at lull 63 A 75hp shaft output motor that has an efficiency

at lull 63 A 75-hp (shaft output) motor that has an efficiency of percent is worn out and is to be replaced by a high-effi ciency motor that has an efficiency of 95.4 percent. The motor operates 4368 hours a year at a load factor of 0.75. Taking the cost of electricity to be $0.08/kWh, determine the amount of energy and money saved as a result of installing the high-efficiency motor instead of the standard motor. Also, determine the simple payback period if the purchase prices of the standard and high-efficiency motors are $5449 and $5520, respectively.

Solution

Physics involved: A low efficiency motor consumes more power for the same out put than a high efficiency motor. Hence when you replace the old motor by a more efficient motor, you need less power for the same power output. This saves input energy costs. This saved cost should cover the difference in the prices of the motors by properly utilising the new one.

Solution: Efficiency of new motor = 95.4% = 0.954 , efficiency of old motor = 91.0% = 0.91

1 hp = 0.7457 KW = 0.7457 KJ/s

Maximum Out put power = Maximum possible load = P₀ = 75 hp = 0.7457 * 75 = 56 KW

The new motor must have the same power output as the old one i.e. 56 KW

load factor of new motor= used load / Maximum load = 0.75 => Actual Out put power of new motor = used load = 0.75 * 56 = 42 KW

Total energy output by the new motor used for 4386 h/year = 42 * 4386 = 184212 KWh

Total input energy to the motor = output energy / efficiency = 184212 / 0.954 = 193094 KWh

Total input energy cost for new motor = consumed energy * price per unit energy = 193094 * 0.08 = 15448 $

Total input energy to the old motor for the same use= output energy / efficiency = 184212 / 0.91 = 202431 KWh

Total input energy cost for old motor for the same use = consumed energy * price per unit energy = 202431 * 0.08 = 16194 $

Saved cost = 16194 - 15448 = 746 $ / year (Ans)

Difference in buying cost of motors = 5520 - 5449 = 71 $

To recover this 71 $ we need = 71 / 746 = 0.095 year = 0.095* 365 = 35 days approximately. (Ans)

efficiency of motor = ooutput power / input power