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For the following velocity field, calculate the value of b and c such that it is a non-rotational flow: =(1.35+2.78x+0.754y+4.21z)+(3.45+cx−2.78y+z)+(−4.21x−1.89y)

2 Answers

7 votes

Step-by-step explanation:

To determine the values of b and c such that the velocity field represents a non-rotational flow, we need to ensure that the curl of the velocity field is zero. The curl of a vector field can be calculated using the cross product of the del operator (∇) with the vector field.

Given the velocity field:

V = (1.35 + 2.78x + 0.754y + 4.21z) + (3.45 + cx - 2.78y + z) + (-4.21x - 1.89y)

Let's calculate the curl of V:

∇ × V = (∂/∂y)(-4.21x - 1.89y) - (∂/∂z)(3.45 + cx - 2.78y + z)

+ (∂/∂z)(1.35 + 2.78x + 0.754y + 4.21z) - (∂/∂x)(-4.21x - 1.89y)

+ (∂/∂x)(3.45 + cx - 2.78y + z) - (∂/∂y)(1.35 + 2.78x + 0.754y + 4.21z)

Simplifying the above expression, we have:

∇ × V = (-∂(4.21x)/∂y - ∂(1.89y)/∂y) - (∂(3.45)/∂z + ∂(cx)/∂z - ∂(2.78y)/∂z + ∂(z)/∂z)

+ (∂(1.35)/∂z + ∂(2.78x)/∂z + ∂(0.754y)/∂z + ∂(4.21z)/∂z) - (-∂(4.21x)/∂x - ∂(1.89y)/∂x)

+ (∂(3.45)/∂x + ∂(cx)/∂x - ∂(2.78y)/∂x + ∂(z)/∂x) - (∂(1.35)/∂y - ∂(2.78x)/∂y - ∂(0.754y)/∂y - ∂(4.21z)/∂y)

Now, we can compute each partial derivative term:

∂(4.21x)/∂y = 0

∂(1.89y)/∂y = 1.89

∂(3.45)/∂z = 0

∂(cx)/∂z = 0

∂(2.78y)/∂z = 0

∂(z)/∂z = 1

∂(1.35)/∂z = 0

∂(2.78x)/∂z = 0

∂(0.754y)/∂z = 0.754

∂(4.21z)/∂z = 4.21

∂(4.21x)/∂x = 4.21

∂(1.89y)/∂x = 0

∂(3.45)/∂x = 0

∂(cx)/∂x = c

∂(2.78y)/∂x = 2.78

∂(z)/∂x = 0

User Graeme Wicksted
by
8.4k points
3 votes

Answer and Explanation:

To determine the values of b and c that make the velocity field a non-rotational flow, we need to find the conditions under which the curl of the velocity field is equal to zero.

The curl of a vector field can be calculated using the following formula:

curl(F) = (∂F₃/∂y - ∂F₂/∂z)i + (∂F₁/∂z - ∂F₃/∂x)j + (∂F₂/∂x - ∂F₁/∂y)k

In this case, we have the velocity field F = (1.35 + 2.78x + 0.754y + 4.21z) + (3.45 + cx - 2.78y + z) + (-4.21x - 1.89y).

Let's calculate the curl of this velocity field:

∂F₁/∂x = 2.78

∂F₁/∂y = 0.754 - 2.78c

∂F₁/∂z = 4.21

∂F₂/∂x = -4.21

∂F₂/∂y = -2.78 - 1.89c

∂F₂/∂z = 1

∂F₃/∂x = 0

∂F₃/∂y = -3.45

∂F₃/∂z = 1

Using these partial derivatives, we can calculate the curl of the velocity field:

curl(F) = (0 - (1))i + ((-4.21) - (-3.45))j + ((-2.78 - 1.89c) - 0)k

= -i - 0.76j - (2.78 + 1.89c)k

For the velocity field to be non-rotational, the curl must be equal to zero. Thus, we have the following conditions:

-1 = 0 (from the i-component)

-0.76 = 0 (from the j-component)

2.78 + 1.89c = 0 (from the k-component)

From these conditions, we can see that there is no value of c that satisfies the equation 2.78 + 1.89c = 0. Therefore, there are no values of b and c that make the velocity field a non-rotational flow.

User Narnie
by
8.1k points

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