The figure shows a pump recirculating 200 galmin of water at

The figure shows a pump re-circulating 200 gal/min of water at 70 degree F through a cooling loop made of commercial steel pipe. The reservoir is open to atmosphere. The loss factor for the heat exchanger is 15. Assume the elbows are smooth 90 degree and threaded. Find: Determine the head added by the pump to the fluid at these conditions Determine the pressure just before the inlet to the pump.

Solution

b) The pressure just at the inlet of the pump = atmospheric pressure + pressure due the water column in the suction line – pressure loss due to friction in the pipe

Atmospheric pressure = 214105 lb_f/ft²

Pressure due the water column in the suction line = *g*h = 62.3*32.2*15 = 30091 lb_f/ft²

Diameter of the pipe in the suction line, D₁ = 4 in = 0.33 ft

Cross-section of the discharge line, A₁ = (/4)* D₁² = (/4)* (0.33)² = 0.085 ft²

Rate of flow, Q = 200 gal/min = (200*0.13368)/60 ft³/s = 0.45 ft³/s [1 gal = 0.13368 ft³ ]

Average velocity of flow in the pipe, V = 0.45/0.085 = 5.29 ft/s

Reynolds Number, R_e = (V D₁)/ = (62.3*5.29*0.33)/ (2.03*10^-5)

[viscosity of water = 2.03*10^-5 lb_f-s/ ft²]

= 5357493 > 4000

So, the flow is a turbulent flow.

Friction factor, f = 0.0791 /( R_e)^0.25 = 0.0791 /(5357493)^0.25 = 0.0016

Loss in pressure due to friction = (4fLV²)/2 = (4*62.3*0.0016*24*5.29²)/2 = 133.89 lb_f/ft²

The pressure just at the inlet of the pump = 214105 + 30091– 133.89 = 244062 lb_f/ft²