Calculating Condensate Pressure from Steam Usage

[hmvyoral]

If I have 400# and 180# steam, how can I calculate the pressure of the condensate that is being produced after the steam is used on the various tanks and equipment? I am being asked by one of my process engineers to find this.
Any help would be greatly appreciated.

Thanks

 

[Travis_King]

Do you mean that your 400 lb steam becomes 180 lb steam after some process? Or do you have two steam lines?

 

[Travis_King]

Either way, look at this

 

[hmvyoral]

We have 400, 280, and 180 lines. Thank you for your help.
Another engineer working on this has forwarded me this answering my question.

The steam at 400 psig is superheated steam, meaning instead of being at the saturation temperature, 448.16 F, it is at 600F. Therefore it has ~ 152 degrees of superheat. When the 400 psig steam is condensed, the following happens: First, the temperature is lowered from 600F to 448.16 F. Then, it condenses at 400 psig and 448.16F. The pressure drop is small, ~ 1psi in doing this.

The other situation that exists within the plant are the drop down stations. In this scenario, steam at 400 psig and 600 F undergoes a pressure drop to 180 psig. The resulting temperature drop is down to 578 F.

In the same instance mentioned before, when the 180 psig steam condenses, the following happens. The temperature is dropped from 578F down to 380 F (the saturation temperature of steam at 180 psig). The pressure drop is < 2 psi.

 

[russ_watters]

So it was a trick question, right? Other than piping and fitting losses, 400# steam yields 400# condensate. No calculation required.

 

[hmvyoral]

Apparently it was, but not intended to be. I knew that there would be some due to temperature change, but apparently there is not enough to make a difference. I guess that is why i could not find any such calculations.

 

[Travis_King]

Dropping steam from 400 psi to 180 psi will create measurable condensate.

 

[russ_watters]

Apparently it was, but not intended to be. I knew that there would be some due to temperature change, but apparently there is not enough to make a difference. I guess that is why i could not find any such calculations.

Er, no. There is no pressure drop due to temperature change in a piping system. Zero. Zilch. Nada. Not just "not enough".

 

[russ_watters]

Dropping steam from 400 psi to 180 psi will create measurable condensate.

Why?

 

[Travis_King]

He's using a PRV station. This station will remove latent energy from the steam. Hence condensate return or traps are used on most PRV stations.

Obviously there are differences between recoverable and non-recoverable pressure drops.

Also, maybe I'm not reading it right, but the PRV stations are dropping superheated steam from 400 to 180 psi?

 

[hmvyoral]

Our plant shares resources with another. We receive 400# super heated steam and use that for some processes. We also have let down stations where we use 280# and 180# steam. the plant is over 60 years old and we are in the process of redesigning our steam and condensate lines for better efficiency.

 

[Travis_King]

Just size the steam traps and drip pans at your PRV stations as per vendor suggestions.

Size your condensate lines based on the heat load of the steam services.

 

[russ_watters]

He's using a PRV station. This station will remove latent energy from the steam. Hence condensate return or traps are used on most PRV stations.

I'm still not clear on why (or if?) that would be. When the pressure is reduced, the temperature will drop due to the throttling process, but the lower pressure steam also has a lower saturation/boiling point. The only thing I can think of is that if the loss through the throttling valve is too much, some steam will have to condense.

But typically, the opposite problem exists: when hot, pressurized condensate goes through a throttling valve, some will flash to steam.

Here's a diagram of a PRV station that does not include a condensate connection at the PRV. Certainly it would at the risers, but because it is at the low point of the system. I'd think if condensation was an issue, it would be collected at the valve itself.
http://www.forbesmarshall.com/fm_mi...d=156&s2name=Steam Pressure Reducing Stations

In any case, the OP mentions the pressure after process equipment. Worst case scenario is low flow, where pressure is constant everywhere that doesn't have a valve shutting it off. So any steam trap on the system will have condensate at the same pressure as the steam. During normal operation, valve and fitting losses drop the steam pressure and therefore the condensate pressure a little, but that's not what you design the pressure requirement for.

 

[Chestermiller]

I'm still not clear on why (or if?) that would be. When the pressure is reduced, the temperature will drop due to the throttling process, but the lower pressure steam also has a lower saturation/boiling point. The only thing I can think of is that if the loss through the throttling valve is too much, some steam will have to condense.

But typically, the opposite problem exists: when hot, pressurized condensate goes through a throttling valve, some will flash to steam.

Here's a diagram of a PRV station that does not include a condensate connection at the PRV. Certainly it would at the risers, but because it is at the low point of the system. I'd think if condensation was an issue, it would be collected at the valve itself.
http://www.forbesmarshall.com/fm_mi...d=156&s2name=Steam Pressure Reducing Stations

In any case, the OP mentions the pressure after process equipment. Worst case scenario is low flow, where pressure is constant everywhere that doesn't have a valve shutting it off. So any steam trap on the system will have condensate at the same pressure as the steam. During normal operation, valve and fitting losses drop the steam pressure and therefore the condensate pressure a little, but that's not what you design the pressure requirement for.

For an ideal gas, the Joule-Thompson coefficient is zero, and thus in pressure drop across an adiabatic oriface, the temperature change is close to zero. The change in enthalpy is zero for this steady flow process.

 

[Travis_King]

I don't think the OP is sizing the condensate pressure requirement. I believe he, or she, is looking to size the volume requirement. I could be wrong, though.