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A two-dimensional flow in cylindrical polar coordinates is given by Vr = 2r sin θ cos θ Vθ = −2r sin2 θ Check whether these velocity components represent a physical possible flow field.
- a)Possible for compressible flow only
- b)Possible for incompressible flow only
- c)Not possible for any flow
- d)Possible for both compressible and incompressible flow
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A two-dimensional flow in cylindrical polar coordinates is given by V...
The continuity equation, for a steady, two-dimensional incompressible flow is
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∂ / ∂r (r Vr) + ∂ /∂θ (Vθ) = 0
From the given velocity components,
∂/∂r (r Vr) = Vr + r ∂ /∂r (Vr)
= 2 r sin θ cos θ + r ∂ / ∂r (2r sin θ cos θ )
= 2r sin θ cos θ + 2r sin θ cos θ
= 4 r sin θ cos θ
and ∂ / ∂θ (Vθ) = ∂ / ∂θ (−2r sin2 θ)
= −2r × 2 sin θ cos θ = −4r sin θ cos θ
∴∂ / ∂r (r Vr) + ∂ /∂θ (Vθ)
= 4r sin θ cos θ − 4r sin θ cos θ
= 0
The continuity equation is satisfied so the flow is incompressible.
Most Upvoted Answer
A two-dimensional flow in cylindrical polar coordinates is given by V...
Is the given velocity field physically possible?
To determine whether the given velocity components represent a physically possible flow field, we need to consider the conditions for incompressible flow. Incompressible flow is a condition where the fluid density remains constant throughout the flow.
Conditions for incompressible flow
1. Continuity equation:
The continuity equation states that the mass flow rate into a control volume must be equal to the mass flow rate out of the control volume.
Mathematically, the continuity equation is expressed as:
∇ · V = 0, where ∇ is the del operator and V is the velocity vector.
2. Incompressibility condition:
The incompressibility condition states that the divergence of the velocity vector must be equal to zero.
Mathematically, this condition is expressed as:
∇ · V = ∂Vr/∂r + (1/r) ∂Vθ/∂θ + (1/r) ∂Vφ/∂φ = 0
3. Velocity components:
The given velocity components are:
Vr = 2r sin θ cos θ
Vθ = -2r sin^2 θ
Vφ = 0
Evaluating the given velocity components
1. Continuity equation:
∇ · V = ∂Vr/∂r + (1/r) ∂Vθ/∂θ + (1/r) ∂Vφ/∂φ
= 2sin(θ)cos(θ) + (1/r) (-4sin^2(θ)) + 0
= 2sin(θ)cos(θ) - (4/r)sin^2(θ)
The continuity equation is not satisfied as the divergence of the velocity vector is not equal to zero.
2. Incompressibility condition:
∇ · V = ∂Vr/∂r + (1/r) ∂Vθ/∂θ + (1/r) ∂Vφ/∂φ
= 2sin(θ)cos(θ) + (1/r) (-4sin^2(θ)) + 0
= 2sin(θ)cos(θ) - (4/r)sin^2(θ)
The incompressibility condition is not satisfied as the divergence of the velocity vector is not equal to zero.
Conclusion
Based on the evaluation of the given velocity components, we can conclude that the velocity field does not represent a physically possible flow field. It is only possible for incompressible flow.
To determine whether the given velocity components represent a physically possible flow field, we need to consider the conditions for incompressible flow. Incompressible flow is a condition where the fluid density remains constant throughout the flow.
Conditions for incompressible flow
1. Continuity equation:
The continuity equation states that the mass flow rate into a control volume must be equal to the mass flow rate out of the control volume.
Mathematically, the continuity equation is expressed as:
∇ · V = 0, where ∇ is the del operator and V is the velocity vector.
2. Incompressibility condition:
The incompressibility condition states that the divergence of the velocity vector must be equal to zero.
Mathematically, this condition is expressed as:
∇ · V = ∂Vr/∂r + (1/r) ∂Vθ/∂θ + (1/r) ∂Vφ/∂φ = 0
3. Velocity components:
The given velocity components are:
Vr = 2r sin θ cos θ
Vθ = -2r sin^2 θ
Vφ = 0
Evaluating the given velocity components
1. Continuity equation:
∇ · V = ∂Vr/∂r + (1/r) ∂Vθ/∂θ + (1/r) ∂Vφ/∂φ
= 2sin(θ)cos(θ) + (1/r) (-4sin^2(θ)) + 0
= 2sin(θ)cos(θ) - (4/r)sin^2(θ)
The continuity equation is not satisfied as the divergence of the velocity vector is not equal to zero.
2. Incompressibility condition:
∇ · V = ∂Vr/∂r + (1/r) ∂Vθ/∂θ + (1/r) ∂Vφ/∂φ
= 2sin(θ)cos(θ) + (1/r) (-4sin^2(θ)) + 0
= 2sin(θ)cos(θ) - (4/r)sin^2(θ)
The incompressibility condition is not satisfied as the divergence of the velocity vector is not equal to zero.
Conclusion
Based on the evaluation of the given velocity components, we can conclude that the velocity field does not represent a physically possible flow field. It is only possible for incompressible flow.
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A two-dimensional flow in cylindrical polar coordinates is given by Vr = 2r sin θ cos θ Vθ = −2r sin2 θ Check whether these velocity components represent a physical possible flow field.a) Possible for compressible flow onlyb) Possible for incompressible flow onlyc) Not possible for any flowd) Possible for both compressible and incompressible flowCorrect answer is option 'B'. Can you explain this answer?
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