ENGINEERING FUNDAMENTALS
ENGINEERING FUNDAMENTALS
6th Edition
ISBN: 9781337705011
Author: MOAVENI
Publisher: CENGAGE L
Question
Book Icon
Chapter 20, Problem 14P
To determine

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

Expert Solution & Answer
Check Mark

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A 9,000 gallon storage tank was purchased for $200,000 12 years ago. How much would a 12,000 gallon storage tank cost today if the power-sizing exponent were 0.69 and the cost index increased from 150 to 230 over the last 12 years. Express your answer in $ to the nearest $1,000.
Hello, I need some help with my civil engineer homework. I am stuck on the last question Electricity (See your home electricity bill) 1) You can obtain the daily kilowatt-hour, kWh, use for your home from the electricity bill. This ispresented differently from utility to utility. What is your monthly electricity use?1,186 kWh 2) Use 30 days equal to one month to find the average daily electric use.Daily amount: 39.53 kWh (I divided 1,186/30) 3) How many members are in your household?:  4 people 4) Your personal share is obtained by dividing the daily amount (above) by the number of people (house members) in the household: 9.88 kWh 5) Add 1.6 kWh for electric use at work, in the office, or here at QCC. The total daily electric use foryou equals.personal share + 1.6 kWh =11.48 kWh What is the Crude Oil equivalent? *With electricity, 1 gallon of crude oil produces 15kWh of electricity. Divide your answer intocrude oil conversion factor, 15 kWh, to obtain the crude oil equivalent. Crude…
i need the answer quickly
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Engineering Fundamentals: An Introduction to Engi...
Civil Engineering
ISBN:9781305084766
Author:Saeed Moaveni
Publisher:Cengage Learning
Text book image
Traffic and Highway Engineering
Civil Engineering
ISBN:9781305156241
Author:Garber, Nicholas J.
Publisher:Cengage Learning