Fluids Fluids flow – conform to shape of container –liquids OR gas

Density Defined as: Or: Specific gravity =  /  water at 4°C –  water at 4°C = 1.0 g/cm 3 = 1000 kg/m 3 –Specific gravity has no units density of air (1 atm, 0ºC): g/cm 3 = 1.29 kg/m 3

Example 1 Find the volume of 20,000 N weigh water.

Pressure as Function of Depth Static Fluid (Fluid at rest) P 2 A = P 1 A + mg P 2 = P 1 + ρgh For a given fluid, all points at same depth below surface are at the same pressure.

Example2: Deepest Fish (a) Deepest fish ever sighted was on the floor of the Mariana's trench at 11,500m depth. What is the pressure there? (b) What would be the inward force on a 20cm diameter circular of a submarine wall at that depth?

Which liquid container will have the highest pressure (at the black dot locations)? (1)A, (2) B, (3) C, (4) D, (5) all the same. Clicker!

Consider the four points in the lake. The points are all at the same depth below the surface, but the depth of the bottom of the lake underneath the points varies. At which point is the pressure the greatest? (1)A, (2) B, (3) C, (4) D, (5) all the same. Pressure only depends on depth of point where pressure is being measured. Clicker!

Two beakers are filled with fluid. One is filled with water. The other is filled with a mixture of oil (specific gravity 0.8) and water to the same level. Which beaker has the greatest pressure at the bottom of the beaker. (1) The Water beaker (2) The Oil and Water beaker (3) Both the same Pressure at depth where oil and water meet is lower in the oil and water mixture (oil less dense). Increase in pressure from this level is same, since water in both columns. Clicker!

The following bulbs are connected by a fluid filled tube. The fluid is blue. How do the pressures in the two bulbs, P 1 and P 2, compare? P1P1 P2P2 (1)P 1 > P 2 (2)P 1 = P 2 (3)P 1 < P 2 Pressure at level B greater than at level A in fluid A B Clicker!

Measuring Pressure Using a Column of Fluid Absolute Pressure, P Gauge Pressure = P – P o P o = Atmospheric Pressure = 1.01 x 10 5 Pa PoPo PoPo P P=0 h h Manometer Barometer Height of column and density of fluid determine pressure difference. Pressure sometimes given in inches of Mercury (Hg) 30.2”

Blood Pressure 120 mm of Mercury/80mm of Mercury Gauge pressure (relative to atmosphere)

As you travel up a mountain, your ears pop. The figure below is a pitiful attempt to show the ear drum (in red). As you travel up a mountain, what will the ear drum do between 'pops'? (1) Bow towards the middle ear (2) Hold its shape (3) Bow towards the outside Pressure outside decreases. Clicker!

Example 3 A manometer is made of a solution that has a specific gravity of 4.15 and is used to measure the pressure of gas in a container as shown. The reading in manometer shows 650 mm. (a) What is the gauge pressure? (b) What is the absolute pressure?

The area of the left piston is 10mm 2. The area of the right piston is 10,000mm 2. What force must be exerted on the left piston to keep the 10,000-N car on the right at the same height? (1) 10N (2) 100N (3) 10,000N (4) 10 6 N (5) 10 8 N P = F 1 /A 1 = F 2 /A 2 F 1 = (A 1 /A 2 ) F 2 F 1 = (10 mm 2 / 10,000 mm 2 ) * 10,000N Clicker! Hydraulic Car Lift A1A1 = 10 mm 2 A2A2 = 10,000 mm 2

When an object is submerged in a fluid, there is a buoyant force. An object of 7N weigh submerged in a fluid is hanged by a rope, which has a tension of 5 N. What is the buoyant force? (1) 1 N downward(2) 1 N upward(3) 2 N downward (4) 2 N upward(5) 5 N downward(6) 5 N upward (7) 12 N downward(8) 12 N upward 5N 7N FBFB Just another force Σ F Y = 0 5N – 7N + F B = 0 F B = 7N – 5N Clicker!

Floating versus Completely Submerged Floating: F B = W V DISPLACED < V OBJECT ρ OBJ < ρ FLUID F B = ρ FLUID V DISPLACED g W = ρ OBJ V OBJECT g Completely Submerged V DISPLACED =V OBJECT F B = ρ g V OBJECT FBFB FGFG ρ OBJ > ρ FLUID

Object B is larger than A. If we put both objects into water, which of the following statement is true? (1) Both A, and B will immerse into water. (2) A will completely immerse, B will float. (3) B will completely immerse, A will float. (4) Both will float. (5) None of the above is correct. Density of A > density of water Density of B < density of water Clicker! A 2000 kg/m^3 B 500 kg/m^3

Example 4: If ice has a density of 920 kg/m 3, what percentage of an iceberg sticks out above the waterline? If the mass of the ice is 20 kg, what is the volume of the ice that is sticking out of water? F BUOYANT = Weight ρ w g V DISPLACED = ρ ICE g V ICE ρ w g V SUBMERGED = ρ ICE g V ICE V SUB / V OBJ = ρ ICE / ρ w V SUB / V OBJ = 0.92 So fraction not submerged is or 0.08, 8%

An object of mass 1500 kg with a density of 2000 kg/m 3 is put into water. (a) Will this object float? (b) What will be the weight of the object in water? Example 5

Outline Lecture Couple of Buoyancy Questions No more numerical examples Archimedes principle –Buoyancy – basics –Experiment with cylinders –Rank Buoyant force? Evaluations Demos Bernoulli Experiments Beach Ball and Blower Paper Straw and two bottles Thoughts Need to get across both the simple and the more complex Just another force

Fill two identical glasses with water up to the rim. Now place ice in one (water will overflow into the sink). Now consider the two glasses, one with water, one with water and ice. Which weighs more? (1) The glass without ice cubes (2) The glass with ice cubes (3) The two weigh the same Ice less dense, but occupies more volume. Each cube displaces the weight of the cube in water. Think of the following: two beakers filled to the edge with water. Add an ice cube. The weight of the fluid lost over the edge equals the weight of the ice cube. Clicker!

An incompressible fluid is flowing through a pipe. At which point is the fluid traveling the fastest? (6) All points have the same speed Smallest cross-sectional area, highest speed Clicker!

Bernoulli's Equation – Conservation of Energy One way to look at it: W NC = ΔKE + ΔPE F*d (1/2)mv 2 mgh Divide by volume: ΔPressure (1/2)ρv 2 ρgh P 1 + (1/2)ρv ρgy 1 = P 2 + (1/2)ρv ρgy 2 If height not changing much, higher speed leads to lower pressure

An incompressible fluid flows through a pipe. Compare the speed of the fluid at points 1 and 2. (1) Greater at 1 (2) Greater at 2 (3) Both the same (4) Not enough information Area is smaller at 2, therefore fluid must travel faster in order to conserve the volume flow rate. Clicker!

An incompressible fluid flows through a pipe. Compare the pressure at points 1 and 2. (1) Greater at 1 (2) Greater at 2 (3) Both the same (4) Not enough information P + (1/2) ρv 2 + ρgh = constant (1/2) ρv 2 is larger (higher v due to smaller A) ρgh is larger (higher) Therefore pressure must decrease at point 2 P 1 + (1/2)ρv ρgy 1 = P 2 + (1/2)ρv ρgy 2 Clicker!

Example 6 Calculate the lift force of an airplane wing with a surface area of 12.0m 2. Assume the air above the wing is traveling at 70.0 m/s and the air below the wing is traveling at 60.0m/s. Assume the density of the air to be constant at 1.29 kg/m 3. F LIFT = F UP - F DOWN = P BELOW A – P ABOVE A = (ΔP) A From Bernoulli, find (ΔP) assuming Δy negligible. P A + (1/2)ρv A 2 + ρgy A = P B + (1/2) ρv B 2 + ρgy B (ΔP) = P B – P A = ((1/2)ρv A 2 – (1/2)ρv B 2 ) + (ρgy A – ρgy B ) (assume ΔPE negligible – if y A -y B = 10cm, only 1.26 Pa) ΔP = (1/2) (1.29)(70 2 – 60 2 ) = Pa F LIFT = (838.5Pa)(12m 2 ) = 1.01 x10 4 N

The Water Tank Water is leaving a tank at a speed of 3.0 m/s. The tank is open to air on the top. What is the height of the water level above the spigot? Assume the area of the tank is much larger than the area of the spigot. h = 0.459m P 1 + (1/2)ρv ρgy 1 = P 2 + (1/2)ρv ρgy 2 P 0 + (1/2)ρv = P 0 + (1/2)ρv ρgh Since A 2 >> A 1, v 1 >> v 2 (1/2)ρv 1 2 = (essentially 0) + ρgh