Transcript
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
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