Energy Management: Audits, Electrical Systems, Motors, Lighting, and Thermal and Mechanical Systems

This chapter answers Section XE8.2 of XE8 Energy Science, the sub-heading “Energy management”: “Energy auditing (methodology, analysis of past trends plants data), electrical systems (demand side management, power factor correction), motor efficiency testing, energy efficient motors, lighting (lighting levels, efficient options, fixtures, daylighting, timers), thermal and mechanical systems (insulation, compressors, pumps, boiler, heating and cooling systems).” It belongs to the Engineering Sciences (XE) paper, as Section XE8.2 of XE8 Energy Science, one of the nine optional sections, any two of which a candidate chooses in the exam. The chapter teaches the audit method and trend analysis, load factor, demand-side management and power factor correction, motor losses, testing and efficient motors, the lumen method and daylighting, and the energy saving routes for insulation, boilers, chillers, compressors and pumps. The numericals are specific energy from a baseline line, capacitor kvar, motor-replacement savings, luminaire count, and pump power under the affinity laws, with all data stated.

1. Energy audit: method and trend analysis

An energy audit is a systematic survey that finds where energy enters a facility, where it goes, and which savings are worth buying. A preliminary (walk-through) audit uses bills and a plant tour to rank the obvious opportunities; a detailed audit adds measurement and a full energy balance. The usual sequence: plan and collect past bills and production data; survey and measure (power analyser, flue-gas analyser, lux meter, flow and temperature instruments, thermal imager); build the energy balance (inputs against useful output and losses, which must close); identify measures; rank them by economics (payback, NPV, cost of saved energy); report; implement; and verify the savings against a baseline.

Analysis of past data starts with the specific energy consumption SEC = energy/production (kWh per tonne, per unit). SEC alone misleads, because part of the energy is fixed (lighting, standby, utilities) and part varies with output. Fit the baseline line E = E₀ + k P by regression of monthly energy on production: E₀ is the fixed load and k the marginal energy per unit. Then SEC = E₀/P + k falls as production rises, because the fixed part is spread over more units. With E₀ = 140 MWh and k = 1.0 MWh/t, SEC at 400 t is 540/400 = 1.35 MWh/t, at 200 t it is 1.70. A month whose actual energy exceeds the line is a month to investigate.

  • CUSUM (cumulative sum) is the running total of (actual − expected) energy, month by month; a line drifting upward means a sustained excess, a downward drift a saving, and a change of slope marks when performance changed.
  • Normalise before comparing: adjust for production, weather (degree-days for heating and cooling) and product mix, or two years of bills compare nothing.
⚠️ Falling SEC is not always saving
If production rises and E₀ is large, SEC falls even when nothing was improved. Compare actual energy with the baseline line at the same production, not SEC at different outputs.

2. Electrical systems: demand-side management and power factor correction

A bill for an industrial supply has an energy charge (kWh) and a maximum demand charge on the highest average kW (or kVA) in a billing interval. The load factor = (average demand)/(maximum demand) = kWh/(MD × hours). 360 000 kWh in a 720 h month with MD = 1000 kW gives 360 000/(1000 × 720) = 0.5. A high load factor means the capacity paid for is well used; a spiky load pays much for demand it uses seldom. Demand-side management (DSM) acts on the customer’s side to reshape the load: peak clipping (cut the peak), valley filling (build off-peak load), load shifting (move load from peak to off-peak, helped by time-of-day tariffs and thermal storage), and strategic conservation (lower the whole curve by efficiency).

Power factor cos φ = kW/kVA. Inductive loads (motors, transformers) draw lagging reactive power kvar, which carries no useful work but loads cables, transformers and generators and costs I²R loss. A shunt capacitor supplies kvar locally. To raise the power factor of a load P from cos φ₁ to cos φ₂ at constant kW: Q_c = P (tan φ₁ − tan φ₂). For 500 kW at 0.80 raised to 0.95: tan φ₁ = 0.75, tan φ₂ = 0.3287, so Q_c = 500 × 0.4213 = 210.7 kvar. The kVA falls from 625 to 526, the current falls to 0.80/0.95 = 0.842 of its value, and the I²R loss to (0.842)² = 0.709, a 29 % reduction. The kW drawn by the load does not change.

⚠️ Over-correction and the real benefit
Capacitors larger than the lagging kvar push the power factor leading and can raise voltage, so banks are switched in steps. The saving is in the supply side (lower kVA demand, lower line loss, released capacity, avoided low-power-factor penalty), not in the kWh the load itself uses.

3. Motor losses, efficiency testing and energy-efficient motors

An induction motor’s efficiency η = output/input = 1 − losses/input. Its losses are fixed (core loss in the iron, friction and windage, roughly constant from no load to full load) and variable (stator and rotor copper, rising as I², plus stray load loss). Efficiency is nearly flat from about half to full load and falls steeply at light load, because the fixed loss is then a large share of a small input; the power factor also falls at light load.

  • Direct (input-output) testing measures electrical input and shaft output (by dynamometer) and divides; it is sensitive to instrument error because the two numbers are close. Indirect (loss-segregation) testing measures the individual losses and subtracts them from the input; a no-load test gives the fixed losses (core, friction and windage), a blocked-rotor test and the stator resistance give the copper losses at the test current, and stray load loss is estimated from the test method’s procedure. Accepted test standards define these procedures, so results from different methods are not interchangeable.
  • Load estimation in the field by slip: load % ≈ (N_s − N)/(N_s − N_rated) × 100, with N_s the synchronous speed. For N_s = 1500 rpm, rated 1440 rpm and measured 1455 rpm, load = 45/60 = 75 %. It is approximate and is affected by supply voltage.
  • Energy-efficient motors cut the losses: more and better copper (lower I²R), low-loss thin laminated steel and more of it (lower core loss), a smaller air gap, a better fan and bearings. They cost more and run with lower slip, which is a trap on fans and pumps: a slightly higher speed raises load power as N³, which can eat the saving.

Worked: a motor delivering 30 kW shaft output for 6000 h a year is replaced, efficiency rising from 90 % to 94 %. Input falls from 30/0.90 = 33.33 kW to 30/0.94 = 31.91 kW. Saving = 1.4184 kW × 6000 h = 8511 kWh, or ₹68 085 a year at ₹8/kWh.

4. Lighting: levels, lumen method, efficient options and daylighting

Illuminance E is the luminous flux falling on a surface per unit area, in lux (1 lux = 1 lm/m²); recommended levels depend on the task (a corridor needs far less than precision work). The luminous efficacy of a source = lumens/watt (power input, including ballast or driver for the system efficacy). In the lumen method the number of luminaires is N = E A/(F × UF × MF): A the working-plane area, F the luminaire’s flux, UF the utilisation factor (the fraction of lamp lumens reaching the working plane, set by room shape, reflectances and luminaire) and MF the maintenance factor (allowing for lamp ageing and dirt). For a 20 m × 10 m room at 300 lux with 3000 lm luminaires, UF = 0.6 and MF = 0.8: N = 300 × 200/(3000 × 0.6 × 0.8) = 41.7, so at least 42.

  • Efficient options: replace low-efficacy incandescent lamps with LED or fluorescent sources; use electronic ballasts or drivers in place of magnetic ballasts (lower ballast loss, no flicker); use high-efficiency luminaires with good reflectors and optics so more lumens reach the task; use task lighting and lower general levels; keep lamps and fixtures clean and surfaces light-coloured.
  • Controls: timers for fixed schedules, occupancy sensors for intermittent spaces, and photocell dimming that cuts electric light when daylight is available. A lighting load left on for 4000 h that is cut to 3000 h saves a quarter of its energy at no capital cost.

Daylighting displaces electric light with natural light. The daylight factor DF = (indoor horizontal illuminance at a point)/(simultaneous outdoor horizontal illuminance from an unobstructed sky) × 100 %, so an indoor 150 lux against an outdoor 5000 lux is 3 %. DF falls with depth into the room, so side windows light the first few metres; roof lights and light shelves reach deeper. The design must balance glare and the heat gain that daylight brings, which adds to cooling load.

5. Thermal systems: insulation, boilers, heating and cooling

Insulation adds thermal resistance in series. Through a plane wall of area A, q = ΔT/ΣR″ with R″ = L/k for each layer (m²K/W), neglecting surface films. A wall of thickness 0.2 m and k = 0.8 W/m·K has R″ = 0.25; adding 0.05 m of insulation of k = 0.04 adds 1.25, so for ΔT = 20 K the flux falls from 20/0.25 = 80 W/m² to 20/1.5 = 13.3 W/m². The thickness that is economic is where the saving from one more increment equals its cost, so it is set by fuel price and by the discount rate. On a small pipe or wire there is a trap: for a cylinder with outside coefficient h, insulation up to the critical radius r_c = k_ins/h raises the heat loss, because the larger area outweighs the added resistance.

Boiler efficiency by the direct method is (steam flow × enthalpy gain)/(fuel flow × calorific value): 10 t/h of steam gaining 2400 kJ/kg from 0.8 t/h of fuel of 40 000 kJ/kg is 24 × 10⁶/32 × 10⁶ = 75 %. The indirect method subtracts losses: stack (dry flue gas) loss, which rises with stack temperature and excess air; moisture-in-fuel and hydrogen-combustion loss; unburnt carbon; radiation; and blowdown. The saving routes follow: tune air to the lowest excess air that keeps CO low, add an economiser or air pre-heater, clean heat-transfer surfaces (fouling raises stack temperature), insulate, return condensate, and recover heat from blowdown.

Cooling. A chiller’s performance is COP = refrigeration effect/compressor power, or in kW/TR; 1 TR (ton of refrigeration) = 3.517 kW, so kW/TR = 3.517/COP, and a COP of 4.0 is 0.88 kW/TR. Efficiency improves by raising the evaporator temperature, lowering the condenser temperature (clean condenser, good cooling-tower approach), insulating cold lines, and avoiding the mixing of hot and cold air. Heating savings come from heat recovery between hot waste streams and cold feeds, and from supplying heat at the lowest useful temperature.

6. Mechanical systems: compressors and pumps

Compressed air is an expensive utility, and in a typical plant much of the electricity is lost as heat. The performance measure is specific power, kW per unit of free air delivered (kW per m³/min). The saving routes: repair leaks (a leak flows all the time and costs the compressor power continuously); lower the delivery pressure to what the users need (work per kg rises with the pressure ratio); draw cool intake air (for the same pressure ratio, work per kg is proportional to the inlet temperature T₁); avoid running unloaded; match the number of machines to demand; use drain traps rather than open drains; recover the heat of compression; and do not use compressed air where a blower or electric tool will do.

Pumps and fans. Hydraulic power = ρ g Q H, and shaft power = ρ g Q H/η. For Q = 0.05 m³/s, H = 30 m and η = 0.70, shaft power = 1000 × 9.81 × 0.05 × 30/0.70 = 21.0 kW. The affinity laws for a fixed system at ideal similarity are Q ∝ N, H ∝ N², P ∝ N³. Reducing speed by 20 % cuts power to 0.8³ = 0.512, a 48.8 % saving, whereas throttling a valve to the same flow keeps the speed and the pump runs up its curve with the loss across the valve, saving far less. Hence a variable-speed drive suits variable flow, and a pump chosen with a big margin and throttled wastes energy; trimming the impeller or fitting a smaller pump is the alternative for a fixed duty.

⚠️ Affinity laws are for the fixed system curve
P ∝ N³ assumes the system resistance is purely dynamic (friction), so that head follows N². If much of the head is static lift, flow does not fall in proportion to speed at the same head, and the cubic saving is not obtained; check the system curve first.

Key takeaways

  • An audit measures, balances and ranks: baseline E = E₀ + kP separates fixed and variable energy, SEC = E₀/P + k falls with output, and CUSUM exposes drift; normalise before comparing.
  • Load factor = kWh/(MD × h); DSM reshapes load by peak clipping, valley filling, load shifting and conservation; MD and tariff set the value of each.
  • Power factor correction needs Q_c = P (tan φ₁ − tan φ₂); it cuts kVA, current and I²R loss, but not the load’s own kW, and over-correction must be avoided.
  • Motor losses are fixed (core, friction) and variable (copper); indirect testing segregates them (no-load, blocked-rotor), and efficient motors cut them but run at lower slip, which matters on fans and pumps.
  • Lighting: N = E A/(F × UF × MF); efficacy, electronic ballasts, good fixtures, timers, sensors and daylighting (daylight factor = indoor/outdoor illuminance) cut lighting energy.
  • Thermal and mechanical systems: insulation (economic thickness, critical radius), boiler stack loss and economisers, chiller COP (1 TR = 3.517 kW), compressor leaks and pressure, and pump power ∝ N³ with speed control beating throttling.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A plant’s monthly CUSUM of (actual − expected) energy, plotted against the production-based baseline, shows a line that has been rising steadily for the last six months. This indicates that

    1. energy use has been below the baseline expectation in each of those six months, so the plant is saving energy
    2. production has been rising over those months, and nothing can be concluded about efficiency from the plot
    3. the baseline regression line has a negative slope with respect to production
    4. energy use has been above the baseline in those months, so the cause should be investigated
    Show answer

    Answer: D — energy use has been above the baseline in those months, so the cause should be investigated

    CUSUM adds (actual − expected) month by month, so a steady rise means each month is above expectation. Production changes are already allowed for in the expected value, so they do not explain the rise.

  2. A factory uses 360 000 kWh in a 720 h billing month, and its maximum demand is 1000 kW. Its load factor is

    1. 0.36, from dividing the monthly energy by the maximum demand in kW
    2. 0.50, from the average demand over the maximum demand
    3. 2.00, from the maximum demand divided by the average demand
    4. 0.72, from the hours divided by 1000
    Show answer

    Answer: B — 0.50, from the average demand over the maximum demand

    Average demand = 360 000/720 = 500 kW, so load factor = 500/1000 = 0.5. The inverse (2.0) is impossible for a load factor, which cannot exceed 1.

  3. A shunt capacitor bank raises the power factor of an inductive load from 0.80 to 0.95 lagging, the load kW being unchanged. Which of the following is correct?

    1. The kW drawn by the load falls in proportion to the power factor
    2. The supply current rises, since the capacitor adds a load
    3. The supply current and the I²R loss in the feeder fall
    4. The reactive power drawn by the load itself becomes zero at all times
    Show answer

    Answer: C — The supply current and the I²R loss in the feeder fall

    At fixed kW and voltage, I ∝ 1/cos φ, so the current falls to 0.80/0.95 of its value and I²R falls by about 29 %. The load’s kW and its own reactive demand are unchanged; the capacitor supplies the kvar locally so that less is drawn from the supply.

  4. A capacitor bank rated well above the lagging kvar of the load is left permanently connected, including at light load. The most likely consequence is

    1. the power factor becomes leading and the supply voltage at the load can rise
    2. the kW demand of the load rises in proportion to the excess kvar
    3. the maximum-demand charge in kW falls to zero
    4. the line current falls to zero at all loads
    Show answer

    Answer: A — the power factor becomes leading and the supply voltage at the load can rise

    Excess capacitive kvar over-corrects, so the net reactive power is leading, the power factor is leading, and the voltage can rise; that is why banks are switched in steps with the load. The load’s kW is unaffected and the line current cannot fall to zero while the load draws real power.

  5. In the loss-segregation (indirect) method of testing an induction motor, the no-load test is mainly used to determine

    1. the rotor copper loss at full load, from the slip
    2. the stator copper loss at rated current and rated load
    3. the fixed losses, core loss and friction and windage
    4. the stray load loss at rated load, from the no-load input
    Show answer

    Answer: C — the fixed losses, core loss and friction and windage

    With no shaft load the current is small, so copper loss is small and the input is almost wholly the core and friction-and-windage loss. Copper losses at rated current come from the blocked-rotor test and the winding resistances.

  6. A standard induction motor driving a centrifugal fan is replaced by an energy-efficient motor of the same rating, which runs at a slightly higher speed because its slip is lower. The fan is not otherwise changed. Which statement is correct?

    1. The saving is certain and equals the efficiency gain, because fan power does not depend on the speed of the shaft
    2. The fan power falls, because a higher speed means a lower load
    3. The motor’s losses rise, because a lower slip means more rotor copper loss
    4. The fan power rises roughly as the cube of the speed ratio and can offset the efficiency gain
    Show answer

    Answer: D — The fan power rises roughly as the cube of the speed ratio and can offset the efficiency gain

    For a centrifugal fan P ∝ N³, so a 1.5 % rise in speed raises power by about 4.6 %, comparable to the efficiency gain of a few per cent. A speed-regulating drive or a re-adjusted pulley restores the saving. The slip is lower because rotor copper loss is lower, not higher.

  7. At a point in a room the illuminance measured at the same instant indoors is 150 lux, while the horizontal illuminance outdoors from an unobstructed sky is 5000 lux. The daylight factor at that point is

    1. 3 %, from 150/5000 × 100
    2. 33 %, from 5000/150
    3. 97 %, from (5000 − 150)/5000 × 100
    4. 30 %, from 150/500
    Show answer

    Answer: A — 3 %, from 150/5000 × 100

    DF = (indoor illuminance/outdoor illuminance) × 100 = 150/5000 × 100 = 3 %. The inverse and the fraction lost outdoors are not the definition.

  8. A very thin bare conductor of outside radius 3 mm loses heat to still air with an outside coefficient h = 8 W/m²K. It is covered with a thin layer of insulation of k = 0.05 W/m·K. For the first millimetres of added insulation the heat loss per unit length will

    1. fall, because any added insulation always adds conduction resistance and so must reduce the loss
    2. remain unchanged, because a layer of insulation does not alter the outer surface area
    3. rise, because the critical radius k/h = 6.25 mm exceeds the outer radius
    4. fall to zero as soon as the outer radius passes 3 mm
    Show answer

    Answer: C — rise, because the critical radius k/h = 6.25 mm exceeds the outer radius

    The critical radius of insulation on a cylinder is r_c = k/h = 0.05/8 = 6.25 mm. When the radius is below r_c, the growth in outer surface outweighs the added conduction resistance, so the loss first rises, peaking at r_c, then falls.

  9. For a reciprocating or rotary air compressor at a fixed delivery pressure and a fixed delivered volume flow of free air, drawing intake air at a lower temperature is beneficial because

    1. the work per kg at a given pressure ratio is proportional to the inlet absolute temperature
    2. cold air has a much larger specific heat, so it carries away more of the heat of compression
    3. the pressure ratio of the compressor then falls to unity at all loads
    4. the leaks in the delivery line stop once the intake air is cold
    Show answer

    Answer: A — the work per kg at a given pressure ratio is proportional to the inlet absolute temperature

    The compression work per unit mass of an ideal gas is proportional to T₁ for a fixed pressure ratio and process, so a colder intake reduces the energy per kg. Specific heat varies little with temperature, and cold intake does not change the pressure ratio or stop leaks.

  10. Which of the following measures reduce the stack (dry flue gas) loss of a boiler?

    1. Tuning the air supply to the lowest excess air consistent with low CO
    2. Fitting an economiser that cools the flue gas by heating the feedwater
    3. Cleaning soot and scale from the heat-transfer surfaces
    4. Raising the excess air well above the optimum to ensure complete burnout
    Show answer

    Answer: A — Tuning the air supply to the lowest excess air consistent with low CO; B — Fitting an economiser that cools the flue gas by heating the feedwater; C — Cleaning soot and scale from the heat-transfer surfaces

    Less excess air means less gas to carry heat out, an economiser lowers the stack temperature, and clean surfaces transfer more heat and lower the stack temperature. Excess air far above the optimum increases the flue-gas flow and so increases stack loss.

  11. Which of the following statements about a centrifugal pump on a pure-friction system are correct?

    1. Under the affinity laws the shaft power varies as the cube of the speed
    2. Throttling a valve to reduce flow saves as much energy as reducing speed to the same flow
    3. Hydraulic power equals ρ g Q H
    4. An oversized pump throttled to the required flow wastes energy in the valve
    Show answer

    Answer: A — Under the affinity laws the shaft power varies as the cube of the speed; C — Hydraulic power equals ρ g Q H; D — An oversized pump throttled to the required flow wastes energy in the valve

    P ∝ N³ and the hydraulic power is ρ g Q H; a throttled oversized pump dissipates head across the valve. Speed control removes that loss, so the claim that throttling saves as much as slowing the pump is false.

  12. Which of the following are demand-side management strategies?

    1. Load shifting, moving a process from peak to off-peak hours
    2. Peak clipping, reducing load during the peak period
    3. Valley filling, building load during off-peak hours
    4. Raising the peak so the maximum demand is as large as possible
    Show answer

    Answer: A — Load shifting, moving a process from peak to off-peak hours; B — Peak clipping, reducing load during the peak period; C — Valley filling, building load during off-peak hours

    Load shifting, peak clipping and valley filling are the standard load-shape strategies. Raising the maximum demand is the opposite of the aim, since demand charges and capacity costs rise with the peak.

  13. A plant’s baseline from regression of monthly data is E = E₀ + kP, where E is in MWh and P in tonnes. Two points on the line are 260 MWh at 120 t and 340 MWh at 200 t. The specific energy consumption at a production of 400 t is ____ MWh/t (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 1.35

    k = (340 − 260)/(200 − 120) = 1.0 MWh/t; E₀ = 340 − 1.0 × 200 = 140 MWh. At 400 t, E = 140 + 400 = 540 MWh, so SEC = 540/400 = 1.35 MWh/t. The SEC at 200 t would be 1.70, showing how the fixed energy is spread.

  14. A load of 500 kW operates at a power factor of 0.80 lagging. A capacitor bank is installed to raise it to 0.95 lagging with the kW unchanged. The kvar rating of the bank required is ____ kvar (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 210.7

    tan φ₁ = tan(cos⁻¹ 0.80) = 0.75; tan φ₂ = tan(cos⁻¹ 0.95) = 0.32868. Q_c = 500 × (0.75 − 0.32868) = 210.66 = 210.7 kvar. Using kVA difference 625 − 526 = 99 is wrong, since kVA is not the reactive power.

  15. A motor delivers a constant 30 kW at the shaft for 6000 hours a year. It is replaced by a motor of efficiency 94 % in place of the existing 90 %, the shaft load being unchanged. Electricity costs ₹8 per kWh. The annual saving is ₹ ____ (nearest integer).

    Numerical answer — type the value.

    Show answer

    Answer: 68085

    Input before = 30/0.90 = 33.333 kW; after = 30/0.94 = 31.915 kW; difference = 1.4184 kW. Energy saved = 1.4184 × 6000 = 8510.6 kWh; saving = 8510.6 × 8 = ₹68 085. Using 30 × 0.04 × 6000 × 8 = ₹57 600 is wrong, because the efficiency difference does not apply to the shaft output.

  16. A hall of floor area 20 m × 10 m needs an average illuminance of 300 lux on the working plane. Each luminaire gives 3000 lm. The utilisation factor is 0.6 and the maintenance factor is 0.8. The least whole number of luminaires required by the lumen method is ____ .

    Numerical answer — type the value.

    Show answer

    Answer: 42

    N = E A/(F × UF × MF) = 300 × 200/(3000 × 0.6 × 0.8) = 60 000/1440 = 41.67. A fraction of a luminaire cannot be installed, so the least whole number is 42.

  17. A centrifugal pump on a pure-friction system is slowed by 20 % to 80 % of its original speed, with the system curve unchanged and the affinity laws valid. The reduction in shaft power is ____ % (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 48.8

    P ∝ N³, so P₂/P₁ = 0.8³ = 0.512. Reduction = 1 − 0.512 = 0.488 = 48.8 %. Taking the reduction as 20 % (linear in speed) or 36 % (square) are the common errors.