LMTD for Steam Heated Reboiler 1

[Chance17]

I am a 30+ engineer and am a little embarrased to ask.
But how does one handle Q = U*A*LMTD for a Steam Heated Reboiler?
The main issue seem to be LMTD determination
The process side is a constant boiling temperature.
Neglecting steam superheat & steam subcooling, the steam is a constant condensing temperature.
The greater temperature difference equals the lesser tempeature difference
The arithmetic LMTD = (Greater DT - Lesser DT)/LN [(greater DT)/lesser DT]
= (0) / LN [1]
= (0) / (0)
Any comments would be appreciated

 

[ione]

As far as I know you cannot apply LMTD when you've to deal with change of phase. Try epsilon-NTU method

 

[25362]

http://www.enggcyclopedia.com/2011/08/calculate-lmtd-formula-fail

 

[Latexman]

Just use DT. DT = Greater DT = Lesser DT.

Good luck,
Latexman

Technically, the glass is always full - 1/2 air and 1/2 water.

 

[ione]

How could it be possible to have Greater DT = Lesser DT? You should have a change of state on both sides of the HX. It's a bit strange to me

 

[Christine74]

If both fluids in your exchanger are isothermal then the LMTD, the GTD, and the LTD are all equal. You can prove this by applying L'Hopital's rule to the LMTD equation.

-Christine