Subject: Fluid Mechanics

Transcription

1 Subject: Fluid Mechanics Solution of Board exam question of TU, IOE (New Course) - by Dr. K. N. Dulal 2068, Baishakh (Regular exam) Attempt all questions. 1. Define stream line with its drawing equation. A steady, three dimensional velocity field is given by V = ( x+0.984y+az)i+(2.61+cx+1.91y+bz)j+(-2.73x-3.66y-3.64z)k. Calculate constants a, b, c such that the flow field is irrotational. [2+4] Stream line is an imaginary curve drawn through the flow field in such a way that the tangent to it at any point indicates the direction of velocity at that point. As stream lines join points of equal velocity, these are velocitycontours.it is useful to visualize the flow pattern. Equation of streamline is u = ( x+0.984y+az), v= (2.61+cx+1.91y+bz) and w= (-2.73x-3.66y-3.64z) Rotational components: (Curl of V)/2,, For irrotational flow,, ( )

2 a = 2.73, b = 3.66, c = Define the term pressure and skin friction drag with neat sketch. Describe briefly, with neat sketches, the changes in flow pattern and the drag coefficient with the variation of Reynold s number when a thin circular cylinder of infinite length is placed in a fluid stream. [2+4] Consider an arbitrary shaped solid body placed in a real fluid, which is flowing with a uniform velocity V in a horizontal direction. Consider a small elemental area da on the surface of the body. The forces acting on surface da are: pressure force acting perpendicular surface, shear force acting tangential to the surface. Let θ be the angle made by the pressure force with horizontal direction. PdA F L τ 0 da F R PdA PdAcosθ V θ F D PdAsinθ τ 0 da θ τ 0 dasinθ θ τ 0 dacosθ stationary body Drag force Drag force on elemental area = Total force in the direction of motion Total drag is The first term is known as pressure drag and the second term is known as friction drag.

3 Variation of flow around a cylinder with different Re Re = 2 to 30 Re = 40 to 70 Re = 90 For R e <0.2, the inertia force is negligible and the flow pattern is symmetrical. With the increase of R e,the flow pattern becomes unsymmetrical with respect to the axis perpendicular to the direction of flow. At Re from 2 to 30, very weak vortices are formed on the downstream of the cylinder. It is the initial stage for the development of the wake. At Re from 40 to 70, the wake as well as a pair of vortices become quite distinct. With further increase in the value of Re, the vortices become more and more elongated in the direction of flow. At Re = 90, these vortices become cylindrical, they leave the cylinder and slowly move in the downstream direction. Coefficient of drag (C D ) of cylinder for different Reynold no. (R e ) a. For R e <1, b. For R e between : C D decreases and reaches a minimum value of 0.95 at R e = c. For R e = : C D increases and attains a maximum value of 1.2 at R e = d. For R e = : C D decreases. At R e = , C D = 0.3 e. For R e >300000: C D increases and it becomes equal to 0.7 in the end.

4 3. Derive the expression for center of pressure on a plane inclined surface immersed in liquid. Also show pressure diagram for such surface. [6+2] O θ y p y z F CG CP O z p Consider a plane surface of area A totally immersed in a liquid of specific weight and inclined an angle θ to the free surface. Consider an element of area da at a vertical distance y from the free surface. P = pressure on each element, = vertical distance of CG from free surface, y p = vertical distance of CP from the free surface, F = resultant force, z = distance of element from O-O, = distance of CG from O-O and z p = distance of CP from O-O. Finding resultant force Force on element (df) = Summing the forces, Assuming to be constant Here, = first moment of area about an axis through O-O = Hence, Finding the position center of pressure Taking moment of force on element (dm) about O dm= df x z = Total moment of all forces (M) = = second moment of area about an axis through O-O = moment of inertia = I 0 (a)

5 Moment of resultant force F about O = (b) Equating a and b From parallel axis theorem where I G = M.I. about an axis through CG. Substituting for I 0 ( ) [ ( ) ] Pressure diagram for inclined surface

6 OR The swing check valve in figure covers a 22.86cm diameter circular opening in a slanted wall. The hinge is 15cm from the center line. The valve will open when the hinge moment is 50Nm. Find the value of h for the water to cause this condition. [8] Air 22.86cmdia Water Hinge h B O A cm y cp F 30cmdia Hinge moment = 50 Nm Area (A) = = 0.041m 2 Location of CG ( ) = h Resultant force on the valve is 9810x.041h = h Vertical distance of Center of pressure from the free surface y cp Position of CP from CG (BO) = Location of F from A (AB) = Taking moment about A, FxAB = 50 h = 0.82m 4. Consider a homogeneous right circular cylinder of length L, radius R, and specific gravity SG, floating in water (SG = 1) with its axis vertical. Show that the body is stable is [ ]. [8]

7 Length of cylinder = L Radius of cylinder = R Specific gravity of cylinder = SG h = depth of immersion Weight of cylinder = Weight of water displaced M G B h = SGxL 0 Position of center of buoyancy (OB) = SGxL/2 Position of CG (OG) = L/2 BG = OG-OB = GM = MB-BG = For stable equilibrium, GM 0 [ ] OR A U-tube shown in figure is filled with a liquid of specific gravity 1.25 to a height of 15cm in both the limbs. It is rotated about a vertical axis 15cm from one limb and 30cm from the other. If the speed of rotation is 60rpm, find the difference in the liquid levels in the two limbs. Also find the pressure at points M and N at the base of U-tube. [8] 30cm 15cm Z 1 Z 2 Z 0 N S 60rpm M

8 Distance from the axis of rotation to left side (r 1 ) = 0.3m Distance from the axis of rotation to right side (r 2 ) = 0.15m Speed of rotation (N) = 60 rpm Angular velocity = 6.28 rad/s = 0.181m = 0.045m Difference in level = = 0.136m Z 1 + Z 2 = 0.3 (the total height should be same until no spilling) Z 1 - Z 2 = Solving Z 1 = 0.218m, Z 1 = 0.082m Pressure at N = Pressure at M = = 1.25x9810x0.218 = 2673 Pa = 1.25x9810x0.082 = Pa 5. Write precisely how the boundary layer theory helped in simplifying the complex motion of fluid. Why does the boundary layer increase with distance from the upstream edge? A simplistic laminar boundary layer model is: for ( ) for Does this expression satisfy boundary conditions applicable to the laminar boundary layer velocity profile? [2+2+2] Boundary layer is a thin layer of the flow in the vicinity of the boundary within which the velocity varies from zero at the solid boundary to the free stream velocity in the direction normal to the boundary. In the boundary layer velocity gradient (du/dy) is large and the shear stress acts. Outside the boundary, velocity is constant and velocity gradient is zero and hence shear stress is zero. Hence, there are two regions of flows: one is the boundary layer zone close to the boundary where the effect of viscosity is mostly confined and another region outside theboundary layer zone where the flow is inviscid. Thus, the boundary layer theoryhelped in simplifying the complex motion of fluid by considering the region where the velocity gradient exists. At the leading edge of the fluid, the thickness of the boundary layer is zero. On the downstream, for the fluid in contact with the boundary, the velocity of flow is zero and at some distance from the boundary the velocity is u.hence a velocity gradient is set up which retards the motion of fluid due to the shear

9 resistance. Near the leading edge, the fluid is retarded in thin layer. At subsequent point downstream the leading edge, the boundary layer region increases because the retarded fluid is further retarded. Boundary conditions for laminar flow: (a) At y = 0, u = 0, (b) At, (c) At and in between y = 0 to, velocity gradient exists. The general equation of velocity profile for laminar boundary layer is u = ay+by 2 +cy 3 +dy 4. First equation for For this equation, At y = 0, u = 0 At,. Velocity gradient is constant i.e. the velocity profile is linear. Hence the equation satisfies the boundary conditions for laminar boundary layer velocity profile. Second equation ( ) for At, At and ( ).Velocity gradient is constant i.e. the velocity profile is linear. Hence the equation satisfies the boundary conditions for laminar boundary layer velocity profile. 6. Two water tanks are connected to each other through a mercury manometer with inclined tubes, as shown in the figure below. The pressure difference between two tanks is 20Kpa. Calculate a and θ. [6] Tank A A a 26.8cm X 2a a X B Tank B θ Mercury SG = 13.6

10 P B -P A = 20 Kpa Equating the pressure at XX -20x x9810x2a = 0 a = m 2a = (26.8/100)Sinθ 2x0.075 = 0.268Sinθ θ = A tank of area A is provided with an orifice 40mm in diameter at its bottom. Water flows into the tank at a uniform rate and is discharged through the orifice. It is found that when the head of water over the orifice is 0.68m, the water surface rose at a rate of m/Sec. But, when the head of water is 1.24m, the water surface rose at m/sec. Find the rate of flow and the cross-sectional area of the tank. Take C d = [6] Diameter of orifice (d) = 40mm = 0.04m Area of orifice (a) = = m 2 Coeff. of discharge (C d ) = 0.62 = Inflow rate (Q i ) =? Cross-sectional area of tank (A) =? First case: Head (h) = 0.68m, dh/dt = m/s Second case: Head (h) = 1.24m, dh/dt = m/s General equation for tank with inflow (Q i ) and outflow (Q o ) Substituting values for both cases Q i =0.0014A Q i = A Solving a and b A =1.278 m 2 Q i = m 3 /s (a) (b)

11 OR A tank is in the form of frustum of a cone having top diameter of 2m, a bottom diameter of 0.8m and height 2m and is full of water. Find the time of emptying the tank through an orifice 100mm in diameter provided at the bottom. Take C d = [6] 1m 2m 0.4m H 0 C d = Area of orifice (a) = = m 2 H 1 = 2m, H 2 = 0m, R 1 = 1m, R 0 = 0.4m From similar triangles, H 0 = 1.33m = Time of emptying the tank is [ ] [ ] = 160 Sec

12 8. For the two orifices shown in the figure below, determine Y 2 such that. [6] 2m orifice1 10m orifice2 Y 2 X 2 X 1 H 1 = 2m, Y 1 = 10-2 = 8m, H 2 = 10-Y 2 Y 2 =? Coefficient of velocity for orifice 1 Coefficient of velocity for orifice 2 Since the two orifices are identical C v1 = C v2 Solving for Y 2 Y 2 = 1, 9 As Y 1 = 8m, Y 2 = 9 (>Y 1 ) is not feasible. Hence Y 2 = 1m 9. The diameter of a pipe bend is 30cm at inlet and 15cm at outlet and the flow is turned through in a vertical plane. The axis at inlet is horizontal and the center of the outlet section is 1.5m below the center of the inlet section. Total volume of water in the bend is 0.9m 3. Neglecting friction, calculate the magnitude and direction of the force exerted on the bend by water flowing through it at 250lps and when the inlet pressure is 0.15N/mm 2. [7]

13 Y P 1 V 1 Z 1 W X 1 R R y R x F P 2 V 2 Diameter at section 1 (d 1 ) = 30cm = 0.3m Area at section 1 (A 1 ) = = m 2 Diameter at section 2(d 2 ) = 15cm = 0.15m Area at section 2 (A 2 ) = = m 2 Discharge (Q) = 250 lps = 0.25 m 3 /s Volume of water within control volume (Vol) = 0.9m 3 Weight of water within control volume (W) = = 9810x0.9 = 8829N Velocity at section 1 (V 1 ) =Q/A 1 = 0.25/ = 3.54 m/s Velocity at section 2 (V 2 ) =Q/A 2 = 0.25/ =14.15 m/s Z 2 = 0, Z 1 = 1.5m θ= = 60 0 Pressure at 1 (P 1 ) = 0.15N/mm 2 = 0.15x10 6 N/m 2 Resultant force exerted by the water on the bend =? Applying Bernoulli s equation between 1 and 2 Z 2 2 P 2 = N/m 2 = N

14 = 4681N Resultant force =14650N Resultant force exerted by the water on the bend = 14650N (to the right and downward) Direction of resultant force = = = The pipe flow in the figure is driven by the pump. What gauge pressure is needed to be supplied by the pump to provide water flow rate of Q = 60m 3 /h? Neglect head loss from A to B. Head loss from C to D = ; Head loss from D to E = ; d AB (diameter of pipe AB) = d CD = 5cm; d DE = 2cm. where V CD = velocity in pipe CD and V DE = velocity in pipe DE. [7] D E Q 1 80m 10m A B C Pump Discharge (Q) = 60m 3 /h = 60/3600 m 3 /s = m 3 /s Diameter of pipe AB and CD = 5cm = 0.05m C/s Area of pipe AB and CD (A AB, A CD ) = = m 2 Diameter of pipe DE =2cm = 0.02m

15 C/s Area of pipe DE (A DE ) = = m 2 Velocity of flow through AB and CD (V AB = V CD ) =Q/A AB = / = 8.5m/s Velocity of flow through DE (V DE ) = Q/A DE = / = 53.2m/s Head loss between A and B (hl AB ) = 0 Head loss from C to D (hl CD ) = = 110.5m Head loss from D to E (hl DE ) = = 73.65m Total head loss (h L ) = = m H p = Head supplied by the pump At point 1, V 1 = 0, P 1 = 0 (atm) At point E, P E = 0 (atm) Applying Bernoulli s equation between 1 and E (Taking datum through A) h p =398.4m Applying Bernoulli s equation between B and C V B = V C, Z B = Z C = 9810x398.4 = Pa = Kpa 11. Express the kinematic viscosity in stokes for a liquid with specific gravity 0.95 and dynamic viscosity poise. A U-tube is made up of two capillarities of diameters 1.0mm and 1.5mm respectively. The U-tube is kept vertically and partially filled with water of surface tension kg/m and zero contact angle. Calculate the difference in the level of the miniscii caused by capillarity. [2+3] h d 2 d 1 Specific gravity = 0.95

16 Density of fluid (ρ) = 0.95x1000 = 950 kg/m 3 Dynamic viscosity = poise = 0.1x0.011 = NS/m 2 Kinematic viscosity =? = 1.158x10-6 m 2 /s = 1.158x10-6 x10 4 Stokes = Stokes Diameter of tube 1 (d 1 ) = 1mm = 1x10-3 m Diameter of tube 2 (d 2 ) = 1.5mm = 1.5x10-3 m Surface tension (σ) = kg/m = N/m Contact angle (θ) = 0 Capillary rise in tube 1 (h 1 ) = m Capillary rise in tube 2 (h 2 ) = 0.021m Difference in level (h) = = m 12. The pressure drop in an air duct depends on the length and diameter of the duct, the mass density, viscosity of the fluid and the velocity of the flow. Obtain an expression for the pressure drop in dimensionless form using Buckingham π theorem. Estimate the pressure drop in a 20m long air duct if a model of the duct operating with water produces a pressure drop of 10 KN/m 2 over 10m length. The scale ratio is 1:50. Given, = 1000 kg/m 3, = 1.2kg/m 3 = 0.001NS/m 2, = NS/m 2 [5+4] Total number of variables = 6 No. of fundamental dimensions = 3 No. of π terms = 6-3= 3 (I) Choose D, V and ρ as repeating variables. First π term (II) Writing dimensions Equating the powers of M, L and T c1+1 =0 a1+b1-3c1-1 = 0 -b1-2 = 0

17 c1 = -1, b1 = -2 a =0 a1 = 0 Substituting the values of a1, b1 and c1 in II Second π term Writing dimensions (III) Equating the powers of M, L and T c2 = 0 a2+b2-3c2+1 = 0 b2 = 0 a2 = -1 Substituting the values of a2, b2 and c2 in III Third π term (IV) Writing dimensions Equating the powers of M, L and T c3+1 =0 a3+b3-3c3-1 =0 -b3-1 = 0 c3 = -1 b3 = -1 a = 0 a3 =-1 Substituting the values of a3, b3 and c3 in (IV) Substituting the values of π 1, π 2 and π 3 in (I)

18 Multiplying first πterm by 1/π 2 and 1/π 3 and expressing the product as a function of 1/π 3 Numerical Scale ratio (L r ) = 1/50 Pressure drop in model (P m ) = 10 KN/m 2 = 10x10 3 N/m 2 Density of air (ρ p ) = 1.24 kg/m 3 Density of water (ρ m ) = 1000 kg/m 3 Viscosity of water (μ m ) = 0.001NS/m 2 Viscosity of air (μ p ) = NS/m 2 Pressure drop in prototype (P p ) =? As the problem involves both viscous and pressure force, we have to use both Reynolds and Euler model law. From Reynolds model law Re)model = Re) Prototype L m /L p = 1/50 V m = 0.31V p From Euler model law Eu) model = Eu) prototype P p = 129N/m 2 Pressure drop in prototype corresponding to 10 KPa over 10m length in model = 129N/m 2 Pressure drop for 20m length = 2x129 = 258N/m 2

19 OR How are the repeating variables selected for dimensional analysis? Prove that the scale ratio for discharge for a distorted model is given as where Q p = discharge through prototype, Q m = discharge through model, (L r ) H = horizontal scale ratio, (L r ) V = Vertical scale ratio. [4+5] Repeating variables appear in most of the π groups.they have a large influence on the problem.there is great freedom in choosing these. Some rules which should be followed for selecting repeating variables are as follows: There are n ( = 3) repeating variables. In combination they must contain all of dimensions (M, L, T) The repeating variables must not form a dimensionless group. Dependent variable should not be selected as repeating variable. No two repeating variables should have the same dimensions. They should be measurable in an experiment. They should be of major interest to the designer. Q p = discharge through prototype, Q m = discharge through model, (L r ) H = horizontal scale ratio, (L r ) V = Vertical scale ratio A = C/s Area of flow, V = Velocity, B = width, h = depth of flow (a) (b) where and From Froude s model law (c) From a, b and c

20 2068, Shrawan (Back exam) Attempt all questions. 1. State and prove Newton s law of viscosity. Prove that, where K is Bulk Modulus of elasticity, P is pressure and is density. [1+2+3] Newton s law of viscosity states that the shear stress is proportional to the rate of deformation or velocity gradient. Where = shear stress, du/dy = velocity gradient andthe constant of proportionality = coefficient of viscosity. A dx A D D τ dy dφ Let us consider a fluid confined between two plates, where the bottom plate is stationary and the upper plate is moving. Let ABCD is the fluid at any time t. Due to the application of shear force τ, the fluid deforms to A BCD at time t+dt. Let dy = distance between two layers, AA = dx and shear strain = dφ. For small angle, dx = dφ. dy Also dx = du. dt τ B C Equating Shear stress is proportional to rate of shear strain. From a and b (a) (b)

21 Bulk modulus is (c) Differentiating From c and d, (d) 2. An 8cm diameter piston compresses manometer oil into an inclined 7mm diameter tube, as shown in figure below. When a weight W is added to the top of the piston, the oil rises an additional distance of 10cm up the tube. How large is the weight, in N? [6] W 10cm Piston D = 8cm h X Y Initial level Final level Y X dh Oil, sp gr = D = 7mm 15 0 Diameter of piston (D) = 8cm = 0.08m Diameter of tube (d) = 7mm = 0.007m h = 0.10Sin15 = m Due to compression, the fluid in the container moves down by dh and the fluid in the tube moves up by 10cm. Volume of fluid fallen = Volume of fluid risen

22 dh = m Neglecting dh Equating the pressure at new level (YY) (b) W = 1.05 N 3. A cylinder, 2m in diameter and 3m long weighing 3KN rests on the floor of the tank. It has water to a depth of 0.6m on one side and liquid of sp gr 0.7 to a depth of 1.25m on the other side. Determine the magnitude and direction of the horizontal and vertical components of the force required to hold the cylinder in position. [8] 0.6m C Water E A O D 1.25m Oil B OA= OB = OC = 1m, BD = 0.6m OD = = 0.4m CD = = m OE = = 0.25m = = <AOB = = Weight of cylinder = 3KN = 3000N Net horizontal force = 0.7x9810x1.25x3x1.25/2-9810x0.6x3x0.6/2 = 10797N (left) Net vertical force (F V )= Weight of volume of water vertically above AB + Weight of volume of water vertically above BC = Fv AB (up) + Fv BC (up) =

23 = [ ] = 32917N (up) The components to hold the cylinder in place are N to the right and = 29917N down. 4. An open cylinder tank 0.5m in diameter and 1m height is completely filled with water and rotated about its axis at 240 rpm. Determine the radius up to which the bottom will be exposed and the volume of water spilled out of the tank. [7] r =0.25m A B 1m z C D z1 P r1 Radius (r) = 0.25m N = 240rpm Angular velocity = rad/s = 2.01m Z1 = = 1.01m r1 =? Volume of water spilled =?

24 r 1 = 0.18m Volume of water spilled = Volume of paraboloid (APB-CPD) = m 3 OR Develop an expression for the position of the Metacenter relative to the center of buoyancy. M A E O θ D G B B C F B =W Consider a vessel with water line AC, B as the CB and G as the CGin original position. When the vessel is tilted through a small angle θ, the CB changes from B to B, the position of water line changes to ED and two wedges AOE and COD are formed. M is the metacenter, W is the weight of vessel and F B is the buoyant force. Consider an element of area da at a distance x from the CG of plan of vessel on either side. Volume of element (dv) = da.xθ (xθ = height for small angle) Buoyant force on the element of left side (df B ) = Weight of element = where is the sp wt of water. Similarly, x x Buoyant force on the element of right side (df B ) = Buoyant force on the two elements produces a couple. Moment of couple = = Total moment due to altered displacement (M) = = = I = Second moment of plan of the vessel about an axis through CG (Moment of inertia)

25 (a) Moment due to the movement of CB from B to B (M ) = F B BB = WBB (b) Where V is the volume of water displaced. For equilibrium, M = M GM = BM-BG 5. Derive the continuity equation for cylindrical coordinates. [6] Y R O T V Z S V θ Q dz V r dθ P N θ r M dr X Consider any point P(r, θ, Z) in space. Let dr, dθ and dz be small increments in r, θ and Z direction respectively. Let V r, V θ and V z be the velocity in r, θ and Z direction respectively, and ρ be the density of fluid. Mass of fluid entering face PQNM per unit time = Mass of fluid leaving face RSTO per unit time = Gain of mass in z-direction = [ ] (a) Similarly, Gain of mass in r-direction (b) Gain of mass in θ-direction (c) Net gain of mass

26 [ ] (d) Rate of increase of mass per unit time (e) Equating d and e [ ] [ ] This is the continuity equation in 3D cylindrical polar co-ordinate. For steady flow,. The continuity equation is For incompressible flow,. The continuity equation is 6. Derive Bernoulli s equation from Euler s equation of motion. A venturimeter with a throat diameter of 100mm is fitted in a vertical pipeline of 200mm diameter with oil of sp.gr flowing upwards. The venturimeter coefficient is Two pressure gauges calibrated in KN/m 2 are fitted at tapping points, one at the throat and the other at the inlet pipe 320mm below the throat. The difference between two pressure gauge readings is 28 KN/m 2. Working from Bernoulli s equation, determine (I) the volume rate of oil through the pipe, (II) the difference in level in the two limbs of mercury manometer if it is connected to the tapping points. [4+8] Euler s equation in 1D is Integrating Dividing by g

27 or, This is the Bernoulli s equation. Diameter at inlet (d 1 ) = 20cm = 0.2m C/s area of inlet (A 1 ) = = m 2 Diameter at throat (d 2 ) = 10cm = 0.1m C/s area of throat (A 2 ) = = m 2 Sp.gr. of oil (S 0 ) =0.88 Density of oil (ρ) =0.88x1000 = 880 kg/m 8 Difference in elevation (Z 2 -Z 1 ) = 320cm =0.32m Difference in pressure (P 1 -P 2 )= 28KN/m 2 = 28000N/m 2 Sp.gr. of mercury (S) =13.6 C d =0.96 Discharge of oil (Q) =? Manometer reading (x) =? 2 1 Inlet Throat (I) Applying Bernoulli s equation between inlet (1) and throat (2) (a) According to continuity equation A 1 V 1 =A 2 V 2 = =4V 1 (b) Solving a and b V 1 = 1.95m/s Discharge (Q) =A 1 V 1 = x1.95 = m 3 /s Actual discharge = C d Q = 0.96x0.612 = m 3 /s II. = =2.92

28 x = 0.2m 7. Derive an expression for estimating discharge for partially submerged orifice. [4] In partially sub-merged orifice, the outlet side of the orifice is partially submerged under liquid. This type of orifice has two portions: free discharging orifice at the upper part and submerged orifice at the lower part. H 1 H 2 H Let H 1 = height of liquid above the upper edge of orifice, H 2 =height of liquid above the lower edge of orifice, H= Difference in liquid level between two tanks, b = width of orifice, and C d = coefficient of discharge Discharge through orifice (Q) = Discharge through free portion (Q1) + Discharge through submerged portion (Q2) Free portion is treated as large rectangular orifice. Hence For submerged portion, Q = Q1 + Q2 8. A 45 0 reducing bend is connected in a pipe line carrying water. The diameter at inlet andoutlet of the bend is 400mm and 200mm respectively. Find the force exerted by water on the bend if the intensity of pressure at inlet of the bend is 215.8KN/m 2. The rate of flow of water is 0.5m 3 /s. The loss of head in the bend is 1.25m of oil of sp.gr [7]

29 P 2 +Y P 1 F x +X F y Reactions Diameter at section 1 (d 1 ) = 400mm = 0.6m Area at section 1 (A 1 ) = = m 2 Diameter at section 2(d 2 ) = 200mm = 0.2m Area at section 2 (A 2 ) = = m 2 Discharge (Q) = 0.5 m 3 /s Velocity at section 1 (V 1 ) =Q/A 1 = 3.98 m/s Velocity at section 2 (V 2 ) =Q/A 2 = m/s Pressure at section 1 (P 1 ) = 215.8KN/m 2 = N/m 2 Loss of head = 1.25m of oil of sp gr 0.85, P= 0.85x9810x1.25 N/m 2 Loss of head in terms of water (h L ) = Angle of bend = 45 0 Resultant force (F R ) =? Direction of resultant force =? Applying Bernoulli s equation between 1 and 2 (Z 1 = Z 2 ) =1.0625m P 2 = 86574N/m 2

30 = 21544N = 7551N Resultant force =22829N Resultant force exerted by the water on the bend = 22829N (to the right and downward) Direction of resultant force = = = Define boundary layer with appropriate sketches. How laminar sub-layer is developed in boundary layer. Differentiate between hydraulically smooth and rough boundary. [2+2+2] When a real fluid flows past a solid boundary, the fluid particle on the surface will have the same velocity as that of the surface because of viscosity. This is called no-slip condition. If the boundary is stationary, the velocity of the fluid at the boundary is zero. Further away from the boundary, the velocity gradually increases. U Boundary layer stream velocity (U) Solid body The term boundary layer is defined as the thin layer of the flow on the boundary within which the velocity varies from zero at the solid boundary to the free stream velocity in the direction normal to the boundary.

31 Boundary layer along a thin flat plate U Turbulent Boundary layer Laminar Boundary layer Laminar sublayer Laminar zone Turbulent zone Consider the flow of fluid, having free stream velocity U, over a smooth thin plate which is flat and parallel to the direction of free stream of fluid. At the leading edge of the fluid, the thickness of the boundary layer is zero. As the fluid moves along the plate, the thickness goes on increasing.the zones formed during boundary layer development are: Laminar boundary layer, Transition zone, turbulent boundary layer. If the plate is very smooth, even in the region of turbulent boundary layer, there is a very thin layer just adjacent to the boundary, in which the flow is laminar. This layer is known as laminar sub-layer. Let k is the average height of irregularities projecting from the surface of a boundary and = thickness of laminar sub layer. If k is much less than, the boundary is called hydrodynamically smooth boundary. In this case eddies cannot reach the surface irregularities. If is much less than k, the boundary is called hydrodynamically rough boundary. In this case eddies come in contact with the surface irregularities and lot of energy will be lost. From Nikuradse s experiment: <0.25: Smooth boundary >6: rough boundary 0.25< <6: transition Transition zone 10. Define aerofoil with accepted terminology with neat sketch. A kite, which may be assumed to be a flat plate and mass 1kg, soars at an angle to the horizontal. The tension in the string holding the kite is 60N when the wind velocity is 50 km/h horizontally and the angle of string to the horizontal direction is The density of air is 1.2 kg/m 3. Calculate the drag coefficient for the kite in the given position if the lift coefficient in the same position is Both coefficients have been based on the full area of the kite. [2+6]

32 chord line α L An airfoil is a streamlined body which may be either symmetrical or unsymmetrical. Some of the terminologies used to characterize airfoil are as follows. Chord line: It is the line joining the front and rear edge. Angle of attack: It is the angle between the direction of flowing fluid and chord line. Camber: It is the curvature of an airfoil. Span: The overall length of airfoil is called its span (L). Aspect ratio: The ratio of span (L) to mean chord (C) is called aspect ratio. Stall: An airfoil is said to be in stall condition when the angle of attack of an airfoil is greater than the angle of attack at maximum lift. Kite Lift 35 0 Drag String T W Mass of kite = 1kg Weight of kite (W) = 1x9.81 N = 9.81N Area of kite = A Velocity (V) = 50 km/hr = Density of air = 1.2kg/m 3 Tension (T) = 60N Coefficient of lift (C L ) = 0.45 Coefficient of drag (C D ) =? Forces in X-dir (Drag) F D = Tcos35 = 60cos35 =49.14N =13.88/s

33 Force in Y-dir (Lift) F L = Tsin = 60sin =44.22N A = 0.85 m 2 C D = Determine the relationship between model and prototype kinematic viscosity if both Reynold s number and Froude number are to be satisfied. [4] From Reynolds model law. (Re) model = (Re) prototype (a) From Froude model law (Fr) model = (Fr) prototype (b) From a and b 12. Show by dimensional analysis that the power P required to operate a test tunnel is given by where is density of fluid, is viscosity, V is fluid mean velocity, P is the power required and L is the characteristics tunnel length. [6] Total number of variables = 5 No. of fundamental dimensions = 3

34 No. of π terms = 5-3= 2 (I) Choose ρ, L and V as repeating variables. First π term Writing dimensions (II) Equating the powers of M, L and T a1+1 =0-3a1+b1+c1+2 = 0 -c1-3 = 0 Solving a1 = -1, c1 = -3-3x-1+b1-3+2 = 0 b1 = -2 Substituting the values of a1, b1 and c1 in II Second π term (III) Writing dimensions Equating the powers of M, L and T a2+1 = 0-3a2+b2+c2-1=0 -c2-1 = 0 Solving a2 = -1, c2 = -1-3x-1+b2-1-1 = 0 b2 = -1 Substituting the values of a2, b2 and c2 in III Substituting the values of π 1 and π 2 in I

35 2068, Chaitra (Regular exam) Attempt all questions. 1. Define kinematic and dynamic viscosity. A small thin plane surface is pulled through the liquid filled space between two large horizontal planes in the parallel direction. Show that the force required will be minimum if the plate is located midway between the planes. [2+4] According to Newton s law of viscosity, shear stress is proportional to the rate of deformation or velocity gradient. Where = shear stress, du/dy = velocity gradient andthe constant of proportionality = coefficient of viscosity. The constant is also called dynamic viscosity or absolute viscosity. Kinematic viscosity is defined as the ratio of dynamic viscosity to density. Where = Kinematic viscosity and ρ = density h y V = shear stress in the upper part = shear stress in the upper part F = pull required to drag the plate For F to be minimum, df/dy = 0 [ ] [ ] Thus, the plate is located midway between the planes for the force to be minimum.

36 2. Define atmospheric, gauge and pressure. A U-tube shown in figure is 10mm in diameter and contains mercury. If 12ml of water is poured into the right hand leg, what are the ultimate heights in the two legs? [2+4] The pressure exerted by the atmospheric air on the surface with which it is in contact is known as atmospheric pressure. If the pressure is quoted by taking atmospheric pressure as a datum, it is called gauge pressure. The pressure measured above a perfect vacuum is called absolute pressure. Water 120mm y E y F h Mercury With the addition of 12ml of water, Drop in mercury level in right leg = y Rise in mercury level in left leg = y Depth of water column = h h= Vol of water added/cross sectional area = = 0.153m = 153mm P E = P F 13.6x9810x2y = 9810x0.153 y = m = 5.6mm Height of free mercury level in the left leg = = 125.6mm Height of free water level in the right leg = = 267.4mm 3. Prove that in case of force vortex, rise of liquid level at the end is equal to the fall of liquid level at the axis of rotation. The hemispherical dome shown in figure is filled with water and is attached to the floor by two diametrically opposite bolts. What force in either bolt is required to hold the dome down, if the dome weighs 25KN? [4+6]

37 R A B h r h f M C O N D h P Q R = Radius of cylinder MN = Water level at absolute equilibrium (original water level) After rotation, AOB is the profile of the liquid surface. h r = Rise of liquid at end h f = Fall of liquid at the end Volume of liquid before rotation = Volume of liquid rotation = Volume of cylinder ABQP- Volume of paraboloid AOB Volume of liquid before rotation = Volume of liquid after rotation Hence, rise of liquid level at the end = fall of liquid level at the axis of rotation

38 4cm 5m Water 1.5m Bolt Vertical force on the dome due to water (F v ) = Weight of volume of water (imaginary) vertically above dome = [ ] = N = KN (up) Weight of dome = 25 KN (down) Net Vertical force = = (up) Force on either bolt = /2 = KN (up) [ ] 4. Show that if B is the center of buoyancy and M is the metacenter of a partially immersed floating body, then BM=I/V, where I is the second moment of the area of the surface of floatation about the longitudinal axis and V is the immersed volume. [5] A E df B O M θ D G B B df B C F B =W

39 Consider a vessel with water line AC, B as the CB and G as the CGin original position. When the vessel is tilted through a small angle θ, the CB changes from B to B, the position of water line changes to ED and two wedges AOE and COD are formed. M is the metacenter, W is the weight of vessel and F B is the buoyant force. Consider an element of area da at a distance x from the CG of plan of vessel on either side. Volume of element (dv) = da.xθ (xθ = height for small angle) Buoyant force on the element of left side (df B ) = Weight of element = where is the sp wt of water. Similarly, x x Buoyant force on the element of right side (df B ) = Buoyant force on the two elements produces a couple. Moment of couple = = Total moment due to altered displacement (M) = = = I = Second moment of plan of the vessel about an axis through CG (Moment of inertia) (a) Moment due to the movement of CB from B to B (M ) = F B BB = WBB (b) Where V is the volume of water displaced. For equilibrium, M = M 5. Explain Lagrangian and Eulerian methods of describing fluid flow. [2] Lagrangian method deals with the motion of single fluid particle. This approach is more complex and the equations of motion are difficult to solve. Eulerian method deals with the motion of fluid particles at a point in space. This method is simple and widely used as the resulting equations can be solved easily. 6. A flow is described by the stream function. Locate the point at which the velocity vector has a magnitude of 6 unit and makes an angle with the X-axis. [4]

40 Resultant velocity = 6 (a) y = 0.7x (b) Solving a and b x = 1.16, y = Prove that the Cippoletti weir is a trapezoidal weir having side slopes with vertical. Water flows over a rectangular sharp crested weir 2m long (divided into two bays of 1m each by pier). The head over the sill is 0.75m. The weir is fitted at the end of the long rectangular channel 3m wide and flow depth 1.5m. Starting from the first principles, determine the rate of discharge over the weir. Consider the velocity of approach and the effect of end contraction. Coefficient of discharge for the weir is 0.7. [4+8] θ/2 H θ/2 L Discharge through Cippoletti weir is For a rectangular weir with two end contractions, the discharge is given by From the above expression, it is clear that discharge of a rectangular weir is reduced by due to end contractions, which is compensated by discharge through triangular notch.

41 Discharge through triangular notch = loss of dischargedue to end contraction Length (L) = 2m 1 bay = 2 end contractions n = 2x2 = 4 Head over sill (H) = 0.75m C/s of channel (A) = 3mx1m = 3m 2 Coefficient of discharge (C D ) = 0.7 Discharge over weir (Q) =? Discharge over a rectangular weir without velocity of approach and with end contraction is = 2.28 m 3 /s Velocity of approach (V a ) = Q/A = 2.28/3 = 0.76 m/s Velocity head = 0.03m Discharge by considering velocity of approach ( ) [ ] ( [ ]= 2.38 m 3 /s 8. A pipe bend placed in a horizontal plane tapers from 50cm diameter at inlet to 25cm diameter at outlet. An oil of density 850kg/m 3 enters the reducing bend horizontal and gets turned by 45 0 clockwise direction. The discharge is 0.45 m 3 /s and pressure at inlet of 40 KN/m 2 drops to 23 KN/m 2 at the outlet due to frictional effect. Find out the magnitude and direction of resultant force on the bend. [8] +Y 1 P F y F x Reactions +X 2 P 2

42 Diameter at section 1 (d 1 ) = 0.5m Area at section 1 (A 1 ) = = m 2 Diameter at section 2(d 2 ) = 0.25m Area at section 2 (A 2 ) = = m 2 Discharge (Q) = 0.45 m 3 /s Velocity at section 1 (V 1 ) =Q/A 1 = 2.3 m/s Velocity at section 2 (V 2 ) =Q/A 2 = 9.16m/s Pressure at section 1 (P 1 ) = 40KN/m 2 = 40000N/m 2 Pressure at section 2 (P 2 ) =23KN/m 2 =23000N/m 2 Angle of bend = 45 0 Resultant force (F R ) =? Direction of resultant force =? = 5458N = -3275N Resultant force =6365N Resultant force exerted by the water on the bend = 6365N (to the right and upward) Direction of resultant force = = = What is hydrodynamic boundary layer? Workout the shape factor parameter for the velocity distribution given by. [3+4]

43 When a real fluid flows past a solid boundary, the fluid particle on the surface will have the same velocity as that of the surface because of viscosity. This is called no-slip condition. If the boundary is stationary, the velocity of the fluid at the boundary is zero. Further away from the boundary, the velocity gradually increases. U Boundary layer stream velocity (U) Solid body The term boundary layer is defined as the thin layer of the flow on the boundary within which the velocity varies from zero at the solid boundary to the free stream velocity in the direction normal to the boundary. In the boundary layer velocity gradient (du/dy) is large and the shear stress exerted by the fluid is given by Newton s law of viscosity as. Outside the boundary, velocity is constant and velocity gradient is zero and hence shear stress is zero. Displacement thickness [ ] Momentum thickness [ ] [ ]

44 10. A 2mm diameter spherical ball of unit weight KN/m 3 is dropped in a mass of fluid of viscosity 15 poise and sp.gr Find out drag force, pressure drag, skin drag and terminal velocity of ball. [8] Diameter of ball (D) = 2mm = 0.002m Sp.wt. of fluid = 0.95x9810 N/m 3 Viscosity of fluid = 15 poise = 15/10 = 1.5 PaS Sp.wt. of ball = kn/m 3 Drag force (F D ) =? Pressure drag and friction drag =? Terminal velocity (V) =? Weight of ball (W) = Drag force (F D ) + Buoyant force on ball (F B ) F D = W F B = N Pressure drag = = N Friction drag = = N From Stoke s law, V = m/s Checking the Reynold s no. = 0.02 As Re<0.2, above expression for F D is valid. 11. Discuss types of model studies. If gravity, viscosity and surface tension are equally important in a model, show that for dynamic similarity the relationship between viscosity ratio, surface tension ratio and model scale ratio L r is given by. [2+6]

45 There are two types of models. a. Undistorted model: An undistorted model is that model which is geometrically similar to its prototype. Result from such model can be directly applied to the prototype. b. Distorted model: A distorted model is that model in which one or more terms of the model are not identical with their counter-parts in the prototype. A distorted model may have either geometrical distortion or material distortion or distortion of hydraulic quantities or a combination of these. Results obtained from the distorted models cannot be directly applied. (Re) model = (Re) prototype (a) (Fr) model = (Fr) prototype (b) (Wb) model = (Wb) prototype (c) Equating b and c, and squaring (d) Substituting V r from b and from d into a

46 12. Derive an expression representing Euler s equation of motion. [4] ds B A θ dz PdA dw Consider a streamline AB in which flow is taking place in S direction. Consider a cylindrical element of length ds and area da. Let P be the pressure on left face and ( pressure force and gravity force Resultant force in S- direction = m a s where a s = acceleration in s direction ) on the right face. Forces acting: For steady flow Dividing by ρ As all are function of S, Or, This is 1D Euler s equation in differential form.

47 2069 Ashadh (Back exam) 1. Give two examples of non-newtonian fluid. If the velocity distribution of fluid over a plate is given by where u is velocity in m/s at a distance y from the plate, determine the shear stress at y=0.2m. Take dynamic viscosity as 8.45x10-4 NS/m 2. Examples of non-newtonian fluid: Blood, slurry Dynamic viscosity = 8.45x10-4 NS/m 2 Shear stress at y=0.2m =? At y = 0.2m, shear stress is = 0 N/m 2 2. Find the pressure difference between the container A and B as shown in the figure. A 130mm Liquid of sp.gr = mm 230mm B 550mm Water Mercury 300mm E F Pressure at E = pressure at F

48 = N/m 2 = Kpa 3. When a 22N weight is placed at the end of the uniform floating wooden beam as shown in figure below, the beam tilts at an angle θ with its upper right corner at the surface, as shown. Determine (a) the angle θ and (b) the specific gravity of wood. Assume that the point of application of 22N weight cuts at point B. 2.75m A θ 22N B 10cmx10cm D C Water A h E θ W= 22N B 10cmx10cm D C F B W wood h = 2.75tanθ Weight of wood Buoyant force (F B )= Weight of water displaced = =

49 = W wood + W = F B (a) Taking moment about B = 0 Solving a and b (b) N/m 3 Sp. Gr. Of wood= 6656/9810 = Find the resultant force on ABCDEF of 10m width and its location due to the liquid as shown in figure Kpa Liquid of sp. gr. = 1.2 2m 2m 5m A 2m B C 2m 4m D 2m E r= 1.6m F 3m G

50 Equivalent head of liquid due to Kpa pressure = = -10m This negative pressure will reduce the fluid surface by 10m. Make calculation by taking imaginary fluid surface (IFL) as free surface Kpa Liquid of sp. gr. = 1.2 2m 2m 5m 10m A 2m IFL B C 2m C 4m y p D D F V 2m E F H x r= 1.6m F 3m G Net Horizontal force on curved surface (F H ) = = 1.2x9810x(8.2x10)x4.1 = N = KN = 5.47m from C Vertical force on D D = N (down) Vertical force on EF = N (up)

51 Net vertical force (F V ) = = N = KN Location of F V Taking moment about C x = 2.65m Resultant force (F R ) = = KN Resultant force acts at a distance of 5.47m down IFL and 2.65m left from C D. 5. Consider the following steady, incompressible two dimensional velocity field: m/s. Generate an analytical expression for the flow streamline and draw several streamlines in the upper right quadrant from x = 0 to 3 and y = 0 to 3. Determine if there are any stagnation points in this flow field and if so, where are they. u = x, v = y Equation of streamline Integrating [ ] (let -K1x0.656 = K2) (Replacing e K2 by K3) [] y+0.8x+0.43xy=K3 0.8(x+0.54xy-3.6y)=K Replacing (K )/8 by c X+0.54xy-3.6y=c. This is the equation of streamline. Plotting of streamlines

52 C=0 C=1 C=2 x y x y x y Streamlines At stagnation point u =0 and v = x = 0 x = y = 0 y = -1.9 Stagnation occurs at (-6.6, -1.9). OR A flow has a potential function given by:. Derive the corresponding stream function and show that some of the stream lines are straight lines passing through the origin of coordinates. Find the inclination of these lines. (a)

53 (b) Integrating eq. (a) w.r.t. x Where c= f(y) Differentiating eq. (c) w.r.t. y Equating eq. (b) and (d) Integrating w.r.t. y (d) (c) Substituting the value of c in eq.(c) For For, this equation (in the form of y=ax) shows that some of the stream lines are straight lines passing through the origin of coordinates. Inclination of the straight lines 6. Describe a cippoletti weir. How does it differ from a rectangular sharp crested weir. A cippoletti weir is a trapezoidal weir having side slopes of 4 vertical to 1 horizontal. θ/2 H θ/2 L Discharge through Cippoletti weir is For a rectangular weir with two end contractions, the discharge is given by

54 From the above expression, it is clear that discharge of a rectangular weir is reduced by due to end contractions, which is compensated by discharge through triangular notch. Discharge through triangular notch = loss of dischargedue to end contraction 7. Find the velocity V1 of the water flowing in a vertical pipe as shown in figure below. The loss of head is two times the velocity head of small pipe. The pipe diameter at section 1 is 15cm and that at section 2 is 6cm. 1 V 1 5m 2 0.5m Hg Applying Bernoulli s equation between 1 and 2 (Taking datum through 2) (a)

55 From continuity equation, A 1 V 1 = A 2 V 2 (b) Writing manomertic equation ( = dynamic head due to stagnation, total head = static +dynamic) (c) From a, b and c 0.02m/s 8. Water flows through a 90 0 reducer bend. The pressure at the inlet is 206 KN/m 2 (gauge) where the cross-sectional area is 0.01m 2. At the exit section, the area is m 2 and the velocity is 15m/s. Assume that the bend does not discharge into the atmosphere but m 2 are conduit continues. Further assume that the head loss in the reducer bend is, where u is the average velocity at the inlet and exit of the reducer. Determine the force exerted by the flow on the reducer bend. P 2 2 +Y 1 +X P 1 F x F y Reactions Area at section 1 (A 1 ) = 0.01m 2 Area at section 2 (A 2 ) = m 2 Velocity at section 2 (V 2 ) =15 m/s

56 Discharge (Q) =A 2 V 2 = x15 = m 3 /s Velocity at section 1 (V 1 ) =Q/A 1 = 3.75 m/s Pressure at section 1 (P 1 ) = 206KN/m 2 = N/m 2 u = ( )/2 = 9.375m/s Head loss (hl) = = 4.5m Applying Bernoulli s equation between 1 and 2 (Z 1 = Z 2 ) = N/m 2 = 2200 N = 703 N Resultant force =2310N Resultant force exerted by the water on the bend = 2310N (to the right and downward) Direction of resultant force = = = Air at 20 0 c flows parallel to a smooth thin flat plate at 4.75m/s. The plate is 3.23m long. Determine whether the boundary layer on the plate is most likely laminar, turbulent or somewhere in between (transitional). Compute the boundary layer thickness at the end of the plate for two cases: (a) The boundary layer is laminar everywhere, and (b) The boundary layer is turbulent everywhere. Velocity of air (V) = 4.75 m/s

57 Length of plate (L) = 3.23m Assume kinematic viscosity of air = 1.5x10-5 m 2 /s Reynold no. (Re) = = 10.2x10 5 As Re>5x10 5, the boundary layer is turbulent. a. For laminar boundary layer, the upper value of Reynold no. is Re = 3x10 5 Thickness of laminar boundary layer b. Turbulent boundary layer Re = 10.2x10 5 Thickness of turbulent boundary layer = 0.029m = 29mm = 0.076m = 76mm 10. Draw a neat sketch of aerofoil and define the terms chord line, camber, angle of attack and aspect ratio associated with aerofoil. A wing of a small aircraft is rectangular in plan having a span of 10m and a chord of 1,2m. In straight and level flight of 240km/hr, the total aerodynamic force acting on the wing is 20KN. If the lift/drag ratio is 10, calculate the coefficient of lift and the total weight the aircraft can carry. Assume air density be 1.2 kg/m 3. chord line α L Chord line: It is the line joining the front and rear edge. Angle of attack: It is the angle between the direction of flowing fluid and chord line. Camber: It is the curvature of an airfoil. Aspect ratio: The ratio of span (L) to mean chord (C) is called aspect ratio. Speed of aircraft (V) = 240km/hr = (240x1000)/3600 = 66.67m/s Plan area (A) = 10x1.2 = 12m 2 Total aerodynamic force (F R ) = 20KN

58 F D = 1.99KN F L = 10x1.99 = 19.9KN Weight the aircraft can carry = F L = 19.9KN = Using Buckingham s π theorem, show that the velocity through a circular orifice is given by [ ] where H = head causing flow, D= diameter of orifice, = specific mass, = coefficient of viscosity, g = acceleration due to gravity. Total no. of variables (n) = 6 No. of fundamental dimensions (m) = 3 No. of π terms = 6-3 =3 (a) Take, g, H as repeating variables. First π term Equating the powers of M, L and T a1 = 0-3a1+b1+c1+1 = 0-2b1-1 = 0 b1 = -1/2-3x0+(-1/2)+c1+1 = 0 c1 = -1/2 Substituting the value of a1, b1 and c1 Second π term

59 Equating the powers of M, L and T a2+1 = 0 a2 = -1-3a2+b2+c2-1 = 0-2b2-1 = 0 b2 = -1/2-3x-1+(-1/2)+c2-1 = 0 c2 = 3/2 Substituting the value of a2, b2 and c2 Third π term Equating the powers of M, L and T a3 = 0-3a3+b3+c3+1 = 0-2b3 = 0 b3 = 0-3x-0+0+c3+1 = 0 c3 = -1 Substituting the value of a3, b3 and c3 Substituting the values of π1, π2, π3 in eq. a ( ) ( ) [ ]

60 12. Write down Euler s equation, Navier-Stoke s equation and Bernoulli s equation with appropriate expression of each component. Euler s equation = density of fluid = change in pressure with space = change in velocity with space = change in elevation with space Navier-Stoke s equation X, Y, Z: body force per unit mass in X, Y and Z direction = pressure gradient in X, Y and Z direction a x, a y, a z : acceleration in X, Y and Z direction T x, T y, T z : shear force per unit mass in X, Y and Z direction Bernoulli s equation = pressure head = velocity head Z = elevation head