Heat Exchanger Question

[SoupEng]

My apologies if this is a simplistic question, but it's been a while since I've actually played with the numbers, mostly a management guy now.

The issue that I have is that I've been presented with a proposal that doesn't seem realistic.

150 ft of Tube-in-Tube heat exchanger:

Our product (Call it Gravy):
Flow Rate - 10 USGPM
T1 (Start Temp) 190 F
T2 (End Temp) 70 F
DeltaP - 150 PSIG
Spec gravity = 1.01
spec heat - c = 0.67 BTU/lb.F
thermal conductivity - k = 0.19 BTU/hr.ft.F
Viscosity - 14,856 cP (Centipoise) @ 127 F

Cooling Medium
50% Glycol, 50% Water
Flow Rate - 50 USGPM
T1 - 25 F
T2 - 48 F
Delta P - 21 PSIG

My question is simply, Is this realistic? The manufacturers claim that they can take the temperature from 190F to 70F seems to clash with my previous food product experience. I'm not confident that 150 ft of tube in tube can take this much heat out of a product.
Is this pressure change also realistic? I wish I still had my Thermo texts, it's frustrating that I can't figure this one out.

Thanks for any help.

 

[ardilesd]

Heat balance seems to be OK (Cp for Glycol-Water was taken as 0,75 Btu/lb.F). But we do need the transfer area to verify if this Heat can be effectively transferred.
Regards

 

[SoupEng]

Thanks for the feedback.

The system I'm speaking of would simply be 150 ft of 2" tube inside 150 ft 3" tube. (Straight piping)

 

[MortenA]

on the rough assumption that they two media have the same heat capacity:

50*(48-25)=1150
10*(190-70)=1730

The heat ballance does not solve - and its the "wrong" way. You want to remove more heat than the receiving media can "receive". This is the most basic equation that must be met. Then come heat rate transfer considerations.

If there is a significant difference the post the two media heat capacities and i could adjust the equation.

Best Regards

Morten

 

[ardilesd]

Hello, Morten .
Check your numbers
The second figure should be 1200 , so heat balance does solve approximately as I said before.
But second , Why do you take the same heat capacity if you know that from the product and can estimate the mixture glycol-water ? At least, you call it rough ....

 

[MortenA]

ah sorry

Well i didnt use much time on it (lack of QA resources

I see my error.

Best regards

Morten

 

[JOM]

I'm more interested in the type of heat exchanger being proposed for a food product. This is 150 feet of a single 2 inch tube inside another tube. That's alot of plumbing.

With a two inch diameter tube and a viscous product like "gravy", will it be turbulent flow? Turbulent flow gives a high heat transfer coefficient.

Surely this isn't a single 150 foot length of tubing? What kind of jointing is used if it is several lengths of tubing - say 15x 10 foot. Can it be disassembled for inspection? How do you clean it? CIP cleaning requires turbulent flow. What happens if a hole develops through corrosion? Will that contaminate the gravy?

All these problems are solved by selecting a plate heat exchanger, and the design will be very precise and guaranteed.

I'm not answering the original question, but to me it seems a strange sort of heat exchanger for such a product. Cheers,
John.

 

[ardilesd]

1) From the point of view of heat tranfer, you need to remove Q = 100 000 kcal/h
Based on the DTML ( 46°C countercurrent) and the transfer area ( ca. 8 m2), you would need a transfer coefficient of 270 kcal/h.m2.°C.
From the properties of your product and the flowrate, you will be in the laminar region and being (very) optimistic, the hi (inner heat transfer coefficient)would never be above 25 kcal/h.m2.°C
So, you could never reach the heat tranfer rate required.
Sorry, but I was in a rush and perhaps there is a mistake in some figure, But I guess that you got the idea.
2)The JOM's remarks are very valuable, as the tube in tube configuration is neither "the effective" nor "the clean" solution for some services.
Regards
See Kern, Chapter 6 (in spanish version) for more details

 

[TD2K]

I echo ardilesd's comment. I worked through this using US units but with a Q of 400,000 BTU/hr, area of 93.3 ft2 (based on outside diameter of 150 feet of 2" pipe) and a LMTD of 84.4F, the required U is 51 BTU/hr/ft2F.

Now, if you really have 21 psi dP on the glycol/water side, you will flow a heck of a lot more than 50 gpm which will increase the LMTD but not by that much since you already have a fairly small temperature rise on that side.

Given the viscosity of this stuff on the other side, I would expect a pretty low heat transfer coefficient, less than the 51 BTU/hr/ft2/F required. Granted, I haven't crunched the numbers but a viscosity of 15,000 cP would give me concern especially given the sensitivity of the viscosity at that range to composition changes, small changes in temperature, etc.

 

[ardilesd]

TD2K ,
the conversion factor English --> metric for heat transfer coefficient is 4.88.-
So, your english U=51 is U=250 in my metric world, close to the U=271 I estimated.
Regards

 

[SoupEng]

JOM, I agree wholeheartedly about this not being the ideal solution. The reason that it is being considered is primarily cost. The current geography dictates the 150ft run of piping, but that is done at 180 degrees F, at which temperature there is significantly less viscosity in the product.
The construction would be 7 x 20' lengths of tubing, with one 10' length.

Holes in the tubing would be answered by a required pressure differential. The lowest product pressure has to be significantly above the highest pressure reached by the cooling medium.

The current proposal for cleaning is indeed a CIP system, a full return loop would need to be installed alongside the current piping run.

The alternative scheme would involve a scraped surface heat exchanger, but the cost is prohibitive and there are products with particulate that we run that could be problematic.

I see other issues in the tube in tube exchanger, including maintenance of seals and gaskets, and product inconsistency and cooling efficiency decrease due to "coring" (when cooler product solidifies near the edge of the tube, and hot product creates a tunnel through the middle).

Thanks all for your help.

 

[SoupEng]

ardilesd - Thank you for the direction, but I need to bother you again.

Using 92,200 kcal/hr requirement, 7.7 m2 surface area, DTML of 46.7C, I followed through the calculations, coming up with 254 kcal/hr-m2-C. (Same order of magnitude as the 271 you came up with and the 250 that TD2K found)


The question is then, how did you come up with the number 25 kcal/hr-m2-C for the possible heat transfer?

Again, my apologies if appear too dense.

Regards,

Chris

 

[ardilesd]

Chris
the Overall Heat Transfer coefficient (U) you calculate is the required U based on the heat load , DTML and Area.

The value of 25 we proposed with TD2K is the "real" U you could have in your system, it is normally estimated from correlations based on Reynolds (fluid dynamics conditions) and Prandtl (heat capacity/conductivity) numbers.
With the high viscosisty of your product, you are in low Reynolds numbers, and very low convection (very low U). If you applied the proposed solution, your product will be cooled only 20 or 30 F, instead of the 120 F you are requiring.

 

[SoupEng]

I calculated Re of 1.065 (obviously laminar flow, as you stated) and Pr of 113507. (Please feel free to correct me if these are out of whack)

I am unfamiliar with a method of translating these numbers into a comparative "real" transfer coefficient to determine the feasibility of the proposal.

Thanks again,

Chris

 

[ardilesd]

The correlation proposed by Kern (Sieder & Tate) is

Nu = 1.86 * (Re*Pr*D/L)^(1/3) * (u/uw)^0.14

where:

Nu: h * D/ k

D: pipe diameter
L: Pipe leght w/o convective mixing
u & uw : viscosity & viscosity evaluated (if you can) at wall temperature
Hope this helps