Final answer:
The problem involves using the Continuity Equation and the principle of conservation of mass to find the density of air in the test section of a wind tunnel. Given sea level density, inlet and test section areas, and their corresponding velocities, the density at the test section is calculated.
Step-by-step explanation:
The question involves determining the density of air in the test section of a wind tunnel, given the inlet and test section areas, as well as the inlet and test section velocities. To find the density in the test section, the Continuity Equation and the conservation of mass principle are used. This involves the equation ρ₁A₁V₁ = ρ₂A₂V₂, where ρ is density, A is area, V is velocity, and subscripts 1 and 2 denote conditions at the inlet and test section, respectively.
On sea level standard day conditions, air at sea level has a density of 1.225 kg/m³ (or using a conversion factor 0.002377 slugs/ft³). Plugging the inlet area (A₁ = 10.5 ft²), inlet velocity (V₁ = 650 ft/s), test section area (A₂ = 2.7 ft²), and test section velocity (V₂ = 2950 ft/s) into the Continuity Equation, we can solve for the density at the test section (ρ₂).
Since we are looking for the density in the test section and the problem states that we can use sea level conditions, ρ₁ can be assumed as 0.002377 slugs/ft³. Therefore, we compute ρ₂ by rearranging the equation to ρ₂ = (ρ₁A₁V₁)/(A₂V₂). By plugging in the values, we find that:
ρ₂ = (0.002377 slugs/ft³ * 10.5 ft² * 650 ft/s) / (2.7 ft² * 2950 ft/s).