1. Chapter 9 Fluid Mechanics

2. Fluids “A nonsolid state of matter in which the atoms or molecules are free to move past each other, as in a gas or liquid.” (p. 318) Solids: definite shape and volume. Liquids: definite volume but no definite shape. Gases: no definite shape or volume; has volume and shape of container.

3. Mass Density “The mass per unit volume of a substance.” (p. 319) Mass density = mass / volume ρ = m / V For mass, we will use grams (g) or kilograms (kg) and for volume we will use cm 3, m 3, liters (L) or milliliters (mL). 1 m 3 = 1 x 10 6 cm 3 1 L = 1000 mL 1 cm 3 = 1 mL Common units of density are g / cm 3, g / mL, and kg / m 3 Solids and liquids are almost incompressible, which means their densities do not change. Gases are compressible; so their densities depend on temperature and pressure.

4. Buoyant Force “A force that acts upward on an object submerged in a liquid or floating on a liquid’s surface.” (p. 319) Archimedes Principle: “any object submerged in a fluid experiences an upward buoyant force equal in magnitude to the weight of the fluid displaced by the object.” (p. 320) Buoyant force = weight of displaced fluid F B = m f g (m f = mass of fluid displaced)

5. Floating Objects If an object floats, then its density is less than the density of the fluid (ρ o < ρ f ; ρ o is the density of object and ρ f is the density of the fluid) Recall that the weight of an object is F g = mg. We will let m o = mass of the floating object. So, the weight of the object in air is F g (object) = m o g For a floating object, F B = F g (object) m f g = m o g m f g = m o g And, since m = ρV, ρ f V f g = ρ o V o g See problem 3, p. 324, Practice 9A

6. Sinking Objects… If an object sinks, then its density is greater than the density of the fluid (ρ o > ρ f ). An object submerged in a fluid has an “apparent weight” that is less then it’s normal weight: F net = F B – F g (object) If an object sinks below the surface of a fluid, then the volume of the fluid displaced equals the volume of the object. So, V f = V o and we can replace both with just V: F net = F B – F g (object) = ρ f V f g– ρ o V o g = ρ f Vg– ρ o Vg A simple relationship results from the above equation F g (object) / F B = ρ o / ρ f See problem #1, p. 324, Practice 9A

7. Fluid Pressure and Temperature Pressure is “the magnitude of the force on a surface per unit of area.” (p. 325) Pressure = Force / Area = F / A P = F / A F = PA A = F / P Area of circle = πr 2 Area of rectangle = LW Surface area of a sphere = 4πr 2 The SI (International Standard) unit for pressure is the Pascal (Pa), which is 1 N / m 2 Atmospheric pressure is 101,000 Pa, which is equal to 1atmosphere (atm). 1 atm = 14.7 psi (pounds per square inch)

8. Pascal’s Principle Pressure applied to a fluid in a closed container is transmitted equally to every point of the fluid and to the walls of the container. If pressure is increased at point 1 in the fluid, it increases by the same amount at point 2. Pressure increase = F 1 / A 1 = F 2 / A 2

9. Pressure as a function of depth Pressure increases with depth: Pressure = F / A = mg / A = ρVg / A = ρAhg / A = ρgh (gauge pressure) Absolute pressure = atmospheric pressure + ρgh = P 0 + ρgh

10. Kinetic Theory of Gases and Temperature Kinetic theory of gases describes gas molecules as tiny particles that are constantly moving and colliding with one another and with the walls of the container. Temperature measures the average kinetic energy of the particles in a substance. As the temperature of a gas increases, so does the kinetic energy and, therefore, velocity of the particles. This results in more collisions, which increases pressure. SI unit for temperature is degrees Celsius (ºC) and Kelvin (K). K = ºC + 273

11. Properties of gases and the Ideal Gas Law When the density of a gas is low enough, the gas molecules are very far apart and have fewer collisions. Scientists consider this an ideal gas, because its temperature, pressure and volume can be related using the Ideal Gas Law: PV = Nk B T k B = Boltzmann’s constant = 1.38 x 10 -23 J / K = 1.38 x 10 -23 J / K N = number of molecules T = temperature in Kelvin

12. Ideal gas in a closed container Dividing both sides of the ideal gas law by T yields the following equation: Nk B = PV / T If the number of molecules of a gas in a closed container is constant and there is a change in P, T, or V, then P 1 V 1 / T 1 = P 2 V 2 / T 2