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A viscous liquid is sheared between two parallel disks; the upper disk rotates and the lower one is fixed. The velocity field between the disks is given by \(\vec{V}=\hat{e}_{0} r \omega z / h\). (The origin of coordinates is located at the center of the lower disk; the upper disk is located at \(z=h .\) ) What are the dimensions of this velocity field? Does this velocity field satisfy appropriate physical boundary conditions? What are they?

Short Answer

Expert verified
The dimensions of the given velocity field are [L^2T^-1], the same as a velocity. The velocity field satisfies physical boundary conditions, as at the boundary with the stationary disk (z=0), the fluid is at rest, and at the boundary with the moving disk (z=h), the fluid moves with the disk, precisely as outlined by the physical boundary conditions for shearing flow.

Step by step solution

01

Understand the Given Velocity Field and its Dimensions

The velocity field is given as \(\vec{V}=\hat{e}_{0} r \omega z / h\). The distint elements are \(\hat{e}_{0}\) is the unit vector in direction of increasing r, \( r \) the radial distance from the center, \(\omega \) the angular velocity, \( z \) the vertical position and \( h \) the height between the two disks. The dimensions of each are-\( r, z, h \) are lengths [L], \(\omega \) is [T^-1]. Thus combined the dimensions of velocity field \(\vec{V}\)are [L^2T^-1], the same as any velocity.
02

Check the Physical Boundary Conditions

The physical boundary conditions state that at the interface between a stationary and a moving surface, the velocity of the fluid should match that of the surface. If we apply this to our problem, at \( z=0 \), the lower disk is stationary, so the fluid velocity should be 0, which is true as per our velocity field. At \( z=h \), the fluid should move with the upper disk which is rotating at \( \omega \). Therefore, at \( z=h \), \( \vec{V} = \hat{e}_{0} r \omega \), which is also satisfied. Hence, the given velocity field satisfies the appropriate physical boundary conditions.

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