Question

Water is pumped from a lake to a storage tank \(15 \mathrm{m}\) above at a rate of \(70 \mathrm{L} / \mathrm{s}\) while consuming \(15.4 \mathrm{kW}\) of electric power. Disregarding any frictional losses in the pipes and any changes in kinetic energy, determine ( \(a\) ) the overall efficiency of the pump-motor unit and ( \(b\) ) the pressure difference between the inlet and the exit of the pump.

Short answer

Question: Calculate the overall efficiency of the pump-motor unit and the pressure difference between the inlet and exit of the pump, considering the information given in the solution. Answer: The overall efficiency of the pump-motor unit is 66.7%, and the pressure difference between the inlet and exit of the pump is 147150 Pa.

Step-by-step solution

Calculate the ideal mechanical power required to pump the water

The ideal mechanical power required to pump the water can be given by the formula: \(P_m = \rho q g h\) where \(P_m\) is the mechanical power, \(\rho\) is the density of water, \(q\) is the flow rate in cubic meters per second, \(g\) is the acceleration due to gravity, and \(h\) is the height difference between the lake and the storage tank. First, we need to convert the flow rate from liters per second to cubic meters per second: \(q = 70\, \text{L/s} = 70 / 1000\, \text{m}^3/\text{s} = 0.07\, \text{m}^3/\text{s}\) Now, plug in the values: \(\rho = 1000\, \text{kg/m}^3\) (density of water) \(g = 9.81\, \text{m/s}^2\) (acceleration due to gravity) \(h= 15\, \text{m}\) (height difference) \(P_m = 1000 \times 0.07 \times 9.81 \times 15 = 10273\, \text{W}\)

Calculate the overall efficiency of the pump-motor unit

The overall efficiency of the pump-motor unit can be calculated using the formula: \(\text{Efficiency} = \frac{\text{Ideal mechanical power}}{\text{Actual electrical power}}\) We know that the actual electrical power consumed is \(15.4\, \text{kW} = 15400\, \text{W}\) and the ideal mechanical power is \(10273\, \text{W}\). So, the efficiency is: \(\text{Efficiency} = \frac{10273}{15400} \times 100\% \approx 66.7\%\)

Calculate the pressure difference between the inlet and the exit of the pump

According to the problem statement, we're disregarding frictional losses in the pipes and any changes in kinetic energy. In this case, the pressure difference can be determined using the formula: \(\Delta P = \rho g h\) where \(\Delta P\) is the pressure difference, \(\rho\) is the density of water, \(g\) is the acceleration due to gravity, and \(h\) is the height difference between the lake and the storage tank. Using the known values: \(\Delta P = 1000 \times 9.81 \times 15 = 147150\, \text{Pa}\) The pressure difference between the inlet and the exit of the pump is \(147150\, \text{Pa}\). To summarize: a) The overall efficiency of the pump-motor unit is \(66.7\%\). b) The pressure difference between the inlet and the exit of the pump is \(147150\, \text{Pa}\).

Key concepts

Mechanical Power Calculation

Understanding the mechanical power calculation in a pumping system is crucial for determining the pump's performance. Mechanical power refers to the rate at which work is done or energy is transferred in mechanical processes. When dealing with pumps, this power is used to move fluids from one point to another.

In the case of pumping water from a lake to a storage tank, the mechanical power, denoted as \(P_m\), is calculated using the formula \( P_m = \rho q g h \), where \(\rho\) represents the fluid's density, \(q\) is the volumetric flow rate, \(g\) the acceleration due to gravity, and \(h\) the vertical height the fluid is being lifted. To get accurate results, it's essential to use consistent units, like cubic meters for volume and seconds for time, converting liters to cubic meters by dividing by 1000, as was demonstrated in the step-by-step solution.

When we apply the known values—1000 kg/m³ for water's density, 9.81 m/s² for gravity, and 0.07 m³/s for the flow rate converted from liters per second—we find that the ideal mechanical power required for this scenario is 10273 W or about 10.3 kW. By understanding these calculations, students can better gauge the energy requirements for various pumping scenarios.

Overall Efficiency of Pump

The overall efficiency of a pump system is a measure of how effectively it converts electrical power into mechanical power to move the fluid. It's a critical determinant of the system's performance and economic value.

The formula to find the efficiency of a pump is given by the ratio of the ideal mechanical power output to the actual electrical power input. It's often expressed as a percentage. So, following the equation \( \text{Efficiency} = \frac{\text{Ideal mechanical power}}{\text{Actual electrical power}} \times 100\% \), we can assess the unit's performance. From our example, with an actual power input of 15.4 kW and the previously calculated ideal mechanical power of 10.273 kW, the efficiency works out to approximately 66.7%.

This indicates that out of the total electrical power supplied to the pumping system, about two-thirds is effectively used for pumping water, while the remainder could be lost as heat or due to other inefficiencies in the motor or pump. Understanding this efficiency helps in selecting the right pump for a given application and also in designing systems for optimal energy utilization.

Pressure Difference Calculation

The pressure difference across a pump is a key factor that influences the flow of fluid within the system. It reflects the capability of the pump to lift a fluid to a certain height or overcome resistance in the piping system.

For situations where frictional losses are negligible and changes in kinetic energy are not considered, like our exercise, the pressure difference \(\Delta P\) can be determined with the simple formula \( \Delta P = \rho g h \). This relationship indicates that the pressure difference is directly proportional to the fluid density (\(\rho\)), the gravitational constant (\(g\)), and the height (\(h\)) to which the fluid is lifted.

Plugging in the values for the density of water, the acceleration due to gravity, and the height of the storage tank above the pump, we obtain a pressure difference of 147150 Pa, or 147.15 kPa. This represents the static pressure needed to raise the water by 15 meters, disregarding other possible losses, and is crucial for sizing equipment and ensuring system integrity. Calculating pressure differences correctly ensures that the pump can handle the requirements of the system without overloading or underperforming.