# Compressible duct flow

Compressible Duct Flow with friction. This article concerns a gas flow with friction due to the boundary walls under consideration. This problem is same as the Moody type pipe-friction problem but with considerably larger changes in friction, kinetic energy and enthalpy of the flow.

## Parameters

Consider a gas flowing through a duct with the following assumptions :
* Constant area, straight duct
* Gas is ideal
* Negligible potential energy changes and low work input
* One dimensional adiabatic flow
* Darcy friction factor used to express wall shear stress due to friction

## Calculation

Consider a small element of area A and length dx as shown in the figure (right). Applying conservation laws on this element,
$\rho$-density
V-velocity of the element
A- cross sectional area
ρV A = G =constant
$\rho V={\frac {G}{A}}$ ...(assumption 1)
${\frac {dp}{p}}+{\frac {dV}{V}}=0$.....(1)
By equating total force on the element to its change in momentum (following Newton’s second law),
$\rho A-(p+dp)A-\tau _{w}\Pi Ddx={\dot {m}}(V+dV-V)$
$dp+{\frac {4\tau _{w}dx}{D}}+\rho VdV=0$....(2)
where $\tau _{w}$ is shear stress due to boundary walls and thus $\tau _{w}\Pi Ddx$ is force due to shear stress.
Total energy of the gas
$h+{\frac {1}{2}}V^{2}+h_{o}=c_{p}T_{o}=c_{p}T+{\frac {1}{2}}V^{2}$
$c_{p}dT+VdV=0$.....(3)
Also by virtue of ideal gas equation
$p=\rho RT$...(i)
${\frac {dp}{p}}={\frac {d\rho }{\rho }}+{\frac {dT}{T}}$...(4)
The relation between wall shear and Darcy friction factor f is given by,
$\tau _{w}={\frac {1}{8}}f\rho V^{2}$
But we know that $a^{2}={\frac {kp}{\rho }}$ where a is the speed of sound in perfect gas. ....(ii)
Therefore the above equation transforms into-
$\tau _{w}={\frac {1}{8}}fkpMa^{2}$....(5)
Where Ma is the Mach number which is defined as the ratio of velocity of sound in that medium to the velocity of the object under consideration i.e V/a.....(iii)
Thus from (i), (ii),(iii)
$V^{2}=Ma^{2}kRT$
${\frac {2dV}{V}}={\frac {2dMa}{Ma}}+{\frac {dT}{T}}$.....(6)
By eliminating variables from eq 1 to 6 we get the following relations:
${\frac {dp}{p}}=-kMa^{2}{\frac {1+(k-1)Ma^{2}}{2(1-Ma^{2})}}f{\frac {dx}{D}}$
${\frac {d\rho }{\rho }}=-{\frac {kMa^{2}}{2(1-Ma^{2})}}f{\frac {dx}{D}}=-{\frac {dV}{V}}$
${\frac {dp_{o}}{p_{o}}}={\frac {d\rho _{o}}{\rho _{o}}}=-{\frac {1}{2}}kMa^{2}f{\frac {dx}{D}}$
${\frac {dT}{T}}={\frac {k(k-1)Ma^{4}}{2(1-Ma^{2})}}f{\frac {dx}{D}}$
${\frac {dMa^{2}}{Ma^{2}}}=kMa^{2}{\frac {1+{\frac {1}{2}}k(k-1)Ma^{2}}{(1-Ma^{2})}}f{\frac {dx}{D}}$
All the equations above have a factor of $1-Ma^{2}$ except ${\frac {dp_{o}}{p_{o}}}$; thus the following deductions could be made :
| Property | Subsonic | Supersonic |
| P | Decrease | Increase |
| $\rho$ | Decrease | Increase |
| V | Increase | Decrease |
| $P_{o},\rho _{o}$ | Decrease | Decrease |
| Ma | Decrease | Increase |
| T | Increase | Decrease |
| Entropy | Increase | Increase |
Irrespective of the flow, Mach number tends downstream towards 1 because entropy increases alongside. Entropy v/s Mach number graph is as shown in the figure (for k=1.4):
Entropy is maximum at Ma=1 so sonic point is eventually achieved by virtue of second law. Since $p_{o}$ and $\rho _{o}$ decrease along the flow in this case, the sonic properties (p*, ῥ*, T*, $p_{o}^{*}$, and $\rho _{o}^{*}$) are considered as references.
Now, By manipulating equation 11 and integrating both sides we get,
$\int _{0}^{L^{*}}{f{\frac {dx}{D}}}=\int _{Ma^{2}}^{1.0}{{\frac {1-Ma^{2}}{kMa^{4}[1+{\frac {1}{2}}(k-1)Ma^{2}]}}dMa^{2}}$
L* is the length from some arbitrary point where Mach number is Ma. At the sonic region, Ma=1. The integration results in the following equation :
${\frac {{\bar {f}}L^{*}}{D}}={\frac {1-Ma^{2}}{kMa^{2}}}+{\frac {k+1}{k}}\ln {\frac {(k+1)Ma^{2}}{2+(k-1)Ma^{2}}}$
Where f- is average friction factor from x=0 to L*. Here D is the diameter of circular duct; in cases where the duct is not of circular cross section D=(4*area)/perimeter.

## Choking due to friction

Choking of duct is basically reduction of duct mass flow. The flow conditions change if the actual length L is greater than the predicted maximum length L*, and there are two classifications :

### Subsonic inlet

If L > L*(Ma1), the flow slows down until an inlet Mach number Ma2 is reached such that L = L*(Ma2). The exit flow is sonic, and the mass flow has been reduced by frictional choking. Further increases in duct length will continue to decrease the inlet Ma and mass flow.

### Supersonic inlet

Friction has a large impact on supersonic duct flow. Even an infinite inlet Mach number will be reduced to sonic conditions in only 41 diameters for f = 0.02. Consider an example inlet Ma = 3.0 and f=0.02. For this condition L*=26 diameters. If L is increased beyond 26D, the flow will not choke but a normal shock will form at just the right place for the subsequent subsonic frictional flow to become sonic exactly at the exit. Eventually, a long duct will cause the feed nozzle throat to become choked, thus reducing the duct mass flow. Thus, supersonic friction changes the flow pattern if L > L* but does not choke the flow until L is much larger than L*.