Plate and frame HX modification

[MatthewL]

I am looking for some confirmation of a few calculations I am working on. Right now, we use a plate and frame heat exchanger with water on both sides to cool process water using either cooling tower water or chilled water. I took some readings on a recent run using tower water, and here were the results:

SIDE TEMP IN TEMP OUT FLOW(GPM)
HOT 109 F 100 F 168 (metered)
COLD 79 F 89 F 151 (calculated)

I trended the results over the last year, and our normal hot in/hot out delta is +/- 1 degree of a 10 degree delta, meaning the plates are likely not going to be any more fouled than they are now. Checking the duty, I come up with ~900,000 BTU/hr (m*cp*delta T). The plate and frame has 65 plates at 1.5 sqft/plate, so back calculating for U gives (by rearranging Q = U * A * delta T):

(900,000 BTU/hr) / (20 F * 65 * 1.5 sqft) = 462 BTU/F sqft hr

We are looking at a new product that will require process water at 65 F and our chilled water is at ~55 F. Since the current approach (hot in/cold out) is 20 F, obviously (assuming constant U) we don't have enough area to force the approach. More likely would be a 75 F hot side out, assuming that the U is constant. Since the approach is half as much, the number of frames needed should be 130, correct? Also, we may be able to drive our chilled water down another 5 degrees, meaning we need a 15 degree apporoach. This would give 65 plates * 20 F / 15 F = 87 plates, assuming again that the U is constant even though we have reduced the flow rate. As a side note, this exchanger was originally designed with 95 plates, but the original plate design is no longer available, and we are now using a higher efficiency plate. I know the plate is higher efficiency because we are still meeting design spec with 65 plates. Here's where things get squirrely. I contact our local heat exchanger rep and give him the same numbers and ask him to check with Alfa Laval what we should do. He comes back with the current setup should be able to meet our needs without adding plates! I don't get it. Is my math off? I know I made some assumptions, but if anything I would have thought I would need MORE plates than I calculated. My current inclination is to run the chiller at 50 F and add 25 plates, for a total of 90 plates. Any thoughts?

Thanks,

Matt

Quality, quantity, cost. Pick two.

 

[Dougt115]

Your math looks sound but I wonder if he is using ideal numbers and not your test numbers. Theory and practice. Your 900,000 was for an existing system he may have used a new system without any fouling, so maybe off by 10% or more.

 

[Latexman]

"Checking the duty, I come up with ~900,000 BTU/hr (m*cp*delta T)." I got 756,000 BTU/hr. 168x500x1x9 = 756,000.

Sales Reps can make mistakes too. Ask him to prove it on paper using your data.

Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.

 

[ione]

MatthewL,

As pointed out by Latexman the duty of your HX is 756,000 Btu/hr.
Moreover have you used the logarithmic temperature difference in your calculations?

 

[MatthewL]

Ione,

I used 20 degree for the temp difference. The average is 20.5, and the delta T LM is 20.49. Even so, this is just for calculating U, which drops out when calculating the final needed area.

Latexman,

Thanks for the double check. The duty is pretty constant, since the process is an underwater pelletization. The polymer flow through the die face is fixed by the pump rate, and the die face temperature is not adjusted much between products. Thus, the constant delta/flow rate on the process side over the last year. I was only using the duty to get an idea of the U and to see if it makes sense (a bullsh** test). Even using the new duty, the result is 388 BTU/F sqft hr which still makes sense for a plate and frame. The important part is I feel that given the same duty, the approach can not get better than ~20 F with the number of plates we have, regardless of what the process conditions are. The process water is in a loop, with the pelletized material being separated by centrifuge and the system temp will come to a steady state fairly quickly once the process starts. That steady state will be the process water exiting will be ~20 F hotter than the cooling water entrance temperature.

Matt

Quality, quantity, cost. Pick two.

 

[MechAg06]

Matt

Since you are using a new water utility you may be able to make up some of the difference by increasing its flow over the current CW supply rate. But if the hotside Delta_T is fixed at 9, you won't be able to get all of the way there. At a 15F cold-end approach, you would need a 27F delta at the hot-end. But the difference between the two exceeds your current temperature change. Even if you could modulate the flow, any benefit caused by reducing the process flowrate to increase the temperature change would probably be offset by a reduction in U-value.