E4.14 Pump Work

Example 4.14: A pump with an overall efficiency of 60 % pumps an acid (SG 1.8) from an open tank to a process column at a rate of 18 lb/s. The column operates at 19.65 psia and the fluid is fed into the column with a velocity of 8 ft/s at a point 60 ft above the acid surface level in the tank. Determine the power required to run the pump if the energy losses are equivalent to 9 ft of water head. Density of acid is given as 112.37 lb/ft³.

Solution:

If acid is stored in a large tank, V₁ can be considered to be zero.

Consider the level of acid in the supply tank Z₁ as datum

Elevation of the discharge point Z₂ = 60 ft.

Frictional losses expressed in terms of acid-head are 9/1.8 = 5 ft·lb_f/lb.

Using these values in the Bernoulli equation, we can get the amount of work required as 72.34 ft·lb_f/lb. The mass flow rate of the fluid, m' = 18 lb/s. Ideal pump work called water horsepower can be calculated as.

This work can be converted into more recognizable units. Its value is 2.37 hp. If the overall efficiency of the pump is η, then the actual power requirement, brake horsepower (BHP), can be calculated as follows: