NTU question
NTU question
(OP)
I have a heat exchanger on site and would like to know the outlet temperatures by playing around with the inlet temperatures. If it is just a liquid-liquid heat exchanger, i have no problem, but with a vapour-liquid HX, where the vapour is condensing if have a problem figuring out the NTU
NTU = U.A/Cmin
how to calculate the heat capacity rate for the condensing vapour
C = flowrate x specific heat
but what about the latent heat? It is not taking into account into NTU?
NTU = U.A/Cmin
how to calculate the heat capacity rate for the condensing vapour
C = flowrate x specific heat
but what about the latent heat? It is not taking into account into NTU?





RE: NTU question
During condensation (being condensation a change-phase process) fluid evolves isothermally along the heat exchanger and so its heat capacity is practically infinite. In this case the interesting thing is that the parameter Cstar ---> 0 as Cmax ----> ∞
Where:
Cstar = Cmin/Cmax
Take a look to this
http://en.wikipedia.org/wiki/NTU_method
RE: NTU question
When Cmax ---> ∞
we have
epsilon ---> 1 – e^ (-NTU)
Where:
epsilon =(actual heat transferred)/(max heat that could be transferred)
epsilon = Q/ [Cmin *(Th in – Tc in)]
Being:
Th in = hot fluid temperature in
Tc in = cold fluid temperature in
RE: NTU question
RE: NTU question
cold side water:
Tin: 35°C
Tout: 46.11°C
Flow: 7151 kg/hr
Hot side glycerine:
Tin: 170°C
Tbubble point: 116.17°C
Tout: 50°C
Flow: 272 kg/hr
C*=0
A= 19,6 m2
U= 86.9 W/m2K
Cmin= 4189*7151/3600=8320
NTU = 19.6x86.9/8320= 0.21
epsilon = 0.19
Qmax = 0.19*8320*(170-35) = 210286.6 W
Delta Tcold = 210286.6/8320 = 25.3°C
So Tcold out should be : 25.3+35 = 60.3
This 60.3 conflicts with the 46.11 from the HX spec sheet. If I take the bubblepoint as T hot in I come closer (50°C).
What is the catch over here?
RE: NTU question
Glycerine has got a boiling point of about 290 °C.
Specific heat of glycerine: cp = 2.43 kJ/(kg K)
Cmin = (2430 * 272)/3600 = 183.6 J/(K s)
RE: NTU question
RE: NTU question
1. In the first glycerine temperature drops from Th,in to Tcondensation
2. In the second glycerine changes its phase Tcondensation = const
3. In the third the condensed glycerine temperature drops from Tcondensation to Th,out
Each unit will have its own Cstar = Cmin/Cmax
RE: NTU question
like you said, I had to consider 3 units.
RE: NTU question
for the NTU of the 3 units, is it correct that not the complete surface area will be used
so for the first unit NTU = U.A1/Cmin
the second unit NTU = U.A2/Cmin
the third unit NTU = U.A3/Cmin
with A1+A2+A3= A
if i take for instance the complete surface area for each of the units, the temperature of the glycerine drops below the condensation temperature in the first phase which is not possible.
Are there any rules on determing the surface area used for the single phase and for the 2-phase units?
RE: NTU question
Please consider that not only the area changes, but also the overall heat transfer coeficcient (U) of each zone.
Take a look to the link below: I hope it could help
http://www.cheresources.com/designexzz.shtml