engineering thermodynamic first law of thermodynamic

1. Shroff S.R. Rotary Institute of Chemical Technology
Principle Supporter & Sponsor-United Phosphorous Ltd(UPL)/Shroff family
Managed By Ankleshwar Rotary Education Society
Approved by AICTE, New Delhi, Govt. of Gujarat & GTU Affiliated
SUBJECT :- ENGINEERING THERMODYNAMICS
TOPIC :- FIRST LAW OF THERMODYNAMICS

2.  First law of thermodynamics for
a closed system undergoing a cycle :
•When system is made to undergo a complete cycle
then net work is done on the system or by the
system. Let us consider a cycle in which net work
is done by the system. As per conservation of
energy principle, energy can not be created, so this
mechanical energy must have been supplied to
system from some source of energy.

3. • The first law of thermodynamics can be stated as
follows.
• 1). “When a system undergoes a thermodynamics
cycle then the net heat added to the system from
the surrounding is equal to net work done by the
system on its surroundings” or §δ Q= §δ W
,where § Cyclic integration
represents the sum for a complete cycle.
• 2) "Heat and work are mutually convertible but
energy can neither be created nor destroyed, the
total energy involved with an energy conversion
remains constant".

4. 3). There is not any machine which can produce
energy without corresponding source of energy".

5. Joule's experiment
• During 1843 to 1848 , Joule conducted several
experiment and formulated first law of
thermodynamics. His experiment’s consists of
two process cycles carried out on a system.
• In process 1- 2, work W12 was done on the system
by means of a paddle wheel. The amount of work
mg falling through height h, caused a rise in the
temperature of the fluid. The system was initially
at temperature t 1, the same as that of atmosphere,
and after work transfer, temperature rise t2 at
constant pressure 1 atm. The process. 1 - 2
undergone by the system is shown in figure.

7. • The system was place in contact with
surroundings by removing insulation as shown
in figure Heat was transferred from the fluid to
the surrounding in process 2-1, until the original
state of the fluid was restored. The amount of
heat transfer Q21 from the fluid to the
surroundings during process 2-1 shown in figure
was estimated. Joule found W12 is always
proportional to the heat Q21

9. • There as system performs a thermodynamic
cycle. If the cycle involved number of heat and
work interaction, same result will be found.
• Where J is the joule's equivalent or mechanical
equivalent of heat.
• The conclusion from the Joule's experiment
leads to the statement of first law of
thermodynamics as "During a cycle, a system
undergoes the cyclic integral of heat added is
equal to the cyclic integral of work done".

10. First law of thermodynamics for a
closed system undergoing a change of
state
• In previous article, expression applies only to
system undergoing cycles, in which there is no
change of state, no change an exciting energy of
system or no energy provided by system itself
.But If a system undergoes a change of state
during which both heat transfer and work
transfer are involved,the net energy transfer will
be stored accumulated within the system.

11. • For example, a system receives 15 kJ Heat while
paddle wheel does 8 kJ work on the system and
system rejects7 kJ heat surrounding as shown in
figure
• The net increase in energy of the system for this
process will be 15+8-7=16kJ. This 16kJ energy will
be stored in the system . The stored energy is
neither heat nor work, and is called internal energy
or the energy of the system.
• The generalized our conclusion as shown in figure
the first law of thermodynamics for a closed system
undergoes a change of state may be expressed as
follows.

12. • [Net energy transferred to or from the system is heat and
work] = [Net increase or decrease in the total energy of
the system]
• The total energy of a system is consists of three parts as
internal energy kinetic energy and potential energy and
Total energy of a system during a process can be
expressed as sum of changes in its U,KE and PE.
• It is very important that the sign convention be observed
heat and work interactions.
• Heat flow to a system, Q is positive
• Heat flow form a system, Q is negative
• Work done by a system, W is positive
• Work done on a system, W is negative

13. Enthalpy
• While analyzing processes we frequently come
across certain combination of properties. for sake of
simplicity and convenience this combination is
defined as a new thermodynamics properties the
enthalpy is combination of two properties as internal
energy and product of pressure and volume. It
express by
• Since u, p and v are all properties there for h also
property it is an intensive property of a system
because that is specific enthalpy multiplying both
side by mass m of equation.

14.  PERPETUAL MOTION MACHINE
OF THE FIRST KIND : 1
• Any device that violates either law is called a
perpetual motion machine.
• A device that violates the first law of
thermodynamics is called perpetual motion
machine of the first kind (PMM1).

16.  NON-FLOW PROCESS AND FLOW
PROCESS :-
Non-flow process : A process undergone by a
closed system of fixed mass is called non flow
process.
• This process does not permits the flow of mass
in or out of the system.
Flow process : In this process , the fluid(mass)
enters the system and leaves after doing work.
• This implies open boundary which permits the
flow of the mass to and from the system.

17.  FLOW PROCESS AND CONTROL
VOLUME :-
• A flow process constitute an open system through
which the working fluid enters and leaves from the
surface of the system.
• There may also be energy interaction in the form
of heat and work, internal energy, can you take
energy, potential energy, etc.
• Most of engineering devices such as internal
combustion engine, steam engines, boilers,
turbines, pumps, compressors, condensers, et
cetera, the working flowed continuously flew in
and out of the plant of the devices.

19. • Analyses of flow process in such devices is done
with the concept of control volume and control
surface.
• For example, consider a portion of steamed
turbine plant in the figure.
• The high pressure and temperature steam enters
the steam turbine at section 1 and after doing work
on the rotor due to expansion of steam, and steam
comes out at low pressure at section 2.
• In these types of flow process, the analysis like in
close system (non-flow process), for finding the
heat and work interactions is not feasible.

20. • Hence the concept of control volume and control
surface is used which is very convenient and
feasible.
• Control volume is an imaginary envelope is
supposed to exist around equipment.
• The space bounded by control volume is called
control surface. But there is a difference between
system boundary and control surface.
• The boundary of the system may change shape,
position and orientation, and it may movable, but
the control surface is fixed and the Mars (matter)
flows across the control surface.

21. • The concept of control volume helps to relate the
energy interaction as work and heat with average
change of state of fluid from entrance to exit
section.

22.  STEADY AND UNSTEADY FLOW
PROCESS :
 STEADY FLOW PROCESS :-It is a flow
process in which fluid parameters at
particular point of the control volume remain
constant during entire process.
• The steady means no change with time.
• This means fluid parameters like velocity ,
pressure, temperature etc., are functions only
of location and do not very with time as
shown in figure.

24.  UNSTEADY FLOW PROCESS :-
• It is a flow process in which fluid parameters at
any point of the control volume is not constant
during entire process.
• This means fluid parameters like velocity,
pressure, temperature etc., are not only functions
of location but also function of time as shown in
figure.

26.  The steady flow process is characterized by the
following conditions in a control volume:
(1) The mass flow rate remains constant within the
system, i.e. mass entering the control volume must
be equal to the mass leaving it and do not ready with
time.
(2) The state of fluid at any fixed point in control
volume is same and do not very with time.
(3) The state of energy of the fluid at the entrance
and the exit of the control volume does not vary
with the time.
(4) Heat and work transfer rate across the control
surface does not vary with time.

27. (5) Chemical composition of the fluid within the
control volume is fixed. So change in the chemical
energy is not involved.
• When any flow process does not satisfy the
above conditions of steady state, and process is
known as unsteady flow process.

28.  STEADY FLOW ENERGY
EQUATION (SFEE) :-
• Consider flow of fluid through a generalized
open system as shown in figure.
• The working fluid enter the system at section 1
and leave the system at section 2 and passing at a
steady rate.
• Let,
m= mass flowing through the control volume, kg/s
Q= heat entering the control volume, kJ
W= work transferred from control volume, kJ
C1,C2= velocity of fluid at entrance and exit, m/s

29. p1,p2 = pressure of fluid at entrance and exit, N/m2
u1,u2 =internal energy per kg of fluid at entrance
and exit, kJ/kg
v1,v2 = specific volume of fluid at entrance and exit
m3/kg
h1,h2 = enthalpy of fluid at entrance and exit, kJ/kg
C.V= Control volume
q= heat entering the control volume per kilogram
of fluid, kJ/kg
w= work transferred from the control volume per
kg of fluid, kJ/kg
z1,z2 = elevation of entrance section and exit
section, m.

31. The energy balance required in open system for
flow process may be written as follows
Energy
entering to
the C.V
+
Heat
entering to
the C.V
=
Energy
leaving the
C.V
+
Work
transferred
from the C.V
+
Increase of stored
energy within the
C.V
For steady flow process,increase of stored energy
within the control volume is zero.

32. Energy
entering to
the C.V
Heat
entering to
the C.V
Energy
leaving the
C.V
Work
transferred
from the C.V
+
+
=∴
∴
Internal
energy
at sec 1
Flow
work
at sec 1
Kinetic
energy
at sec 1
Potential
energy at
sec 1
Heat
entering
to the
C.V
+ + + +
= Internal
energy
at sec 2
Flow
work
at sec 2
Kinetic
energy
at sec 2
Potential
energy at
sec 2
+ + + +
Work
transferred
from the
C.V

33. ∴ m1 u1+p1v1+ +z1g
2
1
2
C + Q =
u2+p2v2+ +z2gm2
2
2
2
C
+ W
It is applicable to compressible and incompressible
fluid is, liquids and gases, ideal and real fluids.
If m1 = m2 = m

34. m u1+p1v1+ +z1g
2
1
2
C
+ Q =
u2+p2v2+ +z2gm
2
2
2
C
+ W
m h1+ +z1g
2
1
2
C
+ Q = h2+ +z2gm
2
2
2
C
+ W
∴
∴
This equation is called steady flow energy
equation(SFEE).

35.  THROTTLING DEVICE :
• Function: It is the flow restricting device as
shown in figure, that cause a large pressure drop in
the fluid.
• Throttling process: It is an irreversible process in
which fluid flows across the restriction in a manner
that cause the reduction in pressure and increase
volume.
• In this process no heat flows from and to the
system, no change in enthalpy of the fluid, no
change in kinetic energy and potential energy and
there is no work transfer.

36. •The throttling process is mostly used to regulate
speed of turbine, to determine the condition of
steam (dryness fraction) and to reduce the pressure
of refrigerant before entry into the evaporator in the
refrigeration plant.

38. We know that SFEE,
m h1+ +z1g
2
1
2
C
+ Q = h2+ +z2gm
2
2
2
C
+ W
For throttling device,
1) ∆KE ≈ 0,(C1 = C2 ),(negligible change in KE)
2) ∆PE = 0,(z1 = z2 ),pipe is horizontal
3) Q = 0, (pipe is thermally insulated)
4) W = 0, (no shaft work involved)
∴mh1 = mh2
∴ h1 = h2 ,i.e., enthalpy of fluid remains constant
during throttling process

39. For ideal gas h=Cp T
∴Cp T1 = Cp T2
∴ T1 = T2
For ideal gas, throttling process takes place at
constant enthalpy , constant temperature and
constant internal energy.
For real gas,h ≠ f (T)
∴T1 ≠ T2
∴ u1 ≠ u2
∴ h1 = h2
Or u1 + p1v1 = u2 + p2v2
Internal energy + Flow energy = constant
Therefore change in internal energy is not zero in
real gas.

40. ENGINEERING
APPLICATION OF STEADY
FLOW ENERGY EQUATION
(1)Nozzle
Function : It is a device that increases the velocity
of a fluid at the expense of its pressure drop.
pressure energy Kinetic energy
(pressure) (velocity)
conversion

41. Nozzle is a passage of varying cross
section by means of which pressure energy of
following fluid is converted into kinetic energy.
We know that SFEE,
For nozzle,
(1)W=0,
(2)Q=0,
(3)Z1 – Z2 = 0
Wgz
c
hmQgz
c
hm  )
2
()
2
( 2
2
2
21
2
1
1

42. Hence, SFEE
0)0
2
(0)0
2
(
2
2
2
2
1
1 
c
hm
c
hm
21
2
2
2
1
2
hh
cc



2
1212 )(2 chhc 
When inlet velocity c1 is small compared to the
exit velocity c2,∴ c1 = 0

43. We get 𝐶2 = 2(ℎ1 − ℎ2)
If the working fluid is a perfect gas, then
∴ ℎ1 − ℎ2 = 𝐶 𝑝(𝑇1 − 𝑇2) and
𝑇2
𝑇1
= (
𝑃2
𝑃1
)
𝛾−1
𝛾
∴ 𝐶2 = 2𝐶 𝑝(𝑇1 − 𝑇2)
= 2𝐶 𝑝 𝑇1(1 −
𝑇2
𝑇1
)
∴ 𝐶2 = 2𝐶 𝑝 𝑇1(1 − (
𝑃2
𝑃1
)
𝛾−1
𝛾
)

44. (2) Diffuser
Function : It is a device that increase pressure of
a fluid at the expense of its velocity drop.
A diffuser is a passage of varying cross section
by mean of which kinetic energy of flowing fluid
is converted into pressure energy.
Kinetic energy Pressure energy
(velocity) (pressure)
The cross section area of diffuser increase in the
Flow direction for subsonic flow as shown in fig.
Decrease for supersonic flow.
We know that SFEE,
conversion

45. Wgz
c
hmQgz
c
hm  )
2
()
2
( 2
2
2
21
2
1
1
For Diffuser,
1) W=0
2) Q =0
3) Z1-Z2=0
Hence,
21
2
2
2
1
2
hh
cc



2
2121 )(2 chhc 

46. In case of diffuser, C2≪ C1
𝐶1 = 2(ℎ2 − ℎ1)
= 2𝐶 𝑝(𝑇2 − 𝑇1)
∴ 𝐶1 = 2𝐶 𝑝 𝑇1 (
𝑃2
𝑃1
)
𝛾−1
𝛾
− 1

47. (3) Boiler
Function : It is a device or equipment in which
heat is produce by combustion of fuel is utilize
to produce steam from water. at desired
temperature and pressure as shown in fig.
Fuel + Air Heat Water
Steam
combustion

48. 1) W=0
2) Z1-Z2=0
Hence, mh1 + Q = mh2
∴ 𝑄 = 𝑚 ℎ2 − ℎ1
Wgz
c
hmQgz
c
hm  )
2
()
2
( 2
2
2
21
2
1
1
0
2
2
2
2
1

 cc

49.  Hydraulic or water turbine
Function : It is hydraulic machine, converts
hydraulic energy into mechanical energy as
shown in fig.
Water (with high P.E) Mechanical
energy
We know that SFEE

50. For Hydraulic turbine,
1) Q=0
2) W=+W
3) U1=U2
h1-h2 = u1 + p1v1 - u2 - p2v2 = p1v1 - p2v2
Hence,
Wgz
c
hmgz
c
hm  )
2
()
2
( 2
2
2
21
2
1
1
Wgz
c
vpmgz
c
vpm  )
2
()
2
( 2
2
2
221
2
1
11

51.  Steam or Gas turbine
Function :- It is device for producing work output
from a flow of fluid which is expanding from high
pressure to low pressure. The work output from the
turbine may be used to run a generator and produce
electric power as shown in fig.

52. We know that SFEE,
For turbine,
1. = , ∆PE = 0
2. , ∆KE = 0
3. Q =0
4. W ≠ 0(shaft work is done by the system,so W is
+ve)
i.e, W = m
Wgz
c
hmQgz
c
hm  )
2
()
2
( 2
2
2
21
2
1
1
1z 2z
21 cc 
)( 21 hh 

53. Compressor
Function:- It is device used to increase the pressure of
a fluid.
1. Rotary compressor:- We know that SFEE,
For rotary compressor,
1.
2.
Wgz
c
hmQgz
c
hm  )
2
()
2
( 2
2
2
21
2
1
1
021  zz
21 cc 

54. 3. Q =0
4. W ≠ 0(shaft work is done on the system,so W is -ve)
W = m , where,)( 21 hh  12 hh 

55. (2) Reciprocating compressor :-
We know that SFEE,
For reciprocating compressor,
1. = , ∆PE = 0
2. , ∆KE = 0
3. Q ≠ 0
4. W ≠ 0, shaft work is done on the system, W is –ve.
i.e,
Wgz
c
hmQgz
c
hm  )
2
()
2
( 2
2
2
21
2
1
1
1z 2z
21 cc 
)( 21 hhmQ 

57.  Centrifugal water pump
Function :- The pump is mechanical device which
conveys liquid from one place to another place. It is
the hydraulic machines which converts the mechanical
energy into hydraulic energy.
We know that SFEE,
Wgz
c
hmgz
c
hm  )
2
()
2
( 2
2
2
21
2
1
1

58. For pump, (1) Q = 0, no heat transfer
(2) W = - W,work supplied
(3) Change of internal energy
i.e,
021 uu
Wgz
c
hmgz
c
hm  )
2
()
2
( 2
2
2
21
2
1
1

59. Heat exchanger
It is a device in which transfers heat from one fluid to
another fluid.In heat exchanger, under steady state
operation, the mass flow rate of each fluid stream
flowing through a heat exchanger remains constant.
The condenser, evaporator and radiator are example of
heat exchangers,

60. 1. Condenser :-
Function:- It is a device used to condense the steam
by rejecting heat from steam to cooling medium.
We know that SFEE,
For condenser,
1. ∆PE = 0,
2. ∆KE = 0,
3. W = 0 ,no shaft work
Wgz
c
hmQgz
c
hm  )
2
()
2
( 2
2
2
21
2
1
1
21 zz 
21 cc 

61. i.e,
2112 ),( hhhhmQ s 

62. 2. Evaporator :-
Function:- It is used to extract heat from the space to
be cooled at low temprature.
- The refrigerant liquid enters the evaporator ,which
absorbs latent heat from the space to be cooled, and it
is converted into vapour.

63. We know that,
For evaporator,
1. ∆PE = 0,
2. ∆KE = 0,
3. W = 0 ,no shaft work
i.e,
Wgz
c
hmQgz
c
hm  )
2
()
2
( 2
2
2
21
2
1
1
)( 12 hhmQ 

64.  UNSTEADY FLOW PROCESS:
• As previous article steady flow analysis, there is
neither a change in the energy of control
volume nor in the properties of fluid at any
section with respect to time.
• However in many of thermodynamic processes
like the filling and emptying process, the
internal energy and mass of tank changes with
respect to time.

65. • The filling and emptying process and variable
flow processes and can be analysed by the
control volume technique.

73.  Bottle or tank emptying process
• The tank emptying process is the reverse if tank
filling process.
• There is flow of fluid from tank to the
surroundings as shown in figure.
• If the surroundings is too much large compared
to the system,energy of fluid leaving the tank of
entering the surroundings is constant.