First law of thermodynamics applied to a closed system

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SUMMARY

The forum discussion revolves around the application of the first law of thermodynamics to a closed system containing water in a rigid steel tank. The initial state (state-1) features a saturated liquid-vapor mixture at 100 kPa, with 75% of the mass in liquid form. The calculations involve determining the average specific volume and total internal energy at state-1, and the specific volume at state-2 after complete vaporization. Key equations used include the average specific volume equation (yavg=yf+x(yfg)) and the internal energy change equation (ΔU=mΔu).

PREREQUISITES
  • Understanding of thermodynamic properties of water, including saturated liquid and vapor states.
  • Familiarity with steam tables for water at various pressures and temperatures.
  • Knowledge of the first law of thermodynamics and its application to closed systems.
  • Ability to perform calculations involving specific volume and internal energy.
NEXT STEPS
  • Study the use of steam tables for calculating thermodynamic properties of water.
  • Learn about the implications of vapor quality in thermodynamic calculations.
  • Explore the first law of thermodynamics in greater detail, particularly in closed systems.
  • Investigate the effects of pressure changes on specific volume and internal energy in phase transitions.
USEFUL FOR

Students and professionals in mechanical engineering, particularly those focusing on thermodynamics, heat transfer, and fluid mechanics. This discussion is beneficial for anyone involved in energy systems or thermal analysis of closed systems.

Luchekv
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Homework Statement


- A well insulated rigid steel tank contains water (only)
- A resistance heating element (240V, 3.67Amp) designed to heat the water as well as inlet and outlet valves for - the water.
- At state-1, the water is in a saturated liquid-vapor mixture state at P(abs)=100kPa, measures 5kg and the portion of it in liquid form is 75%, the rest being vapor
- To reach state-2, more power is supplied to the heater, but just enough to vaporize all the remaining liquid. When all the liquid is vaporized, the heater is switched off.

1.) Calculate the (average) specific volume (units: m3 /kg) for the water at state-1.
2.) Calculate the total internal energy (units: kJ) for the water at state-1.
3.) Calculate the (average) specific volume (units: m3 /kg) for the water at state-2.

Homework Equations


yavg=yf+x(yfg)
ΔU=mΔu

The Attempt at a Solution



1.) [/B]I went to the tables and read of the vf and vg data @ 100kPa. For quality, seeing as 75% is liquid that would make it 0.35. I applied that data to the following equation: yavg=yf+x(yfg) and got:
vavg = 0.001043+0.25(1.6941-0.001043) = 0.4243 m^3/kg

2.) I did the same thing for this part I read the data for uf and ufg from the tables @ 100kPa and applied it to this equation: yavg=yf+x(yfg) and got:
uavg=417+0.25(2088.2) =939.65
Which I then plugged into ΔU=mΔu, seeing as its the same state I only needed uavg:
U= 5kg*1147.87 = 4695.25 kJ

3.) In the statement it says at state 2 all the water is vaporized which would make it a quality of 1 and it would be a saturated vapor..so all I did was read of the value for saturated vapor at 100kPa which is 1.6941m^3/kg

Wasn't 100% on a few of these, any feed back would be great :)
Thank you in advance
 
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Luchekv said:

Homework Statement


- A well insulated rigid steel tank contains water (only)
- A resistance heating element (240V, 3.67Amp) designed to heat the water as well as inlet and outlet valves for - the water.
- At state-1, the water is in a saturated liquid-vapor mixture state at P(abs)=100kPa, measures 5kg and the portion of it in liquid form is 75%, the rest being vapor
- To reach state-2, more power is supplied to the heater, but just enough to vaporize all the remaining liquid. When all the liquid is vaporized, the heater is switched off.

1.) Calculate the (average) specific volume (units: m3 /kg) for the water at state-1.
2.) Calculate the total internal energy (units: kJ) for the water at state-1.
3.) Calculate the (average) specific volume (units: m3 /kg) for the water at state-2.

Homework Equations


yavg=yf+x(yfg)
ΔU=mΔu

The Attempt at a Solution



1.) [/B]I went to the tables and read of the vf and vg data @ 100kPa. For quality, seeing as 75% is liquid that would make it 0.35. I applied that data to the following equation: yavg=yf+x(yfg) and got:
vavg = 0.001043+0.35(1.6941-0.001043) = 0.5936 m^3/kg

If 75% is liquid and 35% is vapor, where did the extra 10% come from?
 
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*sigh* Stupid error... how embarrassing haha. Apart from that is the way I went about it correct?
 
Luchekv said:
*sigh* Stupid error... how embarrassing haha. Apart from that is the way I went about it correct?
You seemed to start off OK, but I stopped checking when I got to the quality mistake.
 
I have gone back and fixed the errors.
 
Part 3 is done incorrectly. The pressure is no longer 100 kPa after the water is heated to the point where all the liquid water has vaporized. But, what actually is still constant?

Chet
 
Luchekv said:

Homework Statement


- A well insulated rigid steel tank contains water (only)
- A resistance heating element (240V, 3.67Amp) designed to heat the water as well as inlet and outlet valves for - the water.
- At state-1, the water is in a saturated liquid-vapor mixture state at P(abs)=100kPa, measures 5kg and the portion of it in liquid form is 75%, the rest being vapor
- To reach state-2, more power is supplied to the heater, but just enough to vaporize all the remaining liquid. When all the liquid is vaporized, the heater is switched off.

1.) Calculate the (average) specific volume (units: m3 /kg) for the water at state-1.
2.) Calculate the total internal energy (units: kJ) for the water at state-1.
3.) Calculate the (average) specific volume (units: m3 /kg) for the water at state-2.

Homework Equations


yavg=yf+x(yfg)
ΔU=mΔu

The Attempt at a Solution



1.) [/B]I went to the tables and read of the vf and vg data @ 100kPa. For quality, seeing as 75% is liquid that would make it 0.35. I applied that data to the following equation: yavg=yf+x(yfg) and got:
vavg = 0.001043+0.25(1.6941-0.001043) = 0.4243 m^3/kg

No, you still aren't applying the concept of vapor quality correctly to calculate the properties of a mixture of vapor and liquid.

If x% is vapor, then (100 - x%) will be liquid, and you must calculate the thermo properties of the mixture accordingly.

https://www.engineersedge.com/thermodynamics/steam_tables.htm
 
SteamKing said:
No, you still aren't applying the concept of vapor quality correctly to calculate the properties of a mixture of vapor and liquid.

If x% is vapor, then (100 - x%) will be liquid, and you must calculate the thermo properties of the mixture accordingly.

https://www.engineersedge.com/thermodynamics/steam_tables.htm
Hmmm. It looks correct to me. 0.25x1.6941+0.75x0.001043
What am I missing?

Chet
 
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Chestermiller said:
Hmmm. It looks correct to me. 0.25x1.6941+0.75x0.001043
What am I missing?

Chet
Below is the OP's calculation, Chet. Notice something different w.r.t. your calculation?
Luchekv said:
vavg = 0.001043+0.25(1.6941-0.001043) = 0.4243 m^3/kg
 
  • #10
I looked on the website you supplied and read this "v=vf+x(vfg)" isn't that what I've done? The way I look at it is..if 75% is liquid its closer to the Sat Liq point (0 quality) on the Tv diagram.

Also, me and Chet get the same answer?
Chestermiller said:
Part 3 is done incorrectly. The pressure is no longer 100 kPa after the water is heated to the point where all the liquid water has vaporized. But, what actually is still constant?

Chet

The mass would be constant so then I could use v = V/m ...but I don't have volume so I would then need to rearrange ρ=m/V but I don't have ρ...would it be correct to assume temperature is constant as well? since its only moving along (horizontally) the Tv diagram...because if that's the case I can use the temperature @ 100kPa in the tables to get the density for water for that temperature
 
  • #11
SteamKing said:
Below is the OP's calculation, Chet. Notice something different w.r.t. your calculation?
It's not really different. It's just doing the factoring differently. The result is the same.

Chet
 
  • #12
Chestermiller said:
It's not really different. It's just doing the factoring differently. The result is the same.

Chet
You and the OP are missing the point.

Take a look at the OP's calculation of internal energy:
Luchekv said:
uavg=417+0.25(2088.2) =939.65

The OP is taking 100% of the internal energy of the liquid phase ...
 
  • #13
I'm really confused now haha... I know x = mVapor/mTotal...if 25% is vapor of the 5kg of mass = 1.25kg/5kg = 0.25?

Thank you for your patience guys
 
  • #14
Luchekv said:
I'm really confused now haha... I know x = mVapor/mTotal...if 25% is vapor of the 5kg of mass = 1.25kg/5kg = 0.25?

Thank you for your patience guys
Yes, 25% of the mass is vapor, and 75% of the mass is liquid. When you figure the thermo properties for a liquid-vapor mixture, you've got to use the proportions of each property to find the property for that combination of liquid and vapor.

For example, if you want to calculate the internal energy of this mixture of liquid water and water vapor, you can't use 100% of the internal energy of the liquid and 25% of the internal energy of the vapor, you can only have 75% of the internal energy of the liquid, because only 75% of the sample is liquid ...

That's why I linked to this article:

https://www.engineersedge.com/thermodynamics/steam_tables.htm
 
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  • #15
SteamKing said:
You and the OP are missing the point.

Take a look at the OP's calculation of internal energy:The OP is taking 100% of the internal energy of the liquid phase ...
Oh. I assumed the 2088 was the difference between the liquid and the vapor.

Chet
 
  • #16
I understand what you're saying, but if I'm to follow this equation "v=vf+x(vfg)" and just above I worked the quality out to be 0.25..I don't see what else I'm supposed to use for x ?

2088 is the difference "ufg" (2505.6-417)
 
  • #17
Luchekv said:
I looked on the website you supplied and read this "v=vf+x(vfg)" isn't that what I've done? The way I look at it is..if 75% is liquid its closer to the Sat Liq point (0 quality) on the Tv diagram.

Also, me and Chet get the same answer?

The mass would be constant so then I could use v = V/m ...but I don't have volume so I would then need to rearrange ρ=m/V but I don't have ρ...would it be correct to assume temperature is constant as well? since its only moving along (horizontally) the Tv diagram...because if that's the case I can use the temperature @ 100kPa in the tables to get the density for water for that temperature
Both the mass of water and the volume of the container are constant. So, if that's the situation, how does the average specific volume in case 1 compare with the specific volume of the pure vapor in case 3?

Chet
 
  • #18
Ok..so I have 5kg of water and my spec volume for state 1 is 0.4243m^3/kg...to get the volume of state one that would be m*v1 = 2.1215m^3/kg

considering volume is constant V1=V2...therefore v2=V2/m ---> v2=2.1215/5= 0.4243...which should be the case since no mass has left the system and the volume hasn't changed.

I'm still confused as to what's wrong with my quality equations...is my quality value wrong? or the numbers I've used from the tables? I've used the equation provided. I don't see what is incorrect. How am I using 100% of the liquid phase for u when I've followed the equation uavg=uf +xufg ? :\ was I supposed to multiply 0.75 to uf? if so..then that's not really following the equation.
 
Last edited:
  • #19
OK. So the specific volume of the vapor for case 3 is 0.4243 m^3/kg. From your steam tables, what saturation pressure, temperature, and specific internal energy of the vapor does this correspond to? For 5 kg of this vapor, what is its internal energy?

Chet
 
  • #20
Apologies for the delay, had a mid semester exam to study for,
(Had to interpolate)
Sat Pressure: 439.229kPa (0.46242-0.41392)/(0.4243-0.41392)=(400-450)/(x-450)
Temp: 146.982 degrees Celsius (0.46242-0.41392)/(0.4243-0.41392)=(143.61-147.9)/(x-147.9)
Internal energy (avg): 2556.24 kJ/kg (0.46242-0.41392)/(0.4243-0.41392)=(2553.1-2557.1)/(x-2557.1)

U = m* uavg = 12781.2 kJ
 
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  • #21
I assume by the like, that it is correct haha?
 
  • #22
Luchekv said:
I assume by the like, that it is correct haha?
Yes. I didn't check your arithmetic, but your approach is correct.

Chet
 
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  • #23
Awesome, thank you! - I will double check the interpolation regardless
 

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