The T-dS Equations & Diagrams

Meeting 9 Section 4-3 & 4-4

Second Law of Thermodynamics



S



S 2



S 1



1



2



Q T A

This can be viewed as a mathematical statement of the second law (for a closed system).

Entropy is a non-conserved property!

Entropy

s 2



s 1



1



2

   

q T

  

int rev

Units are

kJ kg



K or Btu lbm



R

s = S/m : intensive property

The Entropy Change Between Two Specific States

The entropy change between two specific states is the same whether the process is reversible or irreversible

We can write entropy change as an equality by adding a new term:

 S 

entropy

2 1   Q T

entropy

A

change transfer due to heat

 P s

entropy generation or transfer production

Ps



0 ; not a property

Entropy generation

• Ps  0 is an actual irreversible process.

• Ps = 0 is a reversible process.

• Ps  0 is an impossible process.

Entropy Production

• •

Ps quantifies irreversibilities. The larger the irreversibilities, the greater the value of the entropy production, Ps.

•

A reversible process will have no entropy production, Ps = 0.

Ps depends upon the process, and thus it is not a property.

Entropy transfer and production

S 2  S 1  2 1   Q T  Ps

Entropy change of system as it goes from 1 to 2; can be + – or zero depending on the other two terms.

Entropy transfer into/out of the system via heat transfer; can be + – or zero, depending on Q and its direction Entropy production: > 0 when internal irrerversibilities are present; = 0 when no int irr are present; < 0 never.

Entropy transfer and production

S 2  S 1              0    0 0 0 : : : : Q could because if if if Q Q Q is  is 0 negative be P s and  is P or  always negative s  ; and 0 and if  2 1 2 1         ,   2 1  T T     positive Q Q       Q T  P s     P s  P s

Entropy Increase

–

Entropy change in a general system (ΔS

sys

) may be negative (due to heat transfer out of the system)

–

Entropy production cannot be negative

Ps       0 0 Irreversib Reversible le process process 0 Impossible process

Entropy Transfer

•

Entropy change is caused by heat transfer, mass flow, 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.

Some Remarks about Entropy

• Process can occur in a certain direction only, not in any direction such that

Ps



0

.

• Entropy is a non-conserved property, and there is no such thing as conservation of entropy principle. The entropy of the universe is continuously increasing.

• 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.

The Tds Equations

In previous slides, we developed a new property,

entropy

s 2



s 1



1



2

   

q T

  

int rev Units are kJ kg



K or Btu lbm



R

The entropy of a pure substance is determined from the tables, just as for any other property

It’s just like the other properties we’ve encountered in the tables

It is tabulated just like u, v, and h.

Also,

s



s f



x ( s g



s f )

And, for compressed or subcooled liquids,

s ( T , p )



s f ( T )

T-s Diagram for Water

TEAMPLAY

• • • •

Use the tables in your book Find the entropy of water at 50 kPa and 500 °C. Specify the units.

Find the entropy of water at 100 °C and a quality of 50%. Specify the units.

Find the entropy of water at 1 MPa and 120 °C. Specify the units.

T-s diagram

Recall that the P-v diagram was very important in first law analysis, and that

w

 

Pdv

Work was the area under the curve.

For a T-s diagram

dS

    

Q T

  

int rev

Rearrange:



Q



TdS

Integrate:

Q



1



2 TdS

•

On a T-S diagram, the area under the process curve represents the heat transfer for internally reversible processes

d

Derivation of Tds equations:

For a simple closed system:



Q

 

W



dU

The work is given by:



W



PdV

Substituting gives:



Q



dU



PdV

More derivation….

For a reversible process:

TdS

 

Q

Make the substitution for



Q in the energy equation:

TdS = dU + PdV

Or on a per unit mass basis:

Tds = du + Pdv

Tds Equations

Entropy is a property. The Tds expression that we just derived expresses entropy in terms of other properties. The properties are independent of path….We can use the Tds equation we just derived to calculate the entropy change between any two states:

Tds



du



Pdv

Tds Equations

Starting with enthalpy h = u + Pv, it is possible to develop a second Tds equation:

dh

  

d ( u



Pv ) du Tds

 

Pdv vdP



vdP Tds



dh



vdP

Schematic of an h-s Diagram for Water

Tds equations

•

These two Tds relations have many uses in thermodynamics and serve as the starting point in developing entropy change relations for processes.

Entropy change of an Pure substance

•

The entropy-change and isentropic relations for a process can be summarized as follows: 1 .

Pure substances: Any process:

Δs = s

2

- s

1

[kJ/(kg-K)] Isentropic process:

s 2 = s 1

Let’s look at the entropy change for an incompressible substance:

We start with the first Tds equation:

Tds



C v ( T ) dT



Pdv

For incompressible substances, v



const, so dv = 0.

We also know that C v (T) = C(T), so we can write:

ds



C ( T ) dT T

Entropy change of an incompressible substance

Integrating

s 2



s 1



T 1



T 2 C ( T T ) dT

If the specific heat does not vary with temperature:

s 2



s 1



C ln T 2 T 1

Entropy Change for Incompressible Substance

•

The entropy-change and isentropic relations for a process can be summarized as follows: 2. Incompressible substances: Any process:

s 2



s 1



C av ln T 2 T 1

Isentropic process:

T 2 = T 1

Sample Problem

Aluminum at 100 o C is placed in a large, insulated tank having 10 kg of water at a temperature of 30 o C. If the mass of the aluminum is 0.5 kg, find the final temperature of the aluminum and water, the entropy of the aluminum and the water, and the total entropy of the universe because of this process.

Draw Diagram

water Insulated wall AL Closed system including aluminum and water Constant volume, adiabatic, no work done

Conservation of Energy

Apply the first law

Q



W

 

U sys

 

U W

 

U AL m W C W ( T 2



T 1 ) W



m AL C AL ( T 2



T 1 ) AL



0

But (T

2 ) W

= (T

2 ) AL = T 2

at equilibrium



T 2



m W C W ( T 1 ) W m W C W

 

m AL C m AL C AL AL ( T 1 ) AL

Solve for Temperature

m W = 10 kg, C W = 4.177 kJ/kg .

K m AL = 0.5 kg, C AL = 0.941 kJ/kg .

K T 2



( 10 kg )( 4 .

177 kJ kg



K )( 303 ( 10 kg )( 4 .

177 kJ kg



K K )



)



( 0 .

5 kg )( 0 .

941 kg kJ



K ( 0 .

5 kg )( 0 .

941 kJ / kg



)( K ) 373 K ) T 2 = 303.8 K

Entropy Transfer

Entropy change for water and aluminum



S W

 

( 10 kg m W )( C W 4 .

ln T 2 T 1 , W 177 kJ kg



K ) ln( 303 .

8 K 303 K )



0 .

1101 kJ K



S AL



m AL C AL ln T 2 T 1 , AL



( 0 .

5 kg )( 0 .

941 kg kJ



K ) ln 303 .

8 K 373 K

 

0 .

0966 kJ K

Entropy Generation

Entropy production of the universe

S gen

 

S total

 

S W

 

S AL



0 .

1101 kJ



0 .

0135 K kJ K



0 .

0966 kJ K



0 S gen > 0

:

irreversible process

The Entropy Change of an Ideal Gas

Let’s calculate the change in entropy for an ideal gas

Start with 2 nd

Tds

Tds equation



dh



vdP

Remember dh and v for an ideal gas?

dh



C p dT and v



RT/P

Substituting:

Tds



C p dT



RT dP P

Change in entropy for an ideal gas

Dividing through by T,

ds



C p dT T



R dP P

Don’t forget, C

p = C p (T)

….. a function of temperature!

Entropy change of an ideal gas

Integrating yields

s 2



s 1



T 1



T 2 C p ( T ) dT T



R ln P 2 P 1

To evaluate entropy change, we’ll have to evaluate the integral:

T 1 2



T C p dT T

Entropy change of an ideal gas

•

Similarly it can be shown from

Tds = du + Pdv

that

s 2



s 1



T 1



T 2 C v

(

T

)

dT T



R

ln

v 2 v 1

Entropy change of an ideal gas for constant specific heats: Approximation

•

Now, if the temperature range is so limited that C p



constant (and C v



constant),

2 1



C p dT T



C p ln T 2 T 1 s 2



s 1



C p ln T 2 T 1



R ln P 2 P 1

Entropy change of an ideal gas for constant specific heats: Approximation

•

Note

Tds = du + Pdv

•

Therefore

s 2



s 1



C v ln T 2 T 1



R ln v 2 v 1

Tds Tds

 

du dh



Pdv 0



C vdP 0 P

  

v P

 

C v



const const dT



dT



ds v dT



ds ds dT v P



ds



T P T C v C P T v = const.

P = const.

dT/ds s

Summary: Entropy change of an Pure substance

1. Pure substances: Any process:

Δs = s

2



Isentropic process:

s 1

[kJ/(kg-K)] (Table)

s 2 = s 1

Summary: Entropy Change for Incompressible Substance

2. Incompressible substances: Any process:

s 2



s 1



C av ln T 2 T 1

Isentropic process:

T 2 = T 1

Summary: Entropy Change for Ideal gases

3. Ideal gases:

Constant specific heats (approximate treatment):

s 2



s 1



C v

,

av

ln

T 2 T 1 s 2



s 1



C p

,

av

ln

T 2 T 1



R

ln

v 2 v 1



kJ/(kg



K)

 

R

ln

P 2 P 1



kJ/(kg



K)



TEAMPLAY

:

Air is compressed in a piston-cylinder device from 90 kPa and 20 o C to 400 kPa in a reversible isothermal process. Determine: (a) the entropy change of air, (b) the work done and (c) the removed heat.

T v 2 1 v 1 2 293K 90kPa 293K 400kPa 293 K Air is ideal gas, R = 287 Jkg -1 K -1 Q<0 s

Entropy change for an ideal gas with constant heat : s 2



s 1



C p ln T 2 T 1



R ln P 2 P 1

 

287 ln 400 90

 

428

 

J K

 

Ideal gas isothermal compression work: W comp

 

RT ln P 2 P 1

 

287



298



ln 400 90

 

1 25.4

 

kJ kg

 

Rejected heat (1 st law): Q



W comp

 

U



Q



W comp

 

1 25.4

 

kJ kg

 

•

The entropy of the air decreased due to the heat extraction. Consider the heat is rejected on the

•

environment at 15 o C. Evaluate the entropy change of the environment and the air.

•

The environment is a heat reservoir . The entropy change is



s = Q/T = 125400/288 = + 435 JK -1 .

•

The entropy change of the system plus the

•

environment is therefore:



s =



s system +



s environ = - 428+ 435 = 7 JK -1 .

If the environment is at 20 o C,



s = 0 because both are at the same temperature (reversible heat transfer).

Sample Problem

A rigid tank contains 1 lbm of carbon monoxide at 1 atm and 90°F. Heat is added until the pressure reaches 1.5 atm. Compute: (a) The heat transfer in Btu.

(b) The change in entropy in Btu/R.

Draw diagram:

Rigid Tank => volume is constant CO: m= 1 lbm State 1: P = 1 atm T = 90 o F State 2: P = 1.5 atm Heat Transfer

Assumptions

• • • • • •

CO in tank is system Work is zero - rigid tank kinetic energy changes zero potential energy changes zero CO is ideal gas Constant specific heats

Q Apply assumptions to conservation of energy equation



W

0  

U

0 0

+



KE +



PE

For constant specific heats, we get:

Q = mC v



T 2



T 1



Need T

2



How do we get it?

Apply ideal gas equation of state:

P 1 V 1 P 2 V 2



mRT mRT 2 1

Cancel common terms...

Solve for T

2

:

T 2

  

P P 1 2

 

T 1

  

1.5

1.0

  

90



460



R



825 R

Solve for heat transfer

Q



( 1 lbm



)



0 .

18 Btu lbm R

  

825



550



R Q



49 .

5 Btu

Now, let’s get entropy change ...

For constant specific

S 2



S 1



m

 

heats:

C v ln T 2 T 1



Rln

0

v 2 v 1

 

Since v 2 = v 1

S 2



S 1



( 1 S 2



S 1



m C v ln T 2 T 1 lbm )

 

0 .

18 lb Btu m R

 

ln

 

825 550 R R

 

S 2



S 1



0 .

073 Btu/R

Entropy Change in Some Selected Irreversible Processes

Some Irreversible Processes ONE WAY PROCESSES

PROCESSOS IRREVERSÍVEIS: EFEITO JOULE

V Q R i

Sistema Isolado > não há calor cruzando a fronteira e todo trabalho é transformado em energia interna:

U f  U i  Ri   t f  t i  (1a.

lei)

T f

S f  S i  f   Q  P s (2a.

lei) i T

Todo trabalho elétrico é convertido em energia interna do sistema. A variação de S é devido a Ps, pois



Q=0. Como S não depende do caminho, T f dS = dU =Ri



t -> Ps = Ri



t/T f > 0.

Trabalho elétrico convertido em energia interna aumentou a entropia do sistema isolado. Não é possível converter a mesma quantidade de energia interna em trabalho elétrico.

PROCESSOS IRREVERSÍVEIS: EFEITO JOULE Por que não é possível converter a mesma quantidade de energia interna em trabalho elétrico

?

Do ponto de vista microscópico: o sistema deveria resfriar para diminuir a energia interna e transforma la em energia elétrica . Certamente este não será o estado mais provável de encontrá-lo portanto, esta transformação não será espontânea!

Do ponto de vista macroscópico a entropia do sistema isolado deveria diminuir e isto violaria a 2a. Lei! Note que de (i) -> (f),



S>0 Um processo é reversível quando S não varia entre os estados final e inicial!

PROCESSOS IRREVERSÍVEIS: ATRITO

k x 0 m x

Sistema Isolado > não há calor cruzando a fronteira e toda energia Potencial da mola é transformada em energia interna:

U f  U i  k  x 2 0 (1a.

lei) Q S f  S i  f   Q T  P s (2a.

lei) i

Toda energ. pot. mola é convertida em energia interna do sistema. A variação de S é devido a Ps, pois



Q=0. Como S não depende do caminho, TdS = dU =(1/2)kx 0 2 -> Ps = (1/2)kx 0 2 /T > 0 & S f > S i .

Aumentou a entropia do sistema isolado. Não é possível converter a mesma quantidade de energia interna em energia potencial da mola!

PROCESSOS IRREVERSÍVEIS: QUEDA LIVRE

m inicial h 0 final

Sistema Isolado > não há calor cruzando a fronteira e toda energia Potencial é transformada em energia interna após choque com água

U f  U i  m  g  h 0 (1a.

lei) água Q S f  S i 

Como S não depende do caminho,

f   Q T  P s (2a.

lei) i

Toda energ. pot. é convertida em energia interna do sistema. A variação de S é devido a Ps, pois



Q=0. TdS = dU =mgh 0 -> Ps = mgh 0 /T > 0 & S f > S i .

Aumentou a entropia do sistema isolado. Não é possível converter a mesma quantidade de energia interna para elevar a bola na posição inicial h 0 !

PROCESSOS IRREVERSÍVEIS: DIFERENÇA TEMP.

inicial final

Sistema Isolado -> energia interna

Quente Th Frio Tc 

S

 T mix T h  T c 2

C ln

  

T mix T c



permanece constante (blocos com mesma massa e calor específico porém a Th e Tc)

U f  U q  U f  T mix  T q  2 T f (1a.

lei)    

C ln

  

T h T mix

   

C ln

  

T h T c

   

0 pq.

T q



T f (2a.

lei) S do sistema isolado é considerado como a soma de S do sistema quente e frio. Squente diminui mas Sfrio aumenta de modo que a variação total é maior que zero.

Troca de calor com diferença de temperatura é irreversível. O bloco que atinge Tmix não volta expontaneamente para Th e Tc, é necessário trabalho!

PROCESSOS IRREVERSÍVEIS: EXPANSÃO COM DIFERENÇA DE PRESSÃO

P =P1 V1=V T1=T AR P=P2 V2 =2V T2=T AR

Sistema Isolado > na expansão para o vácuo não há calor nem trabalho cruzando a fronteira. A energia interna do gás permanece a mesma. T 1 = T 2 (1 a lei) Processo isotérmico: P 1 V 1 = P 2 V 2 ou P 2 = 0.5P

1 Variação da Entropia:

TdS   0  PdV  dS  PdV T  MR dV V   S  MRLN   V 2 V 1    0

Durante a expansão contra o vácuo a capacidade de realizar trabalho do gás foi perdida (não havia máq. p/ extrair trabalho). A transf. inversa é irreversível pois



S>0

Process Efficiency

A reversible process may have an increase or decrease of Entropy but THERE IS NO ENTROPY GENERATION .

An actual process which does or receive work is often compared against an ideal reversible process.

The process efficiency is often taken as the ratio between the actual work and the ideal reversible work.

Isentropic Process

For a reversible, adiabatic process

 Q REV  0  dS  0 or S 2  S 1

For an ideal gas



ds=0 and vdP = C P dT Using v = RT/P and the fact that R = C P -C V ,



C P



C v



dP P



C P dT T

  

P 2 P 1

    

1

    

T 2 T 1

 

Recovers the adiabatic and reversible ideal gas relations, Pv



= const.

Adiabatic Process: Reversible x Irreversible



Q



0



dS



0 or P S



0 P 2 T

2s 2

P 1

1

s For an adiabatic process (1



2)



S is constant or greater than zero due to entropy generation.

Adiabatic Process: Reversible x Irreversible T Compression, W < 0 P 2 T Expansion, W > 0 P 1

2s 2

P 1

1

P 2

1 2s 2

s s



S is constant or greater than zero due to entropy generation independent if it is an adiabatic expansion or compression.

Available Work in an Adiabatic Process: Reversible x Irreversible

•

The first law states for an adiabatic process that W = U 2 – U 1 where 1 and 2 represent the initial and final states.

•

For compression, T 2S < T 2 , since U ~ to T, then the reversible process requires less work than the irreversible, i.e., W REV < W .

•

For expansion, T 2S > T 2 , since U ~ to T, then the reversible process delivers more work than the irreversible, i.e., W REV > W .

Available Work in a Process: Reversible x Irreversible Paths The 1 st and 2 nd law combined result



eliminating dQ

 

Q TdS

 

dU



Q

  

W REAL



Irr PdV

 

Irr

 

W REAL but PdV is a reversible work mode, then



W REV

 

Irr

 

W REAL

 

W REV

 

W REAL

 

Irr



0 or 1



W REAL W REV



0

  

1 the work delivered in a reversible process is always equal or greater than the one in a irreversible process.

Available Work in a Process: Reversible x Irreversible Path Either for compression or expansion, the irreversibilites do: 1. increase the system entropy due to the entropy generation by the irreversibilities, 2. a fraction of the available work is spent to overcome the irreversibilites which in turn increase the internal energy,

What For Are the Reversible Process?

•

They are useful for establishing references between actual and ‘ideal’ processess.

•

The process efficiency defined as the ratio of the work delivered by an actual and an reversible process compares how close they are.

 

W actual W reversible

•

It must not be confused the process efficiency with the thermal eff. of heat engines! The later operates in a cycle.

TEAMPLAY Um carro com uma potência de 90 kW tem uma eficiência térmica de 28%. Determine a taxa de consumo de combustível se o poder calorífico do mesmo é de 44.000 kJ/kg.

Recommended Exercises

4-18 4-19 4-23 4-25 4.26 4-34

TEAMPLAY

A 50-kg iron block and a 20-kg copper block, both initially at 80 o C, are dropped into a large lake at 15 o C. Thermal equilibrium is established after a while as a result of heat transfer between the blocks and the lake water. Determine the total entropy generation for this process.