Question 8.12: The average friction factor of a 25 mm diameter 12 meter lon......

Given: D=0.025  m ; L=12  m ; f=0.004 ; p_1=2 \text { bar; } T_1=300  K ; M_1=0.25

Refer Figure 5.22. The mass flow rate through the pipe can be calculated from the continuity equation applied at section 1.

\begin{aligned}\dot{m}&=\rho_1 A V_1 \\&=\frac{p_1}{R T_1}\times \frac{\pi}{4}\times D_1^2 \times M_1 \times \sqrt{\gamma R T_1}\\&=\frac{2 \times 10^5}{287 \times 300}\times \frac{\pi}{4}\times(0.025)^2 \times 0.25 \times \sqrt{1.4 \times 287 \times 300}\\&=0.0989  kg / s\end{aligned}

The length of the pipe is given as 12 m. The friction parameter for the pipe length can be calculated from the given data.

4 f \frac{L}{D}=\frac{4 \times 0.004 \times 12}{0.025}=7.68

The friction parameter for the given length is known. The maximum friction parameter
corresponding to the inlet section can be obtained from Fanno tables, The maximum friction parameter corresponding to the exit section can be calculated as follows:

\begin{aligned}4 f \frac{L}{D}&=\left(4 f \frac{L_{\max}}{D}\right)_{M=0.25}-\left(4 f \frac{L_{\max}}{D}\right)_{M=M_2}\\7.68 &=8.537-\left(4 f \frac{L_{\max}}{D}\right)_{M=M_2}\\\left(4 f \frac{L_{\max}}{D}\right)_{M-M_2}&=0.857\end{aligned}

From Fanno tables, corresponding to a friction parameter of 0.857, the Mach number is 0.53

M_2=0.53

The properties at the exit section can be calculated from the property ratios at sections 1 and 2 available in Fanno tables.

\frac{p_2}{p_1}=\frac{p_2 / p^*}{p_1 / p^*}=\frac{\left(p / p^*\right)_{M=0.53}}{\left(p / p^*\right)_{M=0.25}}=\frac{2.012}{4.361}=0.461

\begin{aligned}P_2 &=p_1 \times 0.461=2 \times 0.461 \\&=0.922 bar \\\frac{T_2}{T_1}&=\frac{T_2 / T^*}{T_1 / T^*}=\frac{\left(T / T^*\right)_{M=0.53}}{\left(T / T^*\right)_{M=0.25}}=\frac{1.136}{1.185}=0.959 \\T_2 &=T_1 \times 0.959=300 \times 0.959 \\&=287.7   K .\end{aligned}