Understanding Power and Work in Fluid Systems
To pump water to a certain height, a motor must perform work against gravity. The power required is the rate at which this work is done. However, real-world machines are never 100% efficient, meaning the input power must be higher to compensate for energy losses.
The Problem
Given:
- Volume of water ($V$): $250 \text{ m}^3$
- Height ($h$): $20 \text{ m}$
- Time ($t$): $3 \text{ hours} = 3 \times 3600 \text{ s} = 10800 \text{ s}$
- Efficiency ($\eta$): $70\% = 0.70$
- Density of water ($\rho$): $1000 \text{ kg/m}^3$
- Acceleration due to gravity ($g$): $9.8 \text{ m/s}^2$
Step-by-Step Solution
1. Calculate the Mass of Water
First, find the mass ($m$) of the water to be pumped: $$m = \text{Density} \times \text{Volume} = 1000 \text{ kg/m}^3 \times 250 \text{ m}^3 = 250,000 \text{ kg}$$
2. Calculate Useful Work Done
The work done ($W$) to lift the water is equal to the potential energy gained: $$W = m \times g \times h = 250,000 \text{ kg} \times 9.8 \text{ m/s}^2 \times 20 \text{ m} = 49,000,000 \text{ Joules}$$
3. Calculate Useful Power Output
The power required to do this work ($P_{\text{out}}$) is: $$P_{\text{out}} = \frac{W}{t} = \frac{49,000,000 \text{ J}}{10800 \text{ s}} \approx 4537.04 \text{ Watts}$$
4. Calculate Input Power Required
Taking efficiency into account, the input power ($P_{\text{in}}$) is: $$P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{4537.04}{0.70} \approx 6481.48 \text{ Watts}$$
Conclusion: The electric motor needs a power rating of approximately 6.48 kW.