Question

Water flows through a vertical tube of variable cross section .The area of cross section at a and b are 6 and 3 mm*2 respectively .Of 12 cc of water enters per second through a find the pressure difference [p- p'] ( g= 10 m/s*2 ) the separation between the cross section at a and b is 100 cm

Answer 1

Given,

Volume rate of the flow of water = 1 cc/sec.

= 1 cm³/sec.

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Since, Volume rate of the flow of water is constant throughout the Horizontal pipe, therefore,

(a). At A,

Area × Velocity = 1 cm³/s.

4/100 cm² × Velocity = 1

∴ Velocity = 25 cm/seconds.

(b). At B,  

Area × Velocity = 1 cm³/s.

2/100 × Velocity = 1

Velocity = 50 cm/seconds.

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(c).

Using the Bernoulli's Equation,

P₁ + h₁ρg + 1/2 ρv₁² = P₂ + h₂ρg + 1/2 ρv₂²

∴ P₁ - P₂ = 1/2 ρ(v₂² - v₁²) + ρg(h₂ - h₁)

∴ P₁ - P₂ = 1/2 ρ(v₂² - v₁²)   [ ∵ In case of the Horizontal tube, Height difference is zero.]

∴ P₁ - P₂ = 1/2 × 1 × (50² - 25²)

∴ P₁ - P₂ = 937.5 dyne/cm²

∴ P₁ - P₂ = 93.75 N/m² or 93.75 Pa.

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Hope it helps.