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1 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 volume in this case can be treated as a solid body, with a net force or thrust of Thrust: F body m body a ain bm # V aout bm # V (6 9) acting on the body. This approach can be used to determine the linear acceleration of space vehicles when a rocket is fired (Fig. 6 19). EXAMPLE 6 The Force to Hold a Deflector Elbow in Place A reducing elbow is used to deflect water flow at a rate of 14 kg/s in a horizontal pipe upward 30 while accelerating it (Fig. 6 0). The elbow discharges water into the atmosphere. The cross-sectional area of the elbow is 113 cm at the inlet and 7 cm at the outlet. The elevation difference between the centers of the outlet and the inlet is 30 cm. The weight of the elbow and the water in it is considered to be negligible. Determine (a) the gage pressure at the center of the inlet of the elbow and (b) the anchoring force needed to hold the elbow in place. SOLUTION A reducing elbow deflects water upward and discharges it to the atmosphere. The pressure at the inlet of the elbow and the force needed to hold the elbow in place are to be determined. Assumptions 1 The flow is steady, and the frictional effects are negligible. The weight of the elbow and the water in it is negligible. 3 The water is discharged to the atmosphere, and thus the gage pressure at the outlet is zero. 4 The flow is turbulent and fully developed at both the inlet and outlet of the control volume, and we take the momentum-flux correction factor to be b Properties We take the density of water to be 1000 kg/m 3. Analysis (a) We take the elbow as the control volume and designate the inlet by 1 and the outlet by. We also take the x- and z-coordinates as shown. The continuity equation for this one-inlet, one-outlet, steady-flow system is m. 1 m. m. 14 kg/s. Noting that m. rav, the inlet and outlet velocities of water are V 1 m# 14 kg/s 1.4 m/s ra 1 (1000 kg/m 3 )( m ) V m# 14 kg/s 0.0 m/s ra (1000 kg/m 3 )( m ) V 0 = 000 m/s L = m FIGURE 6 19 The thrust needed to lift the space shuttle is generated by the rocket engines as a result of momentum change of the fuel as it is accelerated from about zero to an exit speed of about 000 m/s after combustion. NASA z x mv 1 F Rz FRx P 1,gage mv cm FIGURE 6 0 Schematic for Example 6.

2 40 FLUID MECHANICS We use the Bernoulli equation (Chap. 5) as a first approximation to calculate the pressure. In Chap. 8 we will learn how to account for frictional losses along the walls. Taking the center of the inlet cross section as the reference level (z 1 0) and noting that P, the Bernoulli equation for a streamline going through the center of the elbow is expressed as P 1 rg V 1 g z 1 P rg V g z P 1 P rga V V 1 g z z 1 b P 1 (1000 kg/m 3 )(9.81 m/s ) a (0 m/s) (1.4 m/s) (9.81 m/s ) P 1, gage 0. kn/m 0. kpa (gage) 1 kn 0.3 0ba 1000 kg m/s b (b) The momentum equation for steady one-dimensional flow is a F aout bm # V ain bm # V mv F RzFRx We let the x- and z-components of the anchoring force of the elbow be F Rx and F Rz, and assume them to be in the positive direction. We also use gage pressure since the atmospheric pressure acts on the entire control surface. Then the momentum equations along the x- and z-axes become F Rx P 1, gage A 1 bm # V cos u bm # V 1 F Rz bm # V sin u Solving for F Rx and F Rz, and substituting the given values, F Rx bm # (V cos u V 1 ) P 1, gage A 1 1 N 1.03(14 kg/s)3(0 cos ) m/s4a 1 kg m/s b (0,00 N/m )( m ) N F Rz bm # 1 N V sin u (1.03)(14 kg/s)(0 sin 30 m/s)a 1 kg m/s b 144 N The negative result for F Rx indicates that the assumed direction is wrong, and it should be reversed. Therefore, F Rx acts in the negative x-direction. Discussion There is a nonzero pressure distribution along the inside walls of the elbow, but since the control volume is outside the elbow, these pressures do not appear in our analysis. The actual value of P 1, gage will be higher than that calculated here because of frictional and other irreversible losses in the elbow. 1 P 1,gage mv 1 FIGURE 6 1 Schematic for Example 6 3. EXAMPLE 6 3 The Force to Hold a Reversing Elbow in Place The deflector elbow in Example 6 is replaced by a reversing elbow such that the fluid makes a 180 U-turn before it is discharged, as shown in Fig. 6 1.

3 41 CHAPTER 6 The elevation difference between the centers of the inlet and the exit sections is still 0.3 m. Determine the anchoring force needed to hold the elbow in place. SOLUTION The inlet and the outlet velocities and the pressure at the inlet of the elbow remain the same, but the vertical component of the anchoring force at the connection of the elbow to the pipe is zero in this case (F Rz 0) since there is no other force or momentum flux in the vertical direction (we are neglecting the weight of the elbow and the water). The horizontal component of the anchoring force is determined from the momentum equation written in the x-direction. Noting that the outlet velocity is negative since it is in the negative x-direction, we have F Rx P 1, gage A 1 b m # ( V ) b 1 m # V 1 bm # (V V 1 ) Solving for F Rx and substituting the known values, F Rx bm # (V V 1 ) P 1, gage A 1 1 N (1.03)(14 kg/s)3(0 1.4) m/s4a 1 kg m/s b (0,00 N/m )( m ) N Therefore, the horizontal force on the flange is 591 N acting in the negative x-direction (the elbow is trying to separate from the pipe). This force is equivalent to the weight of about 60 kg mass, and thus the connectors (such as bolts) used must be strong enough to withstand this force. Discussion The reaction force in the x-direction is larger than that of Example 6 since the walls turn the water over a much greater angle. If the reversing elbow is replaced by a straight nozzle (like one used by firefighters) such that water is discharged in the positive x-direction, the momentum equation in the x-direction becomes F Rx P 1, gage A 1 bm # V bm # V 1 F Rx bm # (V V 1 ) P 1, gage A 1 since both V 1 and V are in the positive x-direction. This shows the importance of using the correct sign (positive if in the positive direction and negative if in the opposite direction) for velocities and forces. EXAMPLE 6 4 Water Jet Striking a Stationary Plate Water is accelerated by a nozzle to an average speed of 0 m/s, and strikes a stationary vertical plate at a rate of 10 kg/s with a normal velocity of 0 m/s (Fig. 6 ). After the strike, the water stream splatters off in all directions in the plane of the plate. Determine the force needed to prevent the plate from moving horizontally due to the water stream. SOLUTION A water jet strikes a vertical stationary plate normally. The force needed to hold the plate in place is to be determined. Assumptions 1 The flow of water at nozzle outlet is steady. The water splatters in directions normal to the approach direction of the water jet. 3 The water jet is exposed to the atmosphere, and thus the pressure of the water jet and the splattered water leaving the control volume is atmospheric pressure, which is disregarded since it acts on the entire system. 4 The vertical forces and momentum fluxes are not considered since they have no z V 1 1 In x V Out F R FIGURE 6 Schematic for Example 6 4.

5 43 CHAPTER 6 Properties The density of air is given to be lbm/ft 3. Analysis Kinetic energy is a mechanical form of energy, and thus it can be converted to work entirely. Therefore, the power potential of the wind is proportional to its kinetic energy, which is V / per unit mass, and thus the maximum power is m. V / for a given mass flow rate: ft/s V 1 (7 mph)a b 10.7 ft/s 1 mph m # r 1 V 1 A 1 r 1 V 1 pd 4 (0.076 lbm/ft3 )(10.7 ft/s) W # max m # # V 1 ke 1 m (10.7 ft/s) (551.7 lbm/s) 1.5 kw p(30 ft) lbm/s 4 1 lbf 1 kw a 3. lbm ft/s ba lbf ft/s b Therefore, the available power to the wind turbine is 1.5 kw at the wind velocity of 7 mph. Then the turbine generator efficiency becomes h wind turbine W# act W # max 0.4 kw kw (or 3.7%) (b) The frictional effects are assumed to be negligible, and thus the portion of incoming kinetic energy not converted to electric power leaves the wind turbine as outgoing kinetic energy. Noting that the mass flow rate remains constant, the exit velocity is determined to be or m # ke m # # V ke 1 (1 h wind turbine ) m m # V 1 (1 h wind turbine) V V 1 1 h wind turbine (10.7 ft/s) ft/s We draw a control volume around the wind turbine such that the wind is normal to the control surface at the inlet and the outlet and the entire control surface is at atmospheric pressure. The momentum equation for steady one-dimensional flow is given as a F aout bm # V ain bm # V Writing it along the x-direction and noting that b 1, V 1, x V 1, and V, x V give Substituting the known values gives F R m # 1 lbf (V V 1 ) (551.7 lbm/s)( ft/s) a 3. lbm ft/s b 31.5 lbf F R m # V m # V 1 m # (V V 1 )

6 44 FLUID MECHANICS V 1 Streamline Streamline A Wind turbine V 3 = V 4 FIGURE 6 4 The large and small control volumes for the analysis of a wind turbine bounded by upper and lower streamlines. V The negative sign indicates that the reaction force acts in the negative x- direction, as expected. Then the force exerted by the wind on the mast becomes F mast F R 31.5 lbf. The power generated is proportional to V 3 since the mass flow rate is proportional to V and the kinetic energy to V. Therefore, doubling the wind velocity to 14 mph will increase the power generation by a factor of 3 8 to kw. The force exerted by the wind on the support mast is proportional to V. Therefore, doubling the wind velocity to 14 mph will increase the wind force by a factor of 4 to lbf. Discussion To gain more insight into the operation of devices with propellers or turbines such as helicopters, wind turbines, hydraulic turbines, and turbofan engines, we reconsider the wind turbine and draw two streamlines, as shown in Fig (In the case of power-consuming devices such as a fan and a helicopter, the streamlines converge rather than diverge since the exit velocity will be higher and thus the exit area will be lower.) The upper and lower streamlines can be considered to form an imaginary duct for the flow of air through the turbine. Sections 1 and are sufficiently far from the turbine so that P 1 P. The momentum equation for this large control volume between sections 1 and was obtained to be F R m # (V V 1 ) (1) The smaller control volume between sections 3 and 4 encloses the turbine, and A 3 A 4 A and V 3 V 4 since it is so slim. The turbine is a device that causes a pressure change, and thus the pressures P 3 and P 4 are different. The momentum equation applied to the smaller control volume gives F R P 3 A P 4 A 0 F R (P 4 P 3 )A () The Bernoulli equation is not applicable between sections 1 and since the path crosses a turbine, but it is applicable separately between sections 1 and 3 and sections 4 and : P 1 rg V 1 g z 1 P 3 rg V 3 g z 3 and P 4 rg V 4 g z 4 P rg V g z Adding these two equations and noting that z 1 z z 3 z 4, V 3 V 4, and P 1 P gives V V 1 P 4 P 3 r (3) Substituting m. rav 3 into Eq. 1 and then combining it with Eqs. and 3 gives V 3 V 1 V (4) Thus we conclude that the average velocity of a fluid through a turbine is the arithmetic average of the upstream and downstream velocities. Of course, the validity of this result is limited by the applicability of the Bernoulli equation. Now back to the wind turbine. The velocity through the turbine can be expressed as V 3 V 1 (1 a), where a 1 since V 3 V 1. Combining this expression with Eq. 4 gives V V 1 (1 a). Also, the mass flow rate through the turbine becomes m. rav 3 rav 1 (1 a). When the frictional

7 45 CHAPTER 6 effects and losses are neglected, the power generated by a wind turbine is simply the difference between the incoming and the outgoing kinetic energies: W # m # (ke 1 ke ) m# (V 1 V ) rav 1(1 a)3v 1 V 1(1 a) 4 rav 3 1a(1 a) Dividing this by the available power of the wind W. max m. V 1 / gives the efficiency of the wind turbine in terms of a, h wind turbine W# W # rav 3 1a(1 a) max (rav 1 )V 1/ The value of a that maximizes the efficiency is determined by setting the derivative of h wind turbine with respect to a equal to zero and solving for a. It gives a 1/3. Substituting this value into the efficiency relation just presented gives h wind turbine 16/ , which is the upper limit for the efficiency of wind turbines and propellers. This is known as the Betz limit. The efficiency of actual wind turbines is about half of this ideal value. EXAMPLE 6 6 Repositioning of a Satellite An orbiting satellite has a mass of m sat 5000 kg and is traveling at a constant velocity of V 0. To alter its orbit, an attached rocket discharges m f 100 kg of gases from the reaction of solid fuel at a velocity V f 3000 m/s relative to the satellite in a direction opposite to V 0 (Fig. 6 5). The fuel discharge rate is constant for s. Determine (a) the acceleration of the satellite during this -s period, (b) the change of velocity of the satellite during this time period, and (c) the thrust exerted on the satellite. SOLUTION The rocket of a satellite is fired in the opposite direction to motion. The acceleration, the velocity change, and the thrust are to be determined. Assumptions 1 The flow of combustion gases is steady and one-dimensional during the firing period. There are no external forces acting on the satellite, and the effect of the pressure force at the nozzle exit is negligible. 3 The mass of discharged fuel is negligible relative to the mass of the satellite, and thus the satellite may be treated as a solid body with a constant mass. 4 The nozzle is well-designed such that the effect of the momentumflux correction factor is negligible, and thus b 1. Analysis (a) We choose a reference frame in which the control volume moves with the satellite. Then the velocities of fluid streams become simply their velocities relative to the moving body. We take the direction of motion of the satellite as the positive direction along the x-axis. There are no external forces acting on the satellite and its mass is nearly constant. Therefore, the satellite can be treated as a solid body with constant mass, and the momentum equation in this case is simply Eq. 6 8, 0 d(mv ) CV aout bm # V ain bm # V dv sat m sat m # fv f V f x Satellite m sat CS V 0 FIGURE 6 5 Schematic for Example 6 6.

8 46 FLUID MECHANICS Noting that the motion is on a straight line and the discharged gases move in the negative x-direction, we can write the momentum equation using magnitudes as m dv sat sat m # f V f dv sat m# f V m f m f / t sat m sat V f Substituting, the acceleration of the satellite during the first s is determined to be a sat dv sat m f / t V m f sat (100 kg)/( s) (3000 m/s) 30 m/s 5000 kg (b) Knowing acceleration, which is constant, the velocity change of the satellite during the first s is determined from the definition of acceleration a sat dv sat / to be dv sat a sat V sat a sat t (30 m/s )( s) 60 m/s (c) The thrust exerted on the satellite is, from Eq. 6 9, F sat 0 m # 1 kn f ( V f ) (100/ kg/s)( 3000 m/s)a 1000 kg m/s b 150 kn Discussion Note that if this satellite were attached somewhere, it would exert a force of 150 kn (equivalent to the weight of 15 tons of mass) to its support. This can be verified by taking the satellite as the system and applying the momentum equation. EXAMPLE 6 7 Net Force on a Flange Flange P 1,gage In Spigot CV Water flows at a rate of 18.5 gal/min through a flanged faucet with a partially closed gate valve spigot (Fig. 6 6). The inner diameter of the pipe at the location of the flange is in ( ft), and the pressure at that location is measured to be 13.0 psig. The total weight of the faucet assembly plus the water within it is 1.8 lbf. Calculate the net force on the flange. F R W faucet z W water x Out FIGURE 6 6 Control volume for Example 6 7 with all forces shown; gage pressure is used for convenience. SOLUTION Water flow through a flanged faucet is considered. The net force acting on the flange is to be calculated. Assumptions 1 The flow is steady and incompressible. The flow at the inlet and at the outlet is turbulent and fully developed so that the momentumflux correction factor is about The pipe diameter at the outlet of the faucet is the same as that at the flange. Properties The density of water at room temperature is 6.3 lbm/ft 3. Analysis We choose the faucet and its immediate surroundings as the control volume, as shown in Fig. 6 6 along with all the forces acting on it. These forces include the weight of the water and the weight of the faucet assembly, the gage pressure force at the inlet to the control volume, and the

9 47 CHAPTER 6 net force of the flange on the control volume, which we call F R. We use gage pressure for convenience since the gage pressure on the rest of the control surface is zero (atmospheric pressure). Note that the pressure through the outlet of the control volume is also atmospheric since we are assuming incompressible flow; hence, the gage pressure is also zero through the outlet. We now apply the control volume conservation laws. Conservation of mass is trivial here since there is only one inlet and one outlet; namely, the mass flow rate into the control volume is equal to the mass flow rate out of the control volume. Also, the outflow and inflow average velocities are identical since the inner diameter is constant and the water is incompressible, and are determined to be V V 1 V V# A c Also, V # 18.5 gal/min pd /4 p(0.065 ft) /4 a ft3 1 gal Next we apply the momentum equation for steady flow, ba 1 min b 1.4 ft/s 60 s m # rv # (6.3 lbm/ft ft3 )(18.5 gal/min)a ba 1 min b.568 lbm/s 1 gal 60 s a F aout bm # V ain bm # V We let the x- and z-components of the force acting on the flange be F Rx and F Rz, and assume them to be in the positive directions. The magnitude of the velocity in the x-direction is V 1 at the inlet, but zero at the outlet. The magnitude of the velocity in the z-direction is zero at the inlet, but V at the outlet. Also, the weight of the faucet assembly and the water within it acts in the z-direction as a body force. No pressure or viscous forces act on the chosen control volume in the z-direction. The momentum equations along the x- and z-directions become Solving for F Rx and F Rz, and substituting the given values, F Rx m # V 1 P 1, gage A 1 1 lbf (.568 lbm/s)(1.4 ft/s)a 3. lbm ft/s b (13 lbf/in ) 7.0 lbf F Rz m # V W faucet water F Rx P 1, gage A 1 0 m # ( V 1 ) F Rz W faucet W water m # ( V ) 0 p(0.780 in) 4 1 lbf (.568 lbm/s)(1.4 ft/s)a 3. lbm ft/s b 1.8 lbf 11.8 lbf Then the net force of the flange on the control volume can be expressed in vector form as F R F Rx i F Rz k 7.0 i 11.8 k lbf

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