Physics/A numerical based on power
I've got another doubt yet. Having completed the chapters of Newtonian mechanics and work I have just one doubt remaining. It is this:
"An overhead tank is fixed at the height 10m from the ground and has a capacity of 500 litres. Calculate the time required to fill the tank from a pump of 0.5 H.P. whose efficiency is 60%"
It's not that I am not able to understand it at all but I fear the answer given in our book is erroneous.
I'll be waiting for your help,
The density of water is 1000 kg/m^3, or we could also say it is 1000 kg/(1000 liters). Therefore the mass is 500 kg. When raised 10 m, its gravitational potential energy, GPE, is
GPE = m*g*h = 500 kg*9.8 m/s^2*10 m = 49,000 J
From the principle of conservation of energy, the work done by the pump must have been 49,000 J. At this point it will be useful to convert the motor's power to units of Watts because the Watt is equivalent to J/s.
0.5 hp * 745.7 Watts/1 hp = 373 W
Now, I have a doubt regarding the significance of the efficiency number. It might have no significance to this problem. I don't know if the rated horsepower is the rate it consumes power or the rate it can do work. I will deal with the 2 possibilities in the next 2 paragraphs.
If the power rating is the rate it can do work, the efficiency number has no affect on the time it will take. That only affects how much electrical energy is wasted. The conclusion in this paragraph is that it does work at the rate of 373 W. Therefore
49,000 J/t = 373 W and
t = 49,000 J /(373 J/s) = 131 seconds.
If the power rating is the rate it consumes energy, then the efficiency is a factor in the time required. Since
efficiency = power output / power consumption
power output = power consumption * efficiency
power output = 373 W * 0.6 = 224 W
t = 49,000 J / (224 J/s) = 219 s
I hope this helps,