1. Water is flowing through a large pipe of diameter 5 feet from left to right as shown. The velocity at the inlet is given by: 2

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1 Chapte II Woked-Out Examples 5 feet dia 1 foot dia 1. Wate is flowing though a lage pipe of diamete 5 feet fom left to ight as shown. The velocity at the inlet is given by: V = 6.5 ft/sec What is the aveage velocity of the flow leaving though the smalle pipe of diamete 1 feet? This is a steady flow. Continuity equation simplifies to ρ[ n] Inlet+ Outlet v V ds = 0 i. At the inlet, the unit nomal vecto is i. At the outlet the unit nomal vecto is At the inlet, V = ( 6.5 )i At the exit, the velocity is V exit i We can assume that the inlet suface is made of annula stips of adius, and width d. Thus, ds=πd We get: Inlet.5 feet [ V n] ds == ρ ( 6.5 ) πd = ρπ 6.5* slugs / sec ρ v At the outlet, we get: Outlet 0 ρ V nds = ρv A = ρv Exit exit Exit π ( 0.5) slugs / sec Sum up the integal at the inlet and at the outlet, and set the sum to zeo as equied by continuity ρπ 6.5* /sec+ ( 0.5) /sec = 0 4 slugs ρvexitπ slugs Solve fo V exit. Afte mino aithmetic, the exit velocity is ft/sec 4

2 Fuel tank x Fuel Line Execise: A fighte aicaft is being efueled in flight at the ate of 150 gallons/minute. The specific gavity of the fuel is The inside diamete of the fuel line is 5 inches. The fluid pessue at the entance to the aicaft is 4 psi. What additional thust does the plane need to develop to keep the aicaft at constant velocity duing the efueling pocess? Note: Density of wate = slug/ft 3. 1 gallon/minute = ft 3 /sec Solution: Step1:We choose a coodinate system fist. In this case, we choose the x-axis to be along the thust diection, as shown above. Step : The next step is to choose a contol volume. The contol volume is the fuel tank, and the fuel line o duct that leads fom the fuel tank to the aicaft nose. We need not know the pecise shape of the tank. Since my softwae has touble dawing anything but ectangles and ellipses, I have dawn the tank and the fuel line as two ectangle-like objects. Step 3: Identify what we ae afte. We ae inteested in computing the x- component of the nomal foce exeted by the fuel inside the contol volume on the tank and fuel line walls. This nomal foce is pessue diffeence between the inteio and exteio of the contol volume walls(the exteio is vented to atmosphee) times the aea. At each point on the tank/fuel line wall, the foce will act along the unit nomal vecto n. The total foce along the x-diection is given by p - p nds Fuel Line and tank ( ) i F x, viscous (1) Recall that F x,viscous is the viscous foce exeted by the suface S of the tank and pipes on the contol volume. To evey action, thee is a eaction. Thus -F x,viscous is the viscous foce exeted by the fluid on the walls, and the aicaft. Step 4: Fom the integal fom of momentum equation isolate what we ae afte. Fom the integal fom of the u-momentum equation we know that:

3 ( p p ) i nds + ρuv nds = F x, viscous () Hee the integal is ove all the boundaies of the contol volume the fuel tank, the fuel line and the entance to the fuel line. We know that no fuel o momentum can escape though the walls of the fuel tank o line. Thus, fom the above equation we can wite: and Fuel Tank line ( p p ) i nds Fx, viscous = ( p p ) i nds Entance Entance ρuv nds (3) As we saw ealie in expession (1), the left side of equation (3) epesents the foce exeted by the fluid inside the contol volume on the fuel tank/fuel line and eventually on the aicaft. Step 5: Evaluate! We can evaluate the ight hand side of (3) as follows. Volume ate of flow of the fuel = 150 gallons/minute = 150 * ft 3 /sec = ft 3 /sec Entance adius = 5/ inches = 5/4 feet Entance aea = π * ( 5/4) * (5/4) ft Velocity at the entance = ft = Volume ate of flow/entance aea = 0.334/ ft/sec =.45 ft/sec This x- component of velocity, given the symbol u, will be along the negative x- axis. Theefoe it is given a sign as well. Thus, u V =.45i = -.45 ft/sec At the entance, the unit nomal n is along the +x diection. It theefoe equals + i. At the entance, p - p = 4 psi = 576 lbf/ft At the entance, V n = -.45 ft/sec Density of fuel = 0.68 * slug/ft 3 = slug/ ft 3

4 The ight hand side thus becomes: - {576 * } - { * (-.45) (-.45) * } lbf. = Pefoming the aithmetic, the ight hand side becomes lbf. This is the foce that the fluid inside the tank will exet on the tanks, and eventually on the aicaft. The negative sign tells us that it is diected along the ve x- axis pushing the aicaft back. Thus lbf of exta thust is needed to ovecome it.

5 Poblem 3: Find the velocity that a wate jet will use to popel a ship with a thust foce equal to 3000 lb. The pipe diamete is ft. V=? Solution: This poblem is simila to the aicaft-efueling poblem we studied befoe. In that poblem, fuel was enteing the aicaft. Hee the wate jet is leaving the ship though a pipe of diamete ft. Step 1: We need to choose the x-axis. We will assume that it is pointing to the ight hand side. Then, at the outlet, the nomal vecto is along the x- axis, and equals i. Step 3:As in the aicaft-efueling poblem, we will constuct a contol volume this will be a wate tank and pipe of unknown shape whee the wate is stoed. The only bounday though which wate can ente o leave is the pipe exit. Step 3: Identify what we ae afte. Now, the foce on the ship is due to pessue exeted by the fluid on the tank and pipe walls and the viscous foces. This is given by: Foce on the ship p n i ds F = Tank+ Pipe ( p atm ) x, viscous Step 4: Fom the momentum equation, isolate the tems we need. The steady state fom of the momentum equation states: n i ds + ρuv nds = S p Fx, viscous S (1) Hee S is the bounday that includes the tank walls, the pipe wall and the pipe exit (outlet).

6 We can subtact atmospheic fom the pessue p in the above equation. As we discussed in the class, adding o subtacting a constant fom the pessue field does not change the pessue foce acting on a closed volume. Thus, the above equation may be viewed as: p p n i ds + ρuv nds + p p n i ds + ρuv nds = F ( ) ( ) atm atm Tank+ Pipe Tank+ Pipe Outlet Outlet At the pipe and tank walls, the nomal component of velocity is zeo. Thus, the second tem in equation () is zeo. The thid tem in equation () is also zeo since at the pipe outlet the pessue p equals atmospheic pessue (given). Thus, the foce on the ship educes to: Foce on the ship = ( p patm ) n i ds Fx, viscous = Tank+ Pipe outlet ρuv nds Step 5: Evaluate the quantities. Assuming unifom flow at the outlet, the ight hand side is -ρv A whee A is the pipe coss section aea. Thus, Foce on the ship = - ρv A The negative sign simply indicates that this thust will act along the negative diection, i.e. to the left, pushing the ship towads the left. We ae given the magnitude of F as 3000 lbf. Substitute fo density ρ= 1.938, use A=πR 3000 We can solve fo V, giving V = =. ft / s π 1 x, viscous ()