A possible design for a parallel plate viscometer consists of a vertical, rectan

Question

A possible design for a parallel plate viscometer consists of a vertical, rectangular box with a centrally located plate inside. The fluid to be tested is placed in the box and the force necessary to remove the plate at a fixed speed is measured. The unit is shown in Fig. 2.16. Calculate the viscosity in cP for the following conditions: the weight of the plate is negligible; the plate is located equidistant between the walls; clearance between plate and each wall is 0.5 cm; total area of plate immersed at instant of reading is 70 cm2 on a side; when the plate is moved at a velocity of 1cm s-1, the force required is 5.6 times 10-4N; end effects are negligible. FIGURE 2.16 Plate viscometer (Problem 2.20).

Explanation / Answer

Since the weight is negligible there will be no gravity effect.

Let the shear stress and force on the left side be 1 and F1 and on the right side be 2 and F2

Since the plate is centrally located and thickness of films on both sides is constant. and Area of plate in same on both sides

1 = 2

and F1 = F2

Total shear force, F' on the plate will be the sum of the two side forces

F' = F1 + F2

F' = 2 F1

This force F' will act downwards and try to pull the plate downwards.

For the plate to move with constant velocity V,

the external force F must be equal to total shear force F'

F' = 5.6 x 10^-4 N

F1 = 2.8 x 10^-4 N

We know that, shear force F1 = 1 x Area = A (u / y)

Here u = 1 cm /s = .01 m/s

y = .5 cm = .005m

Area , A = 70 cm^2 = .007 m^2

= .02 Pa s = 20cP

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