Answer:
A) 600.8 m/s
B) (i) 22.5 cm^2 (ii) 7.11 cm^2
Explanation:
Given data :
P1 = 20 bar , T1 = 2808°C
P2 = 7 bar , T2 = 1808°C
mass flow rate = 1.5 kg/s
Using the superheated vapor region in Table A-4
h1 = 2976.4 KJ/kg , v1 = 0.1200 m^3/kg
h2 = 2799.1 KJ/kg , v2 = 0.2847 m^3/kg
A) calculate exit velocity ( m/s )
given that we are to neglect heat transfer and potential energy
V2 ( exit velocity ) = ( V1^2 + 2 (√h1 - h2 )
= [ (80)^2 + 2 ( √ 2976.4 - 2799.1 )
= 600.8 m/s
B) calculate the inlet and exit flow areas ( cm^2 )
i) Inlet flow area
A1 = ( m * v1 ) / V1
= ( 1.5 * 0.1200 )/ 80 = 22.5 cm^2
ii) exit flow area
A2 = ( m * v2 ) / V2
= ( 1.5 * 0.2847 ) / 600.8 = 7.11 cm^2
