Counterflow heat exchanger model
Counterflow heat exchanger model
(OP)
Hi all,
I am looking at a heat exchanger question in my new job for the first time since 3rd year of my Mech Eng Degree (about 9 years ago!). Anyway I have done some reading and the scenario I am dealing with is fairly simple but I'm still unsure on a few things. Here goes at trying to explain it:
- We have a simple hot water delivery device that delivers about 0.5L/min of water at around 99degC. We want however to knock the output temp down to about 95degC.
- Unfortunately the heater element is a constraint in our situation in that it's of the flash boiler type and our design uses the high temp and pressure to push the water up to the outlet (which is at a higher position than the reservoir) - hence we can't just drop the power supplied to it. (Also trying to mix in some of the inlet water isn't an option because of the pressure difference)
So one thing we were looking at was using the inlet water to cool the outlet water using a simple counterflow heat exchanger set up - see the picture at the link. One of the drivers for water cooling is that space is minimal - ie non-existant - and it was thought that just having a section of copper pipe heat exchanging to the surrounding air wouldn't be sufficient because it would be too short in length.
[URL=http:// img136.ima geshack.us /my.php?im age=heatex changersys temfa5.jpg][IMG]URL]
The rest of the tubing is 2.5mm wall thickness silicon.
What I am trying to determine really is the size of the exchanger required to get the 4degC temp drop. To do this I wanted to use Q = U * A * delta T = mass flow * Cp * delta T
to establish the area required of the copper pipe seperating the two different temperature fluids.
To begin I was going to assume that shell side cold water was insulted to the surrounding air and will have heated by a corresponding 4degC - by conservation of energy (and ignoring the small changes in specific heat over the temp range).
Now I know the equation to work out the value for the overall heat transfer co-eff, U, but it involves knowing "h" the heat transfer coefficient for water. "h" in turns depends on Nusselt's, Prandtls and Reynolds numbers!!
Is there any easier way to do this - any tables of given h values for water to water transfer through a copper wall. I have seen some ranges given for water-to-water heat transfer but it seems like a big range: 150-1200 W/m^2K. Another gave it as 340-450 W/m^2K when the seperating medium is copper.
I hope that's sort of clear to understand!
Please feel free to suggest other methods to get the temperature decrease I am after, but I would really like to understand the fundamentals about modeling heat exchange as well.
Thanks heaps in advance.
James
I am looking at a heat exchanger question in my new job for the first time since 3rd year of my Mech Eng Degree (about 9 years ago!). Anyway I have done some reading and the scenario I am dealing with is fairly simple but I'm still unsure on a few things. Here goes at trying to explain it:
- We have a simple hot water delivery device that delivers about 0.5L/min of water at around 99degC. We want however to knock the output temp down to about 95degC.
- Unfortunately the heater element is a constraint in our situation in that it's of the flash boiler type and our design uses the high temp and pressure to push the water up to the outlet (which is at a higher position than the reservoir) - hence we can't just drop the power supplied to it. (Also trying to mix in some of the inlet water isn't an option because of the pressure difference)
So one thing we were looking at was using the inlet water to cool the outlet water using a simple counterflow heat exchanger set up - see the picture at the link. One of the drivers for water cooling is that space is minimal - ie non-existant - and it was thought that just having a section of copper pipe heat exchanging to the surrounding air wouldn't be sufficient because it would be too short in length.
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The rest of the tubing is 2.5mm wall thickness silicon.
What I am trying to determine really is the size of the exchanger required to get the 4degC temp drop. To do this I wanted to use Q = U * A * delta T = mass flow * Cp * delta T
to establish the area required of the copper pipe seperating the two different temperature fluids.
To begin I was going to assume that shell side cold water was insulted to the surrounding air and will have heated by a corresponding 4degC - by conservation of energy (and ignoring the small changes in specific heat over the temp range).
Now I know the equation to work out the value for the overall heat transfer co-eff, U, but it involves knowing "h" the heat transfer coefficient for water. "h" in turns depends on Nusselt's, Prandtls and Reynolds numbers!!
Is there any easier way to do this - any tables of given h values for water to water transfer through a copper wall. I have seen some ranges given for water-to-water heat transfer but it seems like a big range: 150-1200 W/m^2K. Another gave it as 340-450 W/m^2K when the seperating medium is copper.
I hope that's sort of clear to understand!
Please feel free to suggest other methods to get the temperature decrease I am after, but I would really like to understand the fundamentals about modeling heat exchange as well.
Thanks heaps in advance.
James





RE: Counterflow heat exchanger model
The best reference for step-by-step calculations of this type is:
Hewitt, G. F., G. L. Shires, and T. R. Bott: "Process Heat Transfer", pp 275-285 (CRC Press, 1994).
This is an encyclopedic work written by the leading authorities in the field, surely a worthy successor to Kern's magnum opus with the same title (McGraw-Hill, 1950). In Hewitt, you will find up-to-date correlations and numerical examples on virtually all aspects of process heat transfer.
RE: Counterflow heat exchanger model
Elie Abou Jaoudeh
RE: Counterflow heat exchanger model
sorry for the delayed thanks - been a busy week or few!
I will try to get a copy of the suggested reference.
Elie your point seems a fairly valid one! I think the cooling shell will however dissipate more heat than I am assuming and so Tcold,out will be less than 14deg C. It definitely isn't a completely "closed system" as the surrounding air will come into play.
Anyway my collegue is pursing a few mock ups of different solutions as well to see if he can get an answer too.
Thanks
James
RE: Counterflow heat exchanger model
RE: Counterflow heat exchanger model
Nu = h*d/k (you know what h is and need to find it, d is the ID of you copper pipe and k is the k for water at this temp (0.6802W/m-k @ 93Deg C))
If the flow is laminar it is easy Nu=4.364, Solve for h!
If the fluid is Turbulent h can be approximated as
h=1429*(1+0.0146T)(v (in m/s))^2 / d (in m)^0.2
I got this from my Lindeburg Mech Eng ref Manual.
So now you have h inside and you need h outside.
h outside is natural convection (unless you use the fan that marcoh talked about) for natural convection of air and normal ranges of Gr Grashof Number and Pr Prandtl number and a horizontal cylinder:
h = 1.32 *((Ts-Teff)/OD)^0.25 Where Ts is the surface temp and Teff is the temp of the room. The surface temp should be the average of the water temp and the air temp (good approx.)
For a vertical cylinder:
h = 1.37 *((Ts-Teff)/L)^0.25 Where Ts is the surface temp and Teff is the temp of the room and L is the vertical length of the cylinder.
So now we have the convection - conduction - convection that Elei talked about.
So:
1/U = 2PI /(rout*hout) + ln(ro/ri)/kCu + 1/(rin*hin)
and
Q= U * L * (Tin - Tout) = m * Cp * Delta T
When I started to study for my test this type of question really confused me because of all the Prandtl and Reynolds and other numbers. I think HT engineers came up with all those things to try and confuse other Mechanical Engineers, either way once you start to look into it is not that difficult.
Good Luck.
RE: Counterflow heat exchanger model
Elie Abou Jaoudeh
RE: Counterflow heat exchanger model
GSTeng - it's great to know I wasn't the only one confused at first.
I will plug in the numbers and try to get a reference going through those equations.
Cheers
James