Air enters the combustor of a jet engine at P1 = 10 atm, T1 = 1000oR, and M1 = 0.2. Fuel is injected and burned, with a fuel-air ratio (by mass) of 0.06. The heat released during the combustion is 4.5 * * 108 ft-lb/slug of fuel. Assuming one-dimensional frictionless flow with γγ = 1.4 for the fuel-air mixture, calculate M2, P2, and T2 at the exit of the combustor.

M2 = 0.06404

P2 = 2.273

T2 = 5806.45°R

Explanation:

Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.

Therefore from Steam Table, Po1 = (1.028) * (10) = 10.28 atm,

To1 = (1.008) * (1000) = 1008 ºR

R = 1716 ft-lb/slug-ºR cp = 6006 ft-lb/slug-ºR fuel-air ratio (by mass)

F/A = ? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108) FA ft-lb/sluga

For the air q = cp (To2 - To1)

(Exit flow - inlet flow) - choked flow is assumed For M1 = 0.2

Table A. 3 of steam table gives P/P * = 2.273,

T/T * = 0.2066,

To/To * = 0.1736 To * = To2 = To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp (To * - To) = (6006 ft-lb/sluga-ºR) * (5806.45 - 1008) ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga = FA * (4.5 x 108) ft-lb/slugaFA =

F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit