Case: Specific speed and pumping arrangement

Example 4.20: Select the specific speed of the pump or pumps, required to lift 5835 gpm of water 350 ft through 9400 ft of 2.5 ft-diameter pipe (f = 0.0055). The pump rotating speed is to be 1700 rpm. Consider the following cases: single pump, two pumps in series, two pumps in parallel.

Solution:

C Knowing the value of volumetric flow rate, and the diameter of the pipe we can find the velocity of the fluid in the pipe.

C Knowing the additional information about friction factor, and the length of the pipe we can find skin friction losses through the pipe as

C The elevation difference between points A and B is 350 feet so head required is

C Operating speed of the pump is, N, 1700 rpm. Now specific speed of the pump can be calculated as follows.

C For a single pump, specific speed can be found to be 1575 rpm.

C For two pumps in series, H will be reduced to half = 359/2 = 179.5 and N_s in this case can be found to be 2648 rpm.

C For two pumps in parallel, Q will be reduced to half of original value = 2917.5 gpm. N_s can be calculated to be 1113 rpm.