Ahorizontal pipe of 10cm in diameter has a smooth reduction to a pipe of 5cm in diameter if pressure of water in the larger pipe is 80kpa and the pressure in the smaller pipe is 60kpa at what rate the water does flow through the smaller pipe

Now

Yes

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Physics

To determine the rate at which water flows through the smaller pipe, we can use the principle of conservation of mass, given by the equation:

𝐴₁𝑉₁ = 𝐴₂𝑉₂

Where:
- 𝐴₁ and 𝐴₂ are the cross-sectional areas of the larger and smaller pipe, respectively.
- 𝑉₁ and 𝑉₂ are the velocities of water flow through the larger and smaller pipe, respectively.

Given that the diameter of the larger pipe is 10 cm, we can calculate its radius as:

𝑟₁ = 𝐷₁/2
𝑟₁ = 10/2 = 5 cm = 0.05 m

Similarly, the diameter of the smaller pipe is 5 cm, so its radius can be calculated as:

𝑟₂ = 𝐷₂/2
𝑟₂ = 5/2 = 2.5 cm = 0.025 m

Now, let's calculate the cross-sectional areas of both pipes using the formula:

𝐴 = 𝜋𝑟²

For the larger pipe:

𝐴₁ = 𝜋(𝑟₁)²
𝐴₁ = 3.1415 × (0.05)²
𝐴₁ ≈ 0.00785 m²

And for the smaller pipe:

𝐴₂ = 𝜋(𝑟₂)²
𝐴₂ = 3.1415 × (0.025)²
𝐴₂ ≈ 0.00196 m²

Now, we can set up the equation using the given pressures:

𝑃₁𝐴₁𝑉₁ = 𝑃₂𝐴₂𝑉₂

Plugging in the values:

80𝑘𝑃𝑎 × 0.00785𝑚² × 𝑉₁ = 60𝑘𝑃𝑎 × 0.00196𝑚² × 𝑉₂

Simplifying:

0.627𝑉₁ = 0.1176𝑉₂

Finally, we can solve for the ratio of velocities:

𝑉₂/𝑉₁ = 0.627/0.1176

𝑉₂/𝑉₁ ≈ 5.33

Therefore, the water flows approximately 5.33 times faster through the smaller pipe compared to the larger pipe.