Energy Economics

Amit Singh

Training Program on DSM_IIT Bombay_9 Dec 2016

Outline

Basic Terminologies

• Discount Rate• Net Present Value

(NPV)• Capital Recovery factor

(CRF)

Energy Economics: Figure of merits

• Simple Payback Period• Benefit-Cost Ratio• Internal Rate of Return

(IRR)• Cost of Saved Energy

(CSE)

CSE case studies

• Pump sets• LED Lamps

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Simple payback Period (SPP)

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SPP = 200,000/(5000*32)

= 1.25 Years

There’s a problem!

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An energy auditor recommended additional insulation on a boiler. The cost of the insulation is Rs 200,000. It is estimated that installing the insulation will result in savings of 5 kilo-litres of Light Diesel Oil priced at Rs 32/litre.

What’s the problem with SPP?

• Present investment vs. Future Savings.

• Time adds/discounts Value to/from Money (Investment/Savings)

• Time Value of Money.

What’s the Solution?

To discount the future cost and savings to determine their present worth.

Future SavingsPresent Cost

Discount Rate Net Present Value

Discount rate and NPV (Qualitative Explanation)

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Discount Rate (d): It is a measure of how much worth/ value the investor attaches to his money now (present) in comparison to future.

Who has a higher value of discount Rate?

Net Present Value (NPV):

Salaried personBusinessman or

Value

Time

Discounted Value of the Cash Flow

, Current Cash Flows at various instant of Time

Discount rate and NPV (Quantitative Explanation)

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Internal rate of Return (IRR) & Benefit to Cost (B/C) ratio.

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Cost of Saved Energy (CSE)

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Example 1

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Solar Water Heating SystemAn industry wishes to install a solar water heating system to preheat the make-up water supplied to the boiler. The cost of the system is Rs 2 lakhs and the annual savings are expected to be Rs 30,000. If we neglect depreciation, assuming a life of 20 years and a discount rate of 12%, calculate SPP, NPV, B/C and IRR?

SPP NPV B/C IRR

6.7 Years Rs. 24,083 1.12 13.9 %

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t Cash Flows PV

1 30,000 26785.71

2 30,000 23915.82

3 30,000 21353.41

4 30,000 19065.54

5 30,000 17022.81

6 30,000 15198.93

7 30,000 13570.48

8 30,000 12116.5

9 30,000 10818.3

10 30,000 9659.197

11 30,000 8624.283

12 30,000 7700.253

13 30,000 6875.226

14 30,000 6138.594

15 30,000 5480.888

16 30,000 4893.65

17 30,000 4369.33

18 30,000 3901.188

19 30,000 3483.203

20 30,000 3110.003

NPV (Savings) 224083.3

Discount Rate 0.12

life (N) 20

Example 2

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Replacement of Old Refrigerator with Energy efficient Refrigerator

The cost of a standard refrigerator is Rs 10,000 and the expected electricity consumption per year is 450 kWh. The cost of an energy efficient refrigerator of the same capacity (and with the same features) is Rs 10,500. For the same load, the annual electricity consumption is expected to be 400 kWh. What is the cost of saved energy? The life of the refrigerator can be taken as 10 years and discount rate of 12%.

CSE ={(10,500-10,000) *0.177} / (450-400)

= Rs 1.77/kWh.

If the price of electricity is assumed to be Rs. 5/KWh, investment in EE refrigeration seems feasible by a customer having discount rate 12%.

What about CSE at 30% and 60% discount rate? Is the project still viable?

For d=30%, CSE is Rs 3.23/kWh and for d= 60%, CSE is 6.06/kWh.

Case Study 1: Pump sets

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Sub Standard Pump Efficient New pump Existing old pumps

Pump Type Submersible type (3-Phase) Submersible type (3-Phase) Submersible type (3-Phase)

Pump Rating (kW) 3.7 3.7 3.7

Efficiency 0.3 0.5 0.2

Power Requirement

(kW) 12 7 19

Measure Life (Years) 4 10 3

Usage (Hours/Year) 1200 1200 1200

Retail Market price

(Rs.) 25000 35000 2000

Annual Eelctricty use

(kWh/Year) 14400 8400 22800

Discount Rate CRF CSE(Option 1) CSE(Option 2)

0.1 0.162745395 0.372958197 0.271242325

0.15 0.199252063 0.45661931 0.332086771

0.2 0.238522757 0.546614651 0.397537928

0.25 0.280072562 0.641832956 0.466787604

0.3 0.32346344 0.741270382 0.539105733

0.35 0.36831832 0.844062816 0.613863866

0.4 0.414323844 0.949492142 0.690539739

0.45 0.461226231 1.056976779 0.768710385

0.5 0.508823783 1.166054502 0.848039638

n=10

Option 1: Replacing Existing old pumps with Efficient New pump

Option 2: Replacing Sub Standard pumps with Efficient New pump

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

Cost of Saved Energy

CSE(Option 1) CSE(Option 2)

Case Study 2: Domestic Lighting

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ICL CFL LED

Power Rating (W) 60 11 7

Measure Life

(Hrs) 1000 8000 25000

Usage

(Hours/Year) 1050 1050 1050

Lamp Cost (Rs.) 15 90 100

Effective capital

cost (Rs) 375 270 100

Annual Eelctricty

use (kWh/Year) 63 11.55 7.35

Discount Rate CRF CSE(Option 1) CSE(Option 2)

Cost of

Electricity

0.1 0.458089697 -0.934876933 -2.263695717 5

0.15 0.494850327 -1.009898626 -2.44535202 5

0.2 0.532344369 -1.08641708 -2.630632551 5

0.25 0.570520448 -1.164327444 -2.819283435 5

0.3 0.609331298 -1.24353326 -3.011071102 5

0.35 0.648733424 -1.323945763 -3.20578062 5

0.4 0.688686787 -1.405483238 -3.403214129 5

0.45 0.72915451 -1.488070428 -3.603189402 5

0.5 0.770102611 -1.571637982 -3.805538509 5

n= 2.854Option 1: Replacing ICL with CFL

Option 2: Replacing ICL with LED

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-4

-3

-2

-1

0

1

2

3

4

5

6

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

Cost of Saved energy

CSE(Option 1) CSE(Option 2) Cost of Electricity

Thank you

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