Experimental data indicate that in a region downstream of a given louvered supply vent the velocity of the emitted air is defined by τ = 0.18τo/x where τ and x are expressed in m/s and meters, respectively, and τo is the initial discharge velocity of the air. For τo 3.6 m/s. Determine:

a. The acceleration of the air at x = 2m.

b. The time required for the air to flow from x=1 to x = 3m.

a) a = - 0.0524 m/s²

b) t = 6.17 s

Explanation:

Given

τ = 0.18*τ₀ / x

τ₀ = 3.6 m/s

Determine:

a) The acceleration of the air at x = 2m.

Knowing

a = dτ / dt

Multiplying both the numerator and denominator by dx

a = (dτ / dx) (dx / dt)

Substituting τ for dx / dt

a = τ * (dτ / dx)

⇒ a = τ * (-0.18*τ₀ / x²) = (0.18*τ₀ / x) (-0.18*τ₀ / x²)

⇒ a = - 0.18²*τ₀² / x³ = - 0.18² * (3.6) ² / x³

⇒ a = - 0.4199 / x³

If x = 2m

⇒ a = - 0.4199 / (2) ³

⇒ a = - 0.0524 m/s²

b) The time required for the air to flow from x=1 to x = 3m.

If

τ = dx / dt = 0.18*τ₀ / x

⇒ dt = (x / 0.18*τ₀) dx

⇒ t = (1 / 0.18*τ₀) ∫x dx

⇒ t = (1 / 0.18*τ₀) * (1 / 2) * x²

then

t = (1 / (0.18*3.6)) * (1 / 2) * ((3) ² - (1) ²)

⇒ t = 6.17 s