Answer:
b1 = 0.8228, b2 = 1.078, b3 = 0.1772, b5 = 5.472, Φ2 = 1.012 rad, Φ1 = 1.043 rad,
EF= 14 KJ, EFs = 14.6 K,
Explanation:
We are given that the measured engine fuel flow rate is 0.4g/s = 0.004 kh, airflow rate is 5.6 g/s = 0.0056 kg . Therefore the ratio of airflow rate to the engine fuel flow rate = EF = 5.6/0.4 = 14.
We also know that the "fuel is Gasoline with an H/C ratio of 1.87" that is x = 1.87.
The equation for the reaction is given below;
CHx + 1/Φ × (1 + x/ 4) (O2 + 3. 773 N2) ----------------> b1 CO2 + b2 H2O + b3 CO + b4 H2 + b5 N2.
So, if n1 + n2 = 1 ; x = 2 × b2 + 2 × b4.
Note that our EFs = 34.56 [ 4 + x / 12.01 + 1.08 + x ].
Note that Φ1 = EFs/EF = 1.043 radian.
(Where EFs= 14.6).
Then, 1/ Φ2 × [ 1 + x/4 ] = 2 × b1 + b2 + b3. -------------------------------(1).
1/ Φ2 × [1 + x/4] ×2 × 3.773 = 2 × b5. ----------------------(1).
If we solve the above equations we will have b1 = 0.8228, b2 = 1.078, b3 = 0.1772, b5 = 5.472, Φ2 = 1.012 radian.
Pressure of the dry composition = b1 + b3 + b4+ b5 = 6.329 kPa.