Answer:
Given that
Mass flow rate ,m=2.3 kg/s
T₁=450 K
P₁=350 KPa
C₁=3 m/s
T₂=300 K
C₂=460 m/s
Cp=1.011 KJ/kg.k
For ideal gas
P V = m R T
P = ρ RT
[tex]\rho_1=\dfrac{P_1}{RT_1}[/tex]
[tex]\rho_1=\dfrac{350}{0.287\times 450}[/tex]
ρ₁=2.71 kg/m³
mass flow rate
m= ρ₁A₁C₁
2.3 = 2.71 x A₁ x 3
A₁=0.28 m²
Now from first law for open system
[tex]h_1+\dfrac{C_1^2}{200}+Q=h_2+\dfrac{C_2^2}{2000}[/tex]
For ideal gas
Δh = CpΔT
by putting the values
[tex]1.011\times 450+\dfrac{3^2}{200}+Q=1.011\times 300+\dfrac{460^2}{2000}[/tex]
[tex]Q=1.011\times 300+\dfrac{460^2}{2000}-\dfrac{3^2}{200}-1.011\times 450[/tex]
Q= - 45.49 KJ/kg
Q =- m x 45.49 KW
Q= - 104.67 KW
Negative sign indicates that heat transfer from air to surrounding