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(T_{2}/T_{1}) - R
ln(P_{2}/P_{1}) = 0

Substituting T_{1} = 700 C = 973 K, P_{1} = 300
kPa, P_{2} = 200 kPa.

Also, for steam C_{p} = 35.928 J/mol-K and R= 8.314
J/mol-K

We get exit temperature **T _{2} = 885.86 K =
612.86 C**

Also, from steady state steady flow equation,

h_{in} + 0.5u^{2}_{in} = h_{out}
+ 0.5u^{2}_{out}

Or h_{in} - h_{out} = 0.5
(u^{2}_{out} - u^{2}_{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**

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