chapter 1- fluid pressure new compatibility mode madam zakiah

8/2/2019 Chapter 1- Fluid Pressure New Compatibility Mode Madam Zakiah

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Fluid & Pressure . : ens y

1.2: Pressure1.3: Pascal’s Principle

1.4: Fluid In Motion1.5: Continuity Equation



1.1: Mass Density

divided by its volume V :

m= ρ



Example 1.1:mac ne s op wor er recor s e

mass of an aluminum cube as 176 g.

cm, what is the density of the



1.2: Pressureressure: magn u e o orce o e

force acting perpendicular to a

which the force acts:

A P =

SI unit: Pascal or N/m



Example 1.2:g woman a ances on one ee

of a pair of high heeled-shoes. If the

radius 0.5cm, what pressure does she



1.2.1: Pressure & Depth In A StaticFluid

goes, the more strongly the water,

the pressure that he experiences.

1downward force whose magnitude is

1 .



m ar y, on e o om ace, epressure P 2 generates an upward

2 .

e now t at t s n equ r um,therefore

Σ Fy= P 2A- P 1A- mg = 0 orP 2 A= P 1A+ mg



u we now:m= ρ v

Then, P 2= P 1+ ρ gh



Example 1.3:a cu a e e a so u e pressure a an

ocean depth of 1000 m. Assume the

and the air exerts a pressure of . .



10.3: Pascal’s Principle’

pressure applied to a completely

undiminished to all parts of the fluidand the enclosing walls.

P 2 = P 1

Consider two interconnectedcylindrical chambers with differentarea and filled with a liquid.



ma er c am er sea e a e opwith a cap.

arger c am er e w a mova e

piston.



Example 1.4:,

has a radius of r 1= 0.012m and a

has a radius of r 2= 0.15m. Thecombined weight of the car and thep unger s 2= . . at nputforce is needed to support the car and

surfaces of the piston and plunger areat the same level?



F1

A1

F2

2



1.4.1: Laminar Flow

i. the streamline never cross one.

ii. The spacing of the streamlines

the streamlines, the faster velocity

becomes.



. ere e u ayers o no m x orintermingle and maintain the same

.

iv. Laminar conditions are obtained at



1.4.2: Turbulent Flowea ures o ur u en ow:

i. the fluid particles move in asor er y manner, caus ng y

Currents.. e part c es o not ma nta n t e

same relative positions but constantlyw r.



. ur u en may e occur as a resu othe fluid flow itself or else be

,pumping the fluid.



1.5: Continuity Equation

enters one end of a pipe at a certain

the same rate, assuming that thereare no places between the entry andex t po nts to a or remove t e u .The rate of fluid’s mass flows through

w r , := ρ Av



un : g sSince no fluid can cross the sidewallso e u e, e mass ow ra e a

positions 1 and 2 must be equal. rom t e cont nu ty equat on,

1=

2ρ 1A1v 1 = ρ 2A2v 2



u s nce e ens y oincompressible fluid does not change

,

ρ 1= ρ 2 en, t e equat on w e as:

A1v

1= A

2v

2



e quan y v represen s evolume of fluid per second that

referred as the volume flow rate, Q:= =

Unit: m 3 /s



Example 1.5:a er ows w ve oc y m s n a

pipe of diameter 100 mm into which a.

Determine the velocity of water.

chapter 1- fluid pressure new compatibility mode madam zakiah

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