Answer:
Given that
k=1.31
Po=5 bar
To=700 K
P₁=1 bar
T₁=?
A₁=50 cm²
We know that
[tex]\dfrac{T_1}{T_o}=\left(\dfrac{P_1}{P_o}\right)^{\dfrac{k-1}{k}}[/tex]
By putting the values
[tex]\dfrac{T_1}{T_o}=\left(\dfrac{P_1}{P_o}\right)^{\dfrac{k-1}{k}}[/tex]
[tex]\dfrac{T_1}{700}=\left(\dfrac{1}{5}\right)^{\dfrac{1.3-1}{1.3}}[/tex]
T₁=482.3 K
Now from first law for open system
[tex]h_o+\dfrac{V_o^2}{2000}=h_1+\dfrac{V_1^2}{2000}[/tex]
[tex]C_p=R\dfrac{k}{k-1}\ KJ/kg.k[/tex]
[tex]C_p=\dfrac{8.314}{23}\times \dfrac{1.31}{1.31-1}\ KJ/kg.k[/tex]
Cp=1.527 KJ/kg.k
h= Cp T
[tex]h_o+\dfrac{V_o^2}{2000}=h_1+\dfrac{V_1^2}{2000}[/tex]
[tex]1.527\times 700+\dfrac{0^2}{2000}=1.527\times 482.3+\dfrac{V_1^2}{2000}[/tex]
[tex]\dfrac{V_1^2}{2000}=1.527\times 700-1.527\times 482.3[/tex]
V₁=815.37 m/s
Mass flow rate m=ρ A₁V₁
[tex]\rho=\dfrac{P}{RT}[/tex]
[tex]\rho=\dfrac{100}{\dfrac{8.314}{23}\times 482.3}\ kg/m^3[/tex]
ρ=0.57 kg/m³
m=ρ A₁V₁
m= 0.57 x 50 x 10⁻⁴ x 815.37 m/s
m=2.32 kg/s
