PHYSICS 231 Lecture 22: fluids and viscous flow

1. PHYSICS 231Lecture 22: fluids and viscous flow Remco Zegers Walk-in hour: Tue 4-5 pm Helproom PHY 231

2. P0 h B w Pressure at depth h P= P0+ fluidgh h: distance between liquid surface and the point where you measure P h P Buoyant force for submerged object B = fluidVobjectg = Mfluidg = wfluid The buoyant force equals the weight of the amount of water that can be put in the volume taken by the object. If object is not moving: B=wobject object= fluid Buoyant force for floating object The buoyant force equals the weight of the amount of water that can be put in the part of the volume of the object that is under water. objectVobject= waterVdisplaced h= objectVobject/(waterA) PHY 231

3. initially B = fluidVobjectg w=Mblockg B=w lower density liquid: w remains the same, B becomes smaller the block will sink to the bottom quiz (extra credit) A block of weight w is placed in water and found to stay submerged as shown in the picture. The liquid is then replaced by another liquid of lower density. What will happen if the block is placed in the liquid of lower density? • the block will float on the surface of the liquid • the block will be partially submerged and partially above • the liquid • the block will again be submerged as shown in the picture • the block will sink to the bottom PHY 231

4. Bernoulli’s equation P1+½v12+gy1= P2+½v22+gy2 P+½v2+gy=constant The sum of the pressure (P), the kinetic energy per unit volume (½v2) and the potential energy per unit volume (gy) is constant at all points along a path of flow. Note that for an incompressible fluid: A1v1=A2v2 This is called the equation of continuity. PHY 231

5. hole in a tank P0 Pdepth=h =Pdepth=0+ gh h y If h=1m & y=3m what is x? Assume that the holes are small and the water level doesn’t drop noticeably. x PHY 231

6. If h=1 m and y=3 m what is X? P0 B Use Bernoulli’s law h PA+½vA2+gyA= PB+½vB2+gyB At A: PA=P0 vA=? yA=y=3 At B: Pb=P0 vB=0 yB=y+h=4 P0+½vA2+g3=P0+g4 vA=(g/2)=2.2 m/s y A x1 PHY 231

7. Each water element of mass m has the same velocity vA. Let’s look at one element m. vA=(g/2)=2.2 m/s vA In the horizontal direction: x(t)=x0+v0xt+½at2=2.2t In the vertical direction: y(t)=y0+v0yt+½at2=3-0.5gt2 = 0 when the water hits the ground, so t=0.78 s so x(0.78)=2.2*0.78=1.72 m 3m 0 x1 PHY 231

8. Viscosity Viscosity: stickiness of a fluid One layer of fluid feels a large resistive force when sliding along another one or along a surface of for example a tube. PHY 231

9. Viscosity Contact surface A moving F=Av/d =coefficient of viscosity unit: Ns/m2 or poise=0.1 Ns/m2 fixed PHY 231

10. Poiseuille’s Law How fast does a fluid flow through a tube? R4(P1-P2) (unit: m3/s) Rate of flow Q= v/t= 8L PHY 231

11. R4(P1-P2) Rate of flow Q= 8L R=[8QL/((P1-P2))]1/4=0.05 m Example Flow rate Q=0.5 m3/s Tube length: 3 m =1500E-03 Ns/m2 PP=106 Pa P=105 Pa What should the radius of the tube be? PHY 231

12. If time permits, I will do additional problems here. PHY 231

13. Buoyant forces When submerged in water an object weighs 1.6N. At the same time, the water level in the water container (with A=0.01 m2) rises 0.01 m. What is the specific gravity (sg) of the object? (water=1.0x103 kg/m3) Use the fact that the Buoyant force on a submerged object equals the weight of the displaced water. W =Fg-B =Mobjectg-Mwater,displacedg =objectVobjectg-waterVobjectg =Vobjectg (object-water) 1.6N =0.01*0.01*g(object-water)=1.0x10-4*9.8*water(sg-1) sg=2.63 A=0.01 m2 PHY 231

14. Keep it coming. A plastic bag contains a glucose solution. The part of the bag that is not filled is under vacuum. If the pressure in a blood vein is 1.33x104 Pa, how high must one hang the bag to make sure the solution (specific gravity 1.02) enters the body? (w=1.0x103kg/m3) P=P0+gh 1.33x104=0+1.02*1.0x103*9.8*h h=1.33 m PHY 231

15. Titanic: After the Titanic sunk, 10 people manage to seek refuge on a 2x4m wooden raft. It is still 1.0 cm above water. A heavy debate follows when another person (60 kg) wants to board as well. Fortunately, a PHY231 student is among the 10. Can she convince the others that it is safe to pull the person on board without the whole raft sinking? (w=1.0x103 kg/m3) With 10 people: Fg=B (Mraft+M10)g=Vdisplaced,before wg With 11 people: Fg=(Mraft+M10+M1)g B=(Vdisplaced,before+Vextra) wg stationary if Fg=B (Mraft+M10)g+M1g=(Vdisplaced,before+Vextra)wg M1g=Vextra wg so Vextra=(M1/w) Vextra=60/1.0x103=0.06m3 Vextra=LxWxH=8H=0.06 so H=0.75cm so there is still 0.25cm to spare! PHY 231

16. A=5cm2 Bernoulli =1.0x103 kg/m3 2m A=2cm2, P=1 atm Water flows over a height of 2m through an oddly shaped pipe. A) If the fluid velocity is 1 m/s at the bottom, what is it at the top?B) What is the water pressure at the top? A) Use the equation of continuity: A1v1=A2v2 5*vtop=2*1 vtop=0.4 m/s B) Use Bernoulli. Ptop+½vtop2+ghtop= Pbot+½vbot2+ghbot Ptop+0.5*(1E+03)*0.42+(1E+3)*9.8*2=(1E+05)+0.5*(1E+03)12 Ptop=80820 Pa. PHY 231

17. question • Imagine holding two bricks under water. (brick> water) • Brick A is just beneath the surface of the water, Brick • B is at a greater depth. The force needed to hold brick • B in place is: • larger • the same as • smaller • than the force required to hold brick A in place. Grav. force is the same on both. Buoyant force B: weight of water displaced by brick. Also same for both! PHY 231

18. Before the ice melts: B=Fg so wgVdisp= icegVice Vdisp= iceVice/water After it melts: Mmelted water=Mice waterVmelted= iceVice Vmelted=iceVice/water the same! question Two cups are filled to the same level with water. One of the two cups has ice cubes floating in it. When the ice cubes melt, in which cup is the level of water higher? • The cup without ice cubes • The glass with ice cubes • It is the same in both global warming? PHY 231