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 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.

4. PHY 231 4 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

5. PHY 231 5 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.

6. PHY 231 6 Viscosity Contact surface A fixed moving F=  Av/d  =coefficient of viscosity unit: Ns/m 2 or poise=0.1 Ns/m 2

7. PHY 231 7 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)

8. PHY 231 8 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?

9. PHY 231 9 If time permits, I will do additional problems here.