heat exchanger thermal analysis 1

[Nivrah]

Hi all,
I am investigating the analysis of an existing heat exchanger subjected to changed inlet conditions.Somehow its confusing me. So hopefully your input will help.
here is the current config:
water/water shell and tube HX
hot water in = 90 degrees C
hot water out = 40 degrees C
hot water flowrate = 50 kL/hr (13.61 kg/s)
cold water in = 20.4 degrees C
cold water out = 89.7 degrees C
cold water flowrate = 35 kL/hr (9.722 kg/s)

HX specs:
Area: 180 sq metres
LMTD = 4.6
U = 3.436 kW/m2/C
Duty = 2823 kW.
max theoretical heat transfer = Cmin*cp*dtmax = 9.722*4.185*(90-20.4)=2831.8 kW
Eff = 2823/2831.8 = 99.7%

The current configuration works fine and the thermal equations all apply, i.e, m*cp*dt = U*A*LMTD for both shell and tube side.

NOW, due to a change in cold water inlet conditions, the new cold water temperature at inlet is raised from 20.4 to 29 degrees C.Nothing else is changed!!!.

new max heat transfer = Cmin*cp*dtmax = 9.722*4.185*(90-29)=2481 kW
New actual heat transfer = 99.6% * 2481 kw = 2455 kW.
new hot water outlet temp = 90- 2455/(13.61*4.185) = 46.9 degrees C
New cold water outlet temp = 2455/(9.722*4.185) + 29 = 89.4 degrees C

So far Ok with me?

this is the confusing part:
Area is still at 180 sq metres
new LMTD = 5.09
U is still the same at 3.436 kW/m2/C.............??

When I check Q = U*A*LMTD = 3.436*180*5.09=3148 kW

3148 kW > 2455 kW.................how can this be possible?
does the "U" change to adjust to 2455 kW? it shoudnt change much right?

can someone help me solve this issue?
thanks

 

[TD2K]

For a shell and tube exchanger like this when I'm making reasonably small changes, I look to see if I can assume the U should be constant (more or less). For this system as a first pass I would feel okay with that assumption.

You have a flow rate of the cold water and the hot water, those remain constant. You also have the temperature of the hot water inlet (90C) and the cold water inlet (now 29C, used to be 20.4).

Q given up by the hot side has to equal the Q gained by the cold side if you ignore exchanger heat losses to the ambient surroundings which is usually a good assumption.

Rearranging: Q/dTlm = UA (we've assumed the UA to be constant).

For every degree the hot water cools, you know how much the cold water has to warm up based on the ratio of the flow rates and if the fluids aren't the same, heat capacities.

So, assume a Tout for the hot side (which gives you a dT for the hot side and a Q), calculate the cold side dT (13.61/9.722 * hot side dT) and calculate the Tout for the cold side. Calculate dTlm and with your Q, see if you have the same UA. Adjust Tout for the hot side till you have the same UA. It's pretty easy on a spreadsheet.

 

[Nivrah]

TD2K,
many thanks for your insight!
I did what you suggested in a spreadsheet.The new dtlm came out as 3.88 with Tout_coldwater = 89.8 degree C and Tout_hotwater = 46.59 degree C.

Discussion:
the dtlm was smaller than the one I calculated (3.88 v/s 5.09). I infer that given the changes in the cold side inlet temperature, the existing HX actually performs better since the cold side outlet temp becomes closer to the hot side inlet temperature.

So changes in the cold side inlet temperature (keeping everything else the same)alters the efficiency of the existing HX. So the reason why i got the fallacy in my first post was that I kept using the same efficiency for the changes conditions which I shoudn't have.

 

[TD2K]

The heat exchanger efficiency isn't really changing. The efficiency definition you use isn't one I'm familiar with.

The higher the duty exchanger for a constant U and A, the larger the dTlm has to be.

By increasing the cold water inlet temperature, you've reduced the driving force (dT) between the hot and cold fluids and you end up transferring less heat.

 

[rakuday]

Hi,

To start with, since you are cooling the hot water from 90C to 40C, I assume it to be the performance expected out of the cooler irrespective of the changes in the cold side. Basis this, if the cold water inlet temperature is increased, either you will have an increased water temperature rise or increase your cooling water flow rate to maintain the same energy balance.

I did a quick check, fixing the hot water outlet at 40C. With this you are already at the highest water temp rise and for the changed inlet condition, the cold water outlet is beyond 90C (99C for 9.722kg/s flow rate). So the only other option is to increase your cold water flow rate to match the cooling needs. This comes to around 11.21kg/sec (appr) with the cold water outlet at 89.7C.

Thanks.

 

[Nivrah]

the efficiency i used here is the ratio of the actual heat transfer/ max theoretical heat transfer possible.
max theoretical heat transfer possible = m_dot(min)*cp*(Th_in-Tc_in)....

By increasing the cold water inlet temperature, the actual heat transfer gets closer to the maximum theoretical heat transfer possible for a constant UA. So it appears as if its getting closer to an ideal situation.

 

[Nivrah]

To rakuday,
i agree with your analysis but the point I was looking at was to see how the outlet temperature changes with an increase in cold side inlet temperature for a constant UA.

 

[MrBTU]

did you say that this shell and tube heat exchanger works fine - with a .3 deg C approach temp (cold water out to hot water in)?
I'd like to see that. Please provide details of the unit - tube size, shell size, pressure drop, etc. We don't quote anything closer than a 5 deg F approach, and as approach temp approaches zero, area required approaches infinity......

 

[IRstuff]

You're doing the math incorrectly. You must assume that the power has not changed. This means that your increase in cooling water temperature must be compensated for by an increase in mass flow rate and/or an increase in outlet temperature, i.e.,

4.185 (J/kg*K) * (89.7K-20.4K) = 290 W*s/kg compared to 4.185 (J/kg*K) * (89.7K-20.4K) = 209.25 W*s/kg. This ratio is reflected in the difference in flow rates. Given a constant power, this results in mass flow rates stipulated. In order to support the same amount of heat transferred, your coolant flow must increase to 40kl/hr to compensate, which is consistent with the deltaT changing from about 70°C to ~60°C, i.e., 1/7th drop in deltaT requires 1/7th increase in flow rate.



TTFN
faq731-376

 

[Nivrah]

Mr BTU,
this HX was part of a mass & heat balance by an external company. they quoted 0.3 deg C approach. So I continud on that study but just changed the inlet conditions.

 

[zekeman]

"...4.185 (J/kg*K) * (89.7K-20.4K) = 209.25 W*s/kg. .."

Where did this come from??

If you assume constant U ( though it will actually increase somewhat due to the anticipated increase in mass flow rate for the cold water), and the hot water mass flow rate and temperatures in and out maintained at 90 and 40 degree, then the LMTD has to remain constant at
4.6177. Therefore, you can write the LMTD equation
4.6177=[(40-29)-(90-x)]/ln((40-29)/(90-x))
I get x, the cold water outlet temp= approximately 89 degrees and a delta T of 89-29=60 deg requiring an increase in cold water flow rate of 0.15 comparing favorably with IRstuff's 1/7.
You could then iterate for a new increased U which will reduce the required LMTD and repeat the process to get a new reduced value of cold water flow rate.

 

[25362]

IMHO, it is economically viable to achieve an approach temperature of no less than 5^oC in counter-current or single pass shell and tube heat exchangers.

To achieve a closer approach performance, or a temperature crossing, a plate or a double-pipe counter-current exchanger should be specified.

 

[zekeman]

"IMHO, it is economically viable to achieve an approach temperature of no less than 5oC in counter-current or single pass shell and tube heat exchangers."

You may be right, but the OP has an EXISTING design that he is asking for help.

 

[25362]

Zekeman, you are right. I tried to show my doubts about such a low temperature approach in a shell and tube unit, the details of which are still unknown.

IMHO your positive approach to the
solution of the problem needs to add a LMTD correction factor depending on the number of shell passes the HE has.

There are a series of graphs in books, such as HOLMAN's Heat Transfer, that give the LMTD correction factors F as in the general formula for heat exchangers.

Q = UAF(LMTD)​

Anyway, when using LMTD's there are correction factors based on P = (t₂-t₁)/(T₁-t₁)and R = (T₁-T₂)/(t₂-t₁).

Small letters refer to the cool fluid.
Rising t₁ from 20.4 to 29^oC increases P, therefore getting a smaller F.

Being this so, and assuming Q and
UA constant, it would be a matter of carrying out iterations until the new equilibrium is found. Let Nivrah tell us what type of heat exchanger are we speaking about.

 

[zekeman]

25362,

Thanks for the response.

Point well taken.

 

[25362]

Nivrah,

The conventional formula: Q = UA(LMTD) is not correct for S&T heat exchangers, it should better be written: Q = UAΔT_m. Where:

ΔT_m = F×LMTD​

F = 1 is the result of taking several assumptions that are not necessarily true for S&T units, especially since these units deviate from a true countercurrent setup. Resulting in:

Q = UAF(LMTD)​

A value of F = 0.8 is quite common and would be reasonable in this case by the envisaged change in temperatures.

There are graphs to estimate F.
A visit to Google asking for LMTD correction factors would provide you with an answer.

 

[Nivrah]

Ok gents,
the opening thread was from a study done by a consulting company. I was given specs of the HX such as area, U, so I cannot confirm that it is a shell-tube or plate heat exchanger.i take it from the study that such a close approach must imply a plate HX.
What I then did is that I changed inlet temperatures too see how the outlet temps will change given the same UA.

After working it out in my own time using TD2K's idea, I agree with 25362 in that a new equilibrium must be found.

 

[25362]

Plate exchangers also have LMTD correction factors, although F→ 1.0, mainly due to the fact that the fluid flowing through the end channels only passes heat through one of the channel walls while in the core channels heat is transferred through both walls, resulting in a lower heat transfer for the end channels than for the bulk of the exchanger.