Thermodynamics [Constant Vapor Pressure Problem]

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Homework Statement


There are expanded 0.90kg/s of steam at constant pressure from 3MPa and 70% quality to a final state. If the process is nonflow for which W = 121.3kJ/s, find (a) The final temperature, (b) Q, (c) the available part of Q for a sink temperature of to = 27 Celsius

  • Answers : a) 282 celsius, b) 966 kJ/s, c) 414 kJ/s

Homework Equations


Wnf = P(V2-V1)
Q = H2 - H1
but theres mass, so
Wnf = mP(V2-V1)
Q = m(H2 - H1)

The Attempt at a Solution


I used the steam table for Sat. Pressure. And I got V1 = Vf1 + x(Vfg1) = 47.04095 x 10^-3 m^3/kg
And so,
Wnf = mP(V2-V1)
121.3 = (0.90)(3MPa)(V2 - 47.04095)
So i Got 91.96687593 x10^-3 m^3/kg
Checking it in table 3 from the steam table the temperature is way off the mark of the desired answer.

Is there another way around it? Please help me. The answers are written above ... below the problem...
 

Answers and Replies

  • #3
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From my understanding the liquid vapor was at 70% quality before it turns into a steam having 0.9kg/s.
Either way the problem is really hard to understand. If i can figure out what's initially there before turning into steam.
 
  • #4
SteamKing
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From my understanding the liquid vapor was at 70% quality before it turns into a steam having 0.9kg/s.
Either way the problem is really hard to understand. If i can figure out what's initially there before turning into steam.

No, that's not how vapor quality works, and it is not what is described by the problem statement.

http://en.wikipedia.org/wiki/Vapor_quality

According to the OP, you already have steam with a vapor quality of 70% and a pressure of 3 MPa entering the machine, where it then expands while doing work (or more accurately having a power output) of 121.3 kJ/s. This means that the specific volume of the steam entering is low, and the specific volume of the steam coming out is greater, since it has expanded while doing the work.
 
  • #5
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I see. Thanks for clearing that. From the link that you gave. I found out that I can use X = Mass of vapor / Mass total. I have 0.70 as x but 0.90kg/s is mass of vapor or mass total?
 
  • #6
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It's total mass flow. And, even though the problem statement says that it's a nonflow process, it definitely appears to be a flow process.

Chet
 
  • #7
SteamKing
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It's the total mass flow rate.
 
  • #8
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0.7 = Mass of the Vapor / 0.90
Mass flow rate of the vapor = 0.63 kg/s
I got V1 = 0.04704095 m^3/kg
and V2 = 0.09196687593 m^3/kg
By using interpolation from the steam table of 3MPa i get 353 degrees Celsius. Is there something wrong?
And V1 is from the saturated temp of 3MPa which is 233.90 celsius at the steam table.
 
  • #9
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I'm sorry. I still don't understand the problem statement, and whether the problem should be treated as flow or non-flow. So, I'm going to drop out of the discussion.

It does look like you determined the initial temperature correctly.

Chet
 

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