Question 6.18: Problem: A wind tunnel takes air from the atmosphere and use...
Problem: A wind tunnel takes air from the atmosphere and uses a fan to accelerate it through the test section. A pressure transducer shows a pressure difference of 500 Pa between the inside of the wind tunnel and the outside atmosphere. Assuming that the density of air is 1.23 kg / m³ , what is the velocity of air in the wind tunnel?
Find: Velocity of air in the wind tunnel \pmb{V} _2 .
Known: Pressure difference between inside and outside of wind tunnel ∆P = 500 Pa.
Assumptions: Air is an ideal gas with constant specific heat, density of air ρ = 1.23 kg / m³ .
Governing equation:
Bernoulli Equation (inviscid, incompressible) \frac{P_1}{\rho } + \frac{\pmb{V}_1^2}{2} + gz_1= \frac{P_2}{\rho } + \frac{\pmb{V}_2^2}{2} + gz_2
The air velocity in the atmosphere is \pmb{V} _1 = 0. Assuming that elevation differences are negligible, the Bernoulli equation reduce to
\frac{\pmb{V}_2^2}{2} =\frac{P_1-P_2}{\rho } =\frac{\Delta P}{\rho } ,
\pmb{V}_2=\sqrt{\frac{2\Delta P}{\rho } } =\sqrt{\frac{2\times 500 \ Pa}{1.23 \ kg/m^3} } =28.513 \ m/s.
Answer: The air velocity in the wind tunnel is 28.5 m / s.