7 replies to this topic

[Limilicious]

Correct me if I am wrong but this is my understanding of MVR.

You take LP steam and compress it to a higher pressure steam. There is a compression ratio that limits the amount of compression you can do. Its usually somewhere around 2.1 to 2.5.

The steam produced by MVR are superheated steam and usually require some sort of desuperheater before usage in heat exchanger/evaporators.

When steam is compressed, condensate is produced. So there would be an inlet steam in the MVR and two outlets: a higher pressure steam and condensate.

My question is how can it produce condensate while pressurizing steam to superheated steam? Won't the superheated steam just vapourize the condensate?

And how do you calculate the amount of condensate produced by pressurizing steam? Just taking random numbers, 100 kg/hour of steam goes in the MVR at 90 kPa and 96.7129 C (saturated) and leaves at 180 kPa (compression ratio of 2). How much steam is leaving? How much condensate is leaving?

I found an isentropic compression calculator here: http://www.engineeri...e.com/calc4.htm
The Kappa is the compressibility factor, right? If so, I used the steam tables from Spirax Sarco, and it gave me a compressibility factor less than 1. This resulted in a decrease in temperature according to that calculator.

This is rather weird... and I'm totally confused now. =]

Edited by Limilicious, 16 October 2010 - 06:38 PM.

[breizh]

HI ,
Let you read this paper from Art Montemayor.

Hope this helps

Breizh

[breizh]

Hi ,
An other great paper
http://www1.eere.ene...fs/heatpump.pdf

Hope this can support your study
Breizh

[Limilicious]

Thanks, I'll read them =D

Edit:
From my research, you can estimate the superheated steam temperature that leaves the recompressor by looking at the entropy.

So this is what I did, I looked at the steam table for the entropy of the inlet of the compressor; lets say its 100 C and 1 bar. My goal is having it compressed by a factor of 1.7 (1.7 bar), I look at the superheated steam tables and look for the same entropy with 1.7 bar.

Does this method seem valid for estimating the superheated steam temperature?

Edited by Limilicious, 17 October 2010 - 12:02 PM.

[breizh]

Hi ,
Kappa = ratio cp/cv .
If you consider the compression isentropic S=cte, you can get access directly the value of the temperature at the outlet using the relation P*V^k =cte & Perfect law gas .

Hope this helps
Breizh

[Limilicious]

Hello,

Sorry for the disappointment but I've never seen those equations before. I assume S = cte means Entropy equals cofficient of thermal expansive?
And P*V^k = cte, I also have no clue what k is due to the large amount of equations that reuse the letter "k".

Edit: I found out what that equation is; its the Polytropic Process equation...

Edited by Limilicious, 18 October 2010 - 01:59 AM.

[breizh]

HI ,
You need to find a book dealing with thermodynamic.
K = kappa = Cp/Cv


Breizh

[Limilicious]

Hello,

While I do thank you for your input, I must advise you something as well.

Next time, when you are suggesting something to someone especially equations, do explain the variables that aren't commonly known or obvious (such as variables other than "n for moles, m for mass, P for pressure, T for temperature. But if you want to use V for volume, you should specify it because V can stand for specific volume as well).
For example, k could mean kappa but it can also mean Boltzmann's constant. Its misleading depending on the person reading it.
Another example would be when you make up variable names like "cte". "Cte" can be a short form for coefficient of thermal expansion. But in your case, it should have been just a "C" for constant. Or you could've simply wrote constant.

Someone who is unfamiliar with the equation or someone who forgot about it wouldn't be able to "flow" or "think" at the same pace at you. Especially when you're providing insights in the "Student" forum.

Well, I was trying to avoid looking at the thermodynamic textbook because the one that was required when I took the course was rather poorly written. The course was also poorly taught; in general my thermodynamic knowledge is rather poor. I do admit that and I should probably find a new/better textbook.

All in all, thank you for your input as well as your unnecessary confusion.

Edited by Limilicious, 18 October 2010 - 02:35 AM.