Given the velocity field v= 10 X^2Y + 15 YZ - (3XY-25 T) A) FIND THE ACCELERATIO

Question

Given the velocity field v= 10 X^2Y  + 15 YZ - (3XY-25 T)

A) FIND THE ACCELERATION OF A FLUID ELEMENT AT A POINT (1,2,-1) AT TIME T=0.5.

B) IDENTIFY THE VALUE OF LOCAL AND CONVECTIVE COMPONENTS OF ACCELERATION IN 1) X DIRECTION , 2) Y DIRECTION 3) Z DIRECTION .

Explanation / Answer

a). V= 10 X^2Y + 15 YZ - (3XY-25 T) u = 10*X^2*Y v = 15 YZ w = -3 XY + 25 T u(1,2,-1) = 10*(1)^2*2 = 20 v(1,2,-1) = 15*2*(-1) = -30 w(1,2,-1) = -3*1*2 + 25*0.5 = 6.5 a = u*(dV/dx) i + v*(dV/dy) j + w*(dV/dz) k + dV/dt => a = 20*(20*XY) i + (-30)*(15*Z) j + 6.5*(0) k + (25) => a = 20*(20*1*2) i - 30 *(15*-1) j + 0 k + 25 => a = 800 i + 450 j + 0 k + 25 b). convective acceleration in X-direction = u*(dV/dx) = 800 sq. units Y-direction = v*(dV/dy) = 450 sq. units Z- direction = w*(dV/dz) = 0 Total convective acceleration = sqrt[800^2 + 450^2 + 0^2] = 917.88 sq. units local acceleration = dV/dt = 25 sq. units

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