If the pump having the performance curve shown is operated at a speed of 1500 rpm, what will be the maximum possible head developed?

Transcription

1 Spring 00 Problem set # Problem 8.4 A pump has the characteristics given in Fig What discharge and head will be produced at maximum efficiency if the pump size is 50 cm and the angular speed is 30 rps? What power is required when pumping water under these conditions? At maximum efficiency, from Fig. 8-5, C Q = 0.64, C P = 0.60, C H = 0.75 Q = C Q nd s m) 3 =.40 m 3 /s h = C Hn D g = s 1 ) 0.5 m) 9.81 m/s = 17. m. P = C P ρd 5 n 3 = kg/m m) 5 30 s 1 ) 3 = 506 kw. Problem 8.1 If the pump having the performance curve shown is operated at a speed of 1500 rpm, what will be the maximum possible head developed? C H = Hg D n Since C H will be the same for the maximum head condition, then ) 1500 H n or H 1500 = H 1000 = 10 ft.5 = 9.5 ft Problem 8.19 What type of pump should be used to pump water at a rate of 1 cfs and under a head of 5 ft? Assume N = 1500 rpm. 1

2 Spring 00 Problem set # n = 1500 rpm 60 s/min = 5 rps n s = n Q gh) 3 4 = 5 s 1 1 cfs 3. ft/s 5 ft) 3 4 = 0.57 Then from Fig. 8-15, n s < 0.60 so use mixed flow pump. Problem 8.3 You want to pump water at a rate of 1.0 m 3 /s from the lower to the upper reservoir shown in the figure. What type of pump would you use for this operation if the impeller speed is to be 600 rpm? h = z f L ) V D g = 18 m 15 m m ) 1.7 m/s) 1 m 3. m/s = 3.14 m n s = n Q gh) 3 4 = 10 s m 3 /s 9.81 m/s 3.14 m) 3 4 = 0.76 From Fig. 8-15, use axial flow pump. Problem 8.4 The pump used in the system shown has the characteristics given in Fig. 8-6, page 418. What discharge will occur under the conditions shown, and what power is required? D = 35.6 cm = m, n = 11.5 r/s. Writing the energy equation from the reservoir surface to the center of the pipe at the outlet: p 1 γ + V 1 g + z 1 + h p = p γ + V g + z + h L

3 Spring 00 Problem set # or h p = z z 1 = 1 + f L ) D + k Q e + k b 1) ga Assuming L = 6.4 m, f = 0.014, k b = 0.35 and k e = 0.1. C Q = Q nd 3 so we can get C H from Fig h p = C Hn D g Going through this calculations with different values of Q gives us the following: Q C Q C H h p 1) h p ) m 3 /s m m PSfrag replacementsthen plotting the system curve and the pump curve, we obtain the operating conditions: Q = 0. m 3 /s, Power: P = 6.7 kw from Fig. 8-6). 5 ) 4 Pump curve hp [m] 3 1 Operating point Q [m 3 /s] 3

4 Spring 00 Problem set # Problem 8.7 What is the specific speed for the pump operating under the conditions given in Prob. 8-4? Is this a safe operation with respect to the susceptibility to cavitation? n = 690 rpm 60 s/min = 11.5 rps Assume temperature of 10 C, Vapor pressure: p v = 1. kpa so p v γ = 1. kpa 9.81 kn = 0.1 m. Assume atmospheric pressure head of 10.3 m. Neglecting head loss and velocity head, the gauge pressure head on the suction side of the impellor will be approximately 1 m. Then NPSH = 10.3 m + 1 m 0.1 m = 11. m n ss = nq 1 g NPSH) 3 4 = 11.5 s m 3 /s) m/s 11. m) 3 4 = The n ss value of is much less than the critical value of 0.494, therefore, the pump is in the safe operating range. Problem 8.3 Two pumps having the performance curve shown are operated in series in the 18-in. diameter steel pipe. When both are operating, estimate the time to fill the tank from the 150-ft level to to the 00-ft level. Estimate the maximum pressure in the pipe during the filling phase. Where will this maximum pressure occur? What would have been the initial discharge if the pumps had been installed in parallel? First write the energy equation from the lower to upper reservoir: p 1 γ + V 1 g + z 1 + h p = p γ + V g + z + h L 4

5 Spring 00 Problem set # ft + h p = z + K e + K b + K E + f L ) V D g Assuming K e = 0.1, K b = 0., K E = 1.0, k s /D = Fig. 5-5) and f = Fig. 5-4). Then ) 300 ft Q h p = z 95 ft ft ga Q = z 95 ft ft/s π 1.5 ft)4 16 = z 95 ft s /ft 5 Q The performance curve for the two pumps in series is given below. The initial discharge will be obtained by solving the performance curve and the energy equation for z = 150 ft) h p = 55 ft s /ft 5 Q PSfrag replacements Plot this on the same graph and find the intersection Q i = 5.3 cfs Performance Curve hp [ft] System curve Q [cfs] To calculate the time to fill the reservoir consider increments of filling in 5 ft steps. 5

6 Spring 00 Problem set # z z Q V t [ft] [ft] [cfs] [ft 3 ] [s] t = s So the total time will be t = t = s or hours 55 minutes. The discharge Q was obtained by solving the system equation with with the performance curve as done for obtaining Q i. Check f : V i = Q i A 5.3 cfs = = ft/s ft for the initial values and V f = 1.8 ft/s in the final value in our table. So Re i = V id ν = 14.3 ft/s 1.5 ft ft /s = and Re f = For either of this value we find our initial assumption of f = to be valid. 6

7 Spring 00 Problem set # The maximum pressure will occur immediately downstream of the pumps when the 00 foot level is reached in the tank. Write the energy equation from the maximum pressure point to the water surface in the reservoir. or p max γ + V g + z p = p r γ + V r g + z r + h L p max γ + Q ga + z p = z r + K E + f p max = γ z r z p f = 6.4 lbs/ft 3 = 75 lbs/ft ) 80 ft Q D ga ) ) 80 ft Q D ga 00 ft 90 ft ) 1 ft = 50.4 psi 1 in 80 ft 1.5 ft ).7 ft 3 /s) 3. ft/s ft ) PSfrag replacements Consider parallel pump installation. The performance curve for the two pumps would be as shown below. Solving the initial system equation with this performance curve yields Q i = 36.4 cfs. 150 Performance Curve hp [ft] System curve Q [cfs] 7

8 Spring 00 Problem set # Problem 8.33 The pump of Prob. 8-1 is used to pump water from reservoir A to reservoir B. The pump is installed in a -mi long, 1-in pipe joining the two reservoirs. There are two bends in the pipe r/d = 1.0), and two gate valves are open when pumping. When the water surface elevation in reservoir B is 30 ft above the water surface in reservoir A at what rate will water be pumped? Write the energy equation from the water surface of reservoir A to the water surface in reservoir B: p 1 γ + V 1 g + z 1 + h p = p γ + V g + z + h L h p = ft + K e + K E + K b + K V + f L ) V D g where K e = 0.5, K E = 1.0, K b = 0.35 and K V = 0.0 Table 5-3). Also k s /D = Fig. 5-5), assume f = ) mi 580 ft/mi V h p = 30 ft ft g = 30 ft Q ga = 30 ft Q 3. ft/s π 1 ft)4 16 = 30 ft s /ft 5 Q 3.5 s /ft 5 = 30 ft gpm/cfs) Q = 30 ft ft/gpm) Q Plotting the above equation system curve) on the performance curve for problem 8-1 yields a discharge of 1500 gpm or 3.34 cfs V = Q A = 3.34 ft3 /s = 4.6 ft/s ft Re = VD ν 4.6 ft/s 1 ft = ft /s = giving f = With this larger f the system equation becomes h p = 30 ft ft/gpm) Q. 8

9 Spring 00 Problem set # Plotting this new system curve, etc. yields Q = 1450 gpm = 3.3 cfs. Problem 8.34 Work Prob but have two pumps like that of Prob. 8-1 operating in parallel. Assume same system curve as for the solution to Prob. 8-33: h p = 30 ft ft/gpm) Q. PSfrag replacements The parallel pump performance curve is given below: Plotting the system curve on the performance curve yields a solution of Q = 1650 gpm or 3.67 cfs System curve, 1 pipe Performance Curve hp [ft] System curve, 18 pipe Q [gpm] 9

10 Spring 00 Problem set # Problem 8.35 Work Prob but have two pumps like that of Prob. 8-1 operating in parallel and have an 18-in pipe instead of a 1-in pipe. For this pipe k s /D = , assume f = Then the energy equation reduces to h p = ) mi 580 ft/mi Q 30 ft ft 3. ft/s π 1.5 ft)4 16 = 30 ft s /ft 5 Q = 30 ft ft/gpm) Q Plotting the above equation on the graph of solution for problem 8-34 yields Q = 3300 gpm or 7.35 cfs. 10