Economic Analysis for Energy Auditing

Overview of Economic Analysis in Energy Auditing

  • Context of Economic Evaluation: Once an energy management opportunity (EMO) has been identified, the energy manager must determine the cost-effectiveness to recommend implementation to management and justify capital expenditure.

  • Presentational Source: Presented by Anil Thompson of the School of Engineering at the University Technology of Jamaica.

  • Primary Reference: Guide to Energy Management by Barney L. Capehart and Wayne C. Turner.

Capital Investments in Energy Projects

  • Core Characteristics of Capital Investment:     * Relatively Large: These projects usually involve significant financial outlays.     * Long-Term Benefits: Returns are expected over the entire lifetime of the investment.     * Irreversibility: Once the capital is committed and the project started, it is often difficult or impossible to reverse.     * Tax Implications: Capital investments affect the company's tax burden through depreciation and expenses.

  • Categories of Costs:     * Acquisition Costs: Includes the purchase price, installation, training for personnel, and engineering charges. It also covers permits and any necessary facility renovations required before project commencement.     * Utilization Costs: These are routine costs for operating and maintaining the investment, primarily consisting of energy consumption, maintenance, and regular repairs.     * Disposal Costs: Incurred at the end of the project's useful life to retire or remove the asset. If the project has a positive worth at this stage, it is referred to as the Salvage Value.

Cash Flow Diagrams and Simple Payback Period

  • Cash Flow Diagram: A pictorial display used to map all costs and revenues over a project's timeline. Typically, an end-of-year approach is used for simplicity.

  • Simple Payback Period (SPP): The number of years required to recover the initial investment through project returns.     * Formula:       SPP=Initial CostAnnual SavingsSPP = \frac{\text{Initial Cost}}{\text{Annual Savings}}     * Example Case: A heat pump costing 10,000 USD10,000\,USD saves 2,500 USD/year2,500\,USD/year but costs 500 USD/year500\,USD/year to maintain.         * Initial Cost: 10,000 USD10,000\,USD         * Net Annual Savings: 2,500−500=2,000 USD/year2,500 - 500 = 2,000\,USD/year         * Calculation: SPP=10,0002,000=5 yearsSPP = \frac{10,000}{2,000} = 5\,\text{years}     * Limitations of SPP: It fails to consider the Time Value of Money (Interest) and ignores any cash flows that occur after the payback period has been reached. However, it is a useful "first cut" tool for initial assessment.

The Time Value of Money (TVM) and Interest

  • The TVM Principle: Money available today is worth more than the same amount in the future due to its earning potential and inflationary pressures.     * Interest (Opportunity Cost): The return that could be earned on the money if invested elsewhere.     * Inflation: The decrease in the purchasing power of money over time.

  • Discounted Cash Flow Analysis: The method of reducing future cash flows to a common basis (the present) through the use of an interest rate, also known as the Discount Rate.

  • Basic Interest Calculation Variables:     * PP = Present cash flow of money.     * FnF_n = Future cash flow at the end of year nn.     * InI_n = Accumulated interest over nn years.     * nn = Number of years between PP and FF.     * Formula: Fn=P+InF_n = P + I_n

  • Simple Interest Calculation:     * Formula: I=P×n×iI = P \times n \times i     * Example: Borrowing 10,000 USD10,000\,USD for 5 years5\,\text{years} at 18%/year18\%/year.         * I=(10,000)×(5)×(0.18)=9,000 USDI = (10,000) \times (5) \times (0.18) = 9,000\,USD

  • Compound Interest: Used almost exclusively in professional economic analysis because lenders prefer it. Interest is earned on previous interest.

Single Sum and Uniform Series Analysis

  • Single Sum Formulas:     * Future Worth (F given P):         * F=P(1+i)nF = P(1+i)^n         * Factor notation: (F/P,i,n)(F/P, i, n)         * Example: 5,000 USD5,000\,USD deposited at 10%10\% for 5 years5\,\text{years}.         * F=5,000×(F/P,10,5)=5,000×(1.611)=8,055 USDF = 5,000 \times (F/P, 10, 5) = 5,000 \times (1.611) = 8,055\,USD     * Present Worth (P given F):         * P=F(1+i)−nP = F(1+i)^{-n}         * Factor notation: (P/F,i,n)(P/F, i, n)         * Example: A boiler replacement will cost 150,000 USD150,000\,USD in 7 years7\,\text{years}. Amount to deposit now at 10%10\%:         * P=150,000×(P/F,10,7)=150,000×(0.5132)=76,980 USDP = 150,000 \times (P/F, 10, 7) = 150,000 \times (0.5132) = 76,980\,USD

  • Uniform Series (Annuity): A series of equal cash flows occurring at the end of consecutive periods (e.g., car or mortgage payments).     * Conversion Factors:         1. Find A given P: (A/P,i,n)(A/P, i, n). Used to find the annual savings needed to justify a present cost.         2. Find A given F: (A/F,i,n)(A/F, i, n). Used to find required annual deposits to reach a future goal.         3. Find P given A: (P/A,i,n)(P/A, i, n). Converts annual savings into a present value.         4. Find F given A: (F/A,i,n)(F/A, i, n). Converts annual savings into a future total.

Key Economic Metrics

  • Internal Rate of Return (IRR): The discount rate where the present value of project costs equals the present value of project savings (NPV=0NPV = 0).

  • Minimum Acceptable Rate of Return (MARR): The hurdle rate or minimum interest rate a company requires to approve a project.

  • Net Present Value (NPV): The difference between the present value of future net cash flows and the initial investment.

  • Benefit/Cost (B/C) Ratio:     * Formula: BC(i)=Equivalent BenefitsEquivalent Costs\text{BC}(i) = \frac{\text{Equivalent Benefits}}{\text{Equivalent Costs}}     * A project is cost-effective if the ratio is greater than 11.

Comprehensive Cost Effectiveness Example (Retrofit Study)

  • Scenario: Variable Air Volume (VAV) retrofit for 100,000 USD100,000\,USD. Savings: 450,000 kWh/year450,000\,kWh/year. Life: 10 years10\,\text{years}. MARR: 10%10\%. Electricity: 0.06 USD/kWh0.06\,USD/kWh. Salvage Value: 500 USD500\,USD.

  • Annual Savings Calculation: 450,000×0.06=27,000 USD/year450,000 \times 0.06 = 27,000\,USD/year.

  • Analysis Methods:     * Present Worth (PW):         * PW=−100,000+27,000(P/A,10,10)+500(P/F,10,10)PW = -100,000 + 27,000(P/A, 10, 10) + 500(P/F, 10, 10)         * PW=−100,000+27,000(6.1446)+500(0.3856)=66,097 USDPW = -100,000 + 27,000(6.1446) + 500(0.3856) = 66,097\,USD     * Future Worth (FW):         * FW=−100,000(F/P,10,10)+27,000(F/A,10,10)+500FW = -100,000(F/P, 10, 10) + 27,000(F/A, 10, 10) + 500         * FW=−100,000(2.594)+27,000(15.937)+500=171,399 USDFW = -100,000(2.594) + 27,000(15.937) + 500 = 171,399\,USD     * Annual Worth (AW):         * AW=−100,000(A/P,10,10)+27,000+500(A/F,10,10)AW = -100,000(A/P, 10, 10) + 27,000 + 500(A/F, 10, 10)         * AW=−100,000(0.1628)+27,000+500(0.0628)=10,751 USDAW = -100,000(0.1628) + 27,000 + 500(0.0628) = 10,751\,USD     * Benefit/Cost Ratio:         * PW of benefits: 165,900 USD165,900\,USD (savings) + 193 USD193\,USD (salvage PV) = 166,093 USD166,093\,USD.         * Net Costs: 100,000−193=99,807 USD100,000 - 193 = 99,807\,USD.         * Ratio=165,90099,807=1.66Ratio = \frac{165,900}{99,807} = 1.66. (Project is attractive).     * IRR Calculation:         * Solve for ii where 100,000=27,000(P/A,i,10)+500(P/F,i,10)100,000 = 27,000(P/A, i, 10) + 500(P/F, i, 10).         * Approximation: AP=27,000100,000=27%\frac{A}{P} = \frac{27,000}{100,000} = 27\%.         * Actual Solution: 23.8%23.8\%. Since 23.8\% > 10\%, the project is excellent.

Life Cycle Costing (LCC)

  • Definition: The total cost of owning and operating equipment over its entire service life, including acquisition, energy, maintenance, and disposal.

  • Decision Criterion: Organizations often err by choosing the lowest initial cost. Rational LCC analysis selects the alternative with the lowest total Life Cycle Cost.

  • Air Compressor Comparison Example:     * Alternative 1 (Efficient): 30,000 USD30,000\,USD cost; 1,000 USD1,000\,USD maintenance; 6,000 USD6,000\,USD energy. Total annual cost = 7,000 USD7,000\,USD.         * LCC=30,000+7,000(P/A,10,10)=30,000+7,000(6.1446)=73,012 USDLCC = 30,000 + 7,000(P/A, 10, 10) = 30,000 + 7,000(6.1446) = 73,012\,USD     * Alternative 2 (Standard): 25,000 USD25,000\,USD cost; 500 USD500\,USD maintenance; 10,000 USD10,000\,USD energy. Total annual cost = 10,500 USD10,500\,USD.         * LCC=25,000+10,500(P/A,10,10)=25,000+10,500(6.1446)=89,518 USDLCC = 25,000 + 10,500(P/A, 10, 10) = 25,000 + 10,500(6.1446) = 89,518\,USD     * Choice: Alternative 1 because it has the lower LCC.

Taxes and Depreciation

  • Impact: Depreciation is not a cash flow but is a business expense that lowers taxable income, thus affecting life cycle analysis.

  • Depreciation Methods: Straight line, sum of the years digits, declining balance, and Accelerated Cost Recovery System (ACRS).

  • After Tax Savings (ATS) or After Tax Cash Flow (ATCF):     * Formula (Straight Line):         * ATS=S−[(S−D)×TR]ATS = S - [(S - D) \times TR]         * SS = Before-tax annual savings.         * DD = Depreciation per year.         * TRTR = Corporate tax rate.     * Example: Chiller costing 100,000 USD100,000\,USD saves 20,000 USD/year20,000\,USD/year for 10 years10\,\text{years}. Tax rate = 40%40\%.         * Annual Depreciation (D)=100,00010=10,000 USD/year(D) = \frac{100,000}{10} = 10,000\,USD/year.         * ATS=20,000−[(20,000−10,000)×0.40]=20,000−4,000=16,000 USDATS = 20,000 - [(20,000 - 10,000) \times 0.40] = 20,000 - 4,000 = 16,000\,USD.