Problem 4.4.2

A pump with an overall efficiency of 60 percent pumps an acid from an open tank to a process column at the rate of 18 pounds per second. The column operates at 19.65 psi. Fluid (an acid) enters the column at a velocity of 8 foot per second at a point 60 feet above the acid level in the tank. Determine the power required to run the pump if the energy losses are equivalent to 9 water of water head. Density of acid is 112.37 pounds per cubic foot.

Solution:

Consider that the open tank is big enough that velocity of fluid at the surface (point 1) is zero. Point 2 is chosen where acid is being discharged at 8 foot per second. Energy losses are given as 9 foot pound-force per pound-mass of water. Adjusting for density, energy loss is 9 times 62.43 divided by 113.27 or 5 foot pound-force per pound-mass of acid.

Now we can apply Bernoulli’s equation: P₁ is 1 atmosphere or 14.696 psi or 2116.2 psf. Pump work is calculated as 72.34 foot pound-force per pound-mass. Note that 1 horse-power is equal to 550 foot pound-force per second. Water horse power (theoretical requirement) is 2.37 hp. Pump is 60 percent efficient, so brake horse power is 2.37over 0.6 or 3.95 hp.