Given data:

The power of the Motor – X (P1) and Motor – Y (P2) is 1.5 kW,

The pump is expected to run 4200 hours every year,

Average electricity cost of the motor is 11 cents per kWh,

The number of years for the life of Motor – X (n1) and Motor – Y (n2) is 5 years,

Initial cost for Motor – X and Motor – Y is $ 300 and $ 400,

Operating point efficiency of the Motor – X (ε1) and Motor – Y (ε2) is 0.75 and 0.85,

Maintenance cost of Motor – X and Motor – Y is $ 12 per year and $ 10 per year.

Calculation:

Assume the interest rate as 8 %, and the system doesn’t have salvage value.

Case (i): For Motor - X

Initially calculate the operating cost for the Motor- X using the power, efficiency at the operating point, electricity cost and pump life time.

Convert the electricity cost from cent per kWh to dollar per kWh using the following relationship.

1 cents=0.01 dollars

Therefore,

11 cents per kWh=11(0.01 $)  per kWh=$ 0.11 per kWh

The average electricity cost of the motor is $ 0.11 per kWh.

Formula to calculate the operating cost of the Motor – X is,

Operating cost=(P1ε1)(Electricity cost)(Pump life time) (1)

Here,

P1 is the power of the Motor – X,

ε1 s the operating efficiency of the Motor – X.

Substitute 1.5 kW for P1, 0.75 for ε1, $ 0.11 per kWh for Electricity cost, and 4200 hours every year for Pump life time in equation (1) to find Operating cost.

Operating cost=(1.5 kW0.75)($ 0.11kWh)(4200 hoursyear)=$ 924 per year

Therefore, the present worth of the Motor – X can be calculated using the following expression:

PW=−(Initial cost)−(Operating cost+Maintenence cost)(PA, i, n1) (2)

Here,

P is the present cost,

A is the uniform series payment,

i is the interest rate,

n1 is the number of years of life of Motor – X.

Substitute $ 300 for Initial cost, $ 924 for Operating cost, $ 12 for Maintenence cost, 8 % for i, and 5 for n1 in equation (2) to find PW.

PW=−($ 300)−($ 924+$ 12)(PA, 8 %, 5)=−($ 300)−($ 936)(PA, 8 %, 5) (3)

Refer to the Table 20.9 in the textbook, the value for (PA, i, n) with the interest rate of 8 %, and number of years of 5, is 3.9928.

Substitute 3.9928 for (PA, 8 %, 5) in equation (3) to find PW.

PW=−($ 300)−($ 936)(3.9928)=−($ 300)−($ 3737.26)=−$ 4037.26 (4)

Hence, the present worth of the Motor – X is −$ 4037.26.

Case (ii): For Motor - Y

Initially calculate the operating cost for the Motor- Y using the power, efficiency at the operating point, electricity cost and pump life time.

Convert the electricity cost from cent to dollar using the following relationship.

1 cents=0.01 dollars

Therefore,

11 cents per kWh=11(0.01 $)  per kWh=$ 0.11 per kWh

The average electricity cost of the motor is $ 0.11 per kWh.

Formula to calculate the operating cost of the Motor – Y is,

Operating cost=(P2ε2)(Electricity cost)(Pump life time) (5)

Here,

P2 is the power of the Motor – Y,

ε2 s the operating efficiency of the Motor – Y.

Substitute 1.5 kW for P2, 0.85 for ε2, $ 0.11 per kWh for Electricity cost, and 4200 hours every year for Pump life time in equation (5) to find Operating cost.

Operating cost=(1.5 kW0.85)($ 0.11kWh)(4200 hoursyear)=$ 815.3 per year

Therefore, the present worth of the Motor – Y can be calculated using the following expression:

PW=−(Initial cost)−(Operating cost+Maintenence cost)(PA, i, n2) (6)

Here,

n2 is the number of years of life of Motor – Y.

Substitute $ 400 for Initial cost, $ 815.3 for Operating cost, $ 10 for Maintenence cost, 8 % for i, and 5 for n2 in equation (6) to find PW.

PW=−($ 400)−($ 815.3+$ 10)(PA, 8 %, 5)=−($ 400)−($ 825.3)(PA, 8 %, 5) (7)

Refer to the Table 20.9 in the textbook, the value for (PA, i, n) with the interest rate of 8 %, and number of years of 5, is 3.9928.

Substitute 3.9928 for (PA, 8 %, 5) in equation (7) to find PW.

PW=−($ 400)−($ 825.3)(3.9928)=−($ 400)−($ 3295.26)=−$ 3695.26 (8)

Hence, the present worth of the Motor – Y is −$ 3695.26.

Compare the equations (4) and (8), from the analysis it is clear that the Motor – Y provides the least present value than the Motor – X.

Therefore, the Motor – Y should be recommended to purchase.

Conclusion:

Thus, the Motor – Y should be recommended to purchase.