MOISTURE DROP OUT 2

[aberta]

How much condensate would I collect by cooling coke oven gas from 80 degrees C to 30 degrees C. The gas has a molecular weight of 10.7. It is fully saturated at 80 degrees C. The coke oven gas flow is 32,500 NM3/HR, density 0.47 kg/m3, specific heat 0.4Kcal/NM3/C.
I need this information to size a heat exchanger, which will cool this condensate and recirculate it to cool gas by direct contact with gas.
Thanks.

 

[TD2K]

Look up the vapor pressure of water from the steam tables at 80C. That, along with the coke gas pressure will give you the inlet concentration of water (I'm assuming this should be a low pressure system so you can assume ideal behavior).

Since you know the flow rate of gas (is this on a dry basis), you can calculate the flow rate of water.

Do the calculation again at the outlet of your cooler at 30C. The difference is the amount of water condensed.

 

[aberta]

TD2K
Thank you very much. The coke oven gas is at -0.58 psig (14.116 psia). This is 730 mmHg. the vapour pressure of water at 80 degree C is 355 mmHg. Therefore the vapour concentration is (355/730)*the total volume of gas and vapour. This will give me volume of vapour. Alternatively, ratio (355/730)* total weight of gas plau vapour. This should give me the weight. Am I correct?
Thanks.

 

[TD2K]

No, but you're close.

A fixed volume of any gas always contains the same numbers of moles as another gas (at the same T and P) although the mass will obviously vary due to the different molecular weights. That's an important consideration when doing gas calculations since you want to work in moles, not kgs.

Nm3/hr is in reality a way of expressing a molar flow rate, NOT a mass flow rate. For example, 1 lbmole of a gas at standard conditions occupies 379.6 cubic feet. So a certain flow rate in scfm tells me how many moles per minute are flowing. I then multiply it by the MW to get the mass flowrate. But, a flow rate of 1000 scfm is the same number of moles per minute be it hydrogen or any other gas (Off-hand, I don't know what the corresponding volume is for normal conditions or I would have used that. 22.4 litres per gram mole sticks in my mind from high school chemistry but I'm not sure what temperature that is at).

Back to your question.
Let's assume the 32,500 Nm3/hr is on a dry (water free basis) because you don't say. If it's on a wet basis, I think you can see how to handle it.

You know from the above work that the % of water vapor is 355/730 or 48.63% on a volume basis and the dry gas is therefore 51.37%. The TOTAl volumetric flow rate is therefore 32,500/(0.5137) Nm3/hr. Subtract from that the 32,500 Nm3/hr dry gas (assumed) flow rate and you have the Nm3/hr of water vapor. You can work out using the ideal gas law (PV = nRT) how many moles and therefore how much water in kg is in the inlet gas stream.

Repeat for the outlet stream.

There are quicker ways but this will give you a good idea what component (dry gas and water) you have at each point and give you a better understanding what is happening.

 

[patrickraj]

For the given density of 0.147 kg/M3, the pressure works out to 18.71 psia as shown below :

Density = PM/(ZRT)= 0.0293lb/ft3 or 0.147kg/M3.
Pressure = 0.0293x1545x(80x1.8+492)/1x144x10.7 =18.71Psia
How you are getting pressure as 14.116 Psia?

Vapour pressure at 80°C = 6.791 Psia.
Vapour pressure at 30°C = 0.613 Psia
I assumed 32500Nm3/Hr as dry gas flow.
Wat cont at IL=6.791/(14.116-6.791)x 32500/22.414x10.7=14383.8 kg/hr.

Water content at outlet = 0.613 / (14.116 - 0.613) x 32500 / 22.414 x 10.7 = 704.3 kg/hr.

Hence, moisture condensation = 14383.8 - 704.3 = 13679.5 kg/hr.

 

[aberta]

Thanks TD2K & patrickraj

Patrickraj, I was given the density at 0.47 kg/m3, under normal temperature and pressure.

TD2K, I have done the calculations as follows.

Volume og Gas , dry basis is 32,500 NM3/HR

Pressure in the pipe, inlet to the Primary Coolers is minus0.58 psig=14.116psia=97326 pascals=730mmHg.

Vapour pressure at 80 degrees C is 355 mmHg.

Thus, vapour concentration is 355/730=48.63%

Dry gas concentration is the rest = 51.37%

Total volumetric flow is 32,500/51.37%=63,266 NM3/HR

Thus, Total vaour flow rate is =63266-32500=30,733NM3/Hr.

pv=nRT

p=97326; v=30,766; n=moles; R=8.31; T=273+80=353K

Thus, n=97326*30766/[353*8.31]=1,020,761 moles

molecular weight of H2O is =18

therefore, weight of vapour/hour=1,020,761*18/1000 Kg
=18,373 Kg.

Repeat calculations at 30 Degrees C:

vapour pressure is 31.8 mmHg
Gas pressure is=730 mm Hg
vapour concentration=31.8/730=4.36%
dry gas is rest =95.64%
Total volumetric flow is 32,500/95.64%=33,982 NM3/HR
of this 32,500 is dry gas
Thus, vopour flow is=33982-32500=1,482 NM3/HR
T=273+30=303K
n=97326*1482/[303*8.31]=57284 moles.
weight of vapour per hour=57284*18/1000 Kg=1031 Kg.

Net vapour condensed by cooling total gas and vapour from 80 degrees C to 30 degrees C = 18373-1031=17,342 Kg

Does this sound about right?

Please note:
I got vapour pressure of water at different temperatures from the CRC handbook of physics and chemistry.

The gas in the pipe is under negative pressure of 730 mmHg abs.

Thanks again.

 

[patrickraj]

Dear Td2K,

Could you please comment where I am wrong? Why the difference between my calculation and aberta's.

Thanks in advance

 

[TD2K]

Anyone for a 3rd set of numbers?

I get a condensation rate of 24,450 kg/hr of water.

Patrickraj, you are using the MW of the dry gas (10.7) rather than the MW of water (18.016) to calculate the inlet and outlet content of the water. Other than that, your approach is right (you still won't get quite the same number I have as your vapor pressure numbers seem more precise than mine are, my copy of the steam tables isn't precise in these temperature ranges).

Aberta, you are mixing up normal and operating conditions. Your inlet water vapor flow rate up determined of 30,733 Nm3/hr is correct. BUT, when you convert this to kg of water, you use the actual pressure and temperature. Since your flow rate is in normal conditions, you must use 14.696 psia and 32F/0 degC when converting this volumetric flow rate to mass (note, you also use 30,766 Nm3/hr as the flow rate in your PV=nRT calc but the volume you just worked out was 30,733 Nm3/hr. Not a major effect I realize).


Also, you don't need to work in Nm3/hr. You can work in m3/hr at 80C and 14.116 psia. Your 32,500 Nm3/hr becomes 32,500*(14.696/14.116)*((80 + 273)/273). When you calculate the volumetric flow rate of water vapor using the above approach, you would calculate the mass flow rate then using the ideal gas law and 14.116 psia and 80C temperatures.

 

[aberta]

TD2K

I have a feeling you are right but I wish you weren't. It explains our problems with cooling. I am going to workout the third set of numbers.
Thanks.

 

[TD2K]

The water vapor is obviously going to be a significant part of the cooler's duty. If that was not done correctly, then I can also see why you are having cooler problems.

Technically speaking, when you do this calculation Aberta, to estimate the remaining water vapor flow leaving the cooler still in the gas phase, you should include the pressure drop across the exchanger which will reduce the system pressure and increase the amount of water the 32,500 Nm3/hr of dry coke gas holds.

It won't be significant here since the majority of the water condenses simply due to the cooling from 80C to 30C but strictly speaking, the 14.116 psia is only valid at the inlet.

 

[patrickraj]

DEar TD2K,

You are very much correct, there is a fault. I should have used MW of water in lieu of the dry gas.
With MW of water I get the value of 23032.7 as shown below:

Wat cont at IL=6.791/(14.116-6.791)x 32500/22.414x18.016=24218.6 kg/hr.

Water content at outlet = 0.613 / (14.116 - 0.613) x 32500 / 22.414 x 18.016= 1185.9 kg/hr.

Hence, moisture condensation = 24218.6 - 1185.9 = 23032.7 kg/hr.

Your value works out to only 6.15% higher.

 

[aberta]

Thank you very much TD2K and patrickraj. I calculated 24734 kg/hour using TD2K method. I think the difference between your numbers is due to my mistake earlier in calculating the total vapour flow rate.
Thanks again TD2K, you have shown that we should be having a problem with our coolers which were designed for condensate estimated at 10,000 kg/hr. Since then we have added additional cooling and now we can cool from 80C to just under 40C. We need to cool to 30C as at this temperature most of naphthalene and tar mist are removed before the exhauster. After that we have an electostatic precipitator to remove finer tar mist. Then we clean the gas for ammonia, lightoils and napthalene using wash oil. All the above process tanks are at present partially plugged with tar because of our primary cooler's poor operation. This has been a problem since we increased our coke making push rate from 50 ovens per day to 62/64 ovens per day.
Thank you again very much.