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pressure in an ammonia tank 3

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safetystudies

Civil/Environmental
Nov 30, 2006
9
Hello, first of all, many thanks for your help.
My problem deals with an ammonia tank filled up to 85 % with ammonia at 3°C. The tank temperature then reaches the ambiant temperature of 20°C. How can I calculate the final pressure in the tank resulting from this temperature increase?
 
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Assuming you're speaking of anhydrous ammonia in a tank containing just pure ammonia, the tabulated equilibrium vapor pressure at 20oC is 8.463 atm.
 
Just so you understand, as the tank warms up, some NH3 will vaporize and those molecules will compress the existing ones in the vapor space and some of those NH3 molecules will condense to a liquid until equalibrium is reached. If you were to assume NH3 liquid were noncompressible, you would end up with the tank 85% full when it warmed up. Because its not, you'll end up with about 88% full tank.

That's why most industrial tanks have the chart printed on the level gauge stated how full you can fill the tank at a given temperature or pressure.
 
As long as you have vapor space in the tank (our 85% full), it will follow the NH3 saturation pressure data which is published.

ANSI 15, The Refrigeration Safety Code, requires that the high pressure receive not be filled above a certain percentage, either 80 or 90% full. Once the tank is completely full of liquid and the vapor space is gone,the liquid will be have a a hydraulic fluid and pressure will rise rapidly and fail a vessel of pipe

Ken



Ken
KE5DFR
 
Many thanks again for your help. I'm very bad in calculations so could you explain me how you find that the new liquid volume will be 88,3 % ?
 

It's not so complicated. The total mass of liquid + vapor, and the vessel volume are assumed constant. Using the densities [ρ] of liquid and vapor "before and after", and making the total volume = 1, a balance would show:

[•] Before, at 3 Celsius:

The liquid: volume L1 = 0.85, [ρ] = 634.44 kg/m3; its mass: 0.85[×]634.44 = 539.274 units

The vapor: volume V1 = 0.15, [ρ] = 3.8405 kg/m3; its mass: 0.15[×]3.8405 = 0.576 units

Total mass: 539.274+0.576 = 539.850 units

[•] After, at 20 Celsius:

The liquid: volume = L2, [ρ] = 610.20 kg/m3
The vapor: volume = 1-L2, [ρ] = 6.7025 kg/m3

L2[×]610.2 + (1-L2)[×]6.7025 = 539.850​

Thus, L2 = 0.883, namely 88.3%.

I hope there is no error in the above exercise.
 
Hi, I've found the same results. Many thanks to Dcasto, 25362 and TXiceman !
 
Hello, first of all and again, many thanks for your help.
This is still in concern with an ammonia tank. I have done the calculations below in response to the following question and with the help of 25362, dcasto and Txiceman in my last message.

“Demonstrate the mechanical resistance of a NH3 vessel considering it’s filled to the maximum level authorized at -3°C and a temperature increase to the ambiant temperature (I took 50 °C).”
The problem is that my client says that the filling rate is around 65% and I find 82%.

The input data that he gave me are :
Maximum authorized mass = 48899 kg
Vessel volume = 92,283 m³
Filling rate : 65%

My calculations :
We have : Vtotal = 92,263 m³ = Vliq + Vgaz and Mtotl = Mliq + Mgaz = 48899 kg
At -3°C, 1m3 of NH3liquid weights 642,66 kg and 1m3 d'NH3gaz weights 3,1038 kg,
Since
Mliq = Vl x 642,66
Mgaz = Vg x 3,1038
Then Mass total = Vl x 642,66 + Vg x 3,1038
Vg = 92,263 - Vliq
Mass total = Vl x 642,66 + (92,263-Vl) x 3,1038
Vl = (Mass total – 3,1038 x 92,263) / ( 642,66 – 3,1038) = 76,01 m³
Thus Vg = 92,263-76,01 =16,25 m³
And the filling rate equals to 82,38% and the pressure is 3,83 bar.


Using the same method, I find that at 50°C the filling rate is about 94% and the pressure = 20,34 bars.

Is is possible that 48899 kg of ammonia at -3°C corresponds to a filling rate of 65%?
 

I checked both your estimates, at -3 and +50 Celsius, based on equilibrium saturated anhydrous ammonia conditions, of the filling ratio, not rate, and found them correct.

Is it possible that your client using the "rate" wording means something totally different ?
 
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