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Fluids of viscosities \(\mu_{1}=0.1 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\) and \(\mu_{2}=0.15 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\) are contained between two plates (each plate is \(1 \mathrm{m}^{2}\) in area). The thicknesses are \(h_{1}=0.5 \mathrm{mm}\) and \(h_{2}=0.3 \mathrm{mm},\) respectively. Find the force \(F\) to make the upper plate move at a speed of \(1 \mathrm{m} / \mathrm{s}\). What is the fluid velocity at the interface between the two fluids?

Short Answer

Expert verified
The force required to move the upper plate at a speed of \(1 \mathrm{m} / \mathrm{s}\) is 700 N. The fluid velocity at the interface between the two fluids is 0.5 m/s.

Step by step solution

01

Compute the force due to each fluid

Using the formula F=\(\mu A (du/dh) \), force on the upper plate due to fluid 1, \(F_{1} = \mu_{1}A(du/dh_{1}) \), and due to fluid 2, \(F_{2} = \mu_{2}A(du/dh_{2})\). Given that \(\mu_{1}=0.1 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}, h_{1}=0.5 \mathrm{mm}, \mu_{2}=0.15 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}, h_{2}=0.3 \mathrm{mm}\), and the area A = 1 \( \mathrm{m}^{2}\), and \( du/dh\) represents the change in speed with respect to thickness of the fluid, which becomes \(1 \mathrm{m} / \mathrm{s}\) for each fluid thickness.
02

Calculate the forces

Substitute the known values into the force equations, \(F_{1} = (0.1 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2})(1 \mathrm{m}^{2})(1 \mathrm{m} / \mathrm{s}/0.5 \mathrm{mm}) = 200 N\), and \(F_{2} = (0.15 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2})(1 \mathrm{m}^{2})(1 \mathrm{m} / \mathrm{s}/0.3 \mathrm{mm}) = 500 N\). Note that the thickness must be converted from mm to meter.
03

Compute total force

The total force F required to move the upper plate at a speed of \(1 \mathrm{m} / \mathrm{s}\) is the sum of the forces due to each fluid, i.e., \(F = F_{1} + F_{2} = 200 N + 500 N = 700 N\).
04

Determine fluid velocity at the interface

The velocity at the interface \(v_{i}\) can be determined from the velocity profile of each of the fluids. For a Newtonian fluid such as in this problem, the velocity profile is linear. Thus, \(v_{i}\) for fluid 1 can be expressed as \((du/dh_{1})h_{1}\), and for fluid 2 as \((du/dh_{2})h_{2}\). Substituting the known values, we find: \(v_{i} = (1 \mathrm{m} / \mathrm{s}/0.5 \mathrm{mm})0.5 \mathrm{mm} = 0.5 m/s\).

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