ppt conyains different gas Laws, processes.

1. Unit: 1
Topic: Properties of Gas
(Introduction of Gas, Gas Laws, Gas Constant)
Faculty Name: Pratik Kikani
Branch: Mechanical Engineering
Atmiya University, "Yogidham Gurukul", Kalawad Road, Rajkot - 360005
Web: www.atmiyauni.ac.in
: : atmiyauniversity : : atmiya_university
Subject Code: 18BTMECC201
Subject Name: ELEMENTS OF MECHANICAL ENGINEERING

2. Outline of Topics
• Introduction of Gases
• Laws of gases
– Boyle’s Law
– Charle’s Law
– Combined Gas Law
• Gas Constant
• Relation between Cp and Cv
2Pratik Kikani

3. Outline of Topics
• Non Flow Processes
1. Constant volume (Isochoric) process
2. Constant pressure (Isobaric) process
3. Constant Temperature (Isothermal) process
4. Adiabatic process
5. Polytropic process
3Pratik Kikani

4. Introduction of Gases
• It is required to study the behavior of gases because
in engines the gas is changing its behavior in
different conditions.
• Vapor: It s defined as that state of substance in which
the evaporation of substance(liquid) is incomplete.
E.g. Wet steam (steam contains water ).
4Pratik Kikani

5. Introduction of Gases
• Gas: It is defined as the state of a substance in which
the evaporation of substance is complete.
E.g.Co2, O2, N2 etc.
5Pratik Kikani

6. Introduction of Gases
• Perfect Gas: A perfect gas may be considered as the
gas that obeys the laws of boyle and charles under
all conditions of temperature and pressure.
6
There is no perfect, but many gases like H2, N2, O2,
Air etc. may be regarded as perfect gases. They are
known as real gases.
Pratik Kikani

7. Laws of gases
– Boyle’s Law
– Charle’s Law
– Combined Gas Law
7Pratik Kikani

8. Boyle’s Law
• The volume of given mass(v) of a
perfect gas varies inversely as the
absolute pressure(p) when the
temperature(t) is constant. (Robert
Boyle, 1661)
8Pratik Kikani

9. Charle’s Law
• If any gas is heated at
constant pressure(p), the
change in volume(v) varies
directly with the
temperature(t). (1787 A.D.)
V α T
9
cTV /
Pratik Kikani

10. Combined Gas Law
• In practice, pressure, volume and temperature of a
gas changes at the same time.
• As the pressure changes, Charles law cannot be
applied and as the temperature changes, Boyle’s law
cannot be applied. By combining these two laws the
final conditions of the gas can be determined.
10Pratik Kikani

11. • Consider a m kg of mass of a
perfect gas to change its state in
the following two successive
processes(i) Process 1-2 at
constant pressure, and (ii)
Process 2-3 at constant
temperature.
11
For process 1-2, Applying Charles law.
since T2 =T3 , we may write
Combined Gas Law
Pratik Kikani

12. 12
For process 2-3,
Combined Gas Law
Applying Boyle’s law
Substituting the value of V2 in
we get
& T2= T3
Pratik Kikani

13. • The magnitude of this constant depends upon the
particular gas and it is denoted by R , where R is
called the
• characteristics gas constant. For m kg of gas,
13
Combined Gas Law
Equation is called equation of state for a perfect gas.
Pratik Kikani

14. 14
Relation between Two specific heats cp & Cv
p = pressure of gas in N/m2
V1 & V2 = Initial and final volume of gas in m3
T1 & T2 = Initial and final Temperature of gas in K
Cp & Cv= Specific heats of gas at constant pressure & at
constant volume respectively. In kJ/kg K
R = Characteristic gas constant in kJ/kg K
m = mass of gas in kg
Pratik Kikani

15. 15
Relation between Two specific heats cp & Cv
Consider m kg of gas is heated at
According to 1st law of thermodynamics
∆Q = ∆U + W
At constant volume
∆U = m cv ∆T
W= 0 (∵ ∫pdv )
dv = o
∵ constant volume process
∆Q = ∆U
∆Q = m cv ∆T
At constant pressure
∆Q = ∆U + W
∆U = m cv ∆T
W= P(v2 – v1) (∵ ∫pdv )
Qv = m cv ∆T Qp = m cp ∆T
Pratik Kikani

16. 16
Since p v= m R T
P1v1 = m R T1 , P2v2 = m R T2
p(v2 – v1) = m R(T2- T1)
p(v2 – v1) = m R ∆T
m cp ∆T = m cv ∆T + m R ∆T
Cp = cv + R
Cp - cv = R
Relation between Two specific heats cp & Cv
m cp ∆T = m cv ∆T + m R ∆T
∆Q = ∆U + W ∆U = m cv ∆T & W= p(v2 – v1)
m cp ∆T = m cv ∆T + p(v2 – v1)
So, Heat is applied at constant pressure
then,
Pratik Kikani

17. Flow Process
• Such process occur in the
system having open
boundary permitting
mass interaction with
boundary.
Non flow process
• There is no mass
interaction across the
system boundaries during
the occurrence of process.
1. Constant volume (Isochoric)
process
2. Constant pressure (Isobaric)
process
3. Constant Temperature
(Isothermal) process
4. Adiabatic process
5. Polytropic process 17
Processes
Pratik Kikani

18. 18
Non Flow Processes
1. Constant volume (Isochoric) process
2. Constant pressure (Isobaric) process
3. Constant Temperature (Isothermal) process
4. Adiabatic process
5. Polytropic process
Pratik Kikani

19. 19
Non Flow Processes
Constant volume
(Isochoric) process
P,V,T relation
Work done
Heat Transferred
(∆Q)
Change in internal
Energy(∆U)
Change in Enthalpy
(∆H)
Pratik Kikani

20. 20
Constant volume (Isochoric) process
In a constant volume process,
the working substance is
contained in a rigid vessel.
Hence the boundaries of the
system are immovable and no
work can be done on or by the
system. This process is also
known as Isochoric process.
Pratik Kikani

21. 21
P,V,T relation
Constant volume (Isochoric) process
But
V1 =V2 constant
We know that
Pratik Kikani

22. 22
Work done p dV
Since the volume is constant, Change in volume dV = 0 .
work done W = 0
Work done
Constant volume (Isochoric) process
Pratik Kikani

23. 23
Constant volume (Isochoric) process
Heat Transferred (∆Q)
According to first law of thermodynamics,
∆ Q =W + ∆ U
For constant volume process,
dV = 0, so W =0
∆ Q = ∆ U = m Cv (T2- T1)
So, Heat transfer = Change in internal energy
Pratik Kikani

24. 24
Constant volume (Isochoric) process
Change in internal Energy(∆U)
∆ u = m Cv (T2- T1)
Thus, internal energy is a function of temperature
alone.
Pratik Kikani

25. 25
Change in Enthalpy (∆H)
As per definition of
enthalpy, H= U + Pv
For state 1 and 2
H1 = U1 + P1v1
H2 = U2 + P2v2
Constant volume (Isochoric) process
Pratik Kikani

26. 26
Constant pressure
(Isobaric) process
P,V,T relation
Work done
Heat Transferred
(∆Q)
Change in internal
Energy(∆U)
Change in Enthalpy
(∆H)
Constant Pressure (Isobaric) process
Pratik Kikani

27. 27
In an isobaric process, the process of
the gas remains constant.
Constant Pressure (Isobaric) process
The work done by the gas
during constant pressure
process is represented by area
below the line 1 - 2 on p –V
diagram as shown in fig.
Pratik Kikani

28. 28
P,V,T relation
Constant Pressure (Isobaric) process
We know that
But p1 = p2 Constant
Pratik Kikani

29. 29
Work done = pdV . But p is constant,
Work done
Constant Pressure (Isobaric) process
Pratik Kikani

30. 30
Constant Pressure (Isobaric) process
Heat Transferred (∆Q)
∆ Q = p(v2 – v1)+ m Cv (T2- T1)
∆ Q =W + ∆ U
W =p(v2 – v1)
∆ U= m Cv (T2- T1)
∆ Q = m R (T2- T1) + m Cv (T2- T1)
∆ Q =m cp(T2- T1)
∆ Q =m (T2- T1) (R+Cv) Since, R+Cv = Cp
Pratik Kikani

31. 31
Change in internal Energy(∆U)
∆ u = m Cv (T2- T1)
Thus, internal energy is a function of temperature
alone.
Constant Pressure (Isobaric) process
Pratik Kikani

32. 32
Constant Pressure (Isobaric) process
Change in Enthalpy (∆H)
As per definition of
enthalpy, H= U + Pv
For state 1 and 2
H1 = U1 + P1v1
H2 = U2 + P2v2
Pratik Kikani

33. 33
In an isothermal process, the
temperature remains constant during
the process
This process follows Boyle’s law.
Constant temperature (Isothermal) process
Thus the law of expansion or
compression for isothermal process on
p – V diagram is hyperbolic as p is
inversely varies as V.
Thus this process is also known as hyperbolic process
or Constant Internal energy process.
Pratik Kikani

34. 34
P,V,T relation
We know that
But T1 = T2 Constant
Constant temperature (Isothermal) process
Pratik Kikani

35. 35
Work done = pdV .
But p V=c = P1V1 = c ,
Work done
Constant temperature (Isothermal) process
Substituting for p in the integral, we get,
Pratik Kikani

36. 36
Work done
Constant temperature (Isothermal) process
Substituting for C , we get
Pratik Kikani

37. 37
Work done
Constant temperature (Isothermal) process
Pratik Kikani

38. 38
Heat Transferred (∆Q)
∆ Q = W=m R T1 loge r = m R T2 loge r
∆ Q =W + ∆ U
W=m R T1 loge r = m R T2 loge r
∆ U= m Cv (T2- T1) =0
Constant temperature (Isothermal) process
Pratik Kikani

39. 39
Change in internal Energy(∆U)
∆ u = m Cv (T2- T1)
Constant Pressure (Isobaric) process
∆U = 0
Since, T1 =T2
Pratik Kikani

40. 40
Constant Temperature (Isothermal) process
Change in Enthalpy (∆H)
∆H = 0
Since, T1 =T2
Pratik Kikani

41. 41Pratik Kikani