PRINCIPLE OF THE INCREASE
OF ENTROPY
BITS Pilani
Pilani Campus
Second law analysis for a control
volume
M. S. SONI
1
M. S. SONI
2
BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956
PRINCIPLE OF THE INCREASE
OF ENTROPY
PRINCIPLE OF THE I
BITS Pil i
Pilani
Pilani Campus
Second Law of Thermodynamics
S
dL
f Th
d
i
Adiabatically impossible process
The change is spontaneous from left to right What about the
right.
reverse?
Not spo a eous, bu ca it be made to happen? If so, a what
o spontaneous
BITS Pilani
Pilani Campus
Lecture 19 SECOND LAW
PERPETUAL MOTION MACHINE
Any device that violets First law or
y
Second is called as Perpetual
Motion Machine
Perpetual Motion Machine of First
Kind (violets First law )
Perpetual M i
P
l Motion M hi
Machine
BITS Pil i
Pilani
Pilani Campus
Carnot cycle, absolute temperature
C
t
l
b l t t
t
The Carnot Cycle
Reversible cycle:
Reversible isothermal
expansion at TH, QH absorbed
Reversible adiabatic expansion
Reversible isothermal
compression at TL, QL rejected
BITS Pilani
Pilani Campus
Lecture 13 FIRST LAW ANALYSIS FOR
FIRSTA CONTROL VOLUME
Compressor
A compressor receives 0.1 kg/s R-134a at 150
g
kPa, -10oC and delivers it at 1000 kPa, 40oC.
The power input is measured to be 3 kW. The
compressor h h t t
has he
BITS Pilani
Pilani Campus
Lecture 11 FIRST LAW ANALYSIS FOR
FIRSTA CONTROL VOLUME
Recap: Problem 1
A piston/cylinder contains 50 kg
of water at 200 kPa with a
volume of 0.1 m3. Stops in the
cylinder restricts the enclosed
y
volume to 0.5 m3, as shown in
f
BITS Pil i
Pilani
Pilani Campus
Lecture 6 G
L t
Gases, Compressibility Factor
C
ibilit F t
The Ideal Gas
PV = nT or Pv = T
1
1
is the universal gas constant 8.3145 kJ kmole-1 K-1
n = m/M
PV = mRT, where R is the gas constant per unit mass of a
g
given
BITS Pilani
Pilani Campus
Lecture 10 First Law for Control Mass
Problem 1
A piston/cylinder contains 50 kg
of water at 200 kPa with a
volume of 0.1 m3. Stops in the
cylinder restricts the enclosed
y
volume to 0.5 m3, as shown in
fig. The water is now heat
BITS Pilani
Pilani Campus
Lecture 12 FIRST LAW ANALYSIS FOR
FIRSTA CONTROL VOLUME
Throttling valve
Throttling valves are any kind of flow-restricting devices
g
y
g
that cause a significant pressure drop in the fluid.
The pressure drop in the f
fluid is of
4/4/2013
TEMPERATURE ENTROPY PLOT
BITS Pilani
Pilani Campus
T
T
1
1
2
2
s
Lecture 21: ENTROPY Mathematical definition
s
REVERSIBLE
ISOTHERMAL
PROCESS
REVERSIBLE
ISENTROPIC
PROCESS
BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956
TEMPE
4/4/2013
BITS Pilani
Entropy Change for a Control
Mass During an Irreversible Process
Pilani Campus
Lecture 22: Entropy Change For
a Control Mass
Q
T
Q
T A+
1
2
=
Q
=
B
2
1
T
0
BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956
E
4/29/2013
BITS Pilani
Pilani Campus
y
p y
Lec 36 - Thermodynamic Property Relations
1
4/29/2013
Gibbs equation alternate forms
du = Tds Pdv
T = (u/s)v, and P = - (u/v)s
dh = Tds + vdP
T = (h/s)P, and v = (h/P)s
da = Pdv sdT
P = - (a/v)T, and s = - (a/T)v
4/29/2013
BITS Pilani
Pilani Campus
Lec 33 Thermodynamic Property Relations
1
4/29/2013
Change of independent variables
Gibbs equation: du = Tds Pdv
u = u(s, v), and T = (u/s)v, and P = - (u/v)s
Enthalpy h is defined as h = u + Pv = u - v(u/s)v, an
exampl
ENTROPY AS A RATE
EQUATION
BITS Pilani
Pilani Campus
The change of entropy during a process
dS
Q
T
The above equation can be rewritten as
dS =
Second law analysis for a control
volume
M. S. SONI
1
Q
T
+ S gen
M. S. SONI
2
BITS Pilani, Deemed to be Univer
4/11/2013
BITS Pilani
Pilani Campus
Lecture 26 What is entropy?
1
4/11/2013
Principle of entropy increase
Consider a closed system in adiabatic enclosure undergoing
an irreversible process a b. What can we say about Sb
b
Sa?
Return the system to state a
4/11/2013
BITS Pilani
Pilani Campus
Lecture 25 Entropy S revisited
1
4/11/2013
Thermodynamics is a funny subject. The first time
you go through it, you don't understand it at all. The
second time you go through it, you think you
understand it, except for
4/4/2013
BITS Pilani
THE INEQUALITY OF CLAUSIUS
Pilani Campus
It is corollary or a consequence of the
Second Law of thermodynamics.
It is valid for all possible cycles,
including
both
reversible
and
irreversible
heat
engines
and
refrigerators.
Lecture 20:
4/4/2013
BITS Pilani
Problem 1
Pilani Campus
Saturated water vapor at 200 kPa is
compressed to 600 kPa in a reversible
p
adiabatic process. Find the new v and T.
Lecture 23-24: Problems
mssoni@bits-pilani.ac.in
mssoni@bits-pilani.ac.in
BITS Pilani, Deemed
BITS Pil i
Pilani
Pilani Campus
Irreversible and Reversible Processes,
I
ibl
dR
ibl P
Carnot cycle
Heat Engine
Cyclic device that results in
positive W (does work),
accepting heat from a
reservoir at a high T, and
rejecting heat to a reservoir
at a lower
BITS Pilani
Pilani Campus
Lecture 14 FIRST LAW ANALYSIS FOR
FIRSTA CONTROL VOLUME
The Transient Process
Assumptions:
The control volume remains constant relative to the
coordinate frame
The state of the mass within control volume may
change with ti
h
it
BITS Pil i
Pilani
Pilani Campus
Lecture 9 Fi t Law for Control Mass
L t
First L f C t l M
First Law for Control Mass
As seen above, for a closed system
U = U2 U1 = - 1W2 (adiabatic, Q = 0).
U = U2 U1 = 1Q2 (W = 0)
What if both work and heat terms present?
BITS Pilani
Pilani Campus
Properties of a pure substance
Pure Substance
The pure substance is one that has
p
a homogeneous and invariable
chemical composition.
p
A pure substance may exist in many
phases, but the chemical composition
is same in all the ph
8.22 a)ok b)impossible c)possible if reversible d)ok
8.23
8.24 a) impossible b)ok
8.25 impossible
8.26 a) s =4.05 kJ/kg K b) s = 6.5452 kJ/kg K c) s = 1.2369 kJ/kg K
d) s = 0.2966 kJ/kg K e) s = 0.2945 kJ/kg K
8.27
b) x = undefined T 682C
a) T = 64.97C, x
BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE, PILANIKK Birla GOA Campus
INSTRUCTION DIVISION
SECOND SEMESTER 2012-2013
Course Handout (Part II)
Date: 09/01/2013
In addition to Part I (General handout for all course appended to timetable), this portion gives
BIRLA INSTITUTE OF TECHNOLOGY AND SCIENCE PILANI
K K BIRLA GOA CAMPUS
SECOND SEMESTER 2012-2013
Course: THERMODYNAMICS (BITS F111)
MODEL SOLUTION OF TEST- II
1(a)
Solution:
C. V. 1: Turbine (Steady-state) Continuity Eqn.: , 0
80 30 50 kg/s
[2 M]
Energy E
Thermodynamics|Chapter 5 Answers to numerical problems 7th Ed. upto 5.68 (T1 Portion)
(except descriptive/short answers/diagrams)
5.25) a) Compressed liquid, T = 1510C, u = 418.8 kJ/kg, v = 0.001043 m3/kg b) Compressed
liquid, T = 189.10C, v = 0.001137 m3
Thermodynamics|Chapter 3 Answers to numerical problems 7th Ed. (except descriptive/short
answers/diagrams)
3.23) a) Compressed liquid b) Compressed solid
3.24) a) Compressed liquid b) Superheated vapor c) Superheated vapor
3.25) a) Liquid + vapor b) super
Phase Change Substance
Property Tables
in
SI Units
for
ME 201 Section 001
Spring 2012
Craig W. Somerton
Asscoiate Professor
Department of Mechanical Engineeirng
Michigan State University
East Lansing, MI 48824
somerton@egr.msu.edu
1
Table of Contents
Tabl