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In a steady, incompressible, two dimensional flow, one velocity component in the x-direction is given by u = cx2/y2. The velocity component in the y-direction will be:
- a)V = –c(x + y)
- b)v = –cx/y
- c)v = –xy
- d)v = –cy/x
Correct answer is option 'B'. Can you explain this answer?
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In a steady, incompressible, two dimensional flow, one velocity compon...
Ans. (b)
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In a steady, incompressible, two dimensional flow, one velocity compon...
Explanation:
Given: u = cx^2/y^2
To find: v, the velocity component in the y-direction
We know that the continuity equation for incompressible flow is given by ∂u/∂x + ∂v/∂y = 0
Differentiating u with respect to x, we get:
∂u/∂x = 2cx/y^2
Differentiating v with respect to y, we get:
∂v/∂y = -∂u/∂x = -2cx/y^2
Integrating the above expression with respect to y, we get:
v = -2cx/y + f(x)
where f(x) is the function of x only.
Now, to find f(x), we use the given information that the flow is two dimensional and steady, which means that the velocity components do not change with time. Therefore, the flow must be irrotational, i.e., ∂u/∂y = ∂v/∂x.
Differentiating u with respect to y, we get:
∂u/∂y = -2cx/y^3
Equating this with ∂v/∂x, we get:
-2cx/y^3 = f'(x)
Integrating the above expression with respect to x, we get:
f(x) = -cx^2/y^3 + C
where C is the constant of integration.
Substituting this value of f(x) in the expression for v, we get:
v = -2cx/y - cx^2/y^3 + C
Simplifying this expression, we get:
v = cx/y - cx^2/y^3 + C
Comparing this with the given options, we see that the correct answer is v = cx/y, which is option B.
Given: u = cx^2/y^2
To find: v, the velocity component in the y-direction
We know that the continuity equation for incompressible flow is given by ∂u/∂x + ∂v/∂y = 0
Differentiating u with respect to x, we get:
∂u/∂x = 2cx/y^2
Differentiating v with respect to y, we get:
∂v/∂y = -∂u/∂x = -2cx/y^2
Integrating the above expression with respect to y, we get:
v = -2cx/y + f(x)
where f(x) is the function of x only.
Now, to find f(x), we use the given information that the flow is two dimensional and steady, which means that the velocity components do not change with time. Therefore, the flow must be irrotational, i.e., ∂u/∂y = ∂v/∂x.
Differentiating u with respect to y, we get:
∂u/∂y = -2cx/y^3
Equating this with ∂v/∂x, we get:
-2cx/y^3 = f'(x)
Integrating the above expression with respect to x, we get:
f(x) = -cx^2/y^3 + C
where C is the constant of integration.
Substituting this value of f(x) in the expression for v, we get:
v = -2cx/y - cx^2/y^3 + C
Simplifying this expression, we get:
v = cx/y - cx^2/y^3 + C
Comparing this with the given options, we see that the correct answer is v = cx/y, which is option B.
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In a steady, incompressible, two dimensional flow, one velocity component in the x-direction is given by u = cx2/y2. The velocity component in the y-direction will be:a)V = –c(x + y)b)v = –cx/yc)v = –xyd)v = –cy/xCorrect answer is option 'B'. Can you explain this answer?
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