More Examples of Fanno Flow

Solution

First calculate the dimensionless resistance, $\frac{4fL}{D}$ .

$$\frac{4fL}{D} = {4 \times 0.05 \times 4 \over 0.02 } = 40$$

From Figure (9.21) for $P_2 / P_1 = 0.1$ $M_1 \approx 0.13$ etc.

or accurately utilizing the program as in the following table.

$\rule[-0.1in]{0.pt}{0.3 in} \mathbf{M_1}$	$\mathbf{M_2}$	$\mathbf{4fL \over D}$	$\mathbf{\left.{ 4fL \over D }\right\vert _{1}}$	$\mathbf{\left.{ 4fL \over D }\right\vert _{2}}$	$\mathbf{P_2 \over P_1}$
0.12728	1.0000	40.0000	40.0000	0.0	0.11637
0.12420	0.40790	40.0000	42.1697	2.1697	0.30000
0.11392	0.22697	40.0000	50.7569	10.7569	0.50000
0.07975	0.09965	40.0000	107.42	67.4206	0.80000

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Only for the pressure ratio of 0.1 the flow is choked.

$\rule[-0.1in]{0.pt}{0.3 in}\mathbf{M}$	$\mathbf{T \over T_0}$	$\mathbf{\rho \over \rho_0}$	$\mathbf{A \over A^{\star}}$	$\mathbf{P \over P_0}$	$\mathbf{A\times P \over A^{*} \times P_0}$
0.12728	0.99677	0.99195	4.5910	0.98874	4.5393
0.12420	0.99692	0.99233	4.7027	0.98928	4.6523
0.11392	0.99741	0.99354	5.1196	0.99097	5.0733
0.07975	0.99873	0.99683	7.2842	0.99556	7.2519

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Therefore, $T\approx T_0$ and for the same the pressure. Hence, the mass rate is a function of the Mach number. The Mach number is indeed function of the pressure ratio but and therefore mass flow rate is function pressure ratio only through Mach number.

The mass flow rate is

$$\dot{m} = P A M \sqrt{k \over R T} = 300000 \times {\pi \times 0.... ...0.127 \times \sqrt{ 1.4 \over 287 300} \approx 0.48 \left( kg \over sec \right)$$

and for the rest

$$\dot{m} \left( \mathbf{P_2 \over P_1} = 0.3 \right) \sim 0.48 \times {0.1242 \over 0.1273}=0.468 \left(kg\over sec\right)$$

$$\dot{m} \left( \mathbf{P_2 \over P_1} = 0.5 \right) \sim 0.48 \times {0.1139 \over 0.1273}=0.43 \left(kg\over sec\right)$$

$$\dot{m} \left( \mathbf{P_2 \over P_1} = 0.8 \right) \sim 0.48 \times {0.07975 \over 0.1273}=0.30 \left(kg\over sec\right)$$