Question

Air ows steadily in a thermally insulated pipe with a constant diameter of 6.35 cm, and an average friction factor of 0.005. At the pipe entrance, the air has a Mach number of 0.12, stagnation pressure 280 kPa, and stagnation temperature 825 K. Determine:

a. The length of pipe required to reach the sonic state
b. The static pressure and temperature at the exit if the pipe is 25 m long

Answer 1

Solution:

Given :

D = 6.35 cm

$\bar f = 0.005$

$P_s = 280 \ kPa$

$T_s= 825 K$

a). From fanno flow table (γ = 1.4)

At $M_1 = 0.12$ ,    $\left(\frac{4 \bar f L_{max}}{D}\right)_1 = 45.408$

At $M_2 = 1$ ,      $\left(\frac{4 \bar f L_{max}}{D}\right)_2 = 0$

∴  $\left(\frac{4 \bar f L_{max}}{D}\right)_1 - \left(\frac{4 \bar f L_{max}}{D}\right)_2 = \frac{4 \bar f L}{D}$

$45.408 - 0 = \frac{4 \times 0.005 \times L}{0.0635}$

$45.408 = \frac{4 \times 0.005 \times L}{0.0635}$

L = 144.17 m

b). If L = 25 m

$\frac{4 \bar f L}{D}=\frac{4 \times 0.005 \times 25}{0.0635} = 7.874$

From fanno flow , (γ = 1.4)

$At, M_2 = 0.26 , \frac{4 \bar f L}{D} = 7.874$

$\frac{P_s}{P_1}=\frac{T_s}{T_1}^{\frac{\gamma}{\gamma - 1}} = (1+\frac{\gamma-1}{2}M_1^2)^{\frac{\gamma}{\gamma - 1}}$

$\frac{280}{P_1}=\frac{825}{T_1}^{\frac{1.4}{1.4 - 1}} = (1+\frac{1.4-1}{2}(0.12)^2)^{\frac{1.4}{1.4 - 1}}$

$\frac{280}{P_1}=\left(\frac{825}{T_1}\right)^{3.5} =1.0101$

$P_1 = 277.2 \ kPa$

$T_1=822.63 \ K$

$\frac{T_2}{T_1}=\frac{1+\frac{\gamma -1}{2}M_1^2}{1+\frac{\gamma -1}{2}M_2^2}$

$\frac{T_2}{822.63}=\frac{1+\frac{1.4 -1}{2}(0.12)^2}{1+\frac{1.4 -1}{2}(0.26)^2}$

$\frac{T_2}{822.63}=\frac{1.00288}{1.01352}$

$T_2=814 \ K$

$\frac{P_2}{P_1}=\frac{M_1}{M_2}\left(\frac{1+\frac{\gamma -1}{2}M_1^2}{1+\frac{\gamma -1}{2}M_2^2}\right)^{1/2}$

$\frac{P_2}{P_1}=\frac{0.12}{0.26}\left(\frac{1+\frac{1.4 -1}{2}(0.12)^2}{1+\frac{1.4 -1}{2}(0.26)^2}\right)^{1/2}$

$\frac{P_2}{277.2}=0.459$

$P_2=127.27 \ kPa$