Viscosity

Example 1 (continued)

Step 1

Calculate the shear stress in the gap.

The figure below shows an enlarged view of the gap, b, with the linear velocity distribution.

The velocity distribution is V(y) = V_o (1 – y/b) where b is the gap width and V_o is the

constant velocity of the shaft. For a Newtonian fluid, τ = μ dV/dy. Thus

τ = μ ( - V_o /b ) Note: The shear stress in the fluid is constant across gap.

Recall μ = ν ρ where ν is the kinematic viscosity and ρ is the fluid density

specific gravity for lubricant = 0.91, ρ for water is 1000 kg/m³

Step 2

Calculate the shear force exerted by the lubricant on the shaft

The shear force, F_S , is just the shear stress time the area it acts on which is ((2π)(D/2)L).

F_S = μ ( - V_o /b ) (2π)(D/2)L where L is the length of the shaft

Step 3

For equilibrium the applied force must balance the shear force.

Now for equilibrium, → Σ F = 0 μ ( - V_o /b ) (2π)(D/2)L + P = 0

P = μ ( V_o /b ) π D L ,

From the data: μ = 8.0 x 10^-4 (m²/sec) (1000 kg/m³)(0.91), V_o = 3 m/sec,

b = 0.3 mm = 0.0003 m, D = 25 mm = 0.025 m, L = 0.5 m

P = 286 N (result)

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