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Unformatted text preview: ME 309 Fall 2008 Section 2 (Merkle) Homework # 28 Due Wed 5 Nov 2008 Problem 281 . Water is pumped between two reservoirs. The total pipe length is 250 m and its inside diameter is 200 mm . The pipe entrance is two meters below the surface of the lower reservoir while the exit empties two meters above the upper reservoir surface. The height difference between the two water levels is 46 m . The height between the pipe entrance and exit is 50 m . Minor losses for the pipe entrance and exit are described by K = 0.41 and K = 1.00. The piping system contains two elbows with K = 0.4 each. The headflow equation for the pump at 1450 RPM is: 2 1200 45 ] [ Q m H P = . For water properties, use: 3 / 997 m kg = , s m kg e  = / 4 91 . 8 . The pipe roughness is = 0.20 mm . The desired flow rate is Q = 0.0315 m 3 /s . a) Find the head required, H req , (in meters) to provide the desired flow rate through the piping system. b) Find the head produced by the pump, H P , at the desired flow rate at 1450 RPM. Will the operating point for the pump/piping system lie above or below this flow rate at this RPM? c) The pump speed is increased to 1550 RPM, but the required head is unchanged. Find the head produced by the pump at 1550 RPM. Is this speed closer to providing the required flow rate, or further away? 2 m Pump Total pipe length (Horizontal plus two vertical legs): 250m 50 m Flow 2 m Solution: Given : L = 250 m D = 200 mm . = 0.20 mm . z res = 46 m . z pipe = 50 m . K = 0.41 + 1.00 +2*0.4 Q = 0.0315 m 3 /s . Pump head at 1450 RPM: 2 1200 45 Q h = . Water Properties: 3 / 997 m kg = , s m kg e  = / 4 91 . 8 Find : a) Required head to reach Q = 0.0315 m 3 /s . b) Head produced by pump at 1450 RPM at Q = 0.0315 m 3 /s . c) Head produced by pump at 1550 RPM at Q = 0.0315 m 3 /s . Analysis: Part a) Use mechanical energy to find head required: Define: Station 1 at surface of lower reservoir; Station 2 at pipe entrance; Station 3 at pipe exit Head required is work to be done: Represent as P (required pump pressure rise) Required pump head: ( ) [ ] g h z z u u g g p p H tot req + + + = 2 3 2 2 2 3 2 3 2 1 z 1 = 0.; z 2 = 2., z 3 = 48, z 3 z 2 = 50 m p 1 = 101325.; p 2 = Bernoulli., p 3 = 101325 u 1 = 0.; u 2 = Bernoulli., u 3 = u 2 Bernoulli from 1 to 2: ( ) [ ] 1 2 2 1 2 2 1 2 2 1 z z g u u p p + + Solve for stagnation pressure at 2: [ ] 1 2 1 2 2 2 2 1 z z g p u p = + Substitute into mechanical energy to get required head relation: [ ] [ ] g h z z u g z z g p p H tot req + + + + = 2 3 2 3 1 2 1 3 2 1...
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 Spring '08
 MERKLE

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