NTU question 1

[semsagro]

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?

 

[ione]

Heat capacity is an extensive (mass related) properties of a body. The intensive property related to heat capacity is specific heat. Specific heat is the amount of energy required to increase the temperature of a unit quantity of a material by a unit of temperature.
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

 

[ione]

Only to complete my previous post:

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

 

[semsagro]

thanks! makes more sense now

 

[semsagro]

Ok for one condensor this works out, for the other one I still have a problem

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?

 

[ione]

I am a bit puzzled:

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)

 

[semsagro]

We are working under vacuum, that's why the boiling point is lower.

 

[ione]

Just my opinion, but you can’t consider the whole HX as a condenser. You should divide it into three units:

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

 

[semsagro]

ione thanks, the issue is solved

like you said, I had to consider 3 units.

 

[semsagro]

one more question about this

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?

 

[ione]

semsagro,

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