A plate 0.025 mm distant from a fixed plate, moves at 50 cm/s and requires a force of 1.471 N/m² to maintain this speed. Determine the fluid viscosity between the plates in poise.

Fluid Viscosity Calculation

Problem Statement

A plate 0.025 mm distant from a fixed plate, moves at 50 cm/s and requires a force of 1.471 N/m² to maintain this speed. Determine the fluid viscosity between the plates in poise.

Given Data

  • Distance between plates, \(dy = 0.025 \, \text{mm}\)
  • Velocity of moving plate, \(du = 50 \, \text{cm/s}\)
  • Shear Stress (Force per unit area), \(\tau = 1.471 \, \text{N/m}^2\)

Solution

1. Convert All Units to SI

For consistency in the formula, we must convert all measurements to standard SI units (meters and seconds).

$$ dy = 0.025 \, \text{mm} = 0.025 \times 10^{-3} \, \text{m} $$ $$ du = 50 \, \text{cm/s} = 0.50 \, \text{m/s} $$

2. Apply Newton's Law of Viscosity

Newton's Law of Viscosity states that the shear stress (\(\tau\)) between adjacent fluid layers is proportional to the velocity gradient (\(\frac{du}{dy}\)). The constant of proportionality is the dynamic viscosity (\(\mu\)).

$$ \tau = \mu \frac{du}{dy} $$

We can rearrange this formula to solve for the viscosity, \(\mu\):

$$ \mu = \tau \frac{dy}{du} $$

3. Calculate Viscosity in SI Units

Substitute the converted values into the rearranged formula.

$$ \mu = 1.471 \, \text{N/m}^2 \times \frac{0.025 \times 10^{-3} \, \text{m}}{0.50 \, \text{m/s}} $$ $$ \mu = 1.471 \times (0.05 \times 10^{-3}) \, \text{N·s/m}^2 $$ $$ \mu = 0.00007355 \, \text{N·s/m}^2 $$

4. Convert Viscosity to Poise

The final result is required in poise. The conversion factor is 10 poise = 1 N·s/m².

$$ \mu_{\text{poise}} = \mu_{\text{N·s/m}^2} \times 10 $$ $$ \mu = 0.00007355 \times 10 = 0.0007355 \, \text{poise} $$
Final Result:

The fluid viscosity between the plates is \( \mu = 0.0007355 \, \text{poise} \).

Explanation of Viscosity

Viscosity (\(\mu\)) is a measure of a fluid's resistance to flow. It describes the internal friction of a moving fluid. A fluid with high viscosity (like honey) resists motion because its molecular makeup creates a lot of internal friction, while a low-viscosity fluid (like water) flows easily.

Shear Stress (\(\tau\)) is the force per unit area required to move one layer of fluid relative to another. In this problem, it's the force needed to keep the plate moving at a constant speed.

Physical Meaning

The result, a very low viscosity of 0.0007355 poise, indicates that the fluid between the plates offers very little resistance to being sheared. For context, the viscosity of water at 20°C is about 0.01 poise, and the viscosity of air is about 0.00018 poise. This fluid is much less viscous than water, behaving more like a gas.

This principle is fundamental to lubrication. A thin film of a low-viscosity fluid (a lubricant) between two surfaces can dramatically reduce the force required to move them relative to each other, minimizing friction and wear.

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