Question:

Consider the following Fanno flow problem: Flow enters a constant area duct at a temperature of 273 K and a Mach number 0.2 and eventually reaches sonic condition (Mach number = 1) due to friction. Assume $\gamma = 1.4$. The static temperature at the location where sonic condition is reached is ................. K (rounded off to 2 decimal places).

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In Fanno flow, total temperature $T_0$ remains constant (adiabatic). To find static temperature at sonic condition, compute $T_0$ from inlet and then use $T = T_0/(1 + (\gamma-1)/2 \, M^2)$.
Updated On: Aug 22, 2025
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Solution and Explanation

Step 1: Stagnation temperature is constant in Fanno flow.
For Fanno flow (adiabatic, with friction in constant area duct): \[ T_0 = \text{constant} \] Step 2: Compute $T_0$ at inlet.
\[ T_0 = T \left(1 + \frac{\gamma - 1}{2} M^2 \right) \] At inlet: \[ T = 273 \, \text{K}, \quad M = 0.2, \quad \gamma = 1.4 \] \[ T_0 = 273 \left( 1 + \frac{0.4}{2} \times (0.2)^2 \right) \] \[ T_0 = 273 \left( 1 + 0.2 \times 0.04 \right) = 273 (1.008) = 275.184 \, \text{K} \] Step 3: Condition at sonic state.
At Mach = 1: \[ T_0 = T \left( 1 + \frac{\gamma - 1}{2} \times 1^2 \right) \] \[ T_0 = T \left( 1 + 0.2 \right) = 1.2T \] \[ T = \frac{T_0}{1.2} = \frac{275.184}{1.2} \] \[ T = 229.32 \, \text{K} \] Step 4: Recheck consistency.
Wait – mistake! Let's carefully recompute. We found $T_0$ correctly as $275.184 \, \text{K}$. At sonic condition: \[ T = \frac{T_0}{1.2} = \frac{275.184}{1.2} = 229.32 \, \text{K} \] Yes, that is correct. Step 5: Final Answer.
\[ \boxed{229.32 \, \text{K}} \]
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