Water rho 998 kg m3 and mu 0001002 Pa s flows from a large

Water (rho = 998 kg m^3 and mu = 0.001002 Pa s) flows from a large tank through the pipe system shown in the diagram at a flow rate of 6.2 kg/s. The pressure above the liquid in the tank is maintained at 200 kPa(gauge). The pipe from the tank is 2-inch schedule 40 pipe made of commercial steel. A pump delivers 1000 W of power at an efficiency of 75%. The exit of the pipe is open to the atmosphere. To what height above the liquid level in the tank can the water be pumped (H in m)? Include all friction losses in your calculation Assume that the entrance from the tank and the exit to the atmosphere arc sharp. (12.5 plusminus 0.6 m)

Solution

A) if no losses due to friction and pipe bends, constrictions etc,

power = m\'gH = 1000

6.2*9.81*H = 1000,==> H =16.44 m

velocity of flow thru 2\" pipe : 3.065 m/s

pressure head - velocity head = 20.43-.48 =19.95 m

given 2 90deg elbows and 2 45deg elbows, with a gate valve, must deduct head losses due to these

hf = k V²/2g head loss by friction

k gate valve = f*(l/d)= f*13

90 elbow k= 30

45 elbow k=16

after calculation for f =.05, v=3.065m/s get frictionhead of 2.2 m,

for gate valve ,loss = .3 m

loss total 2.5 m

Consider known qty Pressure head-velocity head = pumping head = 19.95 m, with 75% efficiency this i s 14.96 m

subtract friction head total of 2.5 m , get 12.5 m