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Thermodynamic System

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Units

M. Koretsky, Engineering and Chemical Thermodynamics, WILEY 2nd2

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Thermodynamic system and properties

• Closed System vs Open system

• Process

3

System State 1 T₁, P₁, v₁ … Surroundings

Properties Boundary

Heat (Q)

Work(W)

System State 1 T₁, P₁, v₁ … Surroundings

Properties Boundary

Heat (Q)

Work(W)

Mass in out

State 1 T₁, P₁, v₁

…

State 2 T₂, P₂, V₂

…

Process

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Thermodynamic properties

• Extensive properties: size-dependent

– m (kg), V (volume)…

• Intensive properties: size-independent

– T, P, v

(specific volume, m³/kg)

…

• Mole

– A unit of measurement for amount of substance (A kind of relative mass)

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20℃ 20℃ 20℃

1 kg 1kg 2kg

𝑚𝑜𝑙𝑒 = 𝑚𝑎𝑠𝑠

𝑀𝑊(𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑊𝑒𝑖𝑔ℎ𝑡) 𝐻₂ 2g = 2g

2g/mole = 1 mole (𝑜𝑟 gmol) 𝐻₂𝑂 1 mole = 1 mole ⋅ 18g

mole = 18g

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Thermodynamic properties

• State Postulate

– If you know two independent intensive

properties, you can specify the state(all other intensive properties) of a simple compressible system of a pure substance.

• EX) if you specify 25℃, 1 atm for water, its state is also specified. That is, all other intensive properties are also decided. (ex: density is 1g/cm

³

)

5

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Properties and Potential energy

• Some properties are a kind of index of potential energy.

– Height difference Gravitational potential energy difference

Object drops from high height to low height

– Pressure difference Mechanical potential difference

Fluid flows from high pressure to low pressure

– Temperature difference Thermal potential difference

Heat flows from high temperature to low temperature

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Steady-state and equilibrium

• Steady-state: no change with time

• Equilibrium: no potential difference

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System

ሶ

𝑚 𝑚ሶ

𝑃₁, 𝑇₁ 𝑃₂, 𝑇₂ 𝑄

St.-St.

Equilibriu m

Steady-state: The temperature profile of the system does not change with time

Not equilibrium:

Heat is continuously move out from the system

Thermal potential is not same. (T_sys>T_surr) Fluid is continuously move from in to out

mechanical potential is not same (P₁>P₂)

System 𝑃, 𝑇

Steady-state: The temperature profile of the system does not change with time

Equilibrium

No potential difference 𝑇

𝑇 ≠ 𝑇₁

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PT diagram

• Phase

– A region of material that is chemically uniform (homogeneous) and physically distinct

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metastable zone

More than three phases are possible!

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Supercritical fluid

• A distinct liquid and gas phases do not exist

– compressible like a gas, and dissolve materials like a liq

* Matthieumarechal at en.wikipedia, http://www.greener- 9

industry.org.uk/pages/superCO2/3superCO2_coffee.htm

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Supercritical fluid

10

300 bar

50℃ 400℃

0 200 400 600 800 1000 1200

0 50 100 150 200 250 300 350 400 450

Density (kg/m3)

Temperature (℃)

300bar 1 atm

𝑃

𝑣

100°C isotherm

C1 C2

C1

C2

50°C

400°C isotherm

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Phase equilibrium for a pure substance

11

Phase Equilibrium can exist only on the line

P=50 kPa T=20℃

𝑃

𝑇 2.34

20℃

101. 35 𝑘𝑃𝑎 (1 atm)

100℃

P=1 kPa T=20℃

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Phase equilibrium of a pure substance

• Let’s imagine water in a cup in vacuum room at const T

• For pure substance, saturation pressure is

– P at which a pure substance boils at a given T.

– P exerted by the vapor that escapes from the liquid at a given T.

– P at which vaporization rate is same as condensation rate.

– P at phase equilibrium

𝑃

𝑇

2.34 kPa

20℃

P_room≈0 (very low) T=20℃

T=20℃ T=20℃

𝑃_{𝑟𝑜𝑜𝑚}(≈ 0) < 𝑃_𝐻2𝑂^𝑠𝑎𝑡(= 2.34 𝑘𝑃𝑎) 𝑃 < 𝑃_𝐻2𝑂^𝑠𝑎𝑡 = 2.34 𝑘𝑃𝑎 𝑃 = 2 .34 𝑘𝑃𝑎 = 𝑃_𝐻2𝑂^𝑠𝑎𝑡 𝑃_𝑐𝑢𝑝 > 𝑃_𝐻2𝑂^𝑠𝑎𝑡

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P-T diagram of pure substance

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* Matthieumarechal at en.wikipedia

Sub- cooled

liquid

Super- heated

vapor Saturated

vapor

𝑃

𝑇 𝑃₁

34 1

𝑇₁

2

𝑇₂ 3

𝑃₁

𝑇₂ 1

𝑃₁

𝑇₁

5 𝑃₁

𝑇₃

𝑇₃

5

2 𝑃₁

Saturated liquid

Saturated liquid & vapor

𝑇₂ 4 𝑃₁

𝑇₂

Saturation curve

𝑇₂

Boiling water with constant pressure

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How to get the properties?

• Ex) You want to know saturation pressure of water at different temperature

• Do experiments

steam table

• Interpolation

• Make a correlation

Antoine equation (Appendix A)

𝑃 (Bar)

𝑇 ℃

81.3

0.5 1

99.6

2

120.2

Antoine Equation

ln 𝑃^𝑠𝑎𝑡[𝑏𝑎𝑟] = 𝐴 − 𝐵 𝑇[𝐾] + 𝐶

𝑦 𝑦₂

𝑦₁

𝑥₁ 𝑥 𝑥₂ 𝑦 − 𝑦₁

𝑥 − 𝑥₁ = 𝑦₂ − 𝑦₁ 𝑥₂ − 𝑥₁ 𝑦 = 𝑦₂ − 𝑦₁

𝑥₂ − 𝑥₁ 𝑥 − 𝑥₁ + 𝑦₁

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Pv Diagram (Pv phase diagram)

• Ideal Gas

– Molecules have zero-size – No interactive force

– No condensed phase

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𝑃 = 𝑅𝑇 𝑣

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P-v diagram

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𝑃

𝑣 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚 𝑇₂

(등온선)

𝑃₁

𝑣_𝑙𝑖𝑞^𝑠𝑎𝑡 𝑣_𝑣𝑎𝑝^𝑠𝑎𝑡

2 3 4 5

Superheated vapor Saturated liquid /

vapor

𝑃₁ 𝑃₁ 𝑃₁ 𝑃₁

12 3 4

𝑇₂ 𝑇₃ 5

𝑇₂ 𝑇₂

Sub-cooled liquid

1

𝑃₁

𝑇₁ 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚 𝑇₃

Boiling water with constant pressure

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Pv diagram

• Real Gas

* OpenStax college, phase changes, Figure 2 17

(http://cnx.org/content/m42218/latest/?collection=col11406/latest)

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P-T and P-v diagram

𝑃

𝑇

1

𝑃₁ 5

𝑇₁ 𝑇₂ 𝑇₃

𝑃

𝑣 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚 𝑇₂

𝑣_𝑙𝑖𝑞^𝑠𝑎𝑡 𝑣_𝑣𝑎𝑝^𝑠𝑎𝑡

1 2 3 4

2,3,4 5

2 3 4 5

Superheated vapor Saturated liquid /

vapor Sub-cooled

liquid

Boiling water with constant pressure 1

𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚 𝑇₁ 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚 𝑇₃

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Lever Rule

• With 𝑣 𝑙𝑖𝑞 𝑠𝑎𝑡 , 𝑣 𝑣𝑎𝑝 𝑠𝑎𝑡 , you can find v of saturated vapor-liquid system from quality x.

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http://www.rmutphysics.com/charud/scibook/crystal- structure/Solid%20solution.htm

𝑜𝑟 𝑥 = 𝑣 − 𝑣_𝑙𝑖𝑞^𝑠𝑎𝑡 𝑣_𝑣𝑎𝑝^𝑠𝑎𝑡 − 𝑣_𝑙𝑖𝑞^𝑠𝑎𝑡

𝑣 = 𝑥 ∙ 𝑣_𝑣𝑎𝑝^𝑠𝑎𝑡 + (1 − 𝑥) ∙ 𝑣_𝑙𝑖𝑞^𝑠𝑎𝑡

𝑃

𝑣_𝑙𝑖𝑞^𝑠𝑎𝑡 𝑣_𝑣𝑎𝑝^𝑠𝑎𝑡 𝑥: 1 − 𝑥

𝑣 𝑥 = 𝑛_𝑣

𝑛_𝑣 + 𝑛_𝑙 𝑣𝑎𝑝𝑜𝑟 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑟 𝑚_𝑣

𝑚_𝑣 + 𝑚_𝑙 (𝑣𝑎𝑝𝑜𝑟 𝑚𝑎𝑠𝑠 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛)

𝑇₂

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P-v-T diagram (typical)

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http://www.chm.bris.ac.uk/~chdms/Teaching/Chemical_Interactions/page_10.htm

Pv diagram PT diagram

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PvT diagram

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* Greg Holland, Oklahoma city community college (http://occc.edu/gholland/)

Pv diagram

PT diagram

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Gibbs Phase Rule

– How many variables (properties) can you select to define a phase?

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* Matthieumarechal at en.wikipedia

𝐹 = 𝐶 − 𝜋 + 2

= 1 - 3 + 2 =0 𝐹 = 𝐶 − 𝜋 + 2

= 1 - 1 + 2 = 2

𝐹 = 𝐶 − 𝜋 + 2

= 1 - 2 + 2 = 1

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State Postulate / Gibbs Phase rule

• “The state of a simple compressible system is completely specified by two independent,

intensive properties”

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Cengel and Boles,Thermodynamics: an engineering approach, McGraw-Hill, 2008

𝑣 =?

steam

Gibbs Phase rule 𝐹 = 𝐶 − 𝜋 + 2

= 1 - 1 + 2 = 2

Gibbs Phase rule 𝐹 = 𝐶 − 𝜋 + 2

= 1 - 2 + 2 = 1 𝑣 =?

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Example

• Two phase system of water: F=1

• If you know one intensive property, then you can know all other intensive properties of each phase (liquid and vapor)

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𝑃₁

𝐹 = 𝐶 − 𝜋 + 2

= 1 - 2 + 2 = 1

𝑃

𝑇₁ 𝑃₁

𝑃

𝑣 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚 𝑇 = 𝑇₁

𝑣_𝑙𝑖𝑞^𝑠𝑎𝑡 𝑣_𝑣𝑎𝑝^𝑠𝑎𝑡

Gibbs Phase Rule

𝑣 =?

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Example

• Still you don’t know the state of the system.

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𝑃₁

𝑇₁

𝑃

𝑇₁ 𝑃₁

𝑃

𝑣 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚 𝑇 = 𝑇₁

𝑣_𝑙 𝑣_𝑣 𝑣 =?

𝑣_𝑙 𝑣_𝑣

You need one more independent intensive property, for example, mass vapor fraction.

State Postulate

𝑣 =?

𝑣 = 𝑥 ∙ 𝑣_𝑣 + (1 − 𝑥) ∙ 𝑣_𝑙

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Example

• Still you don’t know the total volume (V) or total mass (m)

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𝑃₁

𝑇₁ 𝑃

𝑇₁ 𝑃₁

𝑃

𝑣 𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚 𝑇 = 𝑇₁

𝑣_𝑙 𝑣_𝑣 𝑣

𝑣_𝑙 𝑣_𝑣

𝑣 𝑉 =?

𝑚 =?

It’s OK. The total volume or mass

does not change the state of the

system.

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Ex1.3

• Estimate the specific volume of water at P=1.4MPa and T=333℃

• v=?

– From interpolation

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ASME (2018) ASME/NBS(2019) 0.1943 0.1942

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Y. Lim 28

0.1995 0.1942

Steam table

Interpolation

Experimental value vs. estimated value 0.2741m³/kg

0.1809 m³/kg

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Ex1.5

• (a)

– v1=?

• 0.4 m

³

/kg (textbook)

– x=?

• 0.079 (textbook)

• (b)

– T2=?

• 145.8 C (textbook)

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19 0.4

18-1 18-2 19 0.079 0.07915

18-1 18-2 149.3 149.3

1L

2.5g water 70℃