EGR 334 Thermodynamics Chapter 6: Sections 1-5 Lecture 24: Introduction to Entropy Quiz Today?

EGR 334 ThermodynamicsChapter 6: Sections 1-5

Lecture 24: Introduction to Entropy

Quiz Today?

Today’s main concepts:• Explain key concepts about entropy• Learn how to evaluate entropy using property tables • Learn how to evaluate changes of entropy over reversible

processes• Introduction of the temperature-entropy (T-s) diagram

Reading Assignment:

Homework Assignment:

Read Chapter 6, Sections 6-8

Problems from Chap 6: 1,11,21,28

Introducing Entropy Change and the Entropy Balance

• The entropy change and entropy balance concepts are developed using the Clausius inequality expressed as:

where

scycle = 0 no irreversibilities present within the system

scycle > 0 irreversibilities present within the system

scycle < 0 impossible

cycleb

Q

T

Defining Entropy Change

• Consider two cycles, each composed of two internally reversible processes, process A plus process C and process B plus process C, as shown in the figure.

• Applying Classius Equation to these cycles gives,

where scycle is zero because the cycles are composed of internally reversible processes.

Defining Entropy Change

• Subtracting these equations:

• Since A and B are arbitrary internally reversible processes linking states 1 and 2, it follows that the value of the integral is independent of the particular internally reversible process and depends on the end states only.

Defining Entropy Change

• Recalling (from Sec. 1.3.3) that a quantity is a property if, and only if, its change in value between two states is independent of the process linking the two states, we conclude that the integral represents the change in some property of the system.

• We call this property entropy and represent it by S. The change in entropy is

where the subscript “int rev” signals that the integral is carried out for any internally reversible process linking states 1 and 2.

2

2 1int1rev

QS S

T

Defining Entropy Change

• The equation allows the change in entropy between two states to be determined by thinking of an internally reversible process between the two states. But since entropy is a property, that value of entropy change applies to any process between the states – internally reversible or not.

2

2 1int1rev

QS S

T

Entropy Facts

• Entropy is an extensive property.• Like any other extensive property, the change in

entropy can be positive, negative, or zero:

• Units of entropy, S, are: in SI [kJ/K] in US Customary [Btu/oR]

• Units for specific entropy, s, are: in SI [kJ/kg-K] in US Customary [Btu/lbm-oR]

Finding Entropy: Method 1

►For problem solving, specific entropy values are provided in Tables A-2 through A-18. Values for specific entropy are obtained from these tables using the same procedures as for specific volume, internal energy, and enthalpy.

For two phase liquid/vapor mixtures, quality, x, may be used as

For slightly compressed liquids, the entropy may be approximated as:

f

g f

s sx

s s

( , ) ( )fs T p s T

( )f g fs s x s s or

10

Example 1: find s using property tables

b) Given: Ammonia at p =2.5 bar and v = 0.20 m3/kg find s:

a) Given: H20 at T=520 oC and p= 8 MPa find s:

c) Given: R22 at T = 20 oF and p = 80 psi find s:

From Table A-2…substance is found to be superheated vapor.From Table A-4…..s can be found ad 6.7871 kJ/kg-K

From Table A14…substance is found to be liquid/vapor mixture.the quality can be found as

and then

From Table A-7E or A-8E…substance is found to be compressed liquidthen s can be found using s(T,p)=sf(T) = 0.0356 Btu/lbm-oR

( ) /( ) (0.20 .0015) /(0.4821 0.0015) 0.413f g fx v v v v

( ) 0.4753 0.413(5.5190 0.4753) 2.5583 /f g fs s x s s kJ kg K

11

Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R.

(a) Water p1 = 1000 psi, T1 = 800 °F

p2 = 1000 psi, T2 = 1000 °F

(b) Refrigerant 134a h1 = 47.91 BTU/lb, T1 = -40 °F

saturated vapor, p2 = 40 psi

12

Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R.

(a) Water p1 = 1000 psi, T1 = 800 °F

p2 = 1000 psi, T2 = 100 °F

T

v

545°F

1000 psi

Table A-4E@ T1 = 800 °F, p1 = 1000 psi

s1 = 1.5665 BTU/lb·°R

Table A-5E@ T2 = 100 °F, p2 = 1000 psi

s2 = 0.12901 BTU/lb·°R

s2 - s1 = 0.12901 - 1.5665 = -1.43749 BTU/lb·°R

2

1

13

Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R.

T

v

-40°F

Table A-10E, @ T1 = -40 °F

hf=0, hg=95.82 BTU/lb

Table A-11E, sat. vapor, p2 = 40 psi

s2 - s1 = 0.2197 - 0.11415 = 0.1056 BTU/lb·°R

(b) Refrigerant 134a h1 = 47.91 BTU/lb, T1 = -40 °F

saturated vapor, p2 = 40 psi

2

1

5.0082.95

091.47

x

1 f g fs s x s s sf=0, sg=0.2283 BTU/lb·°R

0 0.5 0.2283 0 0.1142 / mBtu lb R

2 0.2197 / ms Btu lb R

Finding Entropy: Method 2

• For problem solving, states often are shown on property diagrams having specific entropy as a coordinate: the Temperature-Entropy (T-s) diagram and the Enthalpy-Entropy (h-s) diagram

• the h-s diagram is also know as the Mollier diagramThese diagramsare in your appendix as Figures 7 and 8 and Figures 7E and 8E.

T-s diagram:

Vertical axis: TemperatureHorizontal axis: Entropy

Shows other constant property lines: Constant Pressure Constant Quality Constant Enthalpy Constant Specific Volume

h-s diagram:

Vertical axis: EnthalpyHorizontal axis: Entropy

Shows other constant property lines: Constant Temperature Constant Pressure Constant Quality

17

Example 2: Using T-s diagram:

b) Given: H20 at s=1.5 Btu/lbm-R and x=85% find h:

a) Given: H20 at T=900 oF and h = 1400 Btu/lbm find s:

From Table A-7E or A-8E…substance is found to be compressed liquidthen s can be found using s(T,p)=sf(T) = 0.0356 Btu/lbm-oR

From Table A-7E or A-8E…substance is found to be compressed liquidthen s can be found using s(T,p)=sf(T) = 0.0356 Btu/lbm-oR

s ≈ 1.5 Btu/lbm-R

h ≈ 1.5 Btu/lbm

Finding Entropy: Method 3…IT

IT can also be used for working with entropy. Once again it’s recommended that you don’t compare individual values from IT with the values you might find on the Appendix tables from you book, but changes in entropy between two states will be consistent and can be compared.

For H20, find the

difference in entropy between

State 1: T = 500 deg C

and p = 20 bar

State 2: T = 200 deg C

and p = 10 bar

Δs=0.1992 kJ/kg-K

19

How is S related to U?

Sec 6.3 : The TdS Equations

Consider a pure, simple, compressible system undergoing a internally reversible process.

Energy Balance:

WQU

WQUPEKE

then differentiate revint revint WQdU

where,

therefore, dU TdS pdV

TdS dU pdV

int revQ TdS

int revW pdV

1st TdS equation

20

How is S related to H?

Sec 6.3 : The TdS Equations

Recall that: H U pV

and

therefore,

dH dU d pV dU pdV Vdp dU pdV dH Vdp

dU TdS pdV

TdS dH Vdp can also write these on a per mass basis

du Tds pdv dh Tds vdp

Even though this derivation is based on a reversible process, these equations hold for any process… even an irreversible process.

TdS dU pdV

2nd T dS equation

Relationships between p, T, v, u, h, and s

from 1st Tds equation

Finding Entropy: Method 4 for incompressible fluidWhen a substance is compressible (like many liquids),

Recall that for incompressible fluids:

du Tds pdv du pdv

dsT T

du c dT 0dv where c is the specific heat capacity

and

then cdTds

T

2 1

c dTs s

T

if the specific heat is treated as constant

22 1 1

lnTdT

s s c cT T

Δs for incompressible liquids

Finding Entropy: Method 5 for Ideal Gas

Recall that if a substance can be treated as an ideal gas, the following relationships may be applied to the defining the properties:

Combining ideal gas relations with the Tds equations

pv RT vdu c dT

/p vk c c

pdh c dT

p vc c R

give

dh vdpds

T T

du pdvds

T T

v

dT dvds c R

T v p

dT dpds c R

T p

ds = f(T,v) ds = f(T,p)

23

So for Ideal Gas, change in entropy can be evaluated by integratingthese relations

Sec 6.4 Idea Gases

and

vc T dT Rds dv

T v

1

2lnv

vR

T

dTTcs V

Pc T dT Rds dp

T P

2

1

lnPc T dT ps R

T p

24Sec 6.4 Idea Gases

and

1

2lnv

vR

T

dTTcs V

2

1

lnPc T dT ps R

T p

there are several options to evaluate the heat capacity. Remember that cV & cP are functions of temperature.

or constant heat capacities from Tables A20, A22, or A23.

then use or

ii) Use Tabulated values( Table A22):

432 TTTTcP

1v

Rc

k

12 TsTss The s° indicates that it is based on a reference temp

Options to evaluate:

or1P

kRc

k

i) Find values for cp and or cv (Using Table A21)

2 2

1 1

ln lnv

T vs c R

T v

22 1

1

lnp

s s T s T Rp

2 2

1 1

ln lnp

T ps c R

T p

25

Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R.Air as an ideal gas with T1 = 40°F = 500° R , p1 = 2 atm

and T2 = 420°F = 880° R, p2 = 1 atm

where: Tave = 230 F = 690 R

R = 0.06855 Btu/lb-R cp = 0.242 Btu/lb-R

(a) using a constant cv at Tave 2

1

lnPc T dT ps R

T p

2 2

1 1

ln lnP

T ps c R

T p

880 10.242ln 0.06855ln 0.1843 /

500 2os Btu R

26

Example (6.4): Using the appropriate tables determine the change in specific enthalpy between the specified states, in BTU/lb·°R.Air as an ideal gas with T1 = 40°F = 500° R , p1 = 2 atm

and T2 = 420°F = 880° R, p2 = 1 atm

from Table A-22E @ T1 = 500 °R, s° = 0.58233 BTU/lb·°R

@ T2 = 880 °R, s° = 0.71886 BTU/lb·°R

(c) Air as an ideal gas using so

1

212 ln

P

PRTsTss

1 0.71886 0.58233 0.06855ln

2

atms

atm

0.1841BTU

slb R

27

end of lecture 24 slides