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The velocity vector for a steady three - dimension flow field is described as: V = yz²i xy²j (xy-2xyz) k At point (1, 2,3) , what is the approximate value of the magnitude of the velocity?
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The velocity vector for a steady three - dimension flow field is descr...
Calculation of Magnitude of Velocity Vector


To calculate the magnitude of the velocity vector at point (1, 2, 3), we need to use the formula:


|V| = sqrt(Vx² + Vy² + Vz²)


Calculation of Vx


Substituting the values of x, y, and z in the given equation:


Vx = yz² = 2*3² = 18


Calculation of Vy


Substituting the values of x, y, and z in the given equation:


Vy = xy² = 1*2² = 4


Calculation of Vz


Substituting the values of x, y, and z in the given equation:


Vz = xy - 2xyz = 1*2 - 2*1*2*3 = -10


Calculation of Magnitude of Velocity


Substituting the values of Vx, Vy, and Vz in the formula:


|V| = sqrt(18² + 4² + (-10)²) = sqrt(416) ≈ 20.39


Explanation



  • We are given the velocity vector for a steady three-dimensional flow field.

  • The velocity vector is described as V = yz²i + xy²j + (xy-2xyz)k.

  • We are asked to find the magnitude of the velocity vector at point (1, 2, 3).

  • We use the formula |V| = sqrt(Vx² + Vy² + Vz²) to calculate the magnitude of the velocity vector.

  • We substitute the values of x, y, and z in the given equation to calculate Vx, Vy, and Vz.

  • We then substitute the values of Vx, Vy, and Vz in the formula to calculate the magnitude of the velocity vector.

  • The approximate value of the magnitude of the velocity vector at point (1, 2, 3) is 20.39.

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The velocity vector for a steady three - dimension flow field is described as: V = yz²i xy²j (xy-2xyz) k At point (1, 2,3) , what is the approximate value of the magnitude of the velocity?
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