Question

Water flows through a horizontal tube as shown in figure (13-E9). If the difference of heights of water column in the vertical tubes is 2cm, and the areas of cross-section at A and B are 4 cm² and 2 cm² respectively, find the rate of flow of water across any section.

Answer 1

The height difference =2 cm =0.02 m 

The difference of pressure between A and B =Pₐ-Pᵦ =ρgh 

=1000*10*0.02 =200 N/m² 

Let the rate of flow of water be Q cc/s i.e. =Q*10⁻⁶ m³/s 

Area of the cross-section at A =4 cm² =4/10000 m² 

The speed at A = Vₐ =Q*10⁻⁶/(4/10000) m/s =Q/400 m/s 

Area of the cross-section at B =2 cm² =2/10000 m² 

The speed at B = Vᵦ =Q*10⁻⁶/(2/10000) m/s =Q/200 m/s 

Since the height of both the points are the same, from Bernoulli's theorem, 

Pₐ+½ρVₐ² = Pᵦ +½ρVᵦ² 

→Pₐ-Pᵦ = ½ρ(Vᵦ²-Vₐ²) =½*1000*Q²{(1/200)²-(1/400)²} 

→Pₐ-Pᵦ =(1/20)*Q²{1/4 - 1/16} =3Q²/320 

→200 = 3Q²/320 

→Q² = 200*320/3 =21333 

→Q = 146 cc/s

Answer 2

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