Question

Water flows in a horizontal tube (see figure). The pressure of water changes by 700 Nm² between A and B where the area of cross section are 40 cm² and 20 cm², respectively. Find the rate of flow of water through the tube.

(density of water = 1000 kgm⁻³)

पर्याय

• 3020 cm³/s

• 2720 cm³/s

• 2420 cm³/s

• 1810 cm³/s

Answer 1

2720 cm³/s

Explanation:

Area at A (A₁) = 40 cm² = 4 × 10⁻³ m²

Area at B (A₂) = 20 cm² = 2 × 10⁻³ m²

Pressure difference (Δ P) = P_A − P_B = 700 N/m²

Density of water (ρ) = 1000 kg/m³

Continuity equation the flow rate is constant:

A₁V₁ = A₂V₂

`V_2 = (A_1)/(A_2)V_1` 

V₂ = 2V₁

Now, using Bernoulli’s equation

`P_A + 1/2 rhoV_1^2 = P_B + 1/2 rhoV_2^2`

`P_A - P_B =  1/2 rhoV_1^2 - 1/2 rhoV_2^2`

Δ P = `1/2rho(V_1^2 - 2V_1^2)`

700 = `1/2 xx 1000 ((2V_1)^2 - V_1^2)`

700 = `500(4V_1^2 - V_1^2)`

700 = `500(3V_1^2)`

`700/500 = 3V_1^2`

`V_1^2 = 700/(500 xx 3)`

`V_1^2 = 700/1500`

`V_1^2 = 7/15`

`V_1 = sqrt(7/15)`

= 0.68 m/s

Q = A₁V₁

= 4 × 10⁻³ × 0.68

= 2.7 × 10⁻³ m³/s

= 2700 cm³/s ≈ 2720 cm³/s