5 replies to this topic

[Mohsen Khodaee]

Hello every one

I would like to know if it's possible that a throttle valve has exery efficiency greater than 100 or not?

I simulated a throttle valve in HYSYS with these properties:

 

inlet stream:

T=20.17 C

P=840 kPa

inlet stream is in liquid phase.

 

Outlet stream:

T=-10.14 C

P=343 kPa

X=0.2014

Outlet stream is in 2 phase condition, vapor and liquid.

 

Reference environment:

T=43.61 C

P=101 kPa

 

After exergy analysis these results are obtained:

Exergy_in=122.3 kj/kg

Exergy_out=143.4 kj/kg

Exergy Efficiency=117.3% !!!!!!!! and exergy distruction is negative!

I can not understand what is wrong!!!!

Edited by Mohsen Khodaee, 23 November 2016 - 03:10 PM.

[Saml]

I don't know about Hysys. In some versions of Aspen plus, if you don't do a full re-initialisation of the simulation, it does not recalculate the exergy in the already calculated streams after you change the reference conditions. Since they share part of the code, it may be the case in Hysys too.

 

By the way. You wrote Exergy_in twice. I suppose one is _out.

 

Just my two cents.

[Mohsen Khodaee]

I don't know about Hysys. In some versions of Aspen plus, if you don't do a full re-initialisation of the simulation, it does not recalculate the exergy in the already calculated streams after you change the reference conditions. Since they share part of the code, it may be the case in Hysys too.

 

By the way. You wrote Exergy_in twice. I suppose one is _out.

 

Just my two cents.

Thanks Saml.

I edited my question. I've used EES for calculations and Aspen HYSYS and Aspen Plus are very similar, I'm sure that my calculation and HYSYS calculation is right. I guess phase change made this problem. But I dont know how to solve it!

[Saml]

Checking in another simulator DWSIM I get the following
In: 149.92
Out: 142.72
 
Exergy is calculated using the following definition. 
 
 Exergy.png   2.35KB   0 downloads
 
_out value is similar to yours, but _in value is not.
I'd suggest that you add an additional block to calculate the Enthalpy and Entropy of the system at reference conditions and do the Exergy calculation from these two variables.

[breizh]

Hi ,

Consider this resource :

 

Hope this helps

 

Breizh

[Saml]

I've seen basically two ways to calculate the expression "the maximum work that can be extracted from a system" regarding the thermomechanical evolution of the system towards equilibrium with the environment. 

.

One is what Breizh linked. In this case, the expansion work that a system does against the environment is considered  as "extracted work" and is part of the calculated exergy. That portion of the exergy is the term 

 

 PdeltaV.gif   453bytes   0 downloads

 

Here, the thermodynamic properties to use for exergy calculation are internal energy, entropy and volume differences.

 

 

 

Another way is to consider an open system where a stream is continuously flowing. The system volume is fixed. The entrance and exit work are not considered as part of the "maximum work that can be extracted from the system". A better explanation is in the following link:

 

https://www.tbm.tude...y_of_energy.pdf

 

 

For this case, the properties to use to calculate exergy are enthalpy and entropy.

 

This last method it what Aspen is calculating. By the way, I've verified the number in Aspen Plus assuming that you are expanding propane at it's bubble point.

Numbers are similar to those obtained from DWSIM

 

 

Now, regarding your original question: it is not possible to produce an irreversible transformation (the pressure reduction) and be able to obtain more work than before.  This would make a perpetual motion machine of the second class.

 

Let's say that

-  A is your inlet stream condition,

-  B is outlet, and

-  C is at equilibrium with environment.

 

You evolve reversibly from A to C generating 122,3 J per Kg. You can evolve the opposite way buy using the same 122,3 J per Kg (because it is a reversible transformation)

 

You can also evolve from B to C generating 143,4 J

 

So that perpetual motion machine would work using the following cycle:

 

- From B, evolve to C reversibly. Generate 143,4 J/kg  exchanging heat between your system and environment only.

- Evolve reversibly from  C to A using 122,3 J/kg and exchanging heat between your system and environment only.

- Cause an irreversible pressure drop from A to B without exchanging heat.

 

We've got 21,1 J per Kg, exchanging heat against the environment only, with no other hotter/colder source (this is a perpetual motion machine of the second class)

Edited by Saml, 26 November 2016 - 10:04 PM.