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The velocity potential function for a two dimensional flow field is given by ϕ = x2 − y2. The magnitude of velocity at the point (1,1) is
- a)2
- b)4
- c)21/2
- d)2 × 21/2
Correct answer is option 'D'. Can you explain this answer?
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The velocity potential function for a two dimensional flow field is g...
Φ = x2 − y2
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we know u = −∂ϕ / ∂x and v = −∂ϕ / ∂y
∴ u = −(2x), v = −(−2y)
⇒ u = −2x, v = 2y
Thus at (1, 1),⃗V⃗ = −2î + 2ĵ |⃗V⃗ | = √4 + 4 = 2 × 21/2
Most Upvoted Answer
The velocity potential function for a two dimensional flow field is g...
Solution:
Given, ϕ = x^2 − y^2
We know that,
Velocity Potential Function, ϕ = ∂ψ/∂y , where ψ is the stream function.
∴ ∂ψ/∂y = x^2 − y^2
Integrating w.r.t y, we get
ψ = x^2y − (y^3)/3 + f(x) , where f(x) is a function of x only.
Now, Velocity Components,
u = ∂ψ/∂y = x^2 - y^2
v = -∂ψ/∂x = -f'(x)
where f'(x) is the derivative of f(x).
Given, the point (1,1).
Therefore,
u = 1^2 - 1^2 = 0
v = -f'(1)
Now, we need to find the magnitude of velocity at (1,1).
Magnitude of Velocity, V = sqrt(u^2 + v^2)
V = sqrt(0^2 + (-f'(1))^2)
V = f'(1)
Now, we need to find f(x) to find v.
Differentiating ψ w.r.t x, we get
∂ψ/∂x = f'(x)
∴ f'(x) = ∂ψ/∂x = 0
Integrating w.r.t x, we get
f(x) = constant
Therefore, v = -f'(1) = 0
Magnitude of Velocity, V = f'(1)
Now, differentiating ψ w.r.t y, we get
∂ψ/∂y = x^2 − y^2
∴ x^2 − y^2 = 2xy * dy/dx
At (1,1),
x^2 - y^2 = 0
∴ 2xy * dy/dx = 0
Therefore, dy/dx = 0 (as xy ≠ 0)
Now,
V = f'(1) = √(u^2 + v^2)
V = √(0^2 + 0^2) = 0
Therefore, the magnitude of velocity at the point (1,1) is 0.
Hence, the given answer is incorrect.
Given, ϕ = x^2 − y^2
We know that,
Velocity Potential Function, ϕ = ∂ψ/∂y , where ψ is the stream function.
∴ ∂ψ/∂y = x^2 − y^2
Integrating w.r.t y, we get
ψ = x^2y − (y^3)/3 + f(x) , where f(x) is a function of x only.
Now, Velocity Components,
u = ∂ψ/∂y = x^2 - y^2
v = -∂ψ/∂x = -f'(x)
where f'(x) is the derivative of f(x).
Given, the point (1,1).
Therefore,
u = 1^2 - 1^2 = 0
v = -f'(1)
Now, we need to find the magnitude of velocity at (1,1).
Magnitude of Velocity, V = sqrt(u^2 + v^2)
V = sqrt(0^2 + (-f'(1))^2)
V = f'(1)
Now, we need to find f(x) to find v.
Differentiating ψ w.r.t x, we get
∂ψ/∂x = f'(x)
∴ f'(x) = ∂ψ/∂x = 0
Integrating w.r.t x, we get
f(x) = constant
Therefore, v = -f'(1) = 0
Magnitude of Velocity, V = f'(1)
Now, differentiating ψ w.r.t y, we get
∂ψ/∂y = x^2 − y^2
∴ x^2 − y^2 = 2xy * dy/dx
At (1,1),
x^2 - y^2 = 0
∴ 2xy * dy/dx = 0
Therefore, dy/dx = 0 (as xy ≠ 0)
Now,
V = f'(1) = √(u^2 + v^2)
V = √(0^2 + 0^2) = 0
Therefore, the magnitude of velocity at the point (1,1) is 0.
Hence, the given answer is incorrect.
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The velocity potential function for a two dimensional flow field is given by ϕ = x2 − y2. The magnitude of velocity at the point (1,1) isa) 2b) 4c) 21/2d) 2 × 21/2Correct answer is option 'D'. Can you explain this answer?
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