Question

A twin-spool turbofan engine at sea level (\(P_a = 1\ \text{bar},\ T_a = 288\ \text{K}\)) has separate cold and hot nozzles. During static thrust test, the total air mass flow rate is 100 kg/s and the cold exhaust temperature is 288 K. Given:
Fan pressure ratio = 1.6
Overall pressure ratio = 20
Bypass ratio = 3.0
Turbine entry temperature = 1800 K
\(C_p = 1.005\ \text{kJ/kg-K}\), \(\gamma = 1.4\).
Find the static thrust from the cold nozzle (ideal fan and ideal expansion), in kN (round to two decimals).

Hint:

For ideal fan nozzles at static conditions, thrust comes entirely from cold-stream momentum: \(F = \dot{m}V\). Always convert \(C_p\) to J/kg-K.

Answer 1

Correct Answer: 20

Total mass flow rate:
\[ \dot{m}_{total} = 100\ \text{kg/s} \] Bypass ratio \(= 3\):
\[ \dot{m}_{cold} = \frac{3}{4}(100) = 75\ \text{kg/s} \] 1. Temperature rise across ideal fan:
For isentropic compression,
\[ T_{t2} = T_a \left( \text{FPR} \right)^{(\gamma-1)/\gamma} \] \[ T_{t2} = 288 (1.6)^{0.286} = 288(1.140) = 328.3\ \text{K} \] Fan temperature rise:
\[ \Delta T_f = 328.3 - 288 = 40.3\ \text{K} \] 2. Cold exhaust jet velocity:
Fully expanded to ambient → \(\text{exit static } T_e = T_a = 288\ \text{K}\).
Total-to-static jet velocity:
\[ V_j = \sqrt{2 C_p (T_{t2} - T_a)} \] Convert \(C_p\): \(1.005\ \text{kJ/kg-K} = 1005\ \text{J/kg-K}\) \[ V_j = \sqrt{2(1005)(40.3)} = \sqrt{81006} = 284.5\ \text{m/s} \] 3. Static thrust from cold nozzle:
\[ F = \dot{m}_{cold} V_j \] \[ F = 75(284.5) = 21337.5\ \text{N} \] \[ F = 21.34\ \text{kN} \] Thus the static thrust is approximately:
\[ \boxed{21.34\ \text{kN}} \]