The horizontal pump in Fig. P3.178 discharges 20 ° C water

at 57 m ^{3} /h. Neglecting losses, what power in kW is delivered

to the water by the pump?

Community Answer

Step 1Given:The volumetric flow rate, Q = 57 m3/hThe diameter of pipe 1, d1 = 9 cmThe diameter of the pipe 2, d2 = 3 cm  Step 2The expression for velocity of flow through pipe 1,{:[V_(1)=(Q)/(A_(1))],[=(4Q)/(pid_(1)^(2))],[=(4(57m^(3)//h((1m^(3)//s)/(3600m^(3)//h))))/(pi(9(cm)((10^(-2)(m))/(1(cm))))^(2))],[=2.48m//s]:} Step 3The expression for velocity of flow through pipe 2,{:[V_(2)=(Q)/(A_(2))],[=(4Q)/(pid_(2)^(2))],[=(4(57m^(3)//h((1m^(3)//s)/(3600m^(3)//h))))/(pi(3(cm)((10^(-2)(m))/(1(cm))))^(2))],[=22.35m//s]:}Step 4Apply ene ... See the full answer