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

luster is not a good characteristic to use by itself because luster is the property of minerals that indicates how much the surface of a mineral reflects light. the luster of a mineral is affected by the brilliance of the light used to observe the mineral surface.