In fluid mechanics it is important to get units right. It would be easy if everyone used the SI system but they do not. Many items of old o ld equipment and almost anything American uses non-SI units. We will now consider important variables used in fluid mechanics, their SI units and dimensions.
Quantity Symbol
Definition
S.I. Units 3
Dimensions
ρ
Mass per unit volume
kg/m
Pressure
P
Force Force per unit area
N/m
Surface
σ
Force per unit length
N/m
[M]/[T]
Velocity
u
self explanatory
m/s
[L]/[T]
Volumetric
Q
Flow volume past a point
m /s
3
[L] /[T]
[M]/[T]
2
[M]/[L]
3
Density
[M}/[L] [T]
2
Tension
flowrate
in unit time.
Mass flowrate M
Flow mass past a point
kg/s
Viscosity
Shear force experienced
Ns/m
µ
at a unit shear rate. i.e. how “sticky” fluid is!
3
2
[M]/[L][T]
2
Tip: Dimensions are useful. If you have derived an equation and are not sure if it is right, check the dimensions of the left hand side and ri ght hand side. They must be the same or
the equation is wrong. Remember, though it can still be wrong but be dimensionally consistent! Just because the dimensions balance does not necessarily mean that it is correct.
2. Pressure Measurement and Hydrostatics
Pressure measurement is one of the most important features of fluid mechanics. First we have an important definition.
Or ΔP = ρgΔh In this consideration of manometers we will always use 1 for the left hand side and 2 for the right hand side.
A Simple Manometer
me.psu.edu
P1 = ρ1gz1 P2 = ρ2gz2 Difference P1 – P2 = (ρ2 – ρ1)g(h2 – h1) or sometimes they use z instead of h. We sometimes say that if fluid 1 is a gas, its ρ is negligible and therefore:
P2 – P1 ≈ ρ2gΔh
Enlarged Limb or Well Manometer
In this case, one limb is so enlarged that the movement of fluid in it is negligible. Thus, we need only measure one level.
4mechtech.blogspot.com
As with the simple manometer, P 1 – P2 = (ρ2 – ρ1)g(h2 – h1) BUT the volume change in both limbs must be the same. i.e. A1h1 = A2h2 where A is the cross sectional area of the limbs. So, P1 – P2 = (ρ2 – ρ1)gh2(1-A2/A1) A2/A1 is negligible So P1 – P2 = (ρ2 – ρ1)gh2
The Inclined Tube Manometer
This is in order to increase sensitivity. The displacement is read in a sloping tube. It will appear larger than if it was upright.
As before, P1 – P2 = (ρ2 – ρ1)g(h2 – h1) BUT h2 = Lsinθ Therefore P1 – P2 = (ρ2 – ρ1)g(Lsinθ – h1) This is sometimes combined with a well manometer
moodle.sbc.ac.uk
In this case, h 1 is more or less constant so if we make it the datum level h1 ≈ 0.
P1 – P2 = (ρ2 – ρ1)gLsinθ
Typical manometer problem for 20 marks Show that for an inclined tube well manometer, the pressure difference can be given by:
[8 marks]
P1 – P2 = (2 – ρ1)gLsinθ
An inclined tube manometer containing an oil of specific gravity 0.8 reads a length of 20 cm. If P2 is atmospheric pressure, ρ 1 is negligible and the manometer has an angle of 30°,
[4 marks]
Calculate the pressure P 1.
Explain the apparent paradox:
commons.wikimedia.org The two flasks shown contain different amounts of liquid but have the same base area and the same height of fluid, h. Hence the force acting downwards on the base must be the same, i.e. ρgh. But b clearly must have a greater mass than a.
[8 marks]
Why do we tend to use oil or mercury in manometers rather than water?
[2 marks]
Solution:
First, start with basic definitions: P 1 = ρ1gh1 P2 = ρ2gh2 (h or z doesn’t matter)
[1] [2]
P1 - P2 = (ρ2 – ρ1)g(h2 – h1)
It’s a well manometer so A 1h1 = A2h2
P1 – P2 = (ρ2 – ρ1)gh2(1 – A2/A1)
A2/A1 is negligible
[2]
P1 – P2 = (ρ2 – ρ1)gh2
It’s an inclined tube manometer so h 2 = Lsinθ
[2]
Therefore P1 – P2 = (ρ2 – ρ1)gLsinθ
[1] ----[8]
L = 20cm = 0.2m θ = 30° so sin θ = 0.5 2
P2 = 101300N/m
[2]
So P1 = 101300 + (800 – 0)(9.81)(0.2)(0.5)
P1 = 102085N/m2
[2] ----[4]
The apparent hydrostatic paradox can be resolved when we remember that pressure acts equally in all directions in a fluid.
[2]
If we look at container b, the force acting on the base is indeed equal to ρghA.
But when the force of the fluid is considered to act in all directions, a component of the force that is acting on the wall will be in a downward direction.
[2]
This force will add to the force ρghA at the base to give a sum equal to mg.
[2] ----[6]
If we used water in a manometer, and it was open to the atmosphere, it would slowly evaporate. This would regularly need topping up and if you missed it for a few days you would get an inaccurate reading. Also, water might allow the growth of algae.
