Fundamental Chemistry and Hydraulics - Mathematics and Statistics for the Environment - page 372

can have C factors very much lower than this. To determine the C factor, published tables are usually
used. In addition, when the friction losses for fittings are factored in, other published tables are avail-
able to make the proper determinations. It is standard practice to calculate the head loss from fittings
by substituting the equivalent length of pipe , which is also available from published tables.

14.2.2 b asiC p umping C alCulations *
Certain computations used for determining various pumping parameters are important to environ-
mental practitioners responsible for water/wastewater treatment plant operation.

Key Point: The rate of flow produced by a pump is expressed as the volume of water pumped during
a given period.

14.2.2.1 Pumping Rates
The mathematical problems most often encountered by water/wastewater operators with regard to
determining pumping rates are often determined by using either of the following equations:

Pumping rate (gpm) = gallons/minutes (14.20)

Pumping rate (gph) = gallons/hours (14.21)

■ EXAMPLE 14. 31
Problem: The meter on the discharge side of the pump reads in hundreds of gallons. If the meter
shows a reading of 110 at 2:00 p.m. and 320 at 2:30 p.m., what is the pumping rate expressed in
gallons per minute?

Solution: The problem asks for pumping rate in gallons per minute (gpm), so we use Equation 14.20:

Pumping rate (gpm) = gallons/minutes

To solve this problem, we must first find the total gallons pumped (determined from the meter
readings):

32,000 gal - 11,000 gal = 21,000 gal

The volume was pumped between 2:00 p.m. and 2:30 p.m., for a total of 30 minutes. From this
information, calculate the gpm pumping rate:

21,000 gal
30 min

Pumpingrate = = 700 gpm

■ EXAMPLE 14. 32
Problem: During a 15-minute pumping test, 16,400 gal were pumped into an empty rectangular
tank. What is the pumping rate in gallons per minute?

Solution: The problem asks for the pumping rate in gallons per minute, so again we use Equation
14.20:

16,400 gal
15 min

Pumpingrate = = 1093 gpm (rou nded)

* The following examples are adapted from Wahren, U., Practical Introduction to Pumping Technology , Gulf Publishing,
Houston, TX, 1997.