1. U3f – L2 1.Using the chart of information you recorded yesterday, add two new columns onto your chart. 2.Title one column “1/P” and calculate 1/P for each of the pressures in your chart (to 3 rd decimal place). 3.Title the second column “P x V” and calculate P x V for each of the volumes and pressures in your chart (to nearest whole number). April 16, 2008 DRILL

2. U3f – L2 Your chart should now look like this: Principles of Fluid Technology

3. U3f - L1 A substance (as a liquid or a gas) that conforms to the outline of its container Fluid Systems have 2 things in common: –They contain a fluid, either gas (pneumatics) or liquid (hydraulics) –They contain a pressure difference that creates a net force What is a FLUID?

4. U3f - L1 The technology of using fluid, either gas (pneumatics) or liquid (hydraulic) to apply force or to transport. Example applications: Air brakes on a truck, Tires on a car, Airfoils on an airplane, Warm-air heating ducts, Hydraulic jack, Plumbing in a school Hydro-electric dam What is Fluid Technology?

5. U3f - L1

6. 1.Graph your Volume and Pressure data 2.Make volume the independent 3.Make pressure the dependent 4.Start both at zero Principles of Fluid Technology X-axis Y-axis Volume (mL) Pressure (psi) What is the data range? Volume: 9-30 mL Pressure: 17-55 psi 0 5 10 15 20 25 30 60 55 50 45 40 35 30 25 20 15 10 5

7. Principles of Fluid Technology Your graph should like the graph below. What can you say about the relationship between volume and pressure? As volume increases, pressure decreases Is this a proportional or inversely proportional relationship? U3f - L1 As Volume goes up, Pressure goes down As Pressure goes up, Volume goes down

8. U3f - L1 Principles of Fluid Technology How is this represented mathematically? V 1 / P or P 1 / V Now graph Volume vs. 1/P –Data range for volume is same; data range for 1/P is.018 to.058 –Volume is independent –1/P is dependent a a [Volume is proportional to 1/P] [Pressure is proportional to 1/V]

9. Principles of Fluid Technology U3f - L1 Is this representative of a proportional relationship?

10. U3f – L2 Back to your chart. What do you notice about the values for P x V? Under constant temperature, the product of pressure and volume for a fluid is a constant. Boyle’s Law: P V = k Principles of Fluid Technology

11. U3f – L2 Robert Boyle (1627-1691) – from Ireland; chemist, physicist, and inventor Boyle’s Law – the pressure and volume of an ideal gas are inversely proportional Increase Pressure –Decrease Volume Increase Volume –Decrease Pressure PV = k P 1 V 1 = P 2 V 2 Boyle’s Law

12. U3f – L2 PV = k P 1 V 1 = P 2 V 2 Boyle’s Law

13. U3f – L2 Problems (copy these problems): 1.A gas system has initial pressure and volume of 60 psi and 8 L. If the volume changes to 5 L, what will the resultant pressure be in psi? 2.A sample of helium gas is compressed from 200 cm 3 to 0.240 cm 3. Its pressure is now 3 psi. What was the original pressure of the helium? P 1 V 1 = P 2 V 2 Boyle’s Law

14. U3f – L2 #1 Solution: Step 1: Write given information P1 = 60 psi V1 = 8 L P2 = ? V2 = 5 L Step 2: Write the formula and solve P 1 V 1 = P 2 V 2 (60 psi) x (8 L) = P2 x (5 L) 480 psi*L = P2 x (5L) 5L P2 = 96 psi Does the answer make sense? Boyle’s Law

15. U3f – L2 #2 Solution: Step 1: Write given information P1 = ? V1 = 200 cm 3 P2 = 3 psi V2 = 0.240 cm 3 Step 2: Write the formula and solve P 1 V 1 = P 2 V 2 P1 x (200 cm 3 ) = (3 psi) x (0.240 cm 3 ) P1 x (200 cm 3 ) =.72 psi*cm 3 200 cm 3 P1 = 0.0036 psi Does the answer make sense? Boyle’s Law

16. U3f – L2 Problem (copy this problem): 1.The diagram below illustrates how hydraulic brakes in a car work. The pedal must be pressed with a force of 10 lbs. The surface area of the piston connected to the pedal is.5 square inch. If the surface area of the piston connected to the other end of the brake line is 1 square inch, what is the force applied to that piston? P 1 = P 2 Boyle’s Law P1 = F1 / A1P2 = F2 / A2

17. U3f – L2 Problems Solution: Step 1: Write given informationP1 = ? F1 = 10 lbs.F2 = ? A1 = 0.5 in 2 A2 = 1 in 2 Step 2: Write the formula and solve for the unknowns P1 = F1 / A1 P1 = (10 lbs) / (0.5 in 2 ) P1 = 20 psi P1 = P2 = 20 psi F2 = 20 psi x 1 in 2 = 20 lbs. P2 = F2 / A2 F2 = P2 x A2

18. U3f – L2 Problems P1 = 20 psi F1 = 10 lbs A1 = 0.5 in 2 P2 = 20 psi F2 = 20 lbs A2 = 1 in 2 What is the mechanical advantage of this system? MA = L / F MA = (20 lbs)/ (10 lbs) MA = 2

19. Complete the fluid technology problem sheet. U3f – L2