Question

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.

1.1	1.2	1.3	1.4	1.5	1.6
99.35	336.4	648.3	997.4	1361	1724

Hint:

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.

Answer 1

Correct Answer: 4.7

Step 1: Recall what the Fanno flow table gives.
In adiabatic flow with friction in a constant-area duct (Fanno flow), \(L^*\) is the extra length of duct that would be needed, from a given Mach number, to bring the flow all the way to \(M=1\). The table lists the dimensionless group \(4fL^*/D\) as a function of \(M\).

Step 2: Read the two table values needed.
At the duct inlet, \(M_1 = 1.5\), giving \((4fL^*/D)_1 = 1361 \times 10^{-4}\).
At the duct exit, \(M_2 = 1.1\), giving \((4fL^*/D)_2 = 99.35 \times 10^{-4}\).

Step 3: Subtract to get the actual duct's friction group.
Since the flow is supersonic and friction is pushing it toward \(M=1\) as it travels down the duct, the actual physical duct length equals the difference of the two "distance to sonic" lengths: \(L = L_1^{*} - L_2^{*}\). In terms of the tabulated group,
\[ \frac{4fL}{D} = \left(\frac{4fL^*}{D}\right)_1 - \left(\frac{4fL^*}{D}\right)_2 = (1361 - 99.35)\times10^{-4} = 1261.65\times10^{-4} = 0.126165 \]

Step 4: Substitute the given L and D.
\(L = 20\) cm \(= 0.2\) m, \(D = 3\) cm \(= 0.03\) m, so \(L/D = 0.2/0.03 = 6.667\).
\[ 4f(6.667) = 0.126165 \]
\[ f = \frac{0.126165}{4 \times 6.667} = \frac{0.126165}{26.667} = 0.0047312 \]

Final Answer:
\(f \approx 4.73\times10^{-3}\), which rounds to
\[ \boxed{f \approx 4.7\times10^{-3}} \]