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Ammonia Evaporation Rate Inquiry

ammonia

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#1 aslungaard

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Posted 30 March 2024 - 10:42 AM

Hello all,

 

I am having an internal debate that I am hoping the collective wisdom of this community can help clarify.
 
The problem statement is as follows, "what determines the rate of evaporation for a pure substance: is it the amount of heat input or the rate that vapors above the liquid surface are removed?"
 
Background: The substance in question is anhydrous ammonia (NH3). At STP, NH3 is a gas, however it is often stored as a liquid under its own pressure. The pressure inside the storage vessel will be a function of its temperature, in accordance with the Antoine equation for vapor pressure.
 
Situation: let's assume we have a 1000gal vessel at 68degF that is 50% full of liquid. The pressure inside this vessel will be 128.8 psia. If a valve on top of the vessel were to suddenly open, the NH3 vapor in the headspace would almost instantaneously flow through the valve and into the surroundings. One could estimate this flow rate by knowing the differential pressure and Cv of the valve.
 
However, this calculation assumes that there is an infinite source of NH3 vapor upstream of the valve. This may not be totally accurate as converting the liquid to vapor requires energy.
 
Here is where I get hung up. Prior to the valve opening, the vapor and liquid inside the vessel are in equilibrium with each other. In other words, the liquid is at its bubble point; any drop in pressure will form a droplet of vapor. This energy can come from a decrease in temperature of the bulk liquid (i.e. auto-refrigeration). The calculation for this is shown below. Q_latent is interchangeable with latent heat of vaporization (i.e. dH_vap).
 
Q_latent = Q_sensible = m_liquid*Cp*dT
 
At 68degF, dH_vap is 574 Btu/lb and Cp is 0.5Btu/lb*F. For convenience, let's assume that the leak rate is 10 lbs/hr. 50% liquid = ~2500lbs so the overall temperature decrease is - 4degF per hour.
 
But what if the flow rate is 100 lbs/hr or 1000lbs/hr? Wouldn't the liquid cool to a point where its vapor pressure is so low that it cannot overcome atmospheric pressure?
 
Any help is most appreciated.


#2 Pilesar

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Posted 30 March 2024 - 08:21 PM

Is your question an academic exercise? If so, the student section is a better sub-forum. For your scenario, vapor is removed and pressure is reduced on a constant volume vessel containing vapor and liquid in equilibrium. The remaining contents of the vessel are no longer in equilibrium but will seek to establish a new equilibrium. The liquid will continue to evaporate until its vapor pressure equals the component's partial pressure in the vapor phase. Liquid ammonia can also be stored in bulk in enclosed vessels near atmospheric pressure at -28 F.



#3 breizh

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Posted 30 March 2024 - 10:34 PM

Hi,

I was thinking JT effect taking place. JT coefficient 2.101 e-03 K/Pa @ 8.8 bara and 20C

Breizh 



#4 aslungaard

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Posted 31 March 2024 - 10:13 AM

Is your question an academic exercise? If so, the student section is a better sub-forum. For your scenario, vapor is removed and pressure is reduced on a constant volume vessel containing vapor and liquid in equilibrium. The remaining contents of the vessel are no longer in equilibrium but will seek to establish a new equilibrium. The liquid will continue to evaporate until its vapor pressure equals the component's partial pressure in the vapor phase. Liquid ammonia can also be stored in bulk in enclosed vessels near atmospheric pressure at -28 F.

 

This is not an academic exercise; rather a genuine question from real-world experience. I worked with anhydrous ammonia systems extensively but never had that "lightbulb moment" when it came to this scenario. Either way, your response is helpful!

 

I understand that VLE will always be maintained within the vessel. Essentially, as the pressure drops inside the vessel, we are moving along the curve from right to left (see attached). So perhaps my question is better framed as such: could the flow of vapor out of the vessel be greater than the time required for the system to establish a new VLE?

Attached Files



#5 aslungaard

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Posted 31 March 2024 - 10:15 AM

Hi,

I was thinking JT effect taking place. JT coefficient 2.101 e-03 K/Pa @ 8.8 bara and 20C

Breizh 

 

Yes, the JT effect would certainly occur as the vapor expands across the valve. 



#6 Pilesar

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Posted 31 March 2024 - 02:23 PM

VLE will NOT always be maintained within the vessel. Some time after a disturbance, equilibrium will be approached. If vapor is released to reduce pressure, the liquid that was in equilibrium at the surface is now above its boiling point temperature and begins to boil. The surface of the liquid cools. The liquid at the bottom of the vessel is not in contact with the vapor and is not forced to be in equalibrium with the vapor. There can be temperature gradients in the liquid also if there is nothing stirring or circulating the liquid... warmer liquid tends to rise and colder liquid tends to sink but the convection currents are not automatic or necessarily quick. Forced circulation of the liquid helps even out the liquid temperature. The pressure at the bottom of the liquid will be higher due to static head. The temperature of the liquid at the bottom of the vessel is not in contact with the vapor and may be significantly subcooled without forced circulation. Even though the pressure at the bottom of the liquid storage is reduced also when vapor is released, it may not reach its boiling point due to subcooling. It has been many years since I worked with large scale ammonia storage and I don't know if temperature gradients are of concern with anhydrous ammonia. I know that it is a concern for LNG storage that the liquid temperatures can stratify with a colder layer above a warmer layer and it is possible for disturbances in the liquid layers to cause large volumes of vapor to be rapidly generated. I did read your 'rate of evaporation' question but I am unable to help with that.



#7 aslungaard

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Posted 31 March 2024 - 07:52 PM

VLE will NOT always be maintained within the vessel. Some time after a disturbance, equilibrium will be approached. If vapor is released to reduce pressure, the liquid that was in equilibrium at the surface is now above its boiling point temperature and begins to boil. The surface of the liquid cools. The liquid at the bottom of the vessel is not in contact with the vapor and is not forced to be in equalibrium with the vapor. There can be temperature gradients in the liquid also if there is nothing stirring or circulating the liquid... warmer liquid tends to rise and colder liquid tends to sink but the convection currents are not automatic or necessarily quick. Forced circulation of the liquid helps even out the liquid temperature. The pressure at the bottom of the liquid will be higher due to static head. The temperature of the liquid at the bottom of the vessel is not in contact with the vapor and may be significantly subcooled without forced circulation. Even though the pressure at the bottom of the liquid storage is reduced also when vapor is released, it may not reach its boiling point due to subcooling. It has been many years since I worked with large scale ammonia storage and I don't know if temperature gradients are of concern with anhydrous ammonia. I know that it is a concern for LNG storage that the liquid temperatures can stratify with a colder layer above a warmer layer and it is possible for disturbances in the liquid layers to cause large volumes of vapor to be rapidly generated. I did read your 'rate of evaporation' question but I am unable to help with that.

 Yes, all valid points and I appreciate the details.






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