Thermodynamics pure substances problem

Click For Summary

Homework Help Overview

The problem involves a rigid tank containing a saturated liquid-vapor mixture of water at 140 °C, which is heated until it reaches the critical state. Participants are tasked with determining the mass and volume of liquid before the heating process.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of mass before heating, exploring the relationship between mass fractions of vapor and liquid. There are questions about the completeness of the problem statement and the use of specific volume values.

Discussion Status

Some participants have provided calculations and values, while others seek clarification on the relationships between mass fractions and total mass. There is an ongoing exploration of how to apply the mass fraction concept to find the mass of liquid water.

Contextual Notes

Participants note potential missing information and discuss the implications of using specific volume values for calculations. The discussion includes references to saturated water tables and the critical state of the substance.

hazzz
Messages
7
Reaction score
0

Homework Statement


Hellooo, so this is the question that i have:

A 0.5-m3 rigid tank initially contained a saturated liquid-vapor mixture of water at 140 °C is now heated until the mixture reaches the critical state. Determine the mass and the volume of liquid before the heating process

Homework Equations


v = vf + x*vfg
x = mvapor/mtotal
v = V/m

The Attempt at a Solution


I do not understand how to find the mass before the heating process :(
I found the X which is 0,004086
vf at 140C is 0,001080m3/kg
vg at 140C is 0,5089m3/kg

the final answer should be mass = 157,83 kg
 
Physics news on Phys.org
hazzz said:

Homework Statement


Hellooo, so this is the question that i have:

A 0.5-m3 rigid tank initially contained a saturated liquid-vapor mixture of water at 140 °C is now heated until the mixture reaches the critical state. Determine the mass and the volume of liquid before the heating process

Homework Equations


v = vf + x*vfg
x = mvapor/mtotal
v = V/m

The Attempt at a Solution


I do not understand how to find the mass before the heating process :(
I found the X which is 0,004086
vf at 140C is 0,001080m3/kg
vg at 140C is 0,5089m3/kg

the final answer should be mass = 157,83 kg
There seems to be information missing. Have you provided the complete problem statement?
 
Chestermiller said:
There seems to be information missing. Have you provided the complete problem statement?
yes everything is there, unless if you do not have access to the saturated water tables! :O For the critical state, the value of v is 0,003155m3/kg
 
hazzz said:
yes everything is there, unless if you do not have access to the saturated water tables! :O For the critical state, the value of v is 0,003155m3/kg
If it can be heated to exactly the critical state, then you can use this specific volume value for the critical state to calculate the total mass of water in the tank. What is that total mass of water in the tank?

EDIT: Wait a second. If you determined X, then you must already have done this. X is the mass fraction of water vapor. What is the mass fraction of liquid water? Given that you know the total mass of water in the tank and the mass fraction of liquid water, what is the mass of liquid water?

Chet
 
Chestermiller said:
If it can be heated to exactly the critical state, then you can use this specific volume value for the critical state to calculate the total mass of water in the tank. What is that total mass of water in the tank?

Chet
Yes i got 158,48kg :) but what i am suppose to do after finding this value? Is this the total mass of water in the tank after or before the heating process ?
 
Chestermiller said:
If it can be heated to exactly the critical state, then you can use this specific volume value for the critical state to calculate the total mass of water in the tank. What is that total mass of water in the tank?

EDIT: Wait a second. If you determined X, then you must already have done this. X is the mass fraction of water vapor. What is the mass fraction of liquid water? Given that you know the total mass of water in the tank and the mass fraction of liquid water, what is the mass of liquid water?

Chet
I don't know :( In the solution book they wrote:

mf = (1-xf) * mtotal
where total = 158,48 kg but i don't know where they took 1-xf and what is xf
 
hazzz said:
Yes i got 158,48kg :) but what i am suppose to do after finding this value? Is this the total mass of water in the tank after or before the heating process ?
It is both. The mass of water in the tank has not changed.
 
hazzz said:
I don't know :( In the solution book they wrote:

mf = (1-xf) * mtotal
where total = 158,48 kg but i don't know where they took 1-xf and what is xf
xf is what you called X in your original post. It is the mass fraction of water vapor in the tank. So, 1-xf is the mass fraction of liquid water in the tank. So, if you know the total mass of water in the tank and the mass fraction of this water that is liquid, you can find the mass of liquid water.
 
Chestermiller said:
xf is what you called X in your original post. It is the mass fraction of water vapor in the tank. So, 1-xf is the mass fraction of liquid water in the tank. So, if you know the total mass of water in the tank and the mass fraction of this water that is liquid, you can find the mass of liquid water.
but where does the 1 come from? I thought the equation was X = mvapor/mtotal so if my X is equal to 0,004086 and my mtotal is 158,48kg how do they get 157,83kg ?
 
  • #10
hazzz said:
but where does the 1 come from? I thought the equation was X = mvapor/mtotal so if my X is equal to 0,004086 and my mtotal is 158,48kg how do they get 157,83kg ?
The sum of the mass fractions has to add up to 1. $$\frac{mvapor}{mtotal}+\frac{mliquid}{mtotal}= \frac{(mvapor + mliquid)}{mtotal}=1$$If the mass fraction of vapor is 0.004086, then the mass fraction of liquid must be 1-0.004086.
 
  • #11
Chestermiller said:
The sum of the mass fractions has to add up to 1. $$\frac{mvapor}{mtotal}+\frac{mliquid}{mtotal}= \frac{(mvapor + mliquid)}{mtotal}=1$$If the mass fraction of vapor is 0.004086, then the mass fraction of liquid must be 1-0.004086.
OHHHH okay! I understand now! :D Thank you very much! You are the best!
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
1K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
Replies
1
Views
5K
  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 16 ·
Replies
16
Views
32K
  • · Replies 11 ·
Replies
11
Views
10K
  • · Replies 3 ·
Replies
3
Views
6K