Question

If a motor-pump set lifts 100 litres of water through a height of 10 m in 1.5 min, then the output power required by the pump will be (Take 1 lit = 1 kg)
1. 0.109 kW
2. 1.102 kW
3. 0.5 kW
4. 10.9 kW

Answer 1

Correct Answer - Option 1 : 0.109 kW

Concept:

The work done is given by the equation W = F.d

Power is given by the equation

\(P = \frac{W}{t}\) 

Where, F = Force exerted in N

d = distance moved in meters

w = work done in Joules

t = time taken in seconds

Calculation:

Given that h = 10 mm, t = 90 sec, Mass,

\({\rm{M}} = 100{\rm{\;litre}} \times \frac{{1kg}}{{litre}} = 100\;kg\) 

The weight of water is therefore,

\(100\;kg \times 9.8\frac{m}{{{s^2}}} = 980\;kg.\frac{m}{{{s^2}}} = 980\;N\) 

(∵ 1N = 1 kg m/s²)

W = F × d

= 980 N × 10 mm = 980 J

\(P = \frac{W}{t}\) 

\(P = {\frac{{9800\;}}{{90s}}^J} = 108.88\;{\rm{\omega }} = 109{\rm{\omega \;or\;}}0.109{\rm{\;kW\;}}\) 

Alternatively,

Power of the pump that lifts, Q = 100 litres, of water where density

P = 1 kg/litre at a vertical height, h = 1.2 m in time, t = 90 sec

\(P = \frac{{\delta\; \times\; g\; \times \;Q \;\times \;h}}{t}\) 

\( = \frac{{1\; \times\; 9.81\; \times\; 100\; \times \;10}}{{90}}\) 

= 109 W

P = 0.109 kW