Thermodynamics: Air and Steam Mixtures

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Homework Help Overview

The discussion revolves around a thermodynamics problem involving a vessel containing a mixture of air and steam, characterized by a specific dryness fraction and conditions of pressure and temperature. Participants are exploring the implications of the dryness fraction and how it relates to the masses of water and air in the mixture.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the meaning of the term "0.75 dry" and its implications for the mixture's composition. There are attempts to relate the dryness fraction to the partial pressures of air and steam. Questions arise regarding the definitions of saturated vapor pressure and how to derive the masses of water and air from the given conditions.

Discussion Status

The discussion is active, with participants seeking clarification on the definitions and relationships between the variables involved. Some guidance has been offered regarding the interpretation of the dryness fraction and its connection to the saturation conditions of the steam. However, there is no explicit consensus on the methodology to find the remaining masses.

Contextual Notes

Participants note the potential confusion surrounding the term "0.75 dry" and its implications for the calculations. There is also mention of the need for relative humidity to find certain values, which may not be directly provided in the problem statement.

nightingale
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Homework Statement


A vessel of 0.3m3 capacity contains a mixture of air and steam which is 0.75 dry. If the pressure is 7 bar and temperature is 116.9 degree Celsius, find: Mass of water present, mass of dry saturated vapour, mass of air.
Answer key: 0.102kg, 0.307kg, 1.394kg

Homework Equations


Relative humidity = mv/mg = Pv/Pg (pressure of vapour/ pressure of dry saturated vapour)
humidity ratio = mv/ma
Moist Volume V = total volume/mass of dry air

The Attempt at a Solution


I was able to determine the mass of dry saturated vapour. The moist volume (V) of dry saturated vapour from the steam table is 0.9774m3/kg and hence:

0.9974 = volume/mass of dry saturated vapour
0.9974 = 0.3/mg
mg = 0.3069kg

I have absolutely no idea however, on how to find the other two masses. I don't understand the term 0.75 dry, I don't know how to find the mass of water (is this mass of water vapour?) without relative humidity. I'd appreciate any help. Thank you very much.
 
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What does that mean? :

"... a mixture of air and steam which is 0.75 dry ..."

would that mean that 75% of the water is vapor and 25% is liquid ?
If this is the case, you know that the steam is saturated and you can find its partial pressure Pv.
Then you know also the partial pressure of air: Ptot - Pv = 7 - Pv .

The rest of the problem is then a simple conversion of quantities from partial pressure to masses.
 
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Which means that the air is 100% saturated in order for the saturated vapour (Pg) to be equal to Pv. Am I correct?
Because if from the steam table I would only get Pg, not Pv. But since you said I can find Pv, that means Pg is equal to Pv?

Thank you.
 
nightingale said:
Which means that the air is 100% saturated in order for the saturated vapour (Pg) to be equal to Pv. Am I correct?
Because if from the steam table I would only get Pg, not Pv. But since you said I can find Pv, that means Pg is equal to Pv?

Thank you.
Yes.

Chet
 
Thank you, everyone!
 

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