Air enters a horizontal, well-insulated nozzle operating at steady state at 12 bar, 500 K, with a velocity of 50 m/s and exits at 7 bar, 440 K. The mass flow rate is 1 kg/s. Determine the net force, in N, exerted by the air on the duct in the direction offlow. Answer:

The net force is 300.8 N

Explanation:

∆H = Cp (T1 - T2) = 1/2 (V2^2 - V1^2)

Cp is the heat capacity of air at constant pressure = 1005 J/kg

T1 is initial temperature of air = 500 K

T2 is the exit temperature of air = 440 K

V1 is the initial velocity of air = 50 m/s

V2 is the exit velocity of air

1005 (500 - 440) = 1/2 (V2^2 - 50^2)

60,300*2 = V2^2 - 2,500

V2^2 = 120,600 + 2,500 = 123,100

V2 = sqrt (123,100) = 350.8 m/s

Net force (F) = mass flow rate * change in velocity = 1 kg/s * (350.8 - 50) m/s = 1 kg/s * 300.8 m/s = 300.8 kgm/s^2 = 300.8 N