first law of thermodynamics and second law

It will be very benifit for everyone

1. Made by:
NAPHIS AHAMAD(Assistant Professor)
MECHANICAL ENGINEERING
JIT
06/10/17Naphis Ahamad (ME) JIT 1

2. 06/10/17Naphis Ahamad (ME) JIT 2

3. First Law of Thermodynamics
06/10/17JIT 3
The First Law is usually referred to as the Law of Conservation
of Energy, i.e. energy can neither be created nor destroyed, but
rather transformed from one state to another.
The energy balance is maintained within the system being
studied/defined boundary.
The various energies associated are then being observed as
they cross the boundaries of the system.

4. 06/10/17JIT 4
Energy Balance for Closed System
Heat
Work
z
Closed
System
Reference Plane, z = 0

V
or
EE E
in out system
- =D

5. 06/10/17JIT 5
According to classical thermodynamics
QW E
net net system
-=D
The total energy of the system, E
system
, is given as
EInternalenergyKineticenergyPotentialenergy
EUKEPE
= + +
= + +
The change in stored energy for the system is
DDD DEUKEPE=+ +
The first law of thermodynamics for closed systems then can be
written as
QW UKEPE
net net-=+ +DD D

6. 06/10/17JIT 6
If the system does not move with a velocity and has no change in
elevation, the conservation of energy equation is reduced to
QW U
net net
-=D
The first law of thermodynamics can be in the form of
)(
1000
)(
2000
12
2
1
2
2
12
kJ
zzgVV
uumWQ net
net ÷
÷
ø
ö
ç
ç
è
æ -
+
-
+-=-
)/(
1000
)(
2000
12
2
1
2
2
12 kgkJ
zzgVV
uuwq
netnet ÷
÷
ø
ö
ç
ç
è
æ -
+
-
+-=-
For a constant volume process,
÷
÷
ø
ö
ç
ç
è
æ -
+
-
+-=-
1000
)(
2000
12
2
1
2
2
12
zzgVV
uumWQ net
net
÷
÷
ø
ö
ç
ç
è
æ -
+
-
+-=
1000
)(
2000
12
2
1
2
2
12
zzgVV
uumQ
net

7. 06/10/17JIT 7
For a constant pressure process,
÷
÷
ø
ö
ç
ç
è
æ -
+
-
+-=-
1000
)(
2000
12
2
1
2
2
12
zzgVV
uumWQ net
net
÷
÷
ø
ö
ç
ç
è
æ -
+
-
+-=--
1000
)(
2000
)(
12
2
1
2
2
1212
zzgVV
uumVVPQ
net
÷
÷
ø
ö
ç
ç
è
æ -
+
-
+-+-=
1000
)(
2000
)(
12
2
1
2
2
1212
zzgVV
VVPuumQ
net
÷
÷
ø
ö
ç
ç
è
æ -
+
-
+-=
1000
)(
2000
12
2
1
2
2
12
zzgVV
hhmQ
net

8. 06/10/17JIT 8
A closed system of mass 2 kg
undergoes an adiabatic process.
The work done on the system is
30 kJ. The velocity of the system
changes from 3 m/s to 15 m/s.
During the process, the elevation
of the system increases 45 meters.
Determine the change in internal
energy of the system.
Example.1
Solution:
Energy balance,
÷
÷
ø
ö
ç
ç
è
æ -
+
-
+-=-
1000
)(
2000
12
2
1
2
2
12
zzgVV
uumWQ net
net
Rearrange the equation
net
Q
( )
()
2 2
2 1 2 1
2 1
2 2
2 1 2 1
2 1
2 2
( )
2000 1000
( )
2000 1000
9.81 4515 3
30 2 2 2
2000 1000
14.451 ..
net
net
V V g z z
W m u u
V V g z z
W m u u
u
u kJ Ans
æ ö - -
- = - + +ç ¸
è ø
æ ö - -
- = - + +ç ¸
è ø
æ öæ ö-
- - = D + + ç ¸ç ¸
ç ¸
è ø è ø
D =

9. 06/10/17JIT 9
Some thermodynamic cycle composes of processes in which
the working fluid undergoes a series of state changes such
that the final and initial states are identical.
For such system the change in internal energy of the
working fluid is zero.
The first law for a closed system operating in a
thermodynamic cycle becomes
Closed System First Law of a Cycle
QW U
QW
net net cycle
net net
-=
=
D

10. 06/10/17JIT 10
NoValue of nProcess Description Result of IGL
1 ∞ isochoricconstant volume (V
1
= V
2
)
2 0 isobaric constant pressure (P
1
= P
2
)
3 1 isothermalconstant temperature
(T
1
= T
2
)
4 1<n< γpolytropic-none-
5 γ isentropicconstant entropy (S
1
= S
2
)
According to a law of constant=
n
VP
2
2
1
1
T
P
T
P
=
2
2
1
1
T
V
T
V
=
2211 VPVP=
1
2
1
1
2
2
1
-
÷
÷
ø
ö
ç
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
=
n
n
n
T
T
V
V
P
P

11. 06/10/17JIT 11
Boundary Works
3
2
4
5
1
P
V

12. 06/10/17JIT 12
Various forms of work are expressed as follows
Process Boundary Work
isochoric
isobaric
isothermal
polytropic
isentropic
0)(
1212 =-= VVPW
)(
1212
VVPW -=
1
2
1112
ln
V
V
VPW=
n
VPVP
W
-
-
=
1
1122
12

13. 06/10/17JIT 13
Conservation of Mass
Conservation of mass is one of the most fundamental
principles in nature. We are all familiar with this
principle, and it is not difficult to understand it!
For closed system, the conservation of mass principle is
implicitly used since the mass of the system remain
constant during a process.
However, for control volume, mass can cross the
boundaries. So the amount of mass entering and leaving
the control volume must be considered.

14. 06/10/17JIT 14
Mass and Volume Flow Rates
Mass flow through a cross-sectional area per unit time is called the
mass flow rate. Note the dot over the mass symbol indicates a time
rate of change. It is expressed as
ò
= dAVm .r
If the fluid density and velocity are constant over the flow cross-
sectional area, the mass flow rate is
voulmespecificcalledis
where
AV
AVm
n
r
n
n
r
1
=
==

15. 06/10/17JIT 15
Principal of Conservation of Mass
The conservation of mass principle for a control volume can be
expressed as
in out CV
m m m- =  
For a steady state, steady flow process the conservation of mass
principle becomes
(kg/s)
in out
m m= 

16. 06/10/17JIT 16
As the fluid upstream pushes mass across the control volume, work
done on that unit of mass is
flow
flow
flow
A
W FdL FdL PdV Pv m
A
W
w Pv
m
d d
d
d
d
= = = =
= =
Flow Work & The Energy of a Flowing Fluid

17. 06/10/17JIT 17
The total energy carried by a unit of mass as it crosses the control
surface is the sum of the internal energy + flow work + potential
energy + kinetic energy
å ++=+++= gz
V
hgz
V
Puenergy
22
22
n
The first law for a control volume can be written as
åå ÷
÷
ø
ö
ç
ç
è
æ
++-
÷
÷
ø
ö
ç
ç
è
æ
++=-
in
in
in
inin
out
out
out
outout
net
net
gz
V
hmgz
V
hmWQ
22
2
.
2
...
Total Energy of a Flowing Fluid

18. 06/10/17JIT 18
Total Energy of a Flowing Fluid
The steady state, steady flow conservation of mass and first law of
thermodynamics can be expressed in the following forms
)(
1000
)(
2000
12
2
1
2
2
12
...
kW
zzgVV
hhmWQ net
net ÷
÷
ø
ö
ç
ç
è
æ -
+
-
+-=-
)(
1000
)(
2000
12
2
1
2
2
12 kJ
zzgVV
hhmWQ net
net ÷
÷
ø
ö
ç
ç
è
æ -
+
-
+-=-
)/(
1000
)(
2000
12
2
1
2
2
12 kgkJ
zzgVV
hhwq
netnet ÷
÷
ø
ö
ç
ç
è
æ -
+
-
+-=-

19. 06/10/17JIT 19
Nozzle & Diffuser
Nozzle - device that increases
the velocity fluid at the expense
of pressure.
Diffuser - device that increases
pressure of a fluid by slowing it
down.
Commonly utilized in jet
engines, rockets, space-craft
and even garden hoses.
Q = 0 (heat transfer from the
fluid to surroundings very
small
W = 0 and ΔPE = 0

20. 06/10/17JIT 20
Energy balance (nozzle & diffuser):
åå ÷
÷
ø
ö
ç
ç
è
æ
++++=
÷
÷
ø
ö
ç
ç
è
æ
++++
out
out
out
outout
out
out
in
in
in
inin
in
in
gz
V
hmWQgz
V
hmWQ
22
2
...
2
...
÷
÷
ø
ö
ç
ç
è
æ
+=
÷
÷
ø
ö
ç
ç
è
æ
+
22
2
.
2
.
out
outout
in
inin
V
hm
V
hm
÷
÷
ø
ö
ç
ç
è
æ
+=
÷
÷
ø
ö
ç
ç
è
æ
+
22
2
2
2
2
1
1
V
h
V
h

21. 06/10/17JIT 21
Turbine & Compressor
Turbine – a work producing device through the expansion of a
fluid.
Compressor (as well as pump and fan) - device used to increase
pressure of a fluid and involves work input.
Q = 0 (well insulated), ΔPE = 0, ΔKE = 0 (very small compare
to Δenthalpy).

22. 06/10/17JIT 22
Energy balance: for turbine
åå ÷
÷
ø
ö
ç
ç
è
æ
++++=
÷
÷
ø
ö
ç
ç
è
æ
++++
out
out
out
outout
out
out
in
in
in
inin
in
in
gz
V
hmWQgz
V
hmWQ
22
2
...
2
...
() ( )
outout
out
inin hmWhm
...
+=
( )
21
..
hhmWout -=

23. 06/10/17JIT 23
Energy balance: for compressor, pump and fan
åå ÷
÷
ø
ö
ç
ç
è
æ
++++=
÷
÷
ø
ö
ç
ç
è
æ
++++
out
out
out
outout
out
out
in
in
in
inin
in
in
gz
V
hmWQgz
V
hmWQ
22
2
...
2
...
() ( )
outoutinin
in hmhmW
...
=+
( )
12
..
hhmWin -=

24. 06/10/17JIT 24
Throttling Valve
Flow-restricting devices that
cause a significant pressure drop
in the fluid.
Some familiar examples are
ordinary adjustable valves and
capillary tubes.

25. 06/10/17JIT 25
Devices where two moving fluid
streams exchange heat without
mixing.
Heat exchangers typically involve
no work interactions (w = 0) and
negligible kinetic and potential
energy changes for each fluid
stream.
Heat Exchanger

26. 06/10/17JIT 26
Second Law of Thermodynamics
Kelvin-Planck statement
No heat engine can have a
thermal efficiency 100
percent.
As for a power plant to
operate, the working fluid
must exchange heat with the
environment as well as the
furnace.

27. 06/10/17JIT 27
Heat Engines
Work can easily be converted to other forms of energy,
but?
Heat engine differ considerably from one another, but
all can be characterized :
othey receive heat from a high-temperature source
othey convert part of this heat to work
othey reject the remaining waste heat to a low-temperature sink
atmosphere
othey operate on a cycle

28. 06/10/17JIT 28
The work-producing
device that best fit into the
definition of a heat engine
is the steam power plant,
which is an external
combustion engine.

29. 06/10/17JIT 29
Represent the magnitude of the energy wasted in order to
complete the cycle.
A measure of the performance that is called the thermal
efficiency.
Can be expressed in terms of the desired output and the
required input
h
th=
DesiredResult
Required Input
For a heat engine the desired result is the net work done
and the input is the heat supplied to make the cycle
operate.

30. 06/10/17JIT 30
The thermal efficiency is always less than 1 or less than
100 percent.
h
th
netout
in
W
Q
=
,
W WW
QQ
netout out in
in net
,= -
¹
where

31. 06/10/17JIT 31
Applying the first law to the cyclic heat engine
Q W U
W Q
W QQ
netin netout
netout netin
netout in out
, ,
, ,
,
- =
=
=-
D
The cycle thermal efficiency may be written as
h
th
netout
in
in out
in
out
in
W
Q
QQ
Q
Q
Q
=
=
-
=-
,
1

32. 06/10/17JIT 32
A thermodynamic temperature scale related to the heat transfers between a
reversible device and the high and low-temperature reservoirs by
Q
Q
T
T
L
H
L
H
=
The heat engine that operates on the reversible Carnot cycle
is called the Carnot Heat Engine in which its efficiency is
h
threv
L
H
T
T
,=-1

33. 06/10/17JIT 33
Heat Pumps and Refrigerators
A device that transfers heat from a low temperature
medium to a high temperature one is the heat pump.
Refrigerator operates exactly like heat pump except that
the desired output is the amount of heat removed out of
the system
The index of performance of a heat pumps or refrigerators
are expressed in terms of the coefficient of performance.

34. 06/10/17JIT 34

35. 06/10/17JIT 35
COP
Q
W
Q
QQ
HP
H
netin
H
H L
= =
-
,
COP
Q
W
R
L
netin
=
,

36. 06/10/17JIT 36
Process Description
1-2 Reversible isothermal heat addition at
high temperature
2-3 Reversible adiabatic expansion from high
temperature to low temperature
3-4 Reversible isothermal heat rejection at
low temperature
4-1 Reversible adiabatic compression from low
temperature to high temperature
Carnot Cycle

37. 06/10/17JIT 37
Execution of Carnot cycle in a piston cylinder device

38. 06/10/17JIT 38

39. 06/10/17JIT 39
The thermal efficiencies of actual and reversible heat engines
operating between the same temperature limits compare as follows
The coefficients of performance of actual and reversible refrigerators
operating between the same temperature limits compare as follows

40. 06/10/17JIT 40
Example1
A steam power plant
produces 50 MW of net
work while burning fuel
to produce 150 MW of
heat energy at the high
temperature. Determine
the cycle thermal
efficiency and the heat
rejected by the cycle to
the surroundings.
Solution:
h
th
netout
H
W
Q
MW
MW
=
= =
,
.
50
150
0333 or 33.3%
W QQ
QQW
MW MW
MW
netout H L
L H netout
,
,
=-
=-
= -
=
150 50
100

41. 06/10/17JIT 41
A Carnot heat engine receives 500 kJ of heat per cycle from a high-
temperature heat reservoir at 652ºC and rejects heat to a low-
temperature heat reservoir at 30ºC. Determine :
(a) The thermal efficiency of this Carnot engine
(b) The amount of heat rejected to the low-temperature heat
reservoir
Example.2
Q
L
W
OUT
Q
H
T
H
= 652
o
C
T
L
= 30
o
C
HE
h
threv
L
H
T
T
K
K
or
,
( )
( )
. .
=-
=-
+
+
=
1
1
30273
652273
0672 672%
Q
Q
T
T
K
K
Q kJ
kJ
L
H
L
H
L
=
=
+
+
=
=
=
( )
( )
.
(.)
30273
652273
0328
5000328
164
Solution:

42. 06/10/17JIT 42
An inventor claims to have developed a refrigerator that maintains
the refrigerated space at 2ºC while operating in a room where the
temperature is 25ºC and has a COP of 13.5. Is there any truth to his
claim?
Example.3
Solution:
Q
L
W
in
Q
H
T
H
= 25
o
C
T
L
= 2
o
C
R
COP
Q
QQ
T
TT
K
K
R
L
H L
L
H L
=
-
=
-
=
+
-
=
( )
( )
.
2273
252
1196
this claim is also false!

43. 06/10/17JIT 43
• Identifies the direction of a process. (e.g.: Heat can only spontaneously
transfer from a hot object to a cold object, not vice versa)
• Used to determine the “Quality” of energy. (e.g.: A high-temperature energy
source has a higher quality since it is easier to extract energy from it to deliver
useable work.)
• Used to exclude the possibility of constructing 100% efficient heat engine and
perpetual-motion machines. (violates the Kevin-Planck and the Clausius
statements of the second law)
• Used to introduce concepts of reversible processes and irreversibilities.
• Determines the theoretical performance limits of engineering systems. (e.g.:
A Carnot engine is theoretically the most efficient heat engine; its performance
can be used as a standard for other practical engines)

44. 06/10/17JIT 44
• A process can not happen unless it satisfies both the first and second laws of
thermodynamics. The first law characterizes the balance of energy which defines
the “quantity” of energy. The second law defines the direction which the process
can take place and its “quality”.
• Define a “Heat Engine”: A device that converts heat into work while operating
in a cycle.
Heat engine
Q
H
Q
L
T
H
T
L
W
net
DQ-W
net
=DU (since DU=0 for a cycle)
ÞW
net
= Q
H
-Q
L
Thermal efficiency, h
th
is defined as
h
th
=W
net
/Q
H
=(Q
H
-Q
L
)/Q
H
=1-(Q
L
/Q
H
)
Question: Can we produce an 100%
heat engine, i.e. a heat engine where
Q
L
=0?

45. 06/10/17JIT 45
• A steam power plant is a good example of a heat engine where the working
fluid, water, undergoes a thermodynamic cycle
W
net
= W
out
- W
in
= Q
in
-Q
out
Q
in
is the heat transferred from the high temp.
reservoir, and is generally referred to as Q
H
Q
out
is the heat transferred to the low temp.
reservoir, and is generally referred to as Q
L
Thermal efficiency
h
th
= W
net
/Q
H
= (Q
H
-Q
L
)/Q
H
=1-(Q
L
/Q
H
)
Typical Efficiency of a large commercial steam
power plant » 40%
Thermal Reservoir
A hypothetical body with a very large thermal
capacity (relative to the system beig examined)
to/from which heat can be transferred without
changing its temperature. E.g. the ocean,
atmosphere, large lakes.

46. 06/10/17JIT 46
• The Kelvin-Planck Statement is another expression of the second law of
thermodynamics. It states that:
It is impossible for any device that operates on a cycle to receive heat
from a single reservoir and produce net work.
• This statement is without proof, however it has not been violated yet.
• Consequently, it is impossible to built a heat engine that is 100%.
Heat engine
Q
H
T
H
W
net
• A heat engine has to reject some
energy into a lower temperature sink
in order to complete the cycle.
• T
H
>T
L
in order to operate the
engine. Therefore, the higher the
temperature, T
H
, the higher the
quality of the energy source and
more work is produced.
Impossible because it violates the Kelvin-Planck Statement/Second Law

47. 06/10/17JIT 47
• A “heat pump” is defined as a device that transfers heat from a low-temperature
source to a high-temperature one. E.g. a heat pump is used to extract energy from
outside cold outdoor air into the warm indoors.
• A refrigerator performs the same function; the difference between the two is in
the type of heat transfer that needs to be optimized.
• The efficiencies of heat pumps and refrigerators are denoted by the Coefficient
of Performance (COP) where
Heat pump/
Refrigerator
Q
H
Q
L
T
H
T
L
W
net
For a Heat Pump:
COP
HP
=Q
H
/W
net
=Q
H
/(Q
H
-Q
L
) = 1/(1-Q
L
/Q
H
)
For a Refrigerator:
COP
R
=Q
L
/W
net
=Q
L
/(Q
H
-Q
L
) = 1/(Q
H
/Q
L
-1)
Note: COP
HP
= COP
R
+ 1
• COP
HP
>1, ex: a typical heat pump has a COP
in the order of 3
• Question: Can one build a heat pump
operating COP= ¥, that is W
net
= 0 and Q
H
=Q?

48. 06/10/17JIT 48
• The Clausius Statement is another expression of the second law of thermodynamics.
It states that:
It is impossible to construct a device that operates in a cycle and produces
no effect other than the transfer of heat from a lower-temperature body to a
higher-temperature body.
• Similar to the K-P Statement, it is a negative statement and has no proof, it is based
on experimental observations and has yet to be violated.
• Heat can not be transferred from low temperature to higher temperature unless
external work is supplied.
Heat pump
Q
H
Q
L
T
H
T
L
Therefore, it is impossible to build
a heat pump or a refrigerator
without external work input.

49. 06/10/17JIT 49
It can be shown that the violation of one statement leads to a violation of the
other statement, i.e. they are equivalent.
A 100% efficient heat engine; violates K-P Statement
Heat pump
Q
L
Q
L
T
H
T
L
Heat transfer from low-temp body to
high-temp body without work; A
violation of the Clausius statement
Heat pump
Q
H
+Q
L
Q
L
T
H
T
L
W
net
=Q
H
Heat engine
Q
H

50. 06/10/17JIT 50
Imagine that we can extract energy from unlimited low-temperature energy sources
such as the ocean or the atmosphere (both can be thought of as thermal reservoirs).
Heat
engine
Heat
pump
Q
L
Q
H
Q
H
W
in
= Q
H
-Q
L
W
net
=Q
L
T
H
Ocean T
L
It is against the Kevin-Planck
statement: it is impossible to
build an 100% heat engine.
Perpetual Motion Machines, PMM, are classified into two types:
PMM1- Perpetual Motion Machines of the First Kind: They violate the First Law
of Thermodynamics
PMM2 - Perpetual Motion Machines of the Second Kind : Violate the Second
Law of Thermodynamics

51. 06/10/17JIT 51
• A reversible process is one that can be executed in the reverse direction with no net
change in the system or the surroundings.
• At the end of a forwards and backwards reversible process, both system and the
surroundings are returned to their initial states.
• No real processes are reversible.
• However, reversible processes are theoretically the most efficient processes.
• All real processes are irreversible due to irreversibilities. Hence, real processes are
less efficient than reversible processes.
Common Sources of Irreversibility:
• Friction
• Sudden Expansion and compression
• Heat Transfer between bodies with a finite temperature difference.

52. 06/10/17JIT 52
• A work-producing device which employs quasi-equlibrium or reversible
processes produces the maximum amount of work theoretically possible.
•A work-consuming device which employs quasi-equilibrium or reversible
processes requires the minimum amount of work theoretically possible.
• One of the most common idealized cycles that employs all reversible processes
is called the Carnot Cycle proposed in 1824 by Sadi Carnot.
•A quasi-equilibrium process, e.g. very slow, frictionless expansion or
compression is a reversible process.