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.

 

[dextermech]

I'm confused...when I do the LMTD, if I say:

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...

 

[vpl]

Dexter

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.

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.

 

[dextermech]

Yea but no matter what, if our delta T on the hot side is equal to your delta T on the cool side, you will get a divide by zero...

Even if I did:

T1: 250
T2: 200
t1: 100
t2: 150

I will get a divide-by-zero in my LMTD equation (???)

 

[insult2injury]

I think that you are misunderstanding the purpose for using the LMTD equation. It accounts for changes in the local [Δ]T values. If you are assuming the same [Δ]T on both ends then the [Δ]T at any point in between must have the same value as well and therefore calculating an LMTD value is pointless. When making this assumption, also note that you are also assuming a constant heat transfer coefficient and constant properties which is not the case. Among those constant properties, is the specific heat. Since the mass flow rates on each side are equal, the constant properties assumption implies that for each degree in temperature that the hot stream decreases, the cold stream must increase that same amount. If you consult a property table, you will notice that the actual specific heat values differ by only a few percent so this is a decent assumption over limited ranges. However, density, viscosity, and thermal conductivity values change by as much as 30% which tells me that the constant htc assumption is a bad one.

I2I

 

[dextermech]

So should I, for the sake of initial calcs, assume that say the cold side increases in temp less than the hot side lowers in temp???

 

[speco]

Dextermech,

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

 

[dextermech]

Something still doesn't make sense. Maybe I am not plugging into the LMTD properly? When they say:

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...

 

[vpl]

dexter

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=m[sub]dot[/sub]c[sub]p[/sub]delta-T.) Since m[sub]dot[/sub] and c[sub]p[/sub] 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.

 

[dextermech]

The temperature differential? You mean use LMTD but fudge it a bit so that I don't divide by zero, or do you mean just use 10 degrees? Thanks so much for your help.

 

[insult2injury]

"Since the mass flow rates on each side are equal, the constant properties assumption implies that for each degree in temperature that the hot stream decreases, the cold stream must increase that same amount. If you consult a property table, you will notice that the actual specific heat values differ by only a few percent so this is a decent assumption over limited ranges."

This means that assuming a constant local temperature difference should be ok because the m[sub]dot[/sub]C[sub]p[/sub] 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

 

[vpl]

Dexter

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 "T[sub]h,i[/sub]", "T[sub]h,o[/sub]", "T[sub]c,i[/sub]", and "T[sub]c,o[/sub]" 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.

 

[IRstuff]

If the cold side outlet temp equals the hot side inlet temp, you have perfect heat transmission, so that case is not physically possible, and produces an infinite area for a zero delta temp, as it should, since the heat transferred is assumed to be a costant.

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*ft[sup]2[/sup]*ºF), results in an area of 1944 ft[sup]2[/sup]

TTFN

 

[SloFlo]

How did you come up with a 2BTU Overall Heat Transfer Coefficient? This is not something that can just be estimated as it has a significant impact on the heat transfer area. Also, I should point out that it is true that the increase in temperature of the cold stream is equal to the decrease in temperature of the hot stream only because the flows and fluids are the same. Otherwise, this would not be true.

 

[25362]

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.

 

[dextermech]

Aren't single pass plate exchangers truly counter-current?

 

[speco]

Dextermech,

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

 

[yank38]

What am I missing here folks. My intuition tells me this is an impossible problem. 2 streams with the same mass flow and different temps. Delta T is required for heat transfer. As the cool side heats the hot side cools and at some point between the 250 and 70 the temperature will be the same, hence no heat transfer, which is why the theory is suggesting an infinite area. Wouldn't the cooling side require a larger mass flow to be capable of maintaining a delta T to cool all the way to 70?

 

[25362]

I estimate that for a countercurrent unit with an effectiveness [&epsilon;] ~ 0.8, the cool air would heat up to ~220[sup]o[/sup]F, and the hot air would cool down to ~100[sup]o[/sup]F. This would represent ~5.5 NTU.

The estimated heat exchange surface would be about 300 ft[sup]2[/sup] for an OHTC = 2 Btu/(hr.sqft.[sup]o[/sup]F)

Using the LMTD method with an LMTD = 30[sup]o[/sup]F, 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.

 

[yank38]

Again, please explain to me how the heat is tranfered from a cold strem to a hotter stream.

 

[insult2injury]

yank - Heat is not transferred from cold to hot. Do a quick google search of counterflow/counter-current heat exchangers. Heat is always transferred from the hot stream to the cold stream, and the hot stream has a higher temperature at every point in the system.

I2I