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Chapter 10: Problem 1

A well is pumping water from an aquifer with a water table 30 feet below ground surface (bgs) to a tank 150 feet above the well. If the well flows \(1,000 \mathrm{gpm}\), what is the required horsepower? (Assume the wire-to- water efficiency is \(68 \% .)\)

Short Answer

Expert verified
The required horsepower is approximately 66.91 HP.

Step by step solution

01

Calculate the Total Head

The total head is the distance the water is lifted. Add the depth below the ground surface (30 feet) to the height above the well (150 feet): \[ \text{Total Head} = 30 + 150 = 180 \text{ feet} \]
02

Convert Flow Rate to Cubic Feet per Second (cfs)

Convert the flow rate from gallons per minute (gpm) to cubic feet per second (cfs). Use the conversion factor: \[ 1 \text{ gpm} = 0.002228 \text{ cfs} \]\[1,000 \text{ gpm} = 1,000 \times 0.002228 = 2.228 \text{ cfs} \]
03

Calculate Water Horsepower (WHP)

Water horsepower can be calculated using the formula: \[ \text{WHP} = \frac{\text{Flow Rate (cfs)} \times \text{Total Head (feet)} \times \text{Weight of Water (lb/ft}^3)}{550} \]Given the weight of water is 62.4 lb/ft³: \[ \text{WHP} = \frac{2.228 \text{ cfs} \times 180 \text{ feet} \times 62.4 \text{ lb/ft}^3}{550} \] \[ \text{WHP} = \frac{25,018.752}{550} \] \[ \text{WHP} \thickapprox 45.49 \text{ HP} \]
04

Calculate Brake Horsepower (BHP)

To find the brake horsepower, divide the water horsepower by the efficiency (in decimal form). Given the efficiency is 68%, or 0.68: \[ \text{BHP} = \frac{\text{WHP}}{Efficiency} = \frac{45.49}{0.68} \] \[ \text{BHP} \thickapprox 66.91 \text{ HP} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Total Head Calculation
To determine the water pumping horsepower, start with the total head calculation. The total head represents the vertical distance the water needs to travel. This distance includes both the depth of the well and the height to which the water is pumped. In our example, the well is 30 feet deep, and the water is being pumped to a height of 150 feet. So, the total head is calculated as follows: \[ \text{Total Head} = 30 \text{ feet} + 150 \text{ feet} = 180 \text{ feet} \]
Understanding the total head is crucial because it impacts the energy required to pump the water. A larger total head value indicates more energy is needed.
Flow Rate Conversion
The flow rate must be converted into the correct units for calculations. We often measure water flow in gallons per minute (gpm). However, since we use cubic feet per second (cfs) for the horsepower formula, we need to convert units. One gallon per minute equals 0.002228 cubic feet per second. For example, for a flow rate of 1,000 gpm:
\[ 1,000 \text{ gpm} \times 0.002228 = 2.228 \text{ cfs} \]
This conversion ensures that all units match those required in the subsequent formulas.
Water Horsepower Calculation
Water horsepower (WHP) is a measure of the actual power needed to lift water to the desired height at the given flow rate. To find the WHP, use the formula:
\[ \text{WHP} = \frac{\text{Flow Rate (cfs)} \times \text{Total Head (feet)} \times \text{Weight of Water (lb/ft}^3)}{550} \]
In this example, given the flow rate is 2.228 cfs, total head is 180 feet, and weight of water is 62.4 lb/ft³:
\[ \text{WHP} = \frac{2.228 \times 180 \times 62.4}{550} = 45.49 \text{ HP} \]
WHP helps determine the base power needed without considering system losses.
Efficiency in Pump Systems
Pump efficiency accounts for energy losses within the pump system. It’s the ratio of the water horsepower to the power supplied to the pump (brake horsepower). Efficiency is given in percentages but is used in calculations in decimal form. In our example, the pump efficiency is 68%, or 0.68. Efficient systems use less energy for the same pumping task, saving costs.
Brake Horsepower Calculation
Brake horsepower (BHP) is the actual power required, considering the system’s efficiency. It’s the water horsepower divided by the efficiency. For our problem, where WHP is 45.49 HP and efficiency is 0.68:
\[ \text{BHP} = \frac{45.49}{0.68} \approx 66.91 \text{ HP} \]
BHP ensures you select a pump motor that can handle the required load, factoring in system losses. It reflects the real-world power needs of the pump.

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Most popular questions from this chapter

What is the water horsepower needed for a well that pumps 2,050 gpm against a pressure of 150 psi?

A well draws water from an aquifer that has an average water level of \(150 \mathrm{ft}\) bgs and pumps to a tank \(225 \mathrm{ft}\) above it. Friction loss to the tank is approximately 22 psi. If the well pumps at a rate of 2,300 gpm and has a wire-to-water efficiency of \(62 \%\), how much will it cost to run this well 14 hours per day. Assume the electrical rate is \(\$ 0.13\) per \(\mathrm{kW}\) Hr.

What is the motor horsepower needed to pump 2,420 AF of water over a year with an average daily pumping operation of 12 hours? Assume the pump is pumping against 95 psi and has a pump efficiency of \(70 \%\) and a motor efficiency of \(80 \%\).

What is the required water horsepower for 100 gpm and a total head pressure of \(50 \mathrm{ft}\) ?

A utility manager is trying to determine which hp motor to purchase for a pump station. A 400 hp motor with a wire-to-water efficiency of \(65 \%\) can pump 3,000 gpm. Similarly, a 250 hp motor with a wire-to-water efficiency of \(75 \%\) can pump 2,050 gpm. With an electrical rate of \(\$ 0.155\) per \(\mathrm{kW}-\mathrm{Hr}\), how much would it cost to run each motor to achieve a daily flow of 2 MG? Which one is less expensive to run?
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