After finding that the old boiler has an efficiency of only 60 percent whereas a new boiler would have an efficiency of 85 percent, a building owner of Example 3.1 has decided to invest the $10,000 in getting a new boiler. Determine whether this investment is costeffective if the lifetime of the boiler is ten years and the

Question

After finding that the old boiler has an efficiency of only 60 percent whereas a new boiler would have an efficiency of 85 percent, a building owner of Example 3.1 has decided to invest the $10,000 in getting a new boiler. Determine whether this investment is costeffective if the lifetime of the boiler is ten years and the discount rate is 5 percent. The boiler consumes 5,000 gallons per year at a cost of $1.20 per gallon. An annual maintenance fee of $150 is required for the boiler (independently of its age). Use all five methods summarized in Table 3.4 to perform the economic analysis.

TABLE 3.4 Summary of the Basic Criteria for the Various Economic Analysis Methods for Energy Conservation Projects

Criterion	Equation	Evaluation Method
NPW > 0	[latex]NPW=-CF_{0}+\sum\limits_{K=1}^{N}{CF_{k}*SPPW(d,k)} [/latex]	Net present worth (NPW)
d′ > d	[latex]-CF_{0}+\sum\limits_{K=1}^{N}{CF_{k}*SPPW(d′,k)}=0 [/latex]	Rate of return (ROR)
BCR > 1	[latex]BCR=\frac {\sum\limits_{K=0}^{N}{B_{k}SPPW(d,k)}}{\sum\limits_{K=0}^{N}{C_{k}SPPW(d,k)}[/latex]	Benefit–cost ratio (BCR)
Y < N	[latex]CF_{0}=\sum\limits_{K=1}^{Y}{CF_{k}*SPPW(d,k)} [/latex]	Discounted payback period (DPB)
Y <	[latex]Y=\frac {CF_{0}}{A}[/latex]	Simple payback period (SPB)

Accepted Answer

The base case for the economic analysis presented in this example is the case where the boiler is not replaced. Moreover, the salvage value of the boiler is assumed insignificant after ten years. Therefore, the only annual cash flows (A) after the initial investment on a new boiler are the net savings due to higher boiler efficiency as calculated below:

[latex]A= Fuel-Use_{before} *(η_{befor}/η_{fter})*Fuel-cos/gallon[/latex]

[latex]A=5000*(1-0.60/0.85)[/latex]*$1.20=$1.765

The cost-effectiveness of replacing the boiler is evaluated as indicated below:

1. Net Present Worth. For this method [latex]CF_{0}[/latex] = $10,000 and [latex]CF_{1}[/latex] = … = [latex]CF_{10}[/latex] = A, d = 0.05, and N= 10 years. Using Eq. (3.24) with USPW = 7.740 (see Example 3.4):

[latex]NPW = −CF_{0} + A*USPW(d,N)[/latex] (3.24)

NPW = $3,682

Therefore, the investment in purchasing a new boiler is cost-effective.

2. Rate of return. For this method also [latex]CF_{0}[/latex] = $10,000 and [latex]CF_{1}[/latex] = …=[latex]CF_{10}[/latex] = A, whereas SPPW(d′,k) is provided by Eq. (3.17). By trial and error, it can be shown that the solution for d′ is:

[latex]SPPW(d,N) = P/F = (1+d)^{−N}[/latex] (3.17)
d′ = 12.5%

Inasmuch as d′ > d = 5%, the investment in replacing the boiler is cost- effective.

3. Benefit–Cost Ratio. In this case, [latex]B_{0}[/latex] = 0 and [latex]B_{1}[/latex]= … = [latex]B_{10}[/latex] = A whereas [latex]C_{0}[/latex] = $10,000 and [latex]C_{1}[/latex] =… = [latex]C_{10}[/latex] = 0.
Note that because the maintenance fee is applicable to both the old and new boiler, this cost is not accounted for in this evaluation method (only the benefits and costs relative to the base case are considered).
Using Eq. (3.26):

[latex]BCR=\frac {\sum\limits_{K=0}^{N}{B_{k}*SPPW(d,k)}}{\sum\limits_{K=0}^{N}{C_{k}*SPPW(d,k)}}[/latex] (3.26)

BCR = 1.368

Thus, the benefit–cost ratio is greater than unity (BCR > 1) and the project of getting a new boiler is economically feasible.

4. Compounded Payback Period. For this method, [latex]CF_{0}[/latex] = $10,000 and [latex]CF_{1}[/latex]= … = [latex]CF_{10}[/latex] = A. Using Eq. (3.27), Y can be solved: Y = 6.9 years.
Thus the compounded payback period is shorter than the lifetime of the project (Y > N = 10 years) and therefore replacing the boiler is cost-effective.

[latex]CF_{0}=\sum\limits_{K=1}^{Y}{CF_{k}*SPPW(d,k)} [/latex] (3.27)

5. Simple Payback Period. For this method, [latex]CF_{0}[/latex] = $10,000 and A =$1,765. Using Eq. (3.29), Y can be easily determined: Y = 5.7 years.
Thus, the simple payback period method indicates that the boiler retrofit project can be cost-effective.

[latex]Y=\frac {CF_{0}}{A}[/latex] (3.29)

Question 3.5: After finding that the old boiler has an efficiency of only ...

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