Question

The mechanical power output of an electric motor is 2.20 hp. The motor is connected to a 120 V source and is 89.0% efficient in converting power that it takes in by electrical transmission into mechanical power. Determine the following (a) current (in A) delivered to the motor (b) energy (in MJ) delivered to the motor by electrical transmission in 3.80 h of operation MJ (c) the cost to run the motor (in dollars) for 3.80 h, if the electric company charges $0.110/kWh

Answer 1

Answer:

a) $i = 15.367\,A$, b) $E = 25.226\,MJ$, c) $C = 0.771\,USD$

Explanation:

a) Let assume that electric motor is single-phase and operates in DC-mode, so that mechanical power output is:

$\dot W = \eta \cdot i \cdot V$

The current delivered to the motor is:

$i = \frac{\dot W}{\eta \cdot V}$

$i = \frac{(2.20\,hp)\cdot \left(\frac{0.746\,kW}{1\,hp} \right)\cdot \left(\frac{1000\,W}{1\,kW} \right)}{(0.89)\cdot (120\,V)}$

$i = 15.367\,A$

b) The power delivered to the motor is:

$\dot W = i\cdot V$

$\dot W = (15.367\,A)\cdot (120\,V)$

$\dot W = 1844.04\,W$

The energy delivered to the motor during 3.80 hours of operation is:

$E = (1844.04\,W)\cdot (3.80\,h)\cdot \left(\frac{3600\,s}{1\,h}\right)\cdot \left(\frac{1\,MJ}{1000000\,J}\right)$

$E = 25.226\,MJ$

c) The cost to run the motor is:

$C = (0.110\,\frac{USD}{kWh} )\cdot (25.226\,MJ)\cdot \left(\frac{0.278\,kWh}{1\,MJ}\right)$

$C = 0.771\,USD$