Lecture 1: Kinematics, ideal mechanical systems and Bernoulli s equation

Size: px
Start display at page:

Download "Lecture 1: Kinematics, ideal mechanical systems and Bernoulli s equation"

Transcription

1 Lecture 1: Kinematics, ideal mechanical systems and Bernoulli s equation Should be able to: Understand some terms used in flow visualization and kinematics Understand Lagrangian and Eulerian frames of reference Understand the material (or total) derivative Derive Bernoulli s theorem using momentum balance along a streamline Solve simple problems using the Bernoulli theorem

2 Streamlines, Pathlines, Streaklines A streamline is everywhere tangent to the velocity field In steady D flow can integrate dx u dy dx = = dy v v u = dz w = ds V A pathline is the actual path travelled by a fluid particle over some time interval E.g. with a tracer particle A streakline is locus of fluid particles that pass through a given point in the flow Common experimentally: e.g. inject dye in water, smoke in air

3 Natural ways to describe flows? Lagrangian description A fluid is comprised of a large number of fluid particles having mass, momentum, internal energy, and other properties. Mathematical laws can then be written for each fluid particle. Eulerian description In the Eulerian description of fluid motion, we consider how flow properties change at a fluid element that is fixed in space and time (x,y,z,t), rather than following individual fluid particles.

4 Eulerian description = most common Eulerian description of fluid flow: a flow domain or control volume is defined by which fluid flows in and out We define field variables which are functions of space and time Pressure field: P=P(x,y,z,t) Velocity field: V=V(x,y,z,t) = u(x,y,z,t)i + v(x,y,z,t)j + w(x,y,z,t)k Acceleration field, a=a(x,y,z,t) = a x (x,y,z,t)i + a y (x,y,z,t)j + a z (x,y,z,t)k These (and other) field variables define the flow field. Suitable for formulation of initial boundary-value problems (PDE's). Leonhard Euler ( ). Particle s velocity at a point is same as fluid velocity Time derivative of particle position is fluid velocity d d V = Vparticle = x particle particle + dt dt What about acceleration? d dt ( t) + y ( t) z ( t) particle

5 Acceleration in Eulerian coordinates? Acceleration is the time derivative of the particle's velocity: To take the time derivative of the velocity, chain rule must be used V = V a particle V = t ( x ( t), y ( t), z ( t) t) particle particle particle particle, V + x d dt x particle ( t) + y ( t) + z ( t) Advective acceleration term: acceleration of fluid even in steady flow V V V V a particle = + u + v + w t x y z Local acceleration V y d dt Advective acceleration particle V z d dt d a particle = V dt particle particle Advective acceleration is nonlinear: source of many phenomena and main challenge in solving fluid flow equations

6 Material derivative & acceleration: Total derivative operator is called material derivative D Dt = + u + v + w = + t x y z t V Other names for the material derivative include: total, particle, Lagrangian, Eulerian, and substantial derivative d Lagrangian: dt D Eulerian: = + V Dt t Consider advective acceleration tangential & normal to steady streamline:

7 Bernoulli s equation Apply Newton s second law along a streamline Assume no viscous effects Assume no heat transfer Acceleration = material derivative of velocity V along a streamline s mas = Fs = PdA W = ρ dads g m = ρ dads V V as = + V t s ( P + dp) da W sinθ Steady incompressible flow, V 1 P dz V + + g = 0 divide by m and take ds 0 s ρ s ds V P Along a streamline: + + gz = constant ρ Bernoulli s equation

8 Using the Bernoulli equation Frictionless flow with no energy transfer 3 terms in Bernoulli equation Potential energy Flow energy Kinetic energy gz P ρ Alternatively, dividing by g: Static head hydrostatic head (pressure head) Dynamic head Unsteady, variable density version of Bernoulli V P 1 ρ + V 1 + g z 1 = P ρ + V + g z = const P P 1 ρ z P ρg V g 1 V ds + t + V V 1 1 dp + ρ 1 + g( z z 1 )= 0 Bernoulli s equation is an energy equation for an ideal mechanical system ( V V ) + g( z z ) = 0 1 1

9 Example 1 Air flows past a ball, as shown. It is determined that the velocity along the x-axis for x<-a is given by: 3 a V = V x 3 Determine the pressure variation along the x-axis and the stagnation pressure at x = -a

10 Example Water flows from a garden hose. A child places his thumb to cover most of the outlet. A thin stream of water jets upwards. a) If the gage pressure in the hose just upstream of his thumb is 400kPa what is the maximum height the jet can achieve, (assuming no energy losses at position 1)? b) Assuming that soon above position 1 the pressure in the jet is P atm, derive an expression for the velocity V J as a function of height z above 1

11 F=ma: Normal to a streamline? Newton s second law normal to streamline Assume no viscous effects Assume no heat transfer Steady incompressible flow, divide by m and take dn 0 n = in direction of curvature Circular streamlines at same height implies radial increase in pressure ma W a n n = F = ρ da V = R n n = PdA n 1 P dz V g = ρ n dn R ( P + dp ) da dn g, m = ρ da n n n W dn cosθ

12 Example 3: Flows a & b have circular streamlines: Sketch the pressure distributions, assuming p(r 0 ) = p 0

13 Fluids in the news:

14 Lecture : Applications and limitations of Bernoulli s equation Bernoulli states that total pressure is constant P total = P + ρv + ρgz Stagnation pressure is the pressure at any point where the fluid is brought to rest isentropically This can be used with a Pitot tube and piezometer to measure fluid velocity P stag ρv = P +

15 Example 4: A piezometer and a Pitot tube are tapped into a horizontal water pipe as shown. Determine the velocity at the center of the pipe?

16 Example 5: a) What is the pressure at the nose of a torpedo moving in salt water at 100ft/s at a depth of 30ft? b) If the pressure at a point on the side of the torpedo, at the same depth as the nose, is 10.0psi gage, what is the fluid velocity at that point?

17 Free streams and jets: Example 6: Water flows along an undulating path, as shown. What is the pressure variation: a) between points 1 &? b) between points 3 & 4?

18 Example 7: A stream of liquid flows from a hole of diameter 0.01m near the bottom of a reservoir of diameter 0.m, with height as shown. The liquid is replenished continuously. Find the flow rate Q that keeps the level constant

19 Example 8: Water flows under the sluice gate as shown. Estimate the flow rate per unit width Q

20 Energy & Hydraulic Grade Lines Bernoulli: frictionless flow with no energy transfer Energy equation for an ideal z Static head + P ρg Pressure head + V g Dynamic head = constant mechanical system Track these quantities along a flow path EGL = total head HGL = static + pressure head Measure with series of: Pitot tubes (EGL) Piezometric taps (HGL)

21 HGL and EGL Plot of EGL and HGL generally decays along flow path Variation in EGL along real flow paths is a measure of energy losses/gains (see later) Frictionless flow: HGL indicates positive & negative pressure P V EGL = + + ρg g z P HGL = + ρg z

22 Limitations on using the Bernoulli Equation Steady flow: d/dt = 0 Frictionless flow No shaft work: w pump =w turbine =0 Incompressible flow: ρ = constant No heat transfer: q net,in =0 Applied along a streamline

23 Fluids in the news:

24 What we covered Eulerian and Lagrangian frames of reference Converting time derivatives between the two systems Material derivative and acceleration Bernoulli s equation Derivation from momentum balance Interpretation as an energy balance Examples of application V V s V + 1 P + ρ s P + gz ρ + g dz ds = 0 = constant

25 Lecture 3: Integral analyses of fluid flow Systems and control volumes Reynolds transport theorem General principals of conservation laws for a control volume Conservation of mass for a control volume

26 Systems and control volumes Generally in mechanics we apply conservation laws to estimate quantities of interest A (closed) system is a region in space consisting of a fixed quantity of matter No mass crosses system boundary A control volume, is any properly selected region in space Analogous to the Free body diagram in Solid mechanics / dynamics Mass can cross control surfaces 6

27 Reynolds Transport Theorem (RTT) The fundamental conservation laws (conservation of mass, energy &momentum) apply directly to systems However, in most fluid mechanics problems, control volume analysis is preferred over system Control volumes are chosen more relevant to the problem at hand Therefore, we need to transform the conservation laws from a system to a control volume. This is accomplished with the Reynolds transport theorem (RTT) At a time t, we define a system to be the mass in the control volume (CV) bounded by the control surface (CS) Reynolds transport theorem: Rate of change of B in the system is equal to the rate of change of B in the CV plus the net flow of B out of the CV, across the CS n

28 Mathematical statement What is B? Any extensive property Associated intensive property is β β is the amount of B per unit mass B sys β = = ρβ dv V db dm Mass Momentum Energy Angular momentum B, Extensive properties m mv E mr V β, Intensive properties 1 V e r V d dt B sys = d dt CV ρβ dv + CS ρβ ( V n) r da Reynolds transport theorem: Rate of change of B in the system is equal to the rate of change of B in the CV plus the net flow of B out of Mech the CV, 80: across Frigaard the CS

29 Moving / deforming control volume Reynolds transport theorem d dt B sys Is valid for fixed, moving and/or deforming control volumes, The velocity V r in the second term is the relative velocity: V r = V -V CS V is the fluid velocity, (typically given in an inertial system) V CS is velocity of the moving/deforming CS, (in the same system as V) = d dt CV ρβ dv + CS ρβ ( V n) r da

30 Conservation of Mass Conservation of mass is one of the most fundamental principles in nature. Mass, like energy, is a conserved property, and it cannot be created or destroyed during a process. For closed systems mass conservation is implicit since the mass of the system remains constant during a process. For control volumes, mass can cross the boundaries which means that we must keep track of the amount of mass entering and leaving the control volume.

31 Mass Conversvation d dt B sys = d dt CV ρβ dv + CS ρβ ( V n) r da B =m = Mass dm β = =1 dm n Mass conserved : d dt B sys = 0 0 = d dt CV ρ dv + CS ρ ( V n) r da

32 Mass and Volume Flow Rates V Consider a fixed CV and CS, so that V r = V Consider flow through a control surface, with area A c. Flow rates obtained by integration The amount of mass flowing through a control surface per unit time is called the mass flow rate and is denoted m = δm = ρv da A c The dot over a symbol is used to indicate time rate of change. Volume flow rate through A c Q = V n dac = Vn dac = Vavg A Ac Ac 1 Vavg = VndAc A Often density doesn t change much A c ( ) c c A m = ρv A c avg c n c

33 Conservation of Mass Principle For fixed CV of arbitrary shape, rate of change of mass within the CV d dt m CV net mass flow rate m net = CS δm = d = ρ dv dt CV CS ( V n) da = m m Therefore, general conservation of mass for a fixed CV is: m in Same expression for deforming or moving CV, except net mass flow rate in/out are relative to the CS ρ m = out dm dt CV out in

34 Steady Flow Processes For steady flow, total amount of mass contained in CV is constant. Total amount of mass entering must be equal to total amount of mass leaving in m = out m For incompressible flows, in VA = VA n n n n out

35 Example 1: If the average water velocity in a in a 300mm pipe is 0.5m/s, what is the average velocity in a 75mm water jet coming from the pipe?

36 Example : An airplane moves forward at speed 971km/hr. The frontal intake area of the jet engine is 0.8m and the entering air density is 0.736kg/m 3. A stationary observer determines that relative to the Earth, the jet engine exhaust gases move away from the engine with a speed 1050km/hr. The engine exhaust area is 0.558m, and the exhaust gas density is 0.515kg/m 3. Estimate the mass flowrate of fuel into the engine in kg/h

37 Example 3: V J A jet of water leaves the nozzle at mean speed 7 m/s And strikes the turning vane as shown. a) Compute the mean velocity exiting the CV if the vane is stationary b) Repeat the calculation if the vane moves to the right at a steady velocity of m/s. For an observer standing at the c.v. inlet (point 1) V 1 = V J U c = 7 = 5 m/s = ρ 1 V 1 A 1 = 998 kg/m 3 *5 m/s*π*.04 /4 = 6.71 kg/s m Note: 1 The inlet velocity used to specify the mass flow rate is again measured relative to the inlet boundary, but now is given by V J U c. Exit: m = 6.71 kg/s, Again, since ρ and A are cons., V = 5 m/s Again, the exit flow is most easily specified by conservation of mass concepts.

38 Example 4: A room contains dust at a uniform concentration, C=ρ dust /ρ. The room is to be cleaned by flushing air through the room. Find an expression for time rate of change of dust mass in the room. V out

39 What we covered Derived the general Reynolds Transport Theorem (RTT) d d B sys = ρβ dv + ρβ ( Vr n) da dt dt CV CS Velocity is relative to moving / deforming control volumes Used the RTT to derive conservation of mass for a CV d 0 = ρ dv + ρ( Vr n) da dt CV CS Mass flow rates: m net = δm = ρ( V n) da = ρvn da Mass conservation can be written as: CS CS m in m = out CS dm dt CV

40 Lecture 4: Conservation of momentum Should be able to: Use Reynolds transport theorem to derive the conservation of linear momentum relation for a CV Examples for fixed & moving CV Examples coupled with mass conservation CV Last lecture we derived Reynolds transport theorem Applied this to give conservation of mass relation in a CV In steady flow conservation of mass is relation between flow rates What about computing forces?

41 Newton s Laws Newton s laws are relations between motions of bodies and the forces acting on them First law: a body at rest remains at rest, and a body in motion remains in motion at the same velocity in a straight path when the net force acting on it is zero Second law: the acceleration of a body is proportional to the net force acting on it and is inversely proportional to its mass F dv = m a = m = dt d dt ( mv)

42 Momentum Conservation for a CV d dt B sys = d dt B =mv = Momentum d β = mv = dm ( ) V CV Momentum conserved : ρβ dv + d dt B sys CS ρβ = F ( V n) r da n n CV F = d dt CV ρv dv + CS ρv V ( n) r da

43 Orientation of CS s and the Momentum Flux Correction Factor A Momentum is a vector quantity Flux term in momentum balance is more complex than for mass conservation Nearly always we orient CS s to be normal to the flow direction Does not affect mass balance Simplifies flux term ( V n) dac = ρv VρVn dac = m βv c A c avg βv avg = A c A VρV c ρv n n da da c c avg Note that at inlet/outlet, V avg often only has component in normal direction: In an incompressible flow, we have 1 Vn β = da c Ac A c Vn, avg β>1 always and for turbulent flow β in range V = A c VdA A c da c c

44 Choosing a Control Volume CV is chosen by fluid dynamicist and there is a lot of freedom. However, selection of CV can either simplify or complicate analysis Clearly define all boundaries. Analysis is often simplified if CS is normal to flow direction Clearly identify all fluxes crossing the CS Clearly identify forces and torques of interest acting on the CV and CS Fixed, moving, and deforming control volumes: For moving CV, use relative velocity, V r = V - V CS = V -V CV For deforming CV, use velocity relative to each deforming control surfaces V r = V V CS Note that for momentum balance, as given, Newton s laws require an inertial frame of reference For non-inertial frames, add fictitious forces addressed later in course

45 Forces Acting on a CV Forces acting on CV consist of body forces that act throughout the entire body of the CV (such as gravity, electric, and magnetic forces) and surface forces that act on the control surface (such as pressure and viscous forces, and reaction forces at points of contact). n Body forces act on each volumetric portion dv of the CV. Surface forces act on each portion da of the CS. df df surface 45

46 Body Forces The most common body force is gravity, which exerts a downward g force on every differential element of the CV df body = df gravity = ρgdv k df body = df gravity = ρgdv Summing over all differential volume elements, gives total body j force acting on CV i F body = ρg CV dv = m CV g

47 Surface Forces Surface forces include both normal and tangential components Diagonal components σ xx, σ yy, σ zz are called normal stresses and are due to pressure and viscous stresses Off-diagonal components σ xy, σ xz, etc., are called shear stresses and are due solely to viscous stresses Force/area acting on CS with normal n: ( n σ) = j Total surface force acting on CS F surface i= 1,, 3 = σ n CS ij i n σ ds

48 Body and Surface Forces Surface integrals are cumbersome. Careful selection of CV allows expression of total force in terms of more readily available quantities like weight, pressure, and reaction forces. Goal is to choose CV to expose only the forces to be determined and a minimum number of other forces. F = Fgravity + pressure viscous other F + F + F Body Simplification for steady flows: F = Out Surface m βv m βv avg In avg

49 Example 1: For the pipe-flow reducing section shown, D 1 = 8cm, D = 5cm and all fluids are at 0 C. If V 1 = 5m/s and the manometer reading is h = 58cm, estimate the total horizontal force on the flange bolts.

50 Example : Force on an elbow Water flows steadily through a 90 elbow, exiting to atmosphere at position. If D = 5cm & P 1 = 3.43 kpa, what anchoring force is needed to keep the elbow in place?

51 Example 3: Moving control volume A water jet 4 cm in diameter with a velocity 7 m/s is directed towards a turning vane (θ = 40 ) that is moving with velocity, U c = m/s. Determine the force F necessary to hold the vane in steady motion. F b = m e V e m i V i V 1 = V J U c = 7 = 5 m/s and V = 5 m/s inclined 40. -F b = 6.71 kg/s * 5 m/s *cos kg/s * 5 m/s and -F b = kg m/s = 7.34 N ans. or Fb

52 Example 4: Non-uniform pressure A sluice gate across a channel of width b operates in both open and close positions, as shown. Is the anchoring force required to hold the gate in place larger when the gate is open or closed?

53 Fluids in the news

54 What we covered Derived linear momentum equation CV F Momentum flux correction factor Simplified forms for momentum balance Sum of forces include: Pressure, body, shear and mechanical Computed examples of: = d dt Stationary control volume Moving, control volume in an inertial frame CV ρv dv + CS ρv V ( n) r da

55 Lecture 5: Momentum II Examples using the momentum balance to generate thrust Derive the conservation of momentum for a control volume moving in an accelerating frame of reference. Examples of accelerating frames of reference

56 Example 1: A liquid jet of velocity V j and area A j strikes a single 180 bucket on a turbine wheel rotating at angular velocity Ω. a) Find an expression for the power delivered b) At what Ω is the power a maximum

57 Example : A static thrust stand as sketched is to be designed for testing a jet engine. The following conditions are known for a typical test: Intake air velocity: 00m/s Exhaust gas velocity: 500m/s Intake cross-sectional area: 1m Intake static gage pressure: -.5kPa Intake static temperature: 68K Exhaust static gage pressure: 0kPa Estimate the nominal anchoring force for design

58 Non-inertial frames of reference Absolute Acceleration = Non-inertial Acceleration + Relative Acceleration Consider fluid particle, with position: Y r i = y x Velocity: z V i = V = velocity in non-inertial frame X How about acceleration? Z

59 Relative acceleration dv a i = + dt a rel Y y x z X Z

60 Non-inertial frames of reference F = ma i Newton s nd Law dv ma i = m + arel = F F marel = dt m dv dt Effect on control volume analysis is to add fictitious forces to the net force balance t F a = + cv d mcv Vρ d V V d me cv cv A e A i V d m In general, situations can be very complex. In many mechanical systems we restrict motion to specific degrees of freedom. i

61 Example 1: Accelerating control volume t F acv d mcv = Vρ d V + V d m e cv cv A water jet 4cm in diameter with a velocity of 7m/s is directed to a turning vane moving with velocity U c (t), and with angle θ = 40 as shown. Determine the acceleration of the vane as a function of time, assuming initially stationary. A e A i V d m i

62 Example : Rocket acceleration t F acv d mcv = Vρ d V + V d m e cv cv A e A i V d m i Calculate the rocket s angular velocity, given the rocket s initial mass, the propellant exit velocity and the exit mass flow rate (both assumed constant)? ω m, V, p = e e e p atm

63 What we covered: Examples of using the linear momentum balance to calculate thrust, in inertial frames We derived the fictional forces that we experience in a non-inertial frame of reference. F a = + cv d mcv Vρ d V V d m t e cv cv Note that the fictional force (acceleration) is negative. A e A i V d m i Examples Don t do this at home

64 Lectures 7 & 8: Energy balance for CV s Should be able to: Use the Reynolds Transport Theorem (RTT) to derive the conservation of energy for a control volume Understand forms of energy and how work is done on system Understand efficiency Relationship with Bernoulli equation for single stream systems Application examples

65 General Energy Equation One of the most fundamental laws in nature is the 1st law of thermodynamics, which is also known as the conservation of energy principle. It states that energy can be neither created nor destroyed during a process; it can only change forms Falling rock, picks up speed as PE is converted to KE. If air resistance is neglected, PE + KE = constant Enthalpy?

66 General Energy Equation Energy in a system is in 3 forms Kinetic energy: 0.5mV Potential energy: mgz Enthalpy: H The energy content of a closed system can be changed by two mechanisms: heat transfer Q and work transfer W Conservation of energy for a closed system can be expressed in rate form as In open system (CV) work comes from Pressure work Viscous work Shaft work (pump / turbine) Net heat transfer to the system: Net power input to the system:

67 Energy Conservation for a CV (from RTT) d dt B sys = d dt CV ρβ dv + CS ρβ ( V n) r da n B = E sys β = e = u+0.5v +gz = (internal + kinetic + potential) specific energies QQ nnnnnn,iiii + WW nnnnnn,iiii = ddee ssssss dddd General Energy Equation: Q net, in + W net, in = de dt sys = dddd dddd = d dt CV eρ dv + CS eρ ( V n) r da

68 Pressure work Where does expression for pressure work come from? When piston moves down ds under the influence of F=PA, the work done on the system is dw boundary =PAds. If we divide both sides by dt, we have δδ WW pppppppppppppppp = δδ WW bbbbbbbbbbbbbbbb = PPPP dddd dddd = PPPPVV pppppppppppp For generalized control volumes: δδ WW bbbbbbbbbbbbbbbb = PPPPPPVV nn = PPPPPP(VV rr.n) V n Note sign conventions: n is outward pointing normal vector Negative sign ensures that work done is positive when is done on the system. dw dt p = P( V n) da cs r

69 Viscous work Viscous work is the work done on the control volume surface due to shear force Can be a real problem because we don't know velocity distribution, therefore don't know shear stress. Solution! Draw control volume carefully, e.g. Perpendicular to flow Where the velocity is zero, e.g., at a solid boundary Or where the shear stress is zero, (or small, or perpendicular to the velocity) dw dt ν = cs V τ ij n da

70 Energy Analysis of Steady Flows Q net, in Simplifications: P V P V + W = m + u + + gz m shaft, net, in + u + out ρ in ρ Kinetic energy correction factor set to 1 (see later) Rate of change of energy content = 0 Well-chosen CV means rate of viscous work is zero Pressure work rate is included in flux terms Note, could use enthalpy: h=u+p/ρ, but confusing later when h= head. Single inflow/outflow: specific quantities P 1 V1 m + u gz1 ρ 1 P m ρ 1 Out In V + u + + gz Fixed control volume + Q net, in Wshaft, net, in q w w net, in shaft, net, in pump + w shaft, net, in P1 V1 + + ρ P1 V1 + + ρ 1 1 Q = m + gz net, in + gz 1 1 P = ρ W + P = ρ shaft, net, in m V + V + + gz + gz P V = + u + ρ + gz + w + [ u u q ] turbine + e + gz 1 net, in e mech, loss mech, loss 1

71 Example 1: A pump delivers water at a steady rate of 300 gal/min as shown. Upstream [sect. (1)] pipe diameter is 3.5 in. & the pressure is 18 psi. Downstream [sect. ()] the pipe diameter is 1 in. & the pressure is 60 psi. The change in elevation is zero. The rise in internal energy of water, u -u 1, associated with a temperature rise across the pump is 93 ft lb/lbm. Q. Determine the power that the pump requires.

72 Example : A steam turbine generator unit is used to produce electricity. The steam enters the turbine with a velocity of 30 m/s and enthalpy, h 1, of 3348 kj/kg. The steam leaves the turbine as a mixture of vapor and liquid having a velocity of 60 m/s and an enthalpy h of 550 kj/kg. The flow through the turbine is adiabatic, and changes in elevation are negligible. Q. Determine the work output involved, per unit mass of steam through-flow

73 Head form of energy equation (single stream CV) Divide by g to express each term in units of length P V P V ρ z1+ hpump = + + z + hturbine + hl 1g g ρg g Magnitude of each term is expressed as an equivalent height of fluid, i.e. a head Note, with no shaft work or irreversible head losses, similar to Bernoulli s equation

74 Example 3: in + in out out zin + = + zout + + hf hp + ht V P V P g ρg g ρg Given the initial exit velocity what is the pump power & the time taken to empty the tank?

75 Example 4: The pump shown in the figure adds 10 horsepower to the water as it pumps water from the lower lake to the upper lake. The elevation difference between the lake surfaces is 30 ft and the head loss is 15 ft. Q. Determine the flow rate and power loss associated with this flow

76 Efficiency Transfer of e mech is usually accomplished by a rotating shaft: shaft work Pump, fan, propulsion: receives shaft work (e.g., from an electric motor) and transfers it to the fluid as mechanical energy Turbine: converts e mech of a fluid to shaft work. In absence of irreversible losses (e.g., friction), mechanical efficiency of a device or process can be defined as E E η mech = = 1 E E mech, out mech, loss mech, in mech, in If η mech < 100%, losses have occurred during conversion.

77 Pump and Turbine Efficiencies In fluid systems, we are usually interested in increasing the pressure, velocity, and/or elevation of a fluid. In these cases, efficiency is better defined as the ratio of (supplied or extracted work) vs. rate of increase in mechanical energy Overall efficiency must include motor or generator efficiency.

78 Example 5: The pump of a water distribution system is powered by a 15-kW electric motor (efficiency 90%). The water flow rate is 50l/s. The inlet/outlet pipe diameters are the same and there is negligible increase in elevation across the pump. Determine a) the mechanical efficiency of the pump; b) the temperature rise of the water as it flows through the pump, due to inefficiency.

79 Example 6: In a hydroelectric power plant, 100m 3 /s of water flows from an elevation of 10m to a turbine. The total irreversible head loss is determined to be 35m. If the overall efficiency of the turbine generator is 80%, estimate the electric power output.

80 Example 7: An axial-flow ventilating fan driven by a motor that delivers 0.4 kw of power to the fan blades produces a 0.6-m diameter axial stream of air having a speed of 1 m/s. The flow upstream of the fan involves negligible speed. Q. Determine how much of the work to the air actually produces useful effects, that is, fluid motion and a rise in available energy. Estimate the mechanical efficiency of this fan.

81 Kinetic Energy Correction Factor Up to this point, the velocity used in the kinetic energy term of the energy equation has been the mass average velocity obtained from the definition of flow rate: mm = ρρρρ VV However, since V varies over the flow area and the kinetic energy term varies with the square of the velocity, using this definition of V may not result in an accurate evaluation of the kinetic energy term for the flow. This problem can be corrected through the use of the kinetic energy correction factor, defined for incompressible flow from ρ u da Vav A = ρβ Solving for ββ we obtain 3 1 u β = da A V av For fully developed laminar flow, β= and for turbulent pipe flow, a value from 1.04 to 1.11 is common

82 Example 8: The small fan shown moves air at a mass flowrate of 0.1 kg/min. Upstream the pipe diameter is 60 mm, the flow is laminar and the kinetic energy coefficient, is.0. Downstream of the fan, the pipe diameter is 30 mm, the flow is turbulent and the kinetic energy coefficient, is The rise in static pressure across the fan is 0.1 kpa, and the fan motor draws 0.14 W. Q. Compare the value of losses calculated: (a) assuming uniform velocity distributions, (b) considering actual velocity distributions.

83 What we covered Derived conservation of energy for a CV Converted to head form for steady, single stream CV in + in out out zin + = + zout + + hf hp + ht V P V P g ρg g ρg Note: Bernoulli s equation as a frictionless flow with no energy in / out. P 1 ρ + V 1 + g z 1 = P ρ + V + g z = const Efficiency of pumps & turbines Examples using energy equation in different formats

84 Fluids in the news:

Chapter 5: Mass, Bernoulli, and Energy Equations

Chapter 5: Mass, Bernoulli, and Energy Equations Chapter 5: Mass, Bernoulli, and Energy Equations Introduction This chapter deals with 3 equations commonly used in fluid mechanics The mass equation is an expression of the conservation of mass principle.

More information

Chapter 5: Mass, Bernoulli, and

Chapter 5: Mass, Bernoulli, and and Energy Equations 5-1 Introduction 5-2 Conservation of Mass 5-3 Mechanical Energy 5-4 General Energy Equation 5-5 Energy Analysis of Steady Flows 5-6 The Bernoulli Equation 5-1 Introduction This chapter

More information

Objectives. Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation

Objectives. Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation Objectives Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation Conservation of Mass Conservation of Mass Mass, like energy, is a conserved

More information

Chapter Four fluid flow mass, energy, Bernoulli and momentum

Chapter Four fluid flow mass, energy, Bernoulli and momentum 4-1Conservation of Mass Principle Consider a control volume of arbitrary shape, as shown in Fig (4-1). Figure (4-1): the differential control volume and differential control volume (Total mass entering

More information

Chapter 2: Basic Governing Equations

Chapter 2: Basic Governing Equations -1 Reynolds Transport Theorem (RTT) - Continuity Equation -3 The Linear Momentum Equation -4 The First Law of Thermodynamics -5 General Equation in Conservative Form -6 General Equation in Non-Conservative

More information

Chapter 6: Momentum Analysis

Chapter 6: Momentum Analysis 6-1 Introduction 6-2Newton s Law and Conservation of Momentum 6-3 Choosing a Control Volume 6-4 Forces Acting on a Control Volume 6-5Linear Momentum Equation 6-6 Angular Momentum 6-7 The Second Law of

More information

vector H. If O is the point about which moments are desired, the angular moment about O is given:

vector H. If O is the point about which moments are desired, the angular moment about O is given: The angular momentum A control volume analysis can be applied to the angular momentum, by letting B equal to angularmomentum vector H. If O is the point about which moments are desired, the angular moment

More information

CLASS Fourth Units (Second part)

CLASS Fourth Units (Second part) CLASS Fourth Units (Second part) Energy analysis of closed systems Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. MOVING BOUNDARY WORK Moving boundary work (P

More information

Lesson 6 Review of fundamentals: Fluid flow

Lesson 6 Review of fundamentals: Fluid flow Lesson 6 Review of fundamentals: Fluid flow The specific objective of this lesson is to conduct a brief review of the fundamentals of fluid flow and present: A general equation for conservation of mass

More information

Chapter 6: Momentum Analysis of Flow Systems

Chapter 6: Momentum Analysis of Flow Systems Chapter 6: Momentum Analysis of Flow Systems Introduction Fluid flow problems can be analyzed using one of three basic approaches: differential, experimental, and integral (or control volume). In Chap.

More information

ROAD MAP... D-0: Reynolds Transport Theorem D-1: Conservation of Mass D-2: Conservation of Momentum D-3: Conservation of Energy

ROAD MAP... D-0: Reynolds Transport Theorem D-1: Conservation of Mass D-2: Conservation of Momentum D-3: Conservation of Energy ES06 Fluid Mechani UNIT D: Flow Field Analysis ROAD MAP... D-0: Reynolds Transport Theorem D-1: Conservation of Mass D-: Conservation of Momentum D-3: Conservation of Energy ES06 Fluid Mechani Unit D-0:

More information

Lecture 4. Differential Analysis of Fluid Flow Navier-Stockes equation

Lecture 4. Differential Analysis of Fluid Flow Navier-Stockes equation Lecture 4 Differential Analysis of Fluid Flow Navier-Stockes equation Newton second law and conservation of momentum & momentum-of-momentum A jet of fluid deflected by an object puts a force on the object.

More information

The Bernoulli Equation

The Bernoulli Equation The Bernoulli Equation The most used and the most abused equation in fluid mechanics. Newton s Second Law: F = ma In general, most real flows are 3-D, unsteady (x, y, z, t; r,θ, z, t; etc) Let consider

More information

COURSE NUMBER: ME 321 Fluid Mechanics I 3 credit hour. Basic Equations in fluid Dynamics

COURSE NUMBER: ME 321 Fluid Mechanics I 3 credit hour. Basic Equations in fluid Dynamics COURSE NUMBER: ME 321 Fluid Mechanics I 3 credit hour Basic Equations in fluid Dynamics Course teacher Dr. M. Mahbubur Razzaque Professor Department of Mechanical Engineering BUET 1 Description of Fluid

More information

FLUID MECHANICS. Chapter 3 Elementary Fluid Dynamics - The Bernoulli Equation

FLUID MECHANICS. Chapter 3 Elementary Fluid Dynamics - The Bernoulli Equation FLUID MECHANICS Chapter 3 Elementary Fluid Dynamics - The Bernoulli Equation CHAP 3. ELEMENTARY FLUID DYNAMICS - THE BERNOULLI EQUATION CONTENTS 3. Newton s Second Law 3. F = ma along a Streamline 3.3

More information

Chapter 3 Bernoulli Equation

Chapter 3 Bernoulli Equation 1 Bernoulli Equation 3.1 Flow Patterns: Streamlines, Pathlines, Streaklines 1) A streamline, is a line that is everywhere tangent to the velocity vector at a given instant. Examples of streamlines around

More information

Therefore, the control volume in this case can be treated as a solid body, with a net force or thrust of. bm # V

Therefore, the control volume in this case can be treated as a solid body, with a net force or thrust of. bm # V When the mass m of the control volume remains nearly constant, the first term of the Eq. 6 8 simply becomes mass times acceleration since 39 CHAPTER 6 d(mv ) CV m dv CV CV (ma ) CV Therefore, the control

More information

Conservation of Momentum using Control Volumes

Conservation of Momentum using Control Volumes Conservation of Momentum using Control Volumes Conservation of Linear Momentum Recall the conservation of linear momentum law for a system: In order to convert this for use in a control volume, use RTT

More information

SYSTEMS VS. CONTROL VOLUMES. Control volume CV (open system): Arbitrary geometric space, surrounded by control surfaces (CS)

SYSTEMS VS. CONTROL VOLUMES. Control volume CV (open system): Arbitrary geometric space, surrounded by control surfaces (CS) SYSTEMS VS. CONTROL VOLUMES System (closed system): Predefined mass m, surrounded by a system boundary Control volume CV (open system): Arbitrary geometric space, surrounded by control surfaces (CS) Many

More information

MASS, MOMENTUM, AND ENERGY EQUATIONS

MASS, MOMENTUM, AND ENERGY EQUATIONS MASS, MOMENTUM, AND ENERGY EQUATIONS This chapter deals with four equations commonly used in fluid mechanics: the mass, Bernoulli, Momentum and energy equations. The mass equation is an expression of the

More information

where = rate of change of total energy of the system, = rate of heat added to the system, = rate of work done by the system

where = rate of change of total energy of the system, = rate of heat added to the system, = rate of work done by the system The Energy Equation for Control Volumes Recall, the First Law of Thermodynamics: where = rate of change of total energy of the system, = rate of heat added to the system, = rate of work done by the system

More information

CEE 3310 Control Volume Analysis, Oct. 10, = dt. sys

CEE 3310 Control Volume Analysis, Oct. 10, = dt. sys CEE 3310 Control Volume Analysis, Oct. 10, 2018 77 3.16 Review First Law of Thermodynamics ( ) de = dt Q Ẇ sys Sign convention: Work done by the surroundings on the system < 0, example, a pump! Work done

More information

Angular momentum equation

Angular momentum equation Angular momentum equation For angular momentum equation, B =H O the angular momentum vector about point O which moments are desired. Where β is The Reynolds transport equation can be written as follows:

More information

ME3560 Tentative Schedule Spring 2019

ME3560 Tentative Schedule Spring 2019 ME3560 Tentative Schedule Spring 2019 Week Number Date Lecture Topics Covered Prior to Lecture Read Section Assignment Prep Problems for Prep Probs. Must be Solved by 1 Monday 1/7/2019 1 Introduction to

More information

Chapter 7 The Energy Equation

Chapter 7 The Energy Equation Chapter 7 The Energy Equation 7.1 Energy, Work, and Power When matter has energy, the matter can be used to do work. A fluid can have several forms of energy. For example a fluid jet has kinetic energy,

More information

Where does Bernoulli's Equation come from?

Where does Bernoulli's Equation come from? Where does Bernoulli's Equation come from? Introduction By now, you have seen the following equation many times, using it to solve simple fluid problems. P ρ + v + gz = constant (along a streamline) This

More information

Chapter 5. Mass and Energy Analysis of Control Volumes

Chapter 5. Mass and Energy Analysis of Control Volumes Chapter 5 Mass and Energy Analysis of Control Volumes Conservation Principles for Control volumes The conservation of mass and the conservation of energy principles for open systems (or control volumes)

More information

ME3560 Tentative Schedule Fall 2018

ME3560 Tentative Schedule Fall 2018 ME3560 Tentative Schedule Fall 2018 Week Number 1 Wednesday 8/29/2018 1 Date Lecture Topics Covered Introduction to course, syllabus and class policies. Math Review. Differentiation. Prior to Lecture Read

More information

3.8 The First Law of Thermodynamics and the Energy Equation

3.8 The First Law of Thermodynamics and the Energy Equation CEE 3310 Control Volume Analysis, Sep 30, 2011 65 Review Conservation of angular momentum 1-D form ( r F )ext = [ˆ ] ( r v)d + ( r v) out ṁ out ( r v) in ṁ in t CV 3.8 The First Law of Thermodynamics and

More information

4 Mechanics of Fluids (I)

4 Mechanics of Fluids (I) 1. The x and y components of velocity for a two-dimensional flow are u = 3.0 ft/s and v = 9.0x ft/s where x is in feet. Determine the equation for the streamlines and graph representative streamlines in

More information

CHAPTER 3 BASIC EQUATIONS IN FLUID MECHANICS NOOR ALIZA AHMAD

CHAPTER 3 BASIC EQUATIONS IN FLUID MECHANICS NOOR ALIZA AHMAD CHAPTER 3 BASIC EQUATIONS IN FLUID MECHANICS 1 INTRODUCTION Flow often referred as an ideal fluid. We presume that such a fluid has no viscosity. However, this is an idealized situation that does not exist.

More information

FE Exam Fluids Review October 23, Important Concepts

FE Exam Fluids Review October 23, Important Concepts FE Exam Fluids Review October 3, 013 mportant Concepts Density, specific volume, specific weight, specific gravity (Water 1000 kg/m^3, Air 1. kg/m^3) Meaning & Symbols? Stress, Pressure, Viscosity; Meaning

More information

Chapter 4 DYNAMICS OF FLUID FLOW

Chapter 4 DYNAMICS OF FLUID FLOW Faculty Of Engineering at Shobra nd Year Civil - 016 Chapter 4 DYNAMICS OF FLUID FLOW 4-1 Types of Energy 4- Euler s Equation 4-3 Bernoulli s Equation 4-4 Total Energy Line (TEL) and Hydraulic Grade Line

More information

Fluid Mechanics. du dy

Fluid Mechanics. du dy FLUID MECHANICS Technical English - I 1 th week Fluid Mechanics FLUID STATICS FLUID DYNAMICS Fluid Statics or Hydrostatics is the study of fluids at rest. The main equation required for this is Newton's

More information

MOMENTUM PRINCIPLE. Review: Last time, we derived the Reynolds Transport Theorem: Chapter 6. where B is any extensive property (proportional to mass),

MOMENTUM PRINCIPLE. Review: Last time, we derived the Reynolds Transport Theorem: Chapter 6. where B is any extensive property (proportional to mass), Chapter 6 MOMENTUM PRINCIPLE Review: Last time, we derived the Reynolds Transport Theorem: where B is any extensive property (proportional to mass), and b is the corresponding intensive property (B / m

More information

2.The lines that are tangent to the velocity vectors throughout the flow field are called steady flow lines. True or False A. True B.

2.The lines that are tangent to the velocity vectors throughout the flow field are called steady flow lines. True or False A. True B. CHAPTER 03 1. Write Newton's second law of motion. YOUR ANSWER: F = ma 2.The lines that are tangent to the velocity vectors throughout the flow field are called steady flow lines. True or False 3.Streamwise

More information

EGN 3353C Fluid Mechanics

EGN 3353C Fluid Mechanics Lecture 8 Bernoulli s Equation: Limitations and Applications Last time, we derived the steady form of Bernoulli s Equation along a streamline p + ρv + ρgz = P t static hydrostatic total pressure q = dynamic

More information

Aerodynamics. Basic Aerodynamics. Continuity equation (mass conserved) Some thermodynamics. Energy equation (energy conserved)

Aerodynamics. Basic Aerodynamics. Continuity equation (mass conserved) Some thermodynamics. Energy equation (energy conserved) Flow with no friction (inviscid) Aerodynamics Basic Aerodynamics Continuity equation (mass conserved) Flow with friction (viscous) Momentum equation (F = ma) 1. Euler s equation 2. Bernoulli s equation

More information

Unit C-1: List of Subjects

Unit C-1: List of Subjects Unit C-: List of Subjects The elocity Field The Acceleration Field The Material or Substantial Derivative Steady Flow and Streamlines Fluid Particle in a Flow Field F=ma along a Streamline Bernoulli s

More information

Chapter 4: Fluid Kinematics

Chapter 4: Fluid Kinematics Overview Fluid kinematics deals with the motion of fluids without considering the forces and moments which create the motion. Items discussed in this Chapter. Material derivative and its relationship to

More information

CEE 3310 Control Volume Analysis, Oct. 7, D Steady State Head Form of the Energy Equation P. P 2g + z h f + h p h s.

CEE 3310 Control Volume Analysis, Oct. 7, D Steady State Head Form of the Energy Equation P. P 2g + z h f + h p h s. CEE 3310 Control Volume Analysis, Oct. 7, 2015 81 3.21 Review 1-D Steady State Head Form of the Energy Equation ( ) ( ) 2g + z = 2g + z h f + h p h s out where h f is the friction head loss (which combines

More information

Lagrangian description from the perspective of a parcel moving within the flow. Streamline Eulerian, tangent line to instantaneous velocity field.

Lagrangian description from the perspective of a parcel moving within the flow. Streamline Eulerian, tangent line to instantaneous velocity field. Chapter 2 Hydrostatics 2.1 Review Eulerian description from the perspective of fixed points within a reference frame. Lagrangian description from the perspective of a parcel moving within the flow. Streamline

More information

Chapter (6) Energy Equation and Its Applications

Chapter (6) Energy Equation and Its Applications Chapter (6) Energy Equation and Its Applications Bernoulli Equation Bernoulli equation is one of the most useful equations in fluid mechanics and hydraulics. And it s a statement of the principle of conservation

More information

Part A: 1 pts each, 10 pts total, no partial credit.

Part A: 1 pts each, 10 pts total, no partial credit. Part A: 1 pts each, 10 pts total, no partial credit. 1) (Correct: 1 pt/ Wrong: -3 pts). The sum of static, dynamic, and hydrostatic pressures is constant when flow is steady, irrotational, incompressible,

More information

Chapter 5. Mass and Energy Analysis of Control Volumes. by Asst. Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn

Chapter 5. Mass and Energy Analysis of Control Volumes. by Asst. Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Chapter 5 Mass and Energy Analysis of Control Volumes by Asst. Prof. Dr.Woranee Paengjuntuek and Asst. Prof. Dr.Worarattana Pattaraprakorn Reference: Cengel, Yunus A. and Michael A. Boles, Thermodynamics:

More information

CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES

CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES Thermodynamics: An Engineering Approach 8th Edition in SI Units Yunus A. Çengel, Michael A. Boles McGraw-Hill, 2015 CHAPTER 5 MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES Lecture slides by Dr. Fawzi Elfghi

More information

Basic Fluid Mechanics

Basic Fluid Mechanics Basic Fluid Mechanics Chapter 5: Application of Bernoulli Equation 4/16/2018 C5: Application of Bernoulli Equation 1 5.1 Introduction In this chapter we will show that the equation of motion of a particle

More information

Benha University College of Engineering at Benha Questions For Corrective Final Examination Subject: Fluid Mechanics M 201 May 24/ 2016

Benha University College of Engineering at Benha Questions For Corrective Final Examination Subject: Fluid Mechanics M 201 May 24/ 2016 Benha University College of Engineering at Benha Questions For Corrective Final Examination Subject: Fluid Mechanics M 01 May 4/ 016 Second year Mech. Time :180 min. Examiner:Dr.Mohamed Elsharnoby Attempt

More information

Chapter 5 Control Volume Approach and Continuity Equation

Chapter 5 Control Volume Approach and Continuity Equation Chapter 5 Control Volume Approach and Continuity Equation Lagrangian and Eulerian Approach To evaluate the pressure and velocities at arbitrary locations in a flow field. The flow into a sudden contraction,

More information

Chapter 5: The First Law of Thermodynamics: Closed Systems

Chapter 5: The First Law of Thermodynamics: Closed Systems Chapter 5: The First Law of Thermodynamics: Closed Systems The first law of thermodynamics can be simply stated as follows: during an interaction between a system and its surroundings, the amount of energy

More information

first law of ThermodyNamics

first law of ThermodyNamics first law of ThermodyNamics First law of thermodynamics - Principle of conservation of energy - Energy can be neither created nor destroyed Basic statement When any closed system is taken through a cycle,

More information

AEROSPACE ENGINEERING DEPARTMENT. Second Year - Second Term ( ) Fluid Mechanics & Gas Dynamics

AEROSPACE ENGINEERING DEPARTMENT. Second Year - Second Term ( ) Fluid Mechanics & Gas Dynamics AEROSPACE ENGINEERING DEPARTMENT Second Year - Second Term (2008-2009) Fluid Mechanics & Gas Dynamics Similitude,Dimensional Analysis &Modeling (1) [7.2R*] Some common variables in fluid mechanics include:

More information

6.1 Momentum Equation for Frictionless Flow: Euler s Equation The equations of motion for frictionless flow, called Euler s

6.1 Momentum Equation for Frictionless Flow: Euler s Equation The equations of motion for frictionless flow, called Euler s Chapter 6 INCOMPRESSIBLE INVISCID FLOW All real fluids possess viscosity. However in many flow cases it is reasonable to neglect the effects of viscosity. It is useful to investigate the dynamics of an

More information

3.25 Pressure form of Bernoulli Equation

3.25 Pressure form of Bernoulli Equation CEE 3310 Control Volume Analysis, Oct 3, 2012 83 3.24 Review The Energy Equation Q Ẇshaft = d dt CV ) (û + v2 2 + gz ρ d + (û + v2 CS 2 + gz + ) ρ( v n) da ρ where Q is the heat energy transfer rate, Ẇ

More information

Pressure in stationary and moving fluid Lab- Lab On- On Chip: Lecture 2

Pressure in stationary and moving fluid Lab- Lab On- On Chip: Lecture 2 Pressure in stationary and moving fluid Lab-On-Chip: Lecture Lecture plan what is pressure e and how it s distributed in static fluid water pressure in engineering problems buoyancy y and archimedes law;

More information

The Impulse-Momentum Principle

The Impulse-Momentum Principle Chapter 6 /60 The Impulse-Momentum Principle F F Chapter 6 The Impulse-Momentum Principle /60 Contents 6.0 Introduction 6. The Linear Impulse-Momentum Equation 6. Pipe Flow Applications 6.3 Open Channel

More information

FE Fluids Review March 23, 2012 Steve Burian (Civil & Environmental Engineering)

FE Fluids Review March 23, 2012 Steve Burian (Civil & Environmental Engineering) Topic: Fluid Properties 1. If 6 m 3 of oil weighs 47 kn, calculate its specific weight, density, and specific gravity. 2. 10.0 L of an incompressible liquid exert a force of 20 N at the earth s surface.

More information

The online of midterm-tests of Fluid Mechanics 1

The online of midterm-tests of Fluid Mechanics 1 The online of midterm-tests of Fluid Mechanics 1 1) The information on a can of pop indicates that the can contains 460 ml. The mass of a full can of pop is 3.75 lbm while an empty can weights 80.5 lbf.

More information

Rate of Flow Quantity of fluid passing through any section (area) per unit time

Rate of Flow Quantity of fluid passing through any section (area) per unit time Kinematics of Fluid Flow Kinematics is the science which deals with study of motion of liquids without considering the forces causing the motion. Rate of Flow Quantity of fluid passing through any section

More information

equation 4.1 INTRODUCTION

equation 4.1 INTRODUCTION 4 The momentum equation 4.1 INTRODUCTION It is often important to determine the force produced on a solid body by fluid flowing steadily over or through it. For example, there is the force exerted on a

More information

4 Finite Control Volume Analysis Introduction Reynolds Transport Theorem Conservation of Mass

4 Finite Control Volume Analysis Introduction Reynolds Transport Theorem Conservation of Mass iv 2.3.2 Bourdon Gage................................... 92 2.3.3 Pressure Transducer................................ 93 2.3.4 Manometer..................................... 95 2.3.4.1 Piezometer................................

More information

2 Navier-Stokes Equations

2 Navier-Stokes Equations 1 Integral analysis 1. Water enters a pipe bend horizontally with a uniform velocity, u 1 = 5 m/s. The pipe is bended at 90 so that the water leaves it vertically downwards. The input diameter d 1 = 0.1

More information

5 ENERGY EQUATION OF FLUID MOTION

5 ENERGY EQUATION OF FLUID MOTION 5 ENERGY EQUATION OF FLUID MOTION 5.1 Introduction In order to develop the equations that describe a flow, it is assumed that fluids are subject to certain fundamental laws of physics. The pertinent laws

More information

V (r,t) = i ˆ u( x, y,z,t) + ˆ j v( x, y,z,t) + k ˆ w( x, y, z,t)

V (r,t) = i ˆ u( x, y,z,t) + ˆ j v( x, y,z,t) + k ˆ w( x, y, z,t) IV. DIFFERENTIAL RELATIONS FOR A FLUID PARTICLE This chapter presents the development and application of the basic differential equations of fluid motion. Simplifications in the general equations and common

More information

Detailed Outline, M E 320 Fluid Flow, Spring Semester 2015

Detailed Outline, M E 320 Fluid Flow, Spring Semester 2015 Detailed Outline, M E 320 Fluid Flow, Spring Semester 2015 I. Introduction (Chapters 1 and 2) A. What is Fluid Mechanics? 1. What is a fluid? 2. What is mechanics? B. Classification of Fluid Flows 1. Viscous

More information

NPTEL Quiz Hydraulics

NPTEL Quiz Hydraulics Introduction NPTEL Quiz Hydraulics 1. An ideal fluid is a. One which obeys Newton s law of viscosity b. Frictionless and incompressible c. Very viscous d. Frictionless and compressible 2. The unit of kinematic

More information

Consider a control volume in the form of a straight section of a streamtube ABCD.

Consider a control volume in the form of a straight section of a streamtube ABCD. 6 MOMENTUM EQUATION 6.1 Momentum and Fluid Flow In mechanics, the momentum of a particle or object is defined as the product of its mass m and its velocity v: Momentum = mv The particles of a fluid stream

More information

Mass of fluid leaving per unit time

Mass of fluid leaving per unit time 5 ENERGY EQUATION OF FLUID MOTION 5.1 Eulerian Approach & Control Volume In order to develop the equations that describe a flow, it is assumed that fluids are subject to certain fundamental laws of physics.

More information

BERNOULLI EQUATION. The motion of a fluid is usually extremely complex.

BERNOULLI EQUATION. The motion of a fluid is usually extremely complex. BERNOULLI EQUATION The motion of a fluid is usually extremely complex. The study of a fluid at rest, or in relative equilibrium, was simplified by the absence of shear stress, but when a fluid flows over

More information

Useful concepts associated with the Bernoulli equation. Dynamic

Useful concepts associated with the Bernoulli equation. Dynamic Useful concets associated with the Bernoulli equation - Static, Stagnation, and Dynamic Pressures Bernoulli eq. along a streamline + ρ v + γ z = constant (Unit of Pressure Static (Thermodynamic Dynamic

More information

EXPERIMENT No.1 FLOW MEASUREMENT BY ORIFICEMETER

EXPERIMENT No.1 FLOW MEASUREMENT BY ORIFICEMETER EXPERIMENT No.1 FLOW MEASUREMENT BY ORIFICEMETER 1.1 AIM: To determine the co-efficient of discharge of the orifice meter 1.2 EQUIPMENTS REQUIRED: Orifice meter test rig, Stopwatch 1.3 PREPARATION 1.3.1

More information

Introduction to Turbomachinery

Introduction to Turbomachinery 1. Coordinate System Introduction to Turbomachinery Since there are stationary and rotating blades in turbomachines, they tend to form a cylindrical form, represented in three directions; 1. Axial 2. Radial

More information

Fundamentals of Fluid Mechanics

Fundamentals of Fluid Mechanics Sixth Edition Fundamentals of Fluid Mechanics International Student Version BRUCE R. MUNSON DONALD F. YOUNG Department of Aerospace Engineering and Engineering Mechanics THEODORE H. OKIISHI Department

More information

2 Internal Fluid Flow

2 Internal Fluid Flow Internal Fluid Flow.1 Definitions Fluid Dynamics The study of fluids in motion. Static Pressure The pressure at a given point exerted by the static head of the fluid present directly above that point.

More information

10.52 Mechanics of Fluids Spring 2006 Problem Set 3

10.52 Mechanics of Fluids Spring 2006 Problem Set 3 10.52 Mechanics of Fluids Spring 2006 Problem Set 3 Problem 1 Mass transfer studies involving the transport of a solute from a gas to a liquid often involve the use of a laminar jet of liquid. The situation

More information

The First Law of Thermodynamics. By: Yidnekachew Messele

The First Law of Thermodynamics. By: Yidnekachew Messele The First Law of Thermodynamics By: Yidnekachew Messele It is the law that relates the various forms of energies for system of different types. It is simply the expression of the conservation of energy

More information

Chapter 5 Mass, Bernoulli, and Energy Equations Chapter 5 MASS, BERNOULLI, AND ENERGY EQUATIONS

Chapter 5 Mass, Bernoulli, and Energy Equations Chapter 5 MASS, BERNOULLI, AND ENERGY EQUATIONS Chapter 5 MASS, BERNOULLI, AND ENERGY EQUATIONS Conservation of Mass 5-C Mass, energy, momentum, and electric charge are conserved, and volume and entropy are not conserved during a process. 5-C Mass flow

More information

Please welcome for any correction or misprint in the entire manuscript and your valuable suggestions kindly mail us

Please welcome for any correction or misprint in the entire manuscript and your valuable suggestions kindly mail us Problems of Practices Of Fluid Mechanics Compressible Fluid Flow Prepared By Brij Bhooshan Asst. Professor B. S. A. College of Engg. And Technology Mathura, Uttar Pradesh, (India) Supported By: Purvi Bhooshan

More information

Thermodynamics ENGR360-MEP112 LECTURE 7

Thermodynamics ENGR360-MEP112 LECTURE 7 Thermodynamics ENGR360-MEP11 LECTURE 7 Thermodynamics ENGR360/MEP11 Objectives: 1. Conservation of mass principle.. Conservation of energy principle applied to control volumes (first law of thermodynamics).

More information

CLASS SCHEDULE 2013 FALL

CLASS SCHEDULE 2013 FALL CLASS SCHEDULE 2013 FALL Class # or Lab # 1 Date Aug 26 2 28 Important Concepts (Section # in Text Reading, Lecture note) Examples/Lab Activities Definition fluid; continuum hypothesis; fluid properties

More information

FLUID MECHANICS. Dynamics of Viscous Fluid Flow in Closed Pipe: Darcy-Weisbach equation for flow in pipes. Major and minor losses in pipe lines.

FLUID MECHANICS. Dynamics of Viscous Fluid Flow in Closed Pipe: Darcy-Weisbach equation for flow in pipes. Major and minor losses in pipe lines. FLUID MECHANICS Dynamics of iscous Fluid Flow in Closed Pipe: Darcy-Weisbach equation for flow in pipes. Major and minor losses in pipe lines. Dr. Mohsin Siddique Assistant Professor Steady Flow Through

More information

In this section, mathematical description of the motion of fluid elements moving in a flow field is

In this section, mathematical description of the motion of fluid elements moving in a flow field is Jun. 05, 015 Chapter 6. Differential Analysis of Fluid Flow 6.1 Fluid Element Kinematics In this section, mathematical description of the motion of fluid elements moving in a flow field is given. A small

More information

ME332 FLUID MECHANICS LABORATORY (PART II)

ME332 FLUID MECHANICS LABORATORY (PART II) ME332 FLUID MECHANICS LABORATORY (PART II) Mihir Sen Department of Aerospace and Mechanical Engineering University of Notre Dame Notre Dame, IN 46556 Version: April 2, 2002 Contents Unit 5: Momentum transfer

More information

Pressure in stationary and moving fluid. Lab-On-Chip: Lecture 2

Pressure in stationary and moving fluid. Lab-On-Chip: Lecture 2 Pressure in stationary and moving fluid Lab-On-Chip: Lecture Fluid Statics No shearing stress.no relative movement between adjacent fluid particles, i.e. static or moving as a single block Pressure at

More information

NPTEL Course Developer for Fluid Mechanics DYMAMICS OF FLUID FLOW

NPTEL Course Developer for Fluid Mechanics DYMAMICS OF FLUID FLOW 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

More information

Lecture 2 Flow classifications and continuity

Lecture 2 Flow classifications and continuity Lecture 2 Flow classifications and continuity Dr Tim Gough: t.gough@bradford.ac.uk General information 1 No tutorial week 3 3 rd October 2013 this Thursday. Attempt tutorial based on examples from today

More information

2. FLUID-FLOW EQUATIONS SPRING 2019

2. FLUID-FLOW EQUATIONS SPRING 2019 2. FLUID-FLOW EQUATIONS SPRING 2019 2.1 Introduction 2.2 Conservative differential equations 2.3 Non-conservative differential equations 2.4 Non-dimensionalisation Summary Examples 2.1 Introduction Fluid

More information

For example an empty bucket weighs 2.0kg. After 7 seconds of collecting water the bucket weighs 8.0kg, then:

For example an empty bucket weighs 2.0kg. After 7 seconds of collecting water the bucket weighs 8.0kg, then: Hydraulic Coefficient & Flow Measurements ELEMENTARY HYDRAULICS National Certificate in Technology (Civil Engineering) Chapter 3 1. Mass flow rate If we want to measure the rate at which water is flowing

More information

Applied Fluid Mechanics

Applied Fluid Mechanics Applied Fluid Mechanics 1. The Nature of Fluid and the Study of Fluid Mechanics 2. Viscosity of Fluid 3. Pressure Measurement 4. Forces Due to Static Fluid 5. Buoyancy and Stability 6. Flow of Fluid and

More information

PTT 277/3 APPLIED THERMODYNAMICS SEM 1 (2013/2014)

PTT 277/3 APPLIED THERMODYNAMICS SEM 1 (2013/2014) PTT 77/3 APPLIED THERMODYNAMICS SEM 1 (013/014) 1 Energy can exist in numerous forms: Thermal Mechanical Kinetic Potential Electric Magnetic Chemical Nuclear The total energy of a system on a unit mass:

More information

HOMEWORK ASSIGNMENT ON BERNOULLI S EQUATION

HOMEWORK ASSIGNMENT ON BERNOULLI S EQUATION AMEE 0 Introduction to Fluid Mechanics Instructor: Marios M. Fyrillas Email: m.fyrillas@frederick.ac.cy HOMEWORK ASSIGNMENT ON BERNOULLI S EQUATION. Conventional spray-guns operate by achieving a low pressure

More information

ENT 254: Applied Thermodynamics

ENT 254: Applied Thermodynamics ENT 54: Applied Thermodynamics Mr. Azizul bin Mohamad Mechanical Engineering Program School of Mechatronic Engineering Universiti Malaysia Perlis (UniMAP) azizul@unimap.edu.my 019-4747351 04-9798679 Chapter

More information

Homework 6. Solution 1. r ( V jet sin( θ) + ω r) ( ρ Q r) Vjet

Homework 6. Solution 1. r ( V jet sin( θ) + ω r) ( ρ Q r) Vjet Problem 1 Water enters the rotating sprinkler along the axis of rotation and leaves through three nozzles. How large is the resisting torque required to hold the rotor stationary for the angle that produces

More information

AER210 VECTOR CALCULUS and FLUID MECHANICS. Quiz 4 Duration: 70 minutes

AER210 VECTOR CALCULUS and FLUID MECHANICS. Quiz 4 Duration: 70 minutes AER210 VECTOR CALCULUS and FLUID MECHANICS Quiz 4 Duration: 70 minutes 26 November 2012 Closed Book, no aid sheets Non-programmable calculators allowed Instructor: Alis Ekmekci Family Name: Given Name:

More information

Page 1. Neatly print your name: Signature: (Note that unsigned exams will be given a score of zero.)

Page 1. Neatly print your name: Signature: (Note that unsigned exams will be given a score of zero.) Page 1 Neatly print your name: Signature: (Note that unsigned exams will be given a score of zero.) Circle your lecture section (-1 point if not circled, or circled incorrectly): Prof. Vlachos Prof. Ardekani

More information

Thermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am - 10:00am & Monday 8:00am - 9:00am MEP 261 Class ZA

Thermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am - 10:00am & Monday 8:00am - 9:00am MEP 261 Class ZA Thermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am - 10:00am & Monday 8:00am - 9:00am MEP 261 Class ZA Dr. Walid A. Aissa Associate Professor, Mech. Engg. Dept. Faculty of Engineering

More information

!! +! 2!! +!"!! =!! +! 2!! +!"!! +!!"!"!"

!! +! 2!! +!!! =!! +! 2!! +!!! +!!!! Homework 4 Solutions 1. (15 points) Bernoulli s equation can be adapted for use in evaluating unsteady flow conditions, such as those encountered during start- up processes. For example, consider the large

More information

Formulae that you may or may not find useful. E v = V. dy dx = v u. y cp y = I xc/a y. Volume of an entire sphere = 4πr3 = πd3

Formulae that you may or may not find useful. E v = V. dy dx = v u. y cp y = I xc/a y. Volume of an entire sphere = 4πr3 = πd3 CE30 Test 1 Solution Key Date: 26 Sept. 2017 COVER PAGE Write your name on each sheet of paper that you hand in. Read all questions very carefully. If the problem statement is not clear, you should ask

More information

Chapter 6 The Impulse-Momentum Principle

Chapter 6 The Impulse-Momentum Principle Chapter 6 The Impulse-Momentum Principle 6. The Linear Impulse-Momentum Equation 6. Pipe Flow Applications 6.3 Open Channel Flow Applications 6.4 The Angular Impulse-Momentum Principle Objectives: - Develop

More information

Atmospheric pressure. 9 ft. 6 ft

Atmospheric pressure. 9 ft. 6 ft Name CEE 4 Final Exam, Aut 00; Answer all questions; 145 points total. Some information that might be helpful is provided below. A Moody diagram is printed on the last page. For water at 0 o C (68 o F):

More information