Thermodynamics chapter 2

Thermodynamics
Chapter 2
Diploma in Engineering
Mechanical Engineering Division, Ngee Ann Polytechnic

Properties of Liquids & Vapours
o Constant pressure process for a liquid and its vapour
o Saturation temperature
o Quality or dryness fraction of wet steam
o Tabulated properties
o Properties of saturated water and steam
o Properties of compressed water
o Properties of wet steam
o Properties of superheated steam
o Interpolation of values in steam tables

Introduction
• 3 existing forms of matters:
• Thermodynamics only concerns liquid and gaseous
phases.
solid liquid gas
thermodynamics

Constant Pressure Process
• Boiling: a liquid heated at any one constant pressure,
there is one fixed temperature at which bubbles of
vapour form in the liquid.
• The higher the pressure of the liquid, the higher will be
the temperature at which boiling occurs.
• The volume occupied by 1kg of a boiling liquid at a
higher pressure is slightly larger than the volume
occupied by 1kg of the same liquid when its boiling at a
lower pressure.

• p-v diagram of a series of boiling points
PR
PQ
PP
Pressure, p
Specific volume, v
R
Q
P

• When a liquid at boiling point
is heated further at constant
pressure. The additional heat
supplied changes the phase
of the substance from liquid
to vapour.
• During the change of phase,
the pressure and
temperatures remain
constant
PR
PQ
PP
Pressure, p
Specific volume, v
R
Q
P
R’
Q’
P’

• The heat required converting
1kg of liquid at boiling point
to 1kg of saturated vapour at
the same temperature and
pressuer is called the latent
heat of vapourisation.
PR
PQ
PP
Pressure, p
Specific volume, v
R
Q
P
R’
Q’
P’
The higher the pressure the smaller will be
the latent heat of vapourisation
The specific volume of a saturated vapour
decreases as the pressure increases

• The point where the
saturated liquid line and the
saturated vapour line meet is
called the critical point
PR
PQ
PP
Pressure, p
Specific volume, v
R
Q
P
R’
Q’
P’
Critical pt
Critical pressure,
At the critical pressure, the latent heat of
vapourisation is zero
The substance exisiting at any point
inside the loop consists of a mixture of
liquid and dry vapour and is known as
wet vapour
A saturated vapour is usually called dry
saturated vapour.

• When a dry saturated vapour is heated at constant
pressure, its temperature rises above saturation
temperature and it becomes superheated vapour
• The difference between the actual temperature of the
superheated vapour and the saturation temperature is
called the degree of superheat

Isothermals on the
Pressure-volume Diagram
PR
PQ
PP
Pressure, p
Specific volume, v
R
Q
P
R’
Q’
P’
Critical pressure
Tc
Critical point • Isothermal: constant temp.
• The temperature lines
become horizontal between
the saturated liquid line and
the saturated vapour line.
Thus there is a
corresponding saturation
temperature for each
pressure
• The critical temp. line just
touches the top of the loop
at the critical point
T3
T2
T1

Quality or Dryness
Fraction of Wet Vapour
• The saturated liquid exists in equilibrium with dry vapour
• The pressure and temperature alone are not sufficient to
define completely the state of mixture
• At pressure PQ and temperature T2, could be a
saturated liquid, a wet vapour or a dry saturated vapour.
• The state can’t be defined until one other property, e.g.
specific volume is known.

Quality or Dryness
Fraction of Wet Vapour
• The condition or quality of a wet vapour is most
frequently defined by its dryness fraction

Tabulated Properties
• The properties are tabulated as functions of pressure
and temperature for a wide variety of substances to
facilitate computation.
• The water and steam properties are tabulated in the
steam tables.

Properties of Saturated
Water and Steam
• The specific properties of saturated water is tabulated in
Saturated Water and Steam Table from page 8-3 to 8-
6.
• The suffix “f” is used to denote the saturated water
state.
Example
• Determine the temperature, specific enthalpy, specific
internal energy and specific volume of saturated water
at p=0.34 bar.

Water and Steam
1
• For most practical purpose, the value of vf is generally
taken as 0.001 m3/kg.
Subscript “f” for saturated water
Subscript “s” for saturated water and steam
Page 8-4

Water and Steam
• The suffix “g” is used to denote the dry saturated
state (i.e. saturated vapour).
Example
internal energy and specific volume of saturated steam
at p=0.34 bar.

Water and Steam
1
Subscript “g” for saturated steam
Page 8-4

Water and Steam
Properties of compressed water
• Water at a temperature below the boiling point
(ts)corresponding to the pressure.
• Assumption: For normal pressure, it is sufficiently
accurate to take the properties of compressed water, as
being equal to those of saturated water at the same
temperature.

Water and Steam
Example
• Determine the specific internal energy, specific enthalpy
and specific volume of compressed water at p=20 bar
and t=179.9ºC Page 8-5
Same as “f” saturated water at the same temp.

Water and Steam
Properties of wet steam
• The specific properties of wet steam can be obtained
using the dryness fraction, x:
The dryness fraction of the wet steam is x
Then 1 kg of wet steam contains:
X kg of dry saturated steam, and
(1-x) kg of saturated water

Water and Steam
Therefore, the specific enthalpy of the wet stream
hx=xhg+(1-x)hf
Similarly, the specific internal energy of the wet steam
ux=xug+(1-x)uf
And the specific volume of the wet steam
vx=xvg+(1-x)vf
Since (1-x)vf≈0
Hence vx=xvg

5
Water and Steam
Example
internal energy and specific volume of wet steam at
p=10 bar with a dryness fraction of 0.5.

Water and Steam
Applying hx=xhg+(1-x)hf
Hence hx=0.5=0.5×2778+(1-0.5)×763
=1770.5 kJ/kg
Applying ux=xug+(1-x)uf
Hence u0.5=0.5×2584+(1-0.5)×762
=1673 kJ/kg
Applying vx=xvg
Hence v0.5=0.5×0.1944
=0.0972 m3/kg

Water and Steam
Example
• A cylinder contains 0.113 m3 of wet steam at 11 bar. If
the mass of the steam is 0.9 kg, determine its dryness
fraction, enthalpy and internal energy.
Solution:

Water and Steam
Page 8-5

Water and Steam
Applying
Applying hx=xhg+(1-x)hf
Hence hx=0.708=0.708×2781+(1-0.708)×781
=2197 kJ/kg
Hence, the enthalpy of the wet steam

Water and Steam
Hence u0.708=0.708×2586+(1-0.708)×781
=2058.94 kJ/kg
Hence, the internal energy of the wet steam

Properties of Superheated
Steam
• Tabulated separately
• Temperature and pressure are independent properties.
i.e. only when the temperature and pressure are given
for superheated steam, then the state is defined and all
the other properties can be found.

Properties of Superheated
Steam
Example
• Determine the specific enthalpy, specific internal energy
and specific volume of superheated steam at p=2 bar
and t=200°C. Page 8-7

Interpolation of Values in
Steam Tables
• Steam tables for the superheated steam as well as the
saturated water & steam give the values of properties at
selected pressures and temperatures.
• To obtain values of properties at conditions other than
those tabulated, interpolation is used.

Steam Tables
Example
• Determine the saturation temperature of the steam at
pressure p=9.6 bar
Solution: refer to saturated water and steam table
Pressure, P Saturation temperature, ts
9 bar 175.4 °C
10 bar 179.9 °C

Steam Tables
Using straight line
interpolation
The slope of AC
=the slope of AB
A
B
C
( 175.4)t 
(179.9 175.4)
(9.69)
(109)
10
9.6
9
175.4 t 179.9 ts(°C)
p (bar)

Steam Tables
Example
• Superheated steam at p=15 bar and t=380ºC. Determine
its specific volume, enthalpy and internal energy.
Solution: refer to superheated steam table
Pressure, p Temperature, t
350°C 400°C
15 bar
v=0.1865 m3/kg v=0.2029 m3/kg
u=2868 kJ/kg u=2952 kJ/kg
h=3148 kJ/kg h=3256 kJ/kg

Steam Tables
400
380
350
t (°C)
0.1865 v 0.2029 v(m3/kg)
Using straight line
interpolation
The slope of AB
=the slope of AC
(400350)
(380350)
( 0.1865)v 
(0.2029 0.1865)

Steam Tables
400
380
350
t (°C)
2868 u 2952 u(kJ/kg)
Using straight line
interpolation
The slope of AB
=the slope of AC
(400350)
(380350)
( 2868)u 
(2952 2868)

Steam Tables
400
380
350
t (°C)
3148 h 3256 h(kJ/kg)
Using straight line
interpolation
The slope of AB
=the slope of AC
(400350)
(380350)
( 3148)h 
(3256 3148)

Non-flow Processes with Steam
o Introduction
o Constant volume process
o Constant pressure process
o Isothermal process
o Adiabatic non-flow process
o Polytropic process

Introduction
• What steam can do?
expand
compress
receive heat
reject heat
do external work
have work done

Introduction
p1 T1
v1 h1
u1
p2 T2
v2 h2
u2
m kg of steam

Constant Volume Process
• The system is contained in a rigid vessel. Hence the
boundary of the system is immoveable and no work can
be done on or by the system.
• v1=v2
• W12=0
Specific volume, v
Pressure, p
system
boundary

• Applying the non-flow energy equation
• The net heat supplied in a constant volume process
goes to increasing the internal energy of the steam.
2 1 12 12
12
12 2 1
12 2 1
U -U =Q -W
Since W =0
Q =U -U
or Q =m(u -u )


Example
• During a constant volume process, the specific internal
energy of the steam increased from 120 kJ/kg to 180
kJ/kg. Determine the amount of heat energy required to
bring about this increase for 2 kg of steam.
Solution:
2 1 12 12
12
12 2 1
12 2 1
U -U =Q -W
Since W =0
Q =U -U
or Q =m(u -u )=2 (180-120)=120kJ



Example
• Dry saturated steam at pressure 3 bar is contained in a
rigid vessel of volume 0.9 m3. if the vessel is cooled until
the pressure is reduced to 1.2 bar. Determine:
A) the mass of steam in the vessel
B) the final dryness fraction of the steam
C) the amount of heat transferred during the cooling
process

Solution:
• A)
3
1 g at p=3bar
1
1
v =v =0.6057m /kg
0.9
1.486
0.6057
V
v
m
V
m kg
v

   
Q

B) for steam undergoing a constant volume process
3
2 1
3
2 1.2
2
2
2
0.6057 /
1.428 /
0.6057
0.4242
1.428
g g at p bar
x g
g
v v m kg
v v m kg
v xv
v
x
v

 
 

   
Q

• C)
1 g at p=3bar
2 2 2
2 2 2 2 2
2 1 12 12
12
12 2 1
12 2 1
=u =2544kJ/kg
at p 1.2 439 / 2512 /
(1 )
(1 ) 0.4242 251 (1 0.4242) 439 1318.37 /
U -U =Q -W
Since W =0
Q =U -U
or Q =m(u -u )=1.486 (1318.37-2544
f g
x g f
g f
u
bar u kJ kg u kJ kg
u xu x u
u x u x u kJ kg
  
  
         


Q
)=-1821.29kJ

• The boundary of the
system expands as
heat is added to the
system or contracts
as heat is removed
from the system .
Hence work is done
either by the system on
its surrounding or by
the surrounding on the
system.
• p1=p2=p Specific volume, v
Pressure, p
v1 v2
1 2p1= p2

• The shaded area on the p-v diagram represents the work
done by the steam.
Hence
Applying the non-flow energy equation
12 2 1
12 2 1
( )
( )
W p V V
orW mp v v
 
 
2 1 12 12
12 2 1 2 1 2 2 1 1
12 2 1
12 2 1
U -U =Q -W
( ) ( ) ( )
since
( )
Q U U p V V U pV U pV
H U pV
Q H H
or Q m h h
        
 
  
 
The net amount of heat energy supplied to or
taken from the steam during a constant pressure
non-flow process is equal to the change of
enthalpy of the steam during the process

Example
• 2.5 kg of wet steam at pressure 15 bar and dryness
fraction 0.8 receives heat at constant pressure until its
temperature become 250ºC. Determine the heat
received by the steam
Solution:
• At p1=15 bar
1 15
1 15
1 15
198.3
845 /
2792 /
s at p bar
f f at p bar
g g at p bar
t t C
h h kJ kg
h h kJ kg



 
 
 
o

Since t2=250ºC and t2 > ts2 (ts2=ts at p=15bar=198.3°C)
Therefore, the steam at state 2 is in the superheated region
h2=hat p=15bar, t=250ºC=2925 kJ/kg
1 1 1 1 1
(1 )
(1 ) 0.8 2792 (1 0.8) 845 2402.6 /
x g f
g f
h xh x h
h x h x h kJ kg
  
         
Q

• For steam undergoing constant pressure process
Heat received
12 2 1
2 1( )
2.5 (2925 2402.6) 1307.5
Q H H
m h h
kJ
 
 
   

Isothermal Process
• Temperature remains
constant
• An isothermal
process for steam is
shown on the right.
The initial and final
sate has been chosen
in the wet region and
superheated region
respectively.
Specific volume, v
Pressure, p
v1 v2
1 Ap1
p2
2
Constant temp. line

Isothermal Process
• From state 1 to state
A, the pressure
remains at p1, since
the wet region the
pressure and
temperature are the
corresponding
saturation values. It
can be seen therefore
that an isothermal
process for wet steam
is also a constant
pressure process. Specific volume, v
Pressure, p
v1 v2
1 Ap1
p2
2
Constant temp. line

Isothermal Process
• Hence
Q1A=HA-H1
Or Q1A=m(hA-h1)
• From state A to state
2, the steam is in the
superheated region
and the pressure fall
from p1 to p2.
Specific volume, v
Pressure, p
v1 v2
1 Ap1
p2
2
Constant temp. line

Isothermal Process
• When state 1 and
state 2 are fixed, then
the specific internal
energy u1 and u2 may
be obtained from
steam tables. The
work done is
represented by the
shaded area.
Specific volume, v
Pressure, p
v1 v2
1 Ap1
p2
2
Constant temp. line

Isothermal Process
Example
• In a closed system, 3 kg of wet steam at 10 bar has an
initial dryness fraction 0.9 and is expanded isothermally
to a pressure of 2 bar. Determine:
A) the change of internal energy and the change of
enthalpy of the steam
B) if the heat supplied during the process is found to be
1000 kJ, determine the work done by the steam.

Isothermal Process
Page 8-5
Solution:

Isothermal Process
1 1 1 1 1
1 1 1 1 1
(1 )
(1 ) 0.9 2778 (1 0.9) 763 2576.5 /
(1 )
(1 ) 0.9 2584 (1 0.9) 762 2401.8 /
x g f
g f
x g f
g f
h xh x h
h x h x h kJ kg
also
u xu x u
u x u x u kJ kg
  
         
  
         
Q
Q

Isothermal Process
At p2=2 bar
t2=t1=179.9°C
From superheated steam table
Using straight line interpolation
T [°C] u[kJ/kg] h [kJ/kg]
150 2578 2770
200 2655 2871
2
2
2655 2578
2578 (179.9 150) 2624 /
200 150
2871 2770
2770 (179.9 150) 2830 /
200 150
u kJ kg
h kJ kg

   


   


Isothermal Process
Hence
And
B) Applying the non-flow energy equation
2 1 2 1
2 1 2 1
( ) 3 (2624 2401.8) 666.6
( ) 3 (2830 2576.5) 760.5
U U m u u kJ
H H m h h kJ
      
      
2 1 12 12
12 12 2 1
U -U =Q -W
( ) 1000 760.5 239.5W Q U U kJ      

Adiabatic Non-flow
Process
• An adiabatic process is one in which no heat is transferred to
or from the system during the process.
Hence Q12=0
U2-U1=Q12-W12
Since Q12=0
W12= -(U2-U1)
• In an adiabatic expansion process: the work done by the
system is at the expense of a reduction in the internal energy
of the system
• In an adiabatic compression process: the work done on the
system goes into increasing the internal energy of the system.

Adiabatic Non-flow
Process
Example
• In a closed system, 2 kg of steam at initial pressure 50
bar and temperature 400°C undergoes adiabatic process
until the pressure is 15 bar and the steam is then dry
saturated. Determine the work done by the steam .
Solution:
u1=uat p=50bar, t=400˚C=2907 kJ/kg
u2=uat p=15bar=2595 kJ/kg

Adiabatic Non-flow
Process
For steam undergoing an adiabatic process
Q12=0
U2-U1=Q12-W12
Hence W12=-(U2-U1)
=-m(u2-u1)
=-2×(2595-2907)
=624 kJ

Polytropic Process
• The most general type of process, in which both heat
energy and work energy cross the boundary of the
system. It is expressed by means of the equation
pVn=constant
Or pvn=constant
Where n known as the index of expansion or compression
and is constant.

Polytropic Process
hence
p1V1
n=p2V2
n
Or p1v1
n=p2v2
n
And the work energy crossing the boundary (or work
transferred) is given by
Or

Polytropic Process
Or

Polytropic Process
Example
• 2.5 kg of wet steam has an initial pressure 6 bar and a
dryness fraction of 0.85 and is expanded according to
the law pV1.2=constant to a final pressure of 1.2 bar.
Determine:
A) the work done by the steam.
B) the heat energy transferred between the steam and
surroundings stating the direction of transfer.

Polytropic Process
Solution:
At p1=6 bar:
vg1=vg at p=6bar=0.3156 m3/kg
uf1=uf at p=6bar=699 kJ/kg
ug1=ug at p=6bar=2568 kJ/kg
Hence u1=x1ug1+(1-x)uf1
=0.85×2568+(1-0.85)×669
=2182.8 kJ/kg
Applying vx=xvg
Hence v1=x1vg1=0.85×0.3156=0.2683 m3/kg

Polytropic Process
For steam undergoing a polytropic process

Polytropic Process
At p2=1.2 bar:
vg2=vg at p=1.2bar=1.428 m3/kg
uf2=uf at p=1.2bar=439 kJ/kg
ug2=ug at p=1.2bar=2512 kJ/kg
Applying vx=xvg
Hence
Hence u2=x2ug2+(1-x)uf2
=0.72×2512+(1-0.72)×439
=931.56 kJ/kg

Polytropic Process
For steam undergoing a polytropic process
U2-U1=Q12-W12
Hence Q12=(U2-U1)+W12
=m(u2-u1)+W12
=2.5×(1931.56-2182.8)+473.4
=-154.7 kJ
Or Q12=154.7 kJ (heat lost)

Properties of Perfect Gas
• The characteristic gas equation
• Specific heats
• Joule’s law
• Relationships between the specific heats
• Enthalpy of a perfect gas
• Ratio of specific heats

The Characteristic Gas
Equation
• One important type of fluid which has many applications
in thermodynamics is the type in which the working
temperature of the fluid remains well above the critical
temperature of the fluid.
• The vapor of the fluid tends to obey the equation:
p: pressure [N/m2]
v: specific volume [m3/kg]
T: temperature [K]

Equation
• In practice, no gasses obey this law rigidly, but many
tend towards it. An imaginary ideal gas, which obeys
the law is called a perfect gas.
• R is the characteristic gas constant in J/kg∙K, each
perfect gas has a different value of gas constant.
• Or

Equation
• For m kg of the perfect gas, occupying V m3
• Hence
R: the characteristic gas constant [J/kg∙K]
p: pressure [N/m2]
m: mass [kg]
v: specific volume [m3/kg]
V: volume [m3]
T: temperature [K]
Characteristic gas equation

Equation
• Anther form of the characteristic gas equation can be
derived using the kilogram-mole.
• The Kilogram-mole is defined as a quantity of a gas
equal to M kg of the gas where M is the molecular weight
of the gas.

Equation
E.g. 1 kilogram-mole of oxygen is equal to 32 kg of oxygen.
Since the molecular weight of oxygen is 32.
For m kg of a gas
Where: n is the number of moles
Hence,

Equation
• Avogadro’s Hypothesis states that the volume of 1
mole (a simplified term for kg-mole) of any gas is the
same as the volume of 1 mole of any other gas. When
the gases have at the same temperature and pressure.
• i.e.
• The constant is called the universal gas constant, and
is given the symbol, R0

Equation
• Experiment has shown that the volume of 1 mole of any
perfect gas at 1 bar and 0°C is approximately 22.71 m3
• Therefore

Equation
• Hence, the gas constant, R for any gas can be found
when the molecular weight of the gas is known.
• Since the molecular weight of oxygen is 32,
• Hence, the gas constant for oxygen

Specific Heats
• The specific heat of a solid or liquid is usually defined
as the amount of heat energy required to raise the
temperature of unit mass of the substance through one
degree
• c: the specific heat [J/kg∙K]
• m: the mass [kg]
• (T2-T1): the increase in temperature [K]
• Q: the heat required [J]

Specific Heats
In case of gas, there are infinite numbers of ways in which heat
may be added between any two temperatures, and hence a gas
could have infinite number of specific heats. However, only two
specific heats for gasses are defined.
 The specific heat at constant volume, cv. It is an amount of
heat energy required to raise the temperature of one kg mass
of the gas by one degree whilst the volume of the gas remain
constant.
 The specific heat at constant pressure, cp. It is an amount of
heat energy sufficient to raise the temperature of one kg mass
of the gas by one degree whilst the pressure of the gas
remains constant.

Specific Heats
• For a perfect gas, the values of cp and cv are constant for
any given gas at all pressure and temperature.
Noted: cp air=1.005 kJ/kg∙K
cv air=0.718 kJ/kg∙K
Hence
Q12=m∙cv(T2-T1) for constant volume non-flow process
Q12=m∙cp(T2-T1) for constant pressure non-flow process

Joule’s Law
• Joule’s law states that the internal energy of a perfect
gas is a function of the temperature only. i.e. u=f(T)
• To evaluate the function, u=f(T), consider m kg of a
perfect gas be heated at constant volume from state 1 to
state 2
U2-U1=Q12-W12
Since W12=0
Hence Q12=U2-U1

Joule’s Law
For a perfect gas undergoing constant volume non-flow
process
Q12=mcv(T2-T1)
Therefore, U2-U1=mcv(T2-T1)
Hence, it is true to say that for a perfect gas undergoing
any non-flow process between state 1 and state 2
U2-U1=mcv(T2-T1)

Relationships between the
Specific Heats
• Consider m kg of a perfect gas at pressure, p1 and
temperature T1, volume V1 undergoes a constant
pressure process to a final pressure p2, temperature T2
and volume V2
Applying W12=p(V2-V1)
And p1V1=mRT1
p2V2=mRT2
Hence W12=mR(T2-T1)

Relationships between the
Specific Heats
For a perfect gas undergoing constant pressure non-flow
process
Q12=mcp(T2-T1)
And U2-U1=mcv(T2-T1)
U2-U1=Q12-W12
Hence, mcv(T2-T1)=mcp(T2-T1)+mR(T2-T1)
cv=cp-R
Or cp-cv=R

Enthalpy of a Perfect Gas
• The enthalpy of a perfect gas is given by
h=u+pv
Since u=cvT
And pv=RT
Hence, h=cvT+RT
=(cv+R)T
=cpT
For m kg of perfect gas
H=mcpT
Hence, H2-H1=mcp(T2-T1)

Ratio of Specific Heats
• The ratio of the specific heat at constant pressure, cp to
the specific heat at constant volume cv is given the
symbol

Example
• A quantity of certain perfect gas has an initial pressure of
150 kN/m2, volume 0.2 m3 and temperature 30°C. It is
then compressed to a pressure of 600 kN/m2 and
temperature 65°C. Determine the final volume of the
gas.

Solution:
Given: p1=150 kN/m2, V1=0.2 m3, t1=30°C,
p2=600 kN/m2, t2=65°C
Applying

Example
• A certain perfect gas has an initial pressure of 360
kN/m2, volume of 0.004 m3, and temperature of 32°C. if
the value of R=0.29 kJ/kg∙K. Determine:
(a) the mass of the gas
(b) if the pressure of the gas is now increased to 10 bar
while the volume remains constant. Calculate the final
temperature of the gas.

Solution:
(a) Applying the characteristic gas equation
pV=mRT
(b)Applying
And V1=V2
Hence

Non-flow Processes with Perfect Gas
• Constant volume process
• Constant Pressure process
• Isothermal or constant temperature process
• Polytropic process
• Adiabatic process

• Consider m kg of a
perfect gas contained in
a rigid vessel at pressure
p1, temperature T1 and
volume V1, undergoes a
constant volume process
to a final pressure p2,
temperature T2, and
volume V2.
1
2
P
P2
P1
V1=V2 V

• For a perfect gas undergoing constant volume process
• Applying

• For a perfect gas undergoing any process
The net amount of heat energy
supplied to or taken from a perfect gas
during a constant volume process is
equal to the change in internal energy
of the perfect gas during that process

Example
• 0.5 kg of air with an initial temperature of 29°C and a
pressure of 1.5 bar is heated to a final temperature of
350°C at constant volume. Determine the heat energy
supplied to the air and its final pressure. Take cp=0.287
kJ/kg∙K, cv=0.718 kJ/kg∙K.
Solution

• Applying
• Since

• Consider m kg of a
perfect gas at pressure
p1, temperature T1 and
volume V1, undergoes a
constant pressure
process to a final
pressure p2, temperature
T2, and volume V2.
1 2
P
V2V1
P1=P2
V

• For a perfect gas undergoing constant pressure process
• Applying

• For a perfect gas undergoing constant pressure process

• Since
• Hence
• Or
The net amount of heat energy
supplied to or taken from a perfect gas
during a constant pressure process is
equal to the change of enthalpy of the
perfect gas during that process

Example
• A mass of air has an initial pressure, 2.5 bar, volume
0.08 m3 and temperature 175°C. It is then undergoes a
constant pressure process until its final temperature
become 30°C. Take R=0.287 kJ/kg∙K, cv=0.718 kJ/kg∙K
as the mass of the air. Determine:
• A) the work done on the air, and
• B) the heat transferred from the air during the process.

Solution
• Applying
• Applying

• Alternatively,
• And

Isothermal or Constant
Temp. Process
• An isothermal expansion or compression process is
defined as a process carried out while the temperature
remains constant.
• Consider m kg of a perfect
gas at pressure p1, temperature T1
and volume V1 undergoes a
constant temperature process
to a final pressure p2, temperature T2
and volume V2.
1
2
P
p2
p1
V1 VV2
pV=c

Temp. Process
• For a perfect gas undergoing isothermal process
• Applying
• Hence

Temp. Process
• For a perfect gas undergoing isothermal process
• and
• Hence

Temp. Process
• since
• Hence
• or
During an isothermal process expansion all the
heat energy received is converted into external
work. Conversely, during an isothermal
compression, all the work done on the gas is
rejected by the gas as heat energy.

Temp. Process
Example
0.2 kg of air has an initial pressure of 30 bar and a volume
of 0.015 m3. it is then expanded to a pressure of 4 bar while
the temperature remains constant. Determine:
• A) the temperature of the air
• B) the final volume of the air
• C)the work done by the air, and
• D) the heat transferred to the air during the process
Take R=0.287 kJ/kg∙K and cv=0.718 kJ/kg∙K

Temp. Process
Solution:
Applying
For a perfect gas undergoing isothermal process
pV=constant
Hence

Temp. Process
For a perfect gas undergoing isothermal process
For a perfect gas undergoing any process

Polytropic Process
• In polytropic process, the gas is compressed or expanded
according to the law
pVn=constant
• Where n is the index of expansion or compression and both
work and heat energy may be transferred across the
boundary.
and volume V2.
1
2
P
p2
p1
V1 VV2
pVn=c

Polytropic Process
• Since
• Applying
• Hence

Polytropic Process
• Substitute into
• For a perfect gas undergoing polytropic process
• And pV=mRT
• Hence

Polytropic Process

Polytropic Process
Example
0.008 kg of air initially at a pressure of 1.2 bar and a
volume of 0.005 m3 is compressed to a volume of 0.001 m3
according to the law pV1.25=constant. Determine:
• A) the final pressure of the air
• B) the initial and final temperature of the air
• C)the work transferred to the air, and
• D)the heat transferred from the air during the process
Take R=0.287 kJ/kg∙K and cv=0.718 kJ/kg∙K

Polytropic Process
Solution:
• Applying
• Applying

Polytropic Process
• Applying

Polytropic Process

Adiabatic Process
• In an adiabatic process, the gas is compressed or
expanded according to the law
pVγ=constant
• Where
and volume V2.
1
2
P
p2
p1
V1 VV2
pVγ=c
Q12=0

Adiabatic Process
• Since
• Applying
• Hence

Adiabatic Process
• Substitute into
• And pV=mRT
• Hence

Adiabatic Process
Q12=0

Adiabatic Process
Example
In a thermally insulated cylinder 0.01 m3 of air has an initial
pressure of 1.0 bar and temperature of 35°C. It is
compressed adiabatically to a final volume of 0.002 m3.
Determine:
• A) the mass of the air
• B) the final pressure of the air
• C) the work done to the air during the process
Take R=0.287 kJ/kg∙K, cv=0.718 kJ/kg∙K, γ=1.4 and
cp=1.005 kJ/kg∙K,

Adiabatic Process
• Solution:
• Applying pV=mRT
• Applying

Adiabatic Process
• For a perfect gas undergoing adiabatic process

Thermodynamics chapter 2

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