Answer:
(a) The area of the duct exit is 1.67 m²
(b) The air pressure at the exit is 1.195 x 10⁵ N/m²
Explanation:
Given;
inlet area, A₁ = 5 m²
inlet velocity, V₁ = 10 m/s
exit velocity, V₂ = 30 m/s
(a) The area of the duct exit is determined by applying continuity equation;
A₁V₁ = A₂V₂
Where;
A₂ is the area of the duct exit
A₂ = (A₁V₁) / (V₂)
A₂ = (5 x 10) / (30)
A₂ = 1.67 m²
(b) Apply Bernoulli’s equation to determine the pressure at the exit;
[tex]P_1 + \frac{\rho V_1^2}{2} = P_2 + \frac{\rho V_2^2}{2}\\\\P_2 = P_1 + \frac{\rho V_1^2}{2} - \frac{\rho V_2^2}{2}\\\\P_2 = P_1 + \frac{\rho }{2} (V_1^2 - V_2^2)[/tex]
Density of air at 300k = 1.177 kg/m³
[tex]P_2 = P_1 + \frac{\rho }{2} (V_1^2 - V_2^2)\\\\P_2 = 1.2*10^5 \ + \ \frac{1.177}{2} (10^2 - 30^2)\\\\P_2 = 1.2*10^5 \ + \ -470.8\\\\P_2 = 1.195*10^5 \ N/m^2[/tex]
Therefore, the air pressure at the exit is 1.195 x 10⁵ N/m²
