Question:

In a static test, a turbofan engine with bypass ratio of 9 has core hot exhaust speed 1.5 times that of fan exhaust speed. The engine is operated at a fuel–air ratio of $f=0.03$. Both the fan and the core streams have no pressure thrust. The ratio of fan thrust to thrust from the core engine is (round off to one decimal place).

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Fan thrust depends only on air mass flow, while core thrust depends on higher jet velocity and fuel addition. Always include the factor $(1+f)$ in core mass flow.
Updated On: Nov 27, 2025
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Correct Answer: 5.7

Solution and Explanation

Let fan exhaust speed be $V_f$. Core exhaust speed = $1.5V_f$. Bypass ratio = $\beta = 9$. Fuel–air ratio = $f = 0.03$.
Fan mass flow rate:
\[ \dot{m}_f = 9\dot{m}_c. \] Fan thrust:
\[ T_f = \dot{m}_f V_f = 9\dot{m}_c V_f. \] Core total mass flow (air + fuel):
\[ \dot{m}_{core} = (1+f)\dot{m}_c. \] Core thrust in static test:
\[ T_c = (1+f)\dot{m}_c(1.5V_f). \] Substitute $f = 0.03$:
\[ T_c = 1.03\dot{m}_c(1.5V_f) = 1.545\dot{m}_c V_f. \] Thrust ratio:
\[ \frac{T_f}{T_c} = \frac{9\dot{m}_c V_f}{1.545\dot{m}_c V_f} = \frac{9}{1.545} \approx 5.83. \] Rounded to one decimal place:
\[ \boxed{5.8} \]
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