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Module 7: Horsepower and
Efficiency
AWWA
Hydraulics
Introduction
When you complete this lesson, you should be able to:
 Evaluate pumping costs based on efficiencies of the motor and pump. Work and Horsepower
The term work is defined as the operation of a force over a specific distance‐‐for example, lifting a one‐pound object one foot. Thus, the amount of work done is measured in foot‐
pounds (ft‐lb).
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Work and Horsepower
 The rate of doing work‐‐that is, the measure of how much work is done in a given time‐‐is called power. Therefore, to make power calculations, you must know the time required to perform the work. The basic unit for measuring power is foot‐pounds per minute (ft‐lb/min).
 One equation for calculating power in foot‐pounds per minute is You will often work with measurements of power expressed in horsepower (hp), which is related to foot‐pounds per minute by the conversion equation
 We use horsepower because the numbers in ft‐lbs/min become unmanageably large. Calculations are generally more convenient with smaller numbers. Work and Horsepower
Work and Horsepower
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Work and Horsepower
 The conversion equation 1 hp = 33,000 ft‐lb/min is called water horsepower (whp) because it is the amount of horsepower required to lift water. Another equation for determining pumping power requirements is this equation:  Remember, whp is the horsepower required for pumping water.
Horsepower Examples
Efficiency
 Pumps are typically powered by electric motors. As an electric motor converts electrical motor horsepower (mhp) to mechanical brake horsepower (bhp) in driving the pump impeller, friction within the drive motor and other factors cause useful power to be lost. This loss of power is expressed as an inefficiency.
 Inefficiencies also occur from the friction of water against a rotating pump impeller as it imparts flow and pressure energy to the water. Depending on the size and type, motors are typically 80 to 95 percent efficient and pumps are typically 50 to 85 percent efficient. 3
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Efficiency

The inefficiencies associated with the electric drive motor are expressed as a motor efficiency that is less than 100 percent. The inefficiencies associated with the pump are expressed as a pump efficiency less than 100 percent. The motor and pump efficiency may be combined mathematically by multiplication, arriving at what is known as an overall wire‐
to‐water efficiency. 
In reality, motor and pump efficiencies must be accounted for in sizing a pumping system. This assures that the pump and motor are big enough to provide adequate power to move the water at the desired flow rate.
Efficiency
In the calculation of water horsepower (whp), pumping system inefficiencies and fluids with specific gravities different than water influence the outcome. In these cases, you must modify the basic whp equation.
Expanded
Efficiency Examples
Example1
 If a pump is to deliver 480 gpm of water against a total head of 105 ft, and the pump has an efficiency of 80 percent, what brake horsepower must be supplied to the pump?
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Efficiency Examples
Example2
 For the system on the previous page, the pump is to deliver 480 gpm of water against a total head of 105 ft, it has an efficiency of 80 percent, and 15.9 hp of brake horsepower must be supplied to the pump. What is the motor hp required if the same system has a motor efficiency of 85 percent?
Efficiency Examples
Example3
 Now, what is the overall wire‐to‐water efficiency of the same system? Remember that the pump is to deliver 480 gpm of water against a total head of 105 ft, it has an efficiency of 80 percent, 15.9 hp of brake horsepower must be supplied to the pump, the motor efficiency is 85 percent, and the motor horsepower is 18.7 mhp.
Efficiency Examples
Example3
 The wire‐to‐water efficiency of this system can be calculated two ways:
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Efficiency Examples
This animation demonstrates how to calculate the efficiencies when the motor, brake, and water horsepowers are known. It starts with the general equation used on the previous page, which explains the relationship of efficiency across the equipment (wire‐to‐water efficiency) to hp output from the equipment and hp supplied to the equipment.
Efficiency Examples
Efficiency Examples
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Efficiency Examples
Efficiency Examples
Efficiency Examples
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Efficiency Examples
Efficiency Examples
Electrical Power and Pumping Costs
Power input to a motor is usually expressed in terms of electrical power rather than horsepower. Electrical power is usually expressed in units of watts or kilowatts.
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Electrical Power and Pumping Costs
Power is generally sold in units of kilowatt‐hours (kW∙h). If a motor draws one kW of power and runs for one hour, then the electric company will charge for one kW∙h. Therefore, to calculate pumping costs, you will need to know the power requirements (power demand) of the motor and the length of time the motor runs. In most situations, you will know, or be able to determine, the total head and flow rate against which the pump is working, so the water horsepower can be calculated. Then, using the wire‐to‐water efficiency of the motor and pump, you can determine the motor horsepower and the kilowatts of power demand. 9