water flow dowen a sloping pipe.the pipe tapers from 2m diameter at the top (A)to 1.4 diameter at the bottom (B)the volume flow rate is 23m/s.

1)calculate the velocities of the flow at points A and B.
2)if the end A is 12m higher than end B, and the presure A is 200kpa what is the presure at point B?

To calculate the velocities of the flow at points A and B, you can use the principle of continuity, which states that the volume flow rate remains constant along a streamline in a steady flow situation.

1) Velocities of the flow at points A and B:
The volume flow rate (Q) is given as 23 m³/s. We can calculate the velocities at points A and B using the formula:

Q = A * V

where Q is the volume flow rate, A is the cross-sectional area, and V is the velocity.

At point A, the diameter is 2 m. So the radius (r) at point A is 1 m (2 m / 2).

Area at point A (A₁) = π * r₁² = π * 1² = π m²

Using the formula for the volume flow rate, we can rearrange it to solve for the velocity at point A:

V₁ = Q / A₁ = 23 / π ≈ 7.32 m/s

At point B, the diameter is 1.4 m. So the radius (r) at point B is 0.7 m (1.4 m / 2).

Area at point B (A₂) = π * r₂² = π * 0.7² ≈ 1.54 π m²

Similarly, we can find the velocity at point B:

V₂ = Q / A₂ = 23 / (1.54π) ≈ 4.73 m/s

Therefore, the velocity at point A is approximately 7.32 m/s, and the velocity at point B is approximately 4.73 m/s.

2) Pressure at point B:
To calculate the pressure at point B, we can use Bernoulli's equation, which states that the total energy of a fluid along a streamline is constant.

P₁ + 0.5ρV₁² + ρgh₁ = P₂ + 0.5ρV₂² + ρgh₂

where P is the pressure, ρ is the density of water, V is the velocity, g is the acceleration due to gravity, and h is the height difference.

Given:
Pressure at point A (P₁) = 200 kPa
Height difference (h) = 12 m
Density of water (ρ) = 1000 kg/m³ (approximate value)

Using the formula, we can rearrange it to solve for the pressure at point B (P₂):

P₂ = P₁ + 0.5ρ(V₁² - V₂²) + ρg(h₁ - h₂)

Substituting the known values:

P₂ = 200 + 0.5 * 1000 * (7.32² - 4.73²) + 1000 * 9.81 * (12 - 0)

P₂ ≈ 200 + 0.5 * 1000 * (53.53 - 22.42) + 1000 * 9.81 * 12

P₂ ≈ 200 + 11555 + 118296 ≈ 130051 kPa

Therefore, the pressure at point B is approximately 130051 kPa.