egr 334 thermodynamics chapter 6: sections 1-5 lecture 24: introduction to entropy quiz today?
Post on 17-Dec-2015
247 Views
Preview:
TRANSCRIPT
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
top related