Question
During an enrgy audit following data were obtained on a 3 phase induction motor
Rated values : 37 kW, 415 V, 66A, 0.88 pf
Operating values :410 V,49A, 0.76 pf
The plant operates for 7000 Hours per year with the electricity cost of Rs 6.00 per unit.
It is proposed to replace the existing motor by a 30 kW energy efficient motor with 92% efficiency.
- Determine the rated efficiency and the loading of the existing motor.
- Calculate the loading with energy efficient motor.
- if replacing the existing motor with energy efficient motor which costs Rs 75,000, determine the payback period for the investment required for the energy efficient motor over the existing motor. Consider the salvage value of the existing motor as Rs 10,000/-.
Answer
Rated input power = 1.732*0.415*66*0.8= 41.746kW.
Rated efficiency of the motor =37/41.746 =88.63%
Actual input power drawn = 1.732*0.410*49*0.76= 26.44 kW
Loading of the motor=26.44/41.746= 0.633 or 63.3%
Shaft power or motor output =37*0.633 =23.44 kW
Energy efficient motor rating =30 kW
Actual output required = 23.44 kW
% Loading of the motor =23.44/30=78%
Annual energy savings =23.44(1/0.8863-1/0.92)*7000* Rs 6 = Rs. 40,740/-
Payback period =(75,000-10,000)/40740 =1.59 years.
