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Chilled Water Coil – Confusion About Delta T and Heat Transfer

Josef M.

Josef M.

Mechanical
Joined
Sep 16, 2024
Messages
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PT
Hi everyone,

I'm a mechanical engineer currently working as a site manager. From time to time, I need to do some HVAC-related calculations, and this particular one is driving me a bit crazy.
I'm hoping someone can help me understand what I'm missing here.

I have a chilled water coil system with inlet and outlet water temperatures, as well as air-side inlet and outlet temperatures across the evaporator.
Here's what's confusing me:
If the chiller lowers the absolute temperature of the water but keeps the ΔT (temperature difference) the same, the amount of heat removed (Q) stays the same—assuming a constant water flow rate.

For example, with water going from 10°C in to 12°C out or from 20°C in to 22ºC out, the heat transfer comes out the same. That doesn’t seem intuitive to me, especially if the air temperature around the coil is constant.
So my questions are:
  • How is it possible for the same amount of heat to be removed in both cases, assuming the environment and airflow are the same?
  • Why don’t we consider the air temperature around the coil in these calculations?
  • If the surrounding air is, say, 25°C, wouldn’t colder water remove more heat due to a larger temperature difference, even with the same ΔT? Shouldn’t that mean you’d need less water flow to achieve the same Q? How is the same flow, with lower temperatures removing the same heat? if the temperature difference is bigger to outside, and the time that water goes through is the same, how can it be that the Q removed is the same?
Sorry if this is a silly question—it’s more about curiosity than an urgent issue. I'd really appreciate any insights.

Thanks!
 
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  1. How is it possible for the same amount of heat to be removed in both cases, assuming the environment and airflow are the same? Your intuition is correct here, in that not all substances will transfer heat at the same rate when it's cold vs. when it's hot. The generic heat transfer equation that is used usually looks like: Q = mdot*cp*(T1-T2). However, this equation is derived assuming cp (specific heat, or the amount of heat necessary to change a substance's temperature by a single degree) is constant. In your application, water has a pretty stable cp across the temperatures you are working with, which is why a temperature difference between 10°C - 12°C and 20°C - 22°C will give you the same heat transfer from/to the water with a very small error.
  2. Why don't we consider the air temperature around the coil in these calculations? It really all just depends on what you want to keep constant or what the system inherently adjusts. For example, in a chilled water system, your biggest concern is getting that water temperature where you want it. So, you will calculate the heat the water needs to give up in order to drop the temperature to that point. This same amount of heat then goes to your air (through whatever process is used). At this point, this no longer affects your chilled water system per se, but if you have to pick out the chiller that is making this heat transfer or you have to design the chiller itself, then the calculations for air temperature do become important. If you slow down the flow of air, the temperature difference for your air will be larger and vice-versa. (This is assuming dry air.) However, if air is the important temperature for your system, then this process would work the same way, but backwards.
  3. If the surrounding air is, say, 25°C, wouldn't colder water remove more heat due to a larger temperature difference, even with the same ΔT?... Not with the same temperature difference, no. Colder water, by intuition, does remove more heat, but that is only because you have adjusted the temperature difference. If you try to remove heat with water at 20°C and the water gets to 25°C, then you try it with 15°C water, the rate of transfer will be the same, but now you can heat the water 10 degrees instead of 5. So more heat in total can be removed, but the rate stays the same. There are some assumptions made here as well, though. One of which is that the heat exchanging method used doesn't see a change in its heat transfer coefficient because of a change in temperature (kind of similar to the cp thing). Ultimately, if you are talking about Q as in heat flow rate, then no it doesn't change, but if you are talking about Q as in total heat, then without getting into specifics yes, it would change.
Hope this helps!
 
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For example, with water going from 10°C in to 12°C out or from 20°C in to 22ºC out, the heat transfer comes out the same. That doesn’t seem intuitive to me, especially if the air temperature around the coil is constant.
So my questions are:

  • How is it possible for the same amount of heat to be removed in both cases, assuming the environment and airflow are the same?
  • Why don’t we consider the air temperature around the coil in these calculations?
When looking at just the water change in temperature - a given change in temperature of the water will give a given heat added from the air for the same total mass or mass flow rate of water assuming a constant specific heat of the water as mentioned above. However to get the same heat transfer on the air side, for a lower or higher water temperature, then something will have to change with the air. For a higher water temperature versus lower water temperature, the heat transfer rate is less per the below heat transfer equation due to a lower dT when keeping the air flow and inlet temperature constant:

Q = U A dT

This is the heat transfer equation for the coil, where:

Q is heat flowrate in BTU/HR from the air to water
U is overall heat transfer coefficient in BTU/HR-SQFT-deg F consisting of 3 layers - internal water convection film, coil metal wall conduction, external air convection film
A is area in SQFT of coil heat transfer surface
dT is differential temperature in degrees Fahrenheit between the water and air - actually based on the average or log mean temperature difference LMTD.

So if dT decreases due to increase in temperature of water while keeping the air inlet temperature and flowrate CFM constant, to maintain the same heat transfer (same dT of water at same flowrate of water) the following must happen:

The heat transfer area A of coil area must increase – replace coil with larger coil area size.

U overall heat transfer can possibly be changed by increasing the velocity of air over the coil. The higher velocity of air improves the convection heat transfer of the external air film. To increase the air velocity then you must increase the flowrate too for a given coil face area.

An increase air flowrate will also result in a higher leaving air temperature for a constant inlet air temperature increasing the LMTD offsetting the loss in heat transfer loss due to increase in water temperature.

For no change in air flow/velocity, there will be no compensation for the change in heat transfer due to the increase int water temperature. The heat transfer rate will drop and the dT of the air will drop so the air exits at a higher temperature. In this case the water will also exit at a lower differential temperature unless the water flow is reduced so that there is more residence time for the water in the coil then water will still exit at 2C difference at the lower heat transfer rate.

The important thing is that air is supplied to the room to handle the room sensible heat load by the equation:

Q = 1.1 CFM dT

Q = heat removed from room in BTU/HR
CFM is air flow in cubic feet per minute
dT is return air temperature minus supply air temperature (coil exit air temperature)

For a given room heat load and CFM the supply air (coil exit) temperature must be low enough so that when heat is added as the air flows through the room, you still can maintain the minimum design temperature - thermostat set temperature of room. Note that in addition to the sensible heat load on the coil, moisture in the air flow is condensed which also puts a latent heat load on the coil.

  • If the surrounding air is, say, 25°C, wouldn’t colder water remove more heat due to a larger temperature difference, even with the same ΔT? Shouldn’t that mean you’d need less water flow to achieve the same Q? How is the same flow, with lower temperatures removing the same heat? if the temperature difference is bigger to outside, and the time that water goes through is the same, how can it be that the Q removed is the same?
The colder water will remove same heat for same total temperature change in accordance with the following as stated by Btutiful:

Q = m Cp dT

Q = heat removed from water in BTU/hr
m = mass flow rate in lbs/hr
Cp = specific heat of water in BTU/lb-deg F
dT = change in temperature of the water

However as stated more dT of air to water will increase the heat flow rate between the water and air by the coil heat transfer equation:

Q = U A dT

Ultimately U A dT = m Cp dT under steady state equilibrium conditions.

Where on left side dT is difference in temperature of air and water and on right side dT is difference in total water temperature in/out. So if you decrease water temperature you will increase heat transfer rate but in the end the heat transferred to the water from the air equals the same heat that goes into raising the temperature of the water by the above equations.

So if you have constant inlet air flow at temperature of 25C and you have water at 20C to 22C which gives you a certain heat transfer rate, if you then lower to 10C to 12C you will increase the heat transfer rate from the air to the water for the same air flow and inlet air temperature. The water will exit at a temperature higher than 2C differential if water flowrate is maintained same. The air too will exit at a lower temperature since heat transfer rate has increased. Under these conditions, at a constant heat input rate of heat into the room to be cooled, equilibrium will be achieved again when temperature of return air lowers to a point to result in the same dT between air and water as when water was at higher temperature. That is when the heat input to the room equals the heat removed by the coil.

However this will develop very low room temperatures if let it run its own course to equilibrium. With a thermostat the coil fan will cycle with water flow through coil control to match the coil load with the room load to maintain the desired room temperature.
 
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The responses above are totally correct, but maybe a bit long.

Your questions:
  • How is it possible for the same amount of heat to be removed in both cases, assuming the environment and airflow are the same?
It isn't possible but then you're putting forward an unrealistic set of conditions. You are assuming NOTHING changes other than the temperature of the water. Its not possible to change only one element of an equation and have no change in any other number and still get the same answer ( heat loss).
  • Why don’t we consider the air temperature around the coil in these calculations
Designers will consider that.
  • If the surrounding air is, say, 25°C, wouldn’t colder water remove more heat due to a larger temperature difference, even with the same ΔT? Shouldn’t that mean you’d need less water flow to achieve the same Q? How is the same flow, with lower temperatures removing the same heat? if the temperature difference is bigger to outside, and the time that water goes through is the same, how can it be that the Q removed is the same?
No, because you're fixing the heat removal element (delta T x flow x Cp), but then not changing anything else other than temperature. If you lower temperature, keep the flow the same the DT will increase if all other things stay the same and hence the heat loss will increase. The question then is whether your chiller can cope with this increased DT to still give you back water at 10C.

Basically you are trying to make X x Y x Z = X x Y x Z1 where Z1 is not equal to Z. This is a mathematical impossibility.
 
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Thank you everyone for the answers.

[B]LittleInch[/B] to clarify one thing on my last question. The delta would not decrease, the absolute temperatures yes, not the delta (22-20=12-10).

From what I understood from the other answers is that, for the same delta T the Q is the same, but what end up happening on real world situation is that this situation is dynamic, and that the equatation, for the same condition outside to the coil, would end up changing by increasing the return temperature because of more absortion of heating.​

 
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Or a lower flow rate if the control was set at a delta T.

Or a gradually higher absolute temperature from the chiller until heat into the circuit equalled the chiller capacity for heat out.
 
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i think you miss the part where a control valve will throttle (or open) chilled water flow to meet the set discharge air temperature (DAT). So, in your scenarios (with chiller supplying 10°C vs. 12°C), the water flow is different.

You also assume entering air temperature (EAT) and air flowrate were constant during your observation. this isn't a given unless you verified that.
 
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I'm not missing that part—I just wasn't understanding how the same heat (same Q) could be removed with the same but lower delta T, and how the formula doesn’t account for the coil or external conditions where the heat is actually being removed.
From your previous answer, I now realize that it would eventually balance with the coil’s heat transfer, which would lead to an increase in delta T. (Then, the equipment itself would either decrease the water flow rate or increase the airflow over the coil). But at a given instant, the same delta T would remove the same Q.

If I'm still mistaken, please correct me.
 
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Josef M. said:
I'm not missing that part—I just wasn't understanding how the same heat (same Q) could be removed with the same but lower delta T, and how the formula doesn’t account for the coil or external conditions where the heat is actually being removed.
From your previous answer, I now realize that it would eventually balance with the coil’s heat transfer, which would lead to an increase in delta T. (Then, the equipment itself would either decrease the water flow rate or increase the airflow over the coil). But at a given instant, the same delta T would remove the same Q.

If I'm still mistaken, please correct me.
Click to expand...
That is the thing you are missing, you don't know it is the same Q.

In addition, in cooling air, you also transfer latent heat. You are only measuring dry bulb temperature, but not dew point. So you don't have the enthalpy of air.
 
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There are three ways to calculate cooling capacity in a chilled water coil: air-side, water-side, and heat exchanger-side. (English unit)

Air-side: Q = 4.5 * CFM * enthalpy difference (accounts for sensible and latent cooling).

Water-side: Q = gpm*dt/24 (measures heat absorbed by water).

Heat exchanger side: Q = U*A (coil surface area including tube and fins) *LMTD (linking air and water through coil design).

In your scenario, raising the water inlet temperature from 10°C (50°F) to 12°C (53.6°F) reduces the LMTD, which lowers the cooling capacity. The heat exchanger formula Q = U * A * LMTD connects the air and water sides, showing that changes in water temperature affect both. As the inlet water temperature increases, the heat removal capacity decreases, influencing air-side conditions (e.g., air outlet temperature). It's not physically possible to keep the same delta T while raising inlet water temperature.
 
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moideen again, I'm saying raising the delta T not just the inlet temperature. Raising the inlet and on the same degree the outlet, keeping the the same delta T (10-12 to 20-22) so the Q is the same.​

 
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Josef M. said:

moideen again, I'm saying raising the delta T not just the inlet temperature. Raising the inlet and on the same degree the outlet, keeping the the same delta T (10-12 to 20-22) so the Q is the same.​

Click to expand...
What is "Q" in your statement? Normally it is volumetric flowrate. But you are NOT measuring that. So you don't know it is the same. If you mean Q being some sort of energy transfer rate, you also don't know that since you don't know the volumetric flowrate.

You only seem to measure temperature. This isn't enough to derive Power.

It is not magic, it is Thermodynamics. No energy disappears or appears from nothing.
 
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by Q I mean the cooling capacity, the amount of heat/energy removed.

Sorry, if I'm being misunderstood but again, same flow, same volumetric flow rate, mass, whaever. Same deta T too. So same Q.

Q = m Cp dT

The thing is, you change the absolute temperature on delta T, and you still remove the same heat (by keeping the same dT 10-12 to 20-22 for eg). Check my original post. Maybe I'm not making any sense.
 
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If the air flow rate is sufficiently high then the temperature difference might be the same
Qwater=m*Cp*dT
where
m = mass flow rate of water, Cp = Specific heat capacity of water, dT = Temperature difference of water
Qair=m*Cp*dT
where
m = mass flow rate of air, Cp = Specific heat capacity of air, dT = Temperature difference of air

Since dT is same and heat transfer is same, we can use the following to determine mass flow rate of air given that mass flow rate of water is known

Qwater = Qair and dTwater = dTair
mwater*Cpwater =mair*Cpair
mair = mwater/mair*Cpwater
 
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Josef M. said:

moideen again, I'm saying raising the delta T not just the inlet temperature. Raising the inlet and on the same degree the outlet, keeping the the same delta T (10-12 to 20-22) so the Q is the same.​

Click to expand...
I have attached fan coil unit selection as an example.

Air inlet condition: 24°C DB / 17.1°C WB

Air outlet condition: 11°C DB / 10.7°C WB

Chilled water temperatures: 5.5°C / 14.5°C

Specified capacity: 9.24 kW (2.6 Tons)

In this scenario, after the installation of the above-selected unit, if you increase the chilled water inlet temperature by 6.5°C from the selection 5.5c, what would be the impact? Will the capacity decrease or increase? Yes, the capacity will decrease.

The air-off coil temperature will also rise if the chilled water flow rate remains unchanged. Most importantly, the chilled water return temperature will be lower than specified.

It is not physically possible to keep the same delta T while raising the inlet water temperature at the same flow rate. As the inlet water temperature increases, LMTD drops, reducing both delta T and capacity. For example, raising the inlet from 5.5°C to 6.5°C will not produce a 15.5°C return; it will be around 14.5°C or less (lower than design deltaT).

When returning to your original question, 10°C in to 12°C out or 20°C in to 22°C out, you cannot achieve the 22°C when increasing the inlet from 10°C to 12°C; it would be 20c or less.
 
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