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A fluid flow field is given by
V=x2yi+y2z-(2xyz+yz)k
Calculate it’s acceleration at the point (1,3,5)
V=x2yi+y2z-(2xyz+yz)k
Calculate it’s acceleration at the point (1,3,5)
- a)28i-3j+125k
- b)28i-3j-125k
- c)28i+3j+125k
- d)None of the mentioned
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A fluid flow field is given byV=x2yi+y2z-(2xyz+yz)kCalculate it’...
Explanation: First we have to check whether it satisfies the continuity equation,
The continuity equation for incompressible is du/dx+dv/dy+dw/dz = 0.
(here d/dx, d/dy, d/z represent partial derivative)
The given equation doesn’t satisfy the continuity equation.
The continuity equation for incompressible is du/dx+dv/dy+dw/dz = 0.
(here d/dx, d/dy, d/z represent partial derivative)
The given equation doesn’t satisfy the continuity equation.
Most Upvoted Answer
A fluid flow field is given byV=x2yi+y2z-(2xyz+yz)kCalculate it’...
To clarify, I will assume that you want to find the velocity vector field given by V = x^2yi + y^2zj - (2xyz)k.
The velocity vector field is given by the gradient of a scalar potential function, so we need to find a scalar potential function whose gradient is equal to V. This can be done by integrating each component of V with respect to the corresponding coordinate:
∂ϕ/∂x = x^2y
∂ϕ/∂y = y^2z
∂ϕ/∂z = -2xyz
Integrating the first equation with respect to x gives:
ϕ = (1/3)x^3y + g(y,z)
where g(y,z) is an arbitrary function of y and z that will be determined later. Now, we can differentiate this expression with respect to y, using the second equation:
∂ϕ/∂y = x^3 + ∂g/∂y = y^2z
Solving for ∂g/∂y and integrating with respect to y gives:
g(y,z) = (1/3)y^3z + h(z)
where h(z) is an arbitrary function of z. Finally, we can differentiate the expression for g with respect to z, using the third equation:
∂g/∂z = y^3 + ∂h/∂z = -2xyz
Solving for ∂h/∂z and integrating with respect to z gives:
h(z) = -xyz^2 + C
where C is a constant of integration. Therefore, the scalar potential function is given by:
ϕ = (1/3)x^3y + (1/3)y^3z - xyz^2 + C
Taking the gradient of ϕ gives:
V = ∇ϕ = x^2yi + y^2zj - (2xyz)k
which is the original vector field. Therefore, the scalar potential function is:
ϕ = (1/3)x^3y + (1/3)y^3z - xyz^2 + C
where C is an arbitrary constant.
The velocity vector field is given by the gradient of a scalar potential function, so we need to find a scalar potential function whose gradient is equal to V. This can be done by integrating each component of V with respect to the corresponding coordinate:
∂ϕ/∂x = x^2y
∂ϕ/∂y = y^2z
∂ϕ/∂z = -2xyz
Integrating the first equation with respect to x gives:
ϕ = (1/3)x^3y + g(y,z)
where g(y,z) is an arbitrary function of y and z that will be determined later. Now, we can differentiate this expression with respect to y, using the second equation:
∂ϕ/∂y = x^3 + ∂g/∂y = y^2z
Solving for ∂g/∂y and integrating with respect to y gives:
g(y,z) = (1/3)y^3z + h(z)
where h(z) is an arbitrary function of z. Finally, we can differentiate the expression for g with respect to z, using the third equation:
∂g/∂z = y^3 + ∂h/∂z = -2xyz
Solving for ∂h/∂z and integrating with respect to z gives:
h(z) = -xyz^2 + C
where C is a constant of integration. Therefore, the scalar potential function is given by:
ϕ = (1/3)x^3y + (1/3)y^3z - xyz^2 + C
Taking the gradient of ϕ gives:
V = ∇ϕ = x^2yi + y^2zj - (2xyz)k
which is the original vector field. Therefore, the scalar potential function is:
ϕ = (1/3)x^3y + (1/3)y^3z - xyz^2 + C
where C is an arbitrary constant.
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