Glycerin of density 1.25 × 10³ kg m⁻³ is flowing through the conical section of pipe The area of cross-section of the pipe at its ends are 10 cm² and 5 cm² and pressure drop across its length is 3 Nm⁻². The rate of flow of glycerin through the pipe is x × 10⁻⁵ m³ s⁻¹. The value of x is ___________.

Solution

Physics sir · 4 steps

Use continuity to tie the two speeds together, then Bernoulli turns the pressure drop into one speed.

1. Continuity first. The same volume passes every section each second, so A₁ v₁ = A₂ v₂. With A₁ = 10 cm² and A₂ = 5 cm², the narrow end has v₂ = 2v₁.
2. Bernoulli. Treat the pipe as horizontal, so the height terms cancel. The pressure drop then becomes the gain in kinetic energy per unit volume.

   P₁ - P₂ = 1/2ρ(v₂² - v₁²) = 1/2ρ(4v₁² - v₁²) = 3/2ρ v₁²

3. Find the speed. 3 = 3/2 × 1250 × v₁², so v₁² = 3/1875 = 0.0016 and v₁ = 0.04 m s⁻¹.
4. So. Q = A₁ v₁ = (10 × 10⁻⁴ m²)(0.04 m s⁻¹) = 4 × 10⁻⁵ m³ s⁻¹, so x = 4.

The common slip: 2. Multiplying the wide-end speed 0.04 m s⁻¹ by the narrow-end area 5 cm² gives 2. Each speed must go with its own area.

The idea to keep. Pressure drop across a pipe of changing area? Continuity links the speeds, Bernoulli gives one speed, and the flow rate is that area times its own speed.

Answer from the NTA final answer key. Solution by PrepWiser.