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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

Sagot :

Answer:

[tex]v_2=549.2 m/s\\[/tex]

Explanation:

Given:

[tex]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[/tex]

Solution:

For [tex]Co_2[/tex] y=1.4

Since Nozzle is adiababic

So,

[tex]h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}\\\frac{v_2^2}{2}=(h_2-h_2)+\frac{r^2}{2}\\v_2^2=2(h_1-h_2)+v_1^2\\v_2=\sqrt{2(h_1-h_2)+v_1^2}[/tex]

Now,

[tex]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=\sqrt{301600+25}\\ =549.2 m/s[/tex]

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