For a velocity field with the equation V = 3z2i + 6 x2J + 2 y2ft Where V is in m

Question

For a velocity field with the equation V = 3z2i + 6 x2J + 2 y2ft Where V is in m/s and x, y and I are in ft.Find the vector equation for the acceleration field.What is the magnitude of the acceleration at a point (3.2,1)-.For a velocity field with the equation V = xt1! + x(x - 1 )(y + 1)/ + xytk where V is in ft/s and x and y are in ft, and t is in seconds,.Determine the vector equation for the acceleration field.What is the magnitude of the acceleration at a point (2.1). at time t=2 seconds.A flow has velocity field is defined by the following equation:The pressure field for this flow is given by the equation:where V0 is in ft/s, r is In slugs/ft, and x, y, and / are in ftWhat is the material derivative for the pressure field?

Explanation / Answer

1)

V = 3z^2 i + 6x^2 j + 2y^2 k

a)

We have u = 3z^2

v = 6x^2

w = 2y^2

Acceleration vector a = dV/dt = (del V / del t) + u* (del V / del x) + v* (del V / del y) + w* (del V / del z)

We have del V / del t = 0

del V / del x = 6*2x j = 12x j

del V / del y = 2*2y k = 4y k

del V / del z = 3*2z i = 6z i

a = (3z^2)(12x j) + (6x^2)(4y k) + (2y^2)(6z i)

a = 12y^2 *z i + 36xz^2 j + 24x^2 y k

b)

At (3,2,1) we

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