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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 3rd 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

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Your chart should now look like this:

Principles of Fluid Technology

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• 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?

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• 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?

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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)

Pre

ssu

re

(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

Principles of Fluid Technology

Pressure (psi)

0

10

20

30

40

50

60

0 5 10 15 20 25 30 35

Volume (mL)

Pre

ssu

re (

psi

)

• 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?

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• As Volume goes up, Pressure goes down

• As Pressure goes up, Volume goes down

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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]

Principles of Fluid Technology

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Volume vs. 1/P

0.000

0.010

0.020

0.030

0.040

0.050

0.060

0.070

0 5 10 15 20 25 30 35

Volume (mL)

1/P

Is this representative of a proportional relationship?

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• 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

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• 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

• P1V1 = P2V2

Boyle’s Law

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PV = k

P1V1 = P2V2

Boyle’s Law

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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 cm3 to 0.240 cm3. Its pressure is now 3 psi. What was the original pressure of the helium?

P1V1 = P2V2

Boyle’s Law

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#1 Solution:Step 1: Write given information

P1 = 60 psi

V1 = 8 L

P2 = ?

V2 = 5 L

Step 2: Write the formula and solve

P1V1 = P2V2

(60 psi) x (8 L) = P2 x (5 L)

480 psi*L = P2 x (5L)

5L 5L

P2 = 96 psi

Does the answer make sense?

Boyle’s Law

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#2 Solution:Step 1: Write given information

P1 = ?

V1 = 200 cm3

P2 = 3 psi

V2 = 0.240 cm3

Step 2: Write the formula and solve

P1V1 = P2V2

P1 x (200 cm3) = (3 psi) x (0.240 cm3)

P1 x (200 cm3) = .72 psi*cm3

200 cm3 200 cm3

P1 = 0.0036 psi

Does the answer make sense?

Boyle’s Law

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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?

P1 = P2

Boyle’s Law

P1 = F1 / A1 P2 = F2 / A2

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Problems

Solution:Step 1: Write given information

P1 = ? P1 = ?

F1 = 10 lbs. F2 = ?

A1 = 0.5 in2 A2 = 1 in2

Step 2: Write the formula and solve for the unknowns

P1 = F1 / A1

P1 = (10 lbs) / (0.5 in2)

P1 = 20 psi

P1 = P2 = 20 psi

F2 = 20 psi x 1 in2 = 20 lbs.

P2 = F2 / A2 F2 = P2 x A2

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Problems

P1 = 20 psi

F1 = 10 lbs

A1 = 0.5 in2

P2 = 20 psi

F2 = 20 lbs

A2 = 1 in2

What is the mechanical advantage of this system?

MA = L / F MA = (20 lbs)/ (10 lbs) MA = 2

Complete the fluid technology problem sheet.

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