Solutions Manual for Fluid Mechanics Seventh Edition in SI Units Frank M. White Chapter 11 Turbomachinery PROPRIETARY AND CONFIDENTIAL This Manual is the proprietary property of The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and protected by copyright and other state and federal laws. By opening and using this Manual the user agrees to the following restrictions, and if the recipient does not agree to these restrictions, the Manual should be promptly returned unopened to McGraw-Hill: This Manual is being provided only to authorized professors and instructors for use in preparing for the classes using the affiliated textbook. No other use or distribution of this Manual is permitted. This Manual may not be sold and may not be distributed to or used by any student or other third party. No part of this Manual may be reproduced, displayed or distributed in any form or by any means, electronic or otherwise, without the prior written permission of the McGraw-Hill. © 2011 by The McGraw-Hill Companies, Inc. Limited distribution only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. 11.1 Describe the geometry and operation of a human peristaltic PDP which is cherished by every romantic person on earth. How do the two ventricles differ? Solution: Clearly we are speaking of the human heart, driven periodically by travelling compression of the heart walls. One ventricle serves the brain and the rest of one’s extremities, while the other ventricle serves the lungs and promotes oxygenation of the blood. Ans. 11.2 What would be the technical classification of the following turbomachines: (a) (b) (c) (d) (e) (f) (g) a household fan = an axial flow fan. Ans. (a) a windmill = an axial flow turbine. Ans. (b) an aircraft propeller = an axial flow fan. Ans. (c) a fuel pump in a car = a positive displacement pump (PDP). Ans. (d) an eductor = a liquid-jet-pump (special purpose). Ans. (e) a fluid coupling transmission = a double-impeller energy transmission device. Ans. (f) a power plant steam turbine = an axial flow turbine. Ans. (g) 11.3 A PDP can deliver almost any fluid, but there is always a limiting very-high viscosity for which performance will deteriorate. Can you explain the probable reason? Solution: High-viscosity fluids take a long time to enter and fill the inlet cavity of a PDP. Thus a PDP pumping high-viscosity liquid should be run slowly to ensure filling. Ans. 11.4 An interesting turbomachine is the torque converter [58], which combines both a pump and a turbine to change torque between two shafts. Do some research on this concept and describe it, with a report, sketches and performance data, to the class. Solution: As described, for example, in Ref. 58, the torque converter transfers torque T from a pump runner to a turbine runner such that ω pumpΤpump ≈ ω turbineΤturbine. Maximum efficiency occurs when the turbine speed ω turbine is approximately one-half of the pump speed ω pump. Ans. 2 11.5 What type of pump is shown in Fig. P11.5? How does it operate? Solution: This is a flexible-liner pump. The rotating eccentric cylinder acts as a “squeegee.” Ans. Fig. P11.5 11.6 Fig. P11.6 shows two points a halfperiod apart in the operation of a pump. What type of pump is this? How does it work? Sketch your best guess of flow rate versus time for a few cycles. Solution: This is a diaphragm pump. As the center rod moves to the right, opening A and closing B, the check valves allow A to fill and B to discharge. Then, when the rod moves to the left, B fills and A discharges. Depending upon the exact oscillatory motion of the center rod, the flow rate is fairly steady, being higher when the rod is faster. Ans. Fig. P11.6 11.7 A piston PDP has a 12.5-cm diameter and a 5-cm stroke and operates at 750 rpm with 92% volumetric efficiency. (a) What is the delivery, in L/min? (b) If the pump delivers SAE 10W oil at 20°C against a head of 15.2 m, what horsepower is required when the overall efficiency is 84%? Solution: For SAE 10W oil, take ρ ≈ 870 kg/m3. The volume displaced is π (12.5E − 2)2 (5E − 2) = 6.14E − 4 m 3 = 0.614 L, 4 strokes L3 ∴ Q = 0.614 750 (0.92 efficiency) min stroke υ= or: Q ≈ 423 L / min Ans. (a) 0.423 (15.2) 870(9.81) 60 ρgQH Power = = = 1.089 kW ≈ 1.46 hp 0.84 η Ans. (b) 3 3 11.8 A centrifugal pump delivers 2.1 m /min of water at 20°C when the brake horsepower is 22 and the efficiency is 71%. (a) Estimate the head rise in m and the pressure rise in kPa. 3 (b) Also estimate the head rise and horsepower if instead the delivery is 2.1 m /min of gasoline at 20°C. Solution: (a) For water at 20°C, take ρ ≈ 998 kg/m3. The power relation is 2.1 H (998)(9.81) 60 ρgQH Power = 22(746) = 16412 W = = , 0.71 η or H ≈ 34 m Ans. (a) Pressure rise Δp = ρgH = (998)(9.81)(34) = 333 kPa Ans. (a) (b) For gasoline at 20°C, take ρ ≈ 680 kg/m3. If viscosity (Reynolds number) is not important, the operating conditions (flow rate, impeller size and speed) are exactly the same and hence the head is the same and the power scales with the density: H ≈ 34 m (of gasoline); Power = Pwater ρgasoline ρwater 680 ≈ 15 hp = 11.2 kW = 22 998 Ans. (b) 11.9 Figure P11.9 shows the measured performance of the Vickers Inc. Model PVQ40 piston pump when delivering SAE 10W oil at 82.2°C (ρ ≈ 910 kg/m3). Make some general observations about these data vis-à-vis Fig. 11.2 and your intuition about PDP behavior. Solution: The following are observed: (a) The discharge Q is almost linearly proportional to speed Ω and slightly less for the higher heads (H or Δp). (b) The efficiency (volumetric or overall) is nearly independent of speed Ω and again slightly less for high Δp. (c) The power required is linearly proportional to the speed Ω and also to the head H (or Δp). Ans. 4 Fig. P11.9 11.10 Suppose that the pump of Fig. P11.9 is run at 1100 r/min against a pressure rise of 20.68 MPa. (a) Using the measured displacement, estimate the theoretical delivery in gal/min. From the chart, estimate (b) the actual delivery; and (c) the overall efficiency. Solution: (a) From Fig. P11.9, the pump displacement is 41 cm3. The theoretical delivery is cm 3 cm 3 r L 41 = 45100 Q = 1100 = 45 r min min min Ans. (a) (b) From Fig. P11.9, at 1100 r/min and Δp = 20.68 MPa, read Q ≈ 47 L/min. Ans. (b) (c) From Fig. P11.9, at 1100 r/min and Δp = 20.68 MPa, read η overall ≈ 87%. Ans. (c) 5 11.11 A pump delivers 1500 L/min of water at 20°C against a pressure rise of 270 kPa. Kinetic and potential energy changes are negligible. If the driving motor supplies 9 kW, what is the overall efficiency? Solution: With pressure rise given, we don’t need density. Compute “water” power: 11.12 In a test of the pump in the figure, the data are: p1 = 100 mmHg (vacuum), p2 = 500 mmHg (gage), D1 = 12 cm, and D2 = 5 cm. The flow rate is 680 L/min of light oil (SG = 0.91). Estimate (a) the head developed; and (b) the input power at 75% efficiency. Fig. P11.12 Solution: Convert 100 mmHg = 13332 Pa, 500 mmHg = 66661 Pa, 680 L/min = 0.01133 m3/s. Compute V1 = Q/A1 = 0.01133/[(π /4)(0.12)2] = 1.00 m/s. Also, V2 = Q/A2 = 5.79 m/s. Calculate γoil = 0.91(9790) = 8909 N/m3. Then the head is Power = γ QH 8909(0.01133)(11.3) = = 1521 W Ans. (b) 0.75 η P11.13 A 3.5 hp pump delivers 5.07 kN of ethylene glycol at 20°C in 12 seconds, against a head of 5.2 m. Calculate the efficiency of the pump. Solution: From Table A.3 for ethylene glycol, ρ = 1117 kg/m3. The specific weight is thus γ = (1117)(9.81) = 10957.8 N/m3. We then may calculate the volume flow rate: Q = (5.07E3 N / 12s) w = = 0.0386 m3 / s = 38.6 L / s 3 γ 10957.8 N / m This gives us enough information to calculate the pump efficiency: (10957.8 N / m3 )(0.0386 P = (3.5 hp)(746 γ QH W ) = 2611 W = = η η hp 2199.4 Solve for η = = 0.84 or 84% 2611 m3 )(5.2 m) s Ans. 6 11.14 A pump delivers gasoline at 20°C and 12 m3/h. At the inlet, p1 = 100 kPa, z1 = 1 m, and V1 = 2 m/s. At the exit p2 = 500 kPa, z2 = 4 m, and V2 = 3 m/s. How much power is required if the motor efficiency is 75%? Solution: For gasoline, take ρ g ≈ 680(9.81) = 6671 N/m3. Compute head and power: P11.15 An eductor or jet pump is a specially designed pump. The concept is to inject a high-velocity fluid in order to increase kinetic energy and reduce the pressure of the fluid to draw it by suction into the flow stream. Show simple calculations to prove the concept. Solution: Given a simple jet pump operation. Assume that sections (1) and (2) have diameter D-m and injector (nozzle) has diameter d–m. At section (1), fluid flows in with speed V1 and jet stream is injected with speed Vj: Under conservation of mass m in = m out That is ρ1V1 A1 + ρ jV j A j = ρ2V2 A2 But in general, we should use the same fluid to inject into the main stream. Therefore, ρ1 = ρ2 = ρ j , we then have V1 A1 + V j A j = V2 A2 π π π V1 (D – d)2 + V j d 2 = V2 D 2 4 4 4 ∴ V2 = V1 (D – d)2 + V j d 2 D2 7 11.16 A lawn sprinkler can be used as a simple turbine. As shown in Fig. P11.16, flow enters normal to the paper in the center and splits evenly into Q/2 and Vrel leaving each nozzle. The arms rotate at angular velocity ω and do work on a shaft. Draw the velocity diagram for this turbine. Neglecting friction, find an expression for the power delivered to the shaft. Find the rotation rate for which the power is a maximum. Fig. P11.16 Solution: Utilizing the velocity diagram at right, we apply the Euler turbine formula: P11.17 The centrifugal pump in Fig. P11.17 has r1 = 15 cm, r2 = 25 cm, b1 = b2 = 6 cm, and rotates counterclockwise at 600 r/min. A sample blade is shown. Assume α 1 = 90°. Estimate the theoretical flow rate and head produced, for water at 20°C, and comment. Impeller 30° 150° Fig. P11.17 40° Solution: For water, take ρ = 998 kg/m3. For counterclockwise rotation, this is a forwardfacing impeller, that is, looking at the figure, β2 = 150°, not 30°. If α 1 = 90°, then Vt1 = 0. Convert 600 r/min × 2π/60 = 62.8 rad/s. Evaluate u1 = ω r1 = (62.8)(0.15) = 9.425 m/s. Then 8 Evaluate u2 = ω r2 = (62.8)(0.25) = 15.71 m/s. From Eq. (3.18), the head produced is The water power is P = ρgQH = (998)(9.81)(0.447)(38.3) = 168,000 ft-lbf/s ≈ 305 hp. 11.18 A centrifugal pump has d1 = 18 cm, d2 = 33 cm, b1 = 10 cm, b2 = 7.5 cm, β1 = 25°, and β2 = 40° and rotates at 1160 r/min. If the fluid is gasoline at 20°C and the flow enters the 3 blades radially, estimate the theoretical (a) flow rate in m /min, (b) horsepower, and (c) head in m. Solution: For gasoline, take ρ ≈ 680 kg/m3. Compute ω = 1160 rpm = 121.5 rad/s. 0.09 ≈ 10.9 m/s u1 = ω r1 = 121.5 10.9 Vn1 = u1 tan β1 = 10.9 tan 25 ° ≈ 5.1 m/s Q = 2π r1b1Vn1 = 2π ( 0.09 ) ( 0.1) (5.1) ≈ 0.29 Vn2 = m3 s Ans. (a) Q 0.29 = ≈ 3.7 m/s 2π r2 b 2 2π ( 0.165 ) ( 0.075 ) u 2 = ωr2 = 121.5(0.165) ≈ 20 m/s Vt2 = u 2 − Vn2 cot 40° ≈ 15.6 m/s Fig. P11.18 Finally, Pideal = ρ Q u2Vt2 = 680(0.29)(20)(15.6) = 61.5 kW ≈ 82.5 hp. Ans. (b) Theoretical head H = P/( ρ gQ) = 61500/[680(9.81)(0.29)] ≈ 31.8 m. Ans. (c) 9 11.19 A jet of velocity V strikes a vane which moves to the right at speed Vc, as in Fig. P11.19. The vane has a turning angle θ. Derive an expression for the power delivered to the vane by the jet. For what vane speed is the power maximum? Fig. P11.19 Solution: The jet approaches the vane at relative velocity (V − Vc). Then the force is F = ρA(V − Vc)2(1 − cosθ ), and Power = FVc = ρ AVc(V − Vc)2(1 − cosθ ) Ans. (a) Maximum power occurs when 11.20 A centrifugal water pump has r2 = 24 cm, b2 = 5 cm, and β2 = 35° and rotates at 1060 r/min. 3 If it generates a head of 55 m, determine the theoretical (a) flow rate in m /min and (b) horsepower. Assume near-radial entry flow. Solution: For water take ρ = 1000 kg/m3. Convert ω = 1060 rpm = 111 rad/s. Then u 2 = ω r2 = 111( 0.24 ) = 26.64 m/s; P Q cotβ2 , and H = = 55 m Power = ρ Qu 2 u 2 − ρ gQ 2π r2 b 2 or: Q P = 9810QH = 1000Q(26.64) 26.64 – cot 35° with H = 55 2π (0.24)(0.05) Solve for Q = 0.34 m 3/s ≈ 20 m3 / min Ans. (a) With Q and H known, P = ρgQH = 9810(0.34)(55) ÷ 746 ≈ 246 hp. Ans. (b) 10 11.21 Suppose that Prob. 11.20 is reversed into a statement of the theoretical power P = 153 hp. Can you then compute the theoretical (a) flow rate; and (b) head? Explain and resolve the difficulty that arises. Solution: With power known, the basic theory becomes quadratic in flow rate: u 2 = 26.64 m Q , P = ρ Qu 2 u 2 − cot β2 s 2π r2 b 2 = 998Q(26.64)[26.64 − 19.57Q] = 153 × 746.7 Clean up: Q 2 − 1.36Q + 0.219 = 0, two roots: Q1 = 0.187 W s m3 m3 ; Q 2 = 1.173 s s Ans. (a) These correspond to H1 = 62.3 m; H 2 = 9.92 m Ans. (b) So the ideal pump theory admits to two valid combinations of Q and H which, for the given geometry and speed, give the theoretical power of 153 hp. Prob. 11.20 was solution 1. 11.22 The centrifugal pump of Fig. P11.22 3 develops a flow rate of 16 m /min with gasoline at 20°C and near-radial absolute inflow. Estimate the theoretical (a) horsepower; (b) head rise; and (c) appropriate blade angle at the inner radius. Solution: For gasoline take ρ ≈ 680 kg/m3. Convert Fig. P11.22 3 Q = 16 m /min = 0.267 m3/s and ω = 1750 rpm = 183 rad/s. Note r2 = 15 cm and β 2 = 30°. The ideal power is computed as Q P = ρ Qu 2 u 2 − cot β2 , where u 2 = ω r2 = 183( 0.15 ) ≈ 27.45 m/s. Plug in: 2π r2 b 2 0.267 P = 680(0.267)(27.45)27.45 − cot 30• = 104200 / 746 ≈ 140 hp 2 π (0.15)(0.075) H= Ans. (a) P 104200 = ≈ 58.5 m Ans. (b) ρgQ 680(9.81)(0.267) Compute Vn1 = Q/[2π r1b1] = 0.267 ≈ 5.7 m/s, u1 = ω r1 = 183(0.1) ≈ 18.3 m/s. [2π (0.1)(0.075)] For purely radial inflow, β1 = tan −1 (5.7/18.3) ≈ 17•. Ans. (c) 11 P11.23 For the sprinkler turbine of Fig. P11.23, let R = 18 cm with the total flow rate of 14 3 m /h of water at 20°C. If the nozzle exit diameter is 0.8 cm, estimate (a) the maximum power delivered in W and (b) the appropriate rotation rate in r/min. Fig. P11.23 3 3 Solution: For water at 20°C take ρ = 998 kg/m . Each arm takes 7 m /h. At maximum power, Q/2 7 / 3600 , = 38.68 m/s Aexit (π / 4)(8E − 3)2 1 = WR = Vrel = 19.34 m/s = w(0.18 m) 2 Vrel = utip Therefore, w = 107.45 rad/s = 17.1 rps ≈ 1026 rpm (b) 2 = 998(14/3600)(19.34)2 Pmax = ρQu tip = 1452 W (a) 11.24 A 37-cm-diameter centrifugal pump, running at 2140 rev/min with water at 20°C produces the following performance data: Q, m3/s: 0.0 0.05 0.10 0.15 0.20 0.25 0.30 H, m: 105 104 102 100 95 85 67 P, kW: 100 115 135 171 202 228 249 η: 0% 44% 74% 86% 92% 91% 79% (a) Determine the best efficiency point. (b) Plot CH versus CQ. (c) If we desire to use this 3 same pump family to deliver 26.5 m /min of kerosene at 20°C at an input power of 400 kW, what pump speed (in rev/min) and impeller size (in cm) are needed? What head will be developed? Solution: Efficiencies, computed by η = ρgQH/Power, are listed above. The best efficiency point (BEP) is approximately 92% at Q = 0.2 m3/s. Ans. (a) The dimensionless coefficients are CQ = Q/(nD3), where n = 2160/60 = 36 rev/s and D = 0.37 m, plus CH = gH/(n2D2) and CP = P/(ρ n3D5), where ρwater = 998 kg/m3. BEP 12 values are A plot of CH versus CQ is below. The values are similar to Fig. 11.8 of the text. Ans. (b) 3 (c) For kerosene, ρ k = 804 kg/m3. Convert 26.5 m /min = 0.442 m3/s. At BEP, we require the above values of dimensionless parameters: 0.442 P 400000 Q = 0.643; 3 = 3 = 0.111; 3 5 = nD 804n 3 D 5 ρn D nD rev rev Solve n = 26.1 = 1560 ; D = 0.534 m Ans. (c) s min Also, H* = C*H (n 2 D 2 )/g = 5.35(26.1)2 (0.534)2 /9.81 = 106 m Ans. (c) P11.25 At what speed, when pumping water, should the 105-cm pump of Fig. 11.7b be run to realize, at highest efficiency, (a) a head of 30.48 m? For this condition, what are the resulting (b) flow rate, and (c) brake horsepower? Solution: This is the same pump, same fluid, hence ρ1 = ρ2 and D1 = D2. Thus the scaling from Eqs. (11.28) is relatively simple. First read the 105-cm BEP values as best the writer can: η = 89% ; n 1 = 710 r / min ; H 1 ≈ 82 m ; Q 1 ≈ 81 m3 / min ; P 1 ≈ 1650 bhp Then get the new rotation rate from the new head, using Eq.(11.28): H2 n D n 30.48 m 30.48 1/ 2 r = ( 2 )2 ( 2 )2 = = ( 2 )2 , n2 = 710( ) = 433 Ans.(a) H1 n1 D1 710 82 m 82 min The flow rate and power follow in like manner: Q2 Q2 n D 433 433 = = 2 ( 2 )3 = , Q2 = (81)( ) ≈ 49 m3 / min Ans.(b) 3 710 710 Q1 n1 D1 81 m / min P2 ρ P2 n D 433 3 = = ( 2 )( 2 )3 ( 2 )5 = ( ) , P2 = (1650)(0.227) = 374 bhp Ans.(c) 710 P1 1650 bhp ρ1 n1 D1 These numbers are slightly uncertain because of the difficulty of reading Fig. 11.7b. 13 11.26 Figure P11.26 shows performance data for the Taco, Inc., model 4013 pump. Compute the ratios of measured shutoff head to the ideal value U2/g for all seven impeller sizes. Determine the average and standard deviation of this ratio and compare it to the average for the six impellers in Fig. 11.7. Fig. P11.26 Performance data for a centrifugal pump. (Courtesy of Taco. Inc., Cranston, Rhode Island.) Solution: All seven pumps are run at 1160 rpm = 121 rad/s. The 7 diameters are given. Thus we can easily compute U = ω r = ω D/2 and construct the following table: D, cm: U, m/s: Ho, m: Ho /(U2/g): 25.4 15.42 13.4 0.553 26.7 16.18 14.9 0.559 27.9 16.98 16.2 0.551 29.2 17.74 18.6 0.580 30.5 18.5 20.4 0.585 31.8 19.29 22.3 0.587 32.9 19.96 24.4 0.599 The average ratio is 0.573 Ans. (a), and the standard deviation is 0.019. Ans. (b) For the six pumps of Fig. 11.7, the average is 0.611, the standard deviation is 0.025. These results are true of most centrifugal pumps: we can take an average of 0.60 ± 0.04. 11.27 At what speed in rpm should the 89cm-diameter pump of Fig. 11.7(b) be run to produce a head of 122 m at a discharge of 3 75.71 m /min? What brake horsepower will be required? Hint: Fit H(Q) to a formula. Solution: A curve-fit formula for H(Q) for this pump is H(m) ≈ 72 − 10.8Q2, with 3 Q in m /s. Then, for constant diameter, the similarity rules predict Fig. 11.7b 14 n H1 = H 2 1 n2 2 2 710 n 710 and Q1 = Q 2 1 , or: H1 = 122 ≈ 72 – 10.8 1.262 n n2 n 2 2 7.02E7 Solve for n 2 = 72 1/2 2 ≈ 987 rpm Ans. 3 Now backtrack to compute H1 ≈ 63 m and Q1 ≈ 54.8 m /min. From Fig. 11.7(b) we may then read the power P1 ≈ 900 bhp (or compute this from η1 ≈ 0.85). Then, by similarity, P2 = (n2/n1)3 = 900(987/710)3 ≈ 2418 bhp. Ans. P11.28 Would the smallest, or the largest, of the seven Taco Inc. pumps in Fig. P11.26 be 3 better (a) for producing, near best efficiency, a water flow rate of 2.27 m /min and a head of 29 m? (b) At what speed, in r/min, should this pump run? (c) What input power is required? 3 Solution: First try the smallest pump, D = 25.4 cm. Read Fig. P11.26: Q* = 1.48 m /min, H* = 12.5 m, ηmax = 65%. To scale up to 2.27, use the similarity rules for constant D: 2.27 n n2 Q2 = = 2= , solve n2 = 1780 r / min 3 Q1 1.48 m / min n1 1160rpm H2 n 1780 2 H2 = =( 2 )2 = ( ) , solve H 2 = 29.43 m Then H1 12.5 m n1 1160 γ Q2 H 2 (9790)(2.27 / 60)(29.43) = = 16770 W = 22.5 hp Power2 = η 0.65 So the small pump will do the job, using 22.5 hp. Next try the largest pump, D = 32.9 cm. 3 3 Read Fig. P11.24: Q* = 2 m /min, H* = 22 m, ηmax = 80%. Again scale up to 2.27 m /min, using the similarity rules at constant diameter: 2.27 n n2 Q2 = = 2 = , solve n2 = 1317 r / min 3 Q1 2 m / min n1 1160rpm H2 n 1317 2 H2 = = ( 2 )2 = ( ) , solve H 2 = 28.4 m Then H1 22 m n1 1160 (9790)(2.27 / 60)(28.4) W γQ H = 13149 = 17.6 hp Power2 = 2 2 = η 0.8 s So the largest pump is better to do the job, with slower speed and less power. Ans. (a) The proper large-pump speed is 1317 r/min. Ans. (b). The power is 17.6 hp. Ans. (c) [NOTE that if we specify a different head, like 60 m, neither pump can do the job.] 15 11.29 The 30.5-cm pump of Fig. P11.26 is to be scaled up in size to provide a head of 27.4 3 m and a flow rate of 3.79 m /min at BEP. Determine the correct (a) impeller diameter; (b) speed in rpm; and (c) horsepower required. 3 Solution: From the chart, at BEP, Q ≈ 1.82 m /min, H ≈ 18.3 m, and η ≈ 76.5%, from which P= ρgQH (9790)(1.82/60)(18.3) = ≈ 7104 W , η 0.765 C*P = 7104 P = ≈ 0.375 3 5 ρn D 998(19.3)3 (0.305)5 1.82/60 (9.81)(18.3) Q* * gH* C*Q = ≈ 5.18 3 = 3 ≈ 0.0554; C H = 2 2 = (19.3)(0.305) (19.3)2 (0.305)2 nD n D Larger pump, H = 27.4, Q = 3.79: 9.81(27.4) ≈ 5.18, n 2 D2 3.79/60 ≈ 0.0554 nD 3 Solve for D ≈ 40 cm Ans. (a) and n ≈ 17.82 rps ≈ 1069 rpm Ans. (b) P* = C*P ρ n 3D 5 = 0.375(998)(17.82)3 (0.4)5 ≈ 21686 W ≈ 29 bhp Ans. (c) 11.30 Tests by the Byron Jackson Co. of a 37.1-cm centrifugal water pump at 2134 rpm yield the data below. What is the BEP? What is the specific speed? Estimate the maximum discharge possible. Q, L/s: 0 56.63 113.27 169.9 226.54 283.17 H, m: 103.63 103.63 103.63 100.58 91.44 67.06 bhp: 135 160 330 330 205 255 Solution: The efficiencies are computed from η = ρgQH/(746 bhp) and are as follows: Q: 0 56.63 113.27 169.9 226.54 283.17 η: 0 0.48 0.751 0.879 0.824 0.755 Thus the BEP is, even without a plot, close to Q ≈ 169.9 L/s. Ans. The specific speed is Ns ≈ nQ*1/2 2134[(0.1699)]1/2 = ≈ 27.7 H*3/4 (100.58)3/4 Ans. For estimating Qmax, the last three points fit a Power-law to within ±0.5%: H ≈ 103.63 − 16496Q 4.85 = 0 if Q ≈ 352 L / s = Q max Ans. 16 11.31 If the scaling laws are applied to the Byron Jackson pump of Prob. 11.30 for the same impeller diameter, determine (a) the speed for which the shut-off head will be 85 m; (b) the speed for which the BEP flow rate will be 226.5 L/s; and (c) the speed for which the BEP conditions will require 80 horsepower. Solution: From the table in Prob. 11.30, the shut-off head at 2134 rpm is 103.63 m. Thus If D1 = D 2 , n 2 = n1 (H 2 /H1 )1/2 = 2134(85/103.63)1/2 ≈ 1933 rpm Ans. (a) If Q*2 = 8 ft 2 , n 2 = n1 (Q*2 / Q1* ) = 2134(226.5 / 169.9) ≈ 2845 rpm Ans. (b) s Finally, if ρ1 = ρ 2 (water) and the diameters are the same, then Power ∝ n3, or, at BEP, n 2 = n1 (P2* /P1* )1/3 = 2134(80/255)1/3 ≈ 1450 rpm Ans. (c) P11.32 A pump, geometrically similar to the 96.5-cm-diameter model in Fig. 11.7b, has a diameter of 60.96 cm and is to develop 300 bhp at BEP when pumping gasoline (not water). Determine (a) the appropriate speed, in r/min; (b) the BEP head; and (c) the BEP flow rate. 3 Solution: For water, ρ1 = 1000 kg/m . For gasoline, from Table A.3, ρ2 = 680 kg/m3. From Fig. 11.7a, read, as best you can, the BEP properties: P1 ≈ 1250 bhp ; Q1 ≈ 75.7 m3 / min ; H1 ≈ 69 m ; n1 = 710 rpm (a) With the BEP power and the diameter known, the power ratio yields the speed: P2 ρ n D 300 bhp 680 n2 3 60.96 cm 5 = = ( 2 )( 2 )3 ( 2 )5 = ( )( ) ( ) , or : 1000 710 96.5 cm P1 1250 bhp ρ1 n1 D1 (1.91E − 10) n 32 = 0.24 , solve n2 ≈ 1080 r min Ans.(a) (b, c) With n2 known, the head and flow rate at BEP follow from the scaling laws: H2 H n D 1080 2 60.96 2 = 2 = ( 2 )2 ( 2 )2 = ( ) ( ) , H 2 = (69)(0.923) ≈ 64 m 710 96.5 H1 69 n1 D1 Ans.(b) Q2 Q2 n D 1080 60.96 3 = = 2 ( 2 )3 = ( )( ) , Q2 = (75.7)(0.383) ≈ 29 m3 / min Ans.(c) 710 96.5 Q1 75.7 m3 / min n1 D1 These answers have a slight uncertainty due to the difficulty of reading Fig. 11.7b. 17 11.33 A centrifugal pump with backward-curved blades has the following measured performance when tested with water at 20°C: 3 Q, m /min: 0 H, m: 37.5 P, hp: 30 1.51 3.03 4.54 6.06 7.57 9.08 35.05 32.92 30.78 28.35 24.67 18.9 44 47 48 46 36 40 (a) Estimate the best efficiency point and the maximum efficiency. (b) Estimate the most efficient flow rate, and the resulting head and brake horsepower, if the diameter is doubled and the rotation speed increased by 50%. 3 Solution: (a) Convert the data above into efficiency. For example, at Q = 1.51 m /min, η= γQH (9790 N/m 3 )(1.51/60 m 3/s)(35.05 m) = = 0.32 = 32% P (36 × 746 W ) When converted, the efficiency table looks like this: Q, m /min: 3 0 1.51 3.03 4.54 6.06 7.57 9.08 η, %: 0 32% 55% 69% 80% 85% 82% 3 So maximum efficiency of 85% occurs at Q = 7.57 m /min. Ans. (a) , but we can set them equal for conditions 1 (b) We don’t know the values of (the data above) and 2 (the performance when n and D are changed): CQ* = Q1 n1 D13 = Q2 n2 D23 = Q2 (1.5n1 )(2D1 )3 , or: Q2 = 12Q1 = 12(7.57 m 3 /min) = 90.84 m 3 /min C *H = Ans. (b) gH 2 gH 2 gH1 , 2 2 = 2 2 = n1 D1 n2 D2 (1.5n1 )2 (2D1 )2 or: H 2 = 9H1 = 9(24.67 m) = 222 m Ans. (b) 11.34 The data of Prob. 11.33 correspond to a pump speed of 1200 r/min. (Were you able to solve Prob. 11.33 without this knowledge?) (a) Estimate the diameter of the impeller [HINT: See Prob. 11.26 for a clue.]. (b) Using your estimate from part (a), calculate the BEP parameters and compare with Eq. (11.27). (c) For what speed of this pump would the BEP head be 85.34 m? Solution: Yes, we were able to solve Prob. 11.33 by simply using ratios. (a) Prob. 11.26 showed that, for a wide range of centrifugal pumps, the shut-off head Ho ≈ 18 0.6U2/g ± 6%, where U is the impeller blade tip velocity, U = ω D/2. Use this estimate with the shut-off head and speed of the pump in Prob. 11.33: ω = 2π (1200 rpm)/60 = 126 rad/s, H o ≈ 37.5 m = 0.6[(126D/2)2 /9.81 m/s2 ] Solve for D ≈ 39.3 cm Ans. (a) (b) With diameter D ≈ 39.3 cm estimated and speed n = 1200/60 = 20 r/s given, we can calculate: CQ* = 2 (7.57/60) m 3/s * = (9.81 m/s )(24.67 m) ≈ 3.9 C ≈ 0.104; H (20 r/s)(0.393 m)3 (20 r/s)2 (0.393 m)2 CP* = (c) Use the estimate of C *H ≈ 3.9 = (48)(746 W/ hp) (998 kg/m 3 )(20 r/s)3 (0.393)5 ≈ 0.48 Ans. (b) to estimate the speed needed to produce 85.34 m of head: (9.81)(85.34 m) , solve for n = 37.3 r/s ≈ 2237 r / min Ans. (c) n 2 (0.393 m)2 3 11.35 In Prob. P11.33, the pump BEP flow rate is 7.57 m /min, the impeller diameter is 40.64 cm, and the speed is 1200 r/min. Scale this pump with the similarity rules to find (a) 3 the diameter and (b) the speed that will deliver a BEP water flow rate of 15.14 m /min and a head of 54.86 m. (c) What brake horsepower will be required for this new condition? Solution: Take the specific weight of water to be γ = 9790 N/m3. Find the efficiency: η1 ≈ η2 = γQ1 H1 (9790 N / m 3 )(7.57 / 60 m 3 / s)(24.67 m) = = 0.851 or 85.1% P1 (48x746 W ) Meanwhile use the scaling laws for BEP flow rate and head: 7.57 Q 15.14 Q1 = = 23 = 3 3 n1 D1 (1200)(0.406) n2 D2 n2 D23 (9.81)(24.67) gH (9.81)(54.86) gH1 = = 2 22 = 2 2 2 2 n1 D1 (1200) (0.406) n2 D2 n22 D22 Solve simultaneously for D2 = 47 cm Ans.(a) and n2 = 1545 r/min Ans.(b). (c) Finally, determine the required horsepower (which we could have done above, using η1): P2 = γ Q2 H 2 (9790)(15.14 / 60)(54.86) = = 1.59 kW η2 0.853 ÷ 746 = 213 hp Ans.(c) 19 11.36 Consider a pump geometrically similar to the 22.9-cm-diameter pump of Fig. P11.36 3 to deliver 4.54 m /min of kerosene at 1500 rpm. Determine the appropriate (a) impeller diameter; (b) BEP horsepower; (c) shut-off head; and (d) maximum efficiency. Fig. P11.36 Performance data for a family of centrifugal pump impellers. (Courtesy of Taco, Inc., Cranston, Rhode Island.) Solution: For kerosene, take ρ ≈ 804 kg/m3, whereas for water ρ ≈ 1000 kg/m3. From Fig. 3 P11.36, at BEP, read Q* ≈ 2.56 m /min, H* ≈ 23.2 m, and ηmax ≈ 0.77. Then CQ* = 2.56 / 60 4.54 / 60 Q* = ≈ 0.121 = , 3 3 (1760/60)(0.229) (1500/60)D 3 nD Solve for D imp ≈ 29 cm Ans. (a) Shut-off: gH o 9.81(25.6 m) 9.81H o = , 2 2 = 2 2 n D (1760/60) (0.229) (1500/60)2 (0.29)2 Solve H o ≈ 29.8 m Ans. (c) 1 − η2 0.229 Moody: ≈ 1 − 0.77 0.29 1/4 , solve for η2 ≈ 0.783 = 78% Fig. P11.34: Read H* ≈ 23.2 m, whence Ans. (d) (crude estimate) 9.81(23.2) 9.81 H*new = , (29.3)2 (0.229)2 (25)2 (0.29)2 or H*new ≈ 27.1 m * = Then Pnew 804(9.81)(4.54/ 60)(27.1) ≈ 2.07 kW ÷ 746 ≈ 27.7 bhp 0.783 Ans. (b) 20 11.37 A 46-cm-diameter centrifugal pump, running at 880 rev/min with water at 20°C, generates the following performance data: 3 Q, m /min: 0.0 7.57 15.14 22.71 30.28 37.85 H, m: 28.04 27.13 25.60 23.77 20.73 15.24 P, hp: 100 112 130 143 156 163 η: 0% 40% 65% 83% 88% 78% Determine (a) the BEP; (b) the maximum efficiency; and (c) the specific speed. (d) Plot the required input power versus the flow rate. Solution: We have computed the efficiencies and listed them. The BEP is the next3 to-last point: Q = 30.28 m /min, η max = 88%. Ans. (a, b) The specific speed is N′s = nQ*1/2/(gH*)3/4 = (880/60)(30.28/60)1/2/[9.81(20.73)]3/4 ≈ 0.193, or Ns = 498.4 (probably a centrifugal pump). Ans. (c) The plot of input horsepower versus flow rate is shown below—there are no surprises in this plot. Ans. (d) 3 P11.38 The pump of Prob. P11.37 has a maximum efficiency of 88% at 30.28 m /min. (a) Can we use this pump, at the same diameter but a different speed, to generate a BEP head of 3 45.72 m and a BEP flow rate of 37.85 m /min? (b) If not, what diameter is appropriate? 3 Solution: We are still pumping water, ρ = 1000 kg/m . Try scaling laws for head and then for flow: H1 = 20.73m ; Q1 = 30.28 n1 =880 rpm = 14.7 m3 m3 = 0.505 ; P1 = 156 bhp ; D1 = 46 cm ; min s r s D1 = D2 : H 2 45.72 n r r = = ( 2 )2 , n2 = 21.8 = 1307 (from the head) H1 20.73 14.7 s min D1 = D2 : Q2 37.85 n r r = = ( 2 ) , n2 = 18.4 = 1103 (from the flow rate) Q1 30.28 14.7 s min 21 These rotation rates are not the same. Therefore we must change the diameter. (b) Allow for a different diameter in both head and flow rate scaling: Ans.(a) n D n2 D H2 45.72 = = ( 2 )2 ( 2 )2 = ( )2 ( 2 )2 , or : n22 D22 = 100.43 H1 n1 D1 14.67 r / s 0.46 20.73 n D Q2 37.85 n2 D2 3 = = ( ) = ( 2 )( 2 )3 , or : Q1 30.28 n1 D1 14.67 0.46 n2 D23 = 1.785 Solve simultaneously to obtain n2 = 23.7 r / s = 1425 rpm ; D2 = 42.2 cm Ans.(b) The power required increases to 430 bhp. 11.39 Consider the two pumps of Problems P11.30 and P11.37, respectively. If the 3 diameters are not changed, which is more efficient for delivering water at 11.36 m /min and a head of 122 m? What is the appropriate rotation speed for the better pump? Solution: Unless we are brilliant, what can we do but try them both? The scaling law for constant diameter states that Q is proportional to n, and H is proportional to n2. The pump of Prob. P11.30 has BEP at Q* = 169.9 L/s and H* = 100.58 m. Scale from there: 11.36 n n2 Q2 = = 2 = , Q1 0.17 × 60 n1 2134 r / min H2 n 2380 2 H2 = = ( 2 )2 = ( ) , 2134 H1 100.58 n1 Solve n2 ≈ 2380 r / min Solve H 2 ≈ 125 m It seems that the Prob. P11.30 pump can do the job quite well. Answer. 3 Try the Prob. P11.37 pump, whose BEP is at Q* = 30.28 m /min and H* = 20.73 m. Then n2 Q2 11.36 n2 = = = , Solve n2 ≈ 330 r / min Q1 30.28 n1 880 r / min H2 n 330 2 H2 = = ( 2 )2 = ( ) , Solve H 2 ≈ 2.9 m 880 H1 20.73 n1 This is not just worse, it is ridiculous – too slow, almost no head, totally inefficient. 22 11.40 A 17.5-cm pump, running at 3500 rpm, has the measured performance at right for water at 20°C. (a) Estimate the horsepower at BEP. If this pump is rescaled in water to provide 20 bhp at 3000 rpm, determine the appropriate (b) impeller diameter; (c) flow rate; and (d) efficiency for this new condition. Q, L/min: 190 379 568 757 946 1136 1325 1514 1703 H, m: 61.3 61 60.4 59.1 57.6 55.2 51.5 47.5 42.4 η, %: 29 50 64 72 77 80 81 79 74 Solution: The BEP of 81% is at about Q = 1325 L/min and H = 51.5 m. Hence the power is P* = ρgQ*H* 998(9.81)(1.325/ 60)(51.5) = = 13, 746.3 W ÷ 746 ≈ 18.4 bhp Ans. (a) η 0.81 If the new conditions are 20 hp at n = 3000 rpm = 50 rps, we equate power coefficients: C*P = 13746.3 3 ? 5 = 0.423 = 998(3500/60) (0.175) Solve D imp ≈ 19.5 cm 20 × 746 , 998(50)3 D 5 Ans.(b) With diameter known, the flow rate is computed from BEP flow coefficient: 1.325 / 60 Q* Q* ? C*Q = = 0.0706 = 3 = 3, 3 nD 50(0.195) (3500/60)(0.175) Solve Q* = 157.0 L / min Ans. (c) Finally, since D1 ≈ D2, we can assume the same maximum efficiency: 81%. Ans. (d) 3 11.41 The Allis-Chalmers D30LR centrifugal compressor delivers 934.46 m /min of SO2 with a pressure change from 96.53 to 124.11 kPa absolute using an 800-hp motor at 3550 r/min. What is the overall efficiency? What will the flow rate and Δp be at 3000 r/min? Estimate the diameter of the impeller. 2 2 Solution: For SO2, take M = 64.06, hence R = 8314/64.06 ≈ 130 m /(s ·K). Then 934.46 ÷ 746 ≈ 576 hp delivered Δp = 124.11 − 96.53 = 27.58 kPa, Power = QΔp = 27.58 60 n 3000 ≈ 790 m3 / min Ans. (b) If n 2 = 3000 rpm, Q 2 = Q1 2 = 934.46 3550 n1 2 3000 Δp 2 = Δp1 (n 2 /n1 )2 = (27.58) ≈ 19.7 kPa 3550 Ans. (c) To estimate impeller diameter, we have little to go on except the specific speed: 23 ρavg ≈ Ns = 110.32 kPa Δp 27.58 kPa ≈ 2.94 kg/m 3 , H = = ≈ 956 m, 130(289) ρg 2.94(9.81) rpm(m 3 /min)1/2 3550[934.46]1/2 = ≈ 631 : Fig. P11.51: C*Q ≈ 0.45 (H-m)3/4 (956)3/4 934.46 / 60 Crudely, C*Q ≈ 0.45 = , solve for D impeller ≈ 84 cm Ans. (d) (3550/60)D 3 Clearly this last part depends upon the ingenuity and resourcefulness of the student. 11.42 The specific speed Ns, as defined by Eq. (11.30), does not contain the impeller diameter. How then should we size the pump for a given Ns? Logan [7] suggests a parameter called the specific diameter DS, which is a dimensionless combination of Q, (gH), and D. (a) If DS is proportional to D, determine its form. (b) What is the relationship, if any, of DS to CQ*, CH*, and CP*? (c) Estimate DS for the two pumps of Figs. 11.8 and 11.13. Solution: If we combine CQ and CH in such a way as to eliminate speed n, and also to make the result linearly proportional to D, we obtain Logan’s result: (c) For the pumps of Figs. 11.8 and 11.13, we obtain 11.43 It is desired to build a centrifugal pump geometrically similar to Prob. 11.30 (data 3 below) to deliver 24.61 m /min of gasoline at 1060 rpm. Estimate the resulting (a) impeller diameter; (b) head; (c) brake horsepower; and (d) maximum efficiency. Q, m3/min: H, m: bhp: 0 10.2 13.6 17 103.6 103.6 103.6 100.6 91.4 67.1 330 330 135 3.4 160 6.8 205 255 Solution: For gasoline, take ρ ≈ 680 kg/m3. From Prob. 11.30, BEP occurs at Q* ≈ 169.9 L/s, ηmax ≈ 0.88. The data above are for n = 2134 rpm = 35.6 rps and D = 37.1 cm. 0.17 ? 24.61/60 , 3 = 0.0935 = 35.6(0.371) (1060/60)D 3 Solve for D imp ≈ 63 cm Ans. (a) Then C*Q = C*H = 9.81(100.6) 9.81H ? , solve for H ≈ 71 m Ans. (b) 2 2 = 5.66 = (35.6) (0.371) (1060/60)2 (0.63)2 24 Step-up the efficiency with Moody’s correlation, Eq. (11.29a), for D1 = 37.1 cm: D 1 − η2 ≈ 1 1 − 0.88 D2 Then P2 = 1/4 37.1 = 63 1/4 = 0.877, solve for η2 ≈ 0.895 ρgQ 2 H 2 680(9.81)(24.61/60 )(71) = = 217, 057 ÷ 746 ≈ 291 bhp Ans. (c) η2 0.895 3 11.44 A 20-cm model pump delivering water at 82.2°C at 3.03 m /min and 2400 rpm begins to cavitate when the inlet pressure and velocity are 82.74 kPa and 6.1 m/s, respectively. Find the required NPSH of a prototype which is 4 times larger and runs at 1000 rpm. Solution: For water at 82.2°C, take ρ g ≈ 9521 N/m3 and pv ≈ 76.6 kPa. From Eq. 11.19, NPSH model p i − p v Vi2 (82.74 − 76.6)kPa (6.1)2 = + = + = 2.54 m ρg 2g 9521 2(9.81) 2 Similarity: NPSH proto 2 2 2 n p Dp 1000 4 = NPSH m = 2.54 ≈ 7.1 m Ans. 2400 1 n m D m 11.45 The 71-cm-diameter pump in Fig. 11.7a at 1170 r/min is used to pump water at 3 20°C through a piping system at 53 m /min. (a) Determine the required brake horsepower. The average friction factor is 0.018. (b) If there is 19.8 m of 30.5-cm-diameter pipe upstream of the pump, how far below the surface should the pump inlet be placed to avoid cavitation? Fig. 11.7a Solution: For water at 20°C, take ρ g ≈ 9790 N/m3 and pv ≈ 2.35 kPa. From Fig. 11.7a 3 (above), at 71 cm and 53 m /min, read H ≈ 97.5 m, η ≈ 0.81, and P ≈ 1400 bhp. Ans. Or: Required bhp = ρgQH (9790)(53/ 60)(97.5) = = 1040943 ÷ 746 ≈ 1395 bhp Ans. η 0.81 3 From the figure, at 53 m /min, read NPSH ≈ 7.6 m. Assuming pa = 1 atm = 101.3 kPa, Eq.11.20: NPSH = pa − pv 101300 − 2350 L V2 − Z i − h fi = − Z i − h fi ≈ 7.6 m, h fi = f , ρg 9790 D 2g V= Q 53/60 = ≈ 12 m/s, A (π /4)(0.305)2 2 19.8 (12) so: Z i = 10.11 − 7.6 − 0.018 ≈ −5.94 m Ans. 0.305 2(9.81) 25 11.46 The pump of Prob. 11.30 is scaled up to a 45.7-cm-diameter, operating in water at BEP at 1760 rpm. The measured NPSH is 4.88 m, and the friction loss between the inlet and the pump is 6.71 m. Will it be sufficient to avoid cavitation if the pump inlet is placed 2.74 m below the surface of a sea-level reservoir? Solution: For water at 20°C, take ρ g = 9790 N/m3 and pv = 2.35 kPa. Since the NPSH is given, there is no need to use the similarity laws. Merely apply Eq. 11.20: NPSH ≤ pa − pv 101300 − 2350 − Z i − h fi , or: Z i ≤ − 6.71 − 4.88 = −1.48 m, OK, ρg 9790 Z actual = −2.74 m Ans. This works. Putting the inlet 2.74 m below the surface gives 1.26 m of margin against cavitation. 11.47 Determine the specific speeds of the seven Taco, Inc. pump impellers in Fig. P11.26. Are they appropriate for centrifugal designs? Are they approximately equal within experimental uncertainty? If not, why not? Solution: Read the BEP values for each impeller and make a little table for 1160 rpm: D, cm: 25.4 26.7 27.9 29.2 30.5 31.8 32.9 Q*, m /min: 3 1.48 1.59 1.67 1.74 1.82 1.93 2 H*, m: 12.5 13.4 14.9 17.1 18.3 20.1 21.9 Specific speed NS: 212 209 198 182 177 170 162 These are well within the centrifugal-pump range (NS < 600) but they are not equal because they are not geometrically similar (7 different impellers within a single housing). Ans. 11.48 The answer to Prob. 11.42 is that the dimensionless “specific diameter” takes the form Ds = D(gH*)1/4/Q*1/2, evaluated at the BEP. Data collected by the writer for 30 different pumps indicates, in Fig. P11.48, that Ds correlates well with specific speed Ns. Use this figure to estimate the appropriate impeller diameter for a pump which delivers 75.71 3 m /min of water and a head of 122 m running at 1200 rev/min. Suggest a curve-fit formula to the data (Hint: a hyperbola). 26 Fig. P11.48 Solution: We see that the data are very well correlated by a single curve. (NOTE: These are all centrifugal pumps—a slightly different correlation holds for mixed- and axial-flow pumps.) The data are well fit by a hyperbola: Figure P11.48: Ds » Const , where Const ≈ 1170 ± 75 Ns Ans. For the given pump-data example, we compute Ns = Hence Ds ≈ (rpm)(m 3 /min)1/2 1200(75.71)1/2 = = 284.4, (Head−m)3/4 (122)3/4 1170 D[9.81(122)]1/4 ≈ 4.11 = , solve D ≈ 78.5 ± 3 cm Ans. 284.4 (75.71/60)1/2 P11.49 A pump must be designed to deliver 6 m3/s of water against a head of 28 m. The specified shaft speed is 20 r/s. What type of pump do you recommend? Solution: Change units to suit the practicing-engineer version of specific speed: 6 m3/s = 360 m3/min, 20 r/s = 1200 r/min. Then the specific speed is Ns = (rpm)(m3 / min)1/ 2 (H − m)3/ 4 = (1200 rpm)(360 m3 / min)1/ 2 (28 m)3/ 4 From Fig. 11.11a, an axial-flow pump is appropriate for this job. Ans. = 1871 27 11.50 A commercial pump runs at 1750 r/min and delivers, near BEP, a flow of 8.71 3 m /min and a head of 40 m. (a) What type of pump is this? (b) Estimate the impeller diameter using the data of Prob. P11.48. (c) Estimate CQ* and add another data point to Fig. P11.51. Solution: (a) To find the type of pump from Fig. 11.11, we need the head in feet: H = 40 m. Then we can calculate the traditional (dimensional) specific speed: NS = (rpm)(m 3 / min)1/2 (1750 rpm)(8.71m 3 / min)1/2 = ≈ 325 [H (m)} 3/4 (40 m)3/4 From Fig. 11.11, this is a centrifugal pump. Ans. (a) (b) Use Fig. P11.48 to estimate the specific diameter and hence the impeller diameter. The curve-fit recommended in the solution to Prob. P11.48 is 1170 1170 D(gH )1/4 D[(9.81 m / s 2 )(40 m)]1/4 DS ≈ = ≈ 3.60 = = = 11.68 D NS 325 Q1/2 (8.71 / 60 m 3 / s)1/2 Solve for D impeller ≈ 31 cm Ans.(b) The accuracy is unusually good, not typical. The actual impeller diameter was 31 cm! (c) Calculate the flow coefficient, which we are told is near BEP: CQ* = Q* (8.71 / 60 m 3 / s) = = 0.17 nD 3 (1750 / 60 r / s)(0.31 m)3 Add this point to Fig. P11.51 at NS = 325.5 and we have an excellent fit to the other data. 11.51 Data collected by the writer for flow coefficient at BEP for 30 different pumps are plotted at right in Fig. P11.51. Determine if the values of fit this correlation for the pumps of Problems P11.26, P11.30, P11.37, and P11.40. If so, suggest a curve fit formula. Solution: Make a table of these values: Fig. P11.51 Flow coefficient at BEP for 30 commercial pumps. 28 3 Prob. 11.30: Prob. 11.37: Prob. 11.40: Prob. P11.26: Q*, m /min 10.19 30.28 1.32 1.74 D, cm n, rpm NS 37.1 45.7 17.4 29.2 2134 880 3500 1160 214 498 210 182 0.0935 0.361 0.0716 0.0602 Ans. When added to the plot shown below, all four data points seem to fit quite well. The data are useful for predicting general centrifugal-pump behavior and are well fit to either a 2nd-order polynomial or a single-term Power-law slightly less than parabolic: Polynomial: C*Q ≈ 1.31E−4N S + 1.15E − 6N S2 (Correlation R 2 ≈ 0.99) Ans. Power C*Q ≈ 2.58E − 6N1.914 S 11.52 Data collected by the writer for power coefficient at BEP for 30 different pumps are for the FOUR pumps of Prob. plotted at right in Fig. P11.52. Determine if the values of 11.51 above fit this correlation. Solution: Make a table of these values, similar to Prob. 11.52: Fig. P11.52 Power coefficient at BEP for 30 commercial pumps. 29 P*, bhp D, cm n, rpm NS Prob. 11.30: 255 37.1 2134 214 0.602 Prob. 11.37: 156 45.7 880 498 1.85 Prob. 11.40: 18.5 17.4 3500 210 0.436 Prob. P11.26: 8.7 29.2 1160 182 0.423 Ans. When added to the plot shown below, all four of them seem to fit reasonably well. The data are moderately useful for predicting general centrifugal-pump behavior and can be fit to either a 2nd-order polynomial or a single-term Power-law: Polynomial: C*P ≈ 1.41E−3N s + 4.22E−6N s2 (poorer correlation, R 2 ≈ 0.96) Ans. Power-law: C*P ≈ 1.25E−4N1.537 s 11.53 An axial-flow pump delivers 1.13 m3/s of air which enters at 20°C and 1 atm. The flow passage has a 25.4-cm outer radius and a 20.3-cm inner radius. Blade angles are α1 = 60° and β 2 = 70°, and the rotor runs at 1800 rpm. For the first stage, compute (a) the head rise; and (b) the power required. Solution: Assume an average radius of (25.4 + 20.3)/2 = 22.9 cm and compute the blade speed: 2π m Q 1.13 m u avg = ω ravg = 1800 ( 0.229 ) ≈ 43.2 ; Vn = = ≈ 15.4 2 2 60 s A π [(0.254 ) − (0.203) ] s Theory: gH = u2 − uVn(cot α1 + cot β 2) = (43.2)2 − 43.2(15.4)(cot 60° + cot 70°), H ≈ 126.4 m Ans. (a) 101300 Ptheory = ρgQH = (9.81)(1.13)(126.4) = 1688 ÷ 746 ≈ 2.26 hp 287(293) Ans. (b) 30 11.54 An axial-flow fan operates in sealevel air at 1200 r/min and has a blade-tip diameter of 1 m and a root diameter of 80 cm. The inlet angles are α1 = 55° and β1 = 30°, while at the outlet β2 = 60°. Estimate the theoretical values of the (a) flow rate, (b) horse-power, and (c) outlet angle α2. Solution: For air, take ρ ≈ 1.205 kg/m3. The average radius is 0.45 m. Thus Fig. P11.54 11.55 Figure P11.48 is an example of a centrifugal pump correlation, where DS is defined in the problem. Logan [3] suggests the following correlation for axial-flow pumps and fans: DS ≈ 51.8 N S0.485 for N S > 1200 where NS is the dimensional specific speed, Eq. (11.30b). Use this correlation to find the 3 appropriate size for a fan that delivers 11.33 m /s of air at sea-level conditions when running at 1620 r/min with a pressure rise of 2 inches of water. [HINT: Express the fan head in feet of air, not feet of water.] 3 Solution: At sea level, take ρair = 1.2255 kg/m . Convert n = 1620 r/min = 27 r/s. Convert the head H into metre of air: H air Δp ρwater gΔhwater (1000 kg / m 3 )(9.81 m / s 2 )(0.05 cm) = = = = 40.82 m of air ρair g ρair g (1.2255 kg / m 3 )(9.81 m / s 2 ) Now compute the dimensional specific speed of the fan: NS = (rpm)(m 3 / min)1/2 (1620)(11.33 × 60)1/2 = ≈ 2615 (m of head)3/4 (40.82)3/4 (yes, axial flow) Finally, use Logan’s correlation to find the specific diameter and hence the fan size: DS ≈ 51.8 D (gH )1/4 D[(9.81 m / s 2 )(40.82)]1/4 = 1.14 = = (2615)0.485 Q1/2 (11.33 m 3 / s)1/2 Solve for D fan ≈ 86 cm Ans. 31 P11.56 It is desired to pump 1.42 m3/s of water at a speed of 22 r/s, against a head of 24.4 m. (a) What type of pump would you recommend? Estimate (b) the required impeller diameter and (c) the brake horsepower. Solution: Evaluate the specific speed: Q = 1.42 m3 / s ; n = 22 Ns = r = 1320 rpm ; H = 24.4 m (OK ) s (rpm)(m3 / min)1/ 2 (m)3/ 4 = (1320rpm)(1.42 × 60)1/ 2 (24.4 m)3/ 4 = 1110 From Fig. 11.11a, this pump should have a mixed-flow impeller. Ans.(a) (b) The writer estimated impeller diameter from the “specific diameter” correlation in Prob. 11.55: D (gH )1/ 4 BG units : Ds = 1/ 2 Q = D[9.81(24.4 m)]1/ 4 1/ 2 (1.42) Dimpeller = (1.73)( Solve 51.8 ≈ N s0.485 1.192 ) ≈ 52 cm 3.933 = 51.8 (1110)0.485 = 1.73 Ans.(b) (c) According to Fig. 11.11a, a really good mixed-flow impeller, Ns = 1110, would have an efficiency of about 92 per cent. Thus we can estimate the power required: Pinput = (9790 N / m3 )(1.42 m3 / s)(24.4 m) γ QH = = 368,700 W ≈ 494 bhp Ans.(c) η 0.92 P11.57 Suppose that the axial-flow pump of Fig. 11.13, with D = 45.7 cm , runs at 1800 3 r/min. (a) Could it efficiently pump 94.64 m /min of water? (b) If so, what head would result? (c) If a head of 36.6 m is desired, what values of D and n would be better? Solution: Convert n = 1800 r/min = 30 r/s. The flow coefficient at BEP is approximately 0.55. Check this against the data: CQ* = Q n D3 = (94.64 / 60 m3 / s) (30 r / s)(0.457)3 ≈ 0.55 (Yes, this is at BEP) Ans.(a) (b) Since we are at BEP, we can use the BEP head coefficient to get H: C H * ≈ 1.07 = gH 2 n D 2 = (9.81) H (30)2 (0.457)2 ; Solve H ≈ 20.5 m Ans.(b) (c) For D = 45.7 cm, H = 36.6 m is far off BEP, CH ≈ 1.9, efficiency only 40 per cent. So let diameter be unknown and set both head and flow coefficients equal to BEP values: CQ* = 0.55 = (94.64 / 60 m3 / s) n D3 Solve simultaneously for , C H * = 1.07 = n = 46.3 (9.81)(36.6) n2 D 2 r r = 2780 , D = 39.6 cm s min Ans.(c) 32 3 11.58 A pump is needed to deliver 151.4 m /min of gasoline at 20°C against a head of 27.4 m. Find the impeller size, speed, and brake horsepower needed to use the pump families of (a) Fig. 11.8; and (b) Fig. 11.13. Which is the better design? Solution: For gasoline, take ρ ≈ 680 kg/m3. (a) For the centrifugal design, C*Q ≈ 0.115 = 151.4/60 nD 3 and C*H ≈ 5.0 = 9.81(27.4) , n 2 D2 Solve for n ≈ 4.24 rps ≈ 255 rpm and D impeller ≈ 1.73 m Ans. (a) P* = C*P ρ n 3 D 5 = 0.65(680)(4.24)3 (1.73)5 ÷ 746 ≈ 699 bhp Ans. (a) (b) For the axial-flow design, Fig. 11.13, 151.4/ 60 9.81(27.4) , C*H = 1.07 = 3 nD n 2 D2 or: n ≈ 1768 rpm, D ≈ 54 cm Ans. (b) C*Q = 0.55 = P* = C*P ρ n 3 D 3 = 0.70(680)(29.5)3 (0.54)5 ÷ 746 ≈ 752 bhp Ans. (b) The axial-flow design (b) is far better for this system: smaller and faster. Ans. 11.59 Performance data for a 53.3-cm-diameter air blower running at 3550 rpm are shown below. What is the specific speed? How does the performance compare with Fig. 11.13? What are ? Δp, m H2O: Q, m3/min: bhp: 0.74 14.15 6 0.76 28.32 8 0.71 56.63 12 0.53 84.95 18 0.25 113.27 25 Solution: Assume 1-atm air, ρ ≈ 1.2 kg/m3. Convert the data to dimensionless form and put the results into a table: Δp, kPa: 7.26 7.46 6.96 5.2 2.45 Q, m3/min: 14.15 28.32 56.63 84.95 113.27 3 0.236 0.472 0.944 1.416 1.888 617 634 591 442 208 CQ: 0.0263 0.0527 0.105 0.158 0.211 CH: 6.09 6.25 5.83 4.36 2.05 CP: 0.419 0.558 0.837 1.256 1.744 η: 0.382 0.590 0.731 0.548 0.248 Q, m /s: H, m (of air): Close enough without plotting: C*Q ≈ 0.105, C*H ≈ 5.83, C*P ≈ 0.837 Ans. Specific speed N s = (3550 rpm)(56.63)1/2 ≈ 223 Ans. (591 m)3/4 This centrifugal pump is very similar to the dimensionless data of Fig. 11.8. Ans. 33 P11.60 Aircraft propeller specialists claim that dimensionless propeller data, when plotted as (CT/J2) versus (CP/J2), form a nearly straight line, y = mx + b. (a) Test this hypothesis for the data of Fig. 11.16, in the high-efficiency range J = V/(nD) equal to 0.6, 0.7, and 0.8. (b) If successful, use this straight line to predict the rotation rate n, in r/min, for a propeller with D = 1.52 m, P = 30 hp, T = 423 N, and V = 152 km/h, for sea-level standard conditions. Comment. Solution: For sea-level air, take ρ = 1.23 kg/m3. The writer read Fig. 11.16 as best he could and came up with the following data, rewritten in the desired form: J 0.6 0.7 0.8 CT 0.055 0.040 0.023 CP 0.042 0.035 0.023 CP/J^2 0.1167 0.0714 0.0359 CT/J^2 0.1528 0.0816 0.0359 The plot is, sure enough, reasonably linear: The specialists were correct; this propeller data fits the simple formula We have all the data except n. Convert to BG units: 152 km/h = 42.222 m/s and 30 hp = 22,380 W. Put these values into our straight-line formula: 423 2 (1.23)(1.52) (42.222) or : 0.0835 ≈ 2 ≈ 1.453[ 22380 (1.23) n(1.52)3 (42.222)2 ] − 0.0184 , 4.223 r r 2.906 − 0.0184 , n ≈ ≈ 41 ≈ 2480 0.102 s min n Ans. This is reasonably accurate and close to maximum efficiency, V/nD ≈ 0.68. However, beware! This clever correlation has reduced the propeller data from three variables (CT, CP, and V/nD) to only two (CT/J2 and CP/J2), without any theoretical justification. Thus, information is lost. Do not use this straight line as a general problem solver. It will fail miserably if fed with data that does not match the conditions in Fig. 11.16. 34 11.61 Suppose it is desired to deliver 19.82 m3/min of propane gas (molecular weight = 44.06) at 1 atm and 20°C with a single-stage pressure rise of 20.3 cm H2O. Determine the appropriate size and speed for using the pump families of (a) Prob. 11.59 and (b) Fig. 11.13. Which is the better design? 2 2 Solution: For propane, with M = 44.06, the gas constant R = 8314/44.06 ≈ 188.7 m /(s ⋅K). Convert Δp = 20.3 cm H2O = (9790)(0.203) = 1.987 kPa. The propane density and head rise are p 101300 Pa kg ρgas = = ≈ 1.832 3 , RT 188.7(293) m 1987 Hence H pump = ≈ 110.56 m propane 1.832(9.81) (a) Prob. 11.57: C*Q ≈ 0.105 = 19.82/60 nD 3 and C*H ≈ 5.83 = 9.81(110.56) n 2 D2 Solve for n = 28.4 rps ≈ 1704 rpm and D ≈ 48 cm Ans. (a) (centrifugal pump) (b) n ≈ 13900 rpm, D ≈ 13.7 cm Ans. (b) (axial flow) The centrifugal pump (a) is the better design—nice size, nice speed. The axial flow pump is much smaller but runs too fast. Ans. P11.62 Performance curves for a certain free propeller, comparable to Fig. 11.16, can be plotted as shown in Fig. P11.62, for thrust T versus speed V for constant power P. (a) What is striking, at least to the writer, about these curves? (b) Can you deduce this behavior by rearranging, or replotting, the data of Fig. 11.16? Fig. P11.62 Solution: (a) The three curves look exactly alike! In fact, we deduce, for this propeller, that the thrust, at constant speed, is linearly proportional to the horsepower! This implies that the dimensionless thrust, CT, is proportional to the dimensionless power, CP, for this propeller. We can go back and see that this is not the case for Fig. 11.16. 35 (b) Here is a plot of thrust versus speed for the actual data of Fig. 11.16. We see that the curves are similar in shape, but the thrust is not proportional to the horsepower. Ans.(b) 11.63 A mine ventilation fan delivers 500 m3/s of sea-level air at 295 rpm and Δp = 1100 Pa. Is this fan axial, centrifugal, or mixed? Estimate its diameter in feet. If the flow rate is increased 50% for the same diameter, by what percent will Δp change? Solution: For sea-level air, take ρ g ≈ 11.8 N/m3, hence H = Δp/ρ g = 1100/11.8 ≈ 93 m, calculate the specific speed: Ns = rpm(m 3 /min)1/2 295(500 × 60)1/2 = ≈ 1706 (axial-flow pump) Ans. (a) (head)3/4 (93)3/4 Estimate C*Q ≈ 0.55 = 500 m 3 /s , solve D impeller ≈ 5.7 m (295/60)D 3 Ans. (b) At constant D, Q ∝ n and Δp (or H) ∝ n2. Therefore, if Q increases 50%, so does n, and therefore Δp increases as (1.5)2 = 2.25, or a 125% increase. Ans. (c) 11.64 The actual mine-ventilation fan in Prob. 11.63 had a diameter of 6.1 m [Ref. 20, p. 339]. What would be the proper diameter for the pump family of Fig. 11.14 to provide 500 m3/s at 295 rpm and BEP? What would be the resulting pressure rise in Pa? Solution: For sea-level air, take ρ g ≈ 11.8 N/m3. As in Prob. 11.63 above, the specific speed of this fan is 1706, hence an axial-flow fan. Figure 11.14 indicates an efficiency of about 90%, and the only values we know for performance are from Fig. 11.13: N s ≈ 1800: C*Q ≈ 0.55 = 500 , solve D impeller ≈ 5.7 m Ans. (a) (295/60)D 3 (295/60)2 (6.1)2 = 98 m, 9.81 Δp = (11.8)(98) ≈ 1160 Pa Ans. (b) C*H ≈ 1.07, H = 1.07 36 3 P11.65 We want to pump 70°C water at 76 m /min and 1800 r/min. Estimate the type of pump, the horsepower required, and the impeller diameter if the required pressure rise for one stage is (a) 170 kPa and (b) 1350 kPa. Solution: For water at 70°C, take ρ ≈ 978 kg/m 3 . Evaluate the specific speed: (a) Δp = 170 kPa Δp 170000 = = 17.72 m ρg (978)(9.81) H= (rpm)(m3 / min)1/ 2 Ns = [H (m)]3/ 4 = (1800)(76)1/ 2 ≈ 1817 (a) (17.72)3/ 4 From Fig. 11.11(a), we select axial-flow pump. Assume that the pump has efficiency, η = 0.9, we then require (978)(9.81)(76 / 60)(17.72) ρ gQH P= = η 0.9 = 2.4E5W = 321 hp (a) From Fig. 11.13 (N s = 1800), CQ* ≈ 0.55 = (b) ∴ D = 42.5 cm Δp = 1350 kPa H= Ns = m3 / s (rps)D 3 = 76 / 60 (1800 / 60)(D 3 ) (a) (1,350,000) = 140.71 m (978)(9.81) (1800)(76)1/ 2 = 384 (b) (140.71)3/ 4 From Fig. 11.11(a), we select centrifugal pump (978)(9.81)(76 / 60)(140.71) ≈ 1.9 MW = 2547 hp 0.9 76 / 60 From Fig. P11.49 (N s = 743), CQ* ≈ 0.23 = (1800 / 60)D 3 P= (ed. 6th) D = 56.8 cm (b) (b) P11.66 A good curve-fit to the head vs. flow for the 81.3 cm pump in Fig. 11.7a is H (in m) ≈ 152.4 − (2.024E − 2)Q 2 , Q in m 3 /min Assume the same rotation rate, 1170 r/min, and estimate the flow rate this pump will provide to deliver water from a reservoir, through 274.3 m of 30.5-cm pipe, to a point 45.7 m above the reservoir surface. Assume a friction factor f = 0.019. 37 Solution: The system head would be the elevation change, 45.7 m, plus the friction head loss: H pump ≈ 152.4 − (6.17E − 3)Q 2 = H system = Δz + f = 45.7 + (0.019)( L V2 D 2g 274.3 [4Q / π / (0.305)2 / 60]2 ] ){ } 0.305 2(9.81) or : 152.4 − (6.17E − 3)Q 2 = 45.7 + (4.53E − 2)Q 2 Solve Q 2 = 106.7 = 2073 , 5.15E − 2 Q ≈ 45.5 m3 min Ans. This is not a BEP match, but the efficiency is about 75%, not too bad. 11.67 A leaf blower is essentially a centrifugal impeller exiting to a tube. Suppose that the tube is smooth PVC pipe, 1.22 m long, with a diameter of 6.35 cm. The desired exit velocity is 117.5 km/h in sea-level standard air. If we use the pump family of Eq. (11.27) to drive the blower, what approximate (a) diameter and (b) rotation speed are appropriate? (c) Is this a good design? Solution: Recall that Eq. (11.27) gave BEP coefficients for the pumps of Fig. 11.7: Apply these coefficients to the leaf-blower data. Neglect minor losses, that is, let the pump head match the pipe friction loss. For air, take ρ = 1.2 kg/m3 and µ = 1.8E−5 kg/m⋅s. Convert 117.5 km/h = 32.6 m/s. We know the Reynolds number, Red = ρVd/µ = (1.2)(32.6)(0.0635)/(1.8E−5) = 138,000, and for a smooth pipe, from the Moody chart, calculate fsmooth = 0.0168. Then H = hf = 17.5 m, and the two previous equations can then be solved for Dpump ≈ 0.39 m; n ≈ 15 r/s = 900 r / min Ans. (a, b) (c) This blower is too slow and too large, a better (mixed or axial flow) pump can be designed. Ans. (c) 38 3 11.68 A 29.2-cm diameter centrifugal pump, running at 1750 rev/min, delivers 3.22 m /min and a head of 32 m at best efficiency (82%). (a) Can this pump operate efficiently when delivering water at 20°C through 200 m of 10-cm-diameter smooth pipe? Neglect minor losses. (b) If your answer to (a) is negative, can the speed n be changed to operate efficiently? (c) If your answer to (b) is also negative, can the impeller diameter be changed to operate efficiently and still run at 1750 rev/min? 3 Solution: For water at 20°C, take ρ = 998 kg/m and µ = 0.001 kg/m-s. Convert: Q = 3.22 3 m /min = 0.0536 m3/s. Compute ReD and f: (a) Now compute the friction head loss at 850 gal/min and see if it matches the pump head: hf = f L V2 200m (6.83m / s)2 = (0.0125)( )[ ] = 59.2m ≠ h pump = 32 m D 2g 0.1m 2(9.81m / s 2 ) The answer to part (a) is No, the pump will operate far off-design to lower the flow rate and hf. (b) Can we just change the pump speed? This answer is also No. At BEP, (gH/n2D2) is constant, therefore pump head is proportional to n2. But the pipe friction loss is also approximately proportional to n2. Thus, for any speed n, the pipe head will always be about twice as large as the pump head and not operate efficiently. Ans.(b) (c) Can we change the impeller diameter, still at 1750 rev/min, and operate efficiently? This time the answer is Yes. First note the BEP dimensionless parameters, n = 1750/60 = 29.2 rps: Now combine these with the fact that hf is approximately proportional to n2. In SI units, 2 3 h f (m) ≈ 20600Q (in m / s) = H pump = 4.33 n2 D p2 g ≈ 20600[0.0738 n D 3p ]2 n cancels, solve for D p4 ≈ 0.00356 m4 , D p ≈ 0.244 m Ans. (c) 3 This is approximately correct. Check: the pump head is 22.4 m and flow rate is 1.88 m /min. For this flow rate, the pipe head loss is 22.3 m. Good match! Ans.(c) NOTE: Part (c) is almost independent of n. For example, if n = 20 rev/s, the best efficiency 3 converges to an impeller diameter of 0.24 m, with Q ≈ 1.22 m /min and H = hf ≈ 10.2 m. 39 11.69 It is proposed to run the pump of Prob. 11.37 at 880 rpm to pump water at 20°C through the system of Fig. P11.69. The pipe is 20-cm diameter commercial steel. What flow rate in m3/min results? Is this an efficient operation? Fig. P11.69 Solution: For water, take ρ = 998 kg/m3 and µ = 0.0010 kg/m⋅s. For commercial steel, take ε = 0.046 mm. Write the energy equation for the system: H pump = Δz + f LV2 40 [Q/( π (0.2)2 /4)]2 = 11− 4 + f ( ) = 7 +10328 fQ 2 , 2(9.81) d 2g 0.2 Meanwhile, H p = fcn(Q)Prob. 11.35 and m3 Q in s ε f = fcn Red , d Moodychart where ε /d = 0.046/200 = 0.00023. If we guess f as the fully rough value of 0.0141, we find that Hp is about 25 m and Q is about 0.35 m3/s. To do better would require some very careful plotting and interpolating, or: EES is made for this job! Iteration with EES leads to the more accurate solution: m ; Ppump = 139 hp; H p = 24.4 m s Q ≈ 20.44 m3 / min Ans. f = 0.01456; V = 10.8 The efficiency is 78%, considerably off the maximum of 88% - not a very good system fit. Ans. Here is a plot of the pump and system curves: 40 11.70 The pump of Prob. 11.37, running at 880 r/min, is to pump water at 20°C through 75 m of horizontal galvanized-iron pipe (ε = 0.15 mm). All other system losses are neglected. Determine the flow rate and input power for (a) pipe diameter = 20 cm; and (b) the pipe diameter yielding maximum pump efficiency. Solution: (a) There is no elevation change, so the pump head matches the friction: Hp = f L V2 75 [Q/ (π (0.2)2 /4)]2 4ρ Q = f = 19366 fQ 2, Red = , 2(9.81) πµd d 2g 0.2 ε 0.15 = = 0.00075 d 200 But also Hp = fcn(Q) from the data in Prob. 11.37. Guessing f equal to the fully rough value 3 of 0.0183 yields Q of about 26.5 m /min. Use EES to get closer: m ; Red = 2.87E6; H p = 73 m s m3 Ans. (a) Q = 0.452 s f = 0.0185; V = 14.4 The efficiency is 87%, not bad! (b) If we vary the diameter but hold the pump at maximum efficiency (Q* = 8000 gal/min), we obtain a best d = 0.211 m. Ans. (b) 11.71 Suppose that we use the axial-flow pump of Fig. 11.13 to drive the leaf blower of Prob. 11.67. What approximate (a) diameter and (b) rotation speed are appropriate? (c) Is this a good design? Solution: Recall that Fig. 11.13 gave BEP coefficients for an axial-flow pump: Apply these coefficients to the leaf-blower data. Neglect minor losses, that is, let the pump head match the pipe friction loss. For air, take ρ = 1.2 kg/m3 and µ = 1.8E−5 kg/m⋅s. h f = H pump = 1.07 Q= n2 D2 9.81 m/s 2 =f 2 1.22 m (32.6 m/s)2 L Vpipe , f = f (Re d ) = f 0.0635 m 2(9.81 m/s 2 ) d pipe 2g m3 π 2 π d pipeV = (0.0635 m)2 (32.6 m/s) = 0.103 = 0.55nD 3 4 4 s We know the Reynolds number, Red = ρVd/µ = (1.2)(32.6)(0.0635)/(1.8E−5) = 138,000, and for a smooth pipe, from the Moody chart, calculate fsmooth = 0.0168. Then H = hf = 17.5 m, and the two equations above can then be solved for: Dpump ≈ 0.122 m ; n ≈ 104 r/s = 6250 r/min Ans. (a, b) This blower is too fast and too small, a better (mixed flow) pump can be designed. Ans. (c) 41 11.72 The pump of Prob. 11.40, running at 3500 rpm, is used to deliver water at 20°C through 183 m of cast-iron pipe to an elevation 30.5 m higher. Find (a) the proper pipe diameter for BEP operation; and (b) the flow rate that results if the pipe diameter is 7.5 cm. Q, m /min: 3 0.189 0.379 0.568 0.757 0.946 1.136 1.325 1.514 1.703 H, m: 61.26 60.96 60.35 59.13 57.61 55.17 51.51 47.55 42.37 η, %: 29 50 64 72 77 80 81 79 74 2 Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 1.003E−3 N.s/m . For cast iron, 3 take ε ≈ 2.59E–4 m. (a) The data above are for 3500 rpm, with BEP at 1.325 m /min: L V2 183 [Q/ (π d 2 /4)]2 9.18 f 1.325 m 3 = 30.5 + f = 30.5 + , Q = Q* = 2(9.81) d 2g d 60 s d5 ε Iterate, converges to Red = 2.87E5, = 0.00265, f = 0.0258, d d ≈ 9.8 cm Ans. (a) H * = 51.51 = Δz + f (b) If d = 7.5 cm, the above solution changes to a new flow rate: Curve-fit: H ≈ 61.26 − 2.811E5Q 2.7 = 30.5 + 183 [Q/ (π (0.075)2 /4)]2 = 30.5 + 6.372E6Q 2 2(9.81) 0.075 Iterate or EES: f = 0.0277, Re = 2.21E5, H ≈ 59 m, Q = 0.76 m3 min Ans. (b) 11.73 The pump of Prob. 11.30, operating at 2134 rpm, is used with water at 20°C in the system of Fig. P11.73. The diameter is 20.3 cm. (a) If it is operating at BEP, what is the proper elevation z2? (b) If z2 = 68.6 m, what is the flow rate? Fig. P11.73 2 Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 1.003E−3 N·s/m . For cast iron, take ε ≈ 2.59E–4 m, hence ε /d ≈ 0.00128. (a) At BEP from Prob. 11.30, Q* ≈ 169.9 L/s and H* ≈ 100.58 m. Then the pipe head loss can be determined: V= Q 0.17 m (998)(5.25)(0.203) = = 5.25 , Re d = = 1.061E6, fMoody ≈ 0.0211 2 A (π /4)(0.203) s 1.003E−3 42 H syst = Δz + f 457.2 (5.25)2 ? L V2 = z 2 − 30.5 + 0.0211 = z 2 − 30.5 + 66.8 = H* = 100.58 0.203 2(9.81) d 2g Solve for z 2 ≈ 64.3 m Ans. (a) (b) If z2 = 68.6 m, the flow rate will be slightly lower and we will be barely off-design: H = H(Q)Table = Δz + f L V2 457.2 V2 Q = 68.6 − 30.5 + f , V= d 2g 0.203 2(9.81) (π /4)(0.203)2 Converges to f ≈ 0.0211, V ≈ 5.06 m , H ≈ 100.89 m, Q ≈ 0.164 m 3 / s Ans. (b) s 11.74 The pump of Prob. 11.40, running at 3500 r/min, delivers water at 20°C through 2195 m of horizontal 12.7-cm-diameter commercial-steel pipe. There are a sharp entrance, sharp exit, four 90° elbows, and a gate valve. Estimate (a) the flow rate if the valve is wide open and (b) the valve closing percentage which causes the pump to operate at BEP. (c) If the latter condition holds continuously for 1 year, estimate the energy cost at 10 ¢/kWh. Data at 3500 rpm: 3 Q, m /min: H, m: η, %: 0.189 0.379 0.568 0.757 0.946 1.136 1.325 1.514 1.703 61.26 60.96 60.35 59.13 57.61 55.17 51.51 47.55 42.37 29 50 64 72 77 80 81 79 74 2 Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 1.003E−3 N⋅s/m . For commercial 3 steel, take ε ≈ 4.57E–5 m, or ε /d ≈ 0.00036. The data above show BEP at 1.325 m /min. The minor losses are a sharp entrance (K = 0.5), sharp exit (K = 1.0), 4 elbows (4 × 0.28), and an open gate valve (K = 0.1), or ∑K ≈ 2.72. Pump and systems heads are equal: H p = H(Q)Table = H syst = V2 L L 2 f + ∑ K , where V = Q/[(π /4)(0.127) ], ≈ 17280 2g d d The friction factor f ≥ 0.0155 depends slightly upon Q through the Reynolds number. (a) Iterate on Q until both heads are equal. The result is f ≈ 0.0176 Re d ≈ 227,442, H p = H syst ≈ 50.62 m, Q ≈ 1.368 m3 / min Ans. (a) 3 (b) Bring Q down to BEP, ≈ 1.325 m /min, by increasing the gate-valve loss. The result is 3 f ≈ 0.0177, Red ≈ 220,300, H ≈ 51.51 m, Q ≈ 1.325 m /min, Kvalve ≈ 25.12 (42% open) Ans. (b) (c) Continue case (b) for 1 year. What does it cost at 10¢ per kWh? Well, we know the power level is exactly BEP, so just figure the energy: P= ρgQH 9790(1.325/60)(51.51) = ≈ 13.75 kW = 13.8 kW η 0.81 Annual cost = 13.75(365 days*24 hours)($0.1/kWh) ≈ 12, 045 Ans. (c) 43 11.75 Performance data for a small commercial pump are shown below. The pump supplies 20°C water to a horizontal 1.6-cm-diameter garden hose (ε ≈ 0.0254 cm), which is 15.2 m long. Estimate (a) the flow rate; and (b) the hose diameter which would cause the pump to operate at BEP. Q, L/min: 0 37.9 75.7 113.6 151.4 189.3 227.1 265 H, m: 22.9 22.9 22.6 21.9 20.7 18.9 14.3 7.3 2 Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 1.003E−3 N·s/m . Given ε /d = 0.0254/(1.6) ≈ 0.017, so f (fully rough) ≈ 0.045. Pump and hose heads must equate, including the velocity head in the outlet jet: H p = H(Q)Table = H syst V2 L = f + 1 , 2g d L 15.2 Q = ≈ 950, V = d 0.016 (π /4)(0.016)2 (a) Iterate on Q until both heads are equal. The result is: f ≈ 0.0467, Re d ≈ 50,148, V ≈ 3.15 m/s, H ≈ 22.92 m, Q ≈ 0.038 m3 / min Ans. (a) The hose is too small. (b) We don’t know exactly where the BEP is, but it typically lies at about 60% of maximum flow rate (0.6 × 290 ≈ 174.9 L/min). Iterate on hose diameter D to make the flow rate lie between 151.4 and 189.3 L/min. The results are: Q = 151.4 L/min, H = 20.7, f ≈ 0.0362, Re ≈ 103122, Dhose ≈ 3.1 cm 170.3 19.8 0.0354 108966 3.3 cm Ans. (b) 189.3 18.9 0.0347 113876 3.51 cm 11.76 The Cessna 172 aircraft has a wing area of 16.17 m2, an aspect ratio of 7.38, and a basic drag coefficient CD∞ = 0.037. Its propeller, whose data is shown in Fig. 11.16, has a diameter of 1.9 m. If it weighs 10.23 kN and flies at 54.86 m/s at 1500 m standard altitude, estimate (a) the appropriate propeller speed, in rev/min, and (b) the power required. Is the propeller efficient? [HINT: The efficiency is good, but not best.] 3 Solution: At 1500 m altitude, take ρ = 1.0583 kg/m . The lift and drag are CL = W 10230 N = = 0.397 2 (1 / 2)ρV A (1 / 2)(1.0583 kg / m 3 )(54.86 m / s)2 (16.17) C L2 (0.397)2 = 0.037 + 0.0068 = 0.0438 π AR π (7.38) 1.0583 ρ )(54.86)2 (16.17) = 1128 N Drag = Thrust = C D V 2 A = (0.0438)( 2 2 C D = C D∞ + = 0.037 + 44 Since T and n are related (strongly), we have to look for the right system match on the performance chart, Fig. 11.16. The HINT says it is pretty efficient. Try best efficiency: J( for ηmax ) = 0.7 = CT* ≈ 0.040, V 54.86 m / s = , for which n = 41.25 rev / s nD n(1.9 m) hence T * = CT* ρ n 2 D 4 = (0.040)(1.0583)(41.25)2 (1.9)4 = 939 N So the thrust is too small, we need larger CT, hence a smaller J. Iterate to this solution: J ≈ 0.67 yields n ≈ 43rev / s CT = 0.044 and T ≈ 1122 N The appropriate propeller speed is n = 43 rev/s ≈ 2580 rev/min. (OK, a match) Ans.(a) (b) At J = 0.67, read, from Fig. 11.16, CP ≈ 0.029. Then the required power is P = CP ρn 3 D 5 = (0.029)(1.0583)(43)3 (1.9)5 = 60, 420 ≈ 81 hp Ans.(b) P11.77 The 93.35-cm pump in Fig. 11.7a is used at 1170 r/min to pump water at 15°C from a reservoir through 350 m of 30 cm-ID galvanized iron pipe to a point 60 m above the reservoir surface. What flow rate and brake horsepower will result? If there are 15 m of pipe upstream of the pump, how far below the surface should the pump inlet be placed to avoid cavitation? Solution: For galvanized pipe, ε ≈ 1.524E–4 m, hence ε /d = 5.08E–4. Assume fully-rough flow, with f ≈ 0.0167. The pipe head loss is approximately: h f = Δz + f L [Q / (π d 2 / 4)]2 2g d 350 [Q / (π (0.3)2 / 4)]2 = 60 + 0.0167 0.3 2(9.81) = 60 + 198.75Q 2 (Q in m 3 /s) Curve-fit Fig. 11.7(a), D = 93.35 cm , H p = 202.7 – 380.12Q 2 Operating point: H p = hf 60 + 198.75Q 2 = 202.7 – 308.12Q 2 ∴ Q = 0.53 m 3 /s (a) at 15°C, ρ = 999 kg.m 3 , µ = 1.155E–3 N·s/m 2 (999)(9.81)(0.53)(115.95) ρ gQH = P= η 0.58 = 1.04 MW = 1392 hp (b) Fig. 11.7a: (Q in m 3 /s) 45 Check: f from Moody Chart (999)(7.5)(0.3) 0.53 = 7.5 m/s, Re = = 1.95E6 V =Q/A= π 1.155E – 3 2 (0.3) 4 6.9 ε / d 1.11 1 = –1.8 log + , f = 0.017 (Ok) f 1/2 Re 3.7 From Fig. 11.7a, at 0.53 m 3 /s, NPSH ≈ 7.6 m. Calculate head loss upstream: 15 (7.5)2 L V2 = 0.017 = 2.44 m 0.3 2(9.81) d 2g Pa – Pv – zi – h fi NPSH = 7.6 ≤ ρg 101300 – 1782 = – zi – 2.44 999(9.81) zi ≤ 0.11 m (c) h fi = f 11.78 The 81.3-cm-diameter pump in Fig. 11.7a is used at 1170 rpm in a system whose 2 3 head curve is Hs(m) ≈ 30.5 + 3.19E–2Q , with Q in m /min. Find the discharge and brake horsepower required for (a) one pump; (b) two pumps in parallel; and (c) two pumps in series. Which configuration is best? Solution: Assume plain old water, ρg ≈ 9810 N/m3. A reasonable curve-fit to the pump 3 head is taken from Fig. 11.7a: Hp(m) ≈ 152.4 − 6.38E–3Q2, with Q in m /min. Try each case: 3 (a) One pump: Hp = 152.4 − 6.38E–3Q2 = Hs = 30.5 + 3.19E–2Q2: Q ≈ 56.43 m /min Ans. (a) (b) Two pumps in parallel: 152.4 − 6.38E–3(Q/2)2 = 30.5 + 3.19E–2Q2, Q ≈ 60.33 3 m /min Ans. (b) 3 (c) Two pumps in series: 2(152.4 − 6.38E–3Q2) = 30.5 + 3.19E–2Q2: Q ≈ 78.37 m /min Ans. (c) Clearly case(c) is best, because it is very near the BEP of the pump. Ans. 46 11.79 Two 88.9-cm pumps from Fig. 11.7b are installed in parallel for the system of Fig. P11.79. Neglect minor losses. For water at 20°C, estimate the flow rate and power required if (a) both pumps are running; and (b) one pump is shut off and isolated. Fig. P11.79 2 Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 1.003E−3 N·s/m . For cast iron, ε ≈ 2.59E–4, or ε /d ≈ 0.000425. The 88.9-cm pump has the curve-fit head relation 3 Hp(m) ≈ 71.63 − 3.372E–5Q3, with Q in m /min. In parallel, each pump takes Q/2: 1 L V2 H p = fcn Q = H syst = Δz + f , 2 curve-fit d 2g and V= ( Q m3 / s ) (π / 4 ) (0.61 m ) 2 L = 2640 d , Δz = 30.5 m (a) Two pumps: Iterate on Q (for Q/2 each) until both heads are equal. The results are: f ≈ 0.0163, Re d ≈ 2.56E6, H p = H s = 69.8 m Q total ≈ 74 m3 / min (37 m 3 /min for each pump) Ans. (a) 9790(37/60)(69.8) P = 2Peach = 2 = 1,154,505 ÷ 746 ≈ 1548 bhp 0.73 Ans. (a) (b) One pump: Iterate on Q alone until both heads are equal. The results are: f ≈ 0.0164, Re ≈ 2.3E6, H ≈ 61.8 m, Q ≈ 66 m3 / min Ans. (b) P = 9790(66/60)(61.8)/0.87 ÷ 746 ≈ 1025 bhp Ans. (b) The pumps in parallel give 12% more flow at the expense of 50% more power. 47 P11.80 Suppose that the two pumps in Fig. P11.79 are modified to be in series at 710 r/min. What pipe diameter is required for BEP operation? 2 Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 1.003E−3 N·s/m . For cast iron ε = 2.59E–4 m. The 88.9 cm pump has the BEP values Q* = 68 m3/min, H* = 60 m. In series, each pump take H/2, so a BEP series operation would match H syst = 2H * = 2(60) = Δz + f 2 2 L V 2 1600 [68 / 60 / (π d / 4)] = 30.5 + f d d 2g 2(9.81) 169.8 f f , = 0.5271 5 d d5 2.59E − 4 4 ρQ ε and = f depends on Re = π dµ d d 120 = 30.5 + Guess, f = 0.02, d = 0.52 m Re = 2,761,176 from Moody Chart or 1 f 1/2 6.9 ε / d 1.11 = –1.8 log + Re 3.7 f = 0.0169 new, f = 0.0169, d = 0.50 m Re = 2,857,519 f = 0.0170 new, f = 0.0170, d = 0.50 m Re = 2,852,869 f = 0.017 – Converge ∴ f = 0.017, Re = 2.85E6, d = 0.5 m, V = 5.77 m/s Ans. (998)(9.81)(68 / 60)(60) ρ gQH =2 Power = 2P = 2 η 0.87 = 1.53 MW = 2052 hp Ans. 11.81 Two 81.3-cm pumps are combined in parallel to deliver water at 20°C through 457 m of horizontal pipe. If f = 0.025, what pipe diameter will ensure a flow rate of 132.5 3 m /min at 1170 rpm? 48 2 Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 1.003E−3 N⋅s/m . As in Prob. 11.78, a reasonable curve-fit to the pump head is taken from Fig. 11.7a: Hp(m) ≈ 3 3 152.4 − 6.38E–3Q2, with Q in m /min. Each pump takes half the flow, 66.25 m /min, for which H p ≈ 152.4 − 6.38E–3(66.25)2 ≈ 124.4 m. Then Q pipe = 132.5 m3 = 2.21 and the pipe loss is s 60 2 2 4.604 457 [2.21/(π d /4)] H syst = 0.025 = = 124.4 m, solve for d ≈ 51.7 cm Ans. d 2(9.81) d5 11.82 Two pumps of the type tested in Prob. 11.24 are to be used at 2140 r/min to pump water at 20°C vertically upward through 100 m of commercial-steel pipe. Should they be in series or in parallel? What is the proper pipe diameter for most efficient operation? Solution: For water take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For commercial steel take ε = 0.046 mm. Parallel operation is not feasible, as the pump can barely generate 100 m of head and the friction loss must be added to this. For series operation, assume BEP operation, Q* = 0.2 m3/s, H* = 95 m: H 2 pumps L V2 100 [0.20/ (π d 2 /4)]2 0.3305 f = 2(95) = Δz + f = 100 + f = 100 + 2(9.81) d 2g d d5 Given Red = 4ρ Q/(πµ d) = 4(998)(0.2)/[π(0.001)d] = 254000/d and ε /d = 0.000046/d, we can iterate on f until d is obtained. The EES result is: f = 0.0156; Red = 1.8E6; V = 12.7 m/s, dbest = 0.142 m Ans. P11.83 Consider the axial-flow pump of Fig. 11.13, running at 4200 r/min, with an impeller diameter of 91.4 cm. The fluid is propane gas (molecular weight 44.06). (a) How many pumps in series are needed to increase the gas pressure from 1 atm to 2 atm? (b) Estimate the mass flow of gas. Solution: Find the gas constant and initial density of the propane. Assume T = 20°C = 293 K. R propane = 8314 p 101300 = 188.7 m2 / (s2 ·K ) ; ρ = = = 1.832 kg / m3 44.06 RT (188.7)(293K ) 49 From Fig. 11.13, at BEP, CQ* = 0.55 and CH* = 1.07. Calculate the head rise for one stage: gH n2 D 2 (9.81) H = 1.07 = (70r / s)2 (0.914)2 , or : H = 446.5m Δp1st stage = ρ g H = (1.832)(9.81)(446.5) = 8.02 kPa (a) Crudely, then, to increase pressure by 1 atm = 101.3 kPa, we need 101.3/8.02 = 12.6 round off to13 stages. However, the pressure rises about 8 per cent per stage, so a better estimate would be ln(2.0)/ln(1.08) = 10 stages. Ans.(a) (b) The mass flow follows from the flow coefficient: CQ* = 0.55 = Q nD 3 = Q (70)(0.914) 3 , or : Q = 29.4 m3 / s m propane = ρQ = (1.832 kg / m3 )(29.4 m3 ) = 53.85 kg / s s Ans.(b) 11.84 Two 81.3-cm pumps from Fig. 11.7a are to be used in series at 1170 rpm to lift water through 152 m of vertical cast-iron pipe. What should the pipe diameter be for most efficient operation? Neglect minor losses. 2 Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 1.003E−3 N·s/m . For cast 3 iron, ε ≈ 2.59E–4 m. From Fig. 11.7a, read H* ≈ 117.35 m at Q* ≈ 75.71 m /min. Equate H p = 2H* = 2(117.35) = H syst = Δz + f L V2 d 2g 2 75.71 / (π d 2 / 4) 152 60 = 152 + 20f = 152 + f d 2(9.81) d5 Iterate, guessing f ≈ 0.02 to get d, then get Red and ε /d and repeat. The final result is f ≈ 0.0185, V ≈ 14 m/s, Re d ≈ 4.72E6, d ≈ 0.339 m Ans. 11.85 Determine if either (a) the smallest, or (b) the largest of the seven Taco pumps in Fig.P11.26, running in series at 1160 r/min, can efficiently pump water at 20°C through 1 km of horizontal 12-cm-diameter commercial steel pipe. Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m-s. For commercial steel pipe, from Table 6.1, ε = 0.046 mm. Then ε/D = 0.046 mm/120 mm = 0.000383. (a) For the smallest pump in Fig. P11.26, the BEP is at about 400 gal/min, with a head of about 12.5 m. See what friction head loss results from this flow rate: 50 V= Q (400 × 6.309E − 5)m 3 / s m ρVD (998)(2.23)(0.12) = = 2.23 ; Re D = = = 267, 000 2 µ A s 0.001 (π / 4)(0.12m) ε = 0.000383; Eq.(6.48) : f Moody ≈ 0.0177, D L V2 1000 (2.23)2 = (0.0177)( ) = 37.4 m hf = f D 2g 0.12 2(9.81) So three small pumps in series, each with 12.5 m of head, would be an efficient system. Ans. (a) 3 (b) For the largest pump in Fig. P11.26, the BEP is at about 1.99 m /min, with a head of about 21.95 m. See what friction head loss results from this flow rate: V= Q (1.99 / 60)m 3 / s m ρVD (998)(2.98)(0.12) = ; Re D = = = 357, 000 2 = 2.98 A (π / 4)(0.12m) s µ 0.001 ε = 0.000383; Eq. (6.48) : f Moody ≈ 0.0173, D L V2 1000 (2.98)2 = (0.0173)( ) = 65.4 m hf = f D 2g 0.12 2(9.81) So three large pumps in series, each with 21.6 m of head, would be an efficient system. Ans. (b) The largest pump is a better solution because of its higher efficiency (80% compared to 65%). 11.86 Reconsider the system of Fig. P6.68. Use the Byron Jackson pump of Prob. 11.30 running at 2134 r/min, no scaling, to drive the flow. Determine the resulting flow rate between the reservoirs. What is the pump efficiency? 2 Solution: For water take ρ = 1000 kg/m3 and µ = 1.788E−3 N·s/m . For cast iron take ε = 2.59E–4 m, or ε /d = 2.59E–4/0.152 = 0.0017. The energy equation, written between reservoirs, is the same as in Prob. 6.68: H p = Δz + f L V2 610 [Q/ (π (0.152)2 /4)]2 = 36.6 + f = 36.6 + 6.212E5 fQ 2 2(9.81) d 2g 0.152 where 4ρ Q ε f = f Moody = fcn Red , with Red = d πµd 51 From the data in Prob. 11.30 for the pump, Hp = fcn(Q) and is of the order of 91 m. Guessing f ≈ 0.02, we can estimate a flow rate of about Q ≈ 0.066 m3/s, down in the low range of the Byron-Jackson pump. Get a closer result with EES: H p = 103.6 m; f = 0.0228; Red = 579, 000; V = 3.81 m / s; bhp = 169 hp; Q = 0.069 m3 / s Ans. Interpolating, the pump efficiency is η ≈ 56%. Ans. The flow rate is too low for this particular pump. 11.87 The S-shaped head-versus-flow curve in Fig. P11.87 occurs in some axial-flow pumps. Explain how a fairly flat system-loss curve might cause instabilities in the operation of the pump. How might we avoid instability? Solution: The stability of pump operation is nicely covered in the review article by Greitzer (Ref. 41 of Chap. 11). Generally speaking, there is little danger of instability if the slope of the pump-head curve, dH/dQ, is negative, unless there are two such points. In Fig. P11.87 above, a flat system curve may cross the pump curve at three points (a, b, c). Of these 3, point b is statically unstable and cannot be maintained. Consider a small disturbance near point b: Suppose the flow rate drops slightly—then the system head decreases, but the pump head decreases even more. Then the flow rate will drop still more, etc., and we move away from the operating point, which therefore is unstable. The general rule is: Fig. P11.87 A pump operating point is statically unstable if the (positive) slope of the pump-head curve is greater than the (positive) slope of the system curve. By this criterion, both points a and c above are statically stable. However, if the points are close together or there are large disturbances, a pump can “hunt” or oscillate between points a and c, so this could also be considered unstable to large disturbances. Finally, even a steep system curve (not shown above) which crosses at only a single point b on the positive-slope part of the pump-head curve can be dynamically unstable, that is, it can trigger an energy-feeding oscillation which diverges from point b. See Greitzer’s article for further details of this and other turbomachine instabilities. 52 11.88 The low-shutoff head-versus-flow curve in Fig. P11.88 occurs in some centrifugal pumps. Explain how a fairly flat system-loss curve might cause instabilities in the operation of the pump. What additional vexation occurs when two of these pumps are in parallel? How might we avoid instability? Fig. P11.88 Solution: As discussed, for one pump with a flat system curve, point a is statically unstable, point b is stable. A ‘better’ system curve only passes through b. For two pumps in parallel, both points a and c are unstable (see above). Points b and d are stable but for large disturbances the system can ‘hunt’ between the two points. 11.89 Turbines are to be installed where the net head is 122 m and the flow rate is 946 3 m /min. Discuss the type, number, and size of turbine which might be selected if the generator selected is (a) 48-pole, 60-cycle (n =150 rpm); or (b) 8-pole (n = 900 rpm). Why are at least two turbines desirable from a planning point of view? Solution: We select two turbines, of about half-flow each, so that one is still available for power generation if the other is shut down for maintenance or repairs. Ans. Assume η ≈ 90%: 946 (122) ÷ 746 ≈ 22720 hp (each turbine ≈ 11360 hp) Ptotal = ηρgQH ≈ 0.9(9790) 60 (a) n = 150 rpm: 1/2 N sp = nP1/2 = 150(11360) (122)5/4 H 5/4 Estimate φ ≈ 0.47 = (b) n = 900 rpm: N sp = ≈ 39.4 (select two impulse turbines) Ans. (a) π nD π (150/60)D = , or: D ≈ 2.93 m 2gH 2(9.81)(122) Ans. (a) 900(11360)1/2 ≈ 236.6 (select two Francis turbines) Ans. (b) (122)5/4 Fig. 11.21: C*p ≈ 2.6 = P* 11360 × 746 , solve D ≈ 1 m Ans. (b) 3 5 = 900 3 5 ρn D 998 D 60 53 11.90 For a high-flow site with a head of 13.7 m, it is desired to design a single 2.1-mdiameter turbine that develops 4000 bhp at a speed of 360 r/min and 88% efficiency. It is decided first to test a geometrically similar model of diameter 0.305 m, running at 1180 r/min. (a) What likely type of turbine is the prototype? What are the appropriate (b) head, and (c) flow rate for the model test? (d) Estimate the power expected to be delivered by the model turbine. 3 Solution: For water, take ρ = 1000 kg/m . (a) We have enough information to determine power specific speed: (rpm)(bhp)1/2 (360rpm)(4000bhp)1/2 N sp = = = 864 [H (m)]5 /4 (13.7m)5 /4 For a water-flow application, this number is characteristic of a propeller turbine. Ans.(a) (b) The similarity rules give us the required model head by equating the head coefficients: CH = n12 gH1 (9.81)(13.7) gH (9.81) H 2 = = 2 22 = , Solve H 2 ≈ 3.1 m Ans.(b) 2 2 2 D1 (360) (2.1) n2 D21 (1180)2 (0.305)2 [We didn’t bother changing n to r/s because the factors of “60” would have cancelled out.] (c) To get the model flow rate, first calculate the prototype flow rate from the known power: P = ηρ gHQ = (4000x746 W ) = (0.88)(1000 kg / m 3 )(9.81 m / s 2 )(13.7 m)Q1 Solve for Q1 = 25.2 m 3 / s Q (25.2) Q Q2 = 23 = CQ = 1 3 = 3 n1 D1 (360)(2.1) n2 D2 (1180)(0.305)3 Then Solve Q2 = 0.25 m 3 / s Ans.(c) If we assume the efficiency is unchanged, the model power output will be P2 = ηρ gH 2Q2 = (0.88)(1000)(9.81)(3.1 m)(0.25 m 3 / s) = 6690 W = 8.97 bhp Ans.(d) [The Moody size-effect formula, Eq. (11.29a), suggests η2 ≈ 0.805, or P2 ≈ 7.7 bhp.] 11.91 The Tupperware hydroelectric plant on the Blackstone River has four 91-cm-diameter turbines, each providing 447 kW at 200 rpm and 5.8 m3/s for a head of 9.1 m. What type of turbines are these? How does their performance compare with Fig. 11.22? Solution: Convert P* = 447 kW = 599 hp. Then, for D = 0.91 m, N sp = nP1/2 200(599)1/2 = ≈ 310 (These are Francis turbines) H 5/4 (9.1)5/4 Ans. Use the given data to compute the dimensionless BEP coefficients: C*Q = 5.8 599(746) ≈ 2.3; C*P = ≈ 19 (both are quite different!) 3 (200/60)(0.91) 1000(200/60)3 (0.91)5 C*H = 9.81(9.1 m) ≈ 9.7 (OK); (200/60)2 (0.91)2 ηmax = 599(746) ≈ 86% (OK) 9810(5.8)(9.1) Ans. 54 11.92 An idealized radial turbine is shown in Fig. P11.92. The absolute flow enters at 30° and leaves radially inward. The flow rate is 3.5 m3/s of water at 20°C. The blade thickness is constant at 10 cm. Compute the theoretical power developed at 100% efficiency. Fig. P11.92 Solution: For water, take ρ ≈ 998 kg/m3. With reference to Fig. 11.22 and Eq. 11.35, 11.93 Performance data for a very small (D = 8.25 cm) model water turbine, operating with an available head of 14.9 m, are as follows: Q, m3/h: 18.7 18.7 18.5 rpm: 0 500 η: 0 14% 18.3 15.1 11.5 1000 1500 2000 2500 3000 3500 27% 50% 11% 38% 17.6 16.7 65% 61% (a) What type of turbine is this likely to be? (b) What is so different about this data compared to the dimensionless performance plot in Fig. 11.22d? Suppose it is desired to use a geometrically similar turbine to serve where the available head and flow are 45.72 m and 0.19 m3/s, respectively. Estimate the most efficient (c) turbine diameter; (d) rotation speed; and (e) horsepower. Solution: (a) Convert Q = 16.7 m3/h = 4.639E–3 m3/s. Use BEP data to calculate power specific speed: bhp = ρ gQH η = (9810 N/m 3 )(4.639E – 3 m 3/s)(14.9 m)(0.65) = 441 W = 0.591 hp N sp = rpm(bhp)1/2 (2500 rpm)(0.59 bhp)1/2 = ≈ 66 (Francis turbine) Ans. (a) (H -m)5/4 (14.9 m)5/4 (b) This data is different because it has variable speed. Our other data is at constant speed. Ans. (b) 55 (c, d, e) First establish the BEP coefficients from the small-turbine data: 3 Q (16.7/3600 m /h) = = 0.198; nD 3 (2500/60 r/s)(0.0825 m)3 (9.81)(49 * 0.3048 m) gH C *H = 2 2 = = 12.4 (2500/60)2 (0.0825)2 n D *= CQ CP* = (326 *1.3558 W) P = = 1.60 3 5 3 ρn D (998 kg/m )(2500/60 r/s)3 (0.0825 m )5 Now enter the new data to the flow and head coefficients: CQ* = 2 0.19 m 3/s *H = (9.81 m/s )(45.72 m) = 12.4 = 0.198; C n2 D23 n22 D22 Solve for n2 = 15 r/s = 900 r / min; D2 = 0.4 m Ans. (c, d) Then P2 = CP*ρ n23 D25 = 1.60(1000 kg/m 3 )(15 r/s)3 (0.4 m)5 P2 = 55296 W = 74 hp Ans. (e) Actually, since the new turbine is 4.84 times larger, we could use the Moody step-up formula, Eq. (11.29a), to predict (1 − η2 ) ≈ (1 − 0.65)/(4.84)1/4 = 0.236, or η2 = 0.764 = 76.4%. We thus expect more power from the larger turbine: P2 = (74 hp)(76%/65%) ≈ 87 hp Better Ans. (e) 11.94 A Pelton wheel of 3.66-m pitch diameter operates under a new head of 610 m. Estimate the speed, power output, and flow rate for best efficiency if the nozzle exit diameter is 10 cm. Solution: First get the jet velocity and then assume BEP at φ ≈ 0.47: Cv ≈ 0.94, so: Vjet = Cv 2gH = 0.94 2(9.81)(610) ≈ 102.8 m/s BEP at φ ≈ 0.47 = π nD π n(3.66) = , 2gH 2(9.81)(610) r × 60 ≈ 268 rpm Ans. s m3 π 2 = 102.8 ( 0.1) = 0.81 ≈ 48.6 m3 / min Ans. 4 s Solve n = 4.47 Q = Vjet Anozzle u best = Vjet /2 ≈ 51.4 m/s Ptheory = ρ Qu(Vj − u)(1 − cos β ) ≈ 1000(0.81)(51.4)(102.8 − 51.4)(1 − cos165°) ÷ 746 ≈ 5640 hp100% Pturb ≈ 4500 bhp, N sp ≈ 268(4500)1/2 ≈ 5.93 : Fig. 11.27, read ηmax < 80%, say, 75% (610)5/4 Actual power output ≈ ηP100% = 0.75(5640) ≈ 4230 bhp Ans. 56 11.95 An idealized radial turbine is shown in Fig. P11.95. The absolute flow enters at 25° with the blade angles as shown. The flow rate is 8 m3/s of water at 20°C. The blade thickness is constant at 20 cm. Compute the theoretical power developed at 100% efficiency. Solution: The inlet (2) and outlet (1) velocity vector diagrams are shown at right. The normal velocities are From these we can compute the tangential velocities at each section: Fig. P11.95 57 11.96 The flow through an axial-flow turbine can be idealized by modifying the stator-rotor diagrams of Fig. 11.12 for energy absorption. Sketch a suitable blade and flow arrangement and the associated velocity vector diagrams. For further details, see Chap. 8 of Ref. 25. Solution: Some typical velocity diagrams are shown below, where u = ω r = blade speed. The power delivered to the turbine, at 100% ideal shock-free flow, is Fig. P11.96 11.97 A dam on a river is being sited for a hydraulic turbine. The flow rate is 1500 m3/h, the available head is 24 m, and the turbine speed is to be 480 r/min. Discuss the estimated turbine size and feasibility for (a) a Francis turbine; and (b) a Pelton wheel. Solution: Assume η ≈ 89%, as in Fig. 11.21d. The power generated by the turbine would be P = ηγ QH = (0.89)(9810 N/m3)(0.417 m3/s)(24 m) = 87,379 W = 117 hp. Now compute Nsp = (480 rpm)(117 hp)1/2/(24 m)5/4 ≈ 98, appropriate for a Francis turbine. (a) A Francis ≈ 0.34 = (0.417 m3/s)/ [(480/60 r/s)D3]. Solve turbine, similar to Fig. 11.21d, would have for a turbine diameter of about 0.54 m, which would be excellent for the task. Ans. (a) (b) A Pelton wheel at best efficiency (half the jet velocity) would only be 0.46 m in diameter, with a huge nozzle, d ≈ 0.15 m, which is too large for the wheel. We conclude that a Pelton wheel would be a poor design. Ans. (b) 11.98 Figure P11.98 shown on the following page, shows a crossflow or “Banki” turbine [Ref. 55], which resembles a squirrel cage with slotted curved blades. The flow enters at about 2 o’clock, passes through the center and then again through the blades, leaving at about 8 o’clock. Report to the class on the operation and advantages of this design, including idealized velocity vector diagrams. 58 Brief Discussion (not a “Solution”): The crossflow turbine is ideal for small dam owners, because of its simple, inexpensive design. It can easily be constructed by a novice (such as the writer) from wood and plastic. It is not especially efficient (≈ 60%) but makes good, inexpensive use of a small stream to produce electric power. For details, see Ref. 55 or the paper “Design and Testing of an Inexpensive Crossflow Turbine,” by W. Johnson et al., ASME Symposium on Small Hydropower Fluid Machinery, Phoenix, AZ, Nov. 1982, ASME vol. H00233, pp. 129−133. Fig. P11.98 11.99 A simple crossflow turbine, Fig. P11.98 above, was constructed and tested at the University of Rhode Island. The blades were made of PVC pipe cut lengthwise into three 3 120°-arc pieces. When tested in water at a head of 1.62 m and a flow rate of 2.4 m /min, the measured power output was 0.6 hp. Estimate (a) the efficiency; and (b) the power specific speed if n ≈ 200 rpm. Solution: We have sufficient information to compute the available water power: 2.4 0.6 Pavail = ρgQH = (9810) ≈ 70% Ans. (a) (1.62) = 636 ÷ 746 = 0.852 hp, ∴ η = 60 0.852 At 200 rpm, N sp = rpm(hp)1/2 200(0.6)1/2 = ≈ 85 (head)5/4 (1.62)5/4 Ans. (b) 59 11.100 One can make a theoretical estimate of the proper diameter for a penstock in an impulse turbine installation, as in Fig. P11.100. Let L and H be known, and let the turbine performance be idealized by Eqs. (11.38) and (11.39). Account for friction loss hf in the penstock, but neglect minor losses. Show that (a) the maximum power is generated when hf = H/3, (b) the optimum jet velocity is (4gH/3)1/2, and (c) the best nozzle diameter is Dj = [D5/(2fL)]1/4, where f is the pipe-friction factor. Fig. P11.100 Solution: From Eqs. 11.38 and 39, maximum power is obtained when u = Vj /2, or: Pmax = ρQ Vj Vj 1 − cos β 3 2 Vj − (1 − cos β ) = ρ A j Vj = CVj Vj , C = constant 2 2 4 Now apply the steady-flow energy equation between the reservoir and the outlet jet: Then the correct pipe flow speed is obtained by back-substitution: 60 11.101 Apply the results of Prob. 11.100 to determining the optimum (a) penstock diameter, and (b) nozzle diameter for the data of Prob. 11.97, with a commercial-steel penstock of 3 length 457 m. [H = 244 m, Q = 151 m /min.] 2 Solution: For water, take ρ = 998 kg/m3 and µ = 1.003E−3 N·s/m . For commercial steel, ε ≈ 4.572E–5 m. We can’t find fMoody until we know D, so iteration is required. We can immediately compute Vjet = (4gH/3)1/2 ≈ 56.5 m/s, but this wasn’t asked! Anyway, h f = H / 3 = 81.33 m = f 2 2 L Vp 457 Vp 151/ 60 =f , where Vpipe = (m / s) D 2g D 2(9.81) (π /4)D 2 Iterate to find f ≈ 0.0120, Re D ≈ 6.24E6, and D = D penstock ≈ 0.513 m Ans. (a) Then D jet D5 = 2fL 1/4 1/4 (0.513)5 = 2(0.0120)(457) ≈ 0.24 m Ans. (b) 11.102 Consider the following non-optimum version of Prob. 11.100: H = 450 m, L = 5 km, D = 1.2 m, Dj = 20 cm. The penstock is concrete, ε = 1 mm. The impulse wheel diameter is 3.2 m. Estimate (a) the power generated by the wheel at 80% efficiency; and (b) the best speed of the wheel in r/min. Neglect minor losses. Solution: For water take ρ = 998 kg/m3 and µ = 0.001 kg/m·s. This is a non-optimum condition, so we simply make a standard energy and continuity analysis. Refer to the figure on the next page for the notation: Δz = H = h f + V j2 V j2 2g , V j D 2j = Vpipe D 2 , combine and solve for jet velocity: 4 ρVD ε L Dj , , 1 + f = 2gH , where f = fcn µ D D D ε 0.001 = = 8.33E−4 1.2 D For example, guessing f ≈ 0.02, we estimate Vj ≈ 91.2 m/s. Using EES yields f = 0.0189, Red = 3.03E6, V jet = 91.23 m m3 , Q = 2.87 s s The power generated (at 80% efficiency) and best wheel speed are 1 D Power = ρ Quwheel (V j − uwheel )(1 − cosβ )(0.8), ubest = Vjet = Ωwheel wheel , β ≈ 165° 2 2 rad rev = 272 , s min Ans. (a, b) If Dwheel = 3.2 m, solve for Ωwheel = 28.5 Power = 9.36 MW 61 As shown on the figure below, which varies Dj, the optimum jet diameter is 34 cm, not 20 cm, and the optimum power would be 16 MW, or 70% more! 11.103 Francis and Kaplan (enclosed) turbines are often provided with draft tubes, which lead the exit flow into the tailwater region, as in Fig. P11.103. Explain at least two advantages to using a draft tube. Solution: Draft tubes have two big advantages: (1) They reduce the exit loss, since a draft tube is essentially a diffuser, as in Fig. 6.23 of the text, so more of the water head is converted to power. Fig. P11.103 (2) They reduce total losses downstream of the turbine, so that the turbine runner can be placed higher up without the danger of cavitation. 62 11.104 Like pumps, turbines can also cavitate when the pressure at point 1 in Fig. P11.103 drops too low. With NPSH defined by Eq. 11.20, the empirical criterion given by Wislicenus [Ref. 4] for cavitation is N ss = (rpm)(m 3 / min)1/2 ≥ 1650 [NPSH(m)]3/4 Use this criterion to compute how high (z1 − z2) the impeller eye in Fig. P11.103 can be placed for a Francis turbine, with a head of 91.4 m, Nsp = 177, and pa = 101.3 kPa, before cavitation occurs in 15°C water. Solution: For water at 15°C, take ρg = 9800 N/m3 and pv ≈ 1.782 kPa. Then Eq. 11.20: NPSH = pa − pv (101.3 − 1.782)(1000) − Δz − h fi = − Δz − 0 = 10.15 m − Δz 9800 ρg Now we need the NPSH, which we find between the two specific-speed criteria: N sp = 177 = N ss = 1650 = n(P in hp)1/2 (91.4 m)5/4 with P = ηρg(Q / 60)(91.4 m) ÷ 746 n(Q in m 3 / min)1/2 . Iterate to solve by assuming η ≈ 90% (NPSH)3/4 (We can’t find n or Q, only nQ1/2). Result: NPSH ≈ 14.8 m, Δz = 10.15 − 14.8 ≈ −4.65 m. Ans. P11.105 A large HAWT could have a swept area diameter up to 150 m, and its overall height could reach 200 m (rated power proposed at 10 MW). What would be the problems for such a large turbine, and how would you solve the problems? Solution: We could think of a few problems. Firstly, velocity profile for wind at different elevation could change, in particular a very large wind turbine. Secondly, a large wind turbine may have limited top speed. It could take quite a few seconds to rotate for one revolution. Any change of velocity profile of the wind could also cause unbalanced on the turbine blade. These problems can be solved by changing the blade angle of attack using a complex control strategy to control the blade pitch angle. 11.106 Consider the large wind turbine in the chapter-opener photo. Pretend that you do not know that 5M means five megawatts. Use the meager photo-caption data, plus your knowledge of wind turbine theory and experiment, to estimate the power delivered by the turbine for a rated wind speed of 12 m/s and a rotation rate of 10 rev/min. Solution: Meager data: It is a HAWT with D = 126 m. For sea level air, take ρ = 1.22 kg/m3. Convert ω = 10 rev/min = (10)(2π)/60 = 1.05 rad/s. Then calculate the speed ratio: 63 Now find the experimental power coefficient on Fig. 11.31 of the text. This giant turbine is probably not “high speed”, but use the “high speed HAWT” curve: Cp ≈ 0.43. Compute This is a good estimate, considering that we have no data for the 5MW turbine. 11.107 A Darrieus VAWT in operation in Lumsden, Saskatchewan, that is 9.75 m high and 6.1 m in diameter sweeps out an area of 40 m2. Estimate (a) the maximum power and (b) the rotor speed if it is operating in 25.75 km/h winds. Solution: For air in Saskatchewan (?), take ρ ≈ 1.185 kg/m3. Convert 25.75 km/h = 7.15 m/s. From Fig. 11.34 for the Darrieus VAWT, read optimum CP and speed ratio: CP,max ≈ 0.42 = P Pmax 3 = (1/2)ρ AV1 (1/2)(1.185)(40)(7.15)3 Solve for Pmax ≈ 3.6 kW Ans. (a) At Pmax , rad 60 ωr ω (6.1/2) ≈ 4.1 = , or: ω ≈ 9.61 × ≈ 92 rpm Ans. (b) s 2π V1 7.15 11.108 An American 1.8-m diameter multiblade HAWT is used to pump water to a height of 3 m through 7.5-cm-diameter cast-iron pipe. If the winds are 19.3 km/h, estimate the rate 3 of water flow in m /min. Solution: For air in “America” (?), take ρ ≈ 1.185 kg/m3. Convert 19.3 km/h = 5.36 m/s. 2 For water, take ρ = 998 kg/m3 and µ = 1.003E−3 N·s/m . From Fig. 11.34 for the American multiblade HAWT, read optimum CP and speed ratio: CP,max ≈ 0.29 at 1 ωr π ≈ 0.9: Pmax ≈ 0.29 (1.185) (1.8)2 (5.36)3 ≈ 67.33 W 2 4 V1 If ηpump ≈ 80%, Ppump ≈ 0.8(67.33) = ρwater gQH syst where Vpipe = Q , (π /4)(0.075)2 L 3 = = 40, D 0.075 2 L Vpipe = 9790Q Δz + f D 2g ε 2.59E – 4 = ≈ 0.0034 0.075 D Clean up to: 5.502E – 3 = Q(3 +1.045ESfMoodyQ 2 ), with Q in m 3/s. Iterate to obtain f ≈ 0.0304, Vpipe ≈ 0.407 m 3 /s, Re ≈ 30400, Q ≈ 0.0018 m 3 /s ≈ 0.108 m3 / min Ans. 64 P11.109 The new 3-blade HAWT municipal wind turbine in Portsmouth, RI has a tower 65 m high, with a turbine diameter of 77 m. The newspaper rates the turbine at 1.5 MW. (a) What wind velocity, and turbine rotation speed, would generate this power? (b) Estimate the annual power production for a steady wind speed of 19.3 km/h. Solution: From Fig. 11.32, best efficiency for a HAWT occurs at ωr/V ≈ 5.5, where the power coefficient is about 0.45. Assume sea-level density of 1.22 kg/m3 – all of Rhode Island is pretty much sea-level. Then find the velocity: 1.22 π ρ AV 3 = (0.45)( )[ (77 m)2 ]V 3 2 4 2 1/3 Solve for V = (1173.5) = 10.55 m / s ≈ 40 km / h Ans.(a) rad r ωr ω (38.5) = 5.5 = , Solve ω = 1.51 = 14.4 Ans.(a) Also, s min V 10.55 P = 1.5E6 W = C P (b) Carry out the same calculations for a steady V = 19.3 km/h = 5.36 m/s: 1.22 π ρ AV 3 = (0.45)( )[ (77 m)2 ](5.36 m/s)3 = 197 kW 2 2 4 Annual production = (197kW )(24 h / day)(365day / year) = 1, 730, 000 kWh / year Ans.(b) P = CP That’s a lot of electric power. Of course, Rhode Island isn’t that windy all the time. 11.110 The controversial Cape Wind project proposes 130 large wind turbines in Nantucket Sound, intended to provide 75% of the electric power needs of Cape Cod and the Islands. The turbine diameter is 100 m. For an average wind velocity of 22.53 km/h, what are the best rotation rate (r/min) and total power output (MW) estimates for (a) a HAWT; and (b) a VAWT? Solution: Convert 22.53 km/h = 6.26 m/s. Use Fig. 11.32 of the text to make your estimates. (a) HAWT geometry. Best efficiency is at (ωr/V) ≈ 5.8, where CP ≈ 0.45: rad 60(0.726) r ωr ω (50 m) = 5.8 = , Solve ω = 0.726 , RPM = = 6.9 s 2π min V 6.26 m / s 1.22 π ρ Power = CP AV 3 = 0.45( ) (100)2 (6.26)3 = 529 kW 2 2 4 Total power = (0.529 MW )(130 units) = 68.8 MW Ans.(a) Ans.(a) (b) Darrieus VAWT geometry. Best efficiency is at (ωr/V) ≈ 4.0, where CP ≈ 0.40: rad 60(0.501) r ωr ω (50 m) = 4.0 = , Solve ω = 0.501 , RPM = = 4.8 Ans.(b) s 2π min V 6.26 m / s 1.22 π ρ ) (100)2 (6.26)3 = 470 kW Power = CP AV 3 = 0.40( 2 4 2 Total power = (0.470 MW )(130 units) = 61 MW Ans.(b) Primarily because of the variability of Cape winds, both of these are overestimates. The official power estimate for the Cape Wind (HAWT) project is 42 MW. 65 P11.111 In 2007, a wind-powered-vehicle contest, held in North Holland [64], was won with a design by students at the University of Stuttgart. A schematic of the winning 3wheeler is shown in Fig. P11.111. It is powered by a shrouded wind turbine, not a propeller, and, unlike a sailboat, can move directly into the wind. (a) How does it work? (b) What if the wind is off to the side? (c) Cite some design questions you might have. Fig. P11.111 Solution: (a) The wind turns the turbine. The turbine turns a shaft that is geared to the wheels and drives the vehicle. (b) If the wind is from the side, the entire turbine swivels to face into the wind. (c) Design questions might include: 1. How large to make the turbine? Answer: D = 1.8 m. 2. How to design the drive train? Answer: Two sets of bicycle gears and a chain drive. 3. How to balance the weight to resist the overturning moment of the turbine thrust? Answer: More weight in the front – total weight was about 1.29 kN. 4. How to manage variable winds? Answer: The turbine blades could change pitch if the wind speed changed. 5. How to make light but strong blades? Answer: The students molded the blades from composite materials. ________________________________________________________________________ P11.112 Analyze the wind-powered-vehicle of Fig. P11.111 with the following data: turbine diameter D = 1.8 m, power coefficient (Fig. 11.32) = 0.3, vehicle CDA = 0.42 m2, and turbine rotation 240 r/min. The vehicle moves directly into a head wind, W = 40.2 km/h. The wind backward thrust on the turbine is approximately T ≈ CT(ρ /2)Vrel2 Aturbine, where Vrel is the air velocity relative to the turbine and CT ≈ 0.7. Eighty per cent of the turbine power is delivered by gears to the wheels, to propel the vehicle. Estimate the sea-level vehicle velocity V, in km/h. 66 Solution: At sea level take ρ = 1.22 kg/m3. Recall the definition of power coefficient from Eq. (11.46). The air velocity relative to the moving turbine is (W+V). Hence ρ ρ π 3 Aturbine Vrel = C P ( ) D 2 (W +V )3 , C P ≈ 0.4 (Fig.11.32) 2 2 4 Also, η Pturbine = Pvehicle = Fvehicle V , where η ≈ 0.8 Pturbine = C P The retarding force on the vehicle is the sum of drag on the vehicle and thrust on the turbine: F vehicle = Fdrag + Fthrust = ρ ρ π 2 (C D A)(W + V )2 + CT D (W + V )2 2 2 4 where CT ≈ 0.7. Combine these two equations and we find that both (W+V)2 and ρ cancel, and the turbine area can be divided out. The result is ηC P (W + V ) = V [ or : V = CD A (π / 4)D 2 + CT ] , or : V = W 0.42 1 [ ][ + 0.7]−1 0.8(0.4) (π / 4)(1.8)2 W CD A 1 ( )[ + CT ] − 1 ηC P (π / 4)D 2 = (analytic answer) W 40.2 = = 23.65 km / h 2.70 − 1 1.7 Ans. This is a simplified analysis, independent of density and RPM, in reasonable agreement with the actual performance of the Stuttgart student design, which averaged 24 km/h during the contest. _____________________________________________________________________________ P11.113 A very large Darrieus VAWT was constructed by the U.S. Department of Energy near Sandia, New Mexico. It is 20 m high and 10 m in diameter, with a swept area of 111.5 2 m . If the turbine is constrained to rotate at 90 r/min, use Fig. 11.32 to plot the predicted power output in kW versus wind speed in the range V = 5 to 70 km/h. 3 Solution: For air in New Mexico, take ρ = 1.23 kg/m . Given r = 5 m and w = 90 rpm = 9.43 rad/s, we can select wind speed V, hence ω r/v is known, read Cp, hence the power can be computed. For example Select ∴ 9.43(5) ωr = 3.0 = V1 V1 V1 = 15.72 m/s = 56.6 km/h Read C p from Fig. 11.32, C p ≈ 0.1 = ∴ P= P 1 ρ AV13 2 1 1 ρ AV13 = (1.23)(111.5)(15.72)3 2 2 = 26,638.4 W = 35.71 hp 67 Continue to do this until we can span wind speed from 5 to 70 km/h. Wind Speed km/h (m/s) 5 10 20 30 40* 50 60 70 (1.39) (2.78) (5.56) (8.33) (11.11) (13.89) (16.67) (19.44) Power kW 0 0 0 12.7 38.6 32.2 22.2 20.2 ωr V1 34 16.96 8.4 5.66 4.24 3.39 2.83 2.43 Cp 0 0 0 0.32 0.41 0.175 0.07 0.04 Note that the rotor gives zero power for V ≤ 22.6 km/h and gives the maximum power at 40 km/h. ___________________________________________________________________________ P11.114 Figure 11.32 showed the typical power performance of a wind turbine. The wind also causes a thrust force that must be resisted by the structure. The thrust coefficient CT of a wind turbine may be defined as follows: CT = Thrust force ( ρ / 2) A V 2 = T ( ρ / 2)[(π / 4)D 2 ] V 2 Values of CT for a typical horizontal-axis wind turbine are shown in Fig. P11.114. The abscissa is the same as in Fig. 11.32. Consider the turbine of Prob. P11.109. If the wind is 56.3 km/h and the rotation rate 24 r/min, estimate the bending moment about the tower base. 68 Fig. P11.114 ω r/V1 Solution: Assume sea-level density of 1.22 kg/m3 – all of Rhode Island is pretty much sealevel. Convert 24 r/min x 2π/60 = ω = 2.51 rad/s. Convert wind speed = 56.3 km/h = 15.65 m/s. Now find where we are on the graph: (2.51 / s)(38.5 m) ωr = = 6.2 15.65 m / s V From Fig. P11.114, read CT ≈ 0.8. Then the thrust force is, approximately, 1.22 ρ A V 2 = (0.8)( )[π (38.5 m)2 ](15.65)2 ≈ 556,570 N 2 2 Moment about the base = T h = (556,570 N )(65 m) = 3.62 N·m Ans. Thrust = C ___________________________________________________________________________ 69 COMPREHENSIVE PROBLEMS C11.1 The net head of a little aquarium pump is given by the manufacturer as a func-tion of volume flow rate as listed: Q, m3/s: H, mmH2O: 0 1E−6 2E−6 3E−6 4E−6 5E−6 1.10 1.00 0.80 0.60 0.35 0.0 What is the maximum achievable flow rate if you use this pump to pump water from the lower reservoir to the upper reservoir as shown in the figure? NOTE: The tubing is smooth, with an inner diameter of 5 mm and a total length of 29.8 m. The water is at room temperature and pressure, and minor losses are neglected. Fig. C11.1 Solution: For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m·s. NOTE: The tubing is so small that the flow is laminar, even at the highest pump flow rate: The energy equation shows that the pump must fight both friction and elevation: One can plot the two relations, as at right, or use EES with a look-up table to get the final result for flow rate and head: The EES print-out gives the results Red = 255, H = 0.999 m, Q = 1.004E−6 m3/s. 70 C11.2 Reconsider Prob. 6.68 as an exercise in pump selection. Select an impeller size and rotational speed from the Byron Jackson pump family of Prob. 11.30 which will deliver a flow rate of 85 L/s to the system of Fig. P6.68 at minimum input power. Calculate the horsepower required. 2 Solution: For water take ρ = 998 kg/m3 and µ = 1.003E−3 N·s/m . For cast iron take Fig. C11.2 ε = 2.59E–4 m, or ε /d = 2.59E–4/0.152 = 0.0017. The energy equation, written between reservoirs, is the same as in Prob. 6.62: H p = Δz + f L V2 600 [Q/ (π /4)/ (0.15)2 ]2 = 36 m + f = 36 + 6.529E5 fQ 2 d 2g 0.15 2(9.81) where ε f = fMoody = fcn Red , d with Red = 4 ρQ πµd If, as given, Q = 0.085 m3/s, then, from above, f = 0.0227 and Hp = 143.07 m. Now we have to optimize: From Prob. 11.30, Q* = 0.17 m3/s, H* = 100.58 m, and P* = 255 bhp when n = 2134 rpm (35.57 rps) and D = 37.1 cm. Thus, at BEP: For the system above, (0.085)/[nD3] = 0.0932, or nD3 = 0.912, and (9.81)(100.58)/n2D2 = 5.66, or n2D2 = 174.33. Solve simultaneously: n = 50.1 rps = 3010 rev ; D p = 26.3 cm; Power = 0.599 ρn 3 D p5 = 127 hp min C11.3 Reconsider Prob. 6.84 as an exer-cise in turbine selection. Select an impeller size and rotational speed from the Francis turbine family of Fig. 11.21d which will deliver maximum power generated by the turbine. Calculate the water turbine power output and remark on the practicality of your design. Fig. C11.3 Ans. 71 Solution: For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m·s. For wrought iron take ε = 0.046 m, or ε /d1 = 0.046/60 = 0.000767 and ε /d2 = 0.046/40 = 0.00115. The energy and continuity equations yields 2 d V2 L V2 L V2 Δz = 20 m = 2 + f1 1 1 + f2 2 2 + hturbine ; V2 = V1 1 = 2.25V1 2g d1 2g d2 2g d2 with the friction factors to be determined by the respective Reynolds numbers and roughness ratios. We use the BEP coefficients from Eq. (11.36) for this turbine: CQ* = Q* gH * P* * = 2 2 = 9.03; CP* = 3 5 = 2.70 3 = 0.34; C H nDt n Dt ρn Dt We know from Prob. 6.83 that the friction factors are approximately 0.022. With only one energy equation (above) and two unknowns (n and Dt ), we vary, say, n from 0 to 100 rev/s and find the resulting turbine diameter and power extracted. The power is shown in the graph below, with a maximum of 381 watts at n = 70 rev/s, with a resulting turbine diameter Dt = 5.3 cm. This is a fast, small turbine! Ans. C11.4 The system of Fig. C11.4 is designed to deliver water at 20°C from a sea-level reservoir to another through new cast iron pipe of diameter 38 cm. Minor losses are ∑K1 = 0.5 before the pump entrance and ∑K2 = 7.2 after the pump exit. (a) Select a pump from either Figs. 11.7a or 11.7b, running at the given speeds, which can perform this task at maximum efficiency. Determine (b) the resulting flow rate; (c) the brake horsepower; and (d) whether the pump as presently situated is safe from cavitation. 72 Fig. C11.4 Solution: For water take ρ = 998 kg/m3 and = 0.001 kg/m·s. First establish the system curve of head loss versus flow rate. For cast iron take ε = 0.26 mm, hence ε /d = 0.26/380 = 0.000684. 3 3 The pumps of Figs. 11.7a,b deliver flows of 15.14 m /min to 106 m /min, no doubt turbulent flow. The energy equation, written from the lower free surface to the upper surface, gives 26 3 Take, for example, Q = 83.28 m /min = 1.39 m3/s. Then Vpipe = 12.24 m/s, Red = 4.64E6, f = 0.0180, hence the head loss becomes hf = (12.24)2 2(9.81) 28 m (0.0180) 0.38 m + 0.5 + 7.2 = 68.9 m Thus a match requires H = h f + z2 = 68.9 m + 11 m = 79.9 m No pump in Fig. 11.7 exactly matches this, but the 71.12-cm pump in Fig. 11.7a and the 105-cm pump in Fig. 11.7b are pretty close, especially the latter. We can continue the calculations: 3 Q, m /min: hf, m: 15.14 13.41 30.28 20.12 45.42 31.39 60.57 47.55 75.71 67.97 83.28 79.86 90.85 106 92.96 122.53 (a) The best match seems to be the 96.5-cm pump of Fig. 11.7b at a flow rate of 3 75.71 m /min, near maximum efficiency of 88%. Ans. (a) 3 (b, c) The appropriate flow rate is 75.71 m /min. Ans. (b) The horsepower is 1250 bhp. Ans. (c) 73 (d) Use Eq. (11.20) to check the NPSH for cavitation. For water at 20°C, pv = 2337 Pa. The velocity in the pipe is V = Q/A = 11.13 m/s. The theoretical minimum net positive suction head is: 3 In Fig. 11.7b, at Q = 75.71 m /min, read NPSH ≈ 16 m >> −0.36 m! So this pump, when placed in its present position, will surely cavitate! Ans. (d) A new pump placement is needed. C11.5 In Prob. 11.25, estimate the efficiency of the pump in two ways: (a) read it directly from Fig. 11.7b (for the dynamically similar pump); (b) calculate it from Eq. (11.5) for the actual kerosene flow. Compare your results and discuss any discrepancies. Solution: Problem 11.25 used the 105-cm-pump in Fig. 11.7b to deliver 22000 gal/min of kerosene at 850 rpm. (a) The problem showed that the dynamically similar water pump, at 710 rpm, had a flow rate of 18,400 gal/min. (a) (b) For kerosene, take ρ = 804 kg/m3. The solution to Prob. 11.25 gave a kerosene head of 103.6 m and a brake horsepower of 1600 hp. Together with the known flow rate, we can calculate the kerosene efficiency by definition: ηkerosene = ρ kerosene gQH (804)(9.81)(83.28/60)(103.6) = = 0.95 or 95% Ans. (b) (1600)(746) W Power This is significantly different from 88.5% in part (a) above. The main reason (the author thinks) is the difficulty of reading bhp from Fig. 11.7b. The actual kerosene bhp for Prob. 11.25 is probably about 1700, not 1600. Ans. C11.6 An interesting turbomachine [58] is the fluid coupling of Fig. C11.6, which delivers fluid from a primary pump rotor into a secondary turbine on a separate shaft. Both rotors have radial blades. Couplings are common in all types of vehicle and machine transmissions and drives. The slip of the coupling is defined as the dimension-less difference between shaft rotation rates, s = 1 − ω s /ω p. For a given percentage of fluid filling, the torque T transmitted is a function of s, ρ, ωp, and impeller diameter D. (a) Non-dimensionalize this function into two pi groups, with one pi proportional to T. Tests on a 30.5-cm-diameter coupling at 2500 r/min, filled with hydraulic fluid of density 897 kg/m3, yield the following torque versus slip data: Slip, s: Torque T, N·m: 0% 0 5% 122 10% 372.8 15% 596.6 20% 786.4 25% 921.9 74 (b) If this coupling is run at 3600 r/min, at what slip value will it transmit a torque of 1220 N·m? (c) What is the proper diameter for a geometrically similar coupling to run at 3000 r/min and 5% slip and transmit 813.5 N·m of torque? Fig. C11.6 Solution: (a) List the dimensions of the five variables, from Table 5.1: Variable: T s Dimension: {ML2/T2} {1} ρ {M/L3} ωp {1/T} D {L} Since s is already dimensionless, the other four must form a single pi group. The result is: Now, to work parts (b) and (c), put the data above into this dimensionless form. Convert ρoil = 897 kg/m3. Convert ωp = 2500 r/min = 41.7 r/s. The results are: Slip, s: ( 0% ) T ρωp2 D 5 : 0 5% 10% 0.0296 0.0906 15% 20% 25% 0.145 0.191 0.224 (b) With D = 0.305 m and ω p = 3500 r/min = 58.3 r/s and T =1220 N·m, 1220 N T = 0.152, Estimate s ≈ 15% Ans. (b) 2 5 = 3 (897 kg /m )(58.3 r/s)2 (0.305)5 ρω p D (c) With D unknown and s = 5% and ω p = 3000 r/min = 50 r/s and T = 813.5 N·m, 813.5 N⋅m T = 0.0298, Solve D ≈ 41.4 cm Ans. (c) 2 5 = (897 kg/m 3 )(50 r/s)2 (D)5 ρω p D C11.7 Report to the class on the Cordier method [63] for optimal design of turbomachinery. The method is related to, and greatly expanded from, Prob. P11.48 and uses both software and charts to develop an efficient design for any given pump or compressor application. Solution: The Cordier method, developed in the textbook G. T. Csanady, Theory of Turbomachines, McGraw-Hill, New York, 1964, is an extensive correlation of the best features of various pumps (or turbines) to design an optimal machine for a given set of conditions. Its modern use, with software, is discussed in Ref. 63. 75 C11.8 A pump-turbine is a reversible device that uses a reservoir to generate power in the daytime and then pumps water back up to the reservoir at night. Let us reset Prob. P6.68 as a pump-turbine. Recall that Δz = 36 m, and the water flows through 600 m of 15-cm-diameter cast iron pipe. For simplicity, assume that the pump operates at BEP (92%) with H*p = 61 m and the turbine operates at BEP (89%) with H*t = 30.5 m. Neglect minor losses. Estimate (a) the input power, in watts, required by the pump; and (b) the power, in watts, generated by the turbine. For further technical reading, consult the URL www.usbr.gov/pmts/hydraulics_lab/pubs/EM/EM39.pdf. 2 Solution: For water, take ρ = 998 kg/m3 and µ = 1.003E-5 N·s/m . (a) Write the steady flow energy equation for pump operation, neglecting minor losses: h p = 61 m = Δz + f Qp L V2 600 m V2 = 36 m + f ( )[ ], V = D 2g 0.15 m 2(9.81) (π / 4)(0.15 m)2 For cast iron, ε = 2.59E–4 m, hence ε /D = 2.59E–4 m/0.15 m = 0.0017. Iterate to find the friction factor and complete the pump head equation. EES quickly finds the result: m m3 ρVD (998)V (0.15) = = 344, 623; f = 0.0230; V = 2.31 ; Q = 0.0408 ; H = 61 m 1.003E − 3 s µ s γ QH (9790)(0.0408)(61) = = 26.5 kW Ans.(a) Hence Powerpump = η 0.92 Re D = (b) Write the steady flow energy equation for turbine operation, neglecting minor losses: ht = 30.5 m = Δz − f L V2 600 m V2 = 36 m − f ( )[ ], V = D 2g 0.15 m 2(9.81) Qp (π / 4)(0.15 m)2 Iterate again and obtain a considerably lower flow rate: m m3 ρVD (998)V (0.15) = = 159700; f = 0.0236; V = 1.07 ; Q = 0.0189 ; H = 30.5m µ s 1.003E − 3 s Hence Powerpump = ηγ QH = (0.88)(9790)(0.0189)(30.5) = 4.97 kW Ans.(b) Re D = The pumping power is much greater than the power generated by the turbine, but the pump’s required electricity is very cheap because usage is so small at night.