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PHYSICS 231 Lecture 22: Pressure

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.

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PHYSICS 231 Lecture 22: Pressure

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  1. PHYSICS 231Lecture 22: Pressure Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom PHY 231

  2. Previously Young’s modulus Solids: Shear modulus Bulk modulus Also fluids General: P=F/A (N/m2=Pa) =M/V (kg/m3) PHY 231

  3. Force and pressure Air (P=1.0E+5 Pa) P=0 Vacuum A F What is the force needed to move the lit? Force due to pressure difference: Fpressure=PA If A=0.01 m2 (about 10 by 10 cm) then a force F=(1.0E+5)*0.01=1000N is needed to pull the lit. PHY 231

  4. Magdeburg’s hemispheres Otto von Guericke (Mayor of Magdeburg, 17th century) PHY 231

  5. Pressure vs Depth Horizontal direction: P1=F1/A P2=F2/A F1=F2 (no net force) So, P1=P2 Vertical direction: Ftop=PatmA Fbottom=PbottomA-Mg=PbottomA-gAh Since the column of water is not moving: Ftop-Fbottom=0 PatmA=PbottomA-gAh Pbottom=Patm+ gh PHY 231

  6. Pressure and Depth: Pdepth=h =Pdepth=0+ gh Where: Pdepth=h: the pressure at depth h Pdepth=0: the pressure at depth 0 =density of the liquid g=9.81 m/s2 h=depth Pdepth=0=Patmospheric=1.013x105 Pa = 1 atm =760 Torr Pascal’s principle: If P0 changes then the pressures at all depths changes with the same value. PHY 231

  7. A submarine A submarine is built in such a way that it can stand pressures of up to 3x106 Pa (approx 30 times the atmospheric pressure). How deep can it go? Pdepth=h =Pdepth=0+ gh 3E+06=(1.0E+05)+(1.0E+03)(9.81)h h=296 m. PHY 231

  8. Does the shape of the container matter? NO!! PHY 231

  9. Your homemade pressure difference meter (PART I) Pdepth=h =Pdepth=0+ gh h1 Part 2 on Friday! h2 PHY 231

  10. Pascal’s principle In other words then before: a change in pressure applied to a fluid that is enclosed in transmitted to the whole fluid and all the walls of the container that hold the fluid. P=F1/A1=F2/A2 If A2>>A1 then F2>>F1. So, if we apply a small force F1, we can exert a very large Force F2. Hydraulic press demo PHY 231

  11. F n FMC/AMC=FBC/ABC FBC=44*1.8/6.4=12.4 N Fshoe-drum=Fdrum-shoe=normal force Ffriction=N=0.5*12.4=6.2N =F*R=6.2*0.34=2.1 NM Hydraulic brake What is the frictional torque about the axle exerted by the shoe on the wheel drum when a force of 44N is applied to the pedal? F=44N 6.4cm2 1.8 cm2 R=0.34 m coef of friction: 0.5 PHY 231

  12. Pressure measurement. The open-tube manometer. The pressure at A and B is the same: P=P0+gh so 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 231

  13. Pressure Measurement: the mercury barometer P0= mercurygh mercury=13.6E+03 kg/m3 mercury,specific=13.6 PHY 231

  14. Pressure at different altitudes The pressure in the lecture room equals 1 atm (1.013E+05 Pa). What will the pressure on the 6th floor of the BPS building be (h=20m)? And at the top of mount Everest (h=8500 m)? Just like the case for a fluid, the pressure changes with depth (height in the case of air). Plecture room=P6th floor+ airgh P6th 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) PHY 231

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