8 replies to this topic

[Sylvia]

Can heat from reactor effluent be directly exchanged to heat up reactor inlet? The amount of heat produced is not sufficient to produce steam, however not utilising the excess heat will be a great waste. In cases like this, what should be done?

Thanks

Regards,
Sylvia

[aliadnan]

Hi

I think you can use a heat exchanger before the reactor inlet in which the heat from reactor effluent can be recovered and the reactor inlet stream can be heated upto the desired reaction temperauture.
If reactor inlet temperature need not to be raised, you can exchange the effluent heat to raise the temperature of another stream in the plant. But all of this depends upon the economics. You can use the pinch technology.

Please correct me if I am wrong.
Hope that helps.

Regards,
Ali

[Sylvia]

Good day Ali,

Yes you are right, i have conducted pinch technology and paired up the reactor inlet and reactor effluent. my concern here is does it promote instability in the process in case of any runaway in the reactions, causing the plant to shut down? The safer way would be to recover heat from the effluent to generate steam, and then use the steam to heat up the reactor inlet (indirect exchange). However steam produced is not sufficient to heat up reactor inlet as the steam temp produced is at 55degC but my exchanger outlet is supposed to be heated up to 100degC from 25degC and further heated to 400degC using the fired heater prior to entering the reactor.

Thanks.

Regards,
Sylvia

[abhi_agrawa]

Sylvia,
You are right, this exchange can promote instability in case of a runaway. However, in your case, if I understand correctly, the reactor feed goes from 25 deg C to 100 deg C, it is then heated in the Fired Heater to 400 deg C and then enters the reactor. Now, if there is a sudden increase in the reactor outlet temperature, then this can be taken care of in the fored heater by reducing the firing. In some of the plant that I have seen, when there is no further heating after the reactor feed-effluent heat exchanger, a bypass to this exchanger is put and in case of temperature excursion, the bypass is opened.
Hope this helps,
-abhishek

[Sylvia]

In order to calculate the area for the heat exchanger, i need to calculate the log mean temp difference. What is the formula if both my shell and tube side involve phase change? Shell side is from 400degC (pure gas phase) to 43.55degC (vapor fraction = 0.54) while my tube side inlet is 25.06 degC (pure liquid) to 235.06 degC (pure gas phase)

The common LMTD calculation that i know of is ((T1-t2)-(T2-t1))/ln((T1-t2)/(T2-t1))
where
T1 = shell inlet temp
T2 = shell outlet temp
t1 = tube inlet temp
t2 = tube outlet temp

However this equation applies ONLY where there is NO change in specific heats and the overall heat transfer coefficient is constant and no heat losses. Clearly in my case, i have to segment my temp changes and calculate the area in sections....but how????

Regards,
Sylvia

[aliadnan]

Hi

What your dealing with is a two phase Heat Exchanger Design, for these types we divide the tube side and shell side (since in your case both the sides have phase change) into 2 zones. One zone accounts for the sensible heat transfer (Preheating or Subcooling) and the other zone accounts for latent heat transfer (vaporization or condensation). Please refer to "Process Heat Transfer" by Kern. Check Example 15.2 of this book and it may help you.

Hope that helps.

Regards
Ali

[Sameer]

Can heat from reactor effluent be directly exchanged to heat up reactor inlet? The amount of heat produced is not sufficient to produce steam, however not utilising the excess heat will be a great waste. In cases like this, what should be done?

Thanks

Regards,
Sylvia

Hi Sylvia,

As you are Quoting that your reactor Effluent is to be used to heat up reactor Inlet , I have seen such an arrangement with Pressure Differential controller (Bypass) and in Normal Circumstances Reactor Effluent Exchanges Heat with Reactor Inlet .
Hope that Helps.
Regards.

[Art Montemayor]

Sylvia:

You have already done a lot in getting close to resolving your answer. Yes, the LMTD equation is not applicable where there is a phase change (the fluid's heat capacity is not constant). However, as Ali has recommended there is another way to skin a cat: divide the two services, treating them separately.

I also recommend you use Don Kern's classic as a source not only of basic process heat transfer, but also for its many worked examples of the practical and real applications found in industry.

I'm attaching a workbook with two of his examples all worked out in spreadsheet fashion in order to allow you to see how he calculates the answers. I highly recommend any student reading this to download it and see the manner that the resolution(s) is reached. This is what your professors are expecting of you when you make project or homework calculations. Excel now permits you to generate this type of presentation - detailed, legible, and documented. This is the quality of rigorous logic applied to heat transfer problems that I would expect to receive if I were teaching the course.

I hope this helps you out in reaching your resolution.

Art Montemayor  Kern__s_Propanol_Condensers.xls   123.5KB   138 downloads