Thermodynamic cycles

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(1)

SNU NAOE Y. Lim

Thermodynamic cycles

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SNU NAOE Y. Lim

Carnot cycle on T-s diagram

Temper atur e T

Entropy s

𝛿𝑞

𝑟𝑒𝑣

= 𝑇𝑑𝑠

12 Isothermal Expansion at T

H

23 Adiabatic Expansion T

H

T

C

34 Isothermal Compression at T

C

41 Adiabatic Compression T

C

T

H

Pressur e P

Volume v

𝑑𝑠 = 𝛿𝑞

𝑟𝑒𝑣

𝑇

𝛿𝑤

𝑟𝑒𝑣

= −𝑃𝑑𝑣

(3)

SNU NAOE Y. Lim

Carnot cycle on T-s diagram

3

T

Entropy s 𝑞

𝐶

3 4

𝛿𝑞

𝑟𝑒𝑣

= 𝑇𝑑𝑠

Isothermal (ΔT=0)

12 Isothermal Expansion at T

H

34 Isothermal Compression at T

C

T

Entropy s 𝑞

𝐻

1 2

𝛿𝑞

𝑟𝑒𝑣

= 𝑇𝑑𝑠

TH TC

Pressure P

Volume v

(4)

SNU NAOE Y. Lim

Carnot cycle on T-s diagram

4

Adiabatic and reversible  isentropic (q

rev

=0Δs=0)

23 Adiabatic Expansion T

H

T

C

41 Adiabatic Compression T

C

T

H

T

Entropy s

3 2

4 1

TC TH

Pressure P

Volume v

(5)

SNU NAOE Y. Lim

Carnot cycle on T-s diagram

5

T

Entropy s

3 2

4 1

𝑞

𝑛𝑒𝑡

= 𝑞

𝐻

+ 𝑞

𝐶

𝛿𝑞

𝑟𝑒𝑣

= 𝑇𝑑𝑠

12 Isothermal Expansion at T

H

23 Adiabatic Expansion T

H

T

C

34 Isothermal Compression at T

C

41 Adiabatic Compression T

C

T

H

TC TH

Pressur e P

Volume v

𝛿𝑤

𝑟𝑒𝑣

= −𝑃𝑑𝑣

(6)

SNU NAOE Y. Lim

Practical problems of Carnot cycle

• 12 boiler: sat. liquid to sat. vapor

• 23 turbine: maybe OK but large amount of liquid flowing into the turbine can cause physical damage

• 34 condenser: how to stop at 4?

• 41 compressor: how to compress L/G mixture?

6

T

Entropy s

3 2

4 1

𝑞 = 𝑞

𝐻

+ 𝑞

𝐶

Liquid

Vapor

(7)

SNU NAOE Y. Lim

Rankine cycle

• 12 Isobaric expansion at P

H

(T change): boiler

• 23 Adiabatic expansion: turbine

• 34 Isobaric and isothermal compression:

• 41 Adiabatic compression: pump (for 100% liquid)

7

T

Entropy s

3 2

4 1

cooled Sub- Liquid

Superheate d Vapor

Saturated Liquid

Saturated Vapor with

less liquid 𝑞

𝐻

at P

H

𝑞𝐶 at PC

William JM Rankine (1820 –1872)

http://en.wikipedia.org/wiki/Rankine_cycle

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SNU NAOE Y. Lim

Efficiency of Rankine cycle

• Δ𝑢 𝑐𝑦𝑐𝑙𝑒 = 𝑞 𝑛𝑒𝑡 + 𝑤 𝑛𝑒𝑡

• 𝜂 = 𝑤

𝑛𝑒𝑡

𝑞

𝐻

= 𝑞

𝑛𝑒𝑡

𝑞

𝐻

8

T

Entropy s

3 2

4 1

𝑞

𝐻

T

Entropy s

3 2

4 1

𝑞

𝑛𝑒𝑡

(9)

SNU NAOE Y. Lim

Ideal Rankine cycle example 3.14

• Ideal Rankine cycle

(100% efficiency turbine and pump)

9

(1)

(2) (3)

(4)

(1)

(3) (2) (4)

𝜂𝑅𝑎𝑛𝑘𝑖𝑛𝑒 = 𝑤

𝑞𝐻 = 1108

3196 = 34.7%

600℃

10MPa

99.6℃

0.1MPa

99.6℃

0.1MPa

Saturated water v.f. 0.924 Superheated steam

Subcooled water

100.3℃

10MPa

(10)

SNU NAOE Y. Lim

Ts diagram

10

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SNU NAOE Y. Lim

Real Rankine cycle example 3.15

• Rankine cycle with 85% efficiency turbine and pump

600℃

10MPa

99.6℃

0.1MPa

99.6℃

0.1MPa

Saturated water v.f. 0.999 Superheated steam

Subcooled water

100.8℃

10MPa

𝜂𝑅𝑎𝑛𝑘𝑖𝑛𝑒 = 𝑤

𝑞𝐻 = 938.5

3194 = 29.4%

(1)

(3) (2) (4)

𝜂𝑅𝑎𝑛𝑘𝑖𝑛𝑒 = 𝑤

𝑞𝐻 = 1108

3196 = 34.7%

(12)

SNU NAOE Y. Lim

Carnot cycle and Rankine cycle

• Ex 3.14

12

T

Entropy s

3 2

4 1

𝑞

𝐻

at P

H

𝑞𝐶 at PC

1 4

2

3

𝑞𝐻

𝑤𝑜𝑢𝑡

𝑞𝐶

1 2

4 3

𝑤𝑖𝑛

Carnot cycle

Carnot cycle

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SNU NAOE Y. Lim

Joule-Thomson effect(expansion)

• How does the temperature of gas change when it is forced through a valve or porous plug while kept insulated so that no heat is exchanged with the environment?

13 JP Joule

(1818-1889) W Thomson (Lord Kelvin)

(1824-1907)

𝑃1, 𝑇1 𝑃2, 𝑇2

(throttling process or Joule–Thomson process)

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SNU NAOE Y. Lim

JT process ≈ isenthalpic process

(Assume steady-state, KE&PE small, adiabatic)

• Energy Balance (1 st law, Open system)

𝑑𝑈

𝑑𝑡

= ሶ 𝑄 + ሶ 𝑊

𝑠

+ σ

𝑖𝑛

𝑚 ሶ

𝑖

𝑖

− σ

𝑜𝑢𝑡

𝑚 ሶ

𝑜

𝑜

• Since the gas spends so little time in the valve,

there is no time for heat transfer. Since there is no component, the shaft work is near zero. If the

kinetic energy difference is negligible, for st-st,

• 0 = 0 + 0 + ሶ 𝑚

1

1

− ሶ 𝑚

2

2

• ℎ

1

= ℎ

2

(isenthalpic)

P1, T1, V1, H1 P2, T2, V2, H2

𝑚1 𝑚ሶ2

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SNU NAOE Y. Lim

Refrigeration cycle

15

ሶ𝑄

Air

(Cold:0~4℃)

(Hot:15~30 ℃)

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SNU NAOE Y. Lim

Refrigeration process

• Carnot cycle

16

𝑄𝐻

𝑄𝐶

𝑊𝑖𝑛

𝑊𝑜𝑢𝑡 𝑊𝑠

Boiler Condenser

Turbine

Pump

Compressor

Evaporator (heater) 𝑄𝐶

𝑄𝐻

Expander or valve Condenser

Refrigeration cycle

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SNU NAOE Y. Lim

Refrigeration cycle

17

ሶ𝑄 ሶ𝑄

Cold

(Cold)

Hot Isenthalpic

expansion

(Very Cold)

Heat exchanger

(hot)

Evaporator

(cold vapor)

High P Low P

Low P

Compressor

High P

(Very hot)

Condenser

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SNU NAOE Y. Lim

Condenser Evaporator

ሶ𝑄 ሶ𝑄

ሶ𝑄

ሶ𝑄

ሶ𝑊 High T

High P Vapor

V. Low T Low P Liquid

Low T Low P Vapor

Condenser Evaporator

ሶ𝑄 ሶ𝑄

ሶ𝑊

High T High P Vapor

Low T High P Liquid V. Low T

Low P Liquid

Low T Low P Vapor

(19)

SNU NAOE Y. Lim

Example

• R134a (Refrigerant 134a)

19

1,1,1,2-Tetrafluoroethane Boiling point of −26.3 °C at 1 atm

Condenser

Evaporator

ሶ𝑄𝑒𝑣𝑎𝑝 ሶ𝑄𝑐𝑜𝑛𝑑

ሶ𝑊

High T (70℃) High P (10bar) Vapor

Low T (30℃) High P (10bar) Liquid

V. Low T(-20℃) Low P (1.5bar) Liquid/Vapor

Low T (-20℃) Low P (1.5 bar) Saturated

vapor

(20)

SNU NAOE Y. Lim

Subcooling & Superheating

• Subcooling (Point 44’)

• Superheating (Point 22’)

∴ 𝐶𝑂𝑃 =𝑄𝐶

𝑊𝑐 = 𝑞𝐶 𝑤𝐶

1 2

3 4

1’ 2’

3’

4’ Saturated Liquid Subcooled Liquid

Saturated Vapor

Superheated Vapor

𝑞𝐶

𝑤𝐶

(21)

SNU NAOE Y. Lim

Multi Compression

• Multi Compression

– ∆ℎ

21

> ∆ℎ

21

+ ∆ℎ

43

Condenser

Evaporator

Compressor Throttling (J-T)

Valve

T2 T3

𝑤𝐶 = ∆ℎ

∆ℎ21

∆ℎ21

∆ℎ43

(T3<T2)

(22)

SNU NAOE Y. Lim

Simple liquefaction process

22

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SNU NAOE Y. Lim

Linde-Hampson liquefaction process

• (1895)

23

http://ethesis.nitrkl.ac.in/1466/1/PROCESS_DESIGN.p df

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SNU NAOE Y. Lim

Example 3.16

• n=? (0.73 mol/s in textbook)

• COP=? (3.22 in textbook)

24

PR

0.72mol/s 3.21

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SNU NAOE Y. Lim

Exergy

• Energy is always conserved, but not all

that energy is available to do useful work.

• How can we define an useful energy that we can use? Exergy

25

Hot T

h

Cold T

c

Carnot Cycle

Useful energy

Not useful energy (=heat loss) Energy 100

Steam table Internal energy u

Enthalpy h (kJ/kg) P=100kPa, T=200℃ 2658.0 2875.3

P=10MPa, T=325℃ 2610.4 2809.0 50

50

(26)

SNU NAOE Y. Lim

Exergy

• Exergy

– Available energy to be used.

– The maximum useful work possible during a process that brings the system into equilibrium with its surroundings.

– The useful energy based on the surroundings condition as the reference.

26

Reversible Adiabatic Expansion

𝑃0, 𝑇0

𝑃1, 𝑇1 𝑃0, 𝑇2

𝑃0, 𝑇0

𝑤 Still you can produce more work when the temperature T2>T0 !

Carnot Cycle

𝑤𝑐𝑎𝑟𝑛𝑜𝑡 𝑐𝑦𝑐𝑙𝑒

𝑇0

𝑃0, 𝑇0

Dead state

(no more useful work)

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SNU NAOE Y. Lim

Exergy

• Exergy b at state 1 can be defined as:

• If the macroscopic kinetic energy and potential energy are not negligible:

27

𝑏

1

≡ 𝑢

1

− 𝑢

0

+ 𝑃

0

𝑣

1

− 𝑣

0

− 𝑇

0

𝑠

1

− 𝑠

0

Available internal energy

(comparing to the dead state)

Available PV work Entropy loss or heat loss

(or Available chemical energy)

𝑏

1

= 𝑢

1

− 𝑢

0

+ 𝑃

0

𝑣

1

− 𝑣

0

− 𝑇

0

𝑠

1

− 𝑠

0

+ 𝑉 ത

12

2 − 𝑉 ത

02

2 + 𝑔 𝑧

1

− 𝑧

0

= 𝑢1 − 𝑢0 + 𝑃0 𝑣1 − 𝑣0 − 𝑇0 𝑠1 − 𝑠0 + 𝑉ത12

2 + 𝑔𝑧1

(28)

SNU NAOE Y. Lim

Exergy in an open system

– This exergy in an open system is called as

“exthalpy 𝑏 𝑓 ,” but many people use just exergy for both meaning.

28

𝑏

1

= 𝑢

1

− 𝑢

0

+ 𝑃

0

𝑣

1

− 𝑣

0

− 𝑇

0

𝑠

1

− 𝑠

0

+ 𝑃

1

− 𝑃

0

𝑣

1

= 𝑢

1

+ 𝑃

1

𝑣

1

− 𝑢

0

+ 𝑃

0

𝑣

0

− 𝑇

0

𝑠

1

− 𝑠

0

= ℎ

1

− ℎ

0

− 𝑇

0

𝑠

1

− 𝑠

0

(29)

SNU NAOE Y. Lim

Example 3.18

29

Q

𝑚𝑎𝑖𝑟 = 30 kg/min 𝑇1 = 285 K

𝑃1 = 1 bar 𝑃2 = 1 bar

𝑚𝑠𝑡𝑒𝑎𝑚 = 3 kg/min 𝑥3 = 0.9

𝑃3 = 10 bar 𝑥4 = 0.2

𝑃4 = 10 bar Ex 3.18 (exergy loss)

814.4 kJ/min (SRK)

(30)

SNU NAOE Y. Lim

Example 3.18

30 From h,s: -794 kJ/min

From ex: -809kJ/min From h s:

From ex: -813/-839

(31)

SNU NAOE Y. Lim

Exergy analysis

0

2479.7 910.9

794.4

1568.8 774.4

2479.7 910.9+794.4=1705.3

774.4

Exergy loss through heat exchanger

(32)

SNU NAOE Y. Lim

Exergy Analysis

32 http://exergy.se/goran/thesis/paper1/paper1.html

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