Homework 5 Assignment Print View

1. Award: 25 out of 50.00 points Water flowing in a horizontal 25-cm-diameter pipe at 6 m/s and 300 kPa gage enters a 90° bend reducing section, which connects to a 15cm-diameter vertical pipe. The inlet of the bend is 50 cm above the exit. Neglecting any frictional and gravitational effects, determine the net resultant force exerted on the reducer by the water. Take the momentum-flux correction factor to be 1.04. The net resultant force exerted on the reducer by the water is The direction of the resultant force is 23.4 °. kN. 18.05 References Worksheet Difficulty: Hard Water flowing in a horizontal 25-cm-diameter pipe at 6 m/s and 300 kPa gage enters a 90° bend reducing section, which connects to a 15cm-diameter vertical pipe. The inlet of the bend is 50 cm above the exit. Neglecting any frictional and gravitational effects, determine the net resultant force exerted on the reducer by the water. Take the momentum-flux correction factor to be 1.04. The net resultant force exerted on the reducer by the water is The direction of the resultant force is 26.8 ± 10% °. 18.5 ± 10% kN. Explanation: The following assumptions have been made here: 1. The flow is steady and incompressible. 2. The frictional effects are negligible in the calculation of the pressure drop (so that the Bernoulli equation can be used). 3. The weight of the elbow and the water in it is disregarded since the gravitational effects are negligible. 4. The momentum-flux correction factor for each inlet and outlet is given to be β = 1.04. Take the density of water as 1000 kg/m3. We take the water within the elbow as the control volume, and designate the inlet as 1 and the outlet as 2. We also designate the horizontal coordinate as x (with the direction of flow being positive) and the vertical coordinate as z. The continuity equation for this one-inlet one-outlet steady flow system is ṁ1 = ṁ2 = ṁ The mass flow rate of water and its outlet velocity are ) = (1000 kg/m3 )(6 m/s) [ ṁ = ρV1 A1 = ρV1 ( πD 4 2 V2 = ṁ ( ρπD2 2 4 ) = 294.52 kg/s (1000 kg/m )( 3 π(0.15 m)2 4 ) π(0.25 m) 4 2 ] = 294.52 kg/s = 16.67 m/s The Bernoulli equation for a streamline going through the center of the reducing elbow is expressed as P1 ρg + 2 V1 2g + z1 = P2 ρg + 2 V2 2g + z2 → P2 = P1 + ρg ( V1 2 + z1 − z2 ) − V2 2 2g Using gage pressures and substituting, the gage pressure at the outlet becomes P2 = (300 kPa) + (1000 kg/m3 )(9. 81 m/s2 ) ( 2 (6 m/s) − (16.67 m/s) 2(9.81 m/s ) 2 2 + 0. 5) ( 1 kN 2 1000 kg⋅m/s )( 1 kPa 2 1 kN/m ) = 184.01 kPa The momentum equation for steady flow is ⇀ ⇀ ⇀ ∑ F = ∑out βṁ V − ∑in βṁ V We let the x- and z-components of the resultant of the reaction forces exerted by the reducer elbow on water be FRx and FRz and assume them to be in the positive directions. Noting that the atmospheric pressure acts from all directions and its effect cancels out, the momentum equations along the x- and z-axes become FRx = − βṁV1 − P1,gage A1 FRz = − βṁV2 − P2,gage A2 We should not forget the negative sign for forces (including pressure forces) and velocities in the negative x- or z-direction. Solving for FRx and FRz and substituting the given values, FRx = − 1.04(294.5 kg/s)(6 m/s) ( 1 kN 2 1000 kg⋅m/s FRz = − 1. 04(294.5 kg/s)(17 m/s) ( ) − (300 kN/m2 ) 1 kN 2 1000 kg⋅m/s π(0.25 m) 4 ) − (184.01 kN/m2 ) 2 = -16.54 kN π(0.15 m) 4 Thus, the magnitude of the resultant force exerted by the water on the reducer is expressed as −−−−−−−−−−−−−−−−−−−−−− −−− −−−−−2− 2 FR = √ FRx + FRz = √(-16.54 kN) 2 + (-8.36 kN) 2 = 18.5 kN -8.36 kN θ = tan-1 F Rz = tan-1 -16.54 = 26.8° kN F Rx 2 = -8.36 kN 2. Award: 50 out of 50.00 points A large lawn sprinkler with two identical arms is used 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 5 gal/s and leaves the nozzles in the tangential direction. The sprinkler rotates at a rate of 180 rpm in a horizontal plane. The diameter of each jet is 0.6 in, and the normal distance between the axis of rotation and the center of each nozzle is 2 ft. Now, if the rotating head somehow gets stuck, determine the moment acting on the head. Take the density of water to be 62.4 lbm/ft3. The moment acting on the head is 441 lbf⋅ft. References Worksheet Difficulty: Medium A large lawn sprinkler with two identical arms is used 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 5 gal/s and leaves the nozzles in the tangential direction. The sprinkler rotates at a rate of 180 rpm in a horizontal plane. The diameter of each jet is 0.6 in, and the normal distance between the axis of rotation and the center of each nozzle is 2 ft. Now, if the rotating head somehow gets stuck, determine the moment acting on the head. Take the density of water to be 62.4 lbm/ft3. The moment acting on the head is 441 ± 10% lbf⋅ft. Explanation: The following assumptions have been made here: 1. The flow is uniform and steady. 2. The water is discharged to the atmosphere, and thus the gage pressure at the nozzle outlet is zero. 3. The nozzle diameter is small compared to the moment arm, and thus we use average values of the radius and velocity at the outlet. We take the disk that encloses the sprinkler arms as the control volume, which is a stationary control volume. The conservation of mass equation for this steady flow system is ṁ1 = ṁ2 = ṁ . Noting that the two nozzles are identical, we have ṁnozzle = ṁ 2 V˙total 2 or V˙nozzle = since the density of water is constant. The average jet outlet velocity relative to the nozzle is Vjet = V˙nozzle Ajet = 2.5 gal/s 2 [π(0.6/12 ft) /4] 1 ft ( 7.480 ) = 170.2 ft/s gal 3 The angular momentum equation can be expressed as ∑ M = ∑out rṁV - ∑in rṁV where all moments in the counterclockwise direction are positive, and all in the clockwise direction are negative. Then the angular momentum equation about the axis of rotation becomes −Mshaft = −2rṁnozzle Vjet or Mshaft = rṁtotal Vjet where 5 ṁtotal = ρV˙total = (62.4 lbm/ft3 )( 7.480 ft3 /s) = 41.71 lbm/s Substituting, the torque transmitted through the shaft is determined to be Mshaft = rṁtotal Vjet = (2 ft)(41.71 lbm/s)(170.2 ft/s) ( 1 lbf 2 32.2 lbm⋅ft/s ) = 441 lbf⋅ft