9 replies to this topic

[koby.geo]

What happens if the heat loads from shell side and tube side differ 10%. For example shellside:1000000 kcal/hr, tubeside:900000 kcal/hr

[manish mahale]

I it your process requirement?
Heat load for Shell & Tube exchanger is function of process fluid properties, operating contions & exchanger geometry. System will not provide varying loads on tube & shell side. It will get balanced itself according to parameters mentioned above.
Correct me if you have another view.
thank you
manish mahale

[JEBradley]

Are you saying that one side needs to heat up by 1E+06 kcal/h and the other needs cooling by 9E+05 kcal/h ???!!??

[Zauberberg]

Koby,

As I understood, this is a simply problem to solve. You need 9E+05 kcal/h in order to cool your hot fluid from starting temperature down to desired temperature, and 1E+06 kcal/h to heat your cold fluid from initial temperature up to the target temperature? And you want to combine these streams in a heat exchanger?

[process equipment]

i often get some deviation ( 10 to 40 % ) on the heat load as a result if i fix the temp. and heat load. Then i leave the outlet temperature of the cold side blanc.
If the heat load is correct the temp. will be calculated.

Is this a good approach ?

[joerd]

I don't understand your question. In a heat exchanger, there is only a limited number of variables, and the energy balance equations. Fundamental mathematics tells you that your degrees of freedom (i.e. the number of variables that you can (must) fix, is equal to the number of variables minus the number of equations.
The steady state energy balance says that Q_h (duty hot side) + Q_c (duty cold side) = 0 (no accumulation of heat). For single phase cooling (only sensible heat), Q_h = m_h*c_p,h*(T_h,in - T_h,out). Likewise, the cold side Q_c = m_c*c_p,c*(T_c,in - T_c,out). Here, m is the mass flow rate, and c_p is the mass heat capacity. If you fix the mass flow rates and compositions (thereby fixing c_p), you can only fix three out of the four temperatures. The 4th temperature is fixed by the energy balance requirement. Likewise, you can fix all temperatures, but then you have to leave one mass flow rate to be calculated. If you specify the exchanger area (A) and heat transfer coefficient (U), then the heat load will be fixed via Q = U*A*LMTD and you can only fix two out of four temperatures (i.e. you will usually leave the outlet temperatures blank.
Any violation of this rule leads to an overspecified or underspecified problem, and an error in the heat balance.

[Art Montemayor]

Bravo, Joerd!

What a great, clear, and succinct response to a confusing query.


Koby:

I, too, have been troubled by not being able to understand what your problem really is. As Joerd has so accurately explained, the path to understanding a heat transfer application lies in algorithmic logic and clear math. Computers – especially simulation programs – are based on this principle. Otherwise, there is no solution.

Can you please explain in detail how it is that you present two (2) values for the heat transfer quantity.
How (or where) do you come up with the shell-side 1,000,000 kcal/hr?

And where does the tubeside 900,000 kcal/hr originate?

Await your clear reply.

[aminmeo]

In shell and tube type exchnager the heat transferred by hot fluid should be equal to the heat recievied by cold fluid. So you can simply verify it with the formula Q = mCp deltaT. If you fiund some difference in Qc and Qh there must be some problem in data entry.

[omnibus]

In shell and tube type exchnager the heat transferred by hot fluid should be equal to the heat recievied by cold fluid. So you can simply verify it with the formula Q = mCp deltaT. If you fiund some difference in Qc and Qh there must be some problem in data entry.

My two cent's worth-

Shell and tube side heat loads can differ in case heat from the hot fluid is being lost to atmosphere enroute from inlet to outlet, ie not all the heat from the hot fluid is getting transferred to cold fluid.

[djack77494]

Looks like four cents worth to me.
Omnibus, though you are technically correct, practically the amount of heat loss (or gain for cryogenic processes) is normally thought of as being negligable.
Doug