# ENGR 2213 Thermodynamics

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### Transcript of ENGR 2213 Thermodynamics

ENGR 2213 Thermodynamics

F. C. LaiSchool of Aerospace and Mechanical EngineeringUniversity of Oklahoma

Second Law of ThermodynamicsFirst law of ThermodynamicsUnlike energy, entropy is a non-conserved property.Second law of ThermodynamicsEnergyEntropyClausius InequalityThe cyclic integral of Q/T is alwaysless than or equal to zero.This inequality is valid for all cycles, reversible or irreversible.

Clausius InequalityFor reversible heat engines= 0

Clausius InequalityFor irreversible heat engines< 0From Carnot principleWrev > WirrevQL, rev < QL, irrevQL, irrev = QL, rev+ Qdiff

Example 1A heat engine receives 600 kJ of heat from a high-temperature source at 1000 K during a cycle. Itconverts 150 kJ of this heat to work and rejects theremaining 450 kJ to a low-temperature sink at 300 K. Determine if this heat engine violates the2nd law of thermodynamics on the basis of the Clausius inequality.(b) the Carnot principle.

Example 1 (continued)(a) Clausius inequality= - 0.9 kJ/K< 0(b) Carnot principleth < rev

Second Law of ThermodynamicsClausius InequalityInternally Reversible ProcessesA process is called internally reversible if noirreversibilities occur within the boundaries of thesystem during the process.

EntropyIf a quantity whose integral dependsonly on the end states and not theprocess path, then it is a property.

EntropyIsothermal ProcessesQ = T0 S

Increase-in-Entropy PrincipleClausius Inequality= S1 S2

Increase-in-Entropy PrincipleThe entropy change of a closed system during an irreversible process is greater than the integral ofQ/T evaluated for that process. For an adiabatic process, Q = 0(S)adiabatic 0In the absence of heat transfer, entropy change is due to irreversibilitiesonly, and their effect is always to increase the entropy.

Increase-in-Entropy Principle(S)adiabatic 0A system plus its surroundings constitutes an adiabatic system, assuming both can be enclosed by a sufficiently large boundary across which there is no heat or mass transfer.(S)total = (S)system + (S)surroundings 0

Increase-in-Entropy PrincipleSgen = (S)totalCauses of Entropy Change Heat TransferIsentropic Process> 0irreversible processes= 0reversible processes< 0impossible processes IrreversibilitiesA process involves no heat transfer (adiabatic) and no Irreversibilities within the system (internally reversible).

Remarks about Entropy1. Process can occur in a certain direction only. A process must proceed in the direction that complies with the increase-in-entropy principle.2. Entropy is a non-conserved property. There is no such thing as the conservation of entropy principle.3. The quantity of energy is always preserved during an actual process (the first law), but the quality decreases (the second law). The decrease in quality is always accompanied by an increase in entropy.

What is Entropy?Entropt can be viewed as a measure of moleculardisorder, or molecular randomness From a statistical point of view, entropy is a measureof the uncertainty about the position of molecules atany instant.The entropy of a pure crystalline substance at absolute zero temperature is zero since there is nouncertainty about the state of the molecules at that instant.- the 3rd Law of Thermodynamics

Second Law of ThermodynamicsHeat is, in essence, a form of disorganized energyand some disorganization (entropy) will flow with heat.Work instead is an organized form of energy, and is free of disorder or randomness and thus free ofentropy.

Example 1Saturated water at 100 C is contained in a piston-cylinder assembly. The water undergoes aninternally reversible heating process to the corresponding saturated vapor state. Find the work per unit mass for the process.(b) the heat transfer per unit mass for the process.

Example 1 (continued)(a)(b)= (101.4)(1.673 0.001044)= 170 kJ/kg= 2257 kJ/kg= (373.15)(7.3549 1.3069)= 2257 kJ/kg

Example 2Steam at 7 MPa and 450 C is throttled through a vavleto 3 MPa. Find the entropy generation through the process.

Example 2 (continued)sgen = sT1 = 450 Cp1 = 7 MPaTable A-6h1 = 3287.1 kJ/kgs1 = 6.6327 kJ/kg KThrottling Process h2 = h1p2 = 3 MPah2 = 3287.1 kJ/kgTable A-6s2 = 6.9919 kJ/kg K= 6.9919 6.6327= 0.3592 kJ/kg K

Second Law of ThermodynamicsCarnot Cycle in T-S Diagram