Chapter 4
Q. 4.7
A two-spool turbojet engine has the following data:
Fuel heating value is 45,000 kJ/kg.
A. It is required to calculate
1. The temperature and pressure at the inlet and outlet of each module of the engine
2. Fuel-to-air ratio for both the combustion chamber and afterburner
3. The exit area for both operative and inoperative afterburner
4. The specific thrust and the TSFC
5. The propulsive, thermal, and overall efficiency
B. Recalculate as above but when the airplane powered by the above engine is flying at 8 km altitude with a Mach number 2.04. The maximum temperature at the afterburner is increased to 2000 K and the air mass flow rate is increased to 314.64 kg/s.
In both cases the nozzle is unchoked.
| \pmb{\eta_{\text{d}}} | \pmb{\eta_{\text{c}1}} \\ \pmb{\eta_{\text{c}2}} | \pmb{\eta_{\text{cc}}} | \pmb{\eta_{\text{t}1}} \\ \pmb{\eta_{\text{t}2}} | \pmb{\eta_{\text{ab}}} | \pmb{\eta_{\text{n}}} | \pmb{\pi_{{\text{c}}_{1}}} | \pmb{\pi_{{\text{c}}_{2}}} | \pmb{\Delta P_{\text{cc}}} | \pmb{\Delta P_{\text{cc}}} | \pmb{\text{TIT}} \\ \pmb{\text{(K)}} | \pmb{T_\text{max}} \\ \pmb{\text{(K)}} | \pmb{A_{\text{i}}(\text{m}^2)} | \pmb{\dot{\text{m}}} \\ \pmb{\text{(kg/s)}} |
| 0.9 | 0.88 | 0.95 | 0.94 | 0.95 | 0.96 | 3.35 | 4.65 | 0.03 | 0.05 | 1200 | 1500 | 0.9519 | 186 |
Step-by-Step
Verified Solution
Using the previous equations, the results are arranged in the following table:
| Properties | Sea Level | Altitude 8 km (26,264 ft) |
| T_{\text{a}}(\text{K}) | 288.16 | 236.23 |
| P_{\text{a}}(\text{Pa}) | 1.013125 \times 10 ^5 | 3.5651 \times 10^4 |
| M | 0 | 2.04 |
| V(m/s) | 0 | 628.5893 |
| T_{02}(\text{K}) | 288.16 | 432.9772 |
| P_{02}(\text{Pa}) | 1.013125 \times 10 ^5 | 4.862 \times 10^4 |
| P_{03} | 3.3944 \times 10^5 | 1.6288 \times 10^5 |
| T_{03} | 442.8617 | 635.3745 |
| P_{04} | 1.5784 \times 10^6 | 7.5739 \times 10^5 |
| T_{04} | 686.9657 | 1032.2 |
| P_{05} | 1.531 \times 10^6 | 7.3467 \times 10^5 |
| T_{05} (TIT) | 1200 | 1200 |
| P_{06} | 6.0126 \times 10^5 | 1.6603 \times 10^5 |
| T_{06} | 964.9536 | 849.7425 |
| P_{07} | 3.4105 \times 10^5 | 6.0315 \times 10^4 |
| T_{07} | 854.0723 | 671.0998 |
| P_{08} | 3.2399 \times 10^5 | 6.0315 \times 10^4 |
| T_{08} \ (T_{\text{max}}) | \text{Operative A/B} = 1500 \\ \text{Inoperative A/B}=T_{07}=854.0723 | Operative A/B = 2000 \text{Inoperative A/B }= T_{07} = 671.0998 |
| P_9=P_{\text{a}} | 1.013125 \times 10^5 | 3.5651 \times 10^4 |
| T_9 | 1136.9 | 1785.2 |
| V_{\text{e}}(\text{m/s}) | 913.1156 | 702.2306 |
| T (N) | \text{Operative A/B }= 1.7582 \times 10^5 \\ \text{Inoperative A/B} = 1.7266 \times 10^5 | \text{Operative A/B }= 3.3388 \times 10^4 \\ \text{Inoperative A/B} = 2.4987 \times 10^4 |
| \eta_{\text{p}} | 0 | 0.9222 |
| \eta_{\text{th}} | 0.5977 | 0.1922 |
| \eta_{\text{0}} | 0 | 0.1802 |
| ƒ | 0.0166 | 0.0082 |
| f_{\text{ab}} | 0.0186 | 0.038 |
| A_{\text{e}}(\text{m}^2) | Operative A/B = 0.4381 Inoperative A/B = 0.4032 |
Operative A/B = 1.0424 Inoperative A/B = 1.0045 |
