LMTD in the presence of noncondensables.

[patrickraj]

Dear friends,

How to go about calculation of LMTD(corrected) for STHE in the presence of noncondensables.I have no problem without the presence of noncondensables.

Thanks in advance to all.

 

[tickle]

LMTD does not change for the presence of non condensables.

(T1-t2)-(T2-t1)/ln(T1-t2)/(T2-t1)

1 = in, 2 = out, T = shell side, t = tubside

If you have isothermal condensation then

LMTD = t1-t2/ln(T-t2)/(T-t1)

assuming that condensation occurs on the shell side.

What may change is the outlet temperature that you need to achieve to condense all of the condensables. The best way to design the HE is to do it in stages and add up all of the required areas.

 

[quark]

It is very difficult to estimate heat transfer in the presence of noncondensable gases. For example, a layer of 0.04 inches thick air offers same resistance to flow of heat as a layer of 1 inch thick water, iron of 4.3 feet thick. (I dare not to hint on copper)

It is always better to take care of them rather than having redundant designs.

Regards,

 

[phex]

just a shot in the dark:

can you estimate the amount of noncondensibles in your hx at a time? one could try to get to the point by reducing the available area for heat transfer. i'd try to guess the amount, calculate volume, then look at the geometry of hx. as gravity tends to effect that liquid is collecting below the nc's, you could guess the part of heat transfer area still submerged in liquids. this would give you a better guess on your hx.

the non-condensibles have an effect on the heat transfer coefficient, like quark has said, not on LMTD or LMTD(c). better to cut open the hx and calculate the heat transfer for liquid submerged area separate from the heat transfer for the area in contact with nc's.

best would be to calculate the hx with a software able to cope with two phase heat transfer and nc's.

hth,
chris

 

[DickRussell]

As suggested by others, the heat transfer coefficient will vary (get worse, mainly) as more stuff is condensed and the remaining vapor gets progressively more concentrated in noncondensables. And, yes, calculation of that part is best left to a good HX program.

Still, if it is just the weighted MTD you want, for comparing different scenarios and the feasibility of something different that you want, the weighted MTD for an exchanger can be done by breaking up the exchanger into equal duty increments and doing a series of isenthalpic flash calculations for both sides, starting at one end of the exchanger, to get the corresponding hot and cold temperatures. The number of sections needed depends on the extent of curvature of the condensing curve, but you won't gain much by using more than 8-10 sections because of uncertainties in heat transfer coefficients. The weighted MTD is then #sections/sum(1/LMTD), where the LMTD is calculated for each of the sections. Don't try this on your wristwatch calculator and stylus; use a decent program, because you're talking multicomponent VLE, isenthalpic flash calculations, etc.

Dick Russell

 

[Niq]

LMTD calculations for vapors + noncondensable mixtures are not a big deal but determinations of heat co-efficients cetainly are.
You are recommended to read Kern's Process Heat Transfer, Chapter 13, section 4 for indepth study. Designing of such heat exchanger is truely an interesting assignment.

Regards

 

[patrickraj]

Dear members,
Thanks.Please advise me how the LMTD(C)arrived by the vendor in the following case.

Dry air flow 61600.0 KG/hr
Moisture with air 3174.0 KG/hr
Water outlet 1660.0 KG/hr

Air inlet temp :153.3 Deg C
Air outlet temp :42.o Deg C
CW inlet temp :34.5 Deg C
CW outlet temp :43.5 Deg C

Data sheet LMTD(C):16.2Deg C
My calculated value LMTD(C):30.14Deg C
I have no problem even with Temperature cross condition.

 

[25362]

Dick Russell's approach is the practice used for cases like this one.

 

[DickRussell]

It seems there may be some room for interpretation of the vendor's data sheet, depending on how it labels the MTD. It can report what is basically a weighted MTD, which provides an equivalent MTD that takes into account curvature of the heat release Q vs. T curve. Or, it can be the LMTD as corrected for the exchanger having U-tubes in one or more shells, which takes into account the fact that both inlet and outlet ends of the tube side fluid are in contact with shell side fluid on that end of the exchanger and derates the straight-line LMTD accordingly. Or what is reported for MTD can be both weighted and corrected. In the end, the basic heat transfer equation Q=U*A*MTD must be using the appropriatly average U and MTD, and both of these can vary substantially.

It's OK to give it your best shot for roughing out the feasibility of using an exchanger for something else, but for final design or verification you really ought to have someone well-versed in use of commercial exchanger rating software and heat transfer equipment design put some time on it.

Dick Russell

 

[DickRussell]

I got a few minutes to run some quick calcs on that air/water cooling problem. You didn't give pressure, which of course affects how much water is condensed at 42 C, so I varied the inlet pressure until the water condensed matched your number. Pressure drop didn't have a strong affect; a different value just changed the inlet pressure needed to give the outlet pressure that matched the water. I had it do the weighted MTD calculation; 8 sections is enough, and increasing this to 10 or 12 only dropped the weighted MTD by 0.16 C. Below is the result for the exchanger, as excerpted from the PD-PLUS simulator output. I hope the formatting doesn't get too badly garbled due to word wrap. Let me know if you want the file emailed to you as an attachment (provide an address). As you can see (hopefully), the weighted MTD is 22 C, but multiplying that by the correction factor for one shell (assuming U-tubes) gives 17.5 C, which is close to the vendor's 16.2. The weighted MTD calcs assume mixed air/water side at any point, as though it were happening in the tube side. More likely, the cooling water would be on the tube side, and condensed water would fall off the tubes and flow along the bottom. A different strategy for weighting the MTD over multiple sections of the shell could account for the small difference. What is important to note is that half the duty occurs in just cooling the inlet air to the dew point, a steep part of the Q-T curve, and condensation flattens out the curve somewhat. The Q-T curve is concave-up, so that the weighted MTD is considerably less than for a straight-line heat release.

Hope this helps. Now the output (hope you can rearrange the lines).

HEAT EXCHANGER BLOCK TT-101 Wet air cooler, TT-101
DUTY= 2.7593 MM KCAL/HR
HOT SIDE INLET TEMP= 153.3 C, L/F=0.0000, FLOW= 64774.00 KG/HR
VAPOR DENSITY= 1.716 KG/M3
OUTLET TEMP= 42.0 C, L/F=0.0400
WT FRACTION VAPOR= 0.9744
VAPOR DENSITY= 2.356 KG/M3
LIQUID DENSITY= 991.83 KG/M3
COLD SIDE INLET TEMP= 34.5 C, L/F=1.0000, FLOW= 307460.63 KG/HR
LIQUID DENSITY= 994.50 KG/M3
OUTLET TEMP= 43.5 C, L/F=1.0000
LIQUID DENSITY= 991.26 KG/M3
MTD CORRECTION FACTOR (OVERALL, FOR EVEN TUBE PASS SHELLS):
0.791 FOR 1 SHELL
0.968 FOR 2 SHELLS
0.987 FOR 3 SHELLS

PROFILES (DEG C, FOR EQUAL DUTY SECTIONS):
HOT TEMP 153.3 132.2 111.0 89.8 68.5 54.7 51.1 46.9 42.0
L/F 0.0000 0.0000 0.0000 0.0000 0.0000 0.0051 0.0173 0.0290 0.0400
COLD TEMP 43.5 42.4 41.2 40.1 39.0 37.9 36.7 35.6 34.5
L/F 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
TEMP DIFF 109.8 89.8 69.7 49.6 29.5 16.8 14.3 11.3 7.5
WEIGHTED MTD= 22.2 C (UNCORRECTED)
REQUIRED U*A= 0.1244 MM KCAL/HR-C, BASED ON UNCORRECTED MTD

 

[patrickraj]

Dear DickRussell,

Thanks for your efforts.
Operating pressure is 2.45Ksc(a)
Calculated press. Drop is 0.13ksc
Stright tubes.You are correct,CW in Tube sides only.
Tube-SIX Passes,Shell-Single pass
Please email the spreadshet as an atttachment(Mail ID: patrickraj2000@yahoo.co.in). I shall be of very much grateful to you.