2. First Law of Thermodynamics
As Energy can neither be created nor
destroyed, the net work in the cycle
must be equal to the net heat transfer
in the various processes of a cycle.
When a closed system executes a
complete cycle the sum of heat
interactions is equal to the sum of work
interactions.
Mathematically
or
The summations being over the entire
cycle
X
A
B
Y
1
2
W
Q
0
W
Q
3. First Law (Contd…)
Alternate statement:
When a closed system undergoes a cycle the cyclic
integral of heat is equal to the cyclic integral of work.
Mathematically Q = W
In other words for a two process cycle
QA1-2+QB2-1=WA1-2+WB 2-1
4. First Law (Contd…)
p
v
A
B
C
1
2
1
2
1
2
1
2
1
2
1
2
1
2
2
1
1
2
2
1
1
2
1
1
2
2
1
1
2
tan
)
(
)
(
....
..........
..........
........
).........
2
(
)
1
(
)
2
......(
..........
..........
1
2
1
...
)
1
.....(
..........
........
1
2
1
C
B
B C B C
A C A C
B
A B A B
t
Cons
W
Q
W
Q
W
W
Q
Q
W
W
Q
Q
C
A
for
W
W
Q
Q
B
A
for
W
Q
(1) Corollary 1 : energy is a property.
There exists a property called energy, such that a change in its value is equal to the
difference between the heat supplied and the work done during any change of state.
5. Hence depends on end states and is independent of the path of
process. It is a point function and is exact differential and a
thermodynamic property of mass.
dE=dU+d(P.E.)+d(K.E.)
For non-flow system K.E. and P.E. are zero and hence energy is the
internal energy (U).
Hence
q-w=u2-u1 Non-Flow Energy Equation.
First Law (Contd…)
)
( W
Q
W
Q
dE
1
2 U
U
dU
W
Q
6. (2) Corollary 2: Conservation of Energy
In an isolated system, the energy of the system remains constant.
For an isolated system, Q=0, W=0; U2-U1 = 0
U2=U1
(3) Corollary 2: Perpitual Motion Machine-1 impossible.
The continuous output of work from a system working in a cycle with
zero input of energy(heat) is impossible.
0
.......
........
0 Q
if
W
7. For a process we can have Q=0 or W=0
We can extract work without supplying
heat(during a process) but sacrificing the energy
of the system.
We can add heat to the system without doing
work(in process) which will go to increasing the
energy of the system.
Energy of a system is an extensive property
First Law(contd…)
8. Engineering Implications
When we need to derive some work, we must expend
thermal/internal energy.
Whenever we expend heat energy, we expect to derive
work interaction (or else the heat supplied is wasted or goes
to to change the energy of the system).
If you spend money, either you must have earned it or you
must take it out of your bank balance!!
!!There is nothing called a free lunch!!
9. Reversible non-flow processes
(1) Constant Volume Process (Isochoric/Isometric)
When gas is heated in a fixed enclosed space.
2
2
1
1
1
2
..
tan
..
...
...
.........
.....
0
)
(
0
1
T
p
T
p
volume
t
cons
at
heat
specific
dT
dU
c
dU
dT
mc
Q
u
q
or
dU
Q
U
U
W
U
Q
V
pd
W
V
v
v
A All the heat supplied goes to increase the internal
energy.
10. (2) Constant Pressure Process(Isobaric/Isopiestic)
gas is heated under constant pressure in a cylinder containing a sliding
piston.
If boundary is rigid as in V=Const. , pressure will raise when heat is
added. Hence for p=Const. the boundary must move against an
external resistance as heat is supplied.
2
2
1
1
1
2
1
1
2
2
1
2
1
2
..
tan
..
...
...
)
(
)
(
)
(
)
(
T
V
T
V
pressure
t
cons
at
heat
specific
dT
dH
c
dH
dT
mc
Q
H
H
pV
U
pV
U
V
V
p
U
U
W
U
Q
pdV
W
p
p
p
11. In an isothermal expansion, heat must be
added continuously in order to keep the
temperature constant. Similarly in an
isothermal compression, heat must be
added.
(3) Constant Temperature Process(Isothermal)
2
1 1
2
1
1
2
1
1
2
1
1
1
ln
ln
v
v
v
v
V
V
mRT
V
V
V
p
dV
V
V
p
pdV
W
2
1
1 2
V
p
V
p
14. Cn is negative when γ>n>1.
Heat is rejected by the gas even though its temperature rises or that heat is supplied
to the gas even though its temperature decreases.(cooling during compression and
heating during expansion).
For γ>n>1 system should work more than the heat added or more work should be
done on the system than the heat removed from the system.
For other values of n, c is positive.
Isentropic curve is steeper than an isothermal curve.
15.
16. Steady Flow Energy Equation
Assumptions :
(1)
(2) state of fluid does not vary with time at any given location.
(3) rate of work and heat transfer , crosses the boundary, is constant.
(4) no chemical energy is involved.
Energy entering the control volume = Energy leaving the control
volume
.
.
.
1 2 m
m
m
u
w
q
w
u
q
u
process
nonflow
a
in
w
gz
c
h
q
gz
c
h
w
gz
c
u
v
p
q
gz
c
u
v
p
2
1
2
2
2
2
1
2
1
1
2
2
2
2
2
2
1
2
1
1
1
1
..
...
....
2
2
2
2
17. Application of SFEE
1. Nozzle and Diffusor
Q=0, w=0, z=constant
2
2
2
2
2
2
1
1
c
h
c
h
2. Turbine and Compressor
Assuming adiabatic flow, q=0
)
(
2
2
)
(
2
2
2
2
2
2
1
1
2
2
2
2
1
1
r
Compreesso
c
h
w
c
h
Turbine
w
c
h
c
h
18. Application of SFEE…..
3. Processes involving heat transfer but no work transfer (boiler,
condenser, evaporator)
)
(
)
(
2
1
2
1
2
1
Condenser
q
h
h
Boiler
h
q
h
h
q
h
4. Reciprocating machines(Engines)
w
h
q
h
2
1
5. Throttling process(Isenthalpic Process)
It is an irreversible process ,porous plug, valve.
In this process due to fall in pressure, the gas should come out with
a large velocity, but due to friction between the fluid and
restrictions, the K.E. is converted into heat which warm up the gas
to its initial temperature and the velocity change is not so much.
2
1 h
h
19. Non-steady flow process
2
2
2
2
2
1
1
1
2
2
2
2
2
2
0
0
0
0
0
C
u
m
C
u
m
E
C
v
p
u
m
C
v
p
u
m
E
E
E
W
Q
storage
i
i
i
i
i
flow
storage
flow
20. 1. One kg of ideal gas is heated from 18oC to 93oC. Assuming R=0.264kJ/kgK, γ=1.18 for
the gas and heat transfer 160 kJ. Find (1) the specific heats, (2) the change in internal energy
(3) the change in enthalpy (4) the work done.
2.The initial pressure, volume and temperature of air in a cylinder fitted with a movable
piston are p1 = 10 bar, V1 = 0.04m3 and the temperature 400 K. if the air expands reversibly
to a volume of 0.2 m3 , calculate the work done, change in internal energy and heat absorbed
or rejected for each of the following methods of expansion
The air expands at (1) constant pressure (2) isothermally (3) according to law pv1.2 =
constant (4) adiabatically (5) according to the law pv2=constant.(Cv=.717, Cp=1.005 kJ/kgK)
3. when a system is taken from state 1 to state 2, as shown in Fig. along the path 1-3-2, 100
kJ of heat flows into the system and the system
does 40 kJ of work. (1) how much will the heat that
flows into the system along path 1-4-2 be, if the work
done is 20 kJ? (2) When the system is returned from
2 to 1 along the straight path, calculate the work done
and heat exchanged. (3) if U1=0 and U4=50 kJ, find the
heat absorbed in the processes 1-4 and 4-2.
4. A 3m3 insulated tank having initial charge of helium at 300 K and 100 kPa is charged
from a constant temperature supply line at 300 K and 1000 kPa. If the final tank pressure is
900 kPa, determine the final helium temperature. Take R for helium 2.077 kJ/kg K and γ=
p
v
3
1
2
4
