EBK ENGINEERING FUNDAMENTALS: AN INTROD
EBK ENGINEERING FUNDAMENTALS: AN INTROD
5th Edition
ISBN: 9780100543409
Author: MOAVENI
Publisher: YUZU
Question
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Chapter 20, Problem 14P
To determine

Find the motor which is recommended to purchase based on the given information.

Expert Solution & Answer
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Answer to Problem 14P

Bases on the given information, the Motor – Y is recommended to purchase.

Explanation of Solution

Given data:

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

The pump is expected to run 4200hours every year,

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

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

Initial cost for Motor – X and Motor – Y is $300and $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 yearand $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.

1cents=0.01dollars

Therefore,

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

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

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

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

Here,

P1 is the power of the Motor – X,

ε1 s the operating efficiency of the Motor – X.

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

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

1cents=0.01dollars

Therefore,

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

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

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

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

Here,

P2 is the power of the Motor – Y,

ε2 s the operating efficiency of the Motor – Y.

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

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

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