5-31 Air enters an adiabatic nozzle steadily at 300 kPa, 200 degree Cesius, and

Question

5-31 Air enters an adiabatic nozzle steadily at 300 kPa, 200 degree Cesius, and 30 m/s and leaves at 100 kPa and 180 m/s. The inlet area of the 80 cm^2. Determine (a) the mass flow rate through the nozzle, (b) the exit temperature exit temperature of the air, and (c) the exit area of the nozzle. Note: can you include the page for the table used to get values.
Book Answers: (a) 0.5304 kg/s, (b) 184.6 degree Cesium, (c) 38.7 cm^2 . Note: For the Solutions I will rate lifesaver (*****) Please give fully detail explanation along with the steps.

Explanation / Answer

(a) Using steady mass conservation equation:

mass flow rate m = 1A1V1 = 2A2V2

From property tables, At 300 kPa, 200 deg C, density 1 = 2.2117 kg/m3 (http://www.engineering-4e.com/calc1.htm Use P = 2.96 atm (=300 kPa) and T = 473 K (=200 deg C))

So, mass flow rate = 2.2117 * 80*10-4 * 30 = 0.5308 kg/s

(b) Using steady flow energy equation:

Q - W = (h2 + V22/2) - (h1 + V12/2)

For adiabatic nozzle, Q = W = 0.

Hence, h1 + V12/2 = h2 + V22/2

From property tables, At 200 deg C (~ 470 K), specific enthalpy h1 = 472.24 kJ/kg (Page 904 of http://books.google.co.in/books?id=6W_epDC0i3EC&printsec=frontcover&dq=Thermodynamics:+An+Engineering+Approach+6th+Edition,+Yunus+A.+Cengel,+Michael+A.+Boles&hl=en&sa=X&ei=fV3IT8CWFM7prQfNtaHWDg&ved=0CEYQ6AEwAQ#v=onepage&q&f=false

472.24*103 + 302/2 = h2 + 1802/2

h2 = 456490 J/kg = 456.49 kJ/kg

Now, looking back in the same table for h = 456.49 kJ/kg, we get, T2 ~ 457 K = 184 deg C.

(c) We have, 2 = P2/RT2 = 100*103/(287*457) = 0.7624 kg/m3

m = 2A2V2

0.5308 = 0.7624*A2*180

A2 = 0.0038677 m2 = 38.677 cm2

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