Fanno Curves Fanno Flow Equations Solution of Fanno Flow Equations Integration of Fanno Flow Equations

1. FANNO CURVE & FANNO EQUATION
Dhaval Chauhan
Mechanical Engineer

2. Fanno Curve
 Consider a flow of a fluid in a perfectly insulated (Q=0)
constant area duct.
 Continuity equation:
ṁ = ρ ∙ AC
∴ ṁ
A = ρ ∙ C = Constant, G …(1)
Where G = Mass flow density.
 Energy equation:
h + C2
2 = h0 …(2)
Where h0= Stagnation enthalpy

3. But from Equation (1), C=
𝐺
𝜌
∴ h +
G2
2ρ2 = h0 …(3)
 If the properties of ℎ0 and G are known on upstream
side, the values of h and ρ can be obtained at any
section of the duct. Fig. 1 shows the relation defined
by Equation (1) for a single value of ℎ0 and various
values of G on the plot of h Vs V or 1
ρ. These curves
are called Fanno lines.
 According to second law of thermodynamics, the
entropy during the adiabatic process with friction
always increases.

4.  Therefore, the path of states for portion of curves
must be from left to right. It follows that during
subsonic flow, the effect of friction will be to increase
the velocity and Mach number with reduction in
enthalpy and pressure. While, during supersonic flow,
the effect of friction will be to decrease the velocity
and Mach number with increase in enthalpy and
pressure.

5. Fig. (1) Fanno lines on (h-s) diagram

6. Fanno flow equation:
 Continuity equation:
dρ
ρ
+
d∁
∁
= 0
 Perfect gas equation:
dp
p
=
dρ
ρ
+
dT
T
 By definition of Mach number:
1
M2 ∙ dM2 =
1
C2 ∙ dC2 −
dT
T
 Energy equation:
dT
T
+
γ_1
2
M2 ∙
dC2
C2 = 0

7.  Momentum equation
γM2
2
×
dC2
C2
+ 4f ∙
dx
D
×
γM2
2
+
dp
p
= 0
 Stagnation pressure – Mach number relationship
dp0
p0
=
dp
p
+
γM2
2
1
1 +
γ − 1
2
M2
∙
dM2
M2
 Impulse function
dF
F
=
dp
p
+
γM2
1 + γM2
∙ d
M2
M2

8. Solution of Fanno Lines Equations and
Effect of Wall Friction on Fluid Properties
 We have discussed above simultaneous Equations
which relate to following eight different variables:
𝑑ρ
ρ
,
𝑑𝐶2
𝐶2 ,
𝑑𝑀2
𝑀2 ,
𝑑𝑇
𝑇
,
𝑑𝑝
𝑝
,
𝑑𝑝0
𝑝0
,
𝑑𝐹
𝐹
and 4𝑓 ∗
𝑑𝑥
𝐷
 Out of the above, variable 4𝑓 ∗
𝑑𝑥
𝐷
is independent
variable which is responsible for changes in flow
properties during the flow in a duct. Now we can solve
the simultaneous equation discussed in terms of
removing seven variables as follows:
 ∴
𝑑𝑝
𝑝
=
𝑑𝜌
𝜌
+
𝑑𝑇
𝑇
…(1)

9. and,
𝑑𝑇
𝑇
+
𝛾−1
2
𝑀2 ∗
𝑑𝐶2
𝐶2 = 0 …(2)
 On substituting the value of
𝑑𝑇
𝑇
from Eq.(2) in Eq.(1),
𝑑𝑝
𝑝
=
𝑑𝜌
𝜌
-
𝛾−1
2
𝑀2
∗
𝑑𝐶2
𝐶2 …(3)
But,
𝑑𝜌
𝜌
= −
1
2
∗
𝑑𝐶2
𝐶2 from Continuity equation, hence,
𝑑𝑝
𝑝
= −
1+ 𝛾−1 𝑀2
2
∗
𝑑𝐶2
𝐶2 …(4)
 On substituting the value of Eq.(4) in Momentum equation
we get,
𝑑𝑃
𝑝
= −
1+ 𝛾−1 𝑀2
2
∗
4𝑓 ∗𝑑𝑥
𝐷
∗
𝛾𝑀2
1−𝑀2 …(5)

10.  On substituting the value of Eq.(5) in (4) we get,
1
2
𝑑𝐶2
𝐶2 =
𝑑𝐶
𝐶
=
4𝑓 ∗𝑑𝑥
𝐷
∗
𝛾𝑀2
2 1−𝑀2 …(6)
 Since, from Continuity equation, therefore substituting
the value of
𝑑𝐶
𝐶
from Eq.(6) in Continuity equation we
get,
𝑑𝜌
𝜌
= −
4𝑓 ∗𝑑𝑥
𝐷
∗
𝛾𝑀2
2 1−𝑀2 …(7)
 From Perfect gas equation we have,
𝑑𝜌
𝜌
=
𝑑𝜌
𝜌
+
𝑑𝑇
𝑇
…(8)
On substituting the value of Eq.(5) & (7) we get,
𝑑𝑇
𝑇
= −
4𝑓 ∗𝑑𝑥
𝐷
∗
𝛾𝑀4(𝛾−1)
2 1−𝑀2 …(9)

11.  From Equation by definition of Mach number,
1
M2 ∙ dM2 =
1
C2 ∙ dC2 −
dT
T
On substituting the values from Eq.(6) & (9) we get,
𝑑𝑀2
𝑀2 =
4𝑓 ∗𝑑𝑥
𝐷
∗
𝛾𝑀2
1−𝑀2 ∗ 1 +
𝛾−1
2
𝑀2 …(10)
 From Momentum equation,
𝑑𝑝0
𝑝0
=
𝑑𝑝
𝑝
+
𝛾𝑀2
2
∗
1
1+
𝛾−1
2
𝑀2
∗
𝑑𝑀2
𝑀2 …(11)
On substituting the values from Eq.(5) & (10) we get,
𝑑𝑝0
𝑝0
= −4𝑓 ∗
𝑑𝑥
𝐷
∗
𝛾𝑀2
2
…(12)

12.  From equation of Impulse Function,
dF
F
=
dp
p
+
γM2
1+γM2 ∙ d
M2
M2 …(13)
On substituting the values from Eq.(5) & (10) we get,
dF
F
= −
4𝑓∗𝑑𝑥
𝐷
∗
𝛾𝑀2
2 1+𝛾𝑀2 …(14)

13. Integration of Fanno equations
 Variation of Mach number with duct length
f =
1
Lmax 0
Lmax
f ∙ dx
 Temperature ratio
T
T∗
=
γRT
2
γRT∗
2 =
a2
a∗2
 Density ratio
ρ
ρ∗
=
C2
C
=
1
M
2 1 +
γ − 1
2
M2
γ − 1

14.  Velocity ratio
C
C∗
= M ∙
γ + 1
2 1 +
γ − 1
2
M2
1
2
 Pressure ratio
p
p∗
=
1
M
∙
γ + 1
2 1 +
γ − 1
2
M2
1
2
 Stagnation pressure ratio
p0
p0
∗ =
1
M
∙
2 1 +
γ − 1
2
M2
γ + 1
γ−1 2 γ−1

15.  Impulse function ratio
F
F∗
=
1 + γM2
M 2 1 + γ 1 +
γ − 1
2
M2
1 2
 Change in entropy
s − s∗ = −R ∙ ln
2 1 +
γ − 1
2
M2
γ + 1
γ−1 2 γ−1
×
1
M