Question:medium

A Mach \(1.5\) air flow enters a round duct of length \(20\) cm and diameter \(3\) cm. If the flow exits with Mach number \(1.1\), the average Fanning friction factor \(f\) of the duct is _______ \(\times 10^{-3}\) (rounded off to 1 decimal place).
An excerpt from the Fanno flow table for air is given below.
1.11.21.31.41.51.6
99.35336.4648.3997.413611724

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Use the Fanno flow subtraction rule: the actual duct's \(4fL/D\) equals the tabulated \(4fL^*/D\) at the inlet Mach number minus the value at the exit Mach number.
Updated On: Jul 16, 2026
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Correct Answer: 4.7

Solution and Explanation

Step 1: State the Fanno flow relation between two stations.
The Fanno flow function $4fL^*/D$ measures, in a dimensionless way, how far a given Mach number is from the sonic point $M=1$. Between any two stations 1 (inlet) and 2 (exit) of an actual duct of length $L$, the physical length traveled equals the difference between the two "distance-to-sonic" values:
\[ \frac{4fL}{D} = \left(\frac{4fL^*}{D}\right)_{M_1} - \left(\frac{4fL^*}{D}\right)_{M_2} \]

Step 2: Pull the two table entries.
$M_1 = 1.5 \Rightarrow 1361\times10^{-4}$; $M_2 = 1.1 \Rightarrow 99.35\times10^{-4}$. Difference: $(1361-99.35)\times10^{-4} = 1261.65\times10^{-4}$.

Step 3: Keep L and D in centimeters, since only their ratio matters.
$L/D$ is dimensionless, so unit conversion to meters is not actually needed; working directly in cm gives the same ratio:
\[ \frac{L}{D} = \frac{20 \text{ cm}}{3 \text{ cm}} = 6.667 \]

Step 4: Solve for f.
\[ f = \frac{1}{4}\left(\frac{D}{L}\right)\left[\left(\frac{4fL^*}{D}\right)_{M_1} - \left(\frac{4fL^*}{D}\right)_{M_2}\right] = \frac{1}{4}\left(\frac{3}{20}\right)(1261.65\times10^{-4}) \]
\[ f = (0.0375)(0.126165) = 0.0047312 \]

Final Answer:
\[ \boxed{f \approx 4.7\times10^{-3}} \]
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