Question

500 lbm/min hydrocarbon feed is pumped into the feed point halfway to the top of a distillation tower. The tower feed point is 40.0 feet above holding tank for the hydrocarbon. The pressure in the holding tank is 25.0 psia. The pressure in the tower at the feed point is 85.0 psia. Assume the energy lost due to friction is negligible. There is no significant change in velocity. The hydrocarbon density is 55.0 lbm/ft 3 . Calculate the pump power required in hp and watts.

Answer 1

500 lbs/min = 13.5 m³/h

height = 40 feet = 12.192 m

P_h(kW) = q ρ g h / (3.6 10⁶)            

where

P_h(kW) = hydraulic power (kW)

q = flow capacity (m³/h)

ρ = density of fluid (kg/m³)

g = gravity (9.81 m/s²)

h = differential head (m)

The hydraulic Horse Power can be calculated as:

P_h(hp) = P_h(kW) / 0.746   

where

P_h(hp) = hydraulic horsepower (hp)

P_h(KW) = [13.5 m³/h x 881.015 kg/m³ x 9.81 m/s² x 12.192 m] / (3.6 x10⁶)

P_h(KW) = 0.3951 KW = 395.1 W power at the feed of tower

HP = 0.5296 HP

HP = (v x pressure / 1714 ) { v = volumetric flow rate in gallons/min, pressure in psi }

at feed point { pressure 85 psi and v = 59.88 gallon/min}

HP = [59.88 gal/min x 85 psi ] / 1714 = 2.9695 HP

in tank {p = 25 psi}

HP = [59.88 gal/min x 25 psi ] / 1714 = 0.8734 HP

total pressure =  0.8734 HP + 2.9695 HP + 0.5296 HP

Ptotal =4.37 HP

Ptotal = 3.262 KW = 3262 W