Given The figure shows a pump recirculating 200 galmin of wa

Given: 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

Flow rate Q = 200 gal/min = 0.4456 ft³/s

Inlet pipe cross-section A_in = pi/4 * D_in² = 3.14/4 * (4/12)² = 0.0872 ft²

Discharge pipe cross-section A_out = pi/4 * D_out² = 3.14/4 * (3/12)² = 0.0491 ft²

Velocity in inlet pipe V₁ = Q / A_in = 0.4456 / 0.0872 = 5.11 ft/s

Velocity in discharge pipe V₂ = Q / A_out = 0.4456 / 0.0491 = 9.08 ft/s

For water at 70F, kinematic viscosity, neu = 1.052*10^-5 ft²/s

Reynolds number in inlet pipe Re₁ = V₁*D_in / neu = 5.11*(4/12) / (1.052*10^-5) = 161917

Reynolds number in discharge pipe Re₂ = V₂*D_out / neu = 9.08*(3/12) / (1.052*10^-5) = 215779

For commercial steel pipe, roughness e = 0.00015 ft

In inlet pipe, relative roughness e/D_in = 0.00015 / (4/12) = 0.00045

In discharge pipe, relative roughness e/D_out = 0.00015 / (3/12) = 0.0006

From Moody diagram, for Re₁ = 161917 and e/D_in = 0.00045, we get friction factor f₁ = 0.019

From Moody diagram, for Re₂ = 215779 and e/D_out = 0.0006, we get friction factor f₂ = 0.0193

For threaded 90 deg elbow, loss coeff K = 1.5

For rounded pipe entrance, K = 0.04

For pipe exit, K = 1

Total inlet pipe length L₁ = 10 + 14 = 24 ft

Total discharge pipe length L₂ = 14 + 20 + 14 + 14 + (6 - (22-20)/2) = 67 ft

Head at inlet of pump = [15 + V₁² / (2g)] - (0.04 + 1.5 + f₁*L₁/D_in) V₁² / (2g)

Head at inlet of pump = 15 - (0.04 + 1.5 + 0.019*24/(4/12) - 1) 5.11² / (2*32.2)

Head at inlet of pump = 14.226 ft

Pressure at inlet of pump = 14.226*rho = 14.226*62.4 = 887.72 lb/ft² = 6.16 psi (gage)....Ans (b)

For the discharge line, 14.226 + H_p - (3*1.5 + 15 + 1 + f₂*L₂ / D_out)*V₂² / (2g) - 22 + 6 = 0 + V₂² / (2g)

14.226 + H_p - (3*1.5 + 15 + 1 + f₂*L₂ / D_out - 1)*V₂² / (2g) - 22 + 6 = 0

14.226 + H_p - (3*1.5 + 15 + 1 + 0.0193*67 / (3/12) - 1)*9.08² / (2*32.2) - 22 + 6 = 0

H_p = 33.36 ft............Ans (a)