# ME 305 Fluid Mechanics I. Part 4 Integral Formulation of Fluid Flow

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1 ME 305 Fluid Mechanics I Part 4 Integral Formulation of Fluid Flow These presentations are prepared by Dr. Cüneyt Sert Mechanical Engineering Department Middle East Technical University Ankara, Turkey Please ask for permission before using them. You are NOT allowed to modify them. 4-1

2 Reynolds Transport Theorem dn sys dt = ρ η d + ρ η V n da t CV CS Rate of change of property N within the system. Rate of change of = property N in the corresponding CV + (zero for steady flows) Net rate at which property N exits the CV Conservation Eqn N η Left hand side Mass m 1 Zero Linear Momentum P V F Angular Momentum H r V T Energy E e Q + W 4-2

3 Conservation of Mass (Continuity Equation) Taking N = m and η = 1, RTT becomes dm sys dt = 0 = ρ d + ρ V n da t CV CS Rate of change of mass in the CV Net rate of mass exiting through the CS t CV ρ d < 0 : Mass inside the CV decreases in time CV CS ρ V n da > 0 : The amount of mass leaving the CV is in balance with the above term 4-3

4 Mass Flow Rate ( m) The surface integral of the continuity equation is known as the mass flow rate ( m). It has the units of kg/s. V m = ρ V n da CS [kg/s] CS n Usually the CS is separated into parts and the integral over each part is evaluated one by one. n V CV n V > 0 if V n > 0 outflow V m = = 0 if V n = 0 V = 0 or flow is parallel to the CS α n < 0 if V n < 0 inflow V n = V cos(α) 4-4

5 Mass Flow Rate ( m) Consider the following pipe branching into two. Section 1 is an inlet. V n & m are negative. 1 3 n Section 3 is an exit. V n & m are positive. V n V n, V 2 Section 2 is an exit. V n & m are positive. Alternative CS selection at section 3 (prefer this one) V n 4-5

6 Volumetric Flow Rate (Q) and Average Velocity (V) If the density is uniform across a cross section A volumetric flow rate (Q) can be defined as Q = m ρ = A V n da [m 3 /s] Average velocity over a section of a flow field is defined as V = m ρa = Q A (ρ is uniform over the section) m = ρav Q = AV u = U max 1 r2 R 2 V V is defined such that these two cases provide the same flow rate. 4-6

7 Volumetric Flow Rate and Average Velocity (cont d) Exercise: Consider the water exiting a pipe of diameter D = 2 cm. To find the flow rate we measured the time it will take to fill a bucket of volume V = 10 liters. It took 65 s to fill it. Determine a) the volumetric flow rate of water. b) the mass flow rate of water. c) Average velocity at the exit. c) the centerline velocity at the exit if the velocity profile is parabolic as given in the previous slide. 4-7

8 Continuity Equation for Steady Flows For a steady flow (time independent flow) partial time derivative of RTT is zero, and the continuity equation reduces to ρ V n da = 0 CS Continuity equation for a steady flow is inlets m i + outlets m o = 0 For a common case of unidirectional, uniform flow with a single inlet and single exit 1 inlet 2 Continuity : ρ 1 A 1 V 1 + ρ 2 A 2 V 2 = 0 ρ 1 A 1 V 1 = ρ 2 A 2 V 2 4-8

9 Continuity Equation for Incompressible Flow For an incompressible flow (not necessarily steady) density is constant and continuity equation simplifies as Zero since volume of a CV is constant ρ t CV d + ρ CS V n da = 0 ρ CV t + ρ CS V n da = 0 CS V n da = 0 For a common case of unidirectional, uniform flow with a single inlet and single exit 1 2 Continuity : A 1 V 1 + A 2 V 2 = 0 A 1 V 1 = A 2 V 2 (true even if the flow is unsteady) 4-9

10 Mass Conservation Exercises Exercise : Parabolic velocity profile for the steady, fully developed, laminar flow in a pipe is as given below. Calculate the mass flow rates at the inlet and exit of the shown CV and show that they are equal. ρ 60 2R u = U max 1 r2 R 2 Exercise : For the pipe flow given above how much i) the maximum velocity, and ii) wall shear stress will change if the same amount of fluid is forced to pass through a smaller pipe with a radius that is half of the one given above. 4-10

11 Mass Conservation Exercises (cont d) Exercise : A hair dryer can be considered as a duct of constant diameter in which layers of electric resistors are placed. A small fan pulls the air in and forces it through the resistors where it is heated. If the density of air is 1.2 kg/m 3 at the inlet and 1.05 kg/m 3 at the exit, determine the percent increase in the velocity of air as it flows through the dryer (Reference: Çengel s book). 1.2 kg/m kg/m

12 Mass Conservation Exercises (cont d) Exercise : A cylindrical tank, open at the top, has a diameter of 1 m. Water is flowing into the tank through pipe 1, and flowing out through pipe 2. Determine the rate at which the water level is rising / falling in the tank. 1 D 1 = 0.2 m V 1 = 3 m/s Water 2 D 2 = 0.1 m V 2 = 10 m/s Exercise : A tank of 0.05 m 3 volume contains air at 800 kpa (absolute) and 15 o C. At t = 0, air begins escaping from the tank through a valve with a flow area of 65 mm 2. The air passing through the valve has a speed of 300 m/s and a density of 6 kg/m 3. Determine the instantaneous rate of change of density in the tank at t = 0. Air 4-12

13 Moving CV Although we defined a CV to be fixed in space, in some problems a moving CV can be used. In such a case use a reference frame attached to the moving CV so that it becomes stationary. This is the same as using relative velocities at the inlets/exits. Relative velocity (W) is the fluid velocity seen by an observer moving with the CV. Absolute velocity (V) is the fluid velocity seen by a fixed observer. V CV V CV : Velocity of the CV W V W V : Relative velocity : Absolute velocity Exercise : An airplane moves forward in still air at a speed of 971 km/h. The frontal area of the jet engine is 0.8 m 2 and the entering air density is 1.2 kg/m 3. The exhaust area is m 2 and the exhaust gas density is kg/m 3. A stationary observer on the ground observes that the jet engine exhaust gases move with a speed of 1050 km/h. Estimate the mass flow rate of fuel into the engine. 4-13

14 Deforming CV Although a CV is defined to have fixed position and shape, for some problems it may be useful to consider deforming CVs. Exercise: Solve the cylindrical tank problem of Slide 4-12 using a deforming CV such that the CS is attached to the free surface and goes up/down with it. Exercise: A syringe is used to inoculate a cow. The plunger has a face area of 500 mm 2. The liquid in the syringe is to be injected steadily at a rate of 300 cm 3 /min. The leakage rate past the plunger is 10% of the flow rate out of the needle. With what speed should the plunger be advanced? (Reference: Munson s book) 4-14

15 Conservation of Linear Momentum Taking N = P and η = V, RTT becomes dp sys dt = F = ρvd + ρv V n da t CV CS Sum of external forces acting on the CV Rate of change of linear momentum in the CV Net rate of flow of linear momentum through the CS Unlike the mass conservation equation, this one is a vector equation. For steady flows partial time derivative is zero. First term drops F = ρv V n da CS 4-15

16 Conservation of Linear Momentum (cont d) For the common case of steady, unidirectional, uniform flow with a single inlet and single exit 1 x 2 m = ρ 1 A 1 V 1 = ρ 2 A 2 V 2 F x = ρ 2 V 2 2 A 2 ρ 1 V 1 2 A 1 = m (V 2 V 1 ) 4-16

17 Cons. of Linear Momentum Important Points Direction of momentum flow rate at the inlets/exits is always in the direction of the outward normal n. In solving problems you can show them with arrows just like forces. Show the forces on a separately drawn CV (like drawing a free body diagram). 1 3 α 2 With this CV, forces acting on the fluid by the pipe can be calculated. Forces acting on the pipe by the fluid are the reverse of the calculated ones. 4-17

18 Cons. of Linear Momentum Important Points (cont d) Which forces to include depend on the selection of the CV. Work on the previous example, but this time select the CV outside the pipe With this CV, anchoring forces can be calculated. These are usually the desired ones in practice. Atmospheric pressure forces come into the picture. 4-18

19 Cons. of Linear Momentum Important Points (cont d) Effect of p atm can be simlified. For the outside CV of the previous slide, concentrate of the pressure forces. p atm acts p 3 A 3 p atm acts p 1 A 1 p atm acts p 2 A 2 p 3,gage A 3 These two systems are equivalent as far as the pressure forces are concerned. When calculating anchoring forces by selecting outside CVs, it is better to use gage pressures at inlets/exits. When this is done p atm s are gone. p 1,gage A 1 p 2,gage A

20 Cons. of Linear Momentum Important Points (cont d) Liquid jets exiting freely to atmosphere at subsonic speed are considered to be at atmospheric pressure. Exercise: The following vane deflects the liquid jet coming out of a nozzle by an angle of α. The jet exits the nozzle with an average velocity of V j through an exit area of A j. Jet s cross sectional area is assumed to remain constant. Select a proper CV and calculate the anchoring force necessary to hold the vane in its place. Nozzle Jet 2 1 Deflecting vane 4-20

21 Cons. of Linear Momentum Important Points (cont d) Conservation of linear momentum is a vector equation. It can be written in component form. Cartesian components are x component : F x = t CV ρ u d + ρ u V n da CS y component : F y = t CV ρ v d + ρ v V n da CS z component : F z = t CV ρ w d + ρ w V n da CS V remains as it is in the dot product 4-21

22 Cons. of Linear Momentum Important Points (cont d) When the velocity is not uniform at a cross section, the calculations can still be done as if the velocity is uniform and the results can be corrected using the momentum correction factor (β). u = U max 1 r2 R 2 V 1 Part of a CV 1 1 Previously we showed that U max = 2 V β = Momentum flow rate of the actual velocity profile Momentum flow rate of the uniform velocity profile = A ρu 1 V 1 1 n 1 da ρ V 2 1 A 1 For the parabolic velocity profile of laminar pipe flow given above β = 4/3. For a turbulent flow the velocity profile will be closer to uniform and β will be close to 1 (~ ) and usually neglected. 4-22

23 Linear Momentum Exercises Exercise : A jet engine operates on a test stand with the shown inlet and exit conditions. Fuel-to-air mass ratio is 1/30. Calculate the horizontal force to be applied by the stand to hold the engine in its place. Assume that the fuel is injected in a vertical direction. Air Fuel Combustion products V in = 250 m s T in = 20 A in = 0.5 m 2 P in = 1 atm V out = 900 m s A out = 0.4 m 2 P out = 1 atm 4-23

24 Linear Momentum Exercises (cont d) Exercise : Consider the flow of a fluid inside a converging duct. Determine the anchoring force force required to hold the duct in its place by using a) a CV inside the duct b) a CV outside the duct Take the pressure at the nozzle exit to be atmospheric. 1 Chamber Nozzle ρ 1 p 1 A 1 V 1 2 ρ 2 p 2 = p atm A 2 V 2 p atm 4-24

25 Linear Momentum Exercises (cont d) Exercise : Water flow inside a constant diameter pipe develops from uniform profile to parabolic profile. Contact force between the water and the pipe wall is F z. Determine an expression for the pressure drop between the inlet and the exit. x u = U max 1 r2 R 2 g r L R u = U 4-25

26 Linear Momentum Exercises (cont d) Exercise : A circular cylinder is immersed in a steady, incompressible flow. Velocities at the upstream and downstream locations are measured and simplified as follows. Calculate the drag force acting on the cylinder using the two control volumes shown below. Assume constant pressure everywhere (which is a questionable assumption). U U U Streamline U D 4D? D 4D Streamline 4-26

27 Linear Momentum Exercises (cont d) Exercise : Water exiting a stationary nozzle strikes a flat plate as shown. Water leaves the nozzle as a circular jet at 15 m/s. Nozzle exit area is 0.01 m 2. After striking the plate water leaves the plate tangentially as a circular disk. Determine the horizontal component of the anchoring force at the support. Solve the problem using two different CVs as shown. Which one is easier to use? Plate Nozzle 4-27

28 Linear Momentum Exercises (cont d) Exercise : Water flows steadily through the 90 o reducing elbow. At the inlet to the elbow, the absolute pressure is 220 kpa and the cross sectional area is 0.01 m 2. At the outlet, cross sectional area is m 2 and the velocity is 16 m/s. The elbow discharges to the atmosphere. Determine the force required to hold the elbow in its place. 4-28

29 Linear Momentum Exercises (cont d) Exercise : The vane with a turning angle of 60 o moves at a constant speed of U = 10 m/s. It receives a jet of water that leaves a stationary nozzle with speed V = 30 m/s. The nozzle exit area is m 2. Area of the jet leaving the vane is the same as the jet leaving the nozzle. Determine the force components that act on the vane by the ground. Select a CV that moves with the cart, i.e. consider relative velocities at the inlet and the exit (Reference: Munson s book). V 60 o Nozzle U 4-29

30 Conservation of Angular Momentum Taking N = H and η = r V, RTT becomes dh sys dt = T = ρ( r V)d + ρ( r V) V n da t CV CS Sum of external torques acting on the CV Rate of change of angular momentum in the CV Rate of angular momentum exiting through the CS For a steady flow partial time derivative is zero. First term drops T = ρ( r V) V n da CS 4-30

31 Conservation of Angular Momentum Exercises Exercise : Underground water is pumped to the surface through a 10 cm diameter pipe that consists of a 2 m long vertical and 1 m long horizontal section. Water discharges to atmospheric air at an average velocity of 3 m/s. Mass of the horizontal pipe section when filled with water is 12 kg per meter length. The pipe is anchored on the ground by a concrete base. Determine a) the moment acting at the base of the pipe (point A) b) length of the horizontal section that would make the moment at point A zero. 1 m V = 3 m/s 2 m A Diameter : 10 cm 4-31

32 Conservation of Angular Momentum Exercises (cont d) Exercise : A large lawn sprinkler with four identical arms is to be converted into a turbine to generate electric power by attaching a generator to its rotating head. Water enters the sprinkler from the base along the axis of rotation at a rate of 20 L/s and leaves the nozzles in the tangential direction. The sprinkler rotates at 300 rpm in a horizontal plane. The diameter of each jet is 1 cm, and the normal distance between the axis of rotation and the center of each nozzle is 0.6 m. a) Calculate the torque transmitted to the shaft. b) Calculate the generated power. c) Plot a curve of generated power vs. rotation speed. Determine the maximum possible power generation and the speed at which it is obtained. (Reference: Cengel and Cimbala) 4-32

33 Conservation of Energy Taking N = E and η = e, RTT becomes de sys dt = Q + W = ρed + ρe V n da t CV CS Rate of heat transfer into the system and rate of work done on the system Rate of change of energy in the CV Rate of energy exiting through the CS Q > 0 if heat is added into the system. W > 0 if work is done on the system. Specific energy, e, includes internal, kinetic and potential energy terms e = u + V2 2 + gz 4-33

34 Conservation of Energy (cont d) Q + W = ρ u + V2 t CV 2 + gz d + ρ u + V2 CS 2 + gz V n da Most common forms of work in fluid systems are shear work : work done on the CS due to viscous stresses. Can be made negligibly small by a proper CV selection. shaft work : e.g. energy delivered to the fluid by a pump or energy extracted from the fluid by a turbine. flow work : energy necessary to push a certain amount of fluid into the CV or out of the CV. It depends on the pressure and velocity at an inlet or an exit. W f = A p V n da 4-34

35 Conservation of Energy (cont d) W f is treated separately from other types of work and it is added to the last integral of the energy equation. Q + W = ρ u + V2 t CV 2 + gz d + ρ u + p CS ρ + V2 2 + gz V n da Using the definition of enthaply h = u + p ρ energy equation becomes Q + W = ρ u + V2 t CV 2 + gz d + ρ h + V2 CS 2 + gz V n da W no longer includes flow work Flow work is accounted inside the enthalpy term 4-35

36 Conservation of Energy (cont d) For a steady, single inlet - single exit flow with uniform properties at inlet and exit sections exit m inlet z Datum for potential energy Q + W = m h + V2 2 + gz exit h + V2 2 + gz inlet Dividing both sides by the constant mass flow rate q + w = h + V2 2 + gz exit h + V2 2 + gz inlet 4-36

37 Conservation of Energy Exercises Exercise : A water pump is powered by a 15 kw electric motor whose efficiency is 90 %. Water flow rate through the pump is 50 L/s. Diameters of the inlet and outlet pipes are the same. Elevation difference across the pump is negligible. The pump is known to be working at 75 % efficiency. If the inlet pressure is measured as 100 kpa determine the exit pressure. Assume adiabatic conditions. 50 L/s p 2 =? η pump = 75 % η motor = 90 % 100 kpa 15 kw 4-37

38 Conservation of Energy Exercises (cont d) Exercise : A pump delivers water at a steady rate of 1150 L/min. Inlet and exit pipe diameters are 10 cm and 3 cm, respectively. Inlet and exit pressures are 130 kpa and 400 kpa, respectively. There is no significant elevation difference between the inlet and exit of the pump. The rise in internal energy of water, due to the temperature rise across the pump is 250 Nm/kg. Considering adiabatic pumping process determine the power delivered to the fluid by the pump. Pump W shaft =? 1150 L/min Inlet D 1 = 10 cm p 1 = 130 kpa Exit D 1 = 3 cm p 1 = 400 kpa u 2 u 1 = 250 Nm/kg 4-38

39 Conservation of Energy Exercises (cont d) Exercise : A steam turbine is used to produce electricity. Steam enters the turbine with a velocity of 30 m/s and enthalpy of 3500 kj/kg. The steam leaves the turbine as a mixture of vapor and liquid having a velocity of 60 m/s and enthalpy of 2300 kj/kg. Assuming adiabatic conditions and negligible elevation changes, determine the work output of the turbine per unit mass of fluid flowing. Turbine w shaft =? Inlet V 1 = 30 m/s h 1 = 3500 kj/kg Exit V 2 = 60 m/s h 1 = 2300 kj/kg 4-39

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