Read a Building's Electrical Story: A Worked Example | Cobler
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Capstone: Read a Building's Electrical Story
A worked exercise to finish the course. Take one example office tower from a load list to kWh, a motor nameplate to measured input, kW to kVA and current, half-hour data to maximum demand, and demand to the lines on a TNB bill, with a shoplot for contrast.
Every month, a building receives one electricity bill with one total at the bottom. Behind that total are lights, computers, fans and pumps, each drawing power for so many hours, and a meter recording what they do together. Can you follow the trail from one piece of equipment to a line on the bill?
This final part is a worked exercise. It uses our example office tower, a fictional 20-storey building in Kuala Lumpur, plus a short comparison with our example shoplot. You get the same kind of information a facility team works with: a load list, a motor nameplate, meter readings and a TNB (Tenaga Nasional Berhad) bill. You then work through six tasks.
How to use this exercise
Each task gives you some information and asks a few questions. Try the questions first with a calculator and a notepad. The Worked answer comes straight after each task so you can check your working.
All the numbers are fictional but consistent with each other. Tariff figures are the RP4 rates (the TNB tariff structure in force since July 2025) for Peninsular Malaysia, and the fuel surcharge figure is for September 2026.
Three-phase current depends on kW, voltage and PF. From Three-Phase Power Explained. The formula is set out in Task 3.
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A facility engineer has measured some of the loads on one office floor of the tower (level 12, east half). The figures are average electrical input over a normal weekday, taken from the floor's sub-meters and the building management system (BMS).
Load on level 12 east
Average power while running
Hours running per weekday
Lighting
12 kW
10 (8am to 6pm)
Socket outlets (computers, screens, printers)
15 kW
10 (8am to 6pm)
Supply fan of air handling unit AHU-12A
6 kW
11 (7am to 6pm)
Assume 22 working days in a month, and ignore weekends and night-time standby loads.
How many kWh do these three loads use on one weekday?
How many kWh do they use in a month?
The tower is on the MV (medium voltage) non-domestic General tariff. Its energy rate is 29.83 sen/kWh, and the September 2026 Automatic Fuel Adjustment (AFA) surcharge is 3.67 sen/kWh. What do these kWh add to the energy and AFA lines of the bill?
Worked answer
Multiply each load's power by its hours:
Lighting: 12 kW × 10 h = 120 kWh
Sockets: 15 kW × 10 h = 150 kWh
AHU-12A fan: 6 kW × 11 h = 66 kWh
One weekday: 120 + 150 + 66 = 336 kWh.
One month: 336 kWh × 22 days = 7,392 kWh.
Energy line: 7,392 × 29.83 sen = RM2,205.03. AFA line: 7,392 × 3.67 sen = RM271.29. Together they add RM2,476.32, which is the same as 7,392 × 33.50 sen.
Two points to notice. First, the socket figure is a measured average, not the total of every appliance's rating. If you added up the ratings of every computer, screen and printer on the floor, you would get a much bigger number that never occurs in practice. Second, these kWh do not tell you anything about the capacity and network charges. On an MV bill, those two lines depend on maximum demand, which Task 4 covers.
Task 2: Nameplate versus measurement
The tower's chiller plant has condenser water pumps that run at a fixed speed. One pump motor has this nameplate:
Rated output: 30 kW
400 V, three-phase, 50 Hz
Rated current: 54 A
Power factor: 0.86
Efficiency: 93.6%
A meter on the pump's feeder cable shows the motor drawing 27.0 kW of electrical input while it runs.
What does the 30 kW on the nameplate describe?
How much electrical input would the motor draw at full load?
Is the motor running at, above or below full load? Roughly how much power reaches the pump shaft?
To estimate this pump's monthly kWh, should you use 30 kW or 27.0 kW?
Worked answer
The 30 kW is the rated output: the mechanical power the motor can deliver at its shaft, continuously, at full load. It is not the electrical power the motor draws. How Electric Motors Work explains how a motor turns electrical input into shaft output.
The motor loses some energy as heat, so the input must be larger than the output. At full load: input = output ÷ efficiency = 30 ÷ 0.936 = about 32.1 kW.
The meter reads 27.0 kW, which is less than the 32.1 kW full-load input. So the motor is running below full load, at about 27.0 ÷ 32.1 ≈ 84% of it. If we assume the efficiency stays close to the rated 93.6% at this loading, the shaft receives about 27.0 × 0.936 ≈ 25.3 kW. The other 1.7 kW or so becomes heat in the motor.
Use the measured 27.0 kW. The meter records electrical input, and electrical input is what adds up to the kWh on the TNB bill. Using 30 kW would mix up output and input, and it would still be wrong at full load, because the input would then be about 32.1 kW, not 30 kW.
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The meter measures what goes into the motor; the nameplate kW describes what the shaft can deliver at full load.
Task 3: kW, kVA and power factor
The same meter shows the pump motor's power factor as 0.85 at this load. The supply is 400 V three-phase.
What is the motor's apparent power in kVA?
What current flows in each phase of its cable?
Use the same formula to check the nameplate's 54 A.
At the tower's maximum demand of 1,600 kW, the power factor at the main switchboard is 0.92 (after the capacitor banks). What is the building's total kVA, and how does it compare with the two 1,500 kVA transformers?
Worked answer
Apparent power is real power divided by power factor:
$$S = \frac{P}{PF}$$
Here \(S\) is apparent power in kVA, \(P\) is real power in kW and \(PF\) is the power factor.
For a balanced three-phase load, the current in each phase is:
$$I = \frac{P}{\sqrt{3} \times V \times PF}$$
Here \(I\) is the current in amps, \(P\) is the real power in watts, \(V\) is the line-to-line voltage (400 V) and \(\sqrt{3}\) is about 1.732.
Pump kVA: 27.0 ÷ 0.85 ≈ 31.8 kVA.
Pump current: 27,000 ÷ (1.732 × 400 × 0.85) = 27,000 ÷ 588.9 ≈ 45.8 A per phase.
Nameplate check: at full load the input is about 32,050 W (from Task 2) and the PF is 0.86. The current is 32,050 ÷ (1.732 × 400 × 0.86) = 32,050 ÷ 595.8 ≈ 53.8 A, which rounds to the nameplate's 54 A. The nameplate is consistent with itself. The measured 45.8 A is lower because the motor is below full load.
Building kVA: 1,600 ÷ 0.92 ≈ 1,739 kVA. The two transformers can supply 2 × 1,500 = 3,000 kVA together, so at maximum demand they are carrying about 1,739 ÷ 3,000 ≈ 58% of their combined rating. If all 1,600 kW were delivered at 400 V (ignoring the small losses in the transformers), the total current would be 1,600,000 ÷ (1.732 × 400 × 0.92), or about 2,510 A, shared between the two transformers.
The capacitor banks at the main switchboard raise the power factor seen by the transformers and TNB. They do not change the pump's own 0.85 power factor or the 45.8 A in its cable, because the correction happens upstream of the pump. Power Factor Correction: What Capacitor Banks Do explains why.
Task 4: Find the maximum demand from half-hour data
The tower's monitoring system logs the energy used in each half hour. Here is part of Thursday 17 September 2026, a hot weekday afternoon. By the end of the month, this afternoon turns out to contain the month's highest reading.
Half hour
Energy in the half hour (kWh)
12:00 to 12:30
710
12:30 to 13:00
695
13:00 to 13:30
725
13:30 to 14:00
755
14:00 to 14:30
785
14:30 to 15:00
800
15:00 to 15:30
790
15:30 to 16:00
770
16:00 to 16:30
745
16:30 to 17:00
715
Turn each reading into an average demand in kW.
What is the maximum demand, and when was it set?
On MV General, the capacity and network charges together are RM89.27 per kW of MD (RM29.43 + RM59.84). What does this MD cost for the month?
What would change if the tower were on the MV ToU (time of use) tariff instead?
Worked answer
Each reading is energy over half an hour. Average power = energy ÷ time = kWh ÷ 0.5 h, which is the same as kWh × 2. For example, 710 kWh in half an hour is an average of 1,420 kW. The ten averages are 1,420, 1,390, 1,450, 1,510, 1,570, 1,600, 1,580, 1,540, 1,490 and 1,430 kW.
The highest half hour is 2:30pm to 3:00pm at 1,600 kW (800 kWh × 2). Because no other half hour in September is higher, the month's MD is 1,600 kW. (For this exercise, assume the logged half hours line up with the periods the TNB meter uses. The TNB meter's own MD register is what appears on the bill.)
MD cost on MV General: 1,600 kW × RM89.27 = RM142,832. That is capacity RM47,088 (1,600 × RM29.43) plus network RM95,744 (1,600 × RM59.84).
On MV ToU, the MD is charged only for the peak window: 2pm to 10pm, Monday to Friday. Weekends are off-peak all day. Readings between 12:00 and 14:00 would not count towards the charged MD, even if one of them had been the highest. In this data the 2:30pm half hour is inside the peak window, so the charged MD would still be 1,600 kW, at the ToU rate of RM97.06/kW: 1,600 × RM97.06 = RM155,296. The energy line would also change, because ToU charges separate peak and off-peak rates. A ToU bill cannot be compared with a General bill from this afternoon alone; the whole month's kWh, split by time, is needed.
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The tallest half-hour bar sets the maximum demand; on MV ToU only bars inside the shaded 2pm to 10pm weekday window would count.
Task 5: Read the office tower's bill, then the shoplot's
Here is the tower's September 2026 bill on MV General, before the Renewable Energy Fund levy (KWTBB) and tax:
Energy: 500,000 kWh × 29.83 sen = RM149,150
Capacity: 1,600 kW × RM29.43 = RM47,088
Network: 1,600 kW × RM59.84 = RM95,744
Retail: RM200
AFA (September 2026): 500,000 kWh × 3.67 sen = RM18,350
Total: RM310,532
Suppose the facility team lowers the MD by 100 kW, to 1,500 kW, by moving some load out of the hottest half hours. The month's kWh stay the same. Which lines change, by how much, and what is the new total?
Which lines do not change, and why?
Our example shoplot is on LV (low voltage) non-domestic General and uses 8,000 kWh a month. Its bill is energy RM2,162.40, capacity RM706.40, network RM1,185.60, retail RM20.00 and AFA RM293.60, a total of RM4,368.00. Would lowering the shoplot's peak demand reduce this bill?
Worked answer
Only the two per-kW lines change:
Capacity falls by 100 × RM29.43 = RM2,943, to RM44,145.
Network falls by 100 × RM59.84 = RM5,984, to RM89,760.
The saving is RM8,927 (the same as 100 × RM89.27). The new total is RM310,532 − RM8,927 = RM301,605, about 2.9% lower.
Energy, retail and AFA stay the same. Energy and AFA are charged per kWh, and the kWh did not change because the load was moved, not removed. Retail is a fixed RM200 a month. If the team had switched load off instead of moving it, the energy and AFA lines would also fall, by 33.50 sen for every kWh saved.
No. On LV tariffs, capacity and network are charged per kWh (8.83 and 14.82 sen/kWh), and there is no per-kW demand charge. Only using fewer kWh reduces the shoplot's bill. Each kWh saved removes 27.03 + 8.83 + 14.82 + 3.67 = 54.35 sen from the energy, capacity, network and AFA lines. The shoplot's average cost of about 54.6 sen/kWh (RM4,368 ÷ 8,000 kWh) is slightly higher because it includes the fixed RM20 retail charge.
The comparison shows why the same advice does not suit both buildings. In the tower, a lower peak saves money even when kWh stay the same. In the shoplot, only lower kWh saves money.
Task 6: Safety check
While collecting information for this exercise, you walk around level 12 and the chiller plant. Decide which of these you may do yourself, and which need the electrical competent person responsible for the installation.
Read the kWh and kW figures on a sub-meter display through the closed door of a distribution board (DB).
Read the pump motor's nameplate, which sits close to the coupling guard.
You notice a burning smell and a discoloured breaker inside the level 12 DB.
You want to clamp a portable power meter onto the pump's cable to check the 27.0 kW reading.
The floor's residual current device (RCD) is due for its test-button check.
Worked answer
You may do this. Reading meter displays, BMS screens and bills does not involve touching anything live.
You may read it only if you can do so without removing a guard or reaching near moving parts. Otherwise, ask for it to be read while the pump is stopped and made safe by the person responsible.
Do not open the board or touch the breaker. Keep people away and report it immediately to the competent person responsible. Follow the site's emergency procedure if there is smoke or fire.
This needs a competent person. Fitting a clamp means opening a panel and working near live conductors. The person's Suruhanjaya Tenaga (ST) registration and certificate must cover the work concerned.
You may press the test button if the site's procedure assigns this to you, at the interval stated by the manufacturer, after arranging for the power interruption. Operate only a safely accessible button and do not remove a board cover. If it does not trip, report it immediately and have the affected circuit made safe and checked. The test button checks the RCD's built-in test function only. Measured RCD testing and earthing tests are work for the competent person.
The rule from Earthing and RCDs applies throughout: observe, read and report, and leave anything behind a cover to the competent person. Meter readings, like the ones in this exercise, do not replace earthing and protective-device tests.
Where to go next
If one task gave you trouble, go back to the part it draws on:
If all six tasks made sense, you have finished the course. Two other courses on Cobler Learn continue from here. Building Electrical Fundamentals follows the power from the building's substation to the socket in the wall: switchboards, breakers, cables and drawings. Energy Management: The Economics of Saving Energy covers what you pay for, how to prove a saving, and how to judge whether a saving, such as the RM8,927 a month in Task 5, is worth what it costs to achieve.
Check your understanding
A fan motor's nameplate says 15 kW and 93% efficiency. A meter shows it drawing 16 kW. Is the motor overloaded? Not necessarily. The 15 kW is rated shaft output. At full load, the input would be about 15 ÷ 0.93 ≈ 16.1 kW, so a 16 kW input means the motor is running close to full load, not above it. Use the measured 16 kW when estimating kWh.
The shoplot owner hears that the office tower saved RM8,927 by lowering its peak. Would staggering the shoplot's air conditioners to lower its peak save money in the same way? No. The shoplot is on LV General, which has no per-kW demand charge; capacity and network are charged per kWh. Its bill falls only if it uses fewer kWh.
Recap: kWh come from kW × hours, using measured input, not nameplate ratings. A motor nameplate's kW is rated shaft output; the electrical input is larger by the losses. kVA = kW ÷ PF, and three-phase current follows from kW, voltage and PF. Maximum demand is the highest 30-minute average kW in the month; on MV General it counts at any hour, and on MV ToU only in the 2pm to 10pm weekday window. On an MV bill, capacity and network follow MD; on an LV bill, every line except retail follows kWh. Read and report; leave anything behind a cover to a competent person.
Cobler builds CobiNeural, a platform that shows a facility team its building's electricity as live numbers: energy, demand and power quality across the whole site. To see how your building uses electricity, talk to us.
Tan Kok XinElectricity Fundamentals
