Exit Conditions of Fanno Flow in a Duct Air enters a 27-m-long 5-cm-diameter adiabatic duct at V1 = 85 m/s, T1 = 450 K, and P1 = 220 kPa (Fig. 12–67). 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.

Question

Exit Conditions of Fanno Flow in a Duct

Air enters a 27-m-long 5-cm-diameter adiabatic duct at [latex] V_{1}=85 m / s , \quad T_{1} = 450 K [/latex] , and [latex] P_{1}=220 kPa [/latex] (Fig. 12–67). 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.

Accepted Answer

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 [latex] k=1.4, c_{p}=1.005 kJ / kg \cdot K [/latex], and [latex] R=0.287 kJ / kg \cdot K [/latex].
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 [latex] f L^{*} / D_{h} [/latex],

[latex] \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 } ight)}=425 m / s \ Ma _{1} &=\frac{V_{1}}{c_{1}}=\frac{85 m / s }{425 m / s }=0.200 \end{aligned} [/latex]

Corresponding to this Mach number we read, from Table A–16, [latex] \left(f L^{*} / D_{h} ight)_{1} [/latex] = 14.5333. Also, using the actual duct length L, we have

[latex] \frac{f L}{D_{h}}=\frac{(0.023)(27 m )}{0.05 m }=12.42<14.5333 [/latex]

Therefore, flow is not choked and the exit Mach number is less than 1. The function [latex] f L^{*} / D_{h} [/latex] at the exit state is calculated from Eq. 12–91,

[latex] \left(\frac{f L^{*}}{D_{h}} ight)_{2}=\left(\frac{f L^{*}}{D_{h}} ight)_{1}-\frac{f L}{D_{h}}=14.5333-12.42=2.1133 [/latex]

[latex] \frac{f L}{D_{h}}=\left(\frac{f L^{*}}{D_{h}} ight)_{1}-\left(\frac{f L^{*}}{D_{h}} ight)_{2} [/latex] (12–91)

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

[latex] Ma _{2}= 0 . 4 2 [/latex]

The mass flow rate of air is determined from the inlet conditions to be

[latex]\begin{aligned} ho_{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}} ight)=1.703 kg / m ^{3} \\dot{m}_{ ext {air }} &= ho_{1} A_{1} V_{1}=\left(1.703 kg / m ^{3} ight)\left[\pi(0.05 m )^{2} / 4 ight](85 m / s )=0.284 kg / s\end{aligned}[/latex]

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 [latex] f L^{*} / D_{h} [/latex] 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

[latex]\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}[/latex]

[latex] ext { (or, } L_{\max , 2}=L_{\max , 1}-L=31.6-27=4.6 m ext { ) } [/latex]. Therefore, the flow would reach sonic conditions if a 4.6-m-long section were added to the existing duct.

Question 12.17: Exit Conditions of Fanno Flow in a Duct Air enters a 27-m-lo...

Related Answered Questions

Choked Fanno Flow in a Duct Air enters a 3-cm-diameter smooth adiabatic duct at Ma1 = 0.4, T1 = 300 K, and P1 = 150 kPa (Fig. 12–66). If the Mach number at the duct exit is 1, determine the duct length and temperature, pressure, and velocity at the duct exit. Also determine the percentage of ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Gas Flow through a Converging Nozzle Nitrogen enters a duct with varying flow area at T1 = 400 K, P1 = 100 kPa, and Ma1 = 0.3. Assuming steady isentropic flow, determine T2, P2, and Ma2 at a location where the flow area has been reduced by 20 percent. ...

Verified Answer:

Effect of Back Pressure on Mass Flow Rate Air at 1 MPa and 600°C enters a converging nozzle, shown in Fig. 12–24, with a velocity of 150 m/s. Determine the mass flow rate through the nozzle for a nozzle throat area of 50 cm² when the back pressure is (a) 0.7 MPa and (b) 0.4 MPa. ...

Verified Answer:

Critical Temperature and Pressure in Gas Flow Calculate the critical pressure and temperature of carbon dioxide for the flow conditions described in Example 12–3 (Fig. 12–19). ...

Verified Answer:

Verified Answer:

Effect of Heat Transfer on Flow Velocity Starting with the differential form of the energy equation, show that the flow velocity increases with heat addition in subsonic Rayleigh flow, but decreases in supersonic Rayleigh flow. ...

Verified Answer:

Verified Answer: