Question

Find runtime hours that makes these two motors economically equal.

2.14 Two electric motors are available to satisfy a continuous demand of 85 kW. Motor 1 has an initial cost of $5,500 with an operating efficiency of 91%. Motor 2 has an initial cost of $4,000 with an operating efficiency of 89%. Each motor has a life of 10 years with no salvage value. The annual cost of electricity to run the machines is $19.20 per kW of demand plus $0.12 per kWh of energy use. Determine the

Answer 1

Case - Motor 1

Cost of Machine=Co=$5500

MARR=i=22%

Useful life=n=10 years

Operating efficiency=91%

Output =85 KW

Input Power=Output power/Efficiency=85/91%=93.40659 KW

Let motor runs h hours per year.

Total Cost=C=19.2*85+0.12*93.40659h=1632+11.20879h

Let us calculate Present Worth Factor of annuity PWFA for n=10% and i=22%

Present Worth of Cost of Machine=Co+C*PWFa

=5500+(1632+11.20879h)* 3.923184

=5500+6402.637+43.974154h

=11902.63683+43.974154h

Case - Motor 2

Cost of Machine=Co=$4000

MARR=i=22%

Useful life=n=10 years

Operating efficiency=89%

Output =85 KW

Input Power=Output power/Efficiency=85/89%=95.50562 KW

Let motor runs h hours per year.

Total Cost=C=19.2*85+0.12*95.50562 *h=1632+11.46067h

Let us calculate Present Worth Factor of annuity PWFA for n=10% and i=22%

Present Worth of Cost of Machine=Co+C*PWFa

=4000+(1632+11.46067h)* 3.923184

=4000+6402.637+44.9623373h

=10402.63683+44.9623373h

Set Present worth of machines equal for economically equivalence

11902.63683+43.974154h=10402.63683+44.9623373h

h=(11902.63683-10402.63683)/(44.9623373-43.974154)= 1517.937

Machines are economically equivalent at 1518 hours/year