me 201 entropy
TRANSCRIPT
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Entropy
Dr. Md. Zahurul Haq
ProfessorDepartment of Mechanical Engineering
Bangladesh University of Engineering & Technology (BUET)Dhaka-1000, Bangladesh
[email protected]://teacher.buet.ac.bd/zahurul/
ME 201: Basic Thermodynamics
c Dr. Md. Zahurul Haq (BUET) Entropy ME 201 ( 2015) 1 / 27
Entropy
Consequences of Second Law of Thermodynamics
If a system is taken through a cycle and produces work, it must be
exchanging heat with at least two reservoirs at2 different
temperatures.
If a system is taken through a cycle while exchanging heat with a
single reservoir, work must be zero or negative.
Heat can never be converted continuously and completely intowork, but work can always be converted continuously andcompletely into heat.
Work is a more valuable form of energy than heat.
For a cycle and single reservoir, Wnet0.
c Dr. Md. Zahurul Haq (BUET) Entropy ME 201 ( 2015) 2 / 27
Entropy
Clausius Inequity
T141
Clausius InequalityQ
T 0
Wnet(Wrev+Wsys) =QR dU QRQ =
TRT Wnet =TR QT dU Wnet =TR QT
=
Q
T 0 asWnet0
QT =0 reversible process
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Entropy
Entropy Generation
dS QT =
dS QT 0
Entropy produced (generated) by internal irreversibilities.
T173
:
>0 irreversible process
=0 internally reversible process
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Second-Law of Thermodynamics for CV Systems
Second Law Analysis for CV System
at t at t+ t
T320
CM (time t) : region A + CV
Scm,t =SA+Scv,t
CM (t+ t) : CV + region B
Scm,t+t =SB+Scv,t+tScm,t+tScm,t
t = Scv,t+tScv,t
t +SBSA
t : dScm
dt = dScv
dt + mBsB mAsA
dScmdt =
Qi
Ti+
dScvdt
=
QiTi
+
i( ms)i
e( ms)e+ cv
c Dr. Md. Zahurul Haq (BUET) Entropy M E 201 (2015) 25 / 27
Second-Law of Thermodynamics for CV Systems
T042
T175
c Dr. Md. Zahurul Haq (BUET) Entropy M E 201 (2015) 26 / 27
Second-Law of Thermodynamics for CV Systems
dScvdt =
j
QjTj
+
(
m s)i
(
ms)e+ cv
For CM systems:
mi =0, me =0 =
dScm
dt =
j
QjTj
+
For steady-state steady-flow (SSSF) process: dSdv/dt =0.For 1-inlet & 1-outlet SSSF process:
mi = me.
(se si) = j qjTj + cvmFor adiabatic 1-inlet & 1-outlet SSSF process:
(se si) = cvm =se si
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