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10.1: The Power of Water

  • Page ID
    7184
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    Previously in this text, we discussed the theory of pressure in both feet (head pressure) and psi (pounds per square inch.) In this chapter, we will look at the “power” requirements to move water with pumps and motors. How does water get to the customer’s home? Water pressure is typically provided to customers from elevation (above ground tanks, reservoirs, elevated storage tanks, etc.) But, how does the water get to these storage structures? This is where the concept of horsepower comes in. Historically, the definition of horsepower was the ability of a horse to perform heavy tasks such as turning a mill wheel or drawing a load. It wasn’t until James Watt (1736‐ 1819) invented the first efficient steam engine that horsepower was used as a standard to which the power of an engine could be meaningfully compared. Watt's standard of comparing “work” to horsepower (hp) is commonly used for rating engines, turbines, electric motors, and water‐power devices.

    In the water industry, there are three commonly used terms to define the amount of hp needed to move water: Water Horsepower, Brake Horsepower, and Motor Horsepower.

    Water Horsepower is a measure of water power. The falling of 33,000 pounds of water over a distance of one foot in one minute produces one horsepower. It is the actual power of moving water.

    \[\text{Water hp} = \dfrac{(\text{flow rate in gallons per minute})(\text{total head in feet})}{3,960} \label{hp}\]

    The above equation is used to calculate the power needed to move a certain flow of water a certain height. The constant 3,960 is the result of converting the 33,000 ft-lb/min with the weight of water flow. For example, instead of using gallons per minute, pounds per minute would be needed because 33,000 is in foot-pounds.

    Water horsepower is the theoretical power needed to move water. In order to actually perform the work a pump and motor are needed. However, neither the pump nor the motor are 100% efficient. There are friction losses with each. The horsepower required by the pump (brake horsepower) can be calculated, but the actual horsepower needed looks at the efficiencies of both the pump and the motor. This efficiency is termed the wire-to-water efficiency. The formula below shows brake horsepower and motor horsepower which includes the combined pump and motor inefficiencies.

    \[\text{Brake hp} = \dfrac{(\text{flow rate in gallons per minute})(\text{total head in feet})}{3,960}(\text{pump efficiency %}) \label{bhp}\]

    \[\text{Motor hp} = \dfrac{(\text{flow rate in gallons per minute})(\text{total head in feet})}{3,960}(\text{pump efficiency %})(\text{motor efficiency %}) \label{mhp}\]

    If the pump and the motor were both 100% efficient, then the resulting answer would be 100% x 100% or 1. Hence, the actual horsepower would be the water horsepower and the equation is not affected. However, this is never the case. Typically there are inefficiencies with both components.

    • Pump Efficiency = 60%
    • Motor Efficiency = 80%
    • 0.6 x 0.8 = 0.48 or 48% efficient

    As with all water-related math problems, it is important for the numbers being used to be in the correct units. For example, the flow needs to be in gallons per minute (gpm) and the total head in feet (ft). These will not always be the units provided in the questions. The example below demonstrates this concept.

    Example \(\PageIndex{1}\)

    What is the horsepower of a well that pumps 2.16 million gallons per day (MGD) against a head pressure of 100 pounds per square inch (psi)? Assume that the pump has an efficiency of 65% and the motor 85%.

    Solution

    In this example, the flow is given in MGD and the pressure in psi. The appropriate conversions need to take place before the horsepower (hp) is calculated.

    • 2.16 MGD ÷ 1440 min/day = 1,500 gpm
    • 100 psi x 2.31 ft/psi = 231 ft

    Now, these numbers can be plugged into the hp formula (Equation \ref{mhp}).

    \[\text{hp} = \dfrac{(1,500\, gpm)(231\, ft)}{3,960}(65 \%)(85\%) \nonumber\]

    Make sure to convert the efficiency percentages to decimals before solving.

    \[\begin{align*} \text{hp} &= \dfrac{(1,500 \,gpm)(231 ft)}{3,960}(0.65)(0.85) \\[4pt] &= 158 \,hp \end{align*}\]

    Exercises

    Calculate the required horsepower related questions.


    10.1: The Power of Water is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.