Energy audit

An energy audit is a systematic survey and analysis of energy flows in a building, process or plant to find where energy is used, how efficiently, and how much can be saved at what cost. As per the Bureau of Energy Efficiency (BEE) definition, it verifies, monitors and analyses the use of energy and submits a technical report with recommendations.

Preliminary versus detailed audit

Steps of a detailed audit

1. Plan and pre-audit: define scope, form audit team, collect plant details, process flow, past energy bills (12–36 months), and tariff.
2. Walk-through survey: inspect the main users, note obvious wastes and decide where to measure.
3. Measurement and data collection: measure electricity, fuel, steam, air, water, temperature and flow of major equipment.
4. Analysis: prepare energy balance, specific energy consumption (kWh per unit of product), load profiles, efficiency of equipment, and compare with benchmarks.
5. Identify energy conservation measures (ECMs): no-cost (housekeeping), low-cost (maintenance, controls) and high-cost (new equipment).
6. Economic evaluation: simple payback = investment ÷ annual saving, net present value and internal rate of return.
7. Report and presentation with action plan, then implementation and follow-up (monitoring of savings).

Instruments

Energy balance

It is an account of energy input equal to useful output plus losses, from the first law:

$$\text{Energy input} = \text{useful output} + \text{losses}$$

Steps: draw a system boundary; list all inputs (electricity, fuel, steam) and outputs; measure each stream; check that the balance closes within about 5 %; find the largest losses.

Sankey diagram (example: boiler, 100 units of fuel)

 Fuel 100 ===+=========> Steam 82 (useful)
             |==> Flue gas 12
             |==> Radiation/convection 3
             |==> Blowdown, others 3

The width of each band is proportional to the energy it carries, so the largest losses (flue gas here) are seen directly. Boiler efficiency = 82/100 = 82 %, and the audit would address flue gas heat recovery and excess air.

An audit report is the final output and the basis for management decisions, so it must be clear, factual and action-oriented.

Contents of the report

1. Title page and executive summary: one or two pages stating the main findings, total savings (kWh, fuel, money), investment and payback.
2. Introduction: objective, scope, method, audit team, dates.
3. Facility description: products, processes, operating hours, energy sources and tariff.
4. Energy use analysis: monthly bills, load profile, energy balance, specific energy consumption, benchmark comparison.
5. Findings by system: boiler, motors and pumps, compressed air, lighting, HVAC, power factor, with measured data and efficiency.
6. Energy conservation measures (ECMs): each with description, saving, cost, payback; grouped as no-cost, low-cost and high-cost.
7. Action plan: priority, responsibility and time schedule; monitoring and verification plan.
8. Appendices: measurements, calculations, instrument list and calibration, drawings.

Presentation to management

• Begin with the summary: total savings and payback in the currency and units that the management uses.
• Present ECMs in order of payback; no-cost and low-cost measures first, with quick wins showing results.
• Use charts: pie chart of energy use by end-use, bar chart of monthly consumption, Sankey diagram, and a table of savings.
• Be specific and honest about assumptions and uncertainty; state also non-energy benefits (maintenance, comfort, reduced emission).
• End with a decision request and an implementation plan.

(i) Lighting retrofit

Existing load per fitting $= 36 + 8 = 44$ W; proposed $= 18$ W.

$$\begin{aligned} \Delta P &= 100 \times (44 - 18) = 2600\ \text{W} = 2.6\ \text{kW} \\ \text{Operating hours} &= 10 \times 300 = 3000\ \text{h/yr} \\ \Delta E &= 2.6 \times 3000 = 7800\ \text{kWh/yr} \\ \text{Saving} &= 7800 \times 12 = \text{Rs } 93\,600/\text{yr} \end{aligned}$$

Investment $= 100 \times 650 = \text{Rs } 65\,000$.

$$\text{Payback} = \frac{65\,000}{93\,600} = 0.69\ \text{yr} = 8.3\ \text{months}$$

(ii) Motor replacement

Shaft output $= 15 \times 0.80 = 12$ kW.

$$\begin{aligned} P_{in,old} &= \frac{12}{0.88} = 13.64\ \text{kW} \\ P_{in,new} &= \frac{12}{0.93} = 12.90\ \text{kW} \\ \Delta P &= 0.733\ \text{kW} \\ \Delta E &= 0.733 \times 6000 = 4399\ \text{kWh/yr} \\ \text{Saving} &= 4399 \times 12 = \text{Rs } 52\,786/\text{yr} \end{aligned}$$ $$\text{Payback} = \frac{160\,000}{52\,786} = 3.03\ \text{yr} = 36.4\ \text{months}$$

Combined

$$\Delta E = 7800 + 4399 = 12\,199\ \text{kWh/yr}, \quad \text{Saving} = 93\,600 + 52\,786 = \text{Rs } 146\,386/\text{yr}$$ $$\text{Investment} = 65\,000 + 160\,000 = \text{Rs } 225\,000, \quad \text{Payback} = \frac{225\,000}{146\,386} = 1.54\ \text{yr}$$

Answer: lighting saves 7800 kWh (Rs 93 600), payback 8.3 months; motor saves 4399 kWh (Rs 52 786), payback 3.0 years; combined 12 199 kWh (Rs 146 386) per year, payback 1.54 years (about 18 months). The lighting measure should be done first.