Sizing a counterflow air-to-air heat exchanger
Sizing a counterflow air-to-air heat exchanger
(OP)
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.
Also, is a shell and tube design or a plate design best for air? Thanks for any help.





RE: Sizing a counterflow air-to-air heat exchanger
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.
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RE: Sizing a counterflow air-to-air heat exchanger
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 70oF. 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.
RE: Sizing a counterflow air-to-air heat exchanger
I2I
RE: Sizing a counterflow air-to-air heat exchanger
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 (www.stoneprocess.com)
RE: Sizing a counterflow air-to-air heat exchanger
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!
RE: Sizing a counterflow air-to-air heat exchanger
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 (www.stoneprocess.com)
RE: Sizing a counterflow air-to-air heat exchanger
RE: Sizing a counterflow air-to-air heat exchanger
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RE: Sizing a counterflow air-to-air heat exchanger
Thanks for the help and the software guys!
RE: Sizing a counterflow air-to-air heat exchanger
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?
RE: Sizing a counterflow air-to-air heat exchanger
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 ft2 in area. If your coworker used a 15ºF, then you'd get 630 ft2 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
RE: Sizing a counterflow air-to-air heat exchanger
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 = UAFΔTlm in which the lm stands for log mean. The log mean temperature difference method is commonly used for heat exchanger sizing. By using ΔTlm, differences in the local temperature difference values are accounted for. The equation for calculating ΔTlm 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
RE: Sizing a counterflow air-to-air heat exchanger
RE: Sizing a counterflow air-to-air heat exchanger
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.
RE: Sizing a counterflow air-to-air heat exchanger
RE: Sizing a counterflow air-to-air heat exchanger
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
RE: Sizing a counterflow air-to-air heat exchanger
RE: Sizing a counterflow air-to-air heat exchanger
The product of htc*area*deltaT must be the same on both sides of the HX.
TTFN
RE: Sizing a counterflow air-to-air heat exchanger
RE: Sizing a counterflow air-to-air heat exchanger
T1=250
T2=80
t1=70
t2=240
I'm going to be in a divide-by-zero situation...
Does that make sense? This is using the LMTD equation for a countercurrent HX...
RE: Sizing a counterflow air-to-air heat exchanger
How many BTU's are you willing to spend to recover the BTU's in your initial air stream?
How is the hot air being created? What happens to it currently? Where and how will you be using the "newly created" hot air?
Remember there are inefficiences in any process. You will not get ALL the BTU's out -- which is where your LMTD equation is going wrong.
Patricia Lougheed
Please see FAQ731-376 for tips on how to make the best use of the Eng-Tips Forums.
RE: Sizing a counterflow air-to-air heat exchanger
Even if I did:
T1: 250
T2: 200
t1: 100
t2: 150
I will get a divide-by-zero in my LMTD equation (???)
RE: Sizing a counterflow air-to-air heat exchanger
I2I
RE: Sizing a counterflow air-to-air heat exchanger
RE: Sizing a counterflow air-to-air heat exchanger
I should have looked at this LMTD business earlier in this discussion.
Lets say that you are basing your design on a 10 deg F approach:
250 --->70
240 <---60
=== ===
10 10
If the two terminal temperature differences are identical or nearly identical (within 10%) then the LMTD is virtually the same as the arithmetic mean of the two temperature differences.
Since the two flows are the same, and the temperatures are almost the same, the properties are VERY nearly identical.
What you can't do is this:
250 --->70
250 <---70
=== ===
0 0
Now there is NO temperature difference in counterflow, hence NO heat transferred.
Regards,
Speco
RE: Sizing a counterflow air-to-air heat exchanger
T1: Hot stream inlet temp
T2: Hot stream outlet temp
Does "hot stream" mean the flow through ONE side of the exchanger, or is T1 and T2 on the same (inlet) side of the HX?
If I blow 250F air through the "hot" side and 70F air through the "cold" side..I can'e imagine how there would ne NO transfer. I just don't get it...
RE: Sizing a counterflow air-to-air heat exchanger
You're absolutely right. Anytime you set the temperature differences so they're absolutely the same, the LMTD equation falls apart. Realistically, the formula collapses to the temperature differential (10 degrees). You can show this by changing one of the values slightly (I added 0.0001 to the hot outlet temperature, but it works on any of the four temperatures.) This very small difference is enough that a calculator is no longer dividing by 0 and it will spit out a value.
Looking back at your scenario, I realize you've set your flows exactly equal. This forces the temperature differentials to be equal in order for the energy balance to work out (Remember Q=mdotcpdelta-T.) Since mdot and cp are the same for both the hot fluid and the cold fluid given your initial conditions, then the delta-T's have to be the same -- making the LMTD formula unusable mathematically.
I've never run into an actual situation where everything was the same, and, in real life, if you instrumented everything, you'd probably find slight differences due to system inefficiencies or heat losses that didn't occur through the heat exchanger. But since you're in the "design mode", just use the temperature differential.
Patricia Lougheed
Please see FAQ731-376 for tips on how to make the best use of the Eng-Tips Forums.
RE: Sizing a counterflow air-to-air heat exchanger
RE: Sizing a counterflow air-to-air heat exchanger
This means that assuming a constant local temperature difference should be ok because the mdotCp terms for both streams are nearly identical. For symmetrical stream flows, the LMTD converges to the arithmetic average.
speco is not saying that there will be no heat flow in the case [250 -> 70] [70 -> 250]. He is, however, saying that the average heat flux (heat transfer per unit area) will be zero. I stated previously that the only way to transfer ALL the energy between streams is in the case of infinite surface area. The heat flux is then the amount of heat transferred divided by the surface. A number divided by infinity is clearly 0.
This may be about the time that you consult a heat exchanger manufacturer and get their recommendation.
I2I
RE: Sizing a counterflow air-to-air heat exchanger
If you adjust one of the temperatures by a small amount, then the LMTD will equal the temperature differential between the hot and cold fluid, which has been assumed to 10 degrees in your example. So it doesn't matter if you "fudge" the LMTD or just use the temperature differential. In real life, it's quite likely your actual temperature differentials will not be the same because there will be inefficiencies in the system.
Regarding your earlier post about conventions. Normally, the fluid that starts off the hottest is labeled the "hot" fluid and is designated by "T". The fluid that started off at the lower temperature is the "cold" fluid, and is labeled "t". The "1" stands for inlet and the "2" for outlet. Personally, I use the designations "Th,i", "Th,o", "Tc,i", and "Tc,o" where "T" stands for temperature, "h" for hot, "c" for cold, "i" for inlet and "o" for outlet.
LMTD calculates the temperature differential between the two fluids. It takes the difference between hot inlet and the cold outlet (10 degrees, your example) and then the difference between the hot outlet and the cold inlet (again 10 degrees). It then subtracts the difference between the two and divides by the natural log of one over the other. Because of the parameters you've set up, you basically come up with 0 over infinity.
Patricia Lougheed
Please see FAQ731-376 for tips on how to make the best use of the Eng-Tips Forums.
RE: Sizing a counterflow air-to-air heat exchanger
Therefore, the only plausible answer is that there will be a finite temperature difference between the cold outlet and the hot inlet. Conservation of energy requires that the same temperature difference between the cold inlet and the hot outlet.
If you assume that they're both 10ºF and make the assumption that the thermal gradients on the hot and cold sides are linear with respect to distance, you' get that the temperature delta is 10ºF everywhere in the HX.
This then allows you to determine the total heat transferred based on (250ºF-80ºF)*specific_heat*mass_flow, which must equal (240ºF-70ºF)*specific_heat*mass_flow.
Then, the area required is heat/(htc*5ºF). The HX must sit half-way between the hot and cold sides, again because of conservation of energy.
The total power is then, 18.36 kBTU/hr, which for a 5ºF temperature delta and 2 BTU/(hr*ft2*ºF), results in an area of 1944 ft2
TTFN
RE: Sizing a counterflow air-to-air heat exchanger
RE: Sizing a counterflow air-to-air heat exchanger
It seems to me an LMTD of 10 deg F is overly optimistic for this cooling range.
I gather that such a high degree of temperature "crossing" to be attainable would need several "passes" of a totally truely counter-current arrangement because the intervening-metal temperature (due to its good thermal conductivity) would interfere with the "linear" heat transfer between the streams creating undesirable pinch points.
RE: Sizing a counterflow air-to-air heat exchanger
RE: Sizing a counterflow air-to-air heat exchanger
Single-pass exchangers can possibly be counterflow. However, it really depends on how they are constructed.
For example, if you use a long shell and tube exchanger with a single pass on the tube side, and a single pass on the shell side in the opposite direction, then it it pretty much counterflow.
Now think of a shell and tube exchanger with U-tubes (2-pass in this case) and a single-pass shell. In half of the exchanger you have counter flow, but in the other half you have co-current (also called Parallel) flow. This is where things begin to fall apart with a temperature cross.
If you have a rectangular exchanger (a more typical air-to-air configuration), with one side single pass and the other side multi-pass, then it is actually counter-cross flow, and also approximates counterflow.
A simple cross-flow exchanger will not work any time there is a temperature crossover required.
Since I first mentioned an overall heat transfer coefficient in the range of 2 to 3, I need to clarify that it's based on the assumption of fairly low velocities on both sides of the exchanger. In an air-to-air exchanger velocity is everything, since it determines the component heat transfer coefficients on both sides (film coefficients). The higher the velocity, the higher the heat transfer coefficient. However, this is at the cost of pressure drop. That is, higher velocities mean higher pressure drops, but a more compact exchanger.
Also, you should take a look at your process requirements. Remember that the overall size is inversely proportional to the LMTD (log mean temperature difference). If you start with a 10 deg F LMTD, the exchanger will be twice as large as one with a 20 deg. F LMTD (with all other factors being equal, but they never really are).
Regards,
speco
RE: Sizing a counterflow air-to-air heat exchanger
RE: Sizing a counterflow air-to-air heat exchanger
I estimate that for a countercurrent unit with an effectiveness ε ~ 0.8, the cool air would heat up to ~220oF, and the hot air would cool down to ~100oF. This would represent ~5.5 NTU.
The estimated heat exchange surface would be about 300 ft2 for an OHTC = 2 Btu/(hr.sqft.oF)
Using the LMTD method with an LMTD = 30oF, and a correction factor of ~0.9 would show about the same result.
See also the article: Consider the plate heat exchanger by Raju and Chand, Chemical Engineering issue of August 11, 1980.
RE: Sizing a counterflow air-to-air heat exchanger
RE: Sizing a counterflow air-to-air heat exchanger
I2I
RE: Sizing a counterflow air-to-air heat exchanger