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.

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

an energy transformation flow diagram is shown

where is it ?

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