16 replies to this topic

[kumars876]

hai to all ,
iam new to htri 6 can any one please do the below feed water heater by htri 6 process and forward me the files
please do this favour by helping me
iam struggling with this problem from 2 weeks

data given
Inlet steam flow rate =5.37 t/hr
Inlet steam flow temperature=359°C
Outlet steam flow temperature=216°C
Shell side mean temperature = 287.5°C
Cooling water flow rate =113.34t/hr
Cooling water inlet flow temperature=175°C
Cooling water outlet flow temperature=200°C
Tube side mean temperature = 187.5°C
shell side steam
tube side water


please any one forward the htri file after doing the problem(if possible send me the hysys file)

my email kumars876@gmail.com

Edited by kumars876, 11 October 2011 - 08:25 PM.

[pavanayi]

Kumar,
What is it that you are struggling with? Is it the software? or is it the concept?
Have you done a material balance?

[kumars876]

iam struggling with the software...
can you please do the problem with the htri software and send me the file...
thanks...

[S.AHMAD]

What is the steam pressure?

[kumars876]

first thank q for ur reply
the steam pressure is at 2.144 mpa
and i don't know the tube side pressure

Edited by kumars876, 13 October 2011 - 12:08 AM.

[pavanayi]

Kumar,
I will not do the problem for you. Neither will anyone else in this forum. Neither will anyone outside the forum do 'a favour' and send you the file.

Being a student, you have been given this problem so that you get trained in the software yourself, so that in future you will be able to use it in real world situations. Do you expect someone to do it for you then too?

If you are not comfortable with the software, either try and get some tutorials on it to try and learn yourself, or ask someone in your department who is more experienced, to guide you through the process.

If you are finding the concept of the question at hand difficult to understand, then someone here might be able to guide you.
For example, if you do not know the tube side pressure, take a reasonable assumption.
There will be people here who will guide you as to what is a 'reasonable assumption', but still it will stop there.

If you have started to solve the problem, then explain what work you have put in till now. Upload any supporting documents and clearly state where is it that you are stuck at. If you are unable to understand any input into the software, state that too.

Then you might expect to get some reasonable answers.

Hope it helps.

[kkala]

Condensate temperature out is 216 oC, corresponding to saturated steam pressure of 21.47 Bara (~ 21.5 Bara), or 2.147 MPa abs. This is practically same as 2.144 MPa (abs). In the majority of cases steam frictional pressure loss along the exchanger can be considered negligible, so ingoing steam pressure (359 oK) can be also considered as 21.5 Bara at exchanger inlet, downstream of the control valve, if any. This means 143 oC superheat in ingoing steam (rather high, yet not impossible).
Inlet - outlet pressure of cooling water will not affect heat transfer (only its mass flow rate will). We used to put an arbitrary inlet pressure, if the program asked for it, checking ΔPs and inlet - outlet operating pressures of the whole line at the end.

Edited by kkala, 13 October 2011 - 03:59 AM.

[S.AHMAD]

1. I suggest that you begin with manual calculation

2. Once you understand the calculations, then you use HTRI.

3. Compare the outcome between manual and HTRI

4. Post specific problem encounter and attached the result from HTRI for review by members

Edited by S.AHMAD, 13 October 2011 - 03:33 AM.

[kumars876]

i have done the problem in excel file but when i upload it is saying error 407

Edited by kumars876, 13 October 2011 - 04:52 AM.

[kkala]

If my understanding is correct, there must be some inconsistency in exchanger heat balance (indicated in the attached "excbal.xls"). Probably it is good to begin by checking exchangers data (e.g. inlet steam condition), to avoid potential "diagnostics" when the software is run.
Note: Shell side mean temperature is close to that of saturated steam (216 oC), if it has some importance.

Attached Files

•  excbal.xls   22.5KB   165 downloads

Edited by kkala, 13 October 2011 - 06:11 AM.

[kkala]

If my understanding is correct, there must be some inconsistency in exchanger heat balance (indicated in the attached "excbal.xls"). Probably it is good to begin by checking exchangers data (e.g. inlet steam condition), to avoid potential "diagnostics" when the software is run.
Note: Shell side mean temperature is close to that of saturated steam (216 oC), if it has some importance.

kumars876, please explain reason of negative feedback on the above post.

[kumars876]

sorry for the delay sir,
i have made the manual calculation by excel file can you tell is it the right process
thanks

Attached Files

•  final hp heater by kerns method.xlsx   84.06KB   152 downloads

[kkala]

A. Comments on final hp heater by kerns method.xlsx calculations.
A1. Parameter table: Condensate enthalpy=925 kJ/kg, pressure same as of steam. Cooling water temp out = 200 oC, operating pressure does not affect heat transfer. See posts of 13 Oct by kkala.
Nevertheless above issues have been corrected in "My Heat Exchanger" sheet. Just make a heat balance (see excbal.xls of 13 Oct) to check / revise data, so that heat given by steam = heat taken by water (not 3326 kW versus 3495 kW, difference not being a matter of heat losses). If others have given the data, ask for clarifications.
A2. "My heater exchanger" sheet : Steam of 359 oC is first desuperheated (giving 355 kJ/kg), then condensed (giving 2230 kJ/kg). For shell side whole heat transfer, you can assume condensation temperature (216 oC) and heat transfer coefficient (U) of condensing steam. Superheated steam has lower U and higher temperature, more or less offsetting each other. This concept has been applied, may be known to others, is most probably reported in Mc Adam's "Heat transmission" (McGraw-Hill, 1954). Accordingly LMTD = [(216-175)-(216-200)]/ln(41/16) oC = 26.6 oC (instead of 87.1 oC taken). Only if steam were not condensing (no phase change) could LMTD have been 87.1 oC. And "Shell side mean temperature is close to that of saturated steam (216 oC), if it has some importance".
Separating the exchanger into two parts, one to desuperheat steam - rest to condense it, would give more precise results, especially in this case of high steam superheat. But calculation would be much more complex (how to elaborate it on a real exchanger?). Let the software do it (if this is possible) to see the difference.
Other comments on the sheet:
-Check tube side coefficient (water) of 20004 W/m2 oK, usually*~4400 W/m2 oK.
-Check shell side coefficient (steam) of 1084 W/m2 oK, usually*~9500 W/m2 oK (no air in steam). Calculation might not be necessary for condensing steam, see http://www.cheresour...h__1#entry38266, post 19 Apr 2010.
-Consider average U for fouling, not the highest.
-In theory new trials have to be made, until assumed overall U ~ calculated overall U, or assumed area ~ calculated area.
(*) From C. R. Branan "Pocket Guide to Chemical Engineering", Gulf, 1999.
B. Please also explain reason of negative feed back on kkala's second post in 13 Oct 2011, as requested.

Edited by kkala, 15 October 2011 - 11:20 AM.

[kumars876]

thank q for the reply sir,
i have done another problem same but with the different temperatures i just confused with the physical properties to take can any one suggest me to how to take the physical properties..for steam and tell is the water physical properties i have taken are correct...
thanq

Attached Files

•  HEAT EXCHANGER.xlsx   97.05KB   122 downloads

[wangzhaocheng]

Kumar,
I will not do the problem for you. Neither will anyone else in this forum. Neither will anyone outside the forum do 'a favour' and send you the file.

Being a student, you have been given this problem so that you get trained in the software yourself, so that in future you will be able to use it in real world situations. Do you expect someone to do it for you then too?

If you are not comfortable with the software, either try and get some tutorials on it to try and learn yourself, or ask someone in your department who is more experienced, to guide you through the process.

If you are finding the concept of the question at hand difficult to understand, then someone here might be able to guide you.
For example, if you do not know the tube side pressure, take a reasonable assumption.
There will be people here who will guide you as to what is a 'reasonable assumption', but still it will stop there.

If you have started to solve the problem, then explain what work you have put in till now. Upload any supporting documents and clearly state where is it that you are stuck at. If you are unable to understand any input into the software, state that too.

Then you might expect to get some reasonable answers.

Hope it helps.

You are so good.

[kumars876]

I have done the problem with the different inputs with the same fluids by Exchanger Design and Rating V7.1 i got the TEMA sheet is this sheet correct
given data...
data given
shell side steam
tube side water
Inlet steam flow rate = 1.4056 kg/sec
Inlet steam flow temperature=252.8°C
Outlet steam flow temperature=129.9°C
inlet steam pressure 8.669 bar
Cooling water flow rate =16.667 kg/sec
Cooling water inlet flow temperature=121.9°C
Cooling water outlet flow temperature=169°C
Cooling water pressure 82.5 bar
and my next doubt is how to increase the performance of the heat exchanger

Attached Files

•  tema sheet.bmp   1.42MB   82 downloads
•  performance.bmp   1.59MB   63 downloads
•  sheet.bmp   1.07MB   52 downloads