Answer
Given,
Air enter density, ρ₁ = 2.21 kg/m³
speed of entry, v₁ = 40 m/s
Air exit density , ρ₂ = 0.752 kg/m³
speed of exit, v₂ = 180 m/s
Inlet area = 90 cm²
a) mass flow rate through nozzle.
[tex]m = \rho_1 A_1 v_1[/tex]
[tex]m = 2.21\times 90\times 10^{-4}\times 40[/tex]
m = 0.796 kg/s
b) exit area
Using continuity equation
[tex]\rho_1 A_1 v_1=\rho_2A_2 v_2[/tex]
[tex]A_2 = \dfrac{\rho_1 A_1 v_1}{\rho_2 v_2}[/tex]
[tex]A_2 = \dfrac{2.21\times 90\times 40}{0.752\times 180}[/tex]
[tex] A_2 = 58\ cm^2[/tex]
Exit area of nozzle is equal to [tex]58\ cm^2[/tex]
(a) The mass flow rate through the nozzle is 0.796 kg/s.
(b) The exit area of the nozzle is 58 cm².
(a)
Using the formula of flow rate,
M' = DvA..................... Equation 1
Where M' = mass flow rate through the Nozzle, D = Density of air as it enters the nozzle, A = inlet area of the nozzle, v = entering velocity of the air
From the question,
Given: D = 2.21 kg/m³, v = 40 m/s, A = 90 cm³ = 9.0×10⁻³ m²
Substitute these values into equation 1
M' = 2.21(40)(9.0×10⁻³)
M' = 0.7956
M' ≈ 0.796 kg/s
Hence, The mass flow rate through the nozzle is 0.796 kg/s
(b)
Also,
making A the subject of the equation in equation 1
A' = M'/D'v'................. Equation 2
where A' = exit area of the nozzle, D' = exit density of air, v' = exit velocity
From the question,
Given: M' = 0.796 kg/s, D' = 0.752 kg/m³, v = 180 m/s
Substitute these values into equation 2
A' = 0.796/(0.752×180)
A' = 0.00588 m²
A' ≈ 58 cm²
Hence, The exit area of the nozzle is 58 cm²
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