Exit Conditions of Fanno Flow in a Duct
Air enters a 27-m-long 5-cm-diameter adiabatic duct at V_{1}=85 m / s , T_{1}=450 K, and P_{1} = 220 kPa (Fig. 12–63). The average friction factor for the duct is estimated to be 0.023. Determine the Mach number at the duct exit and the mass flow rate of air.
SOLUTION Air enters a constant-area adiabatic duct of given length at a specified state. The exit Mach number and the mass flow rate are to be determined.
Assumptions 1 The assumptions associated with Fanno flow (i.e., steady, frictional flow of an ideal gas with constant properties through a constant cross-sectional area adiabatic duct) are valid. 2 The friction factor is constant along the duct.
Properties We take the properties of air to be k = 1.4, c_{p} = 1.005 kJ/kg·K, and R = 0.287 kJ/kg·K.
Analysis The first thing we need to know is whether the flow is choked at the exit or not. Therefore, we first determine the inlet Mach number and the corresponding value of the function f L^{*} / D_{h},
\begin{aligned}c_{1} &=\sqrt{k R T_{1}}=\sqrt{(1.4)(0.287 kJ / kg \cdot K )(450 K )\left(\frac{1000 m ^{2} / s ^{2}}{1 kJ / kg }\right)}=425 m / s \\Ma _{1} &=\frac{V_{1}}{c_{1}}=\frac{85 m / s }{425 m / s }=0.200\end{aligned}
Corresponding to this Mach number we read, from Table A–16, \left(f L * / D_{h}\right)_{1}= 14.5333. Also, using the actual duct length L, we have
\frac{f L}{D_{h}}=\frac{(0.023)(27 m )}{0.05 m }=12.42<14.5333
Therefore, flow is not choked and the exit Mach number is less than 1. The function f L * / D_{h} at the exit state is calculated from Eq. 12–91,
\frac{f L}{D_{h}}=\left(\frac{f L^{*}}{D_{h}}\right)_{1}-\left(\frac{f L^{*}}{D_{h}}\right)_{2} (Eq. 12-91)
\left(\frac{f L^{*}}{D_{h}}\right)_{2}=\left(\frac{f L^{*}}{D_{h}}\right)_{1}-\frac{f L}{D_{h}}=14.5333-12.42=2.1133
The Mach number corresponding to this value of fL*/D is 0.42, obtained from Table A–16. Therefore, the Mach number at the duct exit is
Ma _{2}= 0 . 4 2 0
The mass flow rate of air is determined from the inlet conditions to be
\begin{gathered}\rho_{1}=\frac{P_{1}}{R T_{1}}=\frac{220 kPa }{(0.287 kJ / kg \cdot K )(450 K )}\left(\frac{1 kJ }{1 kPa \cdot m ^{3}}\right)=1.703 kg / m ^{3} \\\dot{m}_{\text {air }}=\rho_{1} A_{1} V_{1}=\left(1.703 kg / m ^{3}\right)\left[\pi(0.05 m )^{2} / 4\right](85 m / s )=0.284 kg / s\end{gathered}
Discussion Note that it takes a duct length of 27 m for the Mach number to increase from 0.20 to 0.42, but only 4.6 m to increase from 0.42 to 1. Therefore, the Mach number rises at a much higher rate as sonic conditions are approached.
To gain some insight, let’s determine the lengths corresponding to f L * / D_{h} values at the inlet and the exit states. Noting that f is assumed to be constant for the entire duct, the maximum (or sonic) duct lengths at the inlet and exit states are
\begin{aligned}&L_{\max , 1}=L_{1}^{*}=14.5333 \frac{D_{h}}{f}=14.5333 \frac{0.05 m }{0.023}=31.6 m \\&L_{\max , 2}=L_{2}^{*}=2.1133 \frac{D_{h}}{f}=2.1133 \frac{0.05 m }{0.023}=4.59 m\end{aligned}
(or, \left.L_{\max , 2}=L_{\max , 1}-L=31.6-27=4.6 m \right). Therefore, the flow would reach sonic conditions if a 4.6-m-long section were added to the existing duct.