from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa at point B. Friction between the water and the pipe walls is negligible. What is the rate of discharge at point B

1.8 m/s

Explanation:

Here is the complete question

The diameters of a water pipe gradually changes from 5 cm at point A to 15 cm at point B. Point A is 5 m lower than point B. The pressure is 700 kPa at point A and 664 kPa and point B. Friction between the water and the pipe walls is negligible.

What is the rate of discharge at point B?

Solution

Using Bernoulli's equation,

P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂² where P₁ = pressure at point A = 700 kPa, h₁ = height at point A, v₁ = speed at point A and P₂ = pressure at point B = 664 kPa, h₂ = height at point B, v₂ = speed at point B, ρ = density of water = 1000 kg/m³

P₁ - P₂ + ρgh₁ - ρgh₂ = 1/2ρv₂² - 1/2ρv₁²

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

(h₂ - h₁) = 5 m

Substituting the values of the variables, we have

700000 Pa - 664000 Pa - (1000 kg/m³ × 9.8 m/s² × 5 m) = 1/2 × 1000 kg/m³(v₂² - v₁²)

36000 Pa - 49000 Pa = 500 kg/m³(v₂² - v₁²)

- 13000 Pa = 500 kg/m³(v₂² - v₁²)

(v₂² - v₁²) = - 13000 Pa/500 kg/m³

(v₂² - v₁²) = -26 m²/s²

By mass flow rate, A₁v₁ = A₂v₂ where A₁ = cross-sectional area at point A and A₂ = cross-sectional area at point B

πd₁²v₁/4 = d₂²v₂/4 where d₁ = diameter at point A = 5 cm = 0.05 m and d₂ = diameter at point B = 15 cm = 0.15 m

d₁²v₁ = d₂²v₂

v₁  = v₂(d₂/d₁)²

= v₂(0.15/0.05)²

= v₂(3)²

= 9v₂

So, substituting v₁ = 9v₂ into the above equation, we have

(v₂² - v₁²) = -26 m²/s²

v₂² - 9v₂² = -26 m²/s²

- 8v₂² = -26 m²/s²

v₂² = -26 m²/s² ÷ (-8)

v₂² = 3.25 m²/s²

taking square root of both sides,

v₂ = √(3.25 m²/s²)

= 1.8 m/s

So, the rate of discharge at point B is 1.8 m/s

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge