Entropy Change

• Property diagrams (T-s and h-s diagrams)

– From the definition of the entropy, it is known that Q=TdS during a reversible process.

– Hence the total heat transfer during such a process is given by Qreversible = TdS

• Therefore, it is useful to consider the T-S diagram for a reversible process involving heat transfer

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

T

S

Example

Show a Carnot cycle on a T-S diagram and identify the heat transfer at both the high and low temperatures, and the work output from the cycle.

S

TTH

TL

S1=S4 S2=S3

1 2

34

A B

• 1-2, reversible isothermal heat transferQH = TdS = TH(S2-S1) area 1-2-B-A

• 2-3, reversible, adiabatic expansionisentropic process, S=constant (S2=S3)

• 3-4, reversible isothermal heat transfer QL = TdS = TL(S4-S3), area 3-4-A-B

• 4-1, reversible, adiabatic compressionisentropic process, S1=S4

Net work Wnet = QH - QL, is the area enclosed by 1-2-3-4, the shaded area

QH

QL

WoutWin

Mollier Diagram

• Enthalpy-entropy diagram, h-s diagram: it is valuable in analyzing steady-flow devices such as turbines, compressors, etc.

• h: change of enthalpy from energy balance (from the first law of thermodynamics)

• s: change of entropy from the second law ( a measure of the irreversibilities during an adiabatic process)

s

h

h

s

TdS (Gibbs) Equations

• Since, the area under the T-s line is equal to the heat transfer for a reversible process, it would be useful to have a relationship between Temperature and Entropy to obtain the heat transfer,T.

• Such a relationship exists for a closed system containing a pure, compressible substance undergoing a reversible process

dU = Qrev - Wrev = TdS - PdV => TdS = dU + PdV or Tds = du + pdv ( per unit mass)

• This is the well-known Gibbs equation

• Alternatively, eliminate du by using the definition of enthalpy, h = u + pv=> dh = du + pdv + vdp, => du + pdv = dh – vdp

Tds = du + pdv

Tds = dh – vdp

TdS

Comments Regarding theTdS Equations

Equations relate the entropy change of a system to other properties, such as enthalpy (h), internal energy (u), pressure and volume.

• Hence, even though T-ds relations were derived for a reversible process, they are valid for any process, reversible or irreversible.

• They are applicable for closed or open systems.

• They are valid for all pure substances, single or multi-phase.

Example – Application of Tds relations

• Consider steam undergoing a phase transition from liquid to vapor at a constant temperature of 20°C. Determine the entropy change sfg=sg-sf using the Gibbs equations and compare the value to that read directly from the thermodynamic table.

)()(1

,

fgfgfgfg vvT

Puu

Tsss

dvT

P

T

duds

vapor to liquid from change

From table A-4:T=20°C, P = 2.338 kPa, vf = 0.001002(m3/kg), vg=57.79(m3/kg), uf=83.9(kJ/kg), ug=2402.9(kJ/kg)Substituting into the Tds relation:sfg= (1/293)(2402.9-83.9) + (2.338/293)(57.79-0.001002) = 8.375(kJ/kg K)

Compares favorably with the tabulated value sfg=8.3715(kJ/kg K)

1/T ug - uf vg - vfP/T

Entropy Change of an Incompressible Substance (YAC:6-8)

• For most liquids and all solids, the density does not change appreciably as pressure changes, hence dv 0.

• Gibbs equation states that Tds = du+pdv

Tds = du where du = CdT, for an incompressible substance and Cp=Cv=C is a function of temperature only.

• Therefore, ds = du/T = C dT/T

2 22

2 111 1

avg

Integrate to determine the entropy change during a process

s ( ) ln( )

where C is the averaged specific heat of the substance

over the given temperature range

avg

dT Ts ds C T C

T T

• Specific heats for some common liquids and solids can be found in thermodynamic tables such as Table A-14 to A-18

Note: According to the above equation, an isothermal process for a pure incompressible substance is isentropic.

Example – Entropy Change of Incomp. Substances

A 1-kg metal bar, initially at 1000 K, is removed from an oven and quenched by immersing in a closed tank containing 20 kg of water, initially at 300 K. Assume both substances are incompressible and Cwater= 4(kJ/kg K), Cmetal = 0.4(kJ/kg K). Neglect heat transfer between the tank and its surroundings.

(a) Determine the final temperature of the metal bar, (b) entropy generation during the process.

Tm=1000 K, mm=1kg, cm=0.4 kJ/kg K

Tw=300 K, mw=20 kg, cw=4 kJ/kg K

Solution

water metal bar f

(a) Energy balance from the first law:

U Q-W 0, no heat transfer and no work done

U U 0, both bar and water reach final temperature T

( ) ( ) 0

( / )

( /

w w f w m m m f

w w m w m mf

w w

m c T T m c T T

m c c T m TT

m c c

(20)(10)(300) (1)(1000)303.5( )

) (20)(10) 1m m

Km

)/(451.0477.0928.01000

5.303ln)4.0)(1(

300

5.303ln)4)(20(

lnlns(bar)s(water)

son)s(generatis system theof balance

entropy the0,Q outside hefer with theat trans No (b)

mw

g

KkJs

T

Tcm

T

Tcms

g

m

fm

w

fwg

The total entropy of the system increases, thus satisfy the second law

Entropy Change of Ideal Gases (YAC:6-9)

From Gibbs’ equations, the change of entropy of an ideal gas can be expressed as:

dPT

v

T

dhdv

T

P

T

duds

For an ideal gas, u=u(T) and h=h(T), du=cv(T)dT and dh=cp(T)dT and Pv = RT

2 22 2

2 1 2 11 11 1

p v

( ) , ( )

By integration, the change of the entropy is

( ) ln( ) or ( ) ln( )

we need to know the function c (T) and c (T) in order to compl

v P

v P

dT dv dT dPds c T R and ds c T R

T v T P

dT v dT Ps s c T R s s c T R

T v T P

ete

the integration,

Entropy Change of Ideal Gases – Special CasesThe dependence of specific heats on temperature makes the integration more complex. However, for certain cases, one can make simplifying assumptions.Case 1: If specific heats are assumed constant, integration is simplified:

Note: The above is strictly true for monatomic gases, e.g.He, Ar It is fairly accurate if the temperature difference is small.

Case 2: Calculate the specific heat at an average temperature, Tavg, and assume it to be constant. This also allows the specific heat to be taken out of the integral

Note: This approximation is generally fairly accurate if the temperature difference is not too large, usually good if T < few hundred degrees.Strictly speaking, one should look at the temp. dependence of specific heats forthe particular substance to evaluate the validity of this approximation.

2 22 1

1 1

2 22 1

1 1

ln( ) ln( ) or

ln( ) ln( )

v

P

T vs s c R

T v

T Ps s c R

T P

Isentropic Processes for Ideal Gases

If a process is isentropic (that is adiabatic and reversible), ds = 0, s1=s2, then it can be shown that:

1 /12 1 2 2

1 2 1 1

p2 1

1 2 v

k

( ) , and ( )

cand ( ) , where

c

k kk

k

T v T P

T v T P

P v

P v

The above are referred to as the: first, second and third, respectively, isentropic relations for Ideal Gases (assuming constant specific heats).They can also be written as:

Tvk-1 = constant First isentropic relationTP(1-k)/k = constant Second isentropic relationPvk = constant Third isentropic relation

Example

• Air is compressed from an initial state of 100 kPa and 300 K to 500 kPa and 360 K. Determine the entropy change using constant cp=1.003 (kJ/kg K)

2 22 1 P

1 1

2 1

ln( ) ln( ) if c is constant

360 5001.003ln (0.287) ln 0.279( / )

300 100

P

T Ps s c R

T P

s s kJ kg K

• Negative entropy due to heat loss to the surroundings