Question

Question 1 Steam enters a nozzle at 300 kPa and 700ºC with a velocity of 20 m/s. The nozzle exit pressure is 200 kPa. Assuming this process is reversible and adiabatic, determine (a) the exit temperature and (b) the exit velocity.

Answer 1

A reversible, adabatic process is isentropic.

Thus, ΔS = C_P ln(T₂/T₁) - R ln(P₂/P₁) = 0

Substituting T₁ = 700 C = 973 K, P₁ = 300 kPa, P₂ = 200 kPa.

Also, for steam C_p = 35.928 J/mol-K and R= 8.314 J/mol-K

We get exit temperature T₂ = 885.86 K = 612.86 C

Also, from steady state steady flow equation,

h_in + 0.5u²_in = h_out + 0.5u²_out

Or h_in - h_out = 0.5 (u²_out - u²_in ) [ specific values i.e. per unit mass ]

h_in - h_out = C_p (T_in - T_out) = 1.996 kJ/kg-K x ( 700 C - 612.86 C ) = 173.9357 kJ/kg

Substitute u_in = 20 m/s to get

outlet velocity u_out = 590.145 m/s