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

2. PHY 231 2 P=P 0 +  fluid gh h: distance between liquid surface and the point where you measure P P0P0 P h B =  fluid V object g = M fluid g = w fluid 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=w object  object =  fluid Pressure at depth h Buoyant force for submerged object Buoyant force for floating object h B w 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.  object V object =  water V displaced h=  object V object /(  water A)

3. PHY 231 3 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? a)the block will float on the surface of the liquid b)the block will be partially submerged and partially above the liquid c)the block will again be submerged as shown in the picture d)the block will sink to the bottom initially B =  fluid V object g w=M block g B=w lower density liquid: w remains the same, B becomes smaller the block will sink to the bottom

4. PHY 231 4 Bernoulli’s equation P 1 +½  v 1 2 +  gy 1 = P 2 +½  v 2 2 +  gy 2 P+½  v 2 +  gy=constant The sum of the pressure (P), the kinetic energy per unit volume (½  v 2 ) and the potential energy per unit volume (  gy) is constant at all points along a path of flow. Note that for an incompressible fluid: A 1 v 1 =A 2 v 2 This is called the equation of continuity.

5. PHY 231 5 hole in a tank P depth=h =P depth=0 +  gh h x y If h=1m & y=3m what is x? Assume that the holes are small and the water level doesn’t drop noticeably. P0P0

6. PHY 231 6 h x1x1 y If h=1 m and y=3 m what is X? Use Bernoulli’s law P A +½  v A 2 +  gy A = P B +½  v B 2 +  gy B At A: P A =P 0 v A =? y A =y=3 At B: P b =P 0 v B =0 y B =y+h=4 P 0 +½  v A 2 +  g3=P 0 +  g4 v A =  (g/2)=2.2 m/s A P0P0 B

7. PHY 231 7 Each water element of mass m has the same velocity v A. Let’s look at one element m. v A =  (g/2)=2.2 m/s x1x1 3m vAvA In the horizontal direction: x(t)=x 0 +v 0x t+½at 2 =2.2t In the vertical direction: y(t)=y 0 +v 0y t+½at 2 =3-0.5gt 2 = 0 when the water hits the ground, so t=0.78 s so x(0.78)=2.2*0.78=1.72 m 0

8. 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.

9. PHY 231 9 Viscosity Contact surface A fixed moving F=  Av/d  =coefficient of viscosity unit: Ns/m 2 or poise=0.1 Ns/m 2

10. PHY 231 10 Poiseuille’s Law How fast does a fluid flow through a tube? Rate of flow Q=  v/  t=  R 4 (P 1 -P 2 ) 8L8L (unit: m 3 /s)

11. PHY 231 11 Example PP=10 6 Pa P=10 5 Pa Flow rate Q=0.5 m 3 /s Tube length: 3 m  =1500E-03 Ns/m 2 What should the radius of the tube be? Rate of flow Q=  R 4 (P 1 -P 2 ) 8L8L R=[8Q  L/(  (P 1 -P 2 ))] 1/4 =0.05 m

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

13. 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 m 2 ) rises 0.01 m. What is the specific gravity (sg) of the object? (  water =1.0x10 3 kg/m 3 ) A=0.01 m 2 Use the fact that the Buoyant force on a submerged object equals the weight of the displaced water. W=F g -B =M object g-M water,displaced g =  object V object g-  water V object g =V object g (  object -  water ) 1.6N=0.01*0.01*g(  object -  water )=1.0x10 -4 *9.8*  water (sg-1) sg=2.63

14. 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 1.33x10 4 Pa, how high must one hang the bag to make sure the solution (specific gravity 1.02) enters the body? (  w =1.0x10 3 kg/m 3 ) P=P 0 +  gh 1.33x10 4 =0+1.02*1.0x10 3 *9.8*h h=1.33 m

15. 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 =1.0x10 3 kg/m 3 ) With 10 people: F g =B (M raft +M 10 )g=V displaced,before  w g With 11 people: F g =(M raft +M 10 +M 1 )g B=(V displaced,before +V extra )  w g stationary if F g =B (M raft +M 10 )g+M 1 g=(V displaced,before +V extra )  w g M 1 g=V extra  w g so V extra =(M 1 /  w ) V extra =60/1.0x10 3 =0.06m 3 V extra =LxWxH=8H=0.06 so H=0.75cm so there is still 0.25cm to spare!

16. PHY 231 16 Bernoulli A=5cm 2 A=2cm 2, P=1 atm 2m 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: A 1 v 1 =A 2 v 2 5*v top =2*1 v top =0.4 m/s B) Use Bernoulli. P top +½  v top 2 +  gh top = P bot +½  v bot 2 +  gh bot P top +0.5*(1E+03)*0.4 2 +(1E+3)*9.8*2=(1E+05)+0.5*(1E+03)1 2 P top =80820 Pa.  =1.0x10 3 kg/m 3

17. 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: a)larger b)the same as c)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!

18. PHY 231 18 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? a)The cup without ice cubes b)The glass with ice cubes c)It is the same in both Before the ice melts: B=F g so  w gV disp =  ice gV ice V disp =  ice V ice /  water After it melts: M melted water =M ice  water V melted =  ice V ice V melted =  ice V ice /  water the same! global warming?