EP WEBASSIGN FOR MOAVENI'S ENGINEERING
EP WEBASSIGN FOR MOAVENI'S ENGINEERING
6th Edition
ISBN: 9780357126592
Author: MOAVENI
Publisher: CENGAGE CO
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
1:08 Il LTE Individual Assignment CEQ 31... CIVIL ENGINEERING ROANTICLO (Earthworks Quantities) (Due Date: Friday, 28/03/2025) Done Thumbi Irrigation Scheme in Mzimba district is under threat of flooding. In order to mitigate against the problem, authorities have decided to construct a flood protection bund (Dyke). Figure 1 is a cross section of a 300m long proposed dyke; together with its foundation (key). Survey data for the proposed site of the dyke are presented in Table 1. Table 2 provides swelling and shrinkage factors for the fill material that has been proposed. The dyke dimensions that are given are for a compacted fill. (1) Assume you are in the design office, use both the Simpson Rule and Trapezoidal Rule to compute the total volume of earthworks required. (Assume both the dyke and the key will use the same material). (2) If you are a Contractor, how many days will it take to finish hauling the computed earthworks using 3 tippers of 12m³ each? Make appropriate assumptions.…
Thumbi Irrigation Scheme in Mzimba district is under threat of flooding. In order to mitigate against the problem, authorities have decided to construct a flood protection bund (Dyke). Figure 1 is a cross section of a 300m long proposed dyke; together with its foundation (key). Survey data for the proposed site of the dyke are presented in Table 1. Table 2 provides swelling and shrinkage factors for the fill material that has been proposed. The dyke dimensions that are given are for a compacted fill. (1) Assume you are in the design office, use both the Simpson Rule and Trapezoidal Rule to compute the total volume of earthworks required. (Assume both the dyke and the key will use the same material). (2) If you are a Contractor, how many days will it take to finish hauling the computed earthworks using 3 tippers of 12m³ each? Make appropriate assumptions. DIKE CROSS SECTION OGL KEY (FOUNDATION) 2m 1m 2m 8m Figure 1: Cross section of Dyke and its foundation 1.5m from highest OGL 0.5m…
this is from CE-192
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
Text book image
Sustainable Energy
Civil Engineering
ISBN:9781337551663
Author:DUNLAP, Richard A.
Publisher:Cengage,
Text book image
Fundamentals Of Construction Estimating
Civil Engineering
ISBN:9781337399395
Author:Pratt, David J.
Publisher:Cengage,
Text book image
Solid Waste Engineering
Civil Engineering
ISBN:9781305635203
Author:Worrell, William A.
Publisher:Cengage Learning,
Text book image
Residential Construction Academy: House Wiring (M...
Civil Engineering
ISBN:9781285852225
Author:Gregory W Fletcher
Publisher:Cengage Learning