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A ventilation system has intake round opening with 20 cm diameter and square out opening with side of 25.06 cm. If the air exits the system with 100 cm/a what is the speed at the intake?

Sagot :

oladayoademosu

The speed at the intake is 199.89 cm/s

Since the volume flow rate, Q = Av is constant where

  • A = cross-sectional area and
  • v = flow speed,

We have that from the continuity equation,

Continuity equation

A₁v₁ = A₂v₂ where

  • A₁ = cross-sectional area of intake = πd²/4 where
  • d = diameter of intake = 20 cm,
  • v₁ = speed of intake,
  • A₂ = cross-sectional area of outlet = L² where
  • L = length of side of outlet = 25.06 cm and
  • v₂ = speed of outlet = 100 cm/s

So, A₁v₁ = A₂v₂

πd²v₁/4 = L²v₂

Speed at the intake

Making v₁ subject of the formula, we have

v₁ = 4L²v₂/πd²

So, substituting the values of the variables into the equation, we have

v₁ = 4L²v₂/πd²

v₁ = 4(25.06 cm)² × 100 cm/s/[π(20 cm)²]

v₁ = 4 × 628.0036 cm�� × 100 cm/s/[400π cm²]

v₁ = 2512.0144 cm² × 100 cm/s/1256.6371 cm²

v₁ = 1.9989 × 100 cm/s

v₁ = 199.89 cm/s

So, the speed at the intake is 199.89 cm/s

Learn more about flow speed here:

https://brainly.com/question/10822213

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