Chapter Three Working Fluids (A) Ideal Gas (Perfect Gas) :: Statement No. (1)
Chapter Three Working Fluids (A) Ideal Gas (Perfect Gas) :: Statement No. (1)
Chapter Three Working Fluids (A) Ideal Gas (Perfect Gas) :: Statement No. (1)
Working fluids
An ideal gas is a gas which obeys the Boyle's law and Charles's law.
Boyle's law:- If a fixed mass of a gas is kept at constant temperature its volume varies
inversely to its pressure.
1
V , (T=constant)
p
C
V , pV=C
p
p1V1 p 2V2
p1 V1
p 2 V2
Charles's law:-
Statement No.(1):- If a fixed mass of a gas receives or reject heat at constant pressure, the
volume of the gas directly proportional to its absolute temperature.
V T , (p=constant)
V 2
V V1 V2
V CT , C,
T T1 T2
V1 T1
V2 T2 1
T
Statement No.(2):- If a fixed mass of gas receives or rejects heat at a constant volume, the
pressure of the gas is directly proportional to it’s
absolute temperature. p 2
pT
p p1 p 2
p CT , C,
T T1 T2
1
V1 T1
V2 T2 T
Equation of state for ideal gases:
Let 1kg of ideal gas follows Boyle's law from state 1 to state i and then change to state 2 at
constant volume.
pV=C
pi p
2 ……………(2) Vi=V2
Ti T2
p1V1 p1V1
From eq.(1) pi
Vi V2
p1V1 p 2
Substituting pi in eq.(2):
V2Ti T2
p1V1 p 2V2
T1 T2
p
cons tan t R
T
,
p RT pV mRT
Another form of the characteristic equation can be derived using the kilogram-mole as a unit.
The kilogram- mole:- Is defined as a quality of a gas equivalent to M kg of the gas, where
M is the molecular weight of the gas.
Example : (the molecular weight of oxygen is 32, then 1kg mole of oxygen is equivalent to
32kg of oxygen)
m =nM
Where;
pV=nMRT
Example (3.1):
Calculate R for air if the air at atmosphere pressure 1.01325 bar and 0°C has the density
1.293kg/m3.
p RT
p p 1.01325 *105
R 287 J / kg.K 0.287kJ / kg.K
T T 1.293 * 273
Avogadro's hypothesis:- State that "equal volumes of different gases at the same pressure
and temperature, contain equal number of molecules".
paVa pbVb
na M aRaTa nb M bRbTb
M aRa M bRb
MR cons tan t Ro
Ro
R
M
Where;
MR Ro
pV=nRoT
**: 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;
pV 10 5 * 22.71
Ro 8314 J / mole.K
nT 1 * 273
Example (3.2):
A vessel of volume 0.2 m3 contains nitrogen at 1.013 bar and 15°C. If 0.2 kg of nitrogen is
now pumped into the vessel, calculate the new pressure when the vessel has returned to its
initial temperature. The molecular weight of nitrogen is 28, and it may be assumed to be a
perfect gas.
Solution:
V=0.2 m3, p1=1.013 bar, T1=15°C, p2=?, T2=T1, M=28
p1V1 m1 RT1
Ro 8314
R 296.9 J / kg .K
M 28
p2V2 m2 RT2
m2 m1 0.2
Example (3.3):
A certain perfect gas of mass 0.01 kg occupies a volume of 0.003 m3 at a pressure of 7 bar
and a temperature of 131 °C. The gas is allowed to expand until the pressure is 1 bar and the
final volume is 0.02 m3. Calculate:
(i) the molar mass of the gas;
(ii) the final temperature.
Solution:
m=0.01kg, V1=0.003 m3, p1=7 bar, T1=131°C, p2=1bar, V2=0.02 m3, M=?, T2=?
p1V1 mRT1
p2V2 mRT2
The specific heat of a solid or liquid is usually defined as the heat required to raise unite
mass through one degree temperature rise.
Where;
**: Two specific heats for gases are defined, the specific heat at constant volume (cυ), and
the specific heat at constant pressure, (cp).
1 dQ
cp
m dT
1 dQ
c
m dT
Note:- For a perfect gas the values of cp and cυ are constant for any one gas at all pressure
and temperature while for real gases, cp and cυ vary with temperature.
Q mc p (T2 T1 )
Q mc (T2 T1 )
Joule's law:- state's that the internal energy of a perfect gas is a function of the absolute
temperature only.
u =F(T)
du=cυdT ,
du
c
dT
du
c
u=cυT+K
dT
where, (K) : constant
U 2 U 1 mc (T2 T1 )
Q mc (T2 T1 ) W
mRT2
pV2 mRT2 V2
p
mRT2 mRT1
W p mR (T2 T1 )
p p
cp=cυ+R
R=cp-cυ
h =u+pυ
H=mcpT
The ratio of specific heat at constant pressure to the specific heat at constant volume called
the sensible heat ratio (γ).
cp
c
R=cp-cυ
cp R R
1 1
c c c
R
c
1
c p c
R
cp
( 1)
Example (3.4):
A certain perfect gas has a specific heat capacities as follows: cp= 0.846 kJ/kg K and cυ=
0.657 kJ/kg K Calculate the gas constant and the molar mass of the gas.
R=cp-cυ
=0.846-0.657=0.189kJ/kg.K
R R 8314
R o M o 44kg / mole
M R 189
Example (3.5):
A perfect gas has a molar rna ss of 26 kg/kmol and a value of γ= 1.26. Calculate the heat
rejected:
(i) when the gas is contained in a rigid vessel at 3 bar and 315°C, and is then cooled until the
pressure falls to1.5 bar.
(ii) when the gas enters a pipeline at 280°C, and flows steadily to the end of the pipe where
the temperature is 20°C. Neglect changes in velocity of the gas in the pipeline.
Solution:-
(i)/
Heat rejected per kg of gas; Q c (T2 T1 )
R 8314
R o 319.8 J / kg.K
M 26
R 319.8
c 1.229kJ / kg.K
1 0.26
p1V1 mRT1
p1 T1
p 2V2 mRT2 V1 V2
p 2 T2
p2 1 .5
T2 T1 ( ) 588( ) 294 K
p1 3
(ii)/
h1 Q h2 W Q h2 h1
For perfect gas, h=cpT
Q mc p (T2 T1 )
The working fluid:- Is the matter contained within the boundaries of a system. When two
independent properties of the fluid are known then the thermodynamic state of the fluid is
defined.
In thermodynamic systems the working fluid can be in the liquid, vapor or gaseous phase but
the solid phase is not important in engineering thermodynamics.
C: critical point.
Saturation state:- Is defined as state at which a change of phase may occur without change
of pressure or temperature.
The latent heat of vaporization:- Is the amount of heat required to changes the phase of the
substance from liquid to vapor, during this change of phase the pressure and temperature
remain constant.
Critical pressure:- Is the pressure at which the turning point occurs between the saturated
liquid line and the saturated vapor line.
Saturated vapor line:- Is a series of point at which vaporization is complete joined to form
line.
Isothermals line:- Is the line of constant temperature.
Note:- when a dry saturated vapor is heated at constant pressure its temperature rises and it
becomes superheated.
Degree of superheated:- Is the difference between the actual temperature of the superheated
vapor and the saturation temperature at the pressure of the vapor.
For example the vapor at point S is superheated at PQ and T3 and the degree of superheated is
(T3-T1).
Region (1):-
Sub cool liquid (compressed liquid) :- A liquid at a pressure higher than the saturation
pressure corresponding to its temperature. (or, a liquid at a temperature lower than the
saturation temperature corresponding to its pressure).
T Tsat .
Example:-
P=1bar, T=50°C
P=1bar, T=100°C
Region (2):-
T=Tsat., p=psat.
Region (3):-
T Tsat .
Vapor,
Q (u 2 u1 ) W
W ( g f ) p
Q (u g u f ) p( g f )
Q (u g p g ) (u f p f )
Q hg h f h fg
The condition or quality of a wet vapor is most frequently defined by its dryness fraction,(x).
mv mv
x
mt ml mv
Where;
Wetness fraction=1-x
Wetness fraction:- is the mass of liquid in 1kg of the mixture.
mv
**: For a dry saturated vapor; x=1 x2 1
0 mv
0
**: For a saturated liquid; x=0 x1 0
ml 0
Note:-
**: The dryness fraction does not used in the superheated region.
f (1 x) g x
**: The volume of liquid is small compared to the volume of dry saturated vapor therefore is
usually negligibly.
g x
2-The enthalpy:
Is given by the sum of the enthalpy of the liquid plus the enthalpy of the dry vapor.
h (1 x)h f xhg
h h f xh fg
h fg hg h f
Is given by the sum of the internal energy of the liquid plus the internal energy of the dry
vapor.
u (1 x)u f xu g
u u f x(u g u f )
For steam in the superheated region temperature and pressure are independent properties.
When temperature and pressure are given for superheated steam then the state is defined and
all the other properties can be found. For example; steam at 2bar and 300°C.
> .
**: For pressure above 70 bar the internal energy can be found from u h p because it
is not tabulated in table.
Example (3.6):
Calculate the specific volume, specific enthalpy and specific internal energy of wet steam at
18 bar, dryness fraction 0.9.
Solution:
g x
At p=18bar, υg=0.1104m3/kg, hf=885kJ/kg, hfg=1912kJ/kg, uf =883kJ/kg, ug=2598kJ/kg
0.9 * 0.1104 0.0994 m 3 / kg
h h f xh fg 885 0.9 *1912 2605 .8kJ / kg
Example (3.7):
Find the dryness fraction, specific volume and specific internal energy of steam at 7 bar and
specific enthalpy 2600 kJ /kg.
Solution:
At p=7bar, hf=697kJ/kg, hfg=2067kJ/kg, uf =696kJ/kg, ug=2573kJ/kg, υg=0.2728m3/kg
h h f xh fg
2600 697 x (2067 ) x 0.921
g x 0.921(0.2728) 0.251m 3 / kg
Example (3.8):
Steam at 110 bar has a specific volume of 0.0196 m3/kg, calculate the temperature, the
specific enthalpy, and the specific internal energy.
Solution:
At p=110bar, υg=0.01598 m3/kg
g
From superheated tables at 110 bar and specific volume is 0.0196 m3/kg , the temperature;
T=350°C, and h=2889 kJ/kg.
Example (3.9):
Steam at 150 bar has a specific enthalpy of 3309 kJ/kg. Calculate the tempera lure, the
specific volume and the specific internal energy.
Solution:
At p=150 bar, hg=2611kJ/kg ≤ h(3309) the steam is superheated
At p=150 bar and h=3309 kJ/kg T=500°C
υ =0.02078 m3/kg
150 *105 * 0.02078
u h p 3309 2997.3kJ / kg
103
Interpolation:-
For properties which are not tabulated exactly in the tables it is necessary to interpolate
between the values tabulated.
Example (3.10):
Find the temperature, specific volume, specific internal energy and specific enthalpy of dry
saturated steam at 9.8 bar.
10 9 9 .8 9
179 .9 175 .4 T 175 .4
Example (3.11):
Steam at 5 bar and 320°C. Find the specific volume and enthalpy.
Note:- In some cases a double interpolation is necessary for example to find the enthalpy of
superheated steam at 18.5 bar and 432°C.
450 400 432 400
3364 3256 h 3256
450 400 3364 3256
432 400 h 3256
432 400
h (3364 3256) 3256 3325.1kJ / kg
450 400
20 15 18.5 15
3317 .8 3325 .1 h 3325 .1
18.5 15
h( )(3317 .8 3325 .1) 3325 .1
20 15
h 0.7( 7.3) 3325 .1 3320 kJ / kg
Example (3.12):
Sketch a pressure-volume diagram for steam and mark on it the following points. Labeling
clearly the pressure, specific volume and temperature of each point.
solution:
point(a):
at p=20 bar, Ts=212.4°C < T(250°C)
superheated
p=20 bar, υ=0.1115m3/kg, T=250°C
point(b):
at T=212.4°C, υg=0.0995m3/kg=υ
dry saturated
p=20 bar, υ=υg=0.0995m3/kg, T=212.4°C
point(c):
at p=10 bar, h=2650kJ/kg, hg=2778kJ/kg
hg>h wet steam
h h f xh fg
2650 763 x ( 2015 ) x 0.936
x g 0.936 * 0.1944 0.182 m 3 / kg
p=10 bar, υ=0.182 m3/kg , T=Ts =179.9°C
point(d):
at p=6 bar, h=3166 kJ/kg, hg=2757 kJ/kg
h > hg superheated
at p=6 bar and h=3166 kJ/kg, T=350°C and υ=0.4743 m3/kg