. The total included angle in this cone is preferably between (length of the cone ~ 7.5 d ) such that the length of convergent cone to be smaller than the divergent part. The pressure measuring systems (such as manometer) is mounted to measure the pressure difference at the inlet section and the throat i.e. sections and of the venturimeter. Considering the continuity equation, it is obvious that in the convergent cone, the fluid is being accelerated from the inlet section to the throat section. But, in the divergent cone, the fluid is retarded from throat section to the end section 3 of the venturi. The acceleration of the flowing fluid takes place in a relatively smaller length without resulting in appreciable loss of energy. So appreciable pressure drop is noticed in the manometer. The measurement of pressure difference between these sections enables the computation of rate of flow of fluid.

38 Module 04; Lecture 5 Mathematical analysis Let and a be the cross-sectional areas at the inlet section and the throat (i.e. section a and ) of the venturi meter respectively, at which the pressures and velocities are p, p and, V respectively. If the fluid is incompressible with no loss of energy between the V sections and, the Bernoulli s equation can be written as, p V p V γ g γ g + + z = + + z If the venturi meter is connected in a horizontal pipe, then the elevation heads at section and will be equal i.e. z = z or if the datum is assumed to passing through the axis of the venturi meter, then z = z = 0. The above equation reduces to, p V p V + = + γ g γ g or, p p V V = () γ γ g g p p In the above expression, is the difference between the pressure heads at γ γ sections and, which is known as venturi head( h ). Further, if through the pipe, then by continuity equation, th By substituting the values of and V from Eq. () in (), V Q th is the discharge Q = av = av () Q th h = g a a i.e. Q th aa = a gh a (3) The above equation gives only the theoretical discharge because the loss of energy is not considered. But, in actual practice, there is always some loss of energy as the fluid flows and the actual discharge( Q) will be always less than the theoretical discharge. The 3

39 Module 04; Lecture 5 actual discharge may be obtained by multiplying the theoretical discharge by a factor, called coefficient of discharge i.e. C d Q = (4) Q th Also, for a given venturi meter, the cross-sectional areas of the inlet section and the throat i.e. and a are fixed. So, one more constant for a given venturi meter can be expressed as, a C d aa C = a g a (5) Using Eqs. (4) and (5) in Eq. (3), we get aa gh Q= Cd = C C h a a d (6) Discussions: The coefficient of discharge of the venturi meter accounts for the effects of nonuniformity of the velocity distribution at sections and. The coefficient of discharge of the venturi meter varies with the flow rate, viscosity of the fluid and the surface roughness. But, in general, for the fluids of low viscosity, the value falls in the range of 0.95 to The venturi head (i.e the pressure difference between the section and ) is usually measured by a manometer. If S and S are the specific gravities of the liquid in the manometer and liquid flowing in the venturi meter and x is the difference in the levels of two limbs of the manometer, then the expression for the venturi head becomes, p p Sm = h= x ( whensm S γ γ > S ) (7a) p p Sm = h= x ( whensm S γ γ < S ) (7b) Venturi meter can also be used to measure the discharge through pipe, which is laid either in an inclined or in vertical position. The same formula for discharge also holds good. But here, m 4

40 Module 04; Lecture 5 h p γ p γ = + ( ) z z (8) Nozzle A flow nozzle is also a device used for measuring the discharge through pipes. As shown in Fig., the flow nozzle consists of a streamlined convergent nozzle through which the fluid is gradually accelerated. Fig. : Flow nozzle. It is essentially a venturi meter with divergent part omitted and hence the basic equations are the same as those for the venturi meter. Since there is no divergent cone on the downstream of throat of the nozzle, there is a greater dissipation of energy than the venturi meter. The discharge coefficient for a nozzle can be given by the following empirical equation; C d d D 0.4 D = D (9) R e d where d and D are the diameter of the nozzle and the pipe, number based on the diameter of the nozzle. R e d is the Reynolds Cavitation When the pressure at any point in a liquid becomes equal to the vapour pressure of the liquid, the liquid vapourizes and forms bubbles. These bubbles have the tendency to break the continuity of the flow. Formation of vapour bubbles, their transport to regions 5

41 Module 04; Lecture 5 of high pressure and subsequent collapse is known as cavitation. It is quantified by a dimensionless number defined by; where p p Cavitation number, σ = v (0) ( ρu 0 ) p is the absolute pressure at the point under consideration, pressure of the liquid, U0 p v is the vapour is the reference velocity and ρ is the density of the liquid. Example- Water flows through an inclined venturi-meter whose inlet and throat diameters are 0mm and 70mm respectively. The inlet and throat section are 60cm and 90cm high above the datum level. For certain flow rate, the pressure difference between the inlet and throat is measured by a mercury manometer and is found as 5cm of Hg. Estimate the flow rate (i) neglecting friction loss; (ii) when the friction head is 5% of head indicated by the manometer and (iii) discharge coefficient. Solution: The inclined venturiemeter is schematically shown in the following figure. () () z h0 h Z Fig. A: Inclined venturiemeter Applying the Bernoulli s equation for the venturiemeter, p V p V γ g γ g + + z = + + z p p V V or, = + ( z z) (I) ρg g 6

42 Module 04; Lecture 5 Pressure balance for the manometer can be written as, ( ) ρ ( ) p + ρg z h = p + g z h h +ρ gh 0 0 Hg Comparing Eqs. (I) and (II), If head loss due to friction ( h p p ρg f ) ρhg = h+ ( z z ρ V V g ρ ρ = Hg h is included, then Eq. (III) becomes, ) (II) (III) V V g In ideal case i.e. without friction loss, By continuity equation, ρ = h h ρ Hg f (IV) Q= A. V = A V. (V) π π 4 4 = 0. = 0.03m, = 0.07 = m Inlet and throat areas are, A ( ) A ( ) Using Eq. (V) in (III), we get, Q ρhg h = g A A ρ Q th = = With friction loss, Using Eq. (V) in (III), we get, Q actual Discharge coefficient, m s ρ ρ = h h = h 0.05h Q Hg Hg f g A A ρ ρ = = m s C Q actual d = = = Qth 0.9 7

43 Module 04; Lecture 5 Example : A 5cm diameter nozzle is attached to pipe of cm diameter at its discharge end. The rate of discharge of water through the nozzle is 5lits/s and pressure at the base of the nozzle is.6bar. In the discharge end, the pressure is atmospheric. Calculate the coefficient of discharge for the nozzle. Assume that the base of the nozzle and the outlet of the nozzle are at same elevation. Solution: The nozzle attached to the pipe is shown in the following figure. Pipe () Nozzle () p =.6bar V, A p = bar V, A Areas at the section and are Discharge, π π π π A =. D =.( 0.) = 0.03m ; A =. d =.( 0.05) = m Pressure at the nozzle base, Q = 5lits s = m s 3 p =.6bar = 60 0 N m V A By continuity equation, = = = 0.73 V A 0.03 Applying Bernoulli s equation for section and with same datum level, Solving for V 00 0 V + = g g V, we get V = m s Theoretical discharge, Qth = A V = = m s Coefficient of discharge, C d Q 0.05 = = = 0.66 Q 0.03 th 8

44 Module 04; Lecture 5 EXERCISES. A jet pump is fitted to a venturimeter whose inlet and throat diameters are 5cm and 7cm respectively. Water is drawn through inlet section of venturie where the pressure and velocity are.5bar and 5m/s respectively. Also, water is drawn by a lift pipe at the throat from a sump m below the venturimeter. The pump discharges the water through the exit section of venturiemter at bar. Calculate the diameters of the venturimeter outlet and lift pipe if the volumetric lift rate is 5% of the main flow.. A venturimeter is to be fitted in a pipe of 0.m diameter where the pressure head is 0m of the flowing fluid and the maximum flow rate is 8m 3 /min. Find the least diameter of the throat so that the pressure head does not become negative. 3. A venturimeter is provided in a 00mm diameter pipe for measurement of water discharge. For a gage pressure of 00kPa in the pipe, determine the diameter of the throat of venturiemeter to produce cavitation at the throat. The throat is m higher than the venturi inlet. The pipe carries a discharge of 60lits/s. Take atmospheric pressure as 0.35kPa and the vapour pressure as.4kpa (absolute). 4. Referring to the following figure, determine the value h at which the cavitation will just start. Take atmospheric pressure as 0.35kPa and the vapour pressure as.4kpa (absolute). (0).8m 50mm h () () Q 75mm 5. A 5mm diameter nozzle discharges lits/hr of water, when the head is 60m. The diameter of the jet is.cm. Determine (i) the value of the coefficients the loss of head due to fluid resistance. Cc, Cv, C d ; (ii) 9

45 Module 04; Lecture 5 6. A fire hosepipe is connected to a nozzle to discharge 0.9lits/hr of water to atmosphere as shown in the following figure. The diameter of the hosepipe is 0cm and that of the nozzle is 3cm. Assuming the frictionless flow, find the force exerted by the flange bolts to hold the nozzle on the hose. Flange Bolts Water D d Hose pipe Nozzle 7. A fireman intends to reach a window 5m above the ground with a fire hose nozzle with 3cm diameter. The height of the nozzle is.5m above the datum level and it discharges 0lits/s of water. Neglecting the air resistance, determine the greatest distance from the building at which fireman can stand and still play the stream upon the window. 0

46 Module 04; Lecture 6 DYMAMICS OF FLUID FLOW PRACTICAL APPLICATIONS OF BERNOULLI S EQUATION Orifice meter (Contd..) An orifice meter is a simple device used for measuring the discharge through pipes. It works on the same principle as that of venturimeter i.e pressure difference is created by reducing the cross-sectional area of the flow passage and measurement of the pressure difference enables the determination of discharge through the pipe. In most of the fluid flow measurements, an orifice is an opening having closed perimeter, made in the walls or the bottom of a tank or a vessel containing fluid, through which fluid may be discharged. It is a cheaper arrangement compared to venturimeter and finds application where Construction space is limited and the accuracy is not that much important. An orifice meter consists of a flat circular plate with a circular hole called orifice which is concentric with the pipe axis (Fig. ). The orifice diameter( d ) varies from 0.4 to 0.8 times the diameter of the pipe ( D) and thickness of the plate( t ) is less than or equal to 0.05 times the diameter of the pipe. A differential manometer is fixed with the orifice meter to measure the pressure difference. One of the limbs of differential manometer is connected at the section, which is at a distance of about 0.9 to. times the pipe diameter from the orifice plate and the position of other limb is about 0.5 times the diameter of the orifice from the orifice plate. Fig. : Orifice meter.

47 Module 04; Lecture 6 As the fluid flows through orifice, the necessary transverse velocity components imparted to the fluid as it approaches the obstruction carry through downstream side. As a result, the minimum stream section occurs not in the plane of the orifice, but somewhat downstream as shown in the Fig.. The term vena contracta is applied to the location and condition of this minimum stream dimension. This is also the location of minimum pressure. The maximum possible pressure difference exists between the sections and (vena contracta), which is measured by connecting a differential manometer. The jet of the fluid coming out of the orifice gradually expands from vena contracta to again fill the pipe. Since there is an abrupt change in the cross sectional area of the flow passage, so greater loss of energy is experienced compared to venturi meter. Mathematical analysis Let p, p and V, V be the pressures and velocities at sections and respectively. Then, for an incompressible fluid, applying Bernoulli s equation between the sections and and neglecting the losses, we have p p V V + z + z = h= γ γ g g where h is the difference between the piezometric heads at sections and. However, if the orifice meter is connected in a horizontal pipe, then (), in which case h will represent the difference between the pressure heads at section and. From Eq. (), we can obtain Since V ( ) 0.5 z = z V = gh+ V () is the velocity of the liquid approaching the orifice, it is often termed as velocity of approach. If the point is considered to be sufficiently far from the orifice, then the velocity of approach will be small in comparison to V and may be neglected in Eq. (), so that V V = gh (3) The above equation is known as Torricelli s formula, which can be used to measure the velocity of a liquid jet experimentally, emerging from a small orifice.

48 Module 04; Lecture 6 Terminologies for orifices (a) Coefficient of velocity, C v In actual practice, as the real fluid flows through orifice, there is always some loss of energy due to friction and surface tension. Hence, the actual velocity of the jet at venacontracta is slightly less than the ideal velocity and is determined by multiplying a factor called coefficient of velocity. It is defined as the ratio of the actual velocity of the jet at vena-contracta to the ideal (theoretical) velocity of the jet i.e. C v = V V V = gh (4) th Experimentally, it is observed that the value of C v varies from 0.95 to 0.99 depending on the shape and size of the orifice. For a sharp-edged orifice discharging water or liquids with similar viscosity, the average value of is C v (b) Coefficient of contraction, Cc A jet of liquid issuing from an orifice has its cross-sectional area at vena-contracta less than the area of the orifice i.e. the jet of liquid undergoes a contraction. The actual area of the jet at vena-contracta is determined by multiplying the area of the orifice by a factor called coefficient of contraction(. If is the area of the jet at vena-contracta and is the area of the orifice, then ) Cc a a0 C a c = (5) a0 In practice, the value of C c varies from 0.6 to 0.69 depending on the size and shape of the orifice and head of the liquid under which the flow takes place. (c) Coefficient of discharge, Cd Theoretically, the cross-sectional area of the jet of liquid issuing from an orifice will be equal to the area of the orifice, which may be considered as an ideal (or theoretical) cross-sectional area of the jet. The product of ideal cross-sectional area of the jet and ideal velocity of jet given by Eq. (3) will give the theoretical discharge. However, on account of the effect of friction due to which the actual velocity of the jet is reduced and 3

49 Module 04; Lecture 6 due to contraction of the jet, the actual discharge of liquid through an orifice is always less than the theoretical discharge. If Q th discharge emerging from an orifice, then and, C is the theoretical discharge and Q is the actual d Q = Q th ( c ) ( v ) Q= a. V = C. a C. gh 0 Hence, C d ( Cc. a0 ) ( Cv. gh) = = Cc. Cv (6) a. gh 0 Therefore, it is possible to find the value of C by determining C, C and then multiplying together. Its value varies from 0.6 to d c v (d) Coefficient of resistance, Cr It is defined as the ratio of the loss of kinetic energy as the liquid flows through an orifice and the actual kinetic energy possessed by the flowing fluid. The loss of kinetic energy as the liquid flows through orifice is equal to the difference between the theoretical kinetic energy and the actual kinetic energy. Theoretical kinetic energy per unit weight of liquid Actual kinetic energy per unit weight of liquid So, Discharge through orifice C r ( Cv ) ( gh ) = V h g = g = th ( gh) = V Cv hc g = g = h = = hcv Cv (7) Considering the losses into account, the actual velocity at the section can be written from Eq. () as, v 4

50 Module 04; Lecture 6 Further, the area of the jet a v ( ) 0.5 V = C gh+ V (8) at the section (i.e. at vena contracta) may be related to area of the orifice a 0 by the following expression, a = C a c 0 Using continuity equation, the expression for V can be written as, By substituting the value of Solving for, we get V V a 0 V = VCc a in Eq. (8), we get a V C gh V C a 0 = v + c 0.5 V = C gh ( 0 ) v CC v c a a Actual discharge Q through the pipe is given by, Q= av = CcaV 0. Taking CC c v = Cd, 0.5 or, Q= Cda gh ( 0 ) 0 Cd a a 0.5 (9) Q= Ca where C = C gh ( a0 a ) 0 d 0.5 ( a0 a ) d ( 0 ) C a a 0.5 (0) () Eq. (0) gives the discharge through orifice meter. The coefficient of discharge for an orifice meter is much smaller than that for venturimeter. This is because in the case of an orifice meter, there are no gradual converging and diverging flow passages as in the case 5

51 Module 04; Lecture 6 of venturimeter, which results in a greater loss of energy and consequent reduction of the coefficient of discharge for an orifice meter. Pitot tube A pitot tube is a simple device used for measuring the velocity of flow. The basic principle of operation for this device is that if the velocity of flow at a particular point (stagnation point) is reduced to zero, the total energy of the flow is converted to pressure energy. By measuring the increase in pressure energy at this point, the velocity of flow may be determined. French engineer Henri de Pitot (695-77) first used this principle for measuring the velocities in the river Seine in 73. Construction The pitot tube consists of a glass tube of suitable length (for which the capillary effects will be negligible), bent at right angles as shown in Fig. (a). The tube is dipped vertically in the flowing stream of the fluid with its open end A, directed to face the flow and the other end projecting above the fluid surface in the stream as shown in Fig. (a). The fluid enters the tube and the level of the fluid in the tube exceeds that of the fluid surface in the surrounding stream because the end A of the tube is a stagnation point where the fluid is at rest, and the fluid approaching A divides at this point and passes around the tube. At the stagnation point, the kinetic energy is converted into the pressure energy. So, the fluid in the tube rises above the surrounding fluid surface by a height, which corresponds, to velocity of the flow of fluid approaching the end A of the Pitot tube. The pressure at the stagnation point is known as stagnation pressure or impact pressure. When the pitot tube is used for measuring the velocity of flow in a pipe or any other conduit, then it may be inserted as shown in Fig. (b). Since the pitot tube measures the stagnation pressure head (or the total head) at its dipped end, the static pressure head is also required to be measured at the same section where the tip of the pitot tube is held. A differential manometer can be used for this purpose for measuring the static pressure as shown in Fig. (b). 6

52 Module 04; Lecture 6 (a) (b) (c) Fig. : (a) Simple pitot tube; (b) Pitot tube for measuring dynamic pressure head in the pipes; (c) Prandtl pitot tube. The tubes recording the static pressure and stagnation pressure are frequently combined into one instrument known as pitot-static tube. In this instrument, the static tube surrounds the total head tube and two or more small holes are drilled radially 7

53 Module 04; Lecture 6 through the outer wall into the annular space. A major problem in the use of ordinary pitot static tube is to obtain proper alignment of the tube with flow direction. The angle formed between the probe axis and the flow streamline at the pressure opening is called yaw angle. Ideally, this angle should be zero, but in many situations, it may not be constant: i.e. the flow may be fixed neither in magnitude nor in direction. Also, The orientation of both impact and static opening of the probe affects the yaw angle. For accurate results, a standard form of pitot static tube known as Prandtl pitot tube (Fig. (c)) is used to measure both static and stagnation pressure. It has a blunt nose and is so designed that the disturbance caused in the flow by the nose and leg cancel each other. Mathematical analysis Referring to Fig. (a), consider a point slightly upstream of end A, lying along the same horizontal plane in the flowing stream where the velocity of flow is V. If the points h 0 and A are at a vertical depth of below the free surface of fluid in the stream and h is the height of the fluid raised in the pitot tube above the free surface, then applying Bernoulli s equation between points and A (neglecting energy loss), V h0 + = h0 + h () g so that, V = gh (3) In the Eq. (), the expression ( h h) 0 + is the stagnation pressure head at point A, which consists of the static pressure head and dynamic pressure head h. Eq. (3) indicates that the dynamic pressure head is proportional to the square of the velocity flow in the stream close to the end A of the pitot tube. Thus, it is possible to measure the velocity of the flow by dipping the pitot tube at any point in the flowing stream and measuring the height h of the fluid raised in the tube above free surface. However, if the flow is highly turbulent, the pitot tube records a value of h higher than that of corresponding mean velocity value in the direction of tube axis. So, by introducing a factor called pitot tube coefficient C v h 0, the actual velocity of flow can be written as, V = C gh (4) v 8

54 Module 04; Lecture 6 The value of C v is usually determined by calibration. But a probable value for this coefficient of the pitot tube is Relative merits of venturi meter, flow nozzle and orifice meter High accuracy, good pressure recovery and resistance to abrasion are the primary advantages of the venturi. The space requirement and cost of the venturi meter is comparatively higher than that of orifice and flow nozzle. The orifice is inexpensive and may often be installed between existing pipe flanges. However, its pressure recovery is poor and it is especially susceptible to inaccuracies resulting from wear and abrasion. It may also be damaged by pressure transients because of its lower physical strength. The flow nozzle possesses the advantages of the venturi, except that it has lower pressure recovery and it has the added advantage of shorter physical strength. It is inexpensive compared with the orifice and is relatively difficult to install properly. The typical pressure characteristics for venturi meter, flow nozzle and the orifice meter is shown in Fig. 3(a-c). In each case, the basic meter acts as an obstacle placed in the path of flowing fluid, causing localized changes in velocity. Concurrently, there will be pressure changes. At points of maximum restriction (or maximum velocity), minimum pressure is observed. A certain portion of this pressure drop becomes irrecoverable owing to the dissipation of the kinetic energy; therefore the output pressure will be always less than the input pressure. As indicated in the Fig. 3(a), the venturi, with its guided re-expansion, is seen be the most efficient. Loses of about 30-40% of differential pressure occur through orifice meter (Fig. 3(c)). (a) (b) 9

55 Module 04; Lecture 6 Fig. 3: Pressure characteristics of (a) venturi meter, (b) nozzle, (c) orifice meter. (c) Example Water flows from a tank through a small circular orifice of 3cm diameter. The water head in the tank is.5m. It is seen that the horizontal displacement of water coming out of the orifice is.3m while the vertical fall is 0cm. Calculate (i) coefficient of velocity, (ii) volume flow rate of water and (iii) the horizontal thrust on the container due to the issue of water jet. (Take coefficient of contraction as 0.6) Solution: The jet coming out orifice drops down due to action of gravity. It is schematically shown in the following figure. If the actual velocity of water at vena-contracta is u, the distance traveled by the water in horizontal direction is x = ut. and that in vertical direction is, y. = gt. A A H C C x y Eliminating t, we have, u = gx.. y The theoretical velocity of the jet is given by, V = gh Head of water in the tank, H =.5m th 0

56 Module 04; Lecture 6 (i) C v u g. x x.3 = = = = = 0.9 V y.gh 4 yh th π a = 0.03 = m 4 Area of the orifice, ( ) 4 Since C = 0.6; C = C. C = = 0.55 (ii) Volume flow rate, c d c v Q= C a g H = = = 4 3 d m s.8lits s (iii) Horizontal thrust on the tank due to the issue of the jet is, F = ρ. Qu. = ρ. QC.. gh = = 8N x v Example : A submarine moves horizontally in the sea and has its axis much below the surface of seawater. A pitot tube properly placed just in front of the submarine is connected to a differential pressure gauge. The pressure differential between the stagnation pressure and the static pressure was found to be 0kN/m. Find the speed of the submarine if the specific gravity of seawater is.06. Solution: Density of seawater = (specific gravity) (density of water) = =06 kg/m 3. At stagnation point, the velocity of the flow is zero. Net head developed in the pitot tube is, 3 pstag pstat 0 0 h = = =.987 ρg Using Eq. (), the velocity can be calculated as, V = gh = = 6.44 m s

57 Module 04; Lecture 6 EXERCISES. A large tank resting on floor is filled to a depth of 5m. A sharp edged circular orifice is located m above the floor level. Determine the discharge and horizontal distance from the tank where the jet will strike the ground. If C = 0.64, C = 0.95, then determine.. A large closed tank is partially filled with oil to a depth of 3.5m above a circular orifice. The specific gravity of oil is 0.85 and the diameter of the orifice is 8cm. What gauge pressure in the tank will produce a jet of kw power? Take C = 0.64, C = A large circular tank contains water to a height of 5m. It is provided with an orifice at its bottom. If the diameters of the orifice and that of tank are 30cm and of 3m respectively, then find (i) the time taken for the water level to fall down from 5m to m (ii) the time taken to completely emptying the tank. Take C d = A pitot tube is inserted in a pipe of 30cm diameter. The static pressure in the pipe is 00mm of mercury (vacuum). The stagnation pressure at the center of the pipe, recorded by the pitot tube is 0.bar. Calculate the rate of flow of water through the pipe, if the mean velocity of the flow is 0.85 times the central velocity. Take C v = Water flows in a pipe of 30cm diameter. Two pitot tubes are installed in the pipe one on the centerline and the other 75mm away from the centerline. The velocities at these two points are 3m/s and m/s respectively. Calculate the reading on the differential manometer connected to the tubes. c v c v C d

58 Module 04; Lecture 7 DYMAMICS OF FLUID FLOW EXPERIMENTAL TEST RIG VENTURI METER A venturi is a device used for measuring the discharge along a pipe. The fluid flowing in the pipe passes through contraction section to throat followed by a diverging portion. The velocity of the fluid at throat is higher than that of any section in the pipe. This increase in velocity is associated with fall in pressure. Thus, by measuring the magnitude of pressures drop, the discharge can be calculated using continuity equation. The working principle of this instrument is demonstrated by using the apparatus (make: TecQuipment, England) installed in the Fluid Mechanics Laboratory in the Department of Mechanical Engineering at IIT Guwahati. Description of Apparatus The apparatus schematically shown in Fig. is fabricated in aluminum. Water is admitted through a flexible hosepipe into the venturi-meter from a hydraulic bench. The flow at the upstream of the venturi-meter is regulated by a supply valve whereas in the downstream of the meter the water is collected in a measuring tank. Piezometer tubes are drilled into the wall along the entire convergent-divergent passage of the venturi-meter (Fig. ). These tubes are further connected to vertical manometer tubes, which are mounted in front of a scale. The entire assembly is supported on a base that may be adjusted to level the equipment. The photograph of the apparatus is shown in Fig. 3. The usual form of a venturi-meter is intended to measure flow rate for which pressure tapping are required only at the entrance and throat. However, more number of pressure tapping is required if it is intended to show pressure distribution along the entire convergent-divergent passage.

59 Module 04; Lecture 7 Fig. : Schematic arrangement of the apparatus. Fig. : Position of piezometer tubes in a venturi-meter tube.

60 Module 04; Lecture 7 Fig. 3: Photograph of test rig for measuring pressure losses in piping system. Objectives of the experiments To calculate the coefficient of venturi-meter To show pressure distribution along the entire convergent-divergent passage of the venturi-meter To plot the variation flow rate with pressure head difference at throat and inlet section of the venturi-meter Theoretical background Consider the flow of an incompressible fluid through the convergent-divergent pipe as shown in Fig. 4. Let a, aand a n be cross-sectional areas at upstream (section ), throat (section ) and any arbitrary section n, u, uand u n (section ), throat (section ) and any arbitrary section n, be velocities at upstream h, hand h n be pressure 3

61 Module 04; Lecture 7 heads at upstream (section ), throat (section ) and any arbitrary section n and Q be the volume-flow rate or discharge rate Fig. 4: Ideal conditions of venturi-meter. Then, continuity equation and Bernoulli s theorem may be stated as, Now, solving for u, Thus, Q= au = a u = a u u u un h+ = h + = hn + g g g u = n n ( h ) g h Q= a a a ( h ) g h a a Incorporating some energy loss occurring between section and, the above expression is modified as, Q = Ca ( h ) g h a a 4

62 Module 04; Lecture 7 where C is known as coefficient of the meter which is established by experimentation and its value varies in the range 0.9 to 0.99 for venturi-meter. For the purpose of calculation and comparison of experimental results, it is convenient to express the pressure head difference as a fraction of velocity head at the throat. Thus, hn h a a = a a ( u g) n Experimental procedure Place the apparatus on the hydraulic bench. Connect the upstream side of the unit to the bench supply valve and down stream end to the weighing tank with a plastic tube. Set the apparatus flow control and bench supply valve to one third of the fully open position. Switch on the bench supply and allow the water to flow. Close the apparatus flow control valve. Adjust the bench supply and apparatus control valve to obtain full flow. At this condition, the maximum pressure difference between throat and inlet is 40mm. Subsequent flow rates can be obtained by closing the apparatus control valve. The rate of flow is measured by weighing the amount of water collected in the tank. While the experiment is under progress, the values of h and h are noted from the manometer scales. Repeat the above procedure for a 0 different flow rates covering entire flow range. Prepare the data sheet in the specified format as given in Table, and 3. 5

63 Module 04; Lecture 7 Table : Cross-sectional areas along the entire convergent-divergent passage of the venturi-meter Piezometer tube no. ( n) A () B C D () E F G H J K L Diameter of the cross section ( d n ) d d n a a n a a a an Table :Pressure distribution as a fraction of velocity head at throat Piezometer tube no. ( n) A () B C D () E F G H J K L h ( mm ) h n ( mm) u For given Q and g h n h ( u g) 6

64 Module 04; Lecture 7 ( m 3 s) Table 3: Values of C from experimental data. Q h ( ) h ( ) A () B C D () E F G H J K L mm mm h h C Presentation of experimental data For each test, various parameters such as flow rate, manometer head, coefficient of venturi-meter etc. are calculated and are plotted in the graphical form. The trends of the graphical plots are given in Figs Fig. 5: Variation of venturi coefficient with flow rate. 7

65 Module 04; Lecture 7 Fig. 6: Measured and ideal pressure distribution along the venturi-meter. Fig. 7: Variation of manometric head difference at inlet and throat with flow rate for the venturi-meter. Conclusion and general remarks An attempt has been made in this experimental apparatus to demonstrate the flow rate measurement by a venturi-meter. The typical trends are plotted in Figs The loss coefficient of the venturi-meter can be calculated and its variation with flow rate can be shown in the fashion as plotted in Fig. 5. The measured pressure distribution for convergent to throat portion of the venturi-meter can be compared with theoretical values of pressure (Fig. 6). Also, its variation with flow rate can be shown in Fig. 7. The general trends and magnitudes obtained during experiments is an indication of success of the experimentation. 8

66 Module 04; Lecture 8 DYMAMICS OF FLUID FLOW IRROTATIONAL FLOW, STREAM FUNCTION AND VELOCITY POTENTIAL Fluid Element Kinematics The bulk motion of fluid flow is decided by the individual movement of fluid elements. These elements are treated as infinitesimally small cubes and execute four possible motions; namely, translation, linear deformation, rotation and angular deformation. This section will focus on the mathematical description of fluid flows. Velocity and Acceleration Fields The velocity field of a fluid element V ( x, y, z, t) can be described by specifying the velocities at all points and at all times within the field of domain. In rectangular coordinate system, it is conveniently written as, where iˆ, ˆj, kˆ V = uiˆ+ vj ˆ+ wkˆ () are the unit vectors in x, y, and z respectively and u, v, and w are the velocities in the respective direction. The acceleration field is then described from velocity field as, V V V V a = + u + v + w t x y z In the component form, u u u u ax = + u + v + w t x y z v v v v ay = + u + v + w t x y z w w w w az = + u + v + w t x y z () (3, a, b, c) The Eq. () can be more conveniently written as, DV a = (4) Dt where the operator

67 Module 04; Lecture 8 D( ) ( ) ( ) ( ) ( ) = + u + v + w Dt t x y z The first term in the Eq. (5) is called time derivative and the second term is known as material/substantial derivative. In vector notations, D ( ) ( ) Dt The gradient operator in the Eq. (6) is, ( V. )( = + t ( ) ( ) ( ) ˆ ( ) = i ˆ + ˆ j+ k x y z (5) ) (6) (7) Linear Deformation The simplest type of deformation that the fluid element can undergo is translation in which all the points in the element have the same velocity. The element will be deformed if there is a presence of velocity gradient. Let us consider the effect of velocity gradient on a small cube having sides x, yand z. δy B D u u δx C A u+( u/ x)δx u+( u/ x)δx δy B D δx Fig. : Linear deformation of a fluid element. u If the effect of single velocity gradient is considered for the BC and DA, then x the points A and C displaced with a velocity C A C ' A ' u u+ δ x x. This difference in velocity causes stretching of the element (from A to A and C to C ) by an amount u. x. t x δ δ during short time δ t. The corresponding change in volume becomes, u δ V = δx. δy. δz. δt x (8)

68 Module 04; Lecture 8 u The rate of change of δ V per unit volume due to the gradient x is, δ ( δ ) ( ) d V u x δt u = lim = V dt δt 0 δt x (9) If the other velocity gradients becomes, v w and y z ( δ ) d V u v w = + + =. V δ V dt x y z are present, then the general expression This rate of change of volume per unit volume is called volumetric dilatation rate. It is seen that the volume of fluid element changes with the variation of velocities with their respective derivatives, but the shape of the element does not change. This variation is known as linear deformation. However, the cross derivatives (such as the element to rotate and change its shape. (0) u ) will cause y Angular Deformation Consider the motion of the fluid element (as shown in Fig. ) in x-y plane in which there exist the velocity variation to cause rotation and hence angular deformation. During short time interval δ t, the elements DA and DB tend to rotate through an angle δα and δβ respectively. δy B D u+( u/ y)δy v u δx C A v+( v/ x) δx δy B D B ' δβ δx C A A ' δα Fig. : Angular deformation of a fluid element. The angular velocities for these elements can be written as; 3

69 Module 04; Lecture 8 If u y is positive, ω ω DA DB v = x () u = y ω DB will be clockwise and with same analogy ω DA becomes anticlockwise. Hence, the rotation of the element about z-axis ( ωz ) becomes the average of the two components. Then it follows; v u ωz = x y The rotational components about x-axis and y-axis can be written as, w v ωx = y z u w ωy = z x These three components specify the rotation vector ω, in the form of, () (3) (4) ω = ω iˆ+ ω ˆj+ ω kˆ (5) x y z In terms of velocity field, ω is expressed as, Vorticity ( ) ξ iˆ ˆj kˆ ω = V = x y z u v w w v ˆ u w ˆ v u = i j ˆ + + k y z z x x y (6) It is defined as a vector that is twice the rotation vector. The significance of vorticity is to describe the rotational characteristics of the fluid to eliminate the factor (/) associated with the rotation vector. It is observed from Eq. () that the fluid element will rotate about z-axis as an undeformed block only when ω DA u v = ωdb or =. Otherwise the rotation will be y x 4

70 Module 04; Lecture 8 associated with an angular deformation. When ω DA u v = ωdb or =, the rotation of the y x fluid element about z-axis is zero (i.e. ξ = 0or V = 0). The flow field that satisfies this condition is known as irrotational flow. Physical Interpretation of Irrotational Flow The concept of irrotationality is a very strange condition of the flow field. The condition that makes the flow field to be irrotational is ξ = 0or V = 0. In order to satisfy, Eq. (6), there must be some kind of specific relationship among the velocity gradients. They are, u v = y x w v = y z u w = z x However, a general flow field does not satisfy all three equations. However, uniform flows (in which irrotational flow. (7) u = U, v = 0, w=0) do satisfy the Eq. (7) and can be called as Stream Function The simplest type of flow in fluid mechanics application is a steady, incompressible, plane, two-dimensional flow. The continuity equation for such type of flow can be written as, u v + = 0 (8) x y { } Now, we introduce stream function ψ ( x, y) components u and v such that,, which relates the velocity ψ ψ u = and v= y x (9) 5

71 Module 04; Lecture 8 Using the components u and v from Eq. (9), it can be shown that the continuity ( ) equation (Eq. 0) gets satisfied. Hence, we have only one functionψ x, y that describes a steady, incompressible, plane, two-dimensional flow instead of two functions u( x, y) and v( x, y). The other specific advantage of stream function is described as follows: The lines along which ψ is constant, are the streamlines. By definition, streamlines are the lines in the flow field such that the tangent at any point gives the direction of velocities. So, slope at any point along the line as shown in Fig. 3 is given by, y v = x u (0) v V streamline y u x Fig. 3: Velocity components along a streamline. Now coming back to stream function, the change in stream function from one point ( x, y ) to nearby point ( x dx, y dy) + + is given by, ψ ψ dψ = dx + dy = vdx + udy x y Along the streamline dψ = 0sothat dy = v, which is the definition of streamline. dx u Thus, if stream function is known, infinite number of streamlines can be drawn to define a particular flow field pattern. In cylindrical coordinate systems (i.e. r, θ and z), the stream function for a plane incompressible, two-dimensional flow can be written as, vr ψ ψ = and vθ = r θ r () 6

72 Module 04; Lecture 8 where v and v θ are velocity components in radial and tangential direction, ψ ( r, θ ) is r the stream function. It may be noted that ψ ( r, ) cylindrical coordinate system i.e. Velocity Potential ( rv ) r vθ + = 0 r r r θ θ, also satisfy the continuity equation in In case of irrotational flows, the velocity components are related by Eq. (7). This equation can also be satisfied if the a new scalar function ( x, yz, ) that, φ φ u =, v=, w= φ x y z ) () φ is introduced such This scalar function φ ( x, yzt,, is known as velocity potential. In vector form, the above equation for irrotational flow can be written as, For incompressible flows, the conservation of mass equation is, V (3) = φ (4).V = 0 (5) Hence, for incompressible and irrotational flow, the above equation becomes,. V =. φ = φ = φ φ φ + + = 0 x y z This equation is known as Laplace equation. Thus, inviscid, incompressible, irrotational flow fields are governed by Laplace equation and these flows are characterized by potential flow. Lines of constant φ are called equipotential lines of 0 (6) the flow. The Laplace equation in cylindrical coordinates (i.e. r, θ and z), is, φ φ φ r = (7) r r r r θ z where φ = φ( r, θ, z). The velocity components in the radial and tangential direction are, 7

73 Module 04; Lecture 8 φ φ φ vr =, vθ =, vz = r r θ z Relation between Streamline and Potential lines It may be noted that velocity potential is a consequence of the irrotationality of the flow field where as the stream function is a consequence of conservation of mass. For incompressible and irrotational flow there exists on orthogonal relation among streamlines and potential lines. In addition both φ and ψ satisfy Laplace equation. In x-y (8) plane, we have, φ ψ u = = x y ψ φ v = = x y (9) For, line of constant φ, φ φ dφ = dx + dy = udx + vdy = 0 so that dy = u x y dx v ψ ψ Similarly, along the streamlines, dψ = dx + dy = vdx + udy = 0 so that dy = v x y dx u Thus, it may be concluded that dy dx ψ = constant v = = u dy dx φ = constant Eq. (30) is the mathematical condition for lines of constant φ and ψ. However, it is not (30) true at the stagnation point, where both u and v are zero. For any potential flow field, a flow net can be drawn that consists of a family of streamlines and equipotential lines. The flow nets are useful graphical tools in visualizing flow patterns. Examples The flow field of a fluid is given by ˆ ˆ ( ) V = xyi + yzj yz+ z k. Show that it represents a possible three-dimensional incompressible flow. Also verify whether the flow is ˆ 8

74 Module 04; Lecture 8 rotational or irrotational. If rotational, then determine at a point (, 4,6 ) : (a) angular velocity; (b) vorticity; (c) shear strain; (d) linear strains. Solution: Here, ( ) u = xy, v= yz, w= yz+ z u v w = y, = z, = y+ x y y ( z) Thus, by continuity equation, u v w + + = 0 is satisfied. x y z Hence, the given flow field represents a steady incompressible flow. Now, u v u x, 0 i.e. v = = y x y x Therefore, the given flow is rotational. The angular velocity can be written as, iˆ ˆj kˆ At (, 4,6 ), ω = ( 4iˆ + k ˆ ) ω = V = x y z Vorticity at (, 4,6 ), ξ = ω = ( 4iˆ + kˆ ) Shear strains, ( ) xy yz yz + z = ( z y) iˆ ( 0) ˆj+ ( x) kˆ 9

75 Module 04; Lecture 8 Linear strains, ( ) u = xy, v= yz, w= yz+ z τ τ τ xy yz xz v u = + = x y v ω = + = z y u ω = + = 0 z x ( ) u = xy, v= yz, w= yz+ z u ε x = = y = 4 x v ε y = = z = y ω ε z = = ( y+ z) = 6 z EXERCISES. Check whether the following functions represent possible irrotational flow. (i) φ = x y + z ; (ii) φ = sin ( x + y+ z) ; Ax (iii) φ = x + y ; (iv) φ = Ur cosθ. A stream function ψ = 5xy in which ψ is in cm /s and x, y are in mts. At section AB (x=3m), find: (i) velocity at B; (ii) acceleration at B; (iii) flow per unit width across AB. Assume the flow to be incompressible. 3. The velocity components in a two-dimensional flow field for an incompressible fluid 3 3 y x is expressed as, u = + x x y; v= xy y 3 3 (a) Show that it represents a possible case of an irrotational flow. (b) Obtain the expressions for stream function and velocity potential. 0

76 Module 04; Lecture 9 BASIC PLANE POTENTIAL FLOWS DYMAMICS OF FLUID FLOW One of the major advantages of Laplace equation is the linearity of partial differential equation. Arithmetic operations (e.g. addition, subtraction etc.) can be performed for the solutions of these equations. It has lot of practical implications that leads to interesting solutions of complicated flow problems. Some of the basic potential flows are discussed below. Uniform Flow It is the simplest type of flow in which the streamlines are straight and parallel. The magnitude of the velocity is constant. Fig. (a) and (b) shows the uniform flow in the positive x-direction. Mathematically, the flow represented in Fig. (a) can be expressed as, u = U and v = 0 () y U φ = φ φ = φ ψ = ψ y U φ = φ φ = φ ψ = ψ ψ = ψ ψ = ψ ψ = ψ3 ψ = ψ4 α ψ = ψ3 ψ = ψ4 (a) x (b) x Fig. : Schematic of a uniform flow: (a) positive x-direction; (b) any arbitrary direction. In terms of velocity potential and stream function, we can write, φ φ = U = 0 () x y ψ ψ = U = 0 y x (3)

77 Module 04; Lecture 9 The above two equations can be integrated (constants of integration may be discarded as it does not affect the velocities in the flow) to yield, φ = Ux and ψ = Uy (4) If the uniform flow is at an angle α with respect to positive x-direction, then Source and Sink ( cos ysin ) ( cos xsin ) φ = U x α + α ψ = U y α α Source and sink are the hypothetical terms used in fluid flow where it is assumed that the flow takes place radially (inward/outward) from origin. Consider a radial fluid flow outward from a line through origin as shown in Fig.. If unit length) along the radial line from the origin, velocities, then by conservation of mass principle, we can write, vr m (5) is the mass flow rate (per and v θ are the tangential and radial m m= ( π r). vr or vr = (6) π r ψ = constant φ = constant r θ Fig. : Streamline and equipotential lines for source. Since the flow is purely radial, so v θ = 0. By definition of velocity potential in streamline coordinate system, we have φ m φ =, = 0 r πr r θ Integrating Eq. (7) and putting the constant of integration to zero, (7) m φ = ln r (8) π In this expression, if m is positive, then the flow will be radially outward and is treated as source flow. A sink flow will occur when the flow is towards origin ( m is

78 Module 04; Lecture 9 negative). The radial velocity becomes infinite at r = 0 which is practically impossible. Thus, sources and sinks do not really exist in real flow fields rather some real flows can be approximated at points away from the origin by using sources and sinks. The stream function for the source can be defined such that ψ ψ m = 0, = r r θ π r m ψ = θ π It may be inferred from Eqs. (8) & (9) and Fig. that the streamlines the radial lines and the equipotential lines at the origin. ( ψ = constant) (9) are ( φ = constant) are concentric circles centered Vortex Flow The streamline patterns in a vortex flow are the concentric circles and the equipotential lines are along the direction of radial lines (Fig. 3). φ = constant ψ = constant θ r Fig. 3: Streamline and equipotential lines for vortex. Hence, the equation of motion for streamlines and velocity potentials can be written as, φ = Kθ and ψ = Kln r (0) By definition of streamlines and velocity potentials, we have, So, vr φ φ = = 0, vθ = r r θ ; v ψ ψ r = = 0, vθ = r θ r 3

79 Module 04; Lecture 9 v θ φ ψ K = = = r θ r r It indicates that the tangential velocity varies inversely with distance from the origin with becomes infinite at origin ( r = 0). A vortex motion may be rotational or irrotational depending on the orientation of fluid element in the flow field. The irrotational vortex will occur when the fluid element does not rotate about its own axis and is not decided by the path followed by the element. The irrotational vortex is also called as free vortex and is governed by Eq. (0). In case of rotational vortex (also referred to forced vortex ) the fluid element is artificially rotated with certain angular velocity ( ω ) about its axis. So, the constant K in Eq. (0) is replaced by ω. A combined vortex may be defined as a forced vortex with central core and free vortex behavior outside the core. Mathematically, it is written as, where r 0 v = ωr r r v θ θ 0 K = r > r r corresponds to the radius of the central core. 0 () () Circulation ( Γ) A vortex motion is mathematically associated with a term called circulation which is defined as the line integral of the tangential component of the velocity taken around a closed curve in the flow field. It is expressed as, Γ= Vds. (3) where the integration is performed around any arbitrary closed curve C, velocity vector and irrotational flow, Therefore, ds C is the is the differential length along the curve as shown in Fig. 4. For V = φ so that Vds. = φ. ds= dφ V (4) Γ = dφ = 0 (5) C 4

80 Module 04; Lecture 9 Any arbitrary curve C ds V Fig. 4: Circulation around a closed curve C. In general, the circulation is zero for irrortational flow. However, it is not true in case of free vortex defined by φ = Kθ (Eq. ) where the circulation around a circular path can be represented by, π Γ Γ= Kdθ = πk sothat K = (6) π 0 Now, the velocity potential and stream function for free vortex can be expressed in terms of circulation as, Γ Γ φ = θ and ψ = ln r (7) π π Doublet The source and sink of equal strength located along same axis can be combined to form another basic flow known as doublet (Fig. 5). The combined stream function for the pair can be written as, m ψ = ( θ θ ) (8) π y P r r r source a θ θ a sink θ x It follows from Eq. (8) that Fig. 5: Schematic representation of a doublet. 5

81 Module 04; Lecture 9 πψ tanθ tanθ tan tan ( ) = θ θ = m + tanθ tanθ From Fig. 5, we get, rsinθ rsinθ tan θ =, tanθ = rcosθ a rcosθ + a Now, Eq. (9) can be simplified as, For small angles, (9) (0) πψ ar sinθ m ar sinθ tan = or ψ = tan () m r a π r a tan ar sinθ ar sinθ, so from Eq. () r a r a mar sinθ ψ = π ( r a ) In Eq. (), if a 0andm keeping the product Hence, () ma π constant, then r r a r K sinθ ψ = (3) r ma where K = is called the strength of doublet. Thus, a doublet is formed as a source π and sink of equal strength approach one another while increasing their strength. The stream lines are governed by Eq. (3) and the corresponding velocity potential is, K cosθ φ = (4) r 6

82 Module 04; Lecture 9 Example c The tangential velocity variation of velocity in a washbasin is given by, V =. Referring r to the following figure, dtermine the circulation: (a) around a closed curve formed by two streamlines with r = R and r = R and two radius vectors with an angle θ between them; (b) around a closed curve in the form of concentric circle of radius R R C B θ R D A Solution: By definition, circulation is defined by, Γ= Vds. C Referring to above figure, there is no velocity along the radial direction i.e. Γ =Γ = 0 BC DA Circulation around ABCD, Γ =Γ +Γ +Γ +Γ ABCD AB BC CD DA = πk + 0 πk + 0= 0 The circulation in the form of concentric circle of radius R = π K This is a free vortex. 7

83 Module 04; Lecture 9 EXERCISES. Consider the superposition of a source with a uniform flow stream as shown in the following figure. y P U r θ r b x uniform flow source The stagnation point is created at a distance b from the source where velocities for both the flow are equal in magnitude and opposite in direction. If m is volume flow rate emanating from the line of the source, U is the uniform velocity, determine; (a) Location of stagnation point b (b) Radial and tangential velocity at a point P downstream of the flow as shown in the figure. A two-dimensional incompressible flow field described by equation V = Cr in which V is the tangential velocity and C is a constant. Determine circulation: (i) around a circle with radius R ; (ii) around a closed path formed by the arcs of two circles of radii R and R with an angle θ between them. Also, find the vorticity of the flow. 8

84 Module 04; Lecture 30 DYMAMICS OF FLUID FLOW FLOW THROUGH MOUTHPIECE A mouthpiece is a short tube of length not more than two to three times its diameter, which is fitted to a tank for measuring discharge of the flow from the tank. By fitting the mouthpiece, the discharge through an orifice of the tank can be increased. Mouthpieces are classified on the basis of their shape, position and discharge conditions. According to the shape, they may be classified as, cylindrical, convergent, divergent and convergent-divergent. Based on the positions, they may be external or internal mouthpieces with respect to reservoir/tank to which it is connected. An external mouthpiece projects outside the tank whereas the internal mouthpiece projects inside the tank. On the basis of discharge conditions, they may be classified as running full and running free mouthpieces. Flow through an External Cylindrical Mouthpiece Consider a cylindrical mouthpiece of cross-sectional area a, which is attached externally to the tank as shown in Fig.. The tank is filled with a liquid of specific weight γ up to a constant height h above the center of the mouthpiece. The discharge from the tank through the mouthpiece, compared to that of orifice, can be increased by running the mouthpiece full so that the jets of liquid emerging from the mouthpiece will be of same diameter as that of mouthpiece. maintaining a sufficient pressure-head in the tank so as to achieve the mouthpiece running full. A mouthpiece will be running full, if its length is equal to about two to three times its diameter and the pressure head in the tank is maintained at some critical level. Any deviation in full running condition of mouthpiece will result in the formation of venacontracta as in case of orifice. Referring to the Fig., if h and h represents the absolute a c pressure head at section b-b (atmospheric) and at vena-contracta, h L is the head loss through the mouthpiece, V c is the velocity of the jet at vena-contracta, then applying

85 Module 04; Lecture 30 Bernoulli s equation between the free surface of the liquid in the tank and section b-b, we get, Water level h C b C Area ac b Area a Fig. : Flow through an external cylindrical mouthpiece. V ha + h= ha + + hl () g Similarly, between the free surface of the liquid and section c-c Vc ha + h= hc + () g The expression for head loss h L can be written as, where C c = h L V = g Cc ac a is the contraction coefficient. Hence, Eq. () can be written as, V h = + g Cc Substituting Eq. (3) in Eq. () and using the definition of contraction coefficient, we get, (3) hc = ha + Cv h Cc (4)

86 Module 04; Lecture 30 The minimum possible value up to which the pressure at vena-contracta may be reduced is the absolute zero pressure. The limiting value of available pressure head in the tank corresponding to zero pressure head at vena-contracta is given by, ha h = Cv C c The discharge through mouthpiece can be written as, (5) Q= C.. d a gh (6) where, C d is the coefficient of discharge of the mouthpiece. Flow through a Convergent -Divergent Mouthpiece The formation of vena-contracta and subsequent enlargement of the jet causes the loss of energy, which results in reducing the coefficient of discharge of the mouthpiece. The energy loss of the jet can be minimized by designing the shape of the mouthpiece similar to that of the flow pattern of the jet at the entry, vena-contracta and exit of the mouthpiece. So, the mouthpiece is made to conform the shape of the jet up to venacontracta, then making it gradually diverging. Such a mouthpiece is known as convergent-divergent mouthpiece (Fig. ). In such a mouthpiece, the velocity of the jet increases in the convergent portion at the expense of pressure up to vena-contracta where the jet experiences minimum area. Beyond the point of vena-contracta, the pressure increases, as a result velocity of the jet drops down. However, the pressure at venacontracta cannot be reduced below absolute zero that limits the maximum divergence to be provided for the mouthpiece. Referring to Fig. and applying Bernoulli s equation between free liquid surface in the reservoir, the vena-contracta (section-cc) and exit of the mouthpiece (section-bb), Vc V h + a h = h + c ha g = + g (7) where h is the atmospheric pressure head in terms of flowing liquid, h is the free liquid is the absolute pressure head at venaa surface above the center of the mouthpiece, h c 3

87 Module 04; Lecture 30 contracta, V c and V are the velocities of flow at vena-contracta and exit of the mouthpiece respectively. Water level h C b C Area ac b Area a From the Eq. (7), Fig. : Convergent-divergent mouthpiece. V Vc = h and = ha + h h g g If a and a are the cross-sectional areas at vena-contracta and the outlet end of the c mouthpiece, then by continuity equation, av. = av. c c c a Vc ha h or, = = + c (8) a V h c Flow through an Internal Mouthpiece An internal mouthpiece fitted in to a tank or reservoir projects in to the tank and is generally of cylindrical shape only. It is also called re-entrant or Borda s mouthpiece and can be operated in running free or running full. When the internal mouthpiece runs full, the flow pattern is same as in the case of external cylindrical mouthpiece because the mouthpiece is sufficiently long enough to expand the jet of liquid completely (Fig. 3-a). 4

88 Module 04; Lecture 30 Water level Water level h h a b Area a a b Vc (a) (b) Fig. 3: Internal mouthpiece; (a) running full; (b) running free. In the case, if the internal mouthpiece runs free, the length of the mouthpiece is small enough to ensure full expansion of the jet. As a result, a vena-contracta is formed as shown in Fig. 3-b. (a) Free running Borda s mouthpiece Let us consider an internal mouthpiece as shown in Fig. 3-b. It has a cross-sectional area a, discharging the liquid under a constant head h above the mouthpiece. Since the mouthpiece is running free, a vena-contracta will be formed. Let the cross-sectional area and velocity of the emerging jet at vena-contracta be a and V respectively. Now, Static thrust on the fluid for the area a = γ ah γ av c c Rate of change of momentum of the jet by this static thrust = ( γ ah).. V g Equating the above two, by applying Newton s law of motion, we get, ac a c c c gh = (9) V Since, the jet is not expanding, so the loss of the energy can be neglected. Then, applying the energy equation between the free surface of the liquid and outside the mouthpiece, c Vc h + a h = h + a so that h g = V c g By combining Eqs. (9) and (0), (0) 5

89 Module 04; Lecture 30 C a a c c = = i.e. the coefficient of contraction for a Borda s mouthpiece is 0.5. (b) Borda s mouthpiece running full As discussed earlier, the flow pattern in this mouthpiece will be same as that of an external mouthpiece (Fig. 3-a). In this case, Applying Bernoulli s equation, where h L ac V Vc =. V = Cc = 0.5 a 0.5 ( ) V ha + h= ha + + hl () g V = is the head loss due to sudden enlargement. Then, solving for velocity V g from the Eq.(), we get Coefficient of velocity becomes, V = gh () C v = V gh V = gh = = th 6

90 Module 04; Lecture 30 FLOW OVER NOTCHES AND WEIRS A notch may be defined as an obstruction over which the flow of liquid occurs. As the depth of flow above the base of the notch is related to the discharge, the notch forms a useful measuring device. In case of measuring tank or reservoir, the opening is provided at the side of the tank such that the liquid surface in the tank is below the top edge of the opening. In fact, this is a large orifice, which has no upper edge, so that it has a variable area depending upon the level of the free surface. A weir is a notch on a large scale used for measuring the flow of a river. It is a concrete or masonry structure of substantial breadth built across the river in the direction of flow. This allows the excess water to flow over its entire length to the downstream side. Thus a weir is similar to a small dam constructed across the river, with a difference that the excess water flows downstream only through a small portion called spillway and incase of weir, the excess water flows over its entire length. The sheet of water flowing through a notch or over a weir is known as nappe or vein. The bottom edge of the notch or the top of a weir over which water flows is known as sill or crest. The height above the bottom of the tank or channel is known as crest height. Notches and weirs can be classified based on the followings; (a) According to the shape of the openings, notches/weirs may be classified Rectangular type Triangular or V-type Trapezoidal type Stepped type (b) Based on the shape of the crest, weirs are classified as, Sharp-crested weir Broad-crested weir Submerged weir Ogee-shaped weir Elementary Theory of Notches and Weirs There exists a considerable similarity between the patterns of flow over a notch/weir of same shape. As such, the same expressions will hold good for notch/weir of same shape. Moreover, due to this similarity, a notch is often termed as sharp-crested weir. 7

91 Module 04; Lecture 30 The method of determining the theoretical flow through a notch is the same as that adopted for the large orifice. For a notch of any shape (Fig. 4) with height H, consider a horizontal strip of width b at a depth h below the free surface. Fig. 4: Discharge through a notch/weir of any shape. Area of the elementary strip = b. δ h Velocity through the strip = gh Discharge through strip, δq= Area Velocity = b. δh gh Integrating from h = 0 at the free surface to h= H at the bottom of the notch, Total theoretical discharge, H Q = g bh. dh (3) th 0 Before the integration of the Eq. (3) to be carried out, b must be expressed in terms of h. As in the case of orifices, the actual discharge through a notch/weir can be found by multiplying the theoretical discharge by the discharge coefficient can be written for actual discharge as, Special cases 0 ( C d ). Then Eq. (3) H Q= C g bh. dh (4) d For a rectangular notch/weir (Fig. 5-a), b= constant = B (i.e. the width of the crest of the rectangular notch/weir), Eq. (4) can be written as, H 3 Q= BC. d g h dh= Cd. BH. g (5) 3 0 8

92 Module 04; Lecture 30 (a) (b) (c) (d) (e) Fig. 5: (a) Rectangular type; (b) Triangular/V- type; (c) Trapezoidal type; (d) Cipolletti type; (e) Stepped type. 9

93 Module 04; Lecture 30 In case of V-notch with included angle θ (Fig. 5-b), b ( H h) Eq. (4) gives, θ =.tan. So, H 3 5 H ( ) d Q. Cd g tan θ H h h dh. C g tan θ = = H. h. h 3 or, 5 8 θ Q= Cd. H g tan (6) 5 Referring to the Fig. 5-c, a trapezoidal weir/notch is a combination of a rectangular and a V-type weir/notch. The discharge over such notch/weir can be determined by adding the individual discharges over the two different types i.e., θ Q= Cdr. BH. g + Cdv. H g tan (7) 3 5 where C dr and C dv are the discharge coefficients of rectangular and triangular weir or notch respectively. A particular type of trapezoidal weir with included angle of 4 0 is called Cipolletti weir, invented by an Italian engineer Cipolletti in 887. For this weir, the decrease in the discharge over rectangular weir due to end contraction is compensated by the increase in discharge through the two triangular portions. Hence, the total discharge over a Cipolletti weir may be computed using the Eq. (5) as used for rectangular weir (Fig. 5-d). A stepped notch is a combination of rectangular notches shown in Fig. 5-e. The discharge through a stepped notch is equal to sum of the discharges through different rectangular notches. Referring to Fig. 5-e, the total discharge for a stepped notch can be calculated as, Q= Q + Q + Q 3 = g C. B. ( H H ) + C. B. ( H H ) + C. B H d d 3 d3 3 3 (8) 0

94 Module 04; Lecture 30 Example- Determine the discharge through a 00mm diameter internal mouthpiece under a head of m when it flows freely. What will be the discharge and pressure at vena-contracta if it runs full? Take Solution C c = 0.5 and atmospheric pressure as 0.33 m of water Internal mouthpieces running free and full are shown in Fig. 3-a and 3-b respectively. (a) For free flow conditions, a vena-contracta is formed as shown. Applying Bernoulli s equation, between section and at the vena-contracta (section c- c) Vc H = i.e. Vc = gh = 9.8 = 6.7 m s g π ac = Cc a = = m 4 So, ( ) Then, Q = V a = = = c c m s 4.7 lits s (b) For free flow conditions, a vena-contracta is formed but the pressure difference is sufficient to expand the jet at exit. Applying Bernoulli s equation, between section and Further, continuity equation yields, ( V V ) V c H = + g g VCa c c = Va ie.. Vc = V Cc = V Hence, V = gh = 9.8 = 4.43m s Q = V a = = = m s 7.4lits s (c) Applying Bernoulli s equation, between section and section c-c p V V H + H = + and av = av i.e. V = = V = 8.86 m s c c a c c c ρg g Cc where H is the atmospheric pressure head. p 8.86 H a H m i.e. pc 8.7kPa ρ g = = + = = c or, ( )

95 Module 04; Lecture 30 EXERCISES. A tank carrying water is fitted with a cylindrical mouthpiece fitted externally. When water flows in the mouthpiece, a vena-contracta is formed at some section of the mouthpiece. For a given height of water in the tank and diameter of the vena-contracta, determine the diameter of the mouthpiece for which the flow rate is maximum.. A convergent-divergent mouthpiece is fitted to the side of a tank. At constant head of.7m, the discharge is 00lits/min. The head loss in the divergent portion is 0. times the kinetic head at outlet. Find the throat and exit diameters if the separation pressure is.5m and the atmospheric pressure head is 0.33m of water. 3. An external cylindrical mouthpiece of diameter cm is discharging water under a head of 6m. Find (i) Discharge through the mouthpiece (ii) Absolute pressure head of water at vena-contracta Take C = 0.88, C = 0.60 and atmospheric head is 0.33m of water. d c 4. An external mouthpiece converges from inlet up to the vena-contracta to the shape of the jet and than diverges gradually. The diameter of the vena-contracta is cm and the head of the water over the center of the mouthpiece is.5m. The head loss in the contraction is % and that in divergent portion is 5% of total energy before the inlet. The pressure in the system falls up to 8m below atmosphere. Find the maximum discharge that can be drawn through outlet and the corresponding diameter at the outlet. 5. A streamlined nozzle of diameter d is supplied at constant water head whose magnitude is larger compared to d. The nozzle discharges water directly into the atmosphere and is so shaped that the issuing jet is parallel at the nozzle exit. In order to increase the flow rate a shroud of the diameter D is firmly secured to the nozzle as shown in the figure. Free water surface Shroud

96 Module 04; Lecture 30 The jet expands to fill the shroud and the shroud is long enough to ensure that the flow leaving is steady and parallel. Determine, (i) the diameter of the shroud in order to achieve maximum flow rate. (ii) percentage increase in discharge. Neglect shear stresses at the walls of the shroud. 3

97 Module 04; Lecture 3 FLOW OVER NOTCHES AND WEIRS Velocity of Approach DYMAMICS OF FLUID FLOW Continued It is defined as the velocity with which the flow approaches/reaches the notch/weir before it flows past it. The velocity of approach for any horizontal element across the notch depends only on its depth below the free surface. In most of the cases such as flow over a notch/weir in the side of the reservoir, the velocity of approach may be neglected. But, for the notch/weir placed at the end of the narrow channel, the velocity of approach to the weir will be substantial and the head producing the flow will be increased by the kinetic energy of the approaching liquid. Thus, if additional head H a is the velocity of approach, then the due to velocity of approach, acts on the water flowing over the notch or weir. So, the initial and final height of water over the notch/weir will be V a ( H + H a ) and H a respectively. It may be determined by finding the discharge over the notch/weir neglecting the velocity of approach i.e. V a Q = () A where Q is the discharge over the notch/weir and A is the cross-sectional area of channel on the upstream side of the weir/notch. Additional head corresponding to the velocity of approach will be, H a α. Va = () g and α being the kinetic energy correction factor to allow for the non-uniformity of velocity in the cross-section of the channel. For example, the discharge over a rectangular notch/weir of width B 3 Q= Cd. B. g. H 3...without velocity of approach 3 3 = C... ( ) d B g H + Ha Ha 3...with velocity of approach (3)

98 Module 04; Lecture 3 Empirical formula for discharge over rectangular weirs A rectangular weir is frequently used for measuring the rate of flow of water in channels. However, many researchers have conducted number of experimental investigations and proposed some empirical relations commonly used for rectangular weirs. Some of them are described below. (a) Francis formula: It is one of the most commonly used formula for computing the discharge over a sharp or narrow crested weirs with and without end contractions. Based on this formula, the discharge is expressed by, 3 3 Q.84 B 0.n( H H ) = ( ) + a H + Ha H a (4) where n is the number of end contractions. (b) Bazin s formula: Based on this formula, the discharge over a rectangular weir is given by, where m = H + H a ( ) 3 Q= m. g. B. H + Ha (5) is the Bazin coefficient (c) Rehbock s formula: Based on the experiments conducted by Rehbock, the following empirical formula is proposed; where z is the crest height in meters. H 0.00 Q = g.. 3 Z H 3 BH (6) Sharp-Crested Weirs A sharp-crested weir is essentially a vertical sharp-edged flat plate placed across the channel in a way such that the fluid must flow across the sharp edge and drop into the pool downstream of the weir plate as shown in Fig.. The specific shape of the flow area in the plane of the weir plate may be of rectangular/triangular/trapezoidal type. The main forces governing flow over a weir are gravity and inertia. The gravity accelerates the fluid from its free surface elevation upstream of the weir to a larger velocity as it flows down the hill formed by the nappe. Although viscous and surface

99 Module 04; Lecture 3 tension effects are usually of secondary importance, such effects cannot be entirely neglected. Generally, appropriate experimentally determined coefficients such as Francis, Bazin s and Rehbock s formulae are used to account for these effects. Q H Nappe z Weir plate Fig. : Sharp-crested weir geometry. Broad-Crested Weirs Broad-crested weirs differ from thin-plate and narrow-crested weirs by the fact that different flow pattern is developed. Experimental investigations have shown that if the length of the crest of the weir ( ) Lw < 0.65H i.e. H L w >.6, the jet of water touches only the upstream edge and flows clear of the downstream. Weirs falling under these classes are called thin-plate weirs. On the other hand, if ( ) 0.5 H L w.6, the jet of water remains in contact with the entire crest and these weirs are called narrow-crested weirs. In both the cases, the flow pattern is similar corresponding to that of a rectangular notch/weir. Water level () H V z () h B v Fig. : Broad-crested weir geometry. A broad-crested weir is a structure in an open channel that has a crest above which the fluid pressure may be considered hydrostatic. The typical configuration is shown in Lw 3

100 Module 04; Lecture 3 Fig.. Broad-crested weirs are operated in the range, ( ) 0.08 < H L w < 0.5 so that nearly uniform critical flow is achieved in the short reach above the weir block. For long weir blocks ( ) < 0.08 H L w hand, for short weir blocks ( ) > 0.5, head losses across the weir can not be neglected. On the other H L w, the streamlines of the flow over the weir block are not horizontal. Although, broad-crested weirs can be used in channels of any cross-sectional shape, but our attention will be limited to rectangular channels. Consider a broad-crested weir with length, width and height of crest as Lw, B and z. Referring to the Fig., Bernoulli s equation can be applied between sections a-a upstream of the weir and section b-b over the weir i.e. or, if the upstream velocity head is negligible, then The discharge over the broad-crested weir is given by, Va v H + z + h z g = + + g (7) v H h= or v= g( H h) (8) g ( ) Q= C. dlw h g H h (9) In order to measure the discharge over the broad-crested weir, two heads ( i.e. Hand ) h need to be measured. However, experiments have shown that the flow adjusts itself to have maximum discharge for the available head H. The downstream head over the weir can be computed mathematically by differentiating Eq. (9) with respect to h This value of and equating it to zero i.e. h dq h = Cd. Lw. g H h 0 dh = H h (0) or, h= H 3 is known as critical depth. In other words, the discharge over the broadcrested weir is maximum when the critical depth of flow occurs over the surface of the weir crest. The maximum discharge over the weir corresponding to critical depth will be, 4

101 Module 04; Lecture 3 Qmax = CdLw. H. g. H =.7 CdLw. H 3 3 If the velocity of approach ( is considered, then the above equation can be modified as, Submerged Weirs V a ) 3 a 3 () V Qmax =.7 CdLw. H + g () When the water level on the downstream of the weir is above the crest of the weir, then the weir is said to be submerged weir as shown in the Fig. 3. These weirs, constructed across the rivers have larger discharging capacity compared with freely discharging weirs and hence become more useful in discharging water during floods. Water level () Va H H () Lw Fig. 3: Submerged weir geometry. As shown in Fig. 3, the discharge over the submerged weir may be obtained by dividing it into two parts; the portion between the upstream and downstream water surfaces is treated as a free weir( ) Q the portion between the downstream water surface and crest of the weir is treated as drowned orifice ( Q ) If V is the velocity of approach, H and a H are the heads on the upstream and downstream of the weir and L w is the length of the weir, then 5

102 Module 04; Lecture 3 V Q = Cd g. L w ( H H) + 3 g g 3 3 a Va (3) where respectively. C and d C d ( ) ( ) Q = C. L H g H H + V d w a (4) are the discharge coefficients for the free and drowned portion Submergence ratio (SR) and Modular limit Submergence ratio may be defined as the ratio of heads available on the downstream side of the weir to the head available in the upstream side. Mathematically, SR H = Va H + g (5) Both sharp and broad crested weirs are susceptible to submergence depending on the values of SR. The sharp crested weirs behave, as a free weir only up to SR value of 0.66 and the corresponding values for broad crested weirs are This is because of the fact that the flow conditions are such that the downstream water level is held away from the crest and hence it does not affect the upstream flow conditions. The limiting value of SR up to which any submerged weir may behave, as free weir is known as Modular Limit. Ogee spillway, Siphon spillway and Proportional/Sutro weirs Ogee spillway A spillway is a portion of a dam over which the excess water, which cannot be stored in the reservoir formed on the upstream of the dam, flows to the downstream side. The profile of an ogee spillway conforms to the shape of the nappe of the sharp-crested weir of the same height as spillway and under the same head as shown in Fig. 4. The main advantage of providing such a shape for the spillway is that the flowing sheet of water remains in contact with the surface of the spillway and thereby preventing negative pressure being developed on the downstream side. This condition will be fulfilled as long 6

103 Module 04; Lecture 3 as the head over the crest of the spillway is equal to or less than the designed head. In the Fig. 4, if (HW) is less than 0.75, the discharge can be taken as that of a rectangular weir. Siphon spillway It is essentially an Ogee weir provided with an airtight cover and large rectangular section pipe connecting the upstream and downstream water surfaces. It allows the discharge of water at a controlled rate. It has the following advantages over Ogee spillway; Operating head and hence discharge is comparatively more. Since the crest of a siphon spillway can be raised, so it allows a greater amount of water to be stored in the reservoir. H Fig. 4: Ogee spillway and / or Siphon spillway. Proportional/Sutro weir The discharge over a weir is mainly proportional to the pressure head above the crest. For most of the weirs it is expressed as, n Q α H, where n = 3 for rectangular weir and n = 5for a triangular weir. In Proportional/Sutro weir as shown in Fig. 5, the discharge varies linearly with the H. Water level H L Fig. 5: Proportional weir. a 7

104 Module 04; Lecture 3 The discharge through this weir is given by, ( ) a Q C... d L g a = H (6) 3 where Cd is coefficient of discharge (0.6 to 0.65), L and a are the width and height of the rectangular shaped aperture that forms the base of the weir. Example It is proposed to use a notch for measuring the water flow from a reservoir. It is estimated that the error in measuring the head above the bottom of the notch could be.5mm. For a discharge of 0.3m 3 /s, determine the percentage error, which may occur, using a rightangled triangular notch with coefficient of discharge of 0.6. Solution: For a V-notch, 5 8 θ Q= Cd. H g tan 5 0 Taking C = 0.6 and θ = 90, d 8 90 Q= 0.6 H 9.8 tan =.47H When Now Or, Q= = 3 0.3m s H m 3 Q 5.5Q =.47 H = H H Q.5 H = = 00 = 0.7% Q H Example The stream of water from a waterfall of height 40m approaches a weir where the measured head is recorded as 0.3m. The length of the weir is 3m and the velocity of approach is.m/s. Determine, the power available at the waterfall. Use Bazin s formula with α =.5 for the flow over the weir. Solution: Q= m. g. B. H + Ha According to Bazin s formula, ( ) 3 8

105 Module 04; Lecture 3 where m = H + H a Now, H a ( ) α. V.5. a = = = 0.m g 9.8 and H = 0.3m, B = 3m So, m = = Power available at the fall = Example 3 ( ) 3 3 Q = =.7 m s ρ. gqh kw = = 667kW A rectangular channel 6m wide carries 68 lits/min at a depth of 0.9m. What height of a rectangular weir must be installed to double the depth? Discharge coefficient of weir may be taken as Solution: The discharge for a broad crested weir is given by, 3 a V Q=.7 CdLw. H + g 3 3 Here, Q = 68 m min =.8 m s ; L = 6m ; C = 0.85 Then, w a Q V H + = = = 0.47m g.7cdlw The depth of the flow required = 0.9 =.8m The velocity of approach is given by, V h a a Q.8 = = = 0.6m s Va = = m g H = = m Height of the broad crested weir = =.3334m. d 9

106 Module 04; Lecture 3 EXERCISES. A triangular notch is used to measure flow in a channel under a head of 0.m. If the discharge is to be measured within 3% accuracy, what is the maximum velocity of approach that can be neglected?. Water flows through a rectangular channel m wide and 0.5m deep and then over a sharp Cipolletti weir of crest length of 0.6m. If the water level in the channel is 0.5m above the weir crest, calculate the discharge over the weir. Take correction for velocity of approach. C d = 0.6 and make 3. A rectangular notch of crest width 0.4m is used to measure the flow of water in a rectangular channel of 0.6m wide and 0.45m deep. If the water level in the channel is 0.5m above the weir crest, find the discharge in the channel. For the notch, assume C d = 0.6 and take velocity of approach into account. 4. For the stepped notch shown below, find the discharge if C d = 0.6 for all the sections. 0.5m 0.3m 0.5m 0.4m 0.8m.m 5. A discharge of 3.6m 3 /min was measured over a right-angled notch. An error of 0.5cm was mead while measuring the head over the notch. Determine the percentage of error in discharge if the coefficient of discharge for the notch is The head of water over a triangular notch of angle 60 0 is 50cm and the coefficient of discharge is 0.6. The flow measured by it is to be within the accuracy of ±.%. Find the limiting values of the head. 7. Determine the discharge over.5m high sharp-crested weir fixed across m wide rectangular channel when the head over the weir is 0.05m. 8. In a 5m wide rectangular channel with.m depth of flow, a sharp-crested weir of.5m length and 0.6m height is fixed symmetrically across the channel width. If it flows free, determine the discharge. 0

107 Module 04; Lecture 3 9. A sharp crested weir of m height is fixed across 5m wide channel as shown in the following figure. The depth of flow on the upstream and downstream sides of the weir is.5m and.m respectively. Find the discharge over the weir and compare the discharge with that of a submerged weir. Water level H H y =.5m W = m y =.m 0. A river 30m wide and 3m deep has a mean velocity of.m/s. Find the height of a weir to raise the water level by m.. A spillway 40m long having discharge coefficient.8 permits a maximum discharge 90m 3 /s from a storage reservoir. It is proposed to replace the spillway by a siphon spillway of section 0.75m.5m with operating head 8m and discharge coefficient Find the number of siphons required and the amount of extra water stored, if the siphons have a priming depth of 0.5m. Take the average surface area of the reservoir as m.. Determine the discharge over an ogee spillway of 50m length under a head of.5m. What will be the depth of the flow at the toe of the dam if the height of the dam is 50m. 3. What will be the head required to carry a discharge of.75m 3 /s through a m wide gate at 0.3m opening under free flow conditions?

108 Module 04; Lecture 3 IIT Guwahati DYMAMICS OF FLUID FLOW OPEN CHANNEL FLOW In a simple statement, the open channel flow may be defined as the flow of liquid in a conduit, which is partially filled. Thus there always exists a free surface between flowing fluid (usually water) and fluid above (usually atmosphere) it. The main mechanisim for such flow is the fluid weight (i.e. gravity) and friction. This is in contrast to the flow through duct (either gas or liquid running full) driven by a pressure difference. There are many practical examples of open channel flows both natural (streams, rivers etc.) and artificial (flumes, spillways, canals etc.). General characteristics of open channel flow The presence of free surface of open channel flow sometimes makes the analysis simple where the pressure on the surface is equivalent to the hydraulic grade line of the flow. It also complicates the issue because the shape/geometry of the surface is unknown. Moreover, the flow is generally turbulent, sometimes unsteady and three-dimensional. The manner in which the fluid depth ( y ) varies with time ( t) and distance along the channel ( x ) is used to partially classify a flow. An open channel flow is classified as: dy uniform flow (UF) if the depth of flow does not vary along the channel = 0 dx dy non-uniform/varied flow if the depth of flow varies with distance 0 dx The non-uniform flow may be rapidly varying flow (RVF) if the flow depth changes dy considerably in a relatively small distance or gradually varying flow (GVF) dx dy in which the flow depth changes slowly along the channel dx The open channel flow may be laminar, transitional or turbulent depending on the Reynolds number of the flow. As a general rule, if R e < 500, the flow is laminar and the flow goes to turbulent if R >,500. e

109 Module 04; Lecture 3 IIT Guwahati Open channel flows also involve free surface that can deform to form waves and move across the surface. The speed of such waves depends on the size (length and height) and properties (depth, fluid velocity) of the channel. A dimensionless parameter (Froude number, F ) decides the characteristics of the flow in the channel. For F <, the flow is sub-critical (or tranquil); F =, the flow is critical and F >, the flow becomes supercritical (or rapid). r r r r Surface wave The most important feature of the open-channel flow is that the free surface can distort into various shapes as in the case of sea/oceans. These distortions are associated with surface wave. Consider an open channel in which water is stationary at initial time. Now, a single elementary wave of small height δ y produced on the surface of the channel by suddenly moving the initial stationary end wall with a speed δ V as shown in Fig. (a). Moving end wall δv δy δv (a) c y Stationary fluid x y+δy Control surface -c+δv -c () () Fig. : Surface wave in a channel: (a) single elementary wave moving with speed c ; (b) stationary wave. Let this wave travels with a speed c. Hence, for a stationary observer, it will be seen as a single wave moving down the channel with a wave speed c with no fluid motion ahead of the wave and a fluid velocity δ V behind the wave. For another observer moving along the channel with a velocity c, the same phenomena will be observed as if the fluid velocity of c moving right and c+ δv moving left of the observer as shown in Fig. (b). Now, for uniform one dimensional flow, the continuity equation becomes, ( δ )( δ ) cyb = c + V y + y b () (b) x

110 Module 04; Lecture 3 IIT Guwahati where b is the width of the channel. This equation is simplified further with assumption of small amplitude waves with δ y y as, δv c= y () δ y By momentum equation, we can write, γ γ δ ρ δ ( + ) = {( ) } yb y y b bcy c V c where Mass flow rate, m = ρbcy Pressure forces on the channel cross-section () and () are, Simplifying Eq. (3) with assumption ( ) b b F = γ y+ δy F = γ y ( ) ; δ y yδ y, we get (3) δv δ y Combination of Eqs. () and (4) results, g = (4) c c = gy (5) Thus, it is seen that the speed of the small amplitude wave is proportional to the square root of fluid depth and independent of the wave amplitude. Hence, such a wave motion is a balance between inertial effects and hydrostatic pressure effects. Sub-critical and Super-critical flows in open channel In order to study the characteristics of the fluid flow in an open channel, it is desirable to define a dimensionless parameter known as Froude number. It is defined as the ratio of fluid velocity to wave speed and is given by, V V F = r c = gy (6) Now, if the fluid is moving to the left with a velocity ( or ) travel to right with a speed of ( c V) V < c F r <, the wave will relative to the fixed observer; it means the upstream locations are in hydraulic communication with downstream. Such a flow is 3

111 Module 04; Lecture 3 IIT Guwahati known as sub-critical flow. On the other hand, if > ( or > ) ) V c F r, the wave will be washed to the left with speed (V c i.e. no upstream communication with downstream location is possible. This kind of flow is known as supercritical flow. As a special case, when V = c (or F r = ), the upstream propagating wave remains stationary and the flow is called critical. Energy of the flow in an open channel Consider a typical segment of an open channel flow as shown in Fig.. The fluid depth and velocities at two different locations () and () are y, y, Vand V respectively. Velocity head at () Energy line (Slope = Sf ) hl Velocity head at () y V y V z () Channel bottom (Slope = S0) () z The slope of the channel bottom More often, S 0 Horizontal datum Fig. : Schematic section of a open channel. S 0 = ( z z ) l is assumed constant over the segment. is very small and hence the bottom of the channel is nearly treated as horizontal. If there is uniform velocity profile across any section of the channel, the one dimensional energy equation for the flow can be written as, 4

112 Module 04; Lecture 3 IIT Guwahati or, p V p V + + z = + + z +h γ g γ g V V y+ + S0l = y + + Sf l g g L (7) p p where hydrostatic pressures at sections () and () are y = and y γ = γ. The head loss hl is expressed in terms of slope of energy line (or friction slope) as, Eq. (7) reduces to, S f h l L =. Thus, V V y y = + ( Sf S0) l (8) g In a special case, when the bottom of the channel is horizontal with no head loss, Eq. (8) is simplified to, V V y y = (9) g Variation of channel depth in an open channel The depth of the flow in an open channel changes with distance along the channel. By using the concept of energy line and Froude number, it is possible to control the depth of the flow in open channel. Referring to Fig., the total head at any section is given by, V H = + y+ z (0) g Also between section () and (), the energy equation can be written as, H = H+ hl () dz dh dhl Slope of the channel bottom and the energy line are, S0 = ; Sf = = dx dx dx Now, differentiating Eq. (0) with respect to x, So, dh V dv dy dz = + + () dx g dx dx dx 5

113 Module 04; Lecture 3 IIT Guwahati VdV g dx dy + = Sf S0 (3) dx For a given flow rate per unit width and flow rate ( q ) in a rectangular channel; V = q y Eqs. (3) and (4) leads to, dv q dy V dy = = (4) dx y dx y dx dy dx Sf S = (5) F It is seen that the change of depth in an open channel depends on the local slope of the channel bottom, energy line and Froude number. It can be zero, positive or negative depending upon the values of S, and 0 S F. f r 0 r Example Water enters to a rectangular channel of uniform cross section at a velocity of 8m/s. The depth of water is 50cm. Check, if the flow is sub-critical or super-critical and then calculate the critical depth and corresponding velocity. Also, calculate the specific head at the inlet and at critical condition. Solution V Given that, V = 8m/s; h = 0.5m Fr = = 3.6 gh Since F r >, so the flow is supercritical. Total head at the inlet, H V 8 = h + = = 3.76m g 9.8 The channel is of uniform cross-section, so the continuity equation for inlet and critical section can be Vh Vc At critical depth, Fr = = gh 3 = Vh c c = = 4m /s per meter width c 6

114 Module 04; Lecture 3 IIT Guwahati V = gh V = g V h = 4g = 39.4 V = 3.4 m s 3 So, ( ) c c c c c c 4 h c = =.7m 3.4 Example Consider a open channel flow of trapezoidal cross section. If b is the base width of the channel, y is the depth of flow, θ is the angle made by the sides with horizontal and side slope is in N ( vertical to N horizontal), then show that for most economical section, (i) b+ Ny = y N + (ii) Hydraulic radius R = (iii) Best side slope is at 60 0 to the horizontal. y (iv) Length of the sloping side is equal to the base width of the channel. Solution (i) The following figure shows the cross-section of a trapezoidal section. b+ny C D Area of the flow, Wetted perimeter, θ y A B b Ny [ ] AB + CD b+ b+ Ny A = y= y= ( b+ Ny) y A b= Ny y 7

115 Module 04; Lecture 3 IIT Guwahati For most economical section, P= AD+ AB+ AC = AB+ BC = + + = + + b N y y b y N A P= Ny+ y N + y dp dy = 0 d A Ny y N dy + + y A N N y or, 0 or, + = + Substituting the value of A in the above equation and then simplifying, we get (ii) Hydraulic radius is defined as, b+ Ny = y N + = i.e. Half of the top width = Length of the sloping side. ( ) A b+ Ny y R = = P b + y N + Since, b+ Ny = y N + P= ( b+ Ny) So, R = y, i.e. hydraulic radius is equal to half of the flow depth. (iii) Side slope will be the best when the section is most economical and wetted perimeter is minimum i.e. dp dn = 0 d A Ny y N dy + + y or, 0 N = N + or, or, 4 N = N + or, N = 3 So, tanθ = = 3 N 0 Hence, θ = 60 = 8

116 Module 04; Lecture 3 IIT Guwahati Referring to the figure, it can be concluded that best side slope is at 60 0 to the horizontal. (iv) For most economical section, half of the top width = Length of sloping side. i.e. y since N =, so b = 3 3 b+ Ny = y N + Substituting these values wetted perimeter becomes, P= b+ y N + = 3b Thus, for a side slope of 60 0, the length of the sloping side is equal to base width of the trapezoidal section. Example 3 Water flows up a 0.5m tall ramp in a constant width rectangular channel at a rate 0.5 m 3 /s. If the upstream flow depth is 0.7m, determine the elevation of water surface down stream of the ramp. Also predict the flow conditions at the upstream and down stream. Solution Water flow over a ramp is shown in the following figure. For the given conditions, q 0.5 z = 0; z = 0.5m; y = 0.7m; V = = = 0.7m/s y 0.7 Free surface of water V y y V z z=0 Applying Bernoulli s equation; Ramp V V y+ + z = y + + g g z So, 0.7 V = y V y + =

117 Module 04; Lecture 3 IIT Guwahati By continuity equation, yv = yv yv = 0.5 Thus, 0.7 V = y y = y 9.8 y 0.576y = 0 3 Solving the above cubic equation by trial and error, the value of y = 0.56m; y = 0.9m; y = 0.4m It should be noted that the negative solution is meaningless and the other two solutions are realistic one. Hence, the elevation of water surface at the downstream of the flow is, y y + z = 0.666m + z = 0.34m The flow velocity at the downstream may be, For, y = 0.56m V = 0.97m/s y = 0.9m V =.6m/s The flow condition at the upstream and downstream depends on Froude s number i.e. At the upstream of the flow, F V F r = V gy r = = = < gy At the downstream of the flow, Sub-critical flow When, 0.97 y = 0.56m V = 0.97m/s Fr = = 0.43 < Sub-critical flow When,.6 y = 0.9m V =.6m/s Fr = =.9 > Super-critical flow

118 Module 04; Lecture 3 IIT Guwahati EXERCISES. In order to measure the volume flow rate in a rectangular channel flow, the width is reduced gradually from 4m to.5m and the floor is raised by 0.4m at a given section. If the depth of flow approaching the channel is m, then determine the rate of flow indicated by.8m drop in water surface elevation.. A trapezoidal section of base width 6m and side slopes of 7 0 carries a flow 60m 3 /s at a depth of.5m. There is a smooth transition of flow from trapezoidal section to rectangular section of 6m width. (i) If the channel bed is horizontal, determine whether the upstream flow is possible as specified; if not determine the upstream depth of the flow. (ii) Determine the minimum amount by which the bed must be lowered for the upstream flow to be possible as specified. (iii) Determine the depth of flow in the rectangular section and change in the water surface level if the transition is accompanied by gradual lowering of the channel bed by 0.6m. (iv) Determine the amount by which the bed must be lowered if the drop in water surface is to be restricted to 0.3m. 3. Calculate the possible depths of flow in a rectangular channel 3.5m wide with specific energy of 3m for a discharge of 30m 3 /s. 4. Water flows at a velocity of 0.9m/s and a depth of.8m in an open channel of rectangular cross-section of 3m width. At a certain section, the width is reduced to.7m and the bed is raised by 0.5m. To what extent, will the upstream depth be affected? 5. A trapezoidal channel with side slopes of in has to be designed to convey 0m 3 /s at a velocity of m/s so that the amount of concrete lining for the bed and sides is minimum. Calculate the area of the lining required per meter length of the canal. 6. For a most economical circular section of diameter D in an open channel flow; determine, depth of water flow and hydraulic mean depth in terms of D for the following conditions: (i) maximum velocity of flow; (ii) maximum discharge.

119 Module 04; Lecture 3 IIT Guwahati 7. Consider a 4m wide rectangular channel carrying flow of 0m 3 /s at a depth of m. (i) Determine the width to which the channel should be contracted so that flow is critical in the contracted section. (ii) What will be the depth at the contracted section if the width at that section is 4m? (iii)what will be the depths of flow in the upstream and contracted section if the width of the channel is reduced to.5? 8. A 3m wide rectangular channel carries a discharge of m 3 /s at a depth of m. If the width of the channel is reduced to.5m and the bed level is lowered by 0.97m, determine the difference in water elevations between the upstream and contracted section. 9. Water flows in a 4m wide rectangular channel at a depth of m and velocity of.5m/s. If the channel is contracted to m and the bed is raised by 0.5m, what will be the depth in the upstream?

120 Module 04; Lecture 33 DYMAMICS OF FLUID FLOW UNIFORM DEPTH OPEN CHANNEL FLOW dy In an open channel, if the depth of flow is constant = 0, then it is treated as dx uniform depth. In some cases, the uniform depth flow can be accomplished by adjusting the bottom slope so that it precisely equals to the slope of the energy line. Physically, the loss of potential energy of the fluid as it flows downhill is exactly balanced by the dissipation of energy through viscous effects. Consider the flow in an open channel as shown in Fig.. The cross-sectional area is constant in shape and size. If the cross-sectional area is A and the wetted perimeter (i.e. length of perimeter of the cross section in contact with fluid) is P, then a new parameter may be defined as hydraulic radius i.e. R h A = () P Free surface a Free surface Flow area (A) Flow in (Q) a Flow out (Q) Section at a-a Wetted perimeter (P) Fig. : Concept of hydraulic radius. Since the fluid must adhere to the solid surface, the actual velocity distribution is not uniform. The velocity is maximum on the free surface and becomes zero on the wetted perimeter where the wall shear stress ( τ w ) is developed. Chezy and Manning Equations The fundamental relations used to determine the uniform flow rate in open channel are semi-empirical and are governed by Chezy and Manning equations. Consider a control volume flow with volume flow rate of Q and weight W in an open channel as shown in

121 Module 04; Lecture 33 Fig.. Since it is uniform depth flow, so it can be shown from continuity equation that V = V. Then, x -momentum equation for the control volume can be written as, Fx = ρq( V V) = 0 () or, F F τ Pl+ Wsinθ = 0 w Control surface τ = 0 F l y = y V = V () θ W τwpl () F θ x Fig. : Control volume flow in an open channel. As the flow is at uniform depth, so the hydrostatic pressure forces across either end of the control volume balance each other i.e. F = F. Thus, Eq. () becomes, ( γ Al) S0 W sinθ τ w = = =γ RhS0 () Pl Pl where γ is the specific weight of the fluid and sinθ tan θ = S0 (since S0 ). More often the open channel flows are turbulent rather that laminar. So, Reynolds number is quite large and for such flows, the wall shear stress ( τ w ) is proportional to dynamic pressure and may be written as τ w = K ρv (3) where K is a constant that depends on the roughness of the pipe. Now, equating Eqs. () and (3), we get, V = C R S (4) h 0 This equation is known as Chezy equation in which C is called Chezy coefficient. Its value is determined from experiment and also, it has the unit of (m) s This equation is modified by Manning by incorporating dependence of hydraulic radius on the bottom slope and is given by.

122 Module 04; Lecture 33 In this equation, the parameter 3 Rh S0 V = (5) n is the Manning resistance coefficient. Its value depends on the surface material of the channel s wetted perimeter and is obtained from experiment. It is also not dimensionless and has the unit of n are given in Table. n s ( m ) 3 Table : Values for Manning coefficient ( n ) (Ref. ; Table 0.) Wetted perimeter Natural channels Floodplains Excavated earth channel Artificially lined channel n. The typical values of In open channel flows, sometimes it is necessary to determine the best possible hydraulic cross-section (i.e. minimum area) for a given flow rate ( ) roughness coefficient ( n). The flow rate can be written as, Q, slope ( and S 0 ) which can be rearranged as, 3 A A S0 P Q = (6) n nq A= S P (7) For a channel with given flow rate the quantity in the parentheses is constant which means the channel with minimum A is also with minimum P. GRADUALLY AND RAPIDLY VARIED OPEN CHANNEL FLOW The depth of the open channel flow varies (either increases or decreases) in the flow direction depending upon the bottom slope and energy line slope. Physically, the difference between component of weight and shear forces in the direction of flow 3

123 Module 04; Lecture 33 produces a change in fluid momentum. Thus, there is a change in velocity and consequently a change in depth. The shape of the surface y y( x) = can be calculated by solving the governing equation obtained form combination of Manning equation and energy equation. The result will be a nonlinear differential equation, which is beyond the scope of this syllabus. However, some physical interpretation of gradually varied flows can be made from the following equation; dy For 0 dx, the factor Sf S0 Fr dy dx gradually or rapidly varying flow. Now, the sign of dy dx Sf S = (8) F 0 r becomes a non-zero quantity, which is essentially the i.e. whether the flow depth increases or decreases with distance along the channel depends on both numerator and denominator of Eq. (8). The sign of denominator depends on whether the flow is subcritical or super-critical. In fact, for a given channel, there exists a critical slope ( S 0 0c = S ) and a corresponding critical depth ( y= y c ) that leads to F = under conditions of uniform flow. ( ) S 0 The character of a gradually varied flow is classified in terms of actual channel slope compared to that of slope required for producing uniform flow r ( S 0c ). They may be, Mild slope with S < S (i.e. the flow would be sub-critical F <, if it were of uniform depth) 0 0c Steep slope with S > S (the flow would be super-critical F >, if it were of uniform depth) 0 0c Horizontal slope with S 0 = 0 Adverse slope, S 0 < 0 (i.e. flow uphill) Thus, it may be inferred from the above that the determination of whether the flow is sub-critical or super-critical depends solely on whether S0 < S0c or S0 > S0c respectively. For gradually varying flows, the conditions for sub-critical and super-critical flows are r r 4

124 Module 04; Lecture 33 met by S0 and depth of flow. For example, with S0 < S0c, it is possible to have either F < or F > r r depending upon the depth of flow i.e. y < yc or y > y c respectively. dy A rapidly varied flow in an open channel is characterized by i.e. the flow dx depth changes occur over a relatively short distance. One such example of a rapidly varied flow is hydraulic jump in which flow changes from a relatively shallow, highspeed condition into a relatively deep, low-speed condition within a horizontal distance of just a few channel depths. The mathematical analysis of hydraulic jump will be discussed in the subsequent lectures. Many open channel flow-measuring devices are based on the principle associated with rapidly varied flows. These devices include broadcrested weirs, sharp-crested weirs and sluice gate. Example Water flows in an open channel of trapezoidal cross section with a velocity of 0.9m/s at a rate 4m 3 /s. The bed slope and side slopes are :500 and : respectively. Find the depth and bottom width of the channel. Take Chezy s constant as Solution Given that, 3 Discharge, Q = 4m /s ; Velocity, V = 0.9m/s ; Bed slope, S 0 = ; Side slope, N = 500 Area of the flow, Q A = = 5.57m V V 0.9 = h 0 h = = 500 =.4m C S By Chezy s formula, V C R S R ( ) The area of the flow and wetted perimeter for a trapezoidal section is given by, By definition of hydraulic radius, Thus, R h ( ) 5.57m A= b+ Ny y = P= b+ y N + = b+.83y A 5.57 = = =.4 b+.83y =.55 P b+.83y 5

125 Module 04; Lecture 33 ( ) ( ) ( ) A= b+ Ny y = b+ y y = y+ y y = 5.57 =.55y.83y 5.57 y y+ = Solving the above quadratic equation, we get, y =.63m ; Thus, b = 7.9 m. Hence, the required depth and bottom width of the channel are.63m and 7.9m respectively. Example A trapezoidal channel with base width m and side slope of : carries water with a depth of m. The bed slope is in 65. Calculate the discharge and average shear stress at the channel boundary. Take Manning coefficient as n = Solution Given that, b= m; y = m; N = ( / ) ; S 0 = ; Side slope, N = 65 The area of the flow and wetted perimeter for a trapezoidal section is given by, ( ).5m A= b+ Ny y = P= b+ y N + = 4.4m By definition of hydraulic radius, Using Manning s formula, R h A.5 = = = 0.59 P 4.4 Discharge, 3 3 h S ( ) ( ) R V = = = 0.95m/s n Q= A V =.375m /s 0.5 Average shear stress at the channel boundary, ( ) τ = ρgr S0 = / 65 = 9.6 N m h 6

126 Module 04; Lecture 33 Example 3 Water flows in channel (cross-section of the shape of isosceles triangle) of bed width a and sides making an angle 45 0 with the bed. Determine the relation between depth of flow d and the bed width a for the condition of: (i) maximum velocity; (ii) maximum discharge. Use Manning s formula. Solution Water flow in a channel of isosceles triangular cross-section is shown in the following figure for which depth of flow and bed width are d and a, respectively. G H J (a/) d 45 deg. 45 deg. E F d a d R (i) By Manning equation, V = S n 3 h 0 The area of the flow and wetted perimeter for the above cross-section is given by, ( ) EF + JH a d + a d A A = d = d = ( a d) d; = a d d d d P P= EF + FJ + EH = a+ d; = d d For a given bed slope, the velocity will be maximum when, Thus by substitution, ( ) d R h d A P d A 0; 0; P d A P 0 d d = = = dd d d d d.83d + ad a = 0 Solving the above quadratic equation, d = 0.34a (ii) Discharge, 3 5 AR 3 h S0 A. Q= AV = = S n n P 0 7

127 Module 04; Lecture 33 For a given bed slope, the discharge will be maximum when, 5 ( ) d P d A Substituting and, we get, d d d d d A P d A d A = 0; 5P A 0 dd d d d d =.63d +.5ad 5a = 0 Solving the above quadratic equation, d = 0.44a Example 4 A rectangular channel of 5m width and.m deep has a slope of in 000 and is lined with rubber for which the Manning s coefficient is It is desired to increase the discharge to a maximum by changing the section so that the channel has same amount of lining. Find the new dimensions and probable increase in discharge. Solution Using Manning s formula, the discharge through the channel is given by, Q AR S n 3 h 0 = AV. = A S0 = ; A= 5. = 6m ; P = 5 +. = 7.4m; Rh = = P Here, ( ) Substituting the values, Let b and y 3 Q = 9.7 m s be the width and depth of the flow for the new section of the channel. In order to have the same amount of lining, P = b+ y = 7.4 For the discharge to be maximum in a rectangular channel, it can be proved that b = y Solving above two equation, y =.85m; b= 3.7m So, the area of cross-section and hydraulic radius for new channel cross-section becomes, A A= = = P 6.845m ; Rh 0.95m 8

128 Module 04; Lecture 33 3 ARh S0 3 By Manning s formula, new discharge, Q = AV. = =.m s n Q Q Percentage increase in discharge = 00 = 4.45% Q EXERCISES. Design a concrete lined channel (trapezoidal cross section) to carry a discharge of 480m 3 /s at a velocity of.3m/s. The bed slope and the side slope of the channel are :4000 and : respectively. Take Manning s roughness coefficient for lining as A canal of trapezoidal cross-section has a bed width of 8m and bed slope of :4000. If the depth of flow is.4m and side slopes are :3, then determine the average flow velocity and discharge in the channel. Also, compute the average shear stress at the channel boundary. 3. A trapezoidal canal is to carry 50m 3 /s of water with a mean velocity of 0.6m/s. One side of the canal is vertical where as the other side has a slope of :3. Find the minimum hydraulic slope if the Manning s coefficient is Water flows in a channel as shown in the following figure. Find the discharge if Chezys constant is 60 and bed slope is in m 0.7m 0.7m 5. Water flows in canal with bed slope of :300. The cross-section of the canal is shown in the following figure. Estimate the discharge when the depth of water is.5m. Assume Chezy s constant as 40. 9

129 Module 04; Lecture m.5m 5m 40m Side slope = vertical to horizontal 6. A concrete lined circular channel of 0.6m diameter has a bed slope of : 500. Find the depth of flow when the discharge is 0.3m 3 /s. Also, determine the velocity and flow rate for conditions of: (i) maximum velocity; (ii) maximum discharge. Assume Chezy s constant as 40. 0

130 Module 04; Lecture 34 DYMAMICS OF FLUID FLOW SPECIFIC ENERGY AND CRITICAL DEPTH In open channel flows one of the important parameter is specific energy and is defined as, V E = y+ () g where y is the water depth and V is the flow velocity. It is also called as energy grade line (EGL). For a given flow rate, there are two possible states for the same specific energy as shown in Fig.. Horizontal hf EGL E Velocity head y Fig. : Illustrative sketch for specific energy. In a simpler case, consider two possible states of specific energy in a rectangular channel of width Eq. () can now be written as, b. The discharge per unit width for this channel is given by, Q q = = V. y b q E = y+ () gy For a given channel of constant width, the value of q remains constant along the channel although the depth y may vary. The variation of q with y is plotted in specific energy diagram (Fig. ). From this curve, it is clear that specific energy attains to a minimum

131 Module 04; Lecture 34 value at certain depth for a given q. This depth is known as critical depth and it can be de obtained by setting = 0 in Eq. (). dy y Constant q Sub-critical yc Critical Super-critical 0 Emin E Fig. : Depth verses specific energy curve. Minimum specific energy occurs at, 3 3 q Q 3 y = yc = = ; E min = y g b. g The velocity of flow at critical depth is known as critical velocity ( and the corresponding discharge is q V y. c =. c c c V c ) (3) Referring to Fig., for E < E min, no solution exists and thus the flow is unrealistic. For E > E min, there are two possible solutions; Large depth with V < V c sub-critical flow. Small depth with V > V c s called super-critical flow. Frictionless flow over a bump Consider an open channel flow over a bump as shown in Fig. 3. The behavior of the free surface is sharply different based on the approach of the flow; i.e. sub-critical or supercritical. For frictionless two-dimensional flow, sections and are related by continuity and momentum equation;

132 Module 04; Lecture 34 V V yv = yv ; y+ = y+ + h g g Eliminating between the above cubic polynomial equations for water depth y over V the bump: 3. y Ey + V y 0 where E V y h g = = g + (4) Supercritical approach flow Subcritical approach flow y V y V Bump h Fig. 3: Frictionless two-dimensional flow over a bump. y y y yc hmax () h () Sub-critical bump Ec E E Super-critical bump Fig. 4: Specific energy plot for flow over a bump. E Following points may be noted down from the above analysis; The Eq. (4) has one negative and two positive solutions if h is not too large. Its behavior is illustrated in Fig. 4 and depends upon whether the point is on the upper or lower leg of the energy curve. 3

133 Module 04; Lecture 34 In sub-critical approach ( F r < ), the water level will decrease at the bump whereas in super-critical approach flows ( F r > ), the water level will increase over the bump. If the bump height reaches hmax = E Ec, the flow at the crest will be critical ( F r = ). In this case, no physical solution is possible i.e. too large bump will choke the channel and cause frictional effect, called hydraulic jump. HYDRAULIC JUMP A hydraulic jump is a discontinuity when there is a conflict between the upstream and downstream control parameters in an open channel flow. It occurs, when the upstream flow is fast and shallow, and the downstream flow is slow and deep and thus provides a mechanism to make a transition between two types of flow. One such example is flow under the sluice gate where the downstream portion of the gate (upstream of the channel) is super-critical flow, while the flow is sub-critical in upstream side (downstream of the channel). Consider a simplest type of hydraulic jump that occurs in a horizontal, rectangular channel as shown in Fig. 5. Take two sections and in upstream and down streamside where the flow is nearly uniform, steady and one-dimensional. Neglecting the wall shear stress, the momentum equation can be written as, ( ) ρ... ( ) F F = ρq V V = V y b V V (5) where b is the channel width. The pressure force at either side is hydrostatic and acts at the channel cross-sections i.e. Using Eq. (6) in Eq. (5), we get, ρ. gy. ρ. gby.. F p A y b = c. =.(. ) = ρ. gy. ρ. gby.. F p A y b = c. =.(. ) = (6) y y V. y( V V = ) (7) g 4

134 Module 04; Lecture 34 Control volume hl () V Energy line () V F y y Q, F Shear stress = 0 Fig. 5: Hydraulic jump geometry. Similarly, the mass and energy conservation equations can be written for section and respectively i.e. ybv.. = ybv.. = Q (8) V V y+ = y + + hl (9) g g The head loss h L is due to violent turbulent mixing and dissipation that occurs within the jump itself. All Eqs. (7 to 9), are satisfied for y y V V h L = ; = and = 0. It represents a trivial case when there is no jump. The other possible solution can be obtained by combining Eqs. (7 and 8) to eliminate Simplification of this equation gives, V i.e. y y V. y V. y V. y = V = ( y y ) (0) g y g. y y y F r =0 y y V where Fr = is the upstream Froude number. The possible solution for Eq. () is gy y ( 8Fr ) () y = + + () h The dimensionless head loss, L y can then be obtained from Eq. (9) as, 5

135 Module 04; Lecture 34 hl y F r y = + y y y (3) For a given values of F, the depth ratios r y y are obtained from Eq. () and then the head loss h L y is calculated from Eq. (3). It will be negative if F r < (since the negative head loss violates second law of thermodynamics). The flow must be supercritical ( F r > ) to produce the discontinuity called as hydraulic jump and there is a considerable energy loss across the hydraulic jump. This in fact is extremely useful in many situations; e.g. the relatively large amount of energy contained in the fluid flowing down the spillway of dam causes damage to the channel below the dam. By placing suitable flow control objects in the channel downstream spillway, it is possible to produce hydraulic jump on the apron of the spillway and thereby dissipate a considerable portion of the energy of the flow i.e. the dam spillway produces super-critical flow and the channel downstream of the dam requires sub-critical flow. Hence, the hydraulic jump provides a means to change the character of the flow. Classification of hydraulic jump The principal parameter affecting hydraulic-jump performance is Froude number. The Reynolds number and the channel geometry have the secondary effect. Based on the Froude number, the hydraulic jumps are classified as; F r < : Jump is impossible as it violates second law of thermodynamics. F r = to.7 : Standing wave or undular jump; low dissipation less than 5%. F r =.7 to.5 : Smooth surface rise known as weak jump ; dissipation is 5 to 5%. F r =.5 to 4.5 : Unstable, Oscillating jump; each irregular pulsation creates a large wave which can travel downstream for miles, damaging earth banks and other structures. Dissipation is 5 to 45%. 6

136 Module 04; Lecture 34 F r = 4.5 to 9 : Stable, well balanced, steady jump; best performance and action, insensitive to downstream conditions. Dissipation is 45 to 70%. F > : Rough, intermittent strong jump but good performance. Dissipation is 70 r 9 to 85%. UNDERFLOW GATES These gates are typical structures constructed at the crest of an overflow spillway, or at the entrance of an irrigation canal/river for controlling the flow rate. Some of the typical structures are vertical gates (commonly called sluice gate), radial gates, and drum gates. Water level Water level Water level y a y (a) (b) (c) Fig. 6: Underflow gates; (a) vertical gate, (b) radial gate, (c) drum gate. The flow under the gate is said to be free flow when the fluid issues as a jet of supercritical flow with free surface open to atmosphere as shown in the Fig. 6. The discharge per unit width of the gate can be expressed as, The discharge coefficient q = C... d a g y (4) C is a function of the contraction coefficient ( C = y a) and d c depth ratio ( y a). The typical values of discharge coefficient from a vertical sluice gate with free out flow are of the order of 0.55 to 0.6. There are certain situations where the depth downstream of the gate is controlled by some downstream obstacle. The jet of water issuing from underflow gate is overlaid by mass of water that is quite turbulent (Fig. 7). Such a gate is known as drowned/submerged gate. The flow rate can be obtained from the same equation (Eq. 4) with appropriate modification in C d. 7

137 Module 04; Lecture 34 Water level y q y a Fig. 7: Drowned outflow from a sluice gate. Flow under a sluice gate The flow pattern under a sluice gate is shown in Fig. 8. If the flow is allowed free discharge (Fig. 8-a), then it smoothly accelerates from sub-critical (upstream) to critical (near the gap) to super-critical (downstream). For free discharge the friction may be neglected. Applying continuity and momentum equation, Eliminating, we get V V V V. y = V. y and + y = + y (5) g g 3 V V. y y + y y + = 0 g g Thus, for a given sub-critical upstream flow ( V, y ) (6), there is only one real solution i.e. super-critical flow at the same specific energy as shown in Fig. 8-b. The flow rate y varies with the ratio y y and reaches to a maximum when =. When the depth y 3 y contracts to 40% less than the gate s gap height, the flow pattern is similar to that of a free orifice discharge and can be approximated in the range H 0.5 y < as, 8

138 Module 04; Lecture 34 Q= C. H. b gy where C d d 0.6 (7) H y If the tail-water is high as in the case of Fig. 8-c, the free discharge is not possible. The sluice gate is said to be drowned or partially drowned. There will be energy dissipation in exit flow, in the form of drowned hydraulic jump and the downstream flow will return to sub-critical. Hence, Eqs (6 and 7) will not be applicable for such situations and experimental correlations are necessary. Water level y Sub-critical () Water level Dissipation High tail water V, y Vena contracta V, y (a) y y V, y () V, y Super-critical E = E E (b) (c) Fig. 8: Flow under a sluice gate; (a) Free discharge; (b) Specific energy for free discharge; (c) Dissipative flow under a drowned gate. Example A rectangular channel 6m wide carries 68 lits/min at a depth of 0.9m. What is the height of a rectangular weir which must be installed to double the depth? Discharge coefficient of weir may be taken as Solution The discharge for a broad crested weir is given by, 3 a V Q=.7 CdLw. H + g 3 3 Here, Q = 68 m min =.8 m s ; L = 6m ; C = 0.85 Then, w a Q V H + = = = 0.47m g.7cdlw d 9

139 Module 04; Lecture 34 The depth of the flow required = 0.9 =.8m The velocity of approach is given by, V h Q.8 = = = 0.6m s Va = = m g H = = m Height of the broad crested weir = =.3334m. a a Example Water flow in a wide channel approaches a 0cm high bump at.5m/s. and a depth of m. Estimate: (a) the water depth over the bump; (b) the bump height that will cause the crest flow to be critical. Take the head loss as 0.m and the flow is frictionless. Solution (a) First, the Froude number is calculated as, F V.5 r = = = gy 9.8 It means that the flow is sub-critical. Take two sections and in the entire length of the flow (Fig 8). Specific energy of the flow is, Applying continuity and energy equations, Eliminating V, (.5) 0.48 V E = + y = +.0 =.5m g 9.8 E = E h=.5 0. =.05m V V V. y = V. y and + y = + y + h g g 3. y Ey + V y = 0 where E V = + y h g g or, y.05y = 0 3 0

140 Module 04; Lecture 34 Three real roots of the above equation are, y = m, m and -0.9m. The negative root is physically impossible. For sub-critical conditions, super-critical condition, y =+ 0.45m. The surface level has dropped by y y h= = 0.04m. y =+ 0.86m and for V. y.5 The crest velocity, V = = =.745m s y 0.86 V.745 The Froude number at the crest is, Fr = = = 0.44 gy i.e. the flow downstream of the bump is sub-critical. (b) For critical flow, Froude s number is unity at the crest and E c 3 V 3 3. y 3 3 q = yc = = = = 0.98m g g 9.8 Maximum height for the bump, h= E E c = = 0.97m The cubic polynomial equation becomes, The solution is y = y c = 0.6m y V. y + = 0 g 3 Ec y or, y 0.98y = 0 3 The surface level has dropped by y y h= = 0.93m c Example 3 Water flows under a sluice gate on a horizontal bed at the inlet to a flume. The water depth is 50cm in the upstream of the gate and the speed is negligible. At the venacontracta downstream of the gate, the flow streamlines are straight and the depth is 6cm. Determine the flow speed downstream from the gate and discharge per unit width. Solution Referring to the Fig. 3-a, Bernoulli s equation can be applied upstream and at the venacontracta of the flow field as,

141 Module 04; Lecture 34 Solving for V, V V + gz = + gz ( ) ( ) V = g z z + V = =.94 m s The discharge per unit width is given by, Q q = = V D = = b m s EXERCISES. Water flows freely under a sluice gate with upstream depth of 5m and gate opening of.5m. Determine: (i) the discharge per unit width; (ii) water depth just upstream of the gate; (iii) what will be the discharge if the water depth immediately downstream of the gate is m; (iv) compare this value with estimation of discharge under submerged condition assuming the flow immediately downstream of the gate to be unaffected by submergence.. Water in an open channel flows under a sluice gate. The flow is incompressible and uniform at two sections and upstream and downstream of the flow respectively. The depth of water and velocity at the section are.5m and m/s respectively. The corresponding values in section are 0.05m and 5m/s respectively. Determine the direction and magnitude of the hydrostatic force per unit width exerted on the gate by the flow. 3. Water flows at a rate 0m 3 /s.m (per unit width) in a wide channel with upstream depth of.5m. If the water undergoes a hydraulic jump, compute the following parameters in the downstream of the gate: (i) depth of the water; (ii) velocity of the flow; (iii) Froude number; (iv) head loss; (v) percentage dissipation; (vi) the power dissipated per unit width; (vii) temperature rise due to dissipation if c p =4.kJ/kg. K. 4. A rectangular channel with a bottom slope of :50 carries water at a rate 0m 3 /s. Determine the width of the channel when the flow is in critical condition. Take Manning s coefficient as 0.06.

142 Module 04; Lecture In a rectangular canal 3.m wide is laid with a slope of The uniform flow occurs at a depth of m. How much height can the hump be raised without causing transition? If the upstream depth of flow is to be raised to.5m, what should be the height of the hump? Assume Manning s coefficient as Water flows steadily in a rectangular channel laid with a slope of The base width of the channel is 5m and depth of flow is m. It is desired to obtain a critical flow in the channel by providing a hump in the bed. Sketch the flow profile and calculate the height of the hump required. Take Manning s coefficient as 0.06 for channel surface. 7. A 3.6m wide rectangular channel conveys 0m 3 /s of water with a velocity of 6m/s. (i) Is there a condition for hydraulic jump to occur? If so, calculate the height, length and strength of the jump? (ii) What is the loss of energy per kg of water? 8. In a rectangular canal of 0.5m width, a hydraulic jump occurs at a point where the depth of flow is 0.m and Froude number is.5. Determine, (i) the specific energy; (ii) the critical and subsequent depths; (iii) loss of head; (iv) energy dissipated. 9. A hydraulic jump occurs in a V-shaped channel with side slope of Derive the expression for flow rate in terms of upstream and downstream depth. If the depths of flow before and after the jump are 0.4m and 0.8m, determine the flow rate and Froude number before and after the jump. 0. The depth and velocity of the flow in a rectangular channel are 0.9m and.5m/s respectively. If a gate at the downstream of the channel is abruptly closed, what will be the height and absolute velocity of the resulting surge? If the channel is 000m long, how much time will be required for the surge to reach the upstream end of the channel.. A rectangular canal carries a discharge of.8m 3 /s per meter width of the canal. The energy loss due to a hydraulic jump is found to be 3m. Determine the conjugate depths before and after the jump. 3

143 Module 04; Lecture 35 IIT Guwahati VISCOUS FLOW GENERAL CHARACTERISTICS OF PIPE FLOW The general method of transporting fluid (liquid or gas) is the flow through a closed conduit. It is commonly called pipe if it is of round cross-section and duct if the crosssection is not round. The common examples include water pipes, hydraulic hoses, air distribution in a duct in an air conditioning plant etc. The main driving force for the flow to occur is the pressure differential at both ends of the pipe and the walls of the pipe which is designed to withstand this pressure difference without any undue distortion of the shape. The basic governing equations such as mass, momentum and energy conservation can be applied for viscous, incompressible fluids in pipes and ducts. Following assumptions are made for present analysis; Cross-section of the duct is circular unless otherwise specified. The pipe is completely filled with fluid being transported otherwise the flow may be treated as open-channel flow where gravity alone is the driving force. The driving potential is the pressure difference across the pipe. The flow of fluid in a pipe may be laminar, turbulent or transitional depending upon the flow rates. Such a flow in the pipe is characterized by a dimensionless number called Reynolds number and is defined as, where R e ρvd = () µ ρ and µ are the density and viscosity of the flowing fluid, D is the diameter of the pipe and V is the average velocity in the pipe. In case of round pipes; For If R R e e < 00, flow is laminar > 4000, flow is turbulent R e falling between the two limits, the flow may switch between laminar or turbulent conditions in a random fashion and is characterized as transitional flow. The physical interpretation of the flow characteristics is shown in Fig.. It represents the x-component of velocity as a function of time at a point A in the flow. For laminar flow, there is only one component of the velocity, the streak-line is a well defined and

144 Module 04; Lecture 35 IIT Guwahati coincides with streamline ( V = uiˆ components velocity fluctuations relative to pipe axis predominantly unsteady. ). The turbulent flow is accompanied by random ( V = uiˆ+ v ˆj+ wkˆ ) and the flow is ua Turbulent Flow in A x Transitional Laminar t Fig. : Flow characteristics in a pipe. FULLY DEVELOPED FLOW The fluid typically enters to the pipe with nearly uniform velocity at some location. The region of the flow near which the fluid enters is known as entrance region. Referring to Fig. (a), the velocity profile at section () is nearly uniform. As the fluid moves through the pipe, the velocity at the wall approaches to zero due to viscous effect and is commonly called no-slip boundary condition. Thus a boundary layer is produced along the pipe wall such that the initial velocity profile changes with distance along the pipe. At the end of entrance length i.e. beyond the section (), the velocity profile does not vary with the distance. The boundary layer completely grows to fill the pipe where the viscous effects are predominant. For the fluid within the inviscid core and surrounding the centerline from () to (), the viscous effects are negligible. The flow between the section () and (3), is fully developed until there is any change in the character of the pipe. The shape of the velocity profile depends on whether the flow is laminar or turbulent which in turn affects the length of the entrance region ( l e ). The dimensionless entrance lengths le D can be correlated with Reynolds number as,

145 Module 04; Lecture 35 IIT Guwahati le = 0.06 Re (for laminar flow) D ( R ) 6 =4.4 (for turbulent flow) e Pressure differential is the driving potential for the flow through pipe. When the flow is fully developed, the pressure gradient is negative i.e. p p = x l. However, referring to Fig. (b), the pressure drop is more significant at the entrance region (outside the inviscid core) where the viscous effects are predominant. Entrance region flow Fully developed flow (a) Inviscid core Boundary layer r D x le () () (3) Entrance pressure drop Linear pressure drop in fully developed region (b) Pressure le x Fig. : (a) Illustration of fully developed flow in a horizontal pipe; (b) Pressure distribution along the horizontal pipe. Fully Developed Laminar Flow Consider a fluid element of length l and radius pipe of diameter time t D r centered on the axis of a horizontal in a fully developed laminar flow at time t (Fig. 3). After certain + δt, the fluid element moves to a new location and the flat end of the element becomes distorted. Since the flow is fully developed and steady, so the distortion on each end of the fluid element is the same and the convective as well as local acceleration is zero. Thus, every part of fluid element moves along a path line and the velocity varies from one path line to other. This velocity variation combined with fluid viscosity 3

146 Module 04; Lecture 35 IIT Guwahati produces shear stress. In addition to this shear force, the pressure drops ( p = p p) along the length of the pipe. Fluid element at time 't' Velocity profile Fluid element at time 't+δt' (a) r D l () () x (τ)(πrl) (b) p(πr) (p- p)(πr) l Fig. 3: (a) Motion of fluid element in a pipe; (b) Free body diagram of fluid element. Now applying Newton s second law, the force balance equation can be written as, ( ) ( ) ( ) After simplification of above equation, For a fully developed flow, where p πr p p πr τ πrl = 0 () p l p τ = (3) l r is constant and hence, τ = Cr (4) C is a constant. At r = 0; τ = 0 i.e. there is no shear stress at the centerline of the pipe. The shear stress at the pipe wall stress ( τ w ). Hence, τ w r = D is maximum and is known as wall shear C =. So, the shear stress distribution throughout the pipe is a D linear function of radial coordinate and is given by, τ w r τ w τ = = (5) D Substituting the value of τ from Eq. (5) and in Eq. (3), we get 4

147 Module 04; Lecture 35 IIT Guwahati p 4τ w τ = = w (6) l D r By definition, the wall shear stress for a laminar flow of a Newtonian fluid in a pipe is given by, du τ w = µ (7) dr The significance of negative sign is that the velocity decreases from the pipe centerline to the pipe wall. Substitution of the value τ w from Eq. (7) in Eq. (6) yields, du p = r dr µ l Now, integrating Eq. (8), the velocity profile is as follows; or, u p du = µ l p µ l rdr = r +C where C is a constant. Applying no-slip boundary condition, i.e. at r = 0, τ = 0 and u = 0. Hence, C p = D 6µl The velocity profile can now be written as, (8) (9) ( ) u r pd. r Vc r = = 6µl D D (0) where V c p. D = is the centerline velocity. An alternative expression can be written 6µl by using the relationship between wall shear stress and pressure gradient (Eqs. 6 and 0) is given by, ( ) u r τ D w r = 4µ D This velocity profile is parabolic in radial coordinate system. The velocity is maximum at ( ) V c the centerline of the pipe and becomes zero at the pipe wall. Integrating Eq. (), the volume flow rate can be obtained as, () 5

148 Module 04; Lecture 35 IIT Guwahati r= R R r Q = uda = u ( r). ( πr). dr = πvc rdr R r= 0 0 RVc π or, Q = Now, the average velocity in terms of volume flow rate is given by, so that for this flow, () Q Q V = A = π R (3) V Vc p D = = (4) 3µl 4 π D p Q = (5) 8µ l Following inferences about the flow rate can be made for a laminar flow ( R 00) in a horizontal pipe from the above analysis; It is directly proportional to the pressure drop It is inversely proportional to the viscosity It is inversely proportional to pipe length It is proportional to fourth power of pipe diameter This flow properties are established experimentally by two independent scientists (G. Hagen and J. Poiseuille) and is known as Hagen-Poiseuille flow. In case of non-horizontal pipes, the above expressions are slightly modified as; p γ lsinθ τ = (6) l r ( p γlsinθ) D V = (5) 3µ l 4 ( p lsin ) D π γ θ Q = (6) 8µ l where θ is the angle made by the pipe with respect to horizontal and γ is the specific weight of the flowing fluid. e 6

149 Module 04; Lecture 35 IIT Guwahati STEADY, LAMINAR FLOW BETWEEN FIXED PARALLEL PLATES Consider the flow between two horizontal, infinite parallel plates as shown in Fig. 4(a). For this geometry, the fluid particles move in the x-direction parallel to the plates and there is no velocity in the y and z direction i.e. v= w= 0. So the continuity equation can be written as, u = 0 (7) x Further, for steady flow, u = 0 (8) t u u h umax h (a) Both plates fixed (b) U Moving plate u b z y x Fixed plate (c) Fig. 4: Steady laminar flow between parallel plates. For infinite plates, there would be no variation of u in the z-direction i.e. u u( y) these conditions, the Navier-Stokes equation can be written as, =. With 7

150 Module 04; Lecture 35 p u + µ 0 = x y p ρg = 0 y p = 0 y IIT Guwahati (9a-c) Eqs. (9-b and c) can be integrated to yield, At ( ) ( ) p = ρgy + f x (0) y= 0, p= 0 f x = 0, i.e. pressure varies hydrostatically in y-direction only. Thus, Eq. (9-a) can be written as, Integrating Eq. () two times, c du dy p µ x p = µ x u = y + c y+ c The two constants c and can be found from boundary conditions. Referring to Fig. 4, () () since the plates are fixed, so at y = ± h, u = 0 (no-slip conditions for viscous fluids). In order to satisfy this boundary condition, c = 0 and distribution becomes, p u = µ x ( y h c ) p = µ x h. Thus, the velocity (3) The Eq. (3) indicates that the velocity profile between two fixed plates is parabolic and is shown in Fig. 4(c). The volume flow rate per unit width, passing through the plates is given by, h h p h p q = udy = y h dy = µ x 3µ x 3 ( ) (4) h h 8

151 Module 04; Lecture 35 IIT Guwahati Since the pressure decreases along the direction of the flow, so the pressure gradient is negative. Thus, if p represents the pressure drop between two points at a distance l, then p l p =. So, the Eq. (4) can be written as, x 3 h p q = (5) 3µl This expression clearly shows that the flow rate is proportional p directly to pressure gradient l 3 directly with cube of gap width ( h ) inversely to the viscosity ( ) µ In terms of mean velocity V, where V = q h, Eq. (5) becomes, h p V = (6) 3µl Referring to Eq. (3), the maximum velocity occurs midway between the plate, i.e. at y = 0, so that u max h p 3 = = V (7) µ x It may be noted that the flow remains laminar if the Reynolds number remains below 400. For higher Reynolds number flow, the above analysis will no longer be valid because the flow field becomes complex, three-dimensional and unsteady. Couette flow Another simple flow similar to that of fixed parallel plates can be developed by fixing one plate and allowing the other plate to move with a constant velocity U as given in Fig. 4 (c). In this case, the governing equations will remain the same, but the boundary conditions for the moving plates will be different i.e. y = 0, u = 0 and y = b, u = U. Then, the values of constants for Eq. () may be obtained as, 9

152 Module 04; Lecture 35 IIT Guwahati It follows from Eq. () that U b p c = and c = 0 (8) b µ x y p u= U + ( y by) (9) b µ x In dimensionless form, the above equation is written as, = U b µ U x b b u y b p y y This type of flow is known as Couette flow. The simplest type of Couette flow is the one for which the pressure gradient is zero i.e. the fluid motion is caused by the fluid p being dragged along the moving boundary. In such case, = 0, so Eq. (30) simplifies x to, It means that velocity varies linearly between two plates. (30) u y = (3) U b Example An inclined pipe (40 0 with horizontal) of circular cross-section (3cm radius) carries oil having density of 900kg/m 3 and kinematic viscosity of 0.000m /s. Between two section and of pipe 0m apart, the pressure are measured to be 350kPa and 50kpa respectively. Assuming steady laminar flow, (i) Show the direction of flow. (ii) Head loss between section and. (iii) Discharge and velocity (iv) Reynolds number of the flow (v) Is the flow really laminar? Solution The flow through the circular pipe is shown in the following figure. 0

153 Module 04; Lecture 35 IIT Guwahati () Q, V 3cm 0m p = 50kPa () 40 deg. p = 350kPa Q, V The elevation at section is given by, z L θ 0 = sin = 0 sin 40 = 6.43m Dynamic viscosity, µ = ρν = = 0.8kg m.s (i) The flow goes in the direction of falling hydraulic grade line (HGL). Thus, p ρ g HGL = z+ = 0 + = 39.65m p ρg HGL = z + = = 34.75m Since, HGL HGL, so the flow goes from section () to (). < (ii) Head loss between section and = HGL- HGL = 4.9m (iii) The discharge can be computed from, ( ) πd. p πd. ρgh π Q = = = = 8µ l 8µ l Q (iv) Average velocity, V = = =.68m s π D 4 π 0.06 (v) Reynolds number, R e ( ) ( ) ρvd = = = 804 µ m s (vi) Since this value is well below the transitional value for a pipe flow (i.e. 300), so the flow is certainly laminar.

154 Module 04; Lecture 35 IIT Guwahati Example The shaft diameter and bearing internal diameter in a journal bearing system are 00mm and 05mm respectively. The bearing length is 40mm and the viscosity of lubricating oil is 0.09 kg/m.s. Calculate the power required to overcome friction when the shaft is rotated at 500rpm and is concentric with the bearing. Solution: The flow of lubricating oil in a journal bearing can be treated as Couette flow because the film thickness is very small (Fig. ). In this case the gap width = = =.5mm. b r0 r i Thus the shear stress may be approximated as, π N 3 π 500 µ r µ U µ rω τ = = = = = 8.75 N m 3 b b b.5 0 Housing ω Rotating shaft Radius ri Radius ro Lubricating oil Frictional force, 3 3 F = πdlτ = π =.44N Frictional power, πn 500 P = Frω = Fr = = π Watts

155 Module 04; Lecture 35 IIT Guwahati EXERCISES. A liquid of specific weight of 900 N/m 3 flows by gravity from a tank of 0.3m length through a capillary tube of same length and mm diameter at a rate of 5m 3 /s. Neglecting entrance effects, compute the viscosity of the liquid.. A viscous liquid of density 50kg/m 3 and kinematic viscosity of m /s flows at cm 3 /s through a horizontal tube of 5mm diameter: (i) Determine the pressure drop per meter length of the tube; (ii) If a.5 mm diameter rod is placed in the 5 mm diameter tube to form a symmetric annulus, then calculate the pressure drop per meter length of the tube for same flow rate. 3. A viscous oil ( µ = 0.07 kg m.s) flows through a 0mm diameter pipe with a maximum velocity of m/s. calculate the pressure drop over 50m length of the pipe. 4. Lubricating oil with specific gravity of 0.85 and dynamic viscosity of.kg/m.s is pumped at the rate of 0.5m 3 /s through a pipe of 300m-length and 0cm diameter. Calculate the pressure drop, average shear stress at the wall of the pipe and power required to maintain the flow: (i) if the pipe is horizontal, (ii) the pipe is incline at 0 0 with horizontal for upward and downward flow. Also determine the slope of the pipe and direction of flow so that the pressure gradient along the pipe is zero. 5. For laminar flow of an oil of dynamic viscosity 0.07kg/m.s in a 5cm diameter pipe, the velocity distribution is parabolic with maximum velocity of.8m/s at the center. Calculate the shear stress at the pipe wall and 5cm from the pipe wall. 6. Using the basic principle of Navier-Stokes equations, derive the expressions for volume flow rate, average velocity and velocity distribution profile for a steady laminar flow in a circular tube. How these expressions can be modified for similar flow in an annulus. 7. A hydraulic system operates with a viscous fluid (dynamic viscosity 0.0 kg/m.s, specific gravity 0.9) at gauge pressure of MPa and 58 0 C. The control valve consists of a piston 0mm long with 5mm diameter, fitted to a cylinder with mean radial clearance of 0.005mm. Determine the leakage flow rate if the gauge pressure on the low-pressure side of the piston is MPa. 3

156 Module 04; Lecture 35 IIT Guwahati 8. The crankshaft journal bearing in an automobile engine is lubricated by SAE30 oil (dynamic viscosity 0.0 kg/m.s, specific gravity 0.9) at 00 0 C. The bearing has the length of 4cm and diameter of 8cm. The crankshaft rotates at 3000rpm and the diametric clearance between the shaft and the bearing is mm. Calculate the power required to overcome friction when there is no load on the bearing. 9. A liquid having dynamic viscosity of 0.0 kg/m.s, specific gravity 0.9 flows between two horizontal plates 5mm apart. If the average velocity is 0.3m/s, estimate the shear stress at the wall, 5mm and 0mm vertically from the wall. Plot the velocity and shear stress distributions over the channel depth. 0. A rectangular plate of.m.5m face area is drawn over a horizontal layer of fluid 5mm thick and supported by a solid plate at rest. The force required to draw the upper plate at a steady speed of 0.6m/s is 5N. Estimate the value of viscosity of the fluid and also plot the velocity profile. 4

157 Module 04; Lecture 36 IIT Guwahati VISCOUS FLOW TURBULENT FLOW THROUGH PIPES Unlike fully developed laminar flow in pipes, turbulent flow occurs more frequently in many practical situations. However, this phenomenon is more complex to analyze. Hence, many empirical relations are developed to understand the characteristics of common flow problems. Before, going into these solutions and empirical relations, first few concepts and characteristics of turbulent flows are discussed. Transition from Laminar to Turbulent Flow Consider a situation in which a water reservoir is connected to a pipe. The water is initially at rest and is allowed to flow through pipe and the flow rate is regulated by a valve. By opening the valve slowly, the flow velocity and hence the Reynolds number increases from zero to the maximum steady state value. For initial time period, the Reynolds number is small enough for laminar flow to occur. At some time, when the Reynolds number reaches 00, intermittent spots and random fluctuations appear indicating the flow transition to turbulent condition. This process continues till the Reynolds number value reaches 4000 beyond which the flow becomes fully turbulent. Random fluctuations Turbulent u Turbulent spots Re 4000 Transitional 00 Laminar Fig. : Transition phenomena in a pipe flow. t

158 Module 04; Lecture 36 IIT Guwahati This phenomena is typically shown in Fig. where the axial velocity component of flow at given location is given by u u( t) =. The flow characteristics such as pressure drop and heat transfer depends strongly on the nature of fluctuations and randomness. Reynolds Time-Averaging Concept in Turbulent Flow The fundamental difference between laminar and turbulent flow is the chaotic and random behavior of flow properties such as velocity, pressure, shear stress, temperature etc. One way to handle such high Reynolds number flow is to standardize in terms of mean/average value of flow parameters. Such a technique is known as Reynolds Time- Averaging Concept. In this method, the flow parameters are expressed in terms of two quantities; one is the time-average value and the other is the fluctuating part with respect to average value. u u' u(t) Time-Average value of u T t0 t0+t t Fig. : Time averaging concept in turbulent flows. For example, referring to Fig., if u u( x, y, z, t) is defined by, =, then mean value of turbulent function u T = u( x, y, z, t) T dt () 0

159 Module 04; Lecture 36 IIT Guwahati where T then defined by, is the averaging time period. The fluctuation part (or the time varying part) is Turbulent Intensity( ) ϕ u = u u () It is clear from Eqs. () and () that the fluctuation has zero mean value. However, the mean square value of fluctuating is not zero and is the measure of turbulent intensity. It is often defined as the ratio of square root of mean square of fluctuating velocity to the time average velocity. Mathematically, it may be written as; ( u ) avg where the mean square value of fluctuating velocity is, ϕ = (3) u 0 t + T ( u ) = ( u ) avg T t0 dt (4) The larger the turbulent intensity, the more will be fluctuation in parameters. Typical values of ϕ range form to 0.. Turbulent Stresses In order to define stresses in turbulent flows, let us write the x-momentum equation with time averaging and fluctuation terms; i.e. du p u u u ρ = + ρ g+ µ ρ u + µ ρ uv µ ρ uw avg avg + dt x x x y y z z (5) ( ) [ ] [ ] The three correlation terms ρ ( u ), ρ[ uv ] and ρ[ u w ] avg avg avg are called turbulent stresses because they have same dimensions as that of laminar shear stress terms i.e. u u u µ, µ and µ. The turbulent stresses are unknown and must be related to x y z experimental flow conditions and geometry. However, experiments in pipe flows reveal that the stress associated with ρ [ uv ] avg reasonable accuracy, the momentum equation is reduced to, in y-direction is dominant. Hence, with avg 3

160 Module 04; Lecture 36 IIT Guwahati du p τ ρ = + ρg + (6) dt x x where { [ uv ] avg} lam turb u τ = µ + ρ = τ + τ y The typical trend of a turbulent-shear layer for a pipe flow is shown in Fig. 3. It is seen that laminar shear is dominant near the wall whereas turbulent shear dominates in the outer layer. There is an intermediate region called overlap layer where both laminar and turbulent shear are important. (7) y y=δ(x) y U(x) τ(x,y) Outer turbulent layer τlam τturb u(x,y) Overlap layer Wall τw(x) 0 Viscous wall layer u(x) Fig. 3: Typical velocity and shear stress distribution in turbulent flow. An alternate form of shear stress for turbulent flow is given in terms of eddy viscosity ( η ) which is analogous to dynamic viscosity in case of laminar flow. It may be written as; du τ = η (8) dy ( avg ) In order to determine the Reynolds stresses in turbulent flows [ uv] ρ, several empirical theories have been attempted. The most common one is Prandtl s concept of mixing length. He proposed that the turbulent process could be viewed as the random transport of bundles of fluid particles over a certain distance ( l m ) from a region of one 4

161 Module 04; Lecture 36 IIT Guwahati velocity to another region of different velocity. This distance is called mixing length. In this mixing length, the eddy viscosity may be defined as, du η = ρlm dy (9) Thus, turbulent shear stress becomes, du turb = lm τ ρ dy (0) Turbulent Velocity Profile A fully developed turbulent flow in a pipe can be divided into three regions which are characterized by their distances from the wall: the viscous sub-layer very near to the pipe wall, the overlap region and the outer turbulent layer throughout the center portion of the flow. Within the viscous sub-layer the viscous shear stress is dominant compared to that of turbulent (or Reynolds) stress i.e. fluid viscosity plays a major role compared to fluid density. In the outer turbulent layer, Reynolds stresses (i.e. fluid density) are dominant and there is a considerable mixing and randomness to the flow. The character of flow within these two regions is entirely different. Considerable efforts have been made to determine the actual velocity profiles in pipe flows. Some of them are discussed here. In viscous sub-layer, the velocity profile is written as, where u u * * yu = () υ y = R r is the distance measured from the wall, υ is the kinematics viscosity of the flow, u is the time-averaged x-component of velocity and velocity defined by, * u is called friction τ = ρ u * w The friction velocity is not the actual fluid velocity rather it has same dimension as that of velocity. The Eq. () sometimes called as the law of wall. For smooth wall, Eq. () 0.5 () is valid very near to the wall for which * yu 0 5. υ 5

162 Module 04; Lecture 36 IIT Guwahati In case of overlap layer the following expression commonly known as Logarithmic Overlap Law has been proposed; * u yu =.5ln 5 * + (3) u υ The most often used correlation is the empirical power law velocity profile defined by, where u V c n r = R (4) V is the centerline velocity and n = 7 holds good for many practical flow c problems. Moody Chart The fundamental difference between laminar and turbulent flow is that the shear stress for laminar flow depends on the viscosity of the fluid whereas in case of turbulent flow, it is the function of density of the fluid. In general, the pressure drop p, for steady, incompressible turbulent flow in a horizontal round pipe of diameter D can be written in the functional form as, turb (,,,,, ) P = f V D l ε µρ (5) where V is the average velocity, l is the length of the pipe and ε is a measure of the roughness of the pipe wall. Similar expression can also be written for the case of laminar flow in which the ε term will be absent because the pressure drop in laminar flow is found to be independent of pipe roughness i.e. lam (,,,, ) By dimensional analysis treatment, we can found that P = F V D l µ ρ (6) P l ρ. VD. l φ, =. ρ V µ D lam P l ρ. VD. l ε φ,, =. ρ V µ D D turb (7) 6

163 Module 04; Lecture 36 IIT Guwahati The only difference between two expressions in Eq. (7) is that the term ( ε D), which is known as the relative roughness. In commercially available pipes, the roughness is not uniform; so it is correlated with pipe diameter and the contribution ( ε D) forms a significant value in friction factor calculation. From tests with commercial pipes, Moody gave the values for average pipe roughness listed in Table. Table : Average values of roughness for commercial pipes (Table 8.; Ref. ) Material ε (mm) Riveted steel Concrete Wood Cast iron 0.6 Galvanized iron 0.5 Commercial steel Plastic and glass 0 (smooth pipes) Now Eq. (7) can be simplified with reasonable assumption that the pressure drop is proportional to pipe length. It can be done only when, It can be rewritten as, P ρ. V l ε = φ Re, (8) D D l ρ. V P= f D (9) where f is known as friction factor and is defined by, ε f φ = R (0) D e, Now, recalling the energy equation for a steady incompressible flow, p V p V + + z = + + z + h L () γ g γ g 7

164 Module 04; Lecture 36 IIT Guwahati where h L is the head loss between two sections. With assumption of horizontal ( z = z ) constant diameter pipe ( D D or V V ) = = with fully developed flow, p = p p = γ h = ρgh () L L From Eqs. (9) and (), we can determine head loss as, h L l V = f D g (3) This is known as Darcy-Weisbach equation and is valid for fully developed, steady, incompressible horizontal pipe flow. If the flow is laminar, the friction factor will be ε independent on D and simply, 64 f = (4) R e The functional dependence of friction factor on the Reynolds number and relative roughness is rather complex. It is found from exhaustive set of experiments and is usually presented in the form of curve-fitting formula/data. The most common graphical representation of friction factor dependence on surface roughness and Reynolds number is shown in Moody Chart (Fig. 4). This chart is valid universally for all steady, fully developed, incompressible flows. The following inferences may be made from Moody chart (Fig. 4). For laminar flows ( 00 ), R < f ( R ) e = 64 e and is independent of surface roughness At very high Reynolds number( R > 4000), the flow becomes completely turbulent e (wholly turbulent flow) and is independent of Reynolds number. In this case, the laminar sub-layer is so thin that the surface roughness completely dominates the character of flow near the wall. The pressure drop responsible for turbulent shear stress is inertia dominated rather than viscous dominated as found in case of laminar viscous sub-layer. Hence, the friction factor is given by, f = φ( ε D) The friction factor at moderate Reynolds number ( ) dependent on both Reynolds number and relative roughness. < < is indeed R e 8

165 Module 04; Lecture 36 IIT Guwahati Even for smooth pipes ( ε = 0), the friction factor is not zero i.e. there is always head loss, no matter how smooth the pipe surface is. There is always some microscopic surface roughness that produces no-slip behavior are called hydraulically smooth. ( thus f 0) on the molecular level. Such pipes Fig. 4: The Moody chart: - friction factor as a function of Reynolds number and surface roughness for round pipes (Pg. 477; Reference ). It must be noted that Moody chart covers extremely wide range of flow parameters i.e. diameter of the pipes( D ), fluid density ( ρ ), viscosity ( ) non-laminar regions of the flow ( R 0 e ) ( ) µ and velocities V in < < that almost accommodates all applications of pipe flows. In the non-laminar regions of fluid flow, Moody chart can be represented by the empirical equation i.e. ε D.5 = log + f 3.7 Re f This equation is called Colebrook formula and is valid with 0% accuracy with the graphical data. (5) 9

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