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koban [17]
1 year ago
11

Water flows through a horizontal pipe with a cross-sectional area of 3mat a speed of 10 m/s with a pressure of 3 Bars at a Point

A, which is 20m higher than Points B and C. At Point B, the cross-sectional area remains the same as Point A, but it has doubled at Point C. Now, calculate the pressure and speed of the water at Points B and C respectively.​
Physics
1 answer:
Over [174]1 year ago
8 0

(a) The speed of the water at point B  is 10 m/s and the speed at point C is 5 m/s.

(b) The pressure of the water at point B is 4.96 bars and the pressure at point C is 5.335 bars.

<h3>Speed of the water at point B </h3>

The speed of the water at point B is determined from continuity equation as shown below;

Q = AV

A₁V₁ = A₂V₂

Since, A₁ = A₂ =  3 m²

V₁ = V₂ = 10 m/s

Thus, the speed of the water at point B  is 10 m/s.

<h3>Speed of water at point C</h3>

A₁V₁ = A₃V₃

A₁V₁  = (2A₁)V₃

V₃ = (A₁V₁) / 2A₁

V₃ = V₁ / 2

V₃ = 10/2

V₃ = 5 m/s

Thus, the speed of the water at point C  is 5 m/s.

<h3>Pressure of the water at point B </h3>

The pressure of the water at point B is determined by appying Bernoulli's equation as shown below;

P₁ + ¹/₂ρV₁² + ρgh₁ = P₂ + ¹/₂ρV₂² + ρgh₂

where;

  • P₁ = 3 bars = 300,000 Pa
  • ρ density of water = 1000 kg/m³
  • g is acceleration due to gravity = 9.8 m/s²

300,000 + ¹/₂(1000)(10)² + (1000)(9.8)(20) = P₂ + ¹/₂(1000)(10)²

546,000 = P₂ + 50,000

P₂ = 496,000 Pa

P₂ = 4.96 bars

<h3>Pressure of the water at point C</h3>

P₁ + ¹/₂ρV₁² + ρgh₁ = P₃ + ¹/₂ρV₃² + ρgh₃

546,000 = P₃ + ¹/₂(1000)(5)²

546,000 = P₃ + 12,500

P₃ = 533,500 Pa

P₃ = 5.335 bars

Learn more about pressure in horizontal pipe here: brainly.com/question/26761275

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