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Heat Exchanger Client-supplied nformation 1

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SirAdam

Mechanical
Sep 18, 2024
3
Hi All,

New in post to a heat exchanger company; where there's no other engineer to discuss/demonstrate. I come from pressure vessel background, though i've been in other industry [machine design] for several years so in need of a refresher!

Case in point; I'm to run through HTRI thermal software to establish a clients proposed tube length is fit/unfit.

provided information
Tube flow [3710 Nm3/hr] 93% o2, 5% Ar, 2% N2
inlet temp 148c
outlet temp 50.3c
inlet pressure 3.03 bara
pressure drop 6.9kpa
fouling resistance - 0.0005 ft2-hr-F/BT
design pressure 10 bar
design temp 200c

Shell side
propylene glycol 40/60 with water
inlet 42C
outlet -
inlet pressure 3 bara
pressure drop 69kpa
fouling 0.0005
design pressure 10 barg
design temp 93C

I'm correct in saying I cannot convert the Nm3/hr into Kg/S without specific volume or density of gas?
I'm correct in saying I either need to understand the required outlet temp OR available duty, to calculate the remaining factors?

Cheers, any tips appreciated.
 
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You know the composition, temperature, and pressure so you can figure out the density of the gas.
And you can also figure out the specific heat and thermal conductivity of the gas.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed
 
You can solve for the shell-side outlet temperature by calculating the required heat duty based on the tube-side flowrate, specific heat capacity, and the inlet/outlet temperatures (assume that the heat exchanger is adiabatic).


-Christine
 
Hi,
As others said you need to calculate the density of the gas based on information given (composition, temperature, pressure).
Mw = 32.32 g/mol
Perfect law gas to be used (PV=ZRT), Z#1 thus qas flow rate inlet is 1912 m3/h and density inlet is about 2.496 kg/m3 and mass flow rate gas 4772 kg/h
Z#0.9987 using Peng Robinson EOS , assumption Z# is validated.
density =Mw /V; Mw molecular weight and V molar volume =RT/P.
No change state in tubes, Gas phase.
Normal m3( 273.15 K, 1.013 bar)
P1 is a given, P2 =P1-delta P (to be calculated)
For heat duty you need to get access to the specific heat capacity of each element and calculate the mixture properties (apply Kay's rule, same you did for Mw and perhaps Z (?)). A Google search will provide all the data missing. (Cp gas)
cp gas inlet = 30.129 J/mol K or 932.2 J/kg K
assumption: cp gas constant (to be checked)
Duty = m*cp gas *(Tin-Tout) = 4772*932.2*(148-50.3)/1000/3600= 120.23 KW
For liquid side, you need to get access to the properties of the liquid (density, Specific heat capacity). From there you need to equate the heat duty liquid =Gas duty to get access to the mass flowrate and or volumetric flow rate of the liquid.
But data are missing to go further in the calculation (liquid outlet temperature or liquid flow rate)
In other words you are quite correct!
Assuming a perfect Counter Current HX, Temp outlet must be < 50.3 C, let say 49 C
density liquid # 1010 k/m3, Cp liquid 0.895 KJ/Kg C
Mass coolant = Duty / Cp Coolant *(Tout -Tin) = 19.19 kg/s or 69.086 ton/h
Flow rate coolant : # 68 m3/h
Good luck
Pierre
 
@pierre's calcs for backing out a shellside flowrate would be valid for 1 shell side pass, 2 or 4 passes on tube side, for which Ft, the LMTD correction factor would most likely be 1.0. Use a TEMA U type HX to avoid differential temp mechanical stress problems.

By the way, if tubeside design T = 200degC, so should shellside design T.

An all welded plate HX would be more compact, and allow you to use less coolant. Both sides are clean fluids. In this case, shellside exit temp could be even 60-70degC - pure counter current flow.


 
Thanks all for the replies!

@Pierre

I'm sorry, can you take me back a step.

I can calculate the Mw value you have, though i can't arrive at the same figures for flow/density/mass flow etc that you have.

Maybe I'm working in the wrong units!

P = in pascals?
Z = 1 [as you state]
R = gas constant 8.3145
T = Kelvin [so in this case, 421 K]
V = m3

@Christine - how do you calculate the heat duty?
 
Hi,
Better if you share your calculation. Better to use an excel file.
Make sure you convert the Normal flowrate to actual flowrate.
Regarding the set of units you are correct.
I will share with you my excel work.
I cannot offer more.
Pierre
 
 https://files.engineering.com/getfile.aspx?folder=258cdb45-d315-495a-9b14-b4a33374c31b&file=SirAdam.xlsx
Hi Pierreick,

That's incredibly helpful thank you!

I don't understand a few constants entered, but i'm sure i can work them out.

I appreciate your assistance

Best Regards
 
Hi Sir Adam,
I did a mistake on the liquid side, being counter current, the temperature outlet should be less than the temperature inlet gas, <148 C
I let you review the calculation, using T liquid outlet being 87 C for example.
Note: pay attention to the limitation related to design temperature/ pressure on shell side (93 C)
I hope this going to put you in the right direction.
Sorry for that.
All the best.
Pierre
 
If P is in pascals, R = 8314 J/kgmole/degK. If it is in kPa, it would be 8.314kJ/kgmole/degK
 
George,
Your last reply is confusing!
The system is SI, nothing to do with Kpa, KJ even your feedback is correct.
R=8.314 J/mol/K where P is in Pa absolute, V in m3and T in K
Pierre
 
Stick it in an ambient vaporizer and not require any PG/Water at all ;)
 
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