PHYSICS 231 Lecture 22: Pressure

Post on 13-Mar-2016

78 views 8 download

Tags:

description

PHYSICS 231 Lecture 22: Pressure. Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom. Previously. Young’s modulus. Solids:. Shear modulus. Bulk modulus Also fluids. General:. P=F/A (N/m 2 =Pa) =M/V (kg/m 3 ). Force and pressure. Air (P=1.0E+5 Pa). P=0 Vacuum. A. F. - PowerPoint PPT Presentation

Transcript of PHYSICS 231 Lecture 22: Pressure

PHY 2311

PHYSICS 231Lecture 22: Pressure

Remco ZegersWalk-in hour: Thursday 11:30-13:30 am

Helproom

PHY 2312

PreviouslySolids:

LAFL

LLAFY

0

0// Young’s modulus

xAFh

hxAFS

// Shear modulus

pressurePVVP

VVAFB

00 //

/ Bulk modulusAlso fluids

P=F/A (N/m2=Pa)

=M/V (kg/m3)

General:

PHY 2313

Force and pressure

A P=0Vacuum

Air (P=1.0E+5 Pa)

F

Force due to pressure difference: Fpressure=PAIf A=0.01 m2 (about 10 by 10 cm) thena force F=(1.0E+5)*0.01=1000N is needed to pullthe lit.

What is the force needed to move the lit?

PHY 2314

Magdeburg’s hemispheres

Otto von Guericke (Mayor of Magdeburg, 17th century)

PHY 2315

Pressure vs DepthHorizontal direction:P1=F1/A P2=F2/A F1=F2 (no net force)So, P1=P2

Vertical direction:Ftop=PatmAFbottom=PbottomA-Mg=PbottomA-gAh

Since the column of water is not moving:Ftop-Fbottom=0PatmA=PbottomA-gAhPbottom=Patm+ gh

PHY 2316

Pressure and Depth:

Pdepth=h =Pdepth=0+ gh

Where:Pdepth=h: the pressure at depth hPdepth=0: the pressure at depth 0=density of the liquidg=9.81 m/s2

h=depthPdepth=0=Patmospheric=1.013x105 Pa = 1 atm =760 TorrPascal’s principle: If P0 changes then the pressures at all depths changes with the same value.

PHY 2317

A submarineA submarine is built in such a way that it can stand pressuresof up to 3x106 Pa (approx 30 times the atmospheric pressure). How deep can it go?

Pdepth=h =Pdepth=0+ gh3E+06=(1.0E+05)+(1.0E+03)(9.81)hh=296 m.

PHY 2318

Does the shape of the container matter?

NO!!

PHY 2319

Your homemade pressure difference meter (PART I)

Pdepth=h =Pdepth=0+ gh

Part 2 on Friday!

h1

h2

PHY 23110

Pascal’s principleIn other words then before: a change in pressure applied to a fluid that is enclosed in transmitted to the wholefluid and all the walls of the container that hold the fluid.

P=F1/A1=F2/A2If A2>>A1 thenF2>>F1.

So, if we apply a smallforce F1, we can exerta very large Force F2.

Hydraulic press demo

PHY 23111

Hydraulic brake

6.4cm2 1.8 cm2

F=44N

R=0.34 mcoef offriction: 0.5

What is the frictionaltorque about the axle exerted by the shoe on the wheel drum when a force of 44N is applied to the pedal?

FMC/AMC=FBC/ABCFBC=44*1.8/6.4=12.4 NFshoe-drum=Fdrum-shoe=normal forceFfriction=N=0.5*12.4=6.2N=F*R=6.2*0.34=2.1 NM

n

F

PHY 23112

Pressure measurement.

The open-tube manometer.The pressure at A and B isthe same:P=P0+ghso h=(P-P0)/(g)

If the pressure P=1.01 atm, what is h? (the liquid is water)h=(1.01-1)*(1.0E+05)/(1.0E+03*9.81)= =0.1 m

PHY 23113

Pressure Measurement: the mercury barometer

P0= mercurygh

mercury=13.6E+03 kg/m3

mercury,specific=13.6

PHY 23114

Pressure at different altitudesThe pressure in the lecture room equals 1 atm (1.013E+05 Pa). What will the pressure on the 6th floorof the BPS building be (h=20m)? And at the top ofmount Everest (h=8500 m)?

Just like the case for a fluid, the pressure changeswith depth (height in the case of air).Plecture room=P6th floor+ airghP6th floor=Plecture room-airgh=1.013E+05-1.29*9.81*20=

=1.010+05 Pa ( 3 promille change)Pmount everest=1.013E+05-1.29*9.81*8500=-6.3E+03 Pa????The density of air changes with altitude, and so the equation does not hold; it is very easy to compress air(small bulk modulus) compared to e.g. water.(see also next chapter)