Thermodynamics - Internal Energy to Find Heat Added

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Discussion Overview

The discussion revolves around a thermodynamics problem involving the calculation of heat added to a closed system containing water as it transitions from one state to another under quasi-static conditions. Participants explore the relationships between pressure, volume, specific volume, internal energy, and the heat transfer involved in the process.

Discussion Character

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

Main Points Raised

  • Participants present a series of calculations to determine the internal energy at two states based on pressure and specific volume, using steam tables.
  • Some participants propose using the No Flow Energy Equation (NFEE) to find the heat added, expressed as Qin = U2 - U1.
  • One participant questions the calculation of work done during the process, suggesting the need to integrate the pressure-volume relationship to find work accurately.
  • Another participant confirms understanding of vapor quality and its importance in determining internal energy from steam tables.
  • There is a discussion about the specific volume at the critical point and how to determine it using the properties of saturated liquid and vapor.
  • Some participants express uncertainty about their calculations and seek clarification on the work done during the process.

Areas of Agreement / Disagreement

While some participants agree on the approach to use internal energy and vapor quality to find heat added, there is disagreement regarding the calculation of work and the correct application of the integral of PdV. The discussion remains unresolved on the correct method for calculating work.

Contextual Notes

Participants reference steam tables and specific properties of water at various pressures, but there are unresolved assumptions regarding the integration of pressure and volume to calculate work accurately. The discussion also touches on the critical point of water, but specific values and methods remain under debate.

Sirsh
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Homework Statement

[/B]
image.png


Homework Equations


Assumptions[/B]: Closed System and Quasi-static.
p = 90 + 73,386*V -> V = (p-90)/(73,386)
v = V/m
Q - W = U2 - U1
mass balance: m = m1 = m2

The Attempt at a Solution



At state 1:
p1 = 110 kPa
V1 = (110-90)/(73,386) = 2.725*10-4 m3
v1 = V1/m1 = 2.725*10-4 / 0.1 = 2.725*10-3 m3/kg

At state 2:
p2 = 1000 kPa
V2 = (1000-90)/(73,386) = 124.002*10-4 m3
v2 = V2/m2 = 124.002*10-4 / 0.1 = 124.002*10-3 m3/kg

Now I've got two independent properties of the steam at states 1 and 2, pressure and specific volume. I should be able to use these to find their respective internal energies from the steam tables. However, I cannot find where to get these points.. As I'm thinking that by getting the internal energies I can substitute them into the No Flow Energy Equation and find the heat added as there is no work into or out of the system thus:

Qin = U2-U1


Any help in the right direction would be appreciated greatly.
 
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Sirsh said:

Homework Statement

[/B]
image.png


Homework Equations


Assumptions[/B]: Closed System and Quasi-static.
p = 90 + 73,386*V -> V = (p-90)/(73,386)
v = V/m
Q - W = U2 - U1
mass balance: m = m1 = m2

The Attempt at a Solution



At state 1:
p1 = 110 kPa
V1 = (110-90)/(73,386) = 2.725*10-4 m3
v1 = V1/m1 = 2.725*10-4 / 0.1 = 2.725*10-3 m3/kg

At state 2:
p2 = 1000 kPa
V2 = (1000-90)/(73,386) = 124.002*10-4 m3
v2 = V2/m2 = 124.002*10-4 / 0.1 = 124.002*10-3 m3/kg

Now I've got two independent properties of the steam at states 1 and 2, pressure and specific volume. I should be able to use these to find their respective internal energies from the steam tables. However, I cannot find where to get these points.. As I'm thinking that by getting the internal energies I can substitute them into the No Flow Energy Equation and find the heat added as there is no work into or out of the system thus:

Qin = U2-U1Any help in the right direction would be appreciated greatly.

Well, do you have steam tables for saturated water?

If you check these tables for saturated water, you should find a specific volume for 100% vapor and 100% liquid at a given pressure. Where does your v1 fall within these two extremes?

Do you understand what vapor quality means?

You should be able to determine the vapor quality of the steam at the indicated pressures, and then use this information to find the internal energy.
 
SteamKing said:
Well, do you have steam tables for saturated water?

If you check these tables for saturated water, you should find a specific volume for 100% vapor and 100% liquid at a given pressure. Where does your v1 fall within these two extremes?

Do you understand what vapor quality means?

You should be able to determine the vapor quality of the steam at the indicated pressures, and then use this information to find the internal energy.

I do have the steam tables for saturated water, and yes I understand what vapour quality means. I have found the specific volume at 100% liquid and vapour, and respective vapour qualities at both p1 and p2.

At p1:
v1 = 2.725*10-3 m3/kg
vf = 0.1046x10-2 m3/kg
vg = 1.549 m3/kg
x = 0.0011

At p2:
v2 = 124.002x10-3 m3/kg
vf = 0.1127x10-2 m3/kg
vg = 0.1944 m3/kg
x = 0.6357

Am I understanding what you're saying correctly, by knowing the vapour quality at both p1 and p2, I can then use the internal energy at 100% vapour and 100% liquid with the vapour quality to find the amount of internal energy at both points p1 and p2?

As such:

At p1 (110 kPa):
uf = 429 kJ/kg
ug = 2510 kJ/kg
u1 = uf + x*(ug-uf) = 429 + 0.0011(2510-429) = 431.2891 kJ/kg
U1 = u1*m1 = 431.2891 kJ/kg * 0.1 kg = 43.1289 kJ

At p2 (1000 kPa):
uf = 762 kJ/kg
ug = 2584 kJ/kg
u2 = uf + x*(ug-uf) = 762 + 0.6357*(2584-762) = 1920.2452 kJ/kg
U2 = u2*m2 = 1920.2452 kJ/kg * 0.1 kg = 192.0245 kJ

Therefore using the NFEE:

Qin = U2 - U1 = 192.0245 - 43.1289 = 148.8956 kJ or 149 kJ (3 SF).
 
Yes, you understand my point exactly.
 
SteamKing said:
Yes, you understand my point exactly.

Thank you for your help once again.

If you don't mind, would be able to help me understand how I could get the specific volume for steam as it passes through critical point?

The information given is: Sealed system of volume 10-4 m3, containing liq/vap mixture in equilibrium at 100 kPa. It states that "The fraction of liquid and vapour is such that when heated the steam passes through the Critical Point". I need to find this specific volume at CP to find the initial dryness fraction.

I'm assuming that:

v(critical point) = vf + x(vg-vf), x = [(v(critical point)-vf)/(vg-vf)].
 
Sirsh said:
Thank you for your help once again.

If you don't mind, would be able to help me understand how I could get the specific volume for steam as it passes through critical point?

The information given is: Sealed system of volume 10-4 m3, containing liq/vap mixture in equilibrium at 100 kPa. It states that "The fraction of liquid and vapour is such that when heated the steam passes through the Critical Point". I need to find this specific volume at CP to find the initial dryness fraction.

I'm assuming that:

v(critical point) = vf + x(vg-vf), x = [(v(critical point)-vf)/(vg-vf)].

The critical point of water is where liquid and vapor can coexist, but not as a mixture:

https://en.wikipedia.org/wiki/Critical_point_(thermodynamics)

For water, the critical point occurs at the end of the saturation line, where P = 22,064 kPa and T = 373.95° C. If you check your steam tables at this point, you will see that vf = vg
 
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Homework Statement


A cylinder fitted with a piston contains 0.1 kg of H2O at 110 kPa. Heat is added in a quasi-static process, during which pressure p (kPa) is related to volume V (m3) by

p = 96 + 70,968*V

until the pressure is 1000 kPa. Determine the heat added in kJ, to three significant figures.

Homework Equations


Assumptions[/B]: Closed System and Quasi-static.
p = 90 + 70,968*V -> V = (p-90)/(70,968)
v = V/m
Q - W = U2 - U1
mass balance: m = m1 = m2

The Attempt at a Solution



p1 = 110 kPa
V1 = (110-96)/(70968) = 1.97*10-4 m3
v1 = V1/m1 = 1.97*10-4 / 0.1 = 1.97*10-3 m3/kg

p2 = 1000 kPa
V2 = (1000-96)/(70968) = 127*10-4 m3
v2 = V2/m2 = 127*10-4 / 0.1 = 127*10-3 m3/kg

At p1:
v1 = 1.97*10-3 m3/kg
vf = 0.1046x10-2 m3/kg
vg = 1.549 m3/kg
x = 0.0006

At p2:
v2 = 127x10-3 m3/kg
vf = 0.1127x10-2 m3/kg
vg = 0.1944 m3/kg
x = 0.6512

At p1 (110 kPa):
uf = 429 kJ/kg
ug = 2510 kJ/kg
u1 = uf + x*(ug-uf) = 429 + 0.0006(2510-429) = 430.2486 kJ
U1 = u1*m1 = 430.2486kJ/kg * 0.1 kg = 43.0249 kJ

At p2 (1000 kPa):
uf = 762 kJ/kg
ug = 2584 kJ/kg
u2 = uf + x*(ug-uf) = 762 + 0.6512*(2584-762) = 1948.4864 kJ/kg
U2 = u2*m2 = 1948.4864kJ/kg * 0.1 kg = 194.8486 kJ

Qin = U2 - U1 = 194.8486 - 43.0249= 151.8237 kJ or 152 kJ (3 SF).

I have been following Sirsh's step with my own values given but i didnt manage to get it correctly.
 
What about the work? You know, the integral of PdV.

Chet
 
Last edited:
Chet, i still get the wrong answer. Q= (U2-U1) + W
whereas W= integral PdV = P(V2-V1)

Based on Sirsh calculation for W = 110( 124.002*10-4 m3 - 2.725*10-4 m3 ) =1.334 kJ
Is my calculation for W correct?
 
  • #10
JackNoRose said:
Chet, i still get the wrong answer. Q= (U2-U1) + W
whereas W= integral PdV = P(V2-V1)

Based on Sirsh calculation for W = 110( 124.002*10-4 m3 - 2.725*10-4 m3 ) =1.334 kJ
Is my calculation for W correct?

That isn't the integral of PdV.

Integrate the function given in the question to find the volumes for state 1 and 2.
 

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