Question

A 40 hp, 1780 rpm 460 V, 3-phase, 60 Hz drip-proof Baldor Super E premium energy induction motor has a power factor of 86% and an efficiency of 92.1%. The motor, priced at $2243, runs at full load 14 hours per day, 5 days a week. Calculate the cost of driving the motor during a 3-year period, knowing that the cost of energy is $0.075/kWh

Answer 1

Answer:

$26,524.95

Explanation:

In order to figure out how much it costs to drive the motor for 3 years, we must first figure out how many kW there are to the 40hp power of the motor, so we do a conversion:

$40Hp*\frac{0.7457kW}{1Hp}=29.828kW$

Once we got how many kW there are to the 40Hp, we can now figure out how much electric power the motor is consuming, we can do that by using the efficiency of the motor, so we get:

$Power=\frac{29.828kW}{0.921}=32.387kW$

Now we can figure out for how many hours the motor is going to be driven throughout the 3 years.

$3years*\frac{52weeks}{1 year}*\frac{5days}{1week} *\frac{14hr}{1day}=10,920 hrs[\tex]$

now that we got the number of hours the motor will be working, we can now find the number of kWh the motor consumes during the 3 years:

32.387kW*10,920hr=353,666.04kWh

So now we can figure out how much it will cost to drive the motor:

$353,666.04kWh*\frac{\ 0.075}{1kWh}= 26,524.95$