ME 331 Thermodynamics II-lecture 4chainarong.me.engr.tu.ac.th/documents/ME331 Thermodynamics...

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Inequality of Clausius 0 net Q T δ Clausius (1865) defined entropy as int rev Q dS T δ = By integrating Clausius inequality, we obtain 2 2 1 int rev 1 Q S S S T δ Δ = = Entropy

Transcript of ME 331 Thermodynamics II-lecture 4chainarong.me.engr.tu.ac.th/documents/ME331 Thermodynamics...

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Inequality of Clausius

0netQT

δ≤∫

Clausius (1865) defined entropy as

int rev

QdSTδ⎛ ⎞= ⎜ ⎟⎝ ⎠

By integrating Clausius inequality, we obtain

2

2 1int rev1

QS S STδ⎛ ⎞Δ = − = ⎜ ⎟⎝ ⎠∫

Entropy

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Change of entropy

0netQT

δ≤∫

From Clausius inequality

A cycle composed of a reversible and an irreversible process

2 1

int rev1 2

0Q QT Tδ δ⎛ ⎞+ ≤⎜ ⎟

⎝ ⎠∫ ∫

( )2

1 21

0Q S STδ

+ − ≤∫2

2 11

QS STδ

− ≥ ∫ andQdSTδ

2

2 11

QS STδ

− = ∫ represents energy transfer with heatFor a reversible process2

2 11

QS STδ

− > ∫For an irreversible process means that some entropy is generated

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The entropy generated during a process is called entropy generation and is denoted as Sgen.

Sgen is always a positive quantity or zero. Its value depends upon the process and thus it is not a property. Sgen is zero for an internally reversible process.

Entropy generation

We can remove the inequality by noting the following

2

2 11

netsys gen

Q kJS S S ST K

δ ⎛ ⎞Δ = − ≥ + ⎜ ⎟⎝ ⎠∫

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Consider an isolated system composed of several subsystems exchanging energy among themselves. Since the isolated system has no energy transfer across its system boundary, the heat transfer across the system boundary is zero.

Applying the definition of entropy to the isolated system

The total entropy change for the isolated system is

0isolatedSΔ ≥

Isolated system

2

1

netisolated

QST

δΔ ≥ ∫

Adiabatic

The entropy of a process never decreases!

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The principal of entropy:

S S S Sgen total sys surr= = + ≥∑Δ Δ Δ 0

where = holds for the totally reversible process> holds for the irreversible process < impossible process

Consider a general system exchanging mass as well as energy with its surroundings.

The total entropy change of an isolated system during a process always increases or, in the limiting case of a reversible process, remains constant.

Total entropy change in an isolated system

Sum of entropy change = Entropy generation

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The total entropy change for a process is the amount of entropy generated during that process (Sgen), and it is equal to the sum of the entropy changes of the system and the surroundings.

The increase in entropy principle can be summarized as follows:

Entropy change is caused by heat transfer and irreversibilities. Heat transfer to a system increases the entropy, and heat transfer from a system decreases it. The effect of irreversibilities is always to increase the entropy.

The entropy for a given system (important or surroundings) may decrease during a process, but the sum of the entropy changes of the system and its surroundings for an isolated system can never decrease.

Summary in entropy change of an isolated system

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1. Processes can occur in a certain direction only, such that Sgen≥0.

2. Entropy is a nonconserved property, and there is no such thing as the conservation of entropy principle. The entropy of the universe is continuously increasing!

3. The performance of engineering systems is degraded by the presence of irreversibilities, and entropy generation is a measure of the magnitudes of the irreversibilities present during that process.

Some Remarks about Entropy

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Example

A heat source at 800 K loses 2000 kJ of heat to a sink at (a) 500 K and (b) 750 K. Determine which heat transfer process is more irreversible.

(a) For heat transfer process to sink at 500 K

2000 2.5 kJ/K800

sourcesource

source

Q kJST K

−Δ = = = −

2000 4 kJ/K500

sinksink

sink

Q kJST K

Δ = = =

2.5 4 1.5 kJ/Kgen total source sinkS S S S= Δ = Δ + Δ = − + =

Entropy generation

Irreversible process!

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Example (cont.)

(b) For heat transfer process to sink at 750 K

2000 2.5 /800

sourcesource

source

Q kJS kJ KT K

−Δ = = = −

2000 2.7 /750

sinksink

sink

Q kJS kJ KT K

Δ = = =

2.5 2.7 0.2 kJ/Kgen total system surroundingsS S S S= Δ = Δ + Δ = − + =

More less irreversible!

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Example: Entropy change of substanceA rigid tank contains 5 kg of refrigerant 134a initially at 20OC and 140 kPa. The refrigerant is now cooled while being stirred until its pressure drops to 100 kPa. Determine the entropy change of the refrigerant during this process.

Assumptions- Volume of the tank is constant; v1 = v2

State 1: 1 1O 3

1 1

140 kPa 1.0532 kJ/kg.K20 C 0.1652 m /kg

P sT v= =⎫

⎬= =⎭

State 2: 32

31 2

0.0007258 m /kg100 kPa0.1917 m /kg

f

g

vPvv v

== ⎫⎬ == ⎭

22

0.1652 0.0007258 0.8610.1916 0.0007258

f

fg

v vx

v− −

= = =−

quality

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Entropy change of the refrigerant during this process:

( ) ( )( )2 1 5 kg 0.8183 1.0532S m s sΔ = − = −

1.175 kJ/kg.K= −

Example: Entropy change of substance (cont.)

- Negative sign indicates that entropy of the system is decreasing during this process!

( )( )2 2 0.0678 0.861 0.9395 0.0678f fgs s x s= + = + −

0.8183 kJ/kg.K=

State 2:

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A piston-cylinder device initially contain 3 lbm of liquid water at 20 psia and 70 OF the water is now heated at constant pressure by the addition of 3450 Btu of heat. Determine the entropy change of the water during this process.

Assumptions- Tank is stationary → ΔKE = ΔPE = 0- Process is quasi equilibrium- P2 = P1

State 1: O

O

1 @70 F1O

1 1 @70 F

0.07463 Btu/lbm.R20 psia70 F 38.09 Btu/lbm

f

f

s sPT h h

≅ == ⎫⎬= ≅ =⎭

Energy balance:

( ) ( ) ( ), ,in out in out mass in mass out sysQ Q W W E E E− + − + − = Δ

( )2 1inQ m h h= −

( )( )23450 3 38.09 Btu/lbmBtu lbm h= −

Example: Entropy change during constant pressure process

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Example: Entropy change during constant pressure process (cont.)

State 2:

2 2

2

100 kPa 1.7759 Btu/lbm.R1188.1 Btu/lbm

P sh

= =⎫⎬= ⎭

( ) ( )( )2 1 3 lbm 1.7759 0.07463S m s sΔ = − = −

5.104 Btu/lbm.R=

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s Sm

s s

=

=2 1

or on a per unit mass basis

The adiabatic, reversible process is a constant entropy process and is called isentropic!

If the process is adiabatic and reversible, the equality holds and the entropy change is

2 1 0S S SΔ = − =

Adiabatic, Reversible (Isentropic) ProcessFor an adiabatic process, one in which there is no heat transfer, the entropy change is

2 1S S=

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The principle of increase of entropy for a closed system exchanging heat with its surroundings at a constant temperature Tsurr is found by using the equation for the entropy generated for an isolated system.

Qout,

sys

A general closed system (a cup of coffee) exchanging heat with its surroundings

SurroundingsTsurr

SystemBoundary

0gen total sys surrS S S S= Δ = Δ + Δ ≥∑

S S m s sQ

Tgen total sysnet surr

surr

= = − + ≥Δ ( ) ,2 1 0

Q Q Q Qnet surr net sys out sys out sys, , , ,( )= − = − − =0where

Example

2 1( )sys sysS S SΔ = −,net surr

surrsurr

QS

TΔ =∑

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ExampleSteam enters an adiabatic turbine at 5 MPa and 450OC and leaves at a pressure of 1.4 MPa. Determine the work output of the turbine per unit mass of steam if the process is reversible.

Assumption - Steady flow - Reversible process - Neglect KE and PE- No heat transfer

( ) ( ) ( ), ,in out in out mass in mass out sysQ Q W W E E E− + − + − = Δ

Energy balance

1 2outmh W mh= +

( )1 2outW m h h= −

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State 1: 1 1

O1 1

5 MPa 6.8186 kJ/kg.K450 C 3316.2 kJ/kg

P sT h

= =⎫⎬= =⎭

State 2: 2

21 2

1.4 MPa2966.6 kJ/kg

Ph

s s= ⎫

=⎬= ⎭

Work output of turbine per unit mass

1 2 3316.2 2966.6 349.6 kJ/kgoutw h h= − = − =

Example (cont.)

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Consider the transfer of heat from a heat reservoir at temperature T to a heat reservoir at temperature T - ΔT > 0 where ΔT > 0, as shown below.

Q

HRatT

HRatT-ΔT

Two heat reservoirs exchanging heat over a finite temperature difference

The second law for the isolated system composed of the two heat reservoirs is

Effect of Heat Transfer on Entropy

S S S SS S S S

gen total sys surr

gen total HR T HR T T

= = + ≥

= = +∑

Δ Δ Δ

Δ Δ Δ Δ

0

@ @

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S SQT

QT Tgen Total= =

−+

+

−Δ

Δ

Now as ΔT → 0, Sgen → 0 and the process becomes totally reversible.Therefore, for reversible heat transfer ΔT must be small. As ΔT gets large, Sgen increases and the process becomes irreversible.

In general, if the heat reservoirs are internally reversible

Effect of Heat Transfer on Entropy

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Find the total entropy change, or entropy generation, for the transfer of 1000 kJ of heat energy from a heat reservoir at 1000 K to a heat reservoir at 500 K.

Q=1000 kJ

HRatT=1000 K

HRatT-ΔT = 500K 0 1 2 S, kJ/K

1000 K

500 K

T Areas= 1000 kJ

The second law for the isolated system is

gen TotalQ QS ST T T− +

= Δ = +−Δ

Example

What happens when the low-temperature reservoir is at 750 K?

The effect of decreasing the ΔT for heat transfer is to reduce the entropy generation or total entropy change of the universe due to the isolated system and the irreversibilities associated with the heat transfer process.

1000 10001000 500

( 1 2)

1

kJ kJK K

kJK

kJK

−= +

= − +

=

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The heat transfer in an internally reversible process is shown as the area under the process curve plotted on the T-S diagram.

For the reversible process, the equation for dS implies that

Heat Transfer as the Area under a T-S Curve

dS QT

Q TdS

net

net

=

=

δ

δor the incremental heat transfer in a process is the product of the temperature and the differential of the entropy, the differential area under the process curve plotted on the T-S diagram.

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Carnot cycle

Give highest thermal efficiency

Constant entropy

Constant entropy

Constant temperature

Composed of two isentropic processes and two isothermal processes

net H LW Q Q= − 1H L Lthermal

H H

Q Q QQ Q−

η = = −

Constant temperature