cheme 260 the clausius inequality & entropy
DESCRIPTION
ChemE 260 The Clausius Inequality & Entropy. Dr. William Baratuci Senior Lecturer Chemical Engineering Department University of Washington TCD 7: A & B CB 6: 1. May 3, 2005. Hot Reservoir. Q H. HE R. W rev. Q C,rev. Cold Reservoir. The Clausius Inequality. Cyclic Integrals - PowerPoint PPT PresentationTRANSCRIPT
ChemE 260 The Clausius Inequality
& Entropy
May 3, 2005
Dr. William BaratuciSenior Lecturer
Chemical Engineering Department
University of Washington
TCD 7: A & BCB 6: 1
The Clausius Inequality
• Cyclic Integrals– Integrate through all the steps in
a cycle and return to the initial state.
• Inexact Differentials: Q & W– Used for path variables, Q and W
• Evaluating Cyclic Integrals– Example 1: Carnot HE
– Example 2: Carnot HE
BaratuciChemE 260May 3, 2005
Q0
T
Hot Reservoir
Cold Reservoir
HER
QH
QC,rev
Wrev
2 4
12 34 H C1 3
Q Q Q Q Q Q Q 0 42
2 4C,rev3 341 12 H
H C H C H C1 3
QQQQQ QQ Q Q
T T T T T T T T T
Clausius: Int. Rev. and Irrev. Cycles• Reversible Cycle, such as Carnot:
– Kelvin: or:
– Therefore:
• Irreversible Cycles:
C,revH
H C
QQQ
T T T
C,rev C
H H
Q T
Q T C,rev H
C H
Q Q
T T
Q0
T
Hot Reservoir
Cold Reservoir
HEIrr
QH
QC,irr
Wirr
rev irr
rev irrW W
H C,rev H C,irrQ Q Q Q
C,rev C,irrQ Q
C,irr C,rev C,irrirr H
H C C C
Q Q QQ Q0
T T T T T
Q0
T
• All Cycles :
– 1st Law :
– Definition of efficiency :
BaratuciChemE 260May 3, 2005
Entropy• Cycle 1-A-B-2:
2 1
1 2A B
Q Q Q0
T T T
• Cycle 1-A-C-2:2 1
1 2A C
Q Q Q0
T T T
• Subtract Eqns:1 1
2 2C B
Q Q
T T
int rev
Q
T
• Does not depend on path !
• It is a state variable or property !
int rev
QdS
T
• Definitionof Entropy:
BaratuciChemE 260May 3, 2005
S: Int. Rev. & Irrev. Processes• Change in Entropy:
2
2 1rev1
QS S S kJ / K
T
2
2 1 2 1int rev irrevrev1
QS S S S
T
• Problem: Since Int. Rev. processes do not exist, how do we evaluate S ?
• Special Case: Reversible, Isothermal Processes
2 2rev
revrev1 1
QQ 1S Q
T T T
Especially useful for evaluating Sreservoir because Treservoir = constantBaratuci
ChemE 260May 3, 2005
Entropy of a Pure Substance• S = fxn( T, P, phase )• NIST Webbook
– Thermophysical Properties of Fluid Systems
– Specific entropy is listed in the thermodynamic data tables
– Observations: sat vap sat liqˆ ˆS S
S with P
S with T All substances at all T :
– Subcooled Liquids: *subcooled sat liq
ˆ ˆS T,P S T,P
Where P* = vapor pressure = Psat
– Sat’d Mixtures: sat mix sat vap sat liqˆ ˆ ˆS x S 1 x S
IG and many real gases :
BaratuciChemE 260May 3, 2005
T-S Diagram
BaratuciChemE 260May 3, 2005
Carnot Cycle• Steps
– 1-2: Isothermal expansion– 2-3: Adiabatic expansion– 3-4: Isothermal compression– 4-1: Adiabatic compression
• Step 1-2:2
2 1rev1
2H
revH1
QS S
T
Q1Q
T T
3
3 2rev2
QS S 0
T
4 434 C
4 3 revrev C C3 3
Q QQ 1S S Q
T T T T
2
2 1rev1
QS S 0
T
• Step 2-3:
• Step 3-4:
• Step 4-1:
Isentropic !
Isentropic !
BaratuciChemE 260May 3, 2005
Heat, Work & TS Diagrams
• 1st Law Cycle
H H 2 1ˆ ˆ ˆQ T S S
C C 3 4ˆ ˆ ˆQ T S S
H Cˆ ˆˆQ W Q
H Cˆ ˆW Q Q
• QH = area under path for step 1-2
• QC = positive area under path for step 3-4
• W = area enclosed by the cycle !BaratuciChemE 260May 3, 2005
Next Class …
• Principle of Increasing Entropy & Entropy Generation– This will let us express the 2nd Law in terms of Entropy
– This very powerful result will let us perform 2nd Law Analysis on processes as well as cycles.
• Fundamental Property Relationships– These are sometimes called the Gibbs Equations
– They show us how entropy is related to other properties, such as H, U, P, V and T
– This will show us how the specific entropy values in the thermodynamic tables were determined.
BaratuciChemE 260May 3, 2005
Example #1
• A piston-and-cylinder device contains saturated R-134a vapor at -5oC. This vapor is compressed in an internally reversible, adiabatic process until the pressure is 1.0 Mpa. Determine the work per kg of R-134a for this process.
Example #2
• Steam 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 kg of steam flowing through the turbine if the process is reversible and changes in kinetic and potential energies are negligible.
Example #3
• Consider a process in which 1.00 kg of saturated water vapor at 100oC is condensed to a saturated liquid in an isobaric process by heat transfer to the surrounding air, which is at 25oC. What is the change in entropy of the water ? What is the change in entropy of the surroundings ? What is the change in the entropy of the universe ?