Exercise 1
DETERMINATION OF THE ISENTROPE EXPONENT FOR AIR
1.1.
Introduction
An isentropic transformation, in other words a reversible adiabatic transformation, is a process that takes
place without the exchange of heat between a system and environment (dQ = 0), which means that
entropy is constant (s = const). Work of the transformation is used to change internal energy.
The equation describing changeability of the parameters of state in the isentropic transformation is
Poisson’s equation which is presented in three forms:
Pv k = const
(1.1)
Tvk-1 = const.
(1.2)
Tk p1-k = const.
(1.3)
where κ is the exponent of isentrope formulated by the following relation
k
Cp
Cv

cp
(1.4)
cv
where Cp (cp) is the molar (specific) heat at constant pressure and Cv (cv) is the molar (specific) heat at
constant volume. The unit of molar heat is
J
kJ
J
or
, while that of specific heat is
.
mol K
kmol K
kg K
The molar (specific) heat at constant pressure Cp (cp) is the amount of heat required to heat up 1 mol
or 1 kmol (1 kg) of gas under constant pressure by 1 degree in the temperature range from 14.5 to15.5°C.
 I 
 i 
Cp  
 ,cp  

 T  p
 T  p
(1.5a,b)
The molar (specific) heat at constant volume Cv (cv) is the amount of heat required to heat up 1 mol
or 1 kmol (1 kg) of gas of constant volume by 1 degree in the temperature range from 14.5 to 15.5°C.
 U 
 u 
Cv  
 ,cp   
 T  v
 T  v
(1.6a,b)
In equations (1.5a,b) and (1.6a,b) the enthalpy and internal energy are given in J/mol (kJ/kmol) and J/kg,
respectively.
The defining equations (1.5) and (1.6) are justified by the fact that the heat in isobaric and isochoric
transformations, according to two forms of the first law of thermodynamics is
Where: q – heat [J/mol] or [J/kg],
i – enthalpy [J/mol] or [J/kg],
dp  di  vdp  di
(1.7)
dq  du  pdv  du
(1.8)
 m3 
 m3 
v – molar or specific volume 
or

 ,
 mol 
 kg 
p – pressure [Pa].
According to the kinetic theory of gas, heat Cp and Cv (cp and cv) changes for perfect gas, depending on
the number of atoms in the molecule.
In general,
C v (cv) 
i
R ( Ri )
2
(1.9)
C p (c v ) 
i
R( Ri )
2
(1.10)
Where: i – degrees of freedom (for monoatomic gas i = 3, diatomic i = 5, and triatomic i = 6).
Hence, changeability of the isentrope exponent is
-
for monoatomic gases
Cv 
3
5
R, C p  R
2
2
or
cv 
3
5
Ri , c p  Ri , k  1,66
2
2

R  J 
Where: Ri – individual gas constant  R i  
  ,
M
kgK



 J
R – universal gas constant equal to 8.314 
 molK

 ,
 kg 
M – molecular weight 
,
 mol 
-
for diatomic gases
Cv 
-
5
7
R, C p  R , k  1,4
2
2
for triatomic gases
C v  3R, C p  4 R
2
Thus, for each perfect gas the value of exponent κ is constant.
For real gas the values of Cp and Cv depend on temperature and pressure. It should be stressed that the
effect of temperature at moderate pressures is higher than the effect of pressure and should not be
neglected in technical calculations. In general, the dependence on temperature is given for molar heat
(and not for specific heat) at constant pressure Cp in the form of the polynomial
C p  A  BT  CT 2  ...
(1.11)
Constants A,B,C,… are given in tables for various gases.
The effect of pressure on Cp is usually presented in a graphic form as a function of reduced parameters –


p 
 pr   , where:
pk 


– reduced temperature, reduced pressure,
temperature  Tr 
Tr, pr
T
 and pressure
Tk 
Tk, pk – critical temperature [K], critical pressure [Pa],
T, p
– given temperature [K], given pressure [Pa].
1.2.
Aim of the exercise
The aim of this exercise is to determine the exponent of isentrope for air, and to study thermodynamic
transformations on theoretical and experimental way.
1.3.
Experimental system
The main element of the experimental system (Fig. 1.1) is a spherical bottle (1) with a volume
of 26.5 dm3 tightly closed on top with a stopper from organic glass. Through the stopper thermometer (2)
and manometer (3) are introduced into the bottle. The bottom of the bottle is connected to the discharge
pipe equipped with a tap (5). A big diameter of the discharge pipe makes it possible to quickly expand air
so as the transformation meets conditions similar to those of the isentropic transformation. Above tap (5)
there is a connecting pipe with tap (4) through which air is supplied by means of a bellows.
Fig. 1.1. Schematic of the experimental system: 1 – glass bottle, 2 – thermometer, 3 – manometer,
4 – connecting pipe with tap and a bellows to supply air, 5 – discharge tap
3
1.4.
Method of measurement
In the open glass bottle there is air at pressure p1 equal to atmospheric pressure and temperature T1 equal
to ambient temperature. After closing the vessel, the air in it is subjected to three subsequent
transformations (Fig. 1.2).
Transformation 1-2 (isothermal compression). Air pressure in the bottle increases from p1 to p2 by
blowing some air volume to the bottle from the environment by means of the bellows. This operation
should be carried out slowly enough to maintain constant temperature T1. Then, the specific air volume
in the bottle decreases from v1 to v2 (in the same volume the gas weight increases).
The initial and final state of the transformation 1-2 can be written as
p1, v1, T1 and p2, v2, T1.
Fig. 1.2. Thermodynamic transformations on pV and Ts charts
Transformation 2-3 (isentropic expansion). To equalize pressures it is enough to quickly turn on tap (5)
and open the glass bottle to surrounding air for a short time. While expanding, the gas decreases rapidly
its temperature from T1 to T3. If expansion time is very short it can be assumed that there is no heat
exchange with the environment and the transformation is isentropic. The gas volume increases from v2 to
v3 and pressure decreases from p2 to p3= p1. Using Poisson’s equation (1.1), changes of parameters
during the transformation 2-3 are represented by the equation
 v3

 v2
k

p
  2
p1

(1.12)
where p2, v2, T1 – air parameters at the beginning of transformation 2-3,
p3 = p1, v3, T3 – air parameters at the end of transformation 2-3.
Transformation 3-4 (isochoric heating). After closing the bottle, the air heats up from temperature T3 to
ambient temperature T1. At the same time its pressure grows from p3 = p1 to p4. Since the volume of gas
(bottle) does not change, the transformation is isochoric.
Air parameters at the beginning and end of transformation 3-4:
4
p3 = p1, v3, T3 and p4, v3, T1.
According to the isothermal transformation which combines states (2) and (4) (Fig. 1.2a):
v3
p
 2
v2
p4
(1.13)
From equations (1.12) and (1.13) it follows that
p2  p2 
 
p1  p 4 
k
(1.14)
By logarithmizing both sides of equation (1.14) we can calculate the exponent of the isentrope
p2
p1
k
p
log 2
p4
log
(1.15)
The initial and final parameters of particular transformations can be measured in the following way:
1. Read the values of atmospheric pressure and ambient temperature.
2. Turn on tap (5) and wait until the moment when air pressure in the bottle becomes equal to the
atmospheric pressure. The level of liquid in both legs of the manometer (3) should be equal.
3. Turn off tap (5), turn on tap (4) and using the bellows increase slowly gas pressure in the bottle
to the value given by the tutor. Wait a few minutes to equalize the gas temperature (if it has
changed) to the ambient temperature and then read the difference of liquid levels Δh1 in the
manometer.
4. Turn on tap (5) and turn it off when the pressures are equalized. After turning on tap (5) the gas
closed in the bottle and atmospheric air should combine for a short time. Wait 5 to 10 minutes to
equalize gas and ambient temperatures. Read the difference in the levels of manometric fluid
Δh2.
Each measurement for identical pressure value to which air should be compressed in the isothermal
transformation (p2 = p1 + Δh1 [mm Hg]), given by the tutor, should be repeated three times. Experimental
measurements are carried out for several different values of Δh1.
1.5.
Safety note
Pressure in the bottle must not exceed the values given by the tutor – there is a danger of explosion!
Taps should be turned on and off holding the tube with one hand, while pressing the valve knob
with the other hand so that the knob does not slide.
1.6.
Description of results
1. Calculate the isentrope exponent for air for each measurement according to formula (1.15). Pressures
p2 and p4 are calculated on the basis of readings of the differential manometer and barometer
5
p2  p1  h1
(1.16)
p4  p1  h2
(1.17)
Where:
p1 – atmospheric pressure [kPa],
Δh1 – difference of liquid levels in the manometers after the isothermal transformation
[kPa],
Δh2 – difference of liquid levels in the manometers after the isochoric transformation
[kPa].
2. Average the values of κ for identical values of Δh1 [kPa]
3. Calculate changes of entropy in transformations 1-2 and 3-4 using the equations:
s 2  s1  Ri ln
v2
p
 Ri ln 1
v1
p2
(1.18)
and
s 4  s3  cv ln
T4
p
 cv ln 4
T3
p3
(1.19)
4. Draw graphs of thermodynamic transformations in pV and Ts systems for each group of
measurements.
1.7.
Table of measurement and calculation results
Table 1.1
p1
T1
Δh1
h1
Δh2
h2
s2 – s1
s4 – s3
[kPa]
[K]
[kPa]
[kPa]
[kPa]
[kPa]
 J 
 kgK 


 J 
 kgK 


No.
κ
κm
1.
2.
3.
4.
5.
6
6.
7.
8.
9.
Table 1.2
p2
p3
p4
v1
v2
v3
v4
s2
s3
s4
T3
[m3]
[m3]
[m3]
[m3]
 J 
 kgK 


 J 
 kgK 


 J 
 kgK 


[K]
L.p.
[kPa] [kPa] [kPa]
1
2
3
7
1.7.
Table of measurement and calculation results
Table 1.1
No.
p1
T1
Δh1
h1
Δh2
h2
s2 – s1
s4 – s3
[kPa]
[K]
[kPa]
[kPa]
[kPa]
[kPa]
 J 
 kgK 


 J 
 kgK 


κ
κm
1.
2.
3.
4.
5.
6.
7.
8.
9.
10
11
12
8