Question
When a water pump was tested at a rotation rate of 1500 rpm, the following data...
When a water pump was tested at a rotation rate of 1500 rpm, the following data was obtained: Q (I/s) 0 10 20 30 40 50 H(m) 10.0 10.5 10.0 8.5 6.0 2.5 n (%) 0 0.40 0.64 0.72 0.64 0.40 It is proposed to use this pump to draw water from an open well and deliver it to a tank 5.5 m above. The delivery pipe is 20.0 m long and 100 mm diameter, and has a friction factor of 0.005. If operating at 1500 rpm, find: (a) The maximum discharge that the pump can provide (b) The pump efficiency at this discharge (c) The input power required. How to solve the problem: To solve this problem follow the following steps: Using appropriate graph paper 1. Draw the pump characteristics (Q, H and n) 2. Draw the system (load) characteristic (Q, Hı) where Hj = hs + hlosses 3. Identify the duty point of the pump 4. Identify the maximum discharge and efficiency at the duty point 5. Calculate the input power required at the identified efficiency (remember P = PgQH Active Go to
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Question is solved below
Q (l/s) Q (m3/s) H (m) Efficiency (%) H system ( m3/s) 0 0 10 0 5.50 10 0.01 10.5 0.4 5.58 20 0.02 10 0.64 5.83 30 0.03 8.5 0.72 6.25 40 0.04 6 0.64 6.83 50 0.05 2.5 0.4 7.58
= -80042 +48Q - 2x1015 Criven : hs = static head - potential head = 5.5m pipe length L=20m, diameter D = 100mm = 0.1m friction foreton f-0.005 Drawing pump characteritie cun ve using given table data get following equations HE - 50002% + 100 2 +10 We Calwating system Chanaelewitic eyn H= Ahst hross hloss=fLQ² 12DS 12 2 H=5.5+ 833.333 22 3 = 0.005 % 20% 2 12 *[0.115 equating ean and ③ we get hloss = 833.3332 5833.3332-1002-4.510 - 0 Solving this we get mor- dischange amane = 0.037638 m/s = 37.638 445 Rump efficiency at Qonax from equation ③ n = 0.6733 à 67.33% The input power required PgQH =p= n =100049.81X 0.037638 X 6.6805 0.6733 P= 3.663 KWHm -H system (m3/s) Efficiency (%) 12 0.8 0.7 10 Duty point 0.6 8 0.5 Head (m) 6 0.4 Efficiency 0.3 4 0.2 2 0.1 0 0 0 10 20 30 40 50 60 Q(1/5)
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