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
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