- Explain why the viscosity of a liquid
decreases while that of a gas increases with a temperature rise.
The following is a table of measurement for a fluid at constant
temperature.
Determine the dynamic viscosity of the fluid.
| du/dy
(s-1) |
0.0 |
0.2 |
0.4 |
0.6 |
0.8 |
| t (N m-2) |
0.0 |
1.0 |
1.9 |
3.1 |
4.0 |
Using Newton's law of viscocity
where m is the viscosity. So viscosity is
the gradient of a graph of shear stress against vellocity gradient of
the above data, or
Plot the data as a graph:
Calculate the gradient for each section of the line
| du/dy (s-1) |
0.0 |
0.2 |
0.4 |
0.6 |
0.8 |
| t (N m-2) |
0.0 |
1.0 |
1.9 |
3.1 |
4.0 |
| Gradient |
- |
5.0 |
4.75 |
5.17 |
5.0 |
Thus the mean gradient = viscosity = 4.98 N s / m2
- The density of an oil is 850 kg/m3.
Find its relative density and Kinematic viscosity if the dynamic
viscosity is 5 ´ 10-3 kg/ms.
r oil = 850 kg/m3 r water = 1000 kg/m3
g oil = 850 / 1000 = 0.85
Dynamic viscosity = m = 5´ 10-3
kg/ms
Kinematic viscosity = n = m / r

- The velocity distribution of a viscous
liquid (dynamic viscosity m = 0.9 Ns/m2) flowing over a
fixed plate is given by u = 0.68y - y2 (u is velocity in
m/s and y is the distance from the plate in m).
What are the shear stresses at the plate surface and at y=0.34m?

At the plate face y = 0m,

Calculate the shear stress at the plate
face

At y = 0.34m,

As the velocity gradient is zero at
y=0.34 then the shear stress must also be zero.
- 5.6m3 of oil weighs 46 800 N.
Find its mass density, r, and relative density, g.
Weight 46 800 = mg
Mass m = 46 800 / 9.81 = 4770.6 kg
Mass density r = Mass / volume = 4770.6
/ 5.6 = 852 kg/m3
Relative density 
- From table of fluid properties the
viscosity of water is given as 0.01008 poises.
What is this value in Ns/m2 and Pa s units?
m = 0.01008 poise
1 poise = 0.1 Pa s = 0.1 Ns/m2
m = 0.001008 Pa s = 0.001008 Ns/m2
- In a fluid the velocity measured at a
distance of 75mm from the boundary is 1.125m/s. The fluid has absolute
viscosity 0.048 Pa s and relative density 0.913. What is the velocity
gradient and shear stress at the boundary assuming a linear velocity
distribution.
m = 0.048 Pa s
g = 0.913

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