Answer:
a) [tex]i = 15.367\,A[/tex], b) [tex]E = 25.226\,MJ[/tex], c) [tex]C = 0.771\,USD[/tex]
Explanation:
a) Let assume that electric motor is single-phase and operates in DC-mode, so that mechanical power output is:
[tex]\dot W = \eta \cdot i \cdot V[/tex]
The current delivered to the motor is:
[tex]i = \frac{\dot W}{\eta \cdot V}[/tex]
[tex]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)}[/tex]
[tex]i = 15.367\,A[/tex]
b) The power delivered to the motor is:
[tex]\dot W = i\cdot V[/tex]
[tex]\dot W = (15.367\,A)\cdot (120\,V)[/tex]
[tex]\dot W = 1844.04\,W[/tex]
The energy delivered to the motor during 3.80 hours of operation is:
[tex]E = (1844.04\,W)\cdot (3.80\,h)\cdot \left(\frac{3600\,s}{1\,h}\right)\cdot \left(\frac{1\,MJ}{1000000\,J}\right)[/tex]
[tex]E = 25.226\,MJ[/tex]
c) The cost to run the motor is:
[tex]C = (0.110\,\frac{USD}{kWh} )\cdot (25.226\,MJ)\cdot \left(\frac{0.278\,kWh}{1\,MJ}\right)[/tex]
[tex]C = 0.771\,USD[/tex]
