Question

A water pump delivers 6 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the pump is measured to be 1.2 psi when the flow rate is \(15 \mathrm{ft}^{3} / \mathrm{s}\) and the changes in velocity and elevation are negligible, determine the mechanical efficiency of this pump.

Short answer

Answer: The mechanical efficiency of the water pump is approximately 78.53%.

Step-by-step solution

Convert the given information into consistent units

We are given the shaft power \(P_s = 6\,\text{hp}\), the pressure differential \(\Delta p = 1.2\,\text{psi}\), and the flow rate \(Q = 15\,\frac{\text{ft}^3}{\text{s}}\). We need to convert these values into consistent units (SI units) to perform our calculations. 1 hp = 745.7 Watt (W), 1 psi = 6895 Pa (Pascal), 1 ft³/s = 0.0283168 m³/s. So, \(P_s = 6\,\text{hp} \times 745.7\,\frac{\text{W}}{\text{hp}} = 4474.2\,\text{W}\), \(\Delta p = 1.2\,\text{psi} \times 6895\,\frac{\text{Pa}}{\text{psi}} = 8274\,\text{Pa}\), and \(Q = 15\,\frac{\text{ft}^3}{\text{s}} \times 0.0283168\,\frac{\text{m}^3}{\text{ft}^3} = 0.424752\,\frac{\text{m}^3}{\text{s}}\). Now that we have the values in consistent units, we can proceed to calculate the water power.

Calculate the water power

Using the formula \(P_w = \Delta p \times Q\), we can calculate the water power: \(P_w = \Delta p \times Q = (8274\,\text{Pa}) \times (0.424752\,\frac{\text{m}^3}{\text{s}}) = 3515.32\,\text{W}\) The water power is calculated to be \(3515.32\,\text{W}\).

Calculate the mechanical efficiency

Using the formula \(\eta_{mech}=\frac{P_w}{P_s}\), we can now calculate the mechanical efficiency of the pump: \(\eta_{mech}=\frac{P_w}{P_s} = \frac{3515.32\,\text{W}}{4474.2\,\text{W}} = 0.78526\) To get the result in percentage, we multiply by 100: Mechanical efficiency = \(\eta_{mech} \times 100\% = 0.78526 \times 100\% = 78.526\%\) The mechanical efficiency of the pump is approximately \(78.53\%\).

Key concepts

Shaft Power

Shaft power is the amount of power that is provided by the motor or engine to a pump's shaft for it to operate. In our exercise, the water pump is delivering 6 horsepower (hp) of shaft power. Since we often work with the metric system in scientific calculations, it's necessary to convert this power into watts (W), which is a unit of power in the SI system. Understanding how to convert shaft power from one unit to another is crucial because it allows for consistent calculations and comparisons across different systems and equipment.

When contemplating about pumps, remember that the shaft power is the input power that the pump needs to overcome various losses such as friction within the pump's mechanisms and to transfer energy to the fluid being moved.

Pressure Differential

Pressure differential is the difference in fluid pressure between two points, often an outlet and inlet of a pump in our context. In the given problem, the pressure differential is 1.2 pounds per square inch (psi). This pressure differential is important as it is one of the key factors that defines how much work the pump can do in moving the fluid through the system. Since calculations involving fluid dynamics are typically performed in the SI units, we must convert psi to pascals (Pa) to maintain consistency.

Knowing the pressure differential helps us understand the amount of force that is exerted by the pump on the fluid to move it, which is a vital part of determining the pump's overall performance.

Flow Rate

Flow rate refers to the volume of fluid that passes through a given point in a system per unit time. In this exercise, the flow rate is given as 15 cubic feet per second (ft³/s). This measure tells us how fast the pump is moving water through the system. For the purpose of precision in calculations, flow rate is also converted to cubic meters per second (m³/s). Understanding flow rate is fundamental when assessing a pump's capacity and efficiency, as it directly impacts the overall effectiveness of a pump in a fluid system.

Unit Conversion

Unit conversion is an essential step in solving many physics and engineering problems including those involving pumps. Because different regions and fields use different measurement systems, having the ability to convert units like horsepower to watts, psi to pascals, and cubic feet per second to cubic meters per second is crucial. Proper unit conversion ensures accuracy in calculations and allows for a standardized way to communicate and compare results.

Always double-check your conversion factors, as an error in conversion can greatly affect the final outcome of a problem. In our example, careful unit conversion enables us to correctly calculate the water power and mechanical efficiency of the pump.

Pump Efficiency Calculation

To determine the mechanical efficiency of a pump, we compare the power imparted to the fluid (water power) to the power supplied to the pump's shaft. This ratio is expressed as a percentage and gives us the efficiency of the pump. The calculation involves determining the actual work output (water power) and dividing it by the input work (shaft power), which we have after performing the necessary unit conversions.

It is important to be precise in these calculations, as the efficiency rating helps in evaluating the performance of the pump and its suitability for a particular application. High-efficiency pumps save energy, reduce operational costs, and can significantly impact the overall system reliability.