Energy Balance for closed rigid container

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Discussion Overview

The discussion revolves around the energy balance for water in a closed, rigid container that is initially saturated vapor at 4 bar and heated to 400°C. Participants explore the implications of the rigid container on the heating process, the determination of internal energy, and the use of steam tables for interpolation of properties.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that the energy balance can be expressed as Q/m = (u2 - u1), where u1 is the internal energy at the initial state.
  • There is uncertainty about how to determine the pressure at state two, with some participants questioning how to interpolate values for u2 from steam tables.
  • Some participants note that the specific volume remains constant throughout the heating process, which raises questions about how to find properties of the superheated vapor.
  • Participants express confusion regarding the choice of pressure tables for interpolation, with differing suggestions on which pressures to use (5 bar and 10 bar vs. 5 bar and 7 bar).
  • One participant highlights the need to search for the constant specific volume when interpolating the superheated vapor tables at 400°C.
  • There is a discussion about the formula used for linear interpolation, with some participants feeling it differs from what they were taught.

Areas of Agreement / Disagreement

Participants generally express confusion and uncertainty regarding the interpolation process and the appropriate use of steam tables. Multiple competing views on the correct approach to interpolation and the choice of pressure tables remain unresolved.

Contextual Notes

Participants mention the dependence on specific steam tables, which may have different tabulations for superheated properties, leading to confusion in the interpolation process.

Who May Find This Useful

This discussion may be useful for students and practitioners dealing with thermodynamics, particularly in understanding energy balances in closed systems and the use of steam tables for property determination.

jdawg
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Homework Statement


Water, initially saturated vapor at 4 bar, fills a closed, rigid container, The water is heated until its temperature is 400C. For the water, determine the heat transfer in KJ per kg of water .

Homework Equations

The Attempt at a Solution


Assume potential and kinetic energy=0, W=0,Steady state

I got the energy balance right: Q/m=(u2-u1)

Found v1 and u1 from steam tables:
specific volume: v1=v2=0.4625 m3/kg
u1 = 2553.6 KJ/kg

I know that state two is super heated, but how do I know what the pressure is?? What values do I interpolate with to get u2??
 
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jdawg said:

Homework Statement


Water, initially saturated vapor at 4 bar, fills a closed, rigid container, The water is heated until its temperature is 400C. For the water, determine the heat transfer in KJ per kg of water .

Homework Equations

The Attempt at a Solution


Assume potential and kinetic energy=0, W=0,Steady state

I got the energy balance right: Q/m=(u2-u1)

Found v1 and u1 from steam tables:
specific volume: v1=v2=0.4625 m3/kg
u1 = 2553.6 KJ/kg

I know that state two is super heated, but how do I know what the pressure is?? What values do I interpolate with to get u2??
Well, you have a closed, rigid container. What does that tell you about the kind of heating process which is going on when the vapor is heated to 400° C?
 
I'm not really sure, the change in internal energy is what causes the increase in heat? Or is there an increase in temperature because the pressure increases?
 
jdawg said:
I'm not really sure, the change in internal energy is what causes the increase in heat? Or is there an increase in temperature because the pressure increases?
Think about the container.

It's rigid, so does its volume change with increasing pressure inside? Does the amount of water inside change while it is being heated?

So what property must be the same at the start of heating and when it is finished?
 
I know the specific volume is constant, but I'm still confused about how to do the interpolating. I guess I'm supposed to use specific volume and internal energy to interpolate? I'm really lost at this part.
 
jdawg said:
I know the specific volume is constant, but I'm still confused about how to do the interpolating. I guess I'm supposed to use specific volume and internal energy to interpolate? I'm really lost at this part.
How can you use the internal energy to interpolate? Isn't that what you're looking for, the internal energy of the vapor after it has been superheated?

If you know the specific volume is constant, what does this tell you about finding the properties of the superheated vapor?
 
I really don't know :( In the solution I have they used the table p=5bar and p=10 bar to interpolate. I don't understand how they knew to use these tables.
 
But then another solution said to use p=5bar and p=7bar instead...
 
jdawg said:
I really don't know :( In the solution I have they used the table p=5bar and p=10 bar to interpolate. I don't understand how they knew to use these tables.

jdawg said:
But then another solution said to use p=5bar and p=7bar instead...

Different steam tables are tabulated differently. Some may tabulate superheated properties for pressure increments of 5 bar, some for closer increments of pressure.

You've answered your own question, and apparently you don't realize it. If the vapor has a constant specific volume, that's the property you search for when interpolating the superheated vapor tables at T = 400° C.
 
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I think both solutions are using the same tables. What you said makes sense, so I just look for the pressure table that has a value close to my constant specific volume at my given temperature.

What formula are they using here though: (0.4397-0.4625)/(0.4397-0.6173)=(2960.9-u2)/(2960.9-2963.2)
I feel like this is different than the formula I was given to use.
 
  • #11
jdawg said:
I think both solutions are using the same tables. What you said makes sense, so I just look for the pressure table that has a value close to my constant specific volume at my given temperature.

What formula are they using here though: (0.4397-0.4625)/(0.4397-0.6173)=(2960.9-u2)/(2960.9-2963.2)
I feel like this is different than the formula I was given to use.
It looks like someone is using linear interpolation to find the value of u for T = 400 °C and v = 0.4625 m3 / kg

https://en.wikipedia.org/wiki/Linear_interpolation

Check the property values in the superheated vapor table.
 
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  • #12
Thanks so much!
 

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