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Dear All,
I want to know something about Kern's method of determining heat exchanger performance. My question is regarding Log Mean Temp Diff. I want to know why is it always necessary to calculate LMTD in Fahrenheit. I tried it with Centigrade and then converted it to Fahrenheit, I got a shocking difference. Can anyone whrow some light on the significance of English units for this method and what if we want to use SI units?
Regards,
Naveed.
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L M T D !
Started by Guest_Naveed !_*, Nov 07 2005 10:30 AM
3 replies to this topic
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#2
Guest_Guest_*
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Posted 07 November 2005 - 02:08 PM
Naveed,
What is the equation you used to convert from C to F?
Remember that LMTD is a temperature DIFFERENCE. Therefore you should not add 32F to compensate for the zero offset in the two scales.
This means that a difference of 10C is 18F NOT 50F.
That would be my first guess for your problem. If that is not your problem, you need to have a good look at all the units in your equations. Heat transfer co-efficinents and constants all have units that need to match what you are using. Fouling factors have units associated with them.
What is the equation you used to convert from C to F?
Remember that LMTD is a temperature DIFFERENCE. Therefore you should not add 32F to compensate for the zero offset in the two scales.
This means that a difference of 10C is 18F NOT 50F.
That would be my first guess for your problem. If that is not your problem, you need to have a good look at all the units in your equations. Heat transfer co-efficinents and constants all have units that need to match what you are using. Fouling factors have units associated with them.
#3
Art Montemayor
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Posted 07 November 2005 - 04:05 PM
Naveed:
Donald Q. Kern’s method(s) for determining heat exchanger performance is not based on solely using US customary engineering units instead of metric or SI units. The Log Mean Temperature Difference LMTD) is not a Kern invention. Don may have made its use and definition more well-known through his famous and pioneering book, “Process Heat Transfer” (1950), but it was already known in engineering circles before he wrote the book. The derivation and employment of the LMTD equation:
LMTD = (T2 - t1) - (T1 - t2) / ln [(T2 - t1) / (T1 - t2)
where,
T1 = Inlet temperature of hot shellside fluid.
T2 = Outlet temperature of hot shellside fluid.
t1 = Inlet temperature of cold tubeside fluid.
t2 = Outlet temperature of cold tubeside fluid.
is one where the units for temperature can be Fahrenheit or Celsius. It doesn’t make any difference as long as you are consistent in the system you work under : US customary engineering units or metric / SI units.
Why do you assert that it is “necessary to calculate LMTD in Fahrenheit”? Were you taught that or is it something you deduced? Whatever led you to believe that, was not correct. There is a Celsius LMTD just as there is a Fahrenheit LMTD – but you must be consistent in your units and system employed.
There is a factor of 1.8 that must be incorporated when converting Celsius temperature differences to Fahrenheit temperature differences. Maybe that is what is troubling you. Perhaps if you tell us the temperatures you were dealing with and the answers you got, we could help out.
Donald Q. Kern’s method(s) for determining heat exchanger performance is not based on solely using US customary engineering units instead of metric or SI units. The Log Mean Temperature Difference LMTD) is not a Kern invention. Don may have made its use and definition more well-known through his famous and pioneering book, “Process Heat Transfer” (1950), but it was already known in engineering circles before he wrote the book. The derivation and employment of the LMTD equation:
LMTD = (T2 - t1) - (T1 - t2) / ln [(T2 - t1) / (T1 - t2)
where,
T1 = Inlet temperature of hot shellside fluid.
T2 = Outlet temperature of hot shellside fluid.
t1 = Inlet temperature of cold tubeside fluid.
t2 = Outlet temperature of cold tubeside fluid.
is one where the units for temperature can be Fahrenheit or Celsius. It doesn’t make any difference as long as you are consistent in the system you work under : US customary engineering units or metric / SI units.
Why do you assert that it is “necessary to calculate LMTD in Fahrenheit”? Were you taught that or is it something you deduced? Whatever led you to believe that, was not correct. There is a Celsius LMTD just as there is a Fahrenheit LMTD – but you must be consistent in your units and system employed.
There is a factor of 1.8 that must be incorporated when converting Celsius temperature differences to Fahrenheit temperature differences. Maybe that is what is troubling you. Perhaps if you tell us the temperatures you were dealing with and the answers you got, we could help out.
#4
Guest_Naveed._*
Guest_Naveed._*
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Posted 08 November 2005 - 03:39 PM
Thanks all of you. I think that will solve my problem. I will do a few examples and then see if anything different comes up. Thankyou all.
Regards,
Naveed.
Regards,
Naveed.
