Calculation of water flow rate in heat exchanger

[abcxyz]

I am not a mechanical engineer; but I have physics background.

I want to lift about 250 watts of heat from a metal block.
At the same time I have to maintain its temperature at 30 deg Cel.

For this I want to use water-air heat exchanger (HE).
Heat exchanger will have inlet & outlet pipes. Hot water coming from metal
block will enter HE. Fan inside HE will pass the heat from water to ambient air.
The water thus cooled will come out of HE through outlet and will cool the metal block.
This is closed loop.

I want to know what should be water flow rate. I have to pick some ready-made HE
available in the market. Hence, I tried using following equation:

q = f * c * (T_hot - T_cold)
where q= heat to be removed from metal block
f = water flow rate through inlet & outlet of HE
c = specific heat capacity of water
T_hot = temperature of water entering HE = 30 deg C
T_cold = temperature of water coming out of HE

Ambient temperature is 22 deg C.

Questions:

1. Is this correct approach ?
2. In this equation, contact area between water and metal block does not come in the picture. Why is this so?
3. Does working/functioning of HE depend on temperature difference between inlet temperature & ambient temperature ?

Any help will be highly appreciated.

Thanks.

 

[BvU]

Hello alphabet,

1. Yes for the water side
2. You can write the same eqn for the air side. Then there is an equation for the heat transfer $q = UA \Delta T_{\rm LM}$ with $\Delta T_{\rm LM}$ the Logarithmic mean temperature difference (LMTD).

[edit] U is the heat transfer coefficient ( Watt/(m².K) ) , A is the area of the exchanger

3. You bet. See 2.

It will be a tough job to remove 250 W with such a small LMTD -- and I expect the 22 degrees ambient temperature will be very hard to keep down as well. What if the sun shines ?[/SUP]

 

[abcxyz]

Hello alphabet,

1. Yes for the water side
2. You can write the same eqn for the air side. Then there is an equation for the heat transfer $q = UA \Delta T_{\rm LM}$ with $\Delta T_{\rm LM}$ the Logarithmic mean temperature difference (LMTD).

[edit] U is the heat transfer coefficient ( Watt/(m².K) ) , A is the area of the exchanger

3. You bet. See 2.

It will be a tough job to remove 250 W with such a small LMTD -- and I expect the 22 degrees ambient temperature will be very hard to keep down as well. What if the sun shines ?[/SUP]

(1)
This is related to CCD cooling; ambient temperature range is from -5 deg C to 22 deg C over a span of a year in my case.
This is optical telescope application. During night observations of sky, temperature range is as mentioned above.

(2)
From my first post, from first eqn, I will find T_cold for water side.
So I know T_hot & T_cold for water.
Now there is fan inside HE.

So I have to find fan's i.e. air's volumetric flow rate so that water of T_hot is
converted into water of T_cold. Right ?

And you mean to say that same first eqn . [ q = f * c * (T_hot - T_cold) ]
I will use for this air/fan side, with appropriate change in c. q will be still 250 watts. Correct ?

(3)
Where & why to use this equation q=U*A*LMTD then ?

Thanks & Regards

 

[BvU]

(1)
This is related to CCD cooling; ambient temperature range is from -5 deg C to 22 deg C over a span of a year in my case.
This is optical telescope application. During night observations of sky, temperature range is as mentioned above.

I see. Good.

(2)
From my first post, from first eqn, I will find T_cold for water side.
So I know T_hot & T_cold for water.

Your T_cold can not be less than 22 C plus some driving temperature difference (a bit like this LMTD). Note this eqn could also be used to determine the water flow. In practice it's both (see below)

So I have to find fan's i.e. air's volumetric flow rate so that water of T_hot is
converted into water of T_cold. Right ?

And you mean to say that same first eqn . [ q = f * c * (T_hot - T_cold) ]
I will use for this air/fan side, with appropriate change in c. q will be still 250 watts. Correct ?

Yes.

(3)
Where & why to use this equation q=U*A*LMTD then ?

The three equations combined help you design the heat exchanger: you have to transport 250 W of heat from a higher temperature (water in/out) side to a lower temperature (air in/out) side. Through some material (the water pipe walls) that has a material-dependent heat conductivity. The heat flux also has to reach the pipe wall from the inside and be given off to the air on the outside. These three steps have a resistance that determines how much the over-all heat transfer coefficient U is. You use this q=U*A*LMTD as a design equation. More q means higher LMTD needed, or more area, or a better U.

 

[Chestermiller]

You have greatly underestimated the complexity of this problem. What you are dealing with here is conceptually the same thing as an automobile cooling system. You need to consider the heat transfer within both the block and the heat exchanger.

The temperature of the water leaving the block will not be 30 C; it will have to be less than 30 C. You need to design the passageways in the block so that they can transfer the required amount of heat to the coolant at the selected flow rate. Otherwise the temperature of the block will be higher than 30 C. In the heat exchanger, the design will also be very important. You need to maximize the heat transfer to the air, and that might even involve the use to cooling fins, like on a car radiator. And there will be resistance to heat transfer both inside the tubes of the heat exchanger and on the outside to the air. I suggest you hire a heat transfer expert to design this thing properly.

 

[abcxyz]

Thanks BvU & Chestermiller.
I have to pick ready-made HE.

(a)
I have attached the datasheets for off-the-shelf HE model WL500 from Laird technologies.
One of them contains dT vs cooling capacity graph.
Since I want to lift 250 watts, corresponding dT = 8 deg C.
So if ambient temp. is 22 C, water inlet temperature should be 30 deg C. Correct ?

But if ambient temp. drops to 10C, water inlet temperature should be 18C;
i.e. water coming from metal block and entering HE should be at 18 C.
But I have to keep my metal block at 30 C.

Any comment ?

(b)
How do I find flow rate in my case from these datasheets of HE WL500 ?
Flow rate for this HE is rated as 2.3 liters/min for 500 watts.

(b) On the graph, dT = difference between ambient Temp. & water inlet temperature,
while on page 2 of "THR-UM-WL500 0412.pdf", (under "Description"),
dT = difference between ambient temperature and the water outlet temperature.
Is this not a discrepancy ?

 

[BvU]

As Chet said, a 30 degree water outlet at the CCD does NOT keep the CCD at 30 degrees. There is a resistance to heat transfer at that end too, meaning the cooling water has to be considerably colder than 30 degrees. Compare the WL500 that is a big device with a fan and nevertheless needs 8 degrees delta T as a driving force.

(a)
At an ambient temperature of 22 degrees your WL500 would cool the 2.3 l/min from 30 degrees to 28.5 degrees:

( 250 J/s ) divided by ( 2.3 l/min * 1 kg/l / 60 s/min * 4200 J/K/kg = 161 J/s/K )​

This leaves a puny 1.5 degrees which is not enough to cool the CCD.

For lower ambient temperatures you have a better chance. Worry about the worst case first, the rest can be dealt with by implementing some from of control.

(b)
the pump curve shows around 2.3 l/min

(c) (!)
No discrepancy. For 500 W cooling you go from 17 degrees at the water inlet to about 14 degrees at the water outlet. Two different ΔT

--------------

Again: look at the situation at the CCD end first. You don't have inifinite area there to pull out the heat, so you'll need very good thermal contact and a big temperature difference. If that requires 5 l/min at 16 degrees you have to think of something else than a simple air cooler at the other end.