Pall ring pack heat exchanger 1

[ione]

I am trying to find a valid approach for the following problem:

I would like to use a Pall rings pack in a counter-flow heat exchanger. Flue gas from natural gas (CH4) combustion, should rise from the bottom of a column, while sprayed water should move in the opposite sense. Pall rings should contribute to fraction water droplets and increase the surface involved in the heat exchange process.
Water flow rate: 6.6 m3/hr
Temperature rise (deltaT) required: 60 °C (Tin = 5°C; Tout = 65 °C)
Flue gas flow rate: 735 m3/hr (49 m3/hr of CH4 for combustion process and 15 m3 of flue gas per each m3 of CH4).
Power required = 470 kW I have considered a high efficiency for this process (98%, maybe even more). Column (H = 2.5 m) is well insulated and sensible heat loss from stack should be very low (very low flue gas temperature from the stack after the Pall ring pack).
Pall ring specs: as per table attached.

And now my questions:
1) Can in your opinion this heat exchanger work properly?
2) Can anyone help with formulas to size the column width and the height of the Pall ring pack to meet the requirements indicated above?

Any comments very appreciated

 

[siretb]

This should work.
the heat transfer is done by direct evaporation of water.
So calculations can be based on mass transfer corrélations, like Onda's correlation.

Your input does not allow, unless I missed something, the calculations
We would need: the feed pressure and feed temperature for your off gas. If it comes direct from combustion will be above 600°C
we would need the gas composition and flowrate (can be assumed to be
Normal cubic meters of CH4 (temp=0°C, 1 atm abs.) with excess air so as to have the ratio of 19:1. Please confirm of provide off gas rate, in kilograms/hour, kilomole/hour or Normalcubicmeters/hour.

For the diameter the top of the column will be at T lower than 65°C, and you know from enthalpy balance how much you evaporate. This gives a net volume rate. Then you size so as to have not more than 80% flood. (you know L, you know V, yo kwon the packing)
The bottom is more tricky because the gas is much hotter. However it will cool very rapidly. Worst case, overly conservative would be to size the bottom segment by using the inlet temperature. However, ost of the water will shower down from the packed bed, and, by the time the off gas reaches the packing it should be already much cooled. I would, for a first pass of calculations, assume a temperature of some 250°C (if t in=600°C) and size
accordingly.

Please provide more details.

 

[ione]

Great. It looks like I’ve found someone that can help!


I don’t know what Onda’s correlation is, could you please report it and explain?

Below further info, hoping they can help you to help me.

From the combustion of CH4 we can consider CO2, H20 (vap) and N2 as products of the combustion.
Stoichiometric air/fuel ratio for CH4 is 10.57:1. I have assumed 15:1, but let’s consider your 19:1 ratio if it is more correct.

Siretb, sorry for poor precision with units: I confirm the flow rate of CH4 is in Nm3/hr (Normal cubic meter per hour). I calculated the flow rate (49 Nm3/hr as stated in my previous post) considering 34.5 MJ/Nm3 as LHV for CH4.

Hot gas stream comes directly from combustion. I do not have idea about feeding pressure, but I can check with the burner’s supplier.


Regards

 

[StoneCold]

A point to consider is that contaminants in the gas stream are going to be scrubbed into the water stream so if your hot water is going to be used for something you will have to deal with the posiblity of sulfur compounds as well as CO2 and possibly CO trapped in the liquid.

Regards
StoneCold

 

[ione]

StoneCold,

Contaminants are not a concern for my specific applications (I had already take this into acconunt). Anyway using CH4 as fuel, sulfur should not be present (nor CO if combustion take place properly).

Regards

 

[siretb]

Onda's correlation can be found easily on the net. Here is one form. Check the net.

klg (R T /(A D) =5.23( G / ap Mug)**0.7 Sc**0.33 (ap dp)**-2
where kg is the mass transfer coef kmols/atm/m2/s
T the ttemperature in °K
A the specific surface of your packing m2/m3
D the diffusivity in this case of H2O m2/s
G the specific gas rate kg/m2 of cross section / s
Mug the ga sviscosity N s/m2
Sc the Schmidt number (adim)
dp the nominal size of the packing (m)
R the gas constant (metric)

then the heat transfer coefficient is calculate using the Chilton analogy
kg Sc**0.667 = h Pr**0.667 / Cp
where Pr is the Prandtl number
and Cp the molar heat capacity J/kilomol/°K
h the heat transfer J/s/°K

The calculations may be fastidious so a process simulator is of help.
A paper heat transfer engineering, vol10 number 2 1989 pages 19+, by Huang and Fair gives a more straightforward correlation (for Raschig ring, and Pall rings )
hg a = c G**0.7 L**0.07
L and G kilogram/s/m2
a m2/m3
kg mass transfer coef kilomols/s/pascal/m2
and backtrack using the Chilton analogy

I would recommand you get this paper of Fair and Huang. there are graphs for the air water system, usable in your case.

Now what is your air rate stoechiometry 10:1 **** no **** can'nt be. How much oxygen do you have left? 3%? do you know your temperature (inlet). I will do a quick calculation for you (no garantee of course) if you provide flow and temperature of incoming gas. Without calculations I would think that 2.5 m of packing should be enough, and that 0.6-1 m diam would be OK. the 6-7 m3 water looks also Ok to me. Ballpark estimate is that you'll need 1-2m3 of packing.

Hope this helps.

 

[ione]

siretb,

Thanks for your valuable post.

As you would have argued I am absolutely not familiar with mass transfer and distillation tower.

I have found on Perry’s (7th edition) the Strigle’s GPDC chart which shows a correlation for pressure drop through a random packed bed and thought it could be useful.

I thought to follow the procedure below to calculate the column diameter:

1. Enter the GPDC chart with a calculated abscissa value,
2. Intersect with an acceptable value of pressure drop dp (at this purpose what did you mean when you suggested 80% flooding?)
3. Find the ordinate value
4. Calculate the superficial gas velocity U
5. Calculate the cross sectional area (and so the diameter) A = Q/U (where Q is the flue gas flow rate).

The ordinate term is:

U*[(rho gas/(rho liq – rho gas))^0.5] * Fp ^0.5*?n^0.05

Where:
U = superficial gas velocity
Fp = packing factor
rho gas = gas density
rho liq = liquid density
n = liquid kinematic viscosity


Further to the last details you ask for, I have assumed air/fuel ratio 15: 1 (735 Nm3/h flue gas flow rate) and inlet temperature 600 °C.

Regards

 

[RoderickH]

When the water hits the hot gas at the base of the tower, the gas will rapidly cool to the adiabatic saturation temperature which depends on the inlet temperature, water vapour content of the incoming gas, and gas molecular weight. For 600C, assuming 10% inlet water vapour (weight/weight, based on dry gas mass flow), and dry gas MW = 30, this temperature is 72C. With 1% inlet water vapour, the saturation temperature is 65C. This applies around atmospheric pressure, and for water, it cannot exceed 100C.
The calculation involves determining the inlet gas enthalpy, and finding the corresponding saturation temperature. For water/air, the saturation line is shown on the normal psychometric chart. For other gas mixtures, the enthalpy/temperature saturation line needs to be calculated.

 

[ione]

The enthalpy balance is undoubtedly useful to evaluate the gas flow rate necessary to the heat transfer process, but as the process should have a very high efficiency (somehow between 99 and 100% in my opinion) I have already evaluated the gas flow rate. My enquiry was much more focused on tower sizing (diameter and height).