Dear friends,
I am going to do a hand calculation that is like example 10.1 in GPSA. But I encountered a big problem! Please open GPSA book, Chapter 10(air fan cooler design).
There is an example10.1 "A liquid HC stream is going to be cooled from 121º C to 66ºC by air fan cooler"
Every thing in clear but there is some thing strange! In Chapter 10 page9 for calculation required area of air fan cooler in the first step overall heat transfer coefficient was determined through table10-10 that is Ok.
Then in the second stage air temperature rise was calculated through a formula that is unknown. As you know GPSA is a user friendly book. Actually I do not know from what equation air temperature rise was calculated?
I mean Δt (air).
T1=121º C
T2=66ºC
t (inlet air)=38º C
t(outlet air from air fan cooler)=unknown
A=? Unknown
Q=Known
How can I calculate t2 (outlet air from air fan cooler)
Please see chapter 10 page9 (air fan cooler), in step2 how air temperature rise was calculated? (In according what formula)?
Does anybody know it?!!!!!!!
I an eagerly waiting for your answer
Thank you very much
Rosa
Cheers!
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Air Temperature Rise Calculation In Air Fan Cooler
Started by rosa, Aug 01 2008 01:03 AM
5 replies to this topic
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#1
Posted 01 August 2008 - 01:03 AM
#2
Posted 01 August 2008 - 08:12 AM
Rosa:
I deleted your other, identical posting in the Hydrocarbons, Refining Forum because you are not supposed to be multi-posting according to Forum Guidelines. Please do not multi-post the same thread.
I don't know what edition of the GPSA and if you are referring to the hard cover copy or the electronic edition. Therefore, I don't have the exact problem in front of me. However, the problem you describe is fairly easy to visualize. If you want to know the outlet temperature of the air in an air-cooled exchanger, you MUST HAVE the quantity of air being administered to the same air-cooled unit. The way it is normally done is that we know the air capacity of the fans at the running speed and their total head. This is either given to you by the fabricator or it is found in the air-cooled unit's data sheet or performance curves. Either way, you must know the air quantity going through the exchanger. With that data known and the amount of heat transferred to the air also known, then you apply the heat balance:
Q = w Cp (T2 - T1)
where,
Q = heat transferred to the air, Btu/hr
w = air flow, in lb/hr
Cp = air specific heat, Btu/hr-lb-oF
T2, T1 = air temperatures out and into the unit, oF
I am sure - from what you describe - that the above is what is done by the GPSA example. The only other way I know is to apply the enthalpy of air to the air heat balance and that is too complicated. The above is the way I normally do it.
I deleted your other, identical posting in the Hydrocarbons, Refining Forum because you are not supposed to be multi-posting according to Forum Guidelines. Please do not multi-post the same thread.
I don't know what edition of the GPSA and if you are referring to the hard cover copy or the electronic edition. Therefore, I don't have the exact problem in front of me. However, the problem you describe is fairly easy to visualize. If you want to know the outlet temperature of the air in an air-cooled exchanger, you MUST HAVE the quantity of air being administered to the same air-cooled unit. The way it is normally done is that we know the air capacity of the fans at the running speed and their total head. This is either given to you by the fabricator or it is found in the air-cooled unit's data sheet or performance curves. Either way, you must know the air quantity going through the exchanger. With that data known and the amount of heat transferred to the air also known, then you apply the heat balance:
Q = w Cp (T2 - T1)
where,
Q = heat transferred to the air, Btu/hr
w = air flow, in lb/hr
Cp = air specific heat, Btu/hr-lb-oF
T2, T1 = air temperatures out and into the unit, oF
I am sure - from what you describe - that the above is what is done by the GPSA example. The only other way I know is to apply the enthalpy of air to the air heat balance and that is too complicated. The above is the way I normally do it.
#3
Posted 01 August 2008 - 11:15 PM
Dear Mr.Art Montemayor;
I asked in different forum to be sure to give my answer.Yes you ar right.
would you please see page2 of uploaded word file that is highlited with red color.I really want to know
what is the basic source of red highlited formula?
many thanks
rosa!
I asked in different forum to be sure to give my answer.Yes you ar right.
would you please see page2 of uploaded word file that is highlited with red color.I really want to know
what is the basic source of red highlited formula?
many thanks
rosa!
Attached Files
#4
Posted 02 August 2008 - 02:38 AM
Dear Art
I think you really mean: w = air flow and Q = Heat rate.
Regards
I think you really mean: w = air flow and Q = Heat rate.
Regards
#5
Posted 02 August 2008 - 09:38 AM
Fallah:
Thank you very much for pointing out a bad product of my hasty writing. In trying to answer the query before departing for the day, I left out the definition of Q and, as you state, erroneously labled the value of w. I have now corrected the original entry.
Thanks for your keen and strict observation.
Thank you very much for pointing out a bad product of my hasty writing. In trying to answer the query before departing for the day, I left out the definition of Q and, as you state, erroneously labled the value of w. I have now corrected the original entry.
Thanks for your keen and strict observation.
#6
Posted 23 August 2008 - 11:27 PM
Hi rosa
I taked a look on ur doc file, and I guess that the highlited formula come from Qsys=Qsur,
because there are both the temperature differences of 'inlet and outlet' and 'air and surround'.
chenblue
I taked a look on ur doc file, and I guess that the highlited formula come from Qsys=Qsur,
because there are both the temperature differences of 'inlet and outlet' and 'air and surround'.
chenblue
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