Answer :
A. 70.5 kPa should be the rating of the pump used if it is 80% efficient.
- Given Data:
Elevation of reservoir A, [tex]h_A = 5 \, \text{m}[/tex]
Elevation of reservoir B, [tex]h_B = 15 \, \text{m}[/tex]
Flow rate, [tex]Q = 500 \, \text{L/s} = 0.5 \, \text{m}^3/s[/tex]
Head loss in the pipe system, [tex]h_{loss} = 1.5 \, \text{m}[/tex]
Pump efficiency, [tex]\text{Efficiency} = 80\% = 0.80[/tex]
- Total Head Required:
The total head, [tex]H[/tex], that the pump must provide can be calculated using the formula:
[tex]H = (h_B - h_A) + h_{loss}[/tex]
[tex]H = (15 \, \text{m} - 5 \, \text{m}) + 1.5 \, \text{m}[/tex]
[tex]H = 10 \, \text{m} + 1.5 \, \text{m}[/tex]
[tex]H = 11.5 \, \text{m}[/tex]
- The Required Pump Power:
The power required to lift the water is calculated using:
[tex]P = \rho g Q H[/tex]
where:
[tex]\rho[/tex] (density of water) = 1000 kg/m³
[tex]g[/tex] (acceleration due to gravity) = 9.81 m/s²
Plugging in the values:
[tex]P = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.5 \, \text{m}^3/s \times 11.5 \, \text{m}[/tex]
[tex]P = 1000 \times 9.81 \times 0.5 \times 11.5[/tex]
[tex]P = 56407.5 \, \text{W} \text{ or } 56.4 \, \text{kW}[/tex]
- Adjusting for Pump Efficiency:
Considering the pump efficiency:
[tex]P_{actual} = \frac{P}{\text{Efficiency}}[/tex]
[tex]P_{actual} = \frac{56407.5 \, \text{W}}{0.80}[/tex]
[tex]P_{actual} = 70509.38 \, \text{W} \text{ or } 70.51 \, \text{kW}[/tex]
- Converting Pressure to kPa:
To find the required pressure in kPa, we use the formula:
[tex]P_{required} = \frac{P_{actual}}{Q}[/tex]
Since 1 kPa = 1000 Pa, we can calculate:
[tex]P_{required} = \frac{70509.38 \, \text{Pa}}{1000}[/tex]
[tex]P_{required} \approx 70.5 \, \text{kPa}[/tex].