Problem Statement
Two plates are placed at a distance of 0.15 mm apart. The lower plate is fixed while the upper plate having surface area 1.0 m² is pulled at 0.3 m/s. Find the force and power required to maintain this speed, if the fluid separating them is having viscosity 1.5 poise.
Given Data
- Distance between plates, \(dy = 0.15 \, \text{mm}\)
- Surface Area of upper plate, \(A = 1.0 \, \text{m}^2\)
- Velocity of upper plate, \(du = 0.3 \, \text{m/s}\)
- Viscosity of fluid, \(\mu = 1.5 \, \text{poise}\)
Solution
1. Convert All Units to SI
We need to work with a consistent set of units (metres, seconds, Newtons).
Distance:
Viscosity: (10 poise = 1 N·s/m²)
2. Calculate the Shear Stress (\(\tau\))
For a thin film of fluid between parallel plates, we can assume a linear velocity profile. The shear stress is given by Newton's law of viscosity.
3. Calculate the Force Required (\(F\))
The force required to pull the plate is equal to the viscous drag force, which is the shear stress multiplied by the area of the plate.
4. Calculate the Power Required (\(P\))
Power is the rate at which work is done, calculated as the force multiplied by the velocity at which the force is applied.
Force Required: \( F = 300 \, \text{N} \)
Power Required: \( P = 90 \, \text{W} \)
Explanation of the Physics (Couette Flow)
This problem is a classic example of Couette flow, which describes the flow of a viscous fluid in the space between two parallel plates, one of which is moving relative to the other.
1. No-Slip Condition:
The fluid in direct contact with the lower, fixed plate is stationary (\(u=0\)). The fluid in contact with the upper plate moves at the same velocity as the plate (\(u=0.3\) m/s). This is known as the no-slip condition.
2. Linear Velocity Profile:
Because the gap between the plates is very small, we assume the velocity of the fluid increases linearly from zero at the bottom plate to its maximum value at the top plate. This creates a constant velocity gradient (\(du/dy\)) across the entire fluid film.
3. Viscous Drag:
The fluid's viscosity resists this shearing motion, creating a drag force that opposes the movement of the top plate. The force we calculated is the force needed to overcome this viscous drag and maintain a constant speed.
Physical Meaning
The calculated force (300 N) is the continuous effort required to slide the upper plate over the lower one due to the "stickiness" (viscosity) of the fluid in between. The power (90 W) represents the rate of energy that is continuously converted into heat within the fluid due to this internal friction.
This principle is fundamental to understanding friction in lubricated systems. For example, in a clutch or a hydrostatic bearing, the viscosity of the fluid, the gap size, and the relative speed of the components determine the forces and energy losses involved. Minimizing this power loss is a key goal in designing efficient machinery.
