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CO2 enters an adiabatic nozzle, operating at steady state, at 200 kPa, 1500 K, 5 m/s and exits at 100 kPa, 1400 K. The exit area of the nozzle is 10 cm2. Using the PG model, determine the exit velocity

User Prelic
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1 Answer

4 votes

Answer:


v_2=549.2 m/s\\

Step-by-step explanation:

Given:


P_1=2500kPa\\T_1=1500 k\\V_1=5 m/s\\P_2=100 kPa\\T_2=1400 k\\A_2=10 cm^2

Solution:

For
Co_2 y=1.4

Since Nozzle is adiababic

So,


h_1+(V_1^2)/(2)=h_2+(V_2^2)/(2)\\(v_2^2)/(2)=(h_2-h_2)+(r^2)/(2)\\v_2^2=2(h_1-h_2)+v_1^2\\v_2=√(2(h_1-h_2)+v_1^2)

Now,


h_1-h_2=Cp_1T_1-CP_2T_2\\h_1-h_2=(1989-1838.2)*10^3\\ =150.8 * 10^3\\Cp for co_2\\C_(p1)=1.326 kj/kg\\C_(p2)=1.313 kj/kg\\v_2=√(301600+25)\\ =549.2 m/s

User Naveen R Kumar
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