Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Tek-Tips community for having the most helpful posts in the forums last week. Way to Go!

Batch cooling with an internal coil 1

Status
Not open for further replies.

skaboy607

Mechanical
Dec 19, 2010
34
0
0
GB
Hi,

I am looking to size a copper coil to cool a volume of water by 5-10 degC. I have found numerous examples for heat exchanger calculations. The volume of water does have an inlet and an outlet but it is very slow so should this be treated as a batch cooling problem (think this is how they are referred to)?

All the equations I have found require an known value that I don't have. This PDF


was helpful but I want to consider a steady state problem as I desire a constant outlet temperature from the bottom of the tank.

Any help, most appreciated. I think I just need pointing in the right direction.
 
Replies continue below

Recommended for you

Is there a good chance the cooling coils will be needed to cool down that volume of water during a non-steady state condition, like after a prolonged shutdown? If so, I'd recommend sizing the cooling coils for batch cooling within an acceptable time frame and for the steady state. Then, put in the larger area of the two.

Good luck,
Latexman
 
The system is transient:

There will exist a volume of water at a temperature (Ta). Water will flow in at the top of the tank at Tin and out at the bottom of the tank Tout. DeltaT between inlet and outlet will be a max of 10 degC.

If Tin is always constant and I require a constant Tout; can this not be considered steady state?
 
Skaboy607,

I think you could attack this problem two ways:.

1. Cool the hot incoming fluid on its way to the tank. Then you have a classic steady-state heat exchanger problem.

2. Treat it as though you had a steady heat input (heat load), and size the internal cooling coil to remove the required amount of heat based on the tank Tout (as a constant) and the coolant tin and tout.

Regards,

Speco
 
Now I am confused. Perhaps I haven't explained myself very well, problems always seem clearer in your own head so apologies if thats the case.

Why would I cool the fluid before it enters the tank, that is why I have got the fluid passing through the tank and why I need to size the coil to cool that delta T (load).

Am I missing something?
 
Ok I see what you mean. Yes the idea of doing the 2nd method is what I think I have been describing. But...how would I got about this? To obtain an overal U value, I need to know the heat tsf cof in the tube and then to the water in the tank. I can calculate the one in the tube but how do I go about calculating the one in the tank?

This is the stage I got to before becomming stumped.
 
Skaboy,

It's hard to find good data for natural convection cooling. I would suggest using data for an analogous type heat exchanger. I found an average overall coefficient of 70 BTU/(hr-deg F-ft^2) for platecoil in a water-to-water cooling application. The SI equivalent is 397 W/(m^2-deg K). If this were my applcation, I would de-rate that a bit for additional safety factor. I am assuming that you have a reasonable cooling water velocity inside the coil.

This should be enough to come up with a workable solution.

Regards,

Speco
 
Thanks Speco. Is that an overall heat transfer coefficient or the h tsf coef from the coil to water in the tank?

Have you got a reference for that value by any chance?

Thanks

Will
 
Skaboy,

It's an overall heat transfer coefficient for cooling, using a coil in a non-agitated tank, with water on both sides. The actual range given was 70-80, by the way. I got these values from an old company that made plate coils.

Here's an online reference. Note the cast iron material for the water-to-water coefficient. It would be a bit higher for a copper coil, so I think the number I gave you looks reasonable.


Speco
 
Can I do this for my calculations....

Heat transfer = mcp(t2-t1) for the fluid coming into and out of the tank. Ans is 418W.

Then use Q=hA(tfluid-tcoil), rearrange for A:

A=Q/(h(tfluid-tcoil)

to give me the surface area needed to satisfy the load?
 
Did you mean, "Heat transfer = mcp(t2-t1) for the fluid coming into and out of the tank. If so, yes, you can do that.

If you are assuming the tank is well mixed, which is typical, the fluid coming out of the tank should be at the same temperature as the contents of the tank, so t2 = t1, right?

Good luck,
Latexman
 
Not quite I don't think. The coil is cooling water that enters the top of the tank at 25degc. I want it to leave the tank at the bottom at 15 deg c so I used those values as my t2 and t1 to calculate the cooling load.
 
Ha, yea I got confused writing it so why I thought anyone else would understand what I was saying. Hopefully the image below describes the calculations and system that I am working on.

System_tank_calcs.jpg
 
Status
Not open for further replies.
Back
Top