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One Complaint, Three Loops: A Worked Cooling Plant Check

A worked exercise to finish the course. Trace a warm, humid complaint in our example office tower through the air, chilled water and condenser water loops using trend readings: delta-T, kW and RT of cooling, kW/RT and tower approach. Then a short version for a shoplot with split units and cassettes.

Tan Kok XinTan Kok XinCooling Fundamentals
Illustration of an air handling unit, chiller and cooling tower linked as three loops with a clipboard of readings

Part 20 of 20 in Cobler's Cooling Fundamentals course. New here? See the course page.

On a hot Thursday afternoon, the helpdesk in the office tower gets a call from level 12, saying "It's warm and sticky up here." The usual first response is to turn the thermostat down, or to blame the air handling unit (AHU) on that floor. But the air on level 12 is cooled by water from the chillers. And the chillers depend on water from the cooling towers on the roof. So the cause could be in any of those three loops, and it's worth a proper cooling plant check before anyone touches a thermostat.

Back in Part 1, we stood next to the little split unit on the shoplot wall and made a promise. We said that by the end of the course, we'd be able to stand in a 20-storey tower's plant room. And we'd be able to explain why one floor is too warm. Well, here's the floor, and here's the plant room.

So we'll follow this one complaint through the air loop, the chilled water loop and the condenser water loop. We'll use a few trend readings and the tools from the whole course. And at the end, there's a shorter version for the shoplot, which has split units and cassettes instead of a chiller plant.

The cooling plant check

Each task gives you some information and asks a few questions. So try the questions first, with a calculator and a notepad. Then open the worked answer under each task to check your working.

As a reminder, the office tower has three 500 RT chillers, two on duty and one on standby. Its chilled water is designed for 6 °C supply and 12 °C return. It also has three cooling towers on the roof and a building management system (BMS). All the readings below are made up, but they're consistent with each other.

We'll need five tools from earlier parts:

The trend readings

The facility engineer pulls the same set of readings from the BMS for two Thursdays at 3pm. One is last week, when nobody complained, and the other is today. The relative humidity comes from a wall sensor on level 12, and the outdoor wet-bulb temperature from the weather sensor on the roof. The plant's electrical input comes from the chiller plant's sub-meter, which covers the chillers, pumps and tower fans.

Reading at 3pm

Thu 10 Sep 2026 (no complaints)

Thu 17 Sep 2026 (complaint)

Air loop: level 12 east and AHU-12A

Room temperature (return air)

24.0 °C

26.5 °C

Room relative humidity

58%

72%

Supply air temperature (setpoint 14.0 °C)

14.0 °C

17.0 °C

Chilled water valve

70% open

100% open

Supply fan speed

75%

100%

Filter pressure drop (change at 250 Pa)

140 Pa

145 Pa

Chilled water loop: the plant

Chillers running

CH-1 and CH-2

CH-1 and CH-2

Chilled water supply (setpoint 6.0 °C)

6.0 °C

8.0 °C

Chilled water return

11.0 °C

12.5 °C

Chilled water flow through the running chillers

140 L/s

140 L/s

Whole plant electrical input

790 kW

800 kW

Condenser water loop: the towers

Outdoor wet-bulb temperature

26.5 °C

27.0 °C

Condenser water leaving the towers

30.0 °C

33.5 °C

Tower fans CT-1 and CT-2

Both running

CT-1 running, CT-2 stopped (BMS command: run)

The helpdesk log also shows that by 3:30pm on 17 September, levels 14 and 17 have called about warm, humid offices too.

The office tower's three loops on 17 September at 3pm: level 12 at 26.5 °C and 72% RH, AHU-12A supply air 17.0 °C with its valve 100% open, chilled water 8.0 °C out and 12.5 °C back at 140 L/s, condenser water 33.5 °C from the towers against a 27.0 °C wet-bulb, and tower fan CT-2 stopped

The complaint-day readings placed on the three loops: air, chilled water and condenser water.

Task 1: the air loop

We'll start where the complaint is, with the air loop rows for AHU-12A. That's the unit whose load we added up in Part 6, and whose control loop we watched in Part 19. The comfort ranges come from the Industry Code of Practice on Indoor Air Quality 2010 (ICOP IAQ 2010) in Part 4.

  1. Are the room conditions outside the ICOP IAQ 2010 comfort ranges? Those are 23 to 26 °C and 40 to 70% relative humidity.
  2. Is AHU-12A trying to cool harder, or has it given up?
  3. Could a dirty filter explain the complaint?
  4. Why is the room humid as well as warm?
Show the worked answer
  1. Yes. The room is at 26.5 °C and 72% relative humidity, just above both upper limits. On the normal day it was 24.0 °C and 58%, well inside them.
  2. It's trying as hard as it can. Its chilled water valve is 100% open and its supply fan is at 100% speed. Even so, its supply air is 17.0 °C, three degrees warmer than its 14.0 °C setpoint. The control loop from Part 19 is asking for more cooling and not getting it.
  3. No. The filter pressure drop is 145 Pa, almost the same as last week. That's well below the 250 Pa change point, so air is getting through the filter normally.
  4. Moisture is removed when water vapour condenses on a coil colder than the air's dew point (Part 3). If the coil is warmer than usual, for example because the water reaching it is warmer, less moisture condenses. So the air leaving the AHU is both warmer and wetter.

AHU-12A is working at full output, then, and still can't reach its setpoint. It looks like a victim, not the cause. The next question is whether the water reaching its coil is as cold as it should be. And the calls from levels 14 and 17 point the same way, because a problem at one AHU wouldn't usually affect three floors.

Task 2: the chilled water loop

Now let's follow the water back, with the chilled water rows.

  1. What is the chilled water delta-T on each day?
  2. How much cooling did the plant deliver on each day, in kW and in RT?
  3. Is this low delta-T syndrome?
  4. What is the most important change between the two days?
Show the worked answer
  1. Normal day: 11.0 − 6.0 = 5.0 °C. Complaint day: 12.5 − 8.0 = 4.5 °C.
  2. Normal day: 140 × 4.19 × 5.0 = 2,933 kW, and 2,933 ÷ 3.517 = about 834 RT. Complaint day: 140 × 4.19 × 4.5 = 2,639.7 kW, and 2,639.7 ÷ 3.517 = about 751 RT.
  3. Not mainly. The delta-T fell a little, from 5.0 to 4.5 °C, but the flow was the same on both days. And the plant didn't start an extra chiller, so low delta-T isn't what changed. The normal-day 5.0 °C is below the 6 °C design, which is worth a separate look (Part 17). But it didn't cause today's complaint.
  4. The chilled water supply is 8.0 °C instead of its 6.0 °C setpoint. The chillers aren't making the water as cold as they're told to. So every AHU in the building gets warmer water, which explains why several floors are complaining.

The two running chillers are rated 500 RT each, so 1,000 RT together. But on the complaint day they delivered only about 751 RT. That's less than the 834 RT of the normal day, although the building was warmer and asking for more. So something is holding the chillers back.

Task 3: the plant's efficiency

  1. Calculate the plant's kW/RT on each day.
  2. How much more electricity is the plant using for each RT on the complaint day?
  3. How do both figures compare with the plant's design target of 0.85 kW/RT?
Show the worked answer
  1. Normal day: 790 ÷ 834 = about 0.95 kW/RT. Complaint day: 800 ÷ 751 = about 1.07 kW/RT. Using the unrounded RT, 790 ÷ 833.95 = 0.947 and 800 ÷ 750.55 = 1.066.
  2. 1.066 ÷ 0.947 is about 1.13. So the plant is using about 13% more electricity for each RT of cooling, while delivering less cooling.
  3. Both are worse than the 0.85 kW/RT design target. The normal-day 0.95 kW/RT is the office tower's usual measured figure, so there's room to improve on an ordinary day. Today's problem has made it worse again.

The plant's electrical input barely changed, from 790 to 800 kW, but its cooling output fell. So the chillers are working as hard as before and achieving less. That's what happens when a chiller has to push its heat out against hotter condenser water, which leads us to the third loop.

Task 4: the cooling towers

Now look at the condenser water rows.

  1. What is the tower approach on each day?
  2. Did the weather change enough to explain the difference?
  3. What else in the table explains it?
  4. Why does hotter condenser water hold the chillers back?
Show the worked answer
  1. Normal day: 30.0 − 26.5 = 3.5 °C, a good approach. Complaint day: 33.5 − 27.0 = 6.5 °C.
  2. No. The outdoor wet-bulb rose by only 0.5 °C, from 26.5 to 27.0 °C. The water leaving the towers rose by 3.5 °C, from 30.0 to 33.5 °C. So most of that rise came from the towers, not the weather.
  3. Tower fan CT-2 has stopped, although the BMS is commanding it to run. Without its fan, CT-2 moves much less air through its fill. So it evaporates less water and cools the condenser water much less. The mixed water going back to the chillers is hotter.
  4. The chillers have to reject their heat into the condenser water. When that water is hotter, the compressor has to lift the heat to a higher temperature and pressure. That takes more power for each RT, the opposite of the condenser relief in Part 15. And when the condenser water gets hot enough, the chiller's own controls limit how hard it works. That protects the machine, and it's why the chillers couldn't hold 6.0 °C.

So here's the whole chain. A stopped tower fan made the condenser water hotter, and that held the chillers back. That let the chilled water warm to 8.0 °C, which left AHU-12A's coil too warm to cool and dry the air. And that made level 12 warm and humid.

Task 5: the next check

So which of these should the facility team do first?

  1. Replace AHU-12A's filter.
  2. Lower the level 12 room setpoint to 22 °C.
  3. Ask the chiller service contractor to check the refrigerant in CH-1 and CH-2.
  4. Send a technician to cooling tower CT-2 to see why its fan has stopped. The electrically competent person responsible for the plant goes too, to inspect and reset or repair the fan motor's starter.
Show the worked answer

Option 4. It goes straight to the cause the data points to.

  • Option 1 wouldn't help, because the filter pressure drop is normal.
  • Option 2 wouldn't help either. AHU-12A's valve is already fully open and its fan is at full speed, so a lower setpoint can't make it deliver more cooling. And once the fault is fixed, someone would have to remember to put the setpoint back.
  • Option 3 has no support in the data. The chillers' trouble started with the hotter condenser water, and nothing in the readings points to their refrigerant.

At the tower, the technician should work to the site's procedures. The fan must be isolated and locked off before anyone reaches into its housing, and electrical work on its starter is for the competent person. Cooling tower spray can carry Legionella bacteria, so the site's tower hygiene procedure also applies (Legionella in cooling towers). If CT-2 can't be repaired quickly, the team can ask the plant's designer or maintenance contractor about the standby tower. CT-3 might take its place for now.

Task 6: the fix

A failed fan motor bearing had tripped CT-2's starter, and it's repaired on Friday 18 September. On Monday 21 September at 3pm, the BMS shows:

  • Outdoor wet-bulb 26.8 °C, and condenser water leaving the towers at 30.5 °C, with both tower fans running.
  • Chilled water supply 6.0 °C, return 11.2 °C, flow 140 L/s. Whole plant electrical input 815 kW.
  • AHU-12A supply air 14.0 °C, with its valve 75% open. Level 12 east at 24.0 °C and 59% relative humidity.
  1. Calculate the approach, delta-T, cooling in RT and kW/RT.
  2. Is the problem fixed?
  3. Does this prove how much energy the repair saves each month?
Show the worked answer
  1. Approach: 30.5 − 26.8 = 3.7 °C. Delta-T: 11.2 − 6.0 = 5.2 °C. Cooling: 140 × 4.19 × 5.2 = 3,050.3 kW, and 3,050.3 ÷ 3.517 = about 867 RT. The kW/RT is 815 ÷ 867 = about 0.94 kW/RT.
  2. Yes, for this fault. The approach is back to about 3.5 to 4 °C, and the chillers are holding 6.0 °C. AHU-12A is reaching its setpoint with its valve only 75% open, and level 12 is back inside the ICOP ranges. The plant's kW/RT is back to its usual level.
  3. No. One afternoon before and one after had different weather and different loads, so they can't give a monthly saving. Proving a saving needs weeks of data, compared with a baseline that allows for weather and occupancy. Part 16 explained how. What the readings do show is that the fault is cleared.

Two items stay on the list for later. The plant's normal delta-T of about 5 °C is below its 6 °C design (Part 17). And its usual 0.94 to 0.95 kW/RT is above the 0.85 kW/RT target. Neither caused the complaint, but both deserve their own investigation.

And notice that this fault was visible in the BMS before anyone called. CT-2's fan had stopped while the BMS was still commanding it to run. So an alarm when a fan's status doesn't match its command would find the next trip before anyone calls the helpdesk.

Task 7: the shoplot

The shoplot has no chilled water and no cooling tower. Its cooling is direct expansion (DX), with six wall-mounted split units (1 to 1.5 HP class) and two ceiling cassettes (2.5 HP class). That's about 33 kW of cooling in total, or about 9 RT. On Saturday 19 September, the café on the ground floor is warm and sticky by 2pm.

The café is cooled by the two cassettes, C-1 and C-2. They're the same model and age, both set to 24 °C in cooling mode with the fan on high. Both have run without stopping since the café opened. So the owner holds a digital thermometer at each cassette's return grille, where air goes in, and at its supply louvre, where air comes out. Then the owner looks at each outdoor unit in the back lane.

Reading at 2pm

Cassette C-1

Cassette C-2

Air going in

25.5 °C

26.0 °C

Air coming out

13.5 °C

20.0 °C

Outdoor unit

Fins clean, open side lane

Fins grey with dust and grease, beside the kitchen exhaust outlet

  1. Which cassette is moving less heat, and how can you tell without a flow meter?
  2. In what order would you check things, and who should do each check?
  3. Which office tower tasks don't apply here, and where will the shoplot see a saving?
Show the worked answer

Answer 1. C-2 is moving less heat.

C-1 cools its air by 25.5 − 13.5 = 12.0 °C. C-2 cools its air by only 26.0 − 20.0 = 6.0 °C, half as much, with the same fan setting and similar air going in. The two units are the same model on the same fan speed, so they move about the same amount of air. So C-2 is removing much less heat. The comparison is fair only because the units are identical, and one unit's temperature drop on its own is harder to judge.

Answer 2. Here's a sensible order:

  1. The owner or staff check and clean both cassettes' indoor filters, with the units switched off. It's quick and safe, and dust on the filter gets onto the coil. A blocked filter reduces airflow, which usually makes the air coming out colder, not warmer. So C-2's warm air points mainly to the outdoor side.
  2. The air conditioning contractor cleans C-2's outdoor coil. The owner arranges for the kitchen exhaust to be pointed away from it. A dirty condenser coil breathing warm, greasy exhaust air can't release heat well. So the unit cools less and uses more electricity (Part 18).
  3. If C-2 is still weak after cleaning, the contractor checks its refrigerant pressures. A low charge means a leak, which is a fault to find and repair, not something to refill and forget. What Is Refrigerant? is a side reading for this course.

After each step, repeat the in-and-out temperature readings to see whether C-2 has caught up with C-1.

Answer 3. The shoplot has no chilled water delta-T, no tower approach and no plant kW/RT to calculate. Its evidence is simpler. It's the temperature drop across each unit, whether units run without stopping, and the kWh on its electricity bill.

The shoplot is on the LV (low voltage) non-domestic General tariff, which has no charge per kW of demand. So the saving from a cleaner, more efficient unit shows up as fewer kWh. In September 2026, each kWh saved removes 54.35 sen from the energy, capacity, network and Automatic Fuel Adjustment (AFA) lines of its bill.

Try it

Loading the interactive

Back in the plant room

So that's the promise from Part 1 kept. A floor in a 20-storey tower felt warm and sticky. We followed it from the room, through the coil, the chilled water and the chiller, to a stopped fan on the roof. And every step used something we built along the way.

The dew point explained why the room felt sticky, and AHU-12A's open valve showed it was a victim. Flow and delta-T put a number on the cooling, kW per ton showed the plant straining, and the approach pointed at the tower. Underneath it all is the idea from the little split unit on the shoplot wall. Heat only flows downhill on its own, so a machine has to push it uphill and out. So when the last step of that push can't release its heat, every loop before it feels the strain. And in the end, so does someone at a desk on level 12.

Where to go next

If one task gave you trouble, go back to the part it draws on:

If all seven tasks made sense, you've finished the course. The Building Automation Fundamentals course continues from Part 19, with the BMS points, trends and alarms you used here. And there are other courses on Cobler Learn.

Check your understanding

1. A plant's chilled water flow is 120 L/s, with water leaving at 6.0 °C and coming back at 11.0 °C. The whole plant draws 680 kW. What is its kW/RT?

Show the answer

Delta-T = 11.0 − 6.0 = 5.0 °C. Cooling = 120 × 4.19 × 5.0 = 2,514 kW, and 2,514 ÷ 3.517 is about 715 RT. So kW/RT = 680 ÷ 715, about 0.95 kW/RT.

2. An AHU's supply air is too warm, but its valve is only 40% open. Is the problem more likely in the AHU or in the chilled water?

Show the answer

More likely in the AHU or its controls. If the water were too warm, the controller would open the valve fully to get more cooling. So a valve only 40% open while the air is too warm points to the AHU itself. It suggests a control fault, a wrong setpoint, a stuck valve or a sensor error there. Check those before looking at the plant.

Recap

  • Follow a complaint through the loops in order, the air, the chilled water, then the condenser water.
  • At each loop, compare the readings with their setpoints and with a normal day.
  • An AHU with its valve and fan at 100% that still can't reach its setpoint is a victim, not the cause.
  • Delta-T and flow give the cooling in kW and RT. The plant's electrical input divided by RT gives kW/RT, and the water leaving the tower minus the wet-bulb gives the approach.
  • Choose the check that goes to the cause, and work safely. Confirm the fix with fresh readings, and use weeks of data, not one afternoon, to prove a saving.
  • In a DX building, compare identical units' in-and-out air temperatures. Then check filters, outdoor coils and refrigerant, in that order.

This is Part 20 of 20 in Cobler's Cooling Fundamentals course. Previous: What Is a BMS? How Building Controls Run the Cooling Plant.

Cobler builds CobiNeural, a platform that shows a facility team its building's energy, water and indoor air data as live numbers across the whole site. To see how your building performs, talk to us.

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