Answer:
Vâ=43.58 m/s
Explanation:
Given that
Pâ=200 KPa
Tâ=90°C
Vâ=? m/s ( exit speed)
Tâ=30°C
Pâ=100 KPa
Vâ=350 m/s
We know that
Heat capacity for air Cp=1.005 KJ/kg.k
We know that for air change in enthalpy only depends only on temperature
Now from first law for open system
[tex]h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}+W[/tex]
[tex]1.005\times 303+\dfrac{350^2}{2000}=1.005\times 363+\dfrac{V_2^2}{2000}[/tex]
[tex]1.005\times 303+\dfrac{350^2}{2000}-1.005\times 363=\dfrac{V_2^2}{2000}[/tex]
[tex]\dfrac{V_2^2}{2000}=0.95[/tex]
Vâ=43.58 m/s