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Question 1 Steam enters a nozzle at 300 kPa and 700ºC with a velocity of 20...

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.

Homework Answers

Answer #1

A reversible, adabatic process is isentropic.

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

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

Also, for steam Cp = 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,

hin + 0.5u2in = hout + 0.5u2out

Or hin - hout = 0.5 (u2out - u2in ) [ specific values i.e. per unit mass ]

hin - hout = Cp (Tin - Tout) = 1.996 kJ/kg-K x ( 700 C - 612.86 C ) = 173.9357 kJ/kg

Substitute uin = 20 m/s to get

outlet velocity uout = 590.145 m/s

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