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Fluids

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has tab

on 13 January 2014

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Transcript of Fluids


Fluid
: A substance that has no fixed shape
Properties of Fluids
Steady flow
: the velocities of moving paricles stays constant at all points
ex calm stream
Unsteady flow
: the velocity of moving particles is not constant nor is the direction ex rapids
Properties of Fluids
Compressible

Fluids
: fluids that can be compressed ex. gasses
Incompressible Fluids
: Fluids that cannot be compressed ex. water
Viscosity
: The ability of a fluid to resist motion
ex low viscosity - water
high viscosity - honey
Density
Density- Mass per unit volume
p =m/v
p = density
m = mass (kg)
v = volune (L)

Pressure
Pressure - is a constant force exerted on or against an object
Pressure is not considered a vector quantity
The direction pressure exerts its force is always perpendicular to the surface its exerting on
P = F/A
P= Pressure (Pascals)
F= Force (Newtons)
A= Area(m)

Pressure and Depth in Static Fluids
Depth is a variable in pressure
Pressure is directly porportional to Depth
ex. glass of water
not moving
therefore sum of forces is 0
apply pressure formula
F = P2A - P1A - mg P1
F = P2A - P1A -pvg P2
P2A = P1A + pAHg
P2= P1 + pgh
Fluids
Fluids
Pascal's Principle
Archemedes' Principal
Knowing that the pressure on the bottom is greater than the pressure on the top we are left with an upward force on the object know as
Buoyant force


Finding buoyant force:
FB = P2A - P1A
=(P2 - P1)A
=pghA

If we were to analyze this formula we find that pAh is the same as mass, therefore when we are finding the buoyant force we are actually finding the weight of the displaced liquid

Archemedes' Principle
: Any fluid applies a bouyant force to an object that is partially or completely immersed in it, the magnitude of the buoyant force equals the weight of the fluid that the object displaces
The Equation of Continuity
Mass flow Rate
: The mass of fluid that flow through a pipe per second
Area * Velocity = volume
volume * density = mass
mass flow rate = AVp

Volume Flow Rate
: The volume of fliuid that flows thriough a pipe per second
Volume flow rate = Area * Velocity = AV

The Equation of Continuity
: an equation that a fluid which enters a pipe must exit the pipe at the same rate assuming there are no flaws in the pipe

knowing that the rate of never changes we can derive the formula
p1V1A1 = p2V2A2

Properties of Fluids
Ideal Fluid: An incompressable nonviscous fluid

Stream Lines:Lines used to represent the trajectories or path of the fluid particles
Bernouille Equation
This theorem is derived from the work energy theorem and it describes the steady flow of an ideal fluid and the motion of an ideal fluid with a density of p.
W = Ef – Ei
This equation relates pressure, speed and elevation/height of two different points.
P1 +1/2p(v1)^2 + pg(y1) = P2 +1/2p(v2)^2 + pg(y2)

From this equation we can further derive the pressure depth formula
P1 + ½ p(v1)^2 + pg(y1) = P2 + ½ p(v2)^2 + pg(y2)

We know that the speed in both cases is the same s e can cancel ½ pv2
P1 + pgy1 = P2 + pgy2
P2 = P1 + pg(y2 – y1)
P2 = P1 + pgh

Viscous Flow
In an ideal fluid there is no viscosity to hinder the flow of a fluid.
However in a non-ideal fluid the viscosity changes the way a fluid flows.
It observed that in fluids with a high viscosity that the fluid in the center always has the highest velocity and the fluid next to the wall has the smallest velocity.
This is known as laminar flow.





As seen in the picture above the flow at the centre has highest velocity and when the fluid it flow it exerts a force on the layers beside it slowing their velocity.

Therefore to compensate for these forces an external force is required to move the fluid along regularly.
Viscous Flow


F = ᶯnAv/y

F = Force (Newtons)
n = coefficient of viscosity (Pa*s or P(poise))
A = Area of the pipe
v = velocity
y = Is the diameter of pipe

Since most fluids are viscous the volume flow rate cannot be calculated in the same manor

It was found that the flow rate is proportional to the difference of pressure in two points on the pipe, the length of pipe and the coefficient of viscosity.




Delta P = change in pressure in two different points
r = radius radius of pipe
n = coefiient of viscosity
l = length of pipe
Poiseuille's Law
Blood in the arteries is flowing, but as a first approximation, the effects of this flow can be ignored and the blood can be treated as a static fluid. Estimate the amount by which the blood pressure P2 in the anterior tibial artery at the foot exceeds the blood pressure P1 in the aorta at the heart when the body is
(a) reclining horiontally
(b) Standing
(Density of blood is 1060 kg/m^3)
Blood in the arteries is flowing, but as a first approximation, the effects of this flow can be ignored and the blood can be treated as a static fluid. Estimate the amount by which the blood pressure P2 in the anterior tibial artery at the foot exceeds the blood pressure P1 in the aorta at the heart when the body is
(a) reclining horiontally
(b) Standing
(Density of blood is 1060 kg/m^3)
(a) When the body is horizontal, there is little or no vertical separation between the feet and the heart. Since h = 0m
P2-P1 = pgh = 0 Pa
Solution:
(b) When an adult is erect, the vertical separation between the feet and the heart is about 1.35m, as the picture indicates.
P2 - P1 = pgh = (1060kg/m^3)(9.80m/s^2)(1.35m) = 1.40x10^4 Pa
Example
Example
In a hydraulic car lift, the input piston has a radius of r1 = 0.0120m and a negligible weight. The output plunger has a radius of r2 = 0.150m. The combined weight of the car and the plunger is F2 = 20500 N . The lift uses hydraulic oil that has a density of 8.00 x 10^2 kg/m^3. What input force F1 is needed to support the car and the output plunger when the bottom surfaces of the piston and plunger are at :
(a) the same level
(b)the levels shown in picture with h = 1.10m

Solution:
(a)Using A= πR^2 for the circular areas of the piston and plunger, we find that,
F1 = F2(A1/A2) = F2((πR1^2)/(πR2^2)) = (20500 N) (((0.0120m)^2)/((0.150m)^2)) = 131 N
(b)In the picture, the bottom surface of the plunger at the point B is at the same level as point A, which is at a depth
h
beneath the input piston. Therefore, we can apply P2 = P1 + pgh with P2 = F2/(πr2*2) and P1 = F1/(πR1^2):
F2/πR2^2 = F1/πR1^2 + pgh
Solving for F1 gives
F1 = F2(r1^2/r2^2)-pgh(πR1^2)
=(20500N)((0.0120m)^2/(0.150m)^2 )- (8.00 x 10^2 kg/m^3) x (9.80 m/s^2)(1.10m) π(0.0120m)^2 = 127N
Pascal's Principle
Any change in the pressure applied is transmitted to all parts of the fluid and enclosing walls


consider the picture
we can see that that the two pipes may have
a different Area but they have the same depth
therefore, P1 = P2
we also know that P =F/A
therefore
F1/A1 = F2/A2
F2 = F1/A1/A2
F2 = F1* A2/A1

From this we can see that if A2 is significantly larger than A1we achieve a much higher output force for a minor imput

By: Maiwand, Isfandyar, Moody & Hasaan
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