CASE STUDY−−Selecting Electric Motors to Power an Assembly Line A large computer manufacturing company is expanding its operations in response to an increase in demand. To meet the new production goals, the company needs to upgrade its assembly line. This upgrade requires additional electric motors

Question

CASE STUDY−−Selecting Electric Motors to Power an Assembly Line A large computer manufacturing company is expanding its operations in response to an increase in demand. To meet the new production goals, the company needs to upgrade its assembly line. This upgrade requires additional electric motors that will be able to deliver a total of 1,550 horsepower output. There are two types of motors available, induction and synchronous, whose specifications are listed in the accompanying table. Both motor types are three phase, and the quoted voltage is rms line-to-line. If the synchronous motors are operated at a power factor (pf) of 1.0, their efficiency is reduced to 80%, but the local power utility will provide electricity at a discounted rate of $0.025 per kilowatt-hour for the first year

	Type of Motor
Specifications	Induction	Synchronous (pf = 0.9 lag)	Synchronous (pf = 1.0)
Rated horsepower	400	500	500
Efficiency	85%	90%	80%
Line voltage	440	440	440
Capital investment	$17,640	$24,500	$24,500
Maintenance cost (First year)	$1,000	$1,250	$1,250
Electricity cost (First year)	$0.030/kWh	$0.030/kWh	$0.025/kWh

Historical data indicate that maintenance costs increase at the rate of 4% per year and that the cost of electricity increases at the rate of 5% per year. The assembly line must operate two eight-hour shifts per day, six days per week, for 50 weeks per year. An eight-year study period is being used for the project, and the company’s market-based cost of capital (MARR[latex]_{m}[/latex]) is 8% per year. Your engineering project team has been tasked with determining what combination of motors should be purchased to provide the additional power requirements of the upgraded assembly line.

Accepted Answer

Your project team has decided that the overall problem of motor selection can be broken down into the following subproblems to arrive at a recommendation:
(a) Determine whether it is less expensive to operate a synchronous motor at a power factor of 0.9 or 1.0.
(b) Determine the PW of the cost of operating one motor of each type at rated horsepower for the total period of operational time during the eight-year project lifetime.
(c) Determine the combination of motors that should be purchased. The objective is to minimize the PW of the total cost.

Solution to Subproblem (a)
Because the capital investment and maintenance costs of the synchronous motor are independent of the power factor employed, only the cost of electricity needs to be considered at this stage of the analysis. The efficiency of the 500-horsepower synchronous motor operating at a power factor of 0.9 is 90%. Therefore, the cost of electricity per operating hour is (recall that 1 hp = 746 watts)
(500 hp/0.9)(0.746 kW/hp)($0.030/kWh) = $12.43/hour.
When operated at a power factor (pf) of 1.0, the efficiency of the motor is reduced to 80%. The electric company, however will provide electricity at a discounted rate. The resulting cost of electricity is
(500 hp/0.8)(0.746 kW/hp)($0.025/kWh) = $11.66/hour.
Therefore, it is less expensive to operate the synchronous motor at pf = 1.0. The discounted electricity rate more than compensates for the loss in efficiency. Given this information, operating the synchronous motor at pf = 0.9 is no longer a feasible alternative.

Solution to Subproblem (b)
The total cost of each motor over the study period consists of the capital investment, maintenance costs, and electricity costs. Specifically, for each motor type,
PW[latex]_{Total costs}[/latex] = Capital investment + PW(maintenance costs) + PW(electricity costs).
Maintenance costs increase 4% annually over the eight year period and can be modeled as a geometric gradient series. Recall from Chapter 4 that the present equivalent of a geometric gradient series is given by:

[latex]P= \begin{cases}\frac{A_{1}[1-(P / F, i \%, N)(F / P, \bar{f} \%, N)]}{} & \bar{f} eq i \ A_{1} N(P / F, i \%, 1) & \bar{f}=i\end{cases}[/latex]

where [latex]\bar{f}[/latex] is the rate of increase per period. Using this relationship,

PW(maintenance costs) =[latex]\frac{A_{1}[1 − (P/F, 8\%, 8)(F/P, 4\%, 8)]}{0.08-0.04} [/latex]

=[latex]\frac{A_{1}}{0.04}[/latex][1 − (0.5403)(1.3686)]

=A[latex]_{1}[/latex](6.5136),

where A[latex]_{1}[/latex] = $1,000 for the induction motor and A1 = $1,250 for the synchronous motor.

Electricity costs are estimated to increase 5% annually over the eight-year study period and can also be modeled as a geometric gradient series.

PW(electricity costs) =[latex]\frac{A_{1}[1 − (P/F, 8\%, 8)(F/P, 4\%, 8)]}{0.08-0.05}[/latex]

=[latex]\frac{A_{1}}{0.03}[/latex][1 − (0.5403)(1.4775)]

=A[latex]_{1}[/latex](6.7236),

Now, the number of hours of motor operation per year is (8 hours/shift)(2 shifts/day)(6 days/week)(50 weeks/year) = 4,800 hours/year. Thus, for the induction motor,
A[latex]_{1}[/latex] = (400 hp/0.85)(0.746 kW/hp)(4,800 hours/year)($0.030/kWh) = $50,552
and
A[latex]_{1}[/latex] = (500 hp/0.8)(0.746 kW/hp)(4,800 hours/year)($0.025/kWh) = $55,950
for the synchronous motor running at pf = 1.0. We can now compute the PW of total costs for each motor type over the entire eight-year study period.
PW[latex]_{Total costs}[/latex](induction motor) = $17,640 + $1,000(6.5136) + $50,552(6.7236)
= $17,640 + $6,514 + $339,891
= $364,045.
PW[latex]_{Total costs}[/latex](synchronous motor) = $24,500 + $1,250(6.5136) + $55,950(6.7236)
= $24,500 + $8,142 + $376,185
= $408,827.

Solution to Subproblem (c)
The last step of the analysis is to determine what combination of motors to use to obtain the required 1,550 horsepower output. The following four options will be considered:
(A) Four induction motors (three at 400 horsepower, one at 350 horsepower)
(B) Four synchronous motors (three at 500 horsepower, one at 50 horsepower)
(C) Three induction motors at 400 horsepower plus one synchronous motor at 350 horsepower
(D) Three synchronous motors at 500 horsepower plus one induction motor at 50 horsepower.
For a motor used at less than rated capacity, the electricity cost is proportional to the fraction of capacity used—all other costs are unaffected.

The PW of total costs for each of the preceding options is computed using the results of subproblem (b):
PW[latex]_{Total costs}[/latex](A) = (3)($364,045) + [$17,640 + $6,514 + (350/400)($339,891)]
= $1,413,694
PW[latex]_{Total costs}[/latex](B) = (3)($408,827) + [$24,500 + $8,142 + (50/500)($376,185)]
= $1,296,742
PW[latex]_{Total costs}[/latex](C) = (3)($364,045) + [$24,500 + $8,142 + (350/500)($376,185)]
= $1,388,107
PW[latex]_{Total costs}[/latex](D) = (3)($408,827) + [$17,640 + $6,514 + (50/400)($339,891)]
= $1,293,121
Option (D) has the lowest PW of total costs. Thus, your project team’s recommendation is to power the assembly line using three 500-horsepower synchronous motors operated at a power factor of 1.0 and one 400-horsepower induction motor. You also include a note to management that there is additional capacity available should future modifications to the assembly line require more power. This case study illustrates the inclusion of price changes in the analysis of engineering projects. The cost of goods and services changes over time and needs to be addressed. Sensitivity analyses on key performance and cost parameters can be used to quantify project risk.

Question 8.7: CASE STUDY−−Selecting Electric Motors to Power an Assembly L...

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