Question

A water pump that consumes \(2 \mathrm{kW}\) of electric power when operating is claimed to take in water from a lake and pump it to a pool whose free surface is \(30 \mathrm{m}\) above the free surface of the lake at a rate of \(50 \mathrm{L} / \mathrm{s}\). Determine if this claim is reasonable.

Short answer

Answer: No, the claim is not reasonable, as the hydraulic power required (14.715 kW) is greater than the electric power consumption of the pump (2 kW).

Step-by-step solution

1. Identify the given information

We are given the following values: - Electric power consumption of the pump: \(P_e = 2 \,\text{kW}\) - Height difference between the lake and the pool: \(h = 30 \,\text{m}\) - Flow rate of water: \(Q = 50 \,\text{L/s}\)

2. Convert the flow rate to \(\text{m}^3/\text{s}\)

To calculate the hydraulic power, we need the flow rate in \(\text{m}^3/\text{s}\). We can convert it from \(\text{L/s}\) to \(\text{m}^3/\text{s}\) using the following conversion: 1 liter = 0.001 cubic meters. $$ Q_\text{m3/s} = Q_\text{L/s} \cdot 0.001 \, \frac{\text{m}^3}{\text{L}} $$ Plug in the given flow rate. $$ Q_\text{m3/s} = 50 \, \frac{\text{L}}{\text{s}} \cdot 0.001 \, \frac{\text{m}^3}{\text{L}} = 0.05 \, \frac{\text{m}^3}{\text{s}} $$

3. Calculate the hydraulic power

To calculate the hydraulic power, we can use the following formula: $$ P_\text{h} = \rho g Q h $$ where \(\rho\) is the density of water (approximately \(1000 \, \text{kg/m}^3\)), \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)), \(Q\) is the flow rate, and \(h\) is the height difference. Plug in the values and compute the hydraulic power. $$ P_\text{h} = 1000 \, \frac{\text{kg}}{\text{m}^3} \cdot 9.81 \, \frac{\text{m}}{\text{s}^2} \cdot 0.05 \, \frac{\text{m}^3}{\text{s}} \cdot 30 \, \text{m} = 14715 \, \frac{\text{kg} \cdot \text{m}^2}{\text{s}^3} $$ Since 1kW = 1000 W, we need to convert the hydraulic power to kiloWatts. $$ P_\text{h,kW} = \frac{14715}{1000} = 14.715 \, \text{kW} $$

4. Compare the hydraulic power with the electric power consumption

Now, we will compare the calculated hydraulic power with the given electric power consumption. The hydraulic power (\(P_\text{h,kW} = 14.715 \, \text{kW}\)) is greater than the electric power consumption (\(P_e = 2 \, \text{kW}\)).

5. Determine if the claim is reasonable

Since the hydraulic power required to pump the water at the given rate and height is greater than the electric power consumption of the pump, the claim is not reasonable. The water pump does not have enough electric power to pump water at the claimed rate and height.