Chapter 1
Introduction: Basic Principles
1.1.(a) Air flows adiabatically through a long straight circular section duct, 0.25 m diameter at
a measured mass flow rate of 40 kg/s. At a particular section along the duct the values of static
temperature T = 150 0 C and static pressure p = 550 kPa. Determine the average velocity of the
airflow and its stagnation temperature.
(b) At another station further along the duct, measurements reveal that the static temperature
has dropped to 147 0C as a consequence of wall friction. Determine the average velocity and the
static pressure of the airflow at this station.
Also determine the change in entropy per unit of mass flow between the two stations.
For air assume that R = 287J/(kgK) and γ = 1.4
Solution
(a) From the gas law
p1
550 × 103
ρ1 =
=
= 4.53 kg/m 3
RT1 287 × 423
From the conservation of mass law
c1 =
mɺ
40
=
= 180 m/s Ans
ρ A 4.53 × 0.04908
π 
where A =   d 2 = 0.04908 m 2
4
From the conservation of energy equation the stagnation temperature is
T01 = T1 +
c12
179.92
= 423 +
= 439.1 K Ans
2C p
2 × 1004.5
(b) T2 = 147 + 273 = 420 K
As the flow is adiabatic, the energy of the air is conserved along the pipe, and h02 = h01
c2
c2
∴ C pT2 + 2 = C pT1 + 1
2
2
∴ c22 = c12 + 2C p ( −3) = 26373
∴ c2 = 162.4 m/s ANS
From the continuity equation
mɺ = ρ1c1 A = ρ 2 c2 A and the gas law ρ = p / ( RT ) we arrive at the
following useful expression:
180 × 420
p2 c1T2
=
=
= 0.9635
p1 c2T1 185.5 × 423
∴ p2 = 530 kPa
(c) The change in entropy between the two stations i:
 γ
T
p
T
p 
∆s = C p ln 2 − R ln 2 = R 
ln 2 − ln 2 
T1
p1
p1 
 γ − 1 T1
= 3.81 J/(kg K)
Ans
N.B.In this type of flow (called a FANNO FLOW) the entropy will always increase.
1.2. Nitrogen gas at a stagnation temperature of 300 K and a static pressure of 2 bar flows
through a pipe duct of 0.3 m diameter. At a particular station along the duct length the Mach
number is 0.6. Assuming the flow is frictionless, determine
(1) the static temperature and stagnation pressure of the flow,
(2) the mass flow of gas if the duct diameter is 0.3 m.
For nitrogen gas take
R = 297 J/ ( kgK ) and γ = 1.4
.
Solution
(1) From eqn. (1.35)
γ
p0  γ − 1 2  γ −1
= 1 +
M  = 1.2755 ∴ p0 = 2.551 bar Ans.
p 
2

and
T0
T
γ −1 2
M = 1.072 ∴ T = 0 = 279.9 K Ans.
= 1+
T
2
1.072
(2) From eqn. (1.38)
mɺ C pT0
An p0
and we have An =
π
4
 γ +1 

−1 
γ
 γ − 1 2  2  γ −1 
=
M 1 +
M 
2
γ −1 

d 2 = 0.07069 m 2 and C p =
γR
= 1040 J/(kg K)
γ −1
Substituting values (and part evaluating) we have, for the LHS of the above equation
mɺ 1040 × 300 ÷ ( 0.07069 × 2.551× 105 ) = 0.03097 mɺ
and for the RHS of the equation
−
1  γ +1 
γ M  γ − 1 2  2  γ −1  1.4 × 0.6
−3
M 
=
(1.072 ) = 1.0782
1 +
2
γ −1 
0.4

∴ mɺ = 27.16 kg/s ANS.
1.3. Air flows adiabatically through a horizontal duct and at a section numbered
(1) the static pressure p1 = 150 kPa, the static temperature T1 = 200 0C and the
velocity c 1 = 100 m/s. At a station further downstream the static pressure p 2 = 50
kPa and the static temperature T 2 =150 0 C. Determine the velocity c 2 and the
change in entropy per unit mass of air.
For air take
γ = 1.4 and R = 287 J/ ( kgK )
Solution
From eqn. (1.12)
h01 = h02
∴ h1 + 12 c12 = h2 + 12 c22
∴ c22 = 2C p (T1 − T2 ) + c12 where C p =
γR
= 3.5 × 287 = 1004.5 J/kgK
γ −1
∴ c22 = 2 ×1004.5 ( 200 − 150 ) + 1002 = 100450
c2 = 316.9 m/s Ans
From eqn.(1.28a)
T
p
150
 50 
∆s = C p ln 2 − R ln 2 = 1.0045ln
− 0.287 ln 

T1
p1
200
 150 
∴∆s = −0.289 + 0.315 = 0.0263 kJ/kg ANS
4
1.5
g
1.7
gg
gg
1.8
gg
1.9
gg
1.6.Steam at a pressure of 80 bar and a temperature of 500 0C is admitted to a turbine
where it expandsto a pressure of 0.15 bar. The expansion through the turbine takes place
adiabatically with an isentropic efficiency of 0.9 and the power output from the turbine is 40
MW. Using a Mollier chart and/or steam tables determine the enthalpy of the steam at exit
from the turbine and the flow rate of the steam.
Use tables to find h1 = 3398 kJ/kg and s1 = 6.723 kJ/kg K at p1 = 80 bar and T1 = 5000 C .
(
)
At the same value of entropy s2s = s1 use the steam tables or, more conveniently, a
Mollier chart we can determine h2s =2180 kJ/kg at p2 =0.15 bar.
Thus the isentropic enthalpy drop h1 –h2s = 1218 kJ/kg and the actual enthalpy drop is
h1 –h2 = 0.9 x 1218 = 1096.2 kJ/kg ∴ h2 = 2301.8 kJ/kg (Ans)
and from the Mollier chart, s2 = 7.115 kJ kg K.
From the conservation of energy equation we have
Power output = mass flow of steam x actual enthalpy drop (in consistent units)
Wɺ = ( h1 − h2 ) mɺ
∴ mɺ =
40 × 106
= 36.5 kg/s
1096.2 × 103
1.10. Carbon dioxide gas (CO 2) flows adiabatically along a duct. At station 1 the static pressure
p1 =120 kPa and the static temperature T 1 = 120 0C. At station 2 further along the duct the static
pressurep2 = 75 kPa and the velocity c 2 = 150 m/s.
Determine
(1) the Mach number M 2
(2) the stagnation pressure p02
(3) stagnationtemperatureT02
(4) the Mach number M 1
For CO 2 take R = 188 J/(kg K) and γ = 1.30.
For adiabatic flow h01 = h02 , ∴ h1 + 12 c12 = h2 + 12 c22 or C pT1 + 12 c12 = C pT2 + 12 c22
Hence, T2 =
c12 − c22
+ T1........ eqn A
2C p
From the perfect gas law and the continuity equation,
c1 =
ρ 2 c2 p2T1c2 75 × 393 × 150
=
=
= 36,844 / T2 eqn.B
ρ1
p1T2
120 × T2
c12 − 1502
or, T2 = 36884 / c1 =
+ T1 eqn.C
2C p
where C p =
γ R 1.3 ×188
=
= 814.7
γ −1
0.3
Eqn. C can be solved iterativly (or as a cubic) which should give c1 = 95.5 m/s .
c12 − 1502
95.52 − 1502
From eqn. A, T2 =
+ T1 =
+ 393 = 384.4 K
1630
1630
Results. (1) M 2 =
c2
150
=
= 0.489
γ RT2
1.3 ×188 × 384.8
γ
4.333
 γ − 1 2  γ −1
(2) p02 = p2 1 +
M 2  = 75 ×103 1 + 0.15 × 0.4892 
= 87.37 kPa
2




(3) T02 = T2 1 +
(4) M 1 = c1 / a1 =
γ −1

M 22  = 385.8 × [1 + 0.15 × 0.4892 ] = 399.6 K
2

95.5
= 0.308
1.3 ×188 × 393
11. Air enters the first stage of an axial flow compressor at a stagnation temperature of 20 0C and
at a stagnation pressure of 1.05 bar and leaves the compressor at a stagnation pressure of 11
bar. The total-to-total efficiency of the compressor is 83 per cent. Determine, the exit stagnation
temperature of the air and the polytropic efficiency of the compressor.
Assume for air that γ = 1.4
Solution
From eqn.(1.46a)
(γ −1) / γ
 p02 
−1


T02 s − T01  p01 
where T01 = 293 K
ηc =
=
T02
T02 − T01
−1
T01
γ −1
Hence,
0.2857
 p02  γ
 11 
−
1
−1




p01 
T02
1.05 


= 1+
= 1+
= 2.1524
T01
ηc
0.83
∴ T02 = 630.6 K Ans
From eqn.(1.50),
  p02  
 ln 

γ − 1   p01  
 2.349 
ηp =
= 0.2857 
γ   T02  
 0.7665 
ln
 

  T01  
∴η p = 0.8756 Ans