Blog

Capstone: Reading a Building's Electricity

The course capstone: walk one office tower from its 11 kV intake to a pantry kettle, then follow its data from CTs and sensors over Modbus to a fault list. Worked numbers for transformer kVA, socket amps, a floor's kWh, motor input versus nameplate, power factor and wasted pump energy.

Tan Kok XinTan Kok XinElectricity Fundamentals
Illustration of an office tower's electrical system with a magnifying glass on one floor

Part 24 of 24 in Cobler's Electricity Fundamentals course. New here? Start with the course map.

With fault detection turning a flood of readings into a ranked list (Part 23), every piece of the course is built. Now we can put them together and read one whole building. So let's do it as a building electricity worked example. We'll walk the fictional office tower we've visited all through the course, in the order the course built it. We'll start at its 11 kV intake and finish at its fault list.

Let's start with something small. When someone switches on the kettle in the tower's level 12 pantry, power stations have to produce a little more electricity almost at once. A second later, a meter on the floor's distribution board has counted it. A few seconds after that, the number has crossed an RS-485 chain to a gateway, and software is comparing it with what it expected. And you've now built every piece of that chain.

So at each stop there are one or two questions. Try them with a calculator first, and then the worked answer follows. The numbers are fictional (illustrative), but they're consistent with each other and with the course. And like any real walk-round, we only look and read. Anything behind a cover is a job for the site's electrical competent person, as in Part 14. Four formulas do most of the work:

$$P = V \times I \qquad E = P \times t \qquad \text{kVA} = \frac{\text{kW}}{\text{PF}} \qquad P_{in} = \frac{P_{out}}{\eta}$$

Power from volts and amps, and energy from power and hours, come from Part 5. Apparent power from power factor (PF) is Part 12. And a motor's input from its output and efficiency (\(\eta\)) is Part 11.

Stop 1: the 11 kV intake and transformers

So we start where the electricity comes from. A generator at a power station makes three-phase AC at 50 Hz (Part 7). A step-up transformer raises it to 275 kV for the National Grid. Then TNB substations step it down to 11 kV at the tower's intake, which is where the TNB meter sits (Part 13). Inside the tower, two 1,500 kVA transformers bring it from 11 kV down to 400 V (Part 9). At its maximum demand of 1,600 kW, the tower's power factor at the main switchboard is 0.92, after the capacitor banks.

Try it. Why raise the voltage to 275 kV, only to lower it again? At maximum demand, what share of the two transformers' rating is the tower using?

Show the worked answer

Well, at a higher voltage the same power needs less current, and line losses go with the current squared (\(I^2 R\), Part 3). So we run long lines at high voltage (Part 8). But a 275 kV line can't supply a building, so transformers step it back down near where it's used. That only works because the supply is AC.

Now the transformers. 1,600 ÷ 0.92 ≈ 1,739 kVA, against 2 × 1,500 = 3,000 kVA, so the tower is using about 58% of their combined rating. We rate transformers in kVA because what heats them up is the current, whatever the kW happens to be.

Stop 2: switchboards and the pantry socket

From the transformers, power runs through the main switchboard and then a riser switchboard for levels 11 to 20. Next comes the level 12 east distribution board, DB-12E (Part 18). DB-12E feeds parallel final circuits (Part 4). A 20 A residual current breaker with overcurrent protection (RCBO) protects the pantry socket circuit, and its residual current function trips at 30 mA. The kettle is rated 3 kW at 230 V.

Try it. The building's supply is 400 V, so why does the kettle get 230 V? How much current does it draw? If a damaged lead leaks current through someone's hand to earth, which part of the RCBO responds?

Show the worked answer

So 400 V is the voltage between any two of the three phases. One phase to neutral is 400 ÷ √3 ≈ 230 V (Part 10), and that's what the kettle gets. It draws 3,000 ÷ 230 ≈ 13 A, which is inside the 20 A rating, though other appliances on the circuit add their own current.

The residual current function catches the leak. Live and neutral pass together through one ring core, which sees nothing while the two currents balance. It trips the circuit when enough current leaks to earth (Part 14). The overcurrent part wouldn't react, because a leak through a person is far too small for it. A residual current device (RCD) adds protection, but it can't prevent every shock, so a damaged lead has to come out of use.

Stop 3: sub-meters

DB-12E's main circuits pass through current transformers that feed sub-meters (Part 19). Let's look at what they show for a normal weekday, as average electrical input:

Load on level 12 east

Average power while running

Hours per weekday

Lighting

12 kW

10 (8 a.m. to 6 p.m.)

Socket outlets (computers, screens, printers)

15 kW

10 (8 a.m. to 6 p.m.)

Supply fan of air handling unit AHU-12A

6 kW

11 (7 a.m. to 6 p.m.)

Try it. How many kWh do these loads use per weekday, and per month of 22 working days? Why use the measured socket average instead of the sum of every computer's rating?

Show the worked answer

12 × 10 = 120 kWh, 15 × 10 = 150 kWh and 6 × 11 = 66 kWh, so that's 336 kWh a weekday. Over a month it's 336 × 22 = 7,392 kWh. Now, adding up the appliance ratings gives a number that never actually happens, because the appliances are never all at full power at once. The meter records what's really drawn, and that's what adds up to kWh.

Stop 4: motors, drives and power factor

The chiller plant's condenser water pumps run at a fixed speed. One pump motor's nameplate reads 30 kW, 400 V three-phase, 54 A, power factor 0.86 and efficiency 93.6%. Its meter shows 27.0 kW of input at a power factor of 0.85.

Try it. What does the 30 kW describe? What would the motor draw at full load, and how loaded is it now? What is its kVA, and do the capacitor banks at the main switchboard change its 0.85?

Show the worked answer

You'd probably think the pump takes 30 kW from the supply. But the 30 kW on the nameplate is the rated shaft output, what the shaft can deliver at full load (Part 11). The input is a different number. At full load the input is 30 ÷ 0.936 ≈ 32.1 kW, so 27.0 kW means about 84% of full load. If the efficiency stays near its rating, about 27.0 × 0.936 ≈ 25.3 kW reaches the shaft. So for kWh we use the measured 27.0 kW.

The pump's apparent power is 27.0 ÷ 0.85 ≈ 31.8 kVA, so the cable carries the current for 31.8 kVA while 27.0 kW does useful work. Now the capacitor banks. They supply reactive current at the main switchboard, so they relieve the transformers and the cables upstream of them. But the current between the switchboard and the pump is unchanged, and its power factor stays at 0.85 (Part 12). That's the trade-off with one central bank. It fixes the power factor the TNB meter sees, and the one at each motor stays where it was.

Diagram showing a meter reading 27.0 kW of electrical input to a motor, about 25.3 kW reaching the pump shaft and about 1.7 kW lost as heat, with the 30 kW nameplate rating shown as rated output

The meter measures what goes into the motor, but the nameplate kW describes what the shaft can deliver at full load.

Now the AHU-12A fan is different, because it runs on a variable speed drive (Part 15). Say its 11 kW motor draws about 12.0 kW at full speed, like the fan in Part 11. Its 6 kW average is half of that, and fan power goes with speed cubed, so it's running at about \(\sqrt[3]{0.5} \approx 0.79\), or 79% speed. The drive's chopped current, added to the computers' power supplies on the same board, distorts the current wave. Triplen harmonics add up in the neutral (Part 16). So if the neutral current on DB-12E is close to the phase currents, that's a clue for the competent person to check.

Stop 5: sensors, controllers and the fault list

So in Stops 1 to 4 we followed the energy the way a person would on a walk-round. But how do we go from that to a building that reports on itself? Well, that's what Parts 20 to 23 built.

The tower also measures what isn't electrical. A carbon dioxide (CO2) sensor in the level 12 meeting room is ranged 0 to 2,000 parts per million (ppm). It sends 4-20 mA to the controller of AHU-12A (Part 20), whose processor turns the reading into a damper position on every scan (Part 21). The DB-12E sub-meters and the plant meters sit on RS-485 chains, read over Modbus RTU by a gateway (Part 22). And fault detection software watches the results (Part 23).

Try it. The CO2 loop reads 12 mA, so how much CO2 is in the room? On a Sunday at 2 a.m., with every chiller off, the condenser pump's meter still shows 27.0 kW. If nobody notices for 30 nights of 6 hours, how much energy is wasted?

Show the worked answer

(12 − 4) ÷ 16 × 2,000 = 1,000 ppm, which is right at the limit in Malaysia's Industry Code of Practice on Indoor Air Quality 2010. So the controller should be calling for full fresh air. For the pump, 27.0 kW × 6 h × 30 = 4,860 kWh, which is about two-thirds of level 12 east's monthly use (Stop 3). And no fixed alarm fires, because 27.0 kW is a normal number for that pump. An anomaly model that expects about 0 kW at that hour flags it on the first night.

Back to the light switch

So you can now read a building's electrical story from end to end. That's the 11 kV intake and its transformers, in kVA, and the switchboard tree down to a protected 230 V socket. It's the meters behind every kWh, and a motor's input against its nameplate, its drive and its harmonics. And it's the sensors, controllers, wires and software that turn all of it into a short list of things to fix. Each piece rests on the one before it, all the way back to the sea of free electrons in copper from Part 1.

Right at the start of the course, at 8 a.m. on the 20th floor, someone flicked on the lights. And a long chain had to answer in less than a second. Now we can name every link in it ourselves.

A generator somewhere turns a little harder (Part 7). Transformers take the voltage up for the journey and back down to 11 kV and then 400/230 V (Parts 9 and 13). The current crosses the main switchboard and a breaker (Part 18). Then it goes through the relay that the lighting controller has just closed (Parts 6 and 21), into the lamp. A meter counts it (Part 19), and software compares it with what it expected (Part 23). It still all happens in less than a second, and nobody sees any of it move.

None of that replaces the chief engineer or the technician. The meters and the software do the watching that no person could, from a pump at 2 a.m. to 17,000 readings a minute. That lets a small team look after a whole tower and spend its time on judgement and repairs.

So where do you go from here? If one stop gave you trouble, go back to the parts it names. If they all made sense, you can choose your next course by your job. Four courses on Cobler Learn build on this one, and each picks up where this course stops:


This is Part 24 of 24 in Cobler's Electricity Fundamentals course. Previous: Fault Detection and Diagnostics. Next: Where to go next, on the course map.

Cobler builds CobiNeural, a platform that reads a building's meters, sensors and BMS. It shows them as live numbers, alerts and anomaly flags across the whole site. To see your own building's story, talk to us.

FAQ

Frequently asked questions

Ask us about this topic

Something in this article you want to dig into — or a situation in your own building it doesn't quite cover? Send us your question. We don't run public comments; the team replies to you directly by email.

0/500
Electricity Fundamentals · Part 24 of 24All 24 parts

Keep reading