Question

Air flows steadily in a pipe with a velocity of 60 ft/s. Surrounding the pipe in an annulus is a second flow of air with a velocity of 40 ft/s. Both flows are exhausted through a 1.5 ft diameter pipe. If the velocity is uniform at the exit, determine the velocity at the exit. Assume constant density.

Answer 1

According to equation of continuity, (mass flown per second) $\Sigma$Q_in = $\Sigma$Q_out

This can be considered as a flow system with two inlets and one exit.

Then, $\Sigma$Q_in   = A₁v₁$\rho$ + A₂ v₂$\rho$      and $\Sigma$Q_out = A₃v₃$\rho$

As $\rho$ is the same, we have: A₁v₁ + A₂ v₂ = A₃v₃ ..................(i)

Here, take d₁ = 1 ft, d₂ = 2 ft and d₃ = 1.5 ft. Also, v₁ = 60 ft/s, v₂ = 40 ft/s, v3 = ?

A₁ = $\pi$(d₁² )/4, A₂ = $\pi$[(d₂² – d₁²)/4] and A₃ = $\pi$(d₃²)/4

[ Note: d stands for diameter.. so area = $\pi$d²/4. A₂ is the annular area. hence the formula].

Putting A1, A2 and A3 in equation (i),

$\pi$(d₁² /4) × v₁ + $\pi$[(d₂² – d₁²)/4] × v₂ = $\pi$(d₃²)/4 × v₃

Equation reduces to

d₁² × v₁ + (d₂² – d₁²) × v₂ = d₃² × v₃  

    ==> 1 × 60 + (4 – 1) × 40 = 1.5² × v₃

    ==> 2.25 v₃ = 180 ==> v₃ = 80 ft/s.....(Required Ans).

Hope it is clear..