Chap. 3 (Sec. 3-5 to End of Chapter)
Mass Flow Rate = r A V
(kg/s)
Volume Flow Rate = A V
(m3 /s)
V = velocity
Conservation of Mass (Continuity Principle)
min
mout
min  mout  msystem
m in  m out  dmsystem / dt
Conservation of Mass (Continuity Principle)
Steady state ?
Change of Any quantity with time = 0
m in  m out  dmsystem / dt  0
m in  m out
(r A V)in = (r A V)out
Incompressible fluid ?
Flow Work
Wflow = PV
wflow = Pv
Total energy of moving fluid
= e + Pv
= u + p.e + k.e + Pv
= h + p.e + k.e
First Law of Thermodynamics (ENERGY BALANCE)
also known as Conservation of Energy Principle
* CLOSED SYSTEMS
* OPEN SYSTEMS (CONTROL VOLUME)
• STEADY FLOW
• UNSTEADY FLOW
* CLOSED SYSTEMS (e.g: piston-cylinder etc.)
Ein – Eout = Esystem
Net Energy transfer by
heat, work, and mass
Change in internal, KE, PE etc
Q – W = E
Qnet, in – Wnet,out = Esystem
Moving boundary work, Shaft work, Paddle Work etc
Q – W = E
Qnet, in – Wnet,out = Esystem
Moving boundary work, Shaft work, Paddle Work etc
How the above equation simplifies for different situations ?
for
for
for
for
stationary systems
constant volume process
constant pressure process
many other situations given in your text book
Q – W = E
Qnet, in – Wnet,out = Esystem
Moving boundary work, Shaft work, Paddle Work etc
For example:
Based on problem statement if you having the following information
Closed System
Stationary
Adiabatic
Constant Volume Process
Paddle Work (wnet, in)
Final Temperature ?
Wnet,in = Usystem
wnet,in = usystem
What will be my approach
R134a
if (a) Substances like Steam,
if (b) ideal gases
if (a) Substances like Steam, R134a
U = m (u2-u1)
As long as you have any two properties,
you can find rest
How?
 State (Saturated liquid , mixture,
superheated)
if (b) ideal gases
U = m (u2-u1) = m Cv (T2 – T1)
PV = mRT
Pv = RT
(PV/T)1 = (PV/T)2
STEADY – FLOW SYSTEMS
E in  E out
Rate of Net Energy transfer in
by heat, work, and mass
=
Rate of Net Energy transfer out
by heat, work, and mass
2
2
V
V
 -W
  m
 e (he  e  gze )   m
 i (hi  i  gzi )
Q
2  
2 


for each exit
Mass balance ?
for each inlet
One inlet (1) and one exit (2)
2
2


V

V
2
1



Q - W  mh2  h1 
 g(z2  z1 )
2


On a unit mass basis


V22  V12
q - w  h2  h1 
 g(z2  z1 )
2


2
2


V

V
2
1



Q - W  mh2  h1 
 g(z2  z1 )
2


In general, how above equation simplifies for
(how to judge - based on
function)
Nozzles
Diffusers
Compressors, Pumps
Turbines
Mixtures, Heat Exchangers
Throttling devices
UNSTEADY – FLOW SYSTEMS
Mass Balance
min – mout = msystem
min – mout = (m2-m1)system
Energy Balance
Ein – Eout = Esystem
(Q in  Win   miθi )  (Q out  Wout   me θ e )  (m2 e 2  m1e1 ) sy stem
θ  h  ke  pe
e  u  ke  pe
(Q in  Win   mihi )  (Q out  Wout   mehe )  (m2u2  m1u1 ) sy stem
Chap. 5
The Second Law of Thermodynamics
(What is second law and how it helps)
Statements (in words or schematic)
Some concepts
Heat Engine, Refrigerator, Heat Pump
Reversible and Irreversible processes
* The Carnot Cycle and it’s importance
1ST LAW
Qnet – Wnet = E
for cyclic devices E = 0
Therefore Wnet = Qnet
Wnet = Qin - Qout
HIGH TEMPERATURE RESERVOIR AT TH
QH

W
HE
net, out
QL

LOW TEMPERATURE SINK AT TL
PERFORMANCE OR EFFICIENCY 
THERMALEFFICIENCY (ηth ) 
ηth 
Wnet,out
QH

DESIRED OUTPUT
REQUIRED INPUT
NET WORK OUTPUT
TOTALHEAT INPUT
Q H  QL
Q
1  L
QH
QH
WARM ENVIRONMENT AT TH>TL
WARM ENVIRONMENT AT TH>TL


QH
W
REFRI.

COPR 

QL
DESIRED OUTPUT
QL

REQUIRED INPUT Wnet,in
QL
1

QH  QL QH / QL  1
QH
W
HEAT
PUMP
net, IN
COLD REFRIGERATED SPACE
AT TEMPERATURE TL
COPR 
(HOME)
net, IN
QL
COLD ENVIRONMENT
AT TEMPERATURE TL
COPHP 
DESIRED OUTPUT
QH

REQUIRED INPUT Wnet,in
COPHP 
QH
1

QH  Q L 1  QL / QH
CARNOT PRINCIPLES
1. The efficiency of an irreversible heat engine is always
less than of a reversible one operating between the
same two reservoirs.
2. The efficiencies of all reversible heat engines
operating between the same two reservoirs are the same
(Independent of working fluid and it’s properties, the way
cycle is executed, or the type of reversible engine)
FOR REVERSIBLE PROCESS ALONE
 QH

 QL

T
  H
 rev TL
TL
ηth, r ev 1 
TH
1
COPR, rev 
TH / TL  1
1
COPHP,rev 
1  TL / TH