8.3: Flow Rate
 Page ID
 7180
Flow rate is the measurement of a volume of liquid (i.e., water) which passes through a given crosssectional area (i.e., pipe) per unit in time. In the waterworks industry, flow rates are expressed in several different units. The most common ones are shown below.
 Flow Rates
 cfs = cubic feet/second
 gpm = gallons/minute
 MGD = million gallons/day
Depending on the application, flow rates are expressed in these or potentially other units. For example, the flow rate from a well or booster pump is commonly expressed as gpm, whereas annual production might be expressed as acrefeet per year (AFY). However, when solving a problem for flow rate the common unit of expression is cfs. The reason for this is in part due to the measurement of unit area of the structure that the water is passing through (i.e., pipe, culvert, aqueduct, etc.) The areas for these structures are typically expressed as square feet (ft^{2}). In addition, the speed (distance over time) at which the water is flowing is commonly expressed as feet per second. The flow rate formula and how the units are expressed are shown in the example below.
 Flow Rate = Area x Velocity
 Flow Rate (Q) = Area (A) x Velocity (V)
 Q = A x V
 Q (cubic feet/sec) = Area (ft^{2}) x Velocity (feet/sec)
Understanding flow rates and velocities can help with the design on pipe sizes for wells, pump stations, and treatment plants. With the understanding that velocities are typically in the range of 2 – 7 feet per second and the known flow rate, pipe diameters can be calculated. For example, if a new well is being drilled and the pump test data determines that the well can produce a specific flow, let’s say 1,500 gpm, and you do not want the velocity to exceed 6.5 fps, the required diameter of the pipe can be determined (see below).
Example
Q = A x V or for this example A = Q/V since the flow rate (Q = 1,500 gpm) and velocity (V = 6.5 fps) are known.
The first step is to make sure the “known” values are in the correct units.
Velocity given at 6.5 fps is in the correct unit. However, the flow rate given in gpm needs to be converted to cfs.
 1,500 gpm ÷ 448.8 = 3.34 cfs
Now that both values are in their correct unit, divide the two to get the unknown value, in this case, the Area (A).
 3.34 cfs/6.5 fps = 0.51 ft^{2}
Knowing that the Area is 0.51 ft^{2} and that the formula for Area is 0.785 x D^{2}, the diameter of the pipe can be calculated.
 0.51 ft^{2} = 0.785 x D^{2}
 D^{2} = 0.51 ft/20.785
 D^{2} = 0.66 ft^{2}
In order to find the value of the diameter (D), you must take the square root of D^{2}.
 D^{2} = 0.66 ft^{2}
 D = 0.81 ft
Since pipe diameters are typically expressed in inches, multiply the answer by 12.
 0.81 ft x 12 in = 9.7 or 10 inches
Exercises
Solve the following problems.

What is the flow rate in MGD of a 24” diameter pipe with a velocity of 3 fps?

What is the velocity through a box culvert that is 3 feet wide and 2 feet deep if the daily flow is 27 AF?

What is the area of a pipe that flows 1.5 MGD and has a velocity of 5 fps?

What is the diameter of a pipe that flows 2,500 gpm with a velocity of approximately 7 fps?

A 15mile aqueduct flows 30,000 AFY at an average velocity of 0.45 fps. If the distance across the top is 13 feet and the depth is 8 feet, what is the distance across the bottom?