Pressure Lecture

Pressure Measurements

Pressure Measurements

Definition:

Pressure is the force acting on a unit area

P = F / A

Force Unit Area

Molecule

Force of a gas molecule colliding with the enclosingchamber or other molecule.

Gas Pressure

Gas Temperature

Gas temperature is a measure of the kinetic energy of the molecules

KE = 1/2 m V2

Ideal Gas Law

Gas pressure increases with temperature for a fixed volume

Gas Pressure increases with the number of molecules

P V = n R TWhere

P = pressure of the gasV = volume of the enclosing containern = number of moleculesR = Rankine Gas constantT = Rankine Temperature

Pressure is always measuredwith respect to a reference.

These references are:

• absolute zero• atmospheric

• relative

Pressure is Measured in 4 Ways

• Absolute Pressure • Gauge Pressure

• Vacuum • Differential Pressure

Patm

Pabsolute zero

14.7 psi

0 psia

Pabsolute

Pgauge

Pvacuum

P2

P1

Pdifferential

Pressure Relationships

Vacuum Pump

Pgauge = Pabs - Patm Pabs > Patm Pgauge >0 (positive)

Pressure Relationship Equations

Pgauge = Pabs - PatmPabs < PatmPgauge < 0 (negative)Vacuum

P2 - P1 = Pdifferential

Some Common Units of Pressure

Psia - absolute pressure, pound per square inch

Psig - gauge pressure, pound per square inch

Pa - Pascal newtons / mm^2

in H2 O- pressure in hydrostatic units with respect to water

Torr - pressure in honor of Torricelli, the inventor of the barometer. Italian physicist

atm - pressure of the atmosphere at sea level

mm_Hg - pressure in hydrostatic units with respect to mercury

Units

Pressure Units

Pascal(Pa)

Bar(bar)

Technicalatmosphere

(at)Atmosphere

(atm)Torr

(mmHg)

Pound persquare inch

(psi)

1 Pa = 1 N/m² 10- 5 10.197×10- 6 9.8692×10- 6 7.5006×10- 3 145.04×10- 6

1 bar 100 000 = 106 dyn/cm² 1.0197 0.98692 750.06 14.504

1 at 98 066.5 0.980665 = 1 kgf/cm² 0.96784 735.56 14.223

1 atm 101 325 1.01325 1.0332= 101 325

Pa760 14.696

1 torr 133.322 1.3332×10- 3 1.3595×10- 3 1.3158×10- 3 = 1 mmHg 19.337×10- 3

1 psi 6 894.76 68.948×10- 3 70.307×10- 3 68.046×10- 3 51.715 torr = 1 lbf/in²

Example reading: 1 Pa = 1 N/m² = 10- 5 bar = 10.197×10- 6 at = 9.8692×10- 6 atm....etc.Note: mmHg is an abbreviation for millimetres of Mercury.

Force = W = volume x weight density

Force = A x H x Wd

Pressure =Force/Area = H x Wd

Pressure Head

Hydrostatic Pressure

Pressure is due to the height (H) of a liquidand its weight density (Wd).

Pressure = H x Wd

WdH

What is the pressure of 1 inch of water ?

Solution:Wd = 62.4 lbf / cu ft

Height = 1 inch

Pressure = =.62.41

..12 12 120.0361 psi

What is the pressure of 100 inches of water ?

Solution:Wd = 62.4 lbf / cu ft

Height = 100 inch

Pressure = =.62.4100

..12 12 123.6111 psi

What is the pressure of 1 mm of Mercury (Hg) ?

psi=..62.413.605

123

1

25.40.019

Solution

Sg= 13.605 Wd = 62.4 x 13.605

H = 1mm = 1 / 25.4 inch

Some Unit Conversions

=.1 Pa 1 106 newton

mm2

=.1 Pa 1.45 104

psi

=.10000 Pa 1.45 psi

=.1 psi 51.715 torr

=.1 psi 2.036 in_Hg

=.1 psi 27.701 in_H20

=.1 atm 14.696 psi

=.1 atm 760.002 torr

Pascal’s Principle

Pressure is transmitted undiminished in an

incompressible medium

ApplicationPascal's law states that when there is an increase in pressure at any point in a confined fluid, there is an equal increase at every other point in the container.

Hydraulic Press

P1 = P2 (since the pressures are equal throughout).

F1/A1 = F2/A2

V1 = V2

A1 D1 = A2 D2

•A = cross sectional area

•D = the distance moved

A1/A2= D2/D1

Pressure Equations

Mechanical Advantage

This system can be thought of as a simple machine (lever), since force is multiplied.The mechanical advantage can be found by rearranging terms in the above equation to

Mechanical Advantage(IMA) = D1/D2 = A2/A1

For the sample problem above, the IMA would be 10:1 (10 inches/ 1 inch or 10 square inches / 1 square inch).

Problem 1

A hydraulic press has an input cylinder 1 inch in diameter and an output cylinder 6 inches in diameter.

a.Assuming 100% efficiency, find the force exerted by the output piston when a force of 10 pounds is applied to the input piston.

b.If the input piston is moved through 4 inches, how far is the output piston moved?

Problem 2

A hydraulic system is said to have a mechanical advantage of 40. Mechanical advantage (MA) is FR (output) / FE (input). If the input piston, with a 12 inch radius, has a force of 65 pounds pushing downward a distance of 20 inches, find

a.the volume of fluid that has been displaced

b.the upward force on the output piston

c.the radius of the output piston

d.the distance the output piston moves

Problem 3

What pressure does a 130 pound woman exert on the floor when she balances on one of her heels? Her heels have an average radius of 0.5 inch.

Problem 4

A car has a weight of 2500 pounds and rests on four tires, each having a surface area of contact with the ground of 14 square inches. What is the pressure the ground experiences beneath the tires that is due to the car?

Problem 5

The input and output pistons of a hydraulic jack are respectively 1 cm and 4 cm in diameter. A lever with a mechanical advantage of 6 is used to apply force to the input piston. How much mass can the jack lift if a force of 180 N is applied to the lever and efficiency is 80%?

1. a. 360 poundsb. 1/9 inch

2.a. 9043 cubic inchesb. 2600 poundsc. 75.9 inchesd. MA = D1 / D2    D2 = D1 / MA    D2 = 20 in / 40    D2 = 0.5 in

3. 165.6 pounds/square inch

4. 44.6 pounds/square inch5. 1410.6 kg