Sizing a counterflow air-to-air heat exchanger 1

[dextermech]

lease help. I have 100 SCFM of 250F air and want to transfer all of it's heat (down to ambient = 70F) to a separate 100 SCFM airflow. How do I calculate the necessary surface area?

Also, is a shell and tube design or a plate design best for air? Thanks for any help.

 

[BigInch]

The overall heat transfer coefficient OHTC between the two air sources through the interfacial pipe or plate will determine the surface area required for any given configuration. You will not be able to transfer "all" the heat, but the more surface area you have, the more heat you will be able to transfer.

Assume a heat exchanger configuration,
Calculate the Nusselt number of each flow against the interfacial surface and shape,
Find the OHTC of the interface shape and material,
Find the heat available to transfer,
Determine the surface area required to transfer the available heat using your assumed configuration,
Try different configurations until you have found the best configuration to give the needed surface area for each one.

Going the Big Inch!

http://virtualpipeline.spaces.msn.com

 

[25362]

Some additional thoughts:

[•] Consider extended (finned) surfaces.
[•] Consider also a cross-flow arrangement (lower friction drops).
[•] The cooling air stream should, of course, be cooler than 70[sup]o[/sup]F. Besides, temperature crossing is not advisable.
[•] Consider a refrigerant as an intermediate cooling factor, finally rejecting the heat to atmospheric air as a final sink. As usual economics will dictate the right approach.

 

[insult2injury]

What is the inlet condition of the secondary airstream? For what process is it required? Where does the primary "hot" airstream come from?

I2I

 

[speco]

Dextermech,

Since you have expressed both flows in SCFM, I presume that both are at approximately atmospheric pressure. If that it the case, the heat transfer coefficient on both sides will be very low, and fins will not make the exchanger any smaller, just more expensive. I would expect the overall heat transfer coefficient to be in the range of 2 to 3 BTU/(hr-ft^2-deg F)with reasonable velocities on both sides. The economics of this exchanger may be awful.

Also, the inlet temperature on the cold side is extremely important here. Since the two stream have equal mass flows, there is a good probability that the temperatures of the two streams will cross. That is, the outlet temperature of the cold stream will be higher than the outlet temperature of the hot stream. If true, then a simple cross-flow exchanger will not work. One of the stream must be arranged in a multiple-pass cofiguration to approximate a counterflow system. In doing so, the pressure drop through the exchanger for this stream could be important and must be accounted for.

Regards,

Speco (

http://www.stoneprocess.com)

 

[dextermech]

Speco,

Assume that one side will be at +20 inches H20 and the other side will be at -20 inches H20. It's a ring compressor blower pushing cold air (70F) through one side and pulling hot air (250F) through the other side. Thanks for your help!

 

[speco]

Dextermech,

Now that you have really defined your design criteria, I would suggest that you contact a vendor who manufactures air-to-air heat exchangers, and give him your data. He/she will also need to know the allowable pressure drop on each side, which is a critical piece of data. This exchanger could be either tubular or plate type. There are probably about 4 or 5 manufacturers in the US who normally make this kind of exchanger.

If you try to design one yourself, you will probably spend a great deal of time to develop a computer program for one-time use. Without a program, the process of calculating the heat transfer coefficients and pressure drops on both sides of the exchanger will be excruciating and reiteriative.

Regards,

Speco (

http://www.stoneprocess.com)

 

[srfish]

Speco is correct about the value of the overall heat transfer coefficient being in the range of 2 to 3 Btu/hr-ft2-F. I ran this problem through our in-house software that estimates the size of shell-and-tube heat exchangers. Using a 10 F. temperature approach where the hot air stream cools to 80 F., the calculated surface is aproximately 600 square feet (wow).

 

[BigInch]

download the software here,

http://www.hudsonproducts.com/products/finfan/program/ACHE2Setup.exe

http://www.hudsonproducts.com/products/tuflite/VTuflite50.exe

BigInch-born in the trenches.

http://virtualpipeline.spaces.msn.com

 

[dextermech]

Not sure what you mean by a 10F temperature approach?

Thanks for the help and the software guys!

 

[dextermech]

I just calculated it like this, correct me if I'm wrong:

assumptions:

airflow: 100 SCFM

air density: .075 lbs. / (cubic foot)

hot air temp: 250F

cold air temp: 70F

specific heat of air: .24 BTU/(lb. * F)

heat transfer coefficient (U): 2 BTU/ (square foot * F * hour)


1. Convert SCFM to hourly mass flow:

100 (cubic foot / minute) * 60 (minute / hour) * .075 (lb. / cubic foot) = 450 (lb. / hour)

2. Calculate the amount of energy that needs to be transferred:

450 lbs * [.24 BTU/(lb.*F)] * (250F – 70F) = 19440 (BTU / hour)

3. Calculate heat exchanger surface area:

A = Q / (U * delta T) = 19440 (BTU / hour) / [2 BTU / (cubic foot * F * hour)]* (250F – 70F)

A = 54 square feet

SOmeone came up with 600 square feet, maybe I am off one decimal place or maybe they are?

 

[IRstuff]

I think both numbers are low, but the higher one is closer, and it's dependent on what he assumed to be the exchanger temperature. You're transfering heat from hot air to metal, then from metal to ambient.

Since you want to remove ALL the heat from the hot air, the output must be close to ambient and the metal must be close to ambient. BUT, that means that the heat moved from the hot air has a harder time getting into the ambient air because the temperature delta is MUCH smaller, and is what is driving the area of the exchanger. If you assume that the metal is at 75ºF, the 5ºF delta causes the ambient side to be at least 1900 ft[sup]2[/sup] in area. If your coworker used a 15ºF, then you'd get 630 ft[sup]2[/sup] area.

I assume that a counter-flow HX will have an additional inefficiency compared to a normal HX, so the area ought to be even larger, or maybe he assumed an even higher delta.

The minimum sized area must be where the metal is approximately halfway between 250ºF and 70ºF.

TTFN

 

[insult2injury]

Several concerns:

1. I wouldn't assume constant properties and heat transfer coefficients with such a large change in temperature (and most likely pressure as well).

2. I would not recommend attempting to cool the hot stream to the inlet temperature of the cold stream. If you take a look at the math, that can only occur in the limit as the area approaches infinity.

3. The correct formula for use in part 3 of your calculations is Q = UAF[Δ]T[sub]lm[/sub] in which the lm stands for log mean. The log mean temperature difference method is commonly used for heat exchanger sizing. By using [Δ]T[sub]lm[/sub], differences in the local temperature difference values are accounted for. The equation for calculating [Δ]T[sub]lm[/sub] depends on the flow conditions of the HEX (parallel, counterflow, crossflow/multipass). Also, F is a correction factor that is read from a graph for the heat exchanger geometry selected.

4. For such large changes in temperature/pressure, it is wise to discretize the HEX into several sections performing the calculations on each section in order to obtain an semi-accurate result.

I2I

 

[dextermech]

I'm finding it hard to beieve, based on my intuition alone, that I need 1900 square feet of aluminum to transfer the heat from a 100 SCFM 250F airflow to 100 SCFM 70F airflow...it just seems off by an order of magnitude...something must be wrong, maybe.

 

[srfish]

dextermech

I am the one who calculated 600 square feet. Our difference is all in the mean temperature difference used. It appears you used the temperature range on the hot side for the MTD. I used an outlet temperature on the hot side of 80 F. instead of 70 F. Then on the cold end you would have a 10 F. difference. That is where the term 10 F. approach comes from. Since this is an estimate, the outlet temperature of the colder stream was assumed to be 240 F. These temperatures calculate a mean temperature of 10 F.

 

[dextermech]

OK, I see how my calculation is totally wrong now...did not use delta T lm, read "F" wrong...I'll re-do it. Thanks for the help. Still don't think my exchanger should be thousands of square feet though...

 

[IRstuff]

Well, it depends on how much heat you willing to leave behind. The minimum area case is about 54 ft[sup]2[/sup], but you'd only get the temperature down to around 160ºF.

The case you came up with would work on the hot side, but the cold side would need to be much, much larger, otherwise, conservation of energy requires that the delta temp on the cold side be the same as on the hot side, e.g., you'd need a -110ºF cold side

TTFN

 

[dextermech]

Remember, it's not that I'm trying to cool the hot air, it's that I want to harvest it's energy, it's energy being any BTU's causing it to be above 70F.

 

[IRstuff]

Either way, the "harvesting" of "all" the heat means that the temperature must be dropped down to lowest available temperature. That heat flow into HX must be exactly balanced by an outgoing heat flow into the cold side air, otherwise, its temperature must go up.

The product of htc*area*deltaT must be the same on both sides of the HX.

TTFN

 

[dextermech]

Wait, the 54 square feet is completely wrong I think, I did the calc completely wrong...