new basic science.pdf

7/25/2019 New Basic science.pdf

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Basic science (Physics and Chemistry)

by Tony R. Kuphaldt (2006)

Unit conversion problems

Work, energy, and power

Simple machines

Static fluids

Dynamic fluids

Elementary thermodynamics Molecular quantities

Balancing chemical reaction equations (stoichiometry)

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Unit conversion problems, by Tony R. Kuphaldt (2006)

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Work, energy, and power



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Metric prefixes and conversion constants

Metric prefixes Yotta = 1024 Symbol: Y Zeta = 1021 Symbol: Z Exa = 1018 Symbol: E

Peta = 1015 Symbol: P

Tera = 1012 Symbol: T Giga = 109 Symbol: G Mega = 106 Symbol: M Kilo = 103 Symbol: k Hecto = 102 Symbol: h Deca = 101 Symbol: da Deci = 101 Symbol: d Centi = 102 Symbol: c Milli = 10

3

Symbol: m Micro = 106 Symbol: Nano = 109 Symbol: n Pico = 1012 Symbol: p Femto = 1015 Symbol: f Atto = 1018 Symbol: a Zepto = 1021 Symbol: z Yocto = 1024 Symbol: y

100

103

106

109

1012

10-3

10-6

10-9

10-12

(none)kilomegagigatera milli micro nano picokMGT m n p

10-2

10-1

101

102

deci centidecahectoh da d c

METRIC PREFIX SCALE

Conversion formulae for temperature oF = (oC)(9/5) + 32 oC = (oF - 32)(5/9) oR = oF + 459.67

K = o

C + 273.15

Conversion equivalencies for distance

1 inch (in) = 2.540000 centimeter (cm)

1 foot (ft) = 12 inches (in)


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Conversion equivalencies for volume

1 gallon (gal) = 231.0 cubic inches (in3) = 4 quarts (qt) = 8 pints (pt) = 128 fluid ounces (fl. oz.)= 3.7854 liters (l)

1 milliliter (ml) = 1 cubic centimeter (cm3)

Conversion equivalencies for velocity

1 mile per hour (mi/h) = 88 feet per minute (ft/m) = 1.46667 feet per second (ft/s) = 1.60934kilometer per hour (km/h) = 0.44704 meter per second (m/s) = 0.868976 knot (knot international)

Conversion equivalencies for mass

1 pound (lbm) = 0.45359 kilogram (kg) = 0.031081 slugs

Conversion equivalencies for force

1 pound-force (lbf) = 4.44822 newton (N)

Conversion equivalencies for area

1 acre = 43560 square feet (ft2) = 4840 square yards (yd2) = 4046.86 square meters (m2)

Conversion equivalencies for common pressure units (either all gauge or all absolute)

1 pound per square inch (PSI) = 2.03602 inches of mercury (in. Hg) = 27.6799 inches of water (in.W.C.) = 6.894757 kilo-pascals (kPa) = 0.06894757 bar

1 bar = 100 kilo-pascals (kPa) = 14.504 pounds per square inch (PSI)

Conversion equivalencies for absolute pressure units (only)

1 atmosphere (Atm) = 14.7 pounds per square inch absolute (PSIA) = 101.325 kilo-pascals absolute(kPaA) = 1.01325 bar (bar) = 760 millimeters of mercury absolute (mmHgA) = 760 torr (torr)

Conversion equivalencies for energy or work

1 british thermal unit (Btu International Table) = 251.996 calories (cal International Table)= 1055.06 joules (J) = 1055.06 watt-seconds (W-s) = 0.293071 watt-hour (W-hr) = 1.05506 x 1010

ergs (erg) = 778.169 foot-pound-force (ft-lbf)

Conversion equivalencies for power

1 horsepower (hp 550 ft-lbf/s) = 745.7 watts (W) = 2544.43 british thermal units per hour(Btu/hr) = 0.0760181 boiler horsepower (hp boiler)

( )


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Physical constants

Speed of light in a vacuum (c) = 2.9979 108 meters per second (m/s) = 186,281 miles per second(mi/s)

Avogadros number (NA) = 6.022 1023 per mole (mol1)

Electronic charge (e) = 1.602

1019 Coulomb (C)

Boltzmanns constant (k) = 1.38 1023 Joules per Kelvin (J/K)

Stefan-Boltzmann constant () = 5.67 108 Watts per square meter-Kelvin4 (W/m2K4)

Molar gas constant (R) = 8.314 Joules per mole-Kelvin (J/mol-K)

Properties of Water

Freezing point at sea level = 32o

F = 0o

CBoiling point at sea level = 212oF = 100oC

Density of water at 4oC = 1000 kg/m3 = 1 g/cm3 = 1 kg/liter = 62.428 lb/ft3 = 1.94 slugs/ft3

Specific heat of water at 14oC = 1.00002 calories/goC = 1 BTU/lboF = 4.1869 Joules/goC

Specific heat of ice 0.5 calories/goC

Specific heat of steam

0.48 calories/goC

Absolute viscosity of water at 20oC = 1.0019 centipoise (cp) = 0.0010019 Pascal-seconds (Pas)

Surface tension of water (in contact with air) at 18oC = 73.05 dynes/cm

pH of pure water at 25o C = 7.0 (pH scale = 0 to 14)

Properties of Dry Air at sea level

Density of dry air at 20o

C and 760 torr = 1.204 mg/cm3

= 1.204 kg/m3

= 0.075 lb/ft3

= 0.00235slugs/ft3

Absolute viscosity of dry air at 20oC and 760 torr = 0.018 centipoise (cp) = 1.8 105 Pascal-seconds (Pas)


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Questions

Question 1

A 120 pound weight is dragged along a level surface with a steady pull of 25 pounds for 10 feet. Howmuch work is done (express your answer in both English and metric units of work)?

120 lb 25 lbpull

file i02589

Question 2

A 3500 pound automobile is hoisted 20 feet in the air by a crane. How much workis done in lifting theautomobile to this height?

file i02621

Question 3

If 40 pounds of books are lifted from floor level to a bookshelf 5 feet above, then later those same booksare taken off the shelf and returned to floor level, what is the total amount of work done by the personmoving the books?

file i02620

Question 4

Suppose a forklift hoists a container weighing 2670 pounds up from ground level, setting it down on aloading dock 4.5 feet above ground level. How much work was done by the forklift in this maneuver?

W = ft-lbs

Now suppose the same forklift moves the container off of the loading dock and into the back of a flatbedtruck 3 feet above ground level. How much work was done by the forklift in this maneuver?

W = ft-lbsfile i04803

Question 5

Suppose a crane picks up a shipping container weighing 32000 pounds, lifting it 17 feet above the ground,moving it horizontally 260 feet to a warehouse, and then setting it down into a pit 5 feet below ground level.Calculate the total (net) amount of work done by the crane on the shipping container from its starting point(on the ground) to its destination (in the pit).

file i02592


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Question 6

Two people are working to move a dead car, one pushing and the other pulling. The person pushingexerts 150 pounds of force, while the one pulling exerts 170 pounds of force. What is the total (resultant)force from the efforts of these two people?

150 lb 170 lb

45 feet of motion

Assuming they are able to drag the car a total distance of 45 feet before collapsing in exhaustion,

calculate the total work done by these two people.file i02614

Question 7

Calculate the amount of work done by this person pushing a lawnmower, given the force and displacementvectors shown:

Lawnmower

28o

F=8N

x= 187 m

file i02616


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Question 8

Danger Dave decides to outfit his car with a solid-fuel rocket mounted in the trunk. Unfortunately, hecant get the rocket motor to fit horizontally in the trunk, and so it sits at a slight angle (15 degrees fromhorizontal). When the rocket is ignited, its thrust is 3400 pounds of force along the centerline of the rocketbody:

15o

Fthrust=3400lb

One fine day, Dave is feeling particularly dangerous, so he ignites the rocket motor to see how fast his

car will perform on a quarter-mile racetrack. Calculate the amount of work done by the rocket on the carduring the quarter-mile run.

Assuming Daves car weighs 3100 pounds (i.e. it has a massof 96.27 slugs) with the new rocket motorinstalled and him in the drivers seat, calculate both his quarter-mile speed and time. Hint: you may needto use one or more of these formulae:

F =ma

x= vt

v= at + v0

x=1

2at2

Where,F= Force applied to objectm= Mass of objectx = Position of object with reference to a starting point (x0, at time t = 0)v = Velocity of objecta = Acceleration of object (i.e. the rate at which velocity changes)t= Elapsed timev0 = Initial velocity of object (at time t = 0)

file i02615


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Question 9

A laborer working on the top of a building uses a manual hoist to lift 20 gallons of water from groundlevel. The height of this lift is 31 feet:

The rope is counterweighted with a mass equal to that of the bucket (empty), so that the bucketsweight does not have to be lifted, only the water inside the bucket. Assuming a vertical lift distance of 31feet, and ignoring the weight of the rope itself, how much work does the laborer do in lifting the water up?Please express your answer in both English and metric units of work.

file i02610Question 10

Suppose a boy working on a farm needs to draw water out of an open well. The water is 23 feet belowground level, and the boy has a 1-gallon bucket with a rope enabling him to manually lift water to thesurface.

Calculate the amount of work the boy must do to lift 1 gallon of water out of the well. Assume thebucket itself weighs 1.5 pounds, but neglect the weight of the rope (for simplicity):

Work = ft-lb

If the boy is able to lift this much water out of the well in 8 seconds, how much is his power output (inunits of horsepower)?

Power = HP

file i04778


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Question 11

A laborer working on the top of a building uses a manual hoist to lift 20 gallons of water 31 feet upfrom ground level. The time required for the laborer to do this job is one minute:

Several hours later, a different laborer does the exact same job of hoisting 20 gallons of water to theroof, but does so in only 40 seconds.

Which laborer performs more work in lifting 20 gallons of water to the roof?file i02611

Question 12

One laborer working on the top of a building uses a manual hoist to lift 10 gallons of water 30 feet upfrom ground level, while a second laborer uses an electric pump to do the same:

Hose

Pump

First, calculate the amount of work needed to lift 10 gallons up to the same roof. Then, calculate thetime required for the pump to do this job, assuming a rating of 1.5 horsepower.

file i02612


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Question 13

How much work is done pumping 3,000 gallons of water from reservoir A to reservoir B over thehill?

Pump

A

B

30o

100ft

55 ft

20 ft

48o

Hill

If the pumps power output is 250 horsepower, how long will it take to pump all 3000 gallons to reservoirB?

Suggestions for Socratic discussion

Calculate the amount of pressure at the discharge port of the pump as it lifts water up to reservoir Bfile i02613

Question 14

Suppose a crate full of bowling balls weighing a total of 5000 newtons (i.e. the crate has a total massof 509.68 kilograms) is lifted 20 meters vertically into the air and then dropped. How fast will its velocitybe just before hitting the ground?


What happens to all the potential energy that was invested in this crate after it impacts the ground?file i02622

Question 15

At the skate park, Riley accidently lets go of his 4 kilogram skateboard at the top of a 5-meter tall

ramp. Assuming a frictionless ride down the ramp, how fast will Rileys skateboard be rolling at the bottomof the ramp?

v = m/s

How much potential energy did Rileys skateboard have when it was 5 meters above ground level (atthe top of the ramp)?


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Question 16

Sam climbs a 130 foot tower with an 8 pound (0.25 slug) textbook. Tony climbs the same tower witha 5 pound (0.16 slug) textbook. Both Sam and Tony drop their textbooks from the top of the tower atexactly the same moment in time. Neglecting the effects of air friction on the books free-fall, calculate thefollowing:

Work done by Sam in lifting his textbook = Kinetic energy of Sams textbook just before it hits the ground = Velocity of Sams textbook just before it hits the ground =

Work done by Tony in lifting his textbook = Kinetic energy of Tonys textbook just before it hits the ground = Velocity of Tonys textbook just before it hits the ground =

file i00430

Question 17A truck weighing 39,000 newtons (N) is traveling at 10 meters per second when its clutch blows out,

disconnecting the engine from the drivetrain (transmission, axle, wheels, etc.). This failure occurs exactlyat the base of a steep incline. How many meters (vertical) will the truck coast up the hill with no enginepower before it stops, neglecting friction of any kind?

How high would the truck have coasted if it had been traveling twice as fast?file i00431

Question 18An automobile weighing 2700 pounds is traveling over level ground at a velocity of 50 miles per hour.

If the driver places the vehicles transmission in neutral (so the engine is disconnected from the wheels) andapplies force to the brake pedal, the vehicle will slow down.

Assuming the braking force on the automobile is a constant 500 pounds (friction opposed to the directionof the automobiles motion), how many feet will the vehicle travel before coming to a rest? Ignore any airfriction, and assume that the braking force is the only decelerating force in this mechanical system.

file i02623

Question 19

Suppose two vehicles are traveling down a highway: a pickup truck and a sports car. The truck weighstwice as much as the sports car, but is traveling only half as fast.

All other factors being equal, which vehicle can stop in the shortest distance?file i02624

Question 20

Suppose a wrecking ball is suspended on the end of a 30 foot cable. If the wrecking ball is drawn tothe side until the cable is at a 20 degree angle (from vertical) and then released to swing, how fast will thewrecking ball be traveling at its maximum velocity?

file i04802


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Question 21

A stupid person fires a gun straight up into the air. The bullet leaves the guns muzzle with a velocityof 1100 feet per second. The bullet has a mass of 150 grains. Ignoring the effects of air friction (I know,aerodynamic friction is no minor effect at supersonic velocities, but just work with me here . . .) how longwill it take for the bullet to return to ground level?

file i02655


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Question 22

Ahorsepoweris defined as 550 ft-lbs of work done in one second of time. An example of this would be a550 pound weight lifted vertically at a speed of one foot per second, or a one pound weight lifted verticallyat a speed of 550 feet per second.

There is a way to relate this to rotary motion, not just linear motion. In the case of rotary motion wemust deal with torque () in lb-ft and angular speed (S) in revolutions per minute (RPM) rather than force

in pounds and linear speed in feet per second.Just as linear power is proportional to the product of force and velocity ( P F v), rotary power is

proportional to torque and rotary speed (P S). What we need to turn this proportionality into anequality is a multiplying constant (k):

P =k S

We may determine the value of this constant by setting up a thought experiment that translatesbetween linear power and rotary power:

Drum

550 lb

weight

1 ft/s

1 ft

In this case, we have a 1 foot radius drum hoisting a 550 pound weight at a linear velocity of 1 footper second, the definition of one horsepower. Translate the linear force and linear velocity to rotary force(torque) and rotary velocity (revolutions per minute), and then calculate the necessary k factor to makeyour own torque/speed/horsepower equation.

file i01430


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Question 23

Calculate energy and/or power for the following scenarios:

One gallon of gasoline stores approximately 121 megajoules (MJ) of energy in chemical form. Ifthis gallon of gasoline is burned completely and steadily over a period of 2 hours, how much poweris output by the fire?

A battery charger connected to a secondary-cell battery applies a constant charging voltage of 14.3volts at a constant charging current of 6.8 amps for 2.5 hours. How much power does this represent(in watts), and how much energy has been delivered to the battery by the charger (in Joules)?

One metric ton of bituminous/anthracite coal contains approximately 28 gigajoules of energy inchemical form. If a coal-burning power plant operating at 35% conversion efficiency (chemical toelectrical energy) is to output 500 MW of power, what rate must the coal be burned in units ofmetric tons per minute?

file i02463


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Answers

Answer 1

In this problem, the figure of 120 lb is of no consequence to our calculations. What matters is the 25pound force in the direction of the weights displacement. Since there is no displacement (motion) in thedirection of the 120 pound force (straight down), there is no work performed by that force.

W=F x cos

W= (25 lb)(10 ft)(cos 0o)

W= 250 ft-lb

We may convert units directly from ft-lb to J in one step, forming a unity fraction with the equivalence:

250 ft-lb

1

1055.06 J778.169 ft-lb

= 338.96 J

Answer 2

W=F x cos

W= (3500 lb)(20 ft)(cos 0

o

)

W= 70000 ft-lb of work

Answer 3

Contrary to intuition, no work has been done. Lifting the 40 pounds of books 5 feet up constitutes 200ft-lb of work done on the books (i.e. potential energy invested in the books), but returning those books backto floor level constitutes 200 ft-lb of energy released(negative work done). Thus, in physics terms, there was

no net work performed.

The person tasked with this pointless exercise, however, may beg to differ.

This same principle of storing and releasing energy is employed in electric vehicles to recover brakingenergy. Instead of converting electrical energy into mechanical potential and visa-versa as happens with theelevator, electric vehicles convert electrical energy into kinetic form (vehicle motion) and visa-versa. Thus,accelerating an electric car from a full stop to some speed and then regeneratively decelerating it back to fullstop is another example of zero (net) work. This energy-recovering capability is what makes electric vehiclesso attractive for stop-and-go travel.

Answer 4

Moving from ground level to loading dock:W = 12015 ft-lbs

Moving from loading dock to flatbed truck:


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Answer 5

The totalamount of work done on the shipping container may be calculated by multiplying its weight(32000 pounds) by the total height change from beginning to end. Assuming it started at ground level, andended up 5 feet below ground level (i.e. the displacement is downwhen the cranes force on the container isup), the work done on the container is:

W=F x cos

W= (32000 lb)(5 ft) cos 180o

W = 160000 ft-lb of work done on the container

Technically, the crane didnt do any work on the shipping container, but rather the shipping containerdid work on the crane!

Note how we are completely ignoring the cranes horizontal motion, because this displacement vector isat right angles (90o) to the containers weight vector as well as the cranes upward force vector, and thereforedoes not count toward or against work done on the container.

Answer 6

When multiple forces act in the same direction, the resultant force will be the simple sum of theindividual forces. In this case, 150 pounds plus 170 pounds is equal to 320 pounds.

Work is calculated by taking this total (net) force of 320 pounds and multiplying by the displacementof 45 feet, since both the force and the displacement vectors are pointed in the same direction (i.e. = 0o):

W=F x cos

W= (320 lb)(45 ft) cos0o

W= 14400 ft-lb of work

Answer 7

W=F x cos

W= (8 N)(187 m) cos 28o

W= 1320

.9 N-m


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Answer 8

The work done by the rocket motor on the car may be calculated with the work formula, taking thedisplacement (x) to be one-quarter of a mile (1320 feet):

W=F x cos

W= (3400 lb)(1320 ft)cos 15o

W= 4335075.1 ft-lb of work done on Daves car

The horizontal component of the rockets 3400 lb thrust will be the 3400 lb times the cosine of 15degrees:

Fx = (3400 lb) cos 15o

= 3284.15 lbThe acceleration of Daves car with this amount of force driving it down the racetrack will be equal to

force divided by mass, following the formula F =ma. In order to calculate a, we will need to know the massof Daves car in slugs (3100 lb / 32.3 ft/s2 = 96.27 slugs):

F =ma

a= F

m

= 3284.15 lb96.27 slugs

= 34.11 ft/s2To put this in perspective, the car would accelerate at 32.2 ft/s2 if pushed off a cliff, so this means

Daves car will rocket down the racetrack at an acceleration faster than free-fall!

The distance traveled (x) by an object under constant acceleration (a) is given by the following formula:

x=1

2at2

In the case of Daves rocket car:

1320 ft =1

2(34.11 ft/s

2)t2

Solving for time (t):

t2 = (2)(1320 ft)

(34.11 ft/s2

)

t=(2)(1320 ft)

(34.11 ft/s2)

t= 8.797 seconds

This, my friends, is a very respectable quarter-mile time!

As for top speed (at the end of the quarter mile run since we assume Dave is still accelerating with the


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As for top speed (at the end of the quarter-mile run, since we assume Dave is still accelerating with therocket motor going at full thrust), this may be calculated by the velocity-acceleration formula v = at:

v= at

v= (34.11 ft/s)(8.797 s) = 300.1 ft/s

This happens to be 204.6 miles per hour, which again is very respectable in a quarter-mile race!

Now, Daves only real problem is how to stop.

Answer 9

To calculate work in this system, we need to know both the displacement (31 feet), and the force exertedin the direction of the displacement. We were not given this force, but we do know it is 20 gallons of waterbeing lifted, and we know the density of water to be 62.4 pounds per cubic foot. So, 20 gallons of waterweighs:

20 gal

1

231 in3

1 gal

1 ft3

1728 in3

62.4 lb

ft3

= 166.83 lb of water in 20 gallons

Now, knowing both the force (166.83 lb) and the displacement (31 ft), we may calculate the work done:

W=F x cos

W= (166.83 lb)(31 ft)(cos 0o) = 5171.83 ft-lb of work

Converting foot-pounds into joules:5171.83 ft-lb

1

1055.06 J

778.169 ft-lb

= 7012.09 J of work

Answer 10

Calculating the weight of the water lifted:

1 gal of water

1

231 in3

1 gal

1 ft3

1728 in3

62.4 lb

1 ft3 of water

= 8.342 lb

Added to the 1.5 pound weight of the bucket, the boys total upward force must be 9.842 pounds.

Since the displacement of the boys effort must be the depth of the water (23 feet), the work done is asimple product of these two figures:

W =F x= (9.842 lb)(23 ft) = 226.36 ft-lb

Power is equal to work done per unit time, so:

P = W

t =

226.36 ft-lb

8 s = 28.29 ft-lb/s

Converting between ft-lb/s and horsepower:


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Answer 11

This is a trick question: both laborers do the exact same amount of work. The second laborer, however,does the same work with greater powerthan the first laborer, since the second laborers work was done in ashorter amount of time.

Answer 12

Calculating the work in raising 10 gallons of water 30 feet up:10 gal

1

231 in3

1 gal

1 ft3

1728 in3

62.4 lb

ft3

= 83.42 lb of water in 10 gallons

Now, knowing both the force (83.42 lb) and the displacement (30 ft), we may calculate the work done:

W=F x cos

W= (83.42 lb)(30 ft)(cos 0o) = 2502.5 ft-lb of work

Now, since we know that 1 horsepower is 550 ft-lbs of work per second of time, we may take this totalamount of work (2502.5 ft-lb) and divide by the pumps power in foot-pounds per second to arrive at ananswer for time in seconds. Since we know the pumps power rating is 1.5 horsepower, it means it is capableof doing 825 ft-lb of work per second:

t=

W

P =

2502.5 ft-lb

825 ft-lb/s

t= 3.033 seconds


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Answer 13

First, lets determine the weight of 3,000 gallons of water:3000 gal

1

231 in3

1 gal

1 ft3

1728 in3

62.4 lb

ft3

= 25025 lb of water in 3000 gallons

This weight of water will have to be lifted to the peak of the hill through the 100 foot pipe, but we willnot use the figure of 100 feet as the displacement, since it is not vertical. Instead, we will use trigonometryto calculate the vertical lift (x):

x= (100 ft)(sin 48o) = 74.31 ft vertical lift

So, the work involved with lifting this much water to that height is:

W=F x cos

W= (25025 lb)(74.31 ft)cos0o

W = 1, 859, 719.9 ft-lb of work

Barring any piping friction, the horizontal section of pipe (55 ft) does not necessitate any work beingdone. The short, 20 foot section of downward-angled pipe, however, actually releasesenergy (performsnegative work) because it lets the water drop in height. This drop is:

drop = (20 ft)(sin 30o

) = 10 ftSince the same amount of water will drop this amount, the negative work done here is:

W=F x cos

W= (25025 lb)(10 ft) cos 180o

W = 250, 250 ft-lb of workThe total (net) work done, then, is the sum of these two figures:

Wnet = 1, 859, 719.9 ft-lb 250, 250 ft-lb = 1, 609, 469.9 ft-lb of work

At a pump power output of 250 HP (137,500 ft-lb per second)

t= W

P

t=1609469.9 ft-lb

137500 ft-lb/s = 11.71 seconds


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Answer 14

The amount of work required to lift this crate of bowling balls 20 meters into the air is 100,000 newton-meters, or 100,000 joules. This is also the amount of potential energy the crate has at its apogee.

When dropped, the crates potential energy converts into kinetic energy, making the crate fall at a fasterand faster velocity as its height approaches ground level (0). If we neglect the effects of air friction, we may

say that the potential energy of the crate at its peak height will be precisely equal to its kinetic energy atthe moment it contacts the ground:

Ep (max.) = Ek (max)

mgh =1

2mv2

Note how the term of mass (m) cancels out of both sides of the equation, letting us know that the massof the crate will be irrelevant to its free-fall velocity:

gh =1

2v2

Solving for v given its initial height of 20 meters, and Earths acceleration of gravity being 9.81 metersper second squared:

2gh = v2

v=

2gh

v=

(2)(9.81 m/s

2)(20 m) = 19.81 m/s

Bowling balls everywhere!!!

Answer 15

v=

2gh = (2)(9.81)(5) = 9.90 m/sEp = mgh = (4)(9.81)(5) = 196.2 Joules


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Answer 16

Work done by Sam in lifting his textbook = 1040 ft-lb Kinetic energy of Sams textbook just before it hits the ground = 1040 ft-lb Velocity of Sams textbook just before it hits the ground = 91.2 ft/s

Work done by Tony in lifting his textbook = 650 ft-lb

Kinetic energy of Tonys textbook just before it hits the ground = 650 ft-lb Velocity of Tonys textbook just before it hits the ground = 91.2 ft/s

Knowing that kinetic energy just before the book hits the ground should be equal to potential energywhen released (assuming zero energy loss due to air friction), we may solve for v quite easily:

Ek =1

2mv2 Ep = mgh

12

mv2 =mgh

1

2v2 =gh

v2 = 2gh

v=

2gh

Answer 17

5.1 meters (measured vertically) at an initial velocity of 10 m/s. At 20 m/s, the truck would have gainedfour timesas much altitude (20.4 meters)!

Knowing that potential energy when the truck reaches its stopping point on the hill should be equal tokinetic energy when the clutch fails (assuming zero energy loss due to friction), we may solve for h quiteeasily:

Ep = mgh E k =1

2mv2

mgh =1

2mv2

gh =1

2v2

h=1

2v2

g

h= v2

2g

N t th t th l f th hill i ifi d b it i i l t t th f h h ti l

A 18


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Answer 18

Because we are going to have to calculate kinetic energy, we know that sooner or later we will need toknow the automobiles mass in units of slugs and the velocity in units of feet per second, so lets get thesecalculations over with now:

F =ma

m= F

a

m= 2700 lb

32.3 ft/s2

= 83.85 slugs

50 mih 1 h

3600 s5280 ft

1 mi = 73.333 ft/s

Now we are all set to calculate the automobiles initial kinetic energy:

Ek =1

2mv2

Ek =1

2(83.85 slugs)(73.33 ft/s)2

Ek = 225, 465.84 ft-lb of kinetic energy at 50 MPH

The automobiles brakes convert this kinetic into heat, through friction. In order to being the velocityto zero, the work done on the automobile by the brakes friction must cancel out the initial kinetic energy at50 MPH. Note that we will use 180o as the angle in the work calculation, because the brakes force vectoropposes the direction of the cars displacement:

0 = W+ Ek

0 = F x cos + Ek

0 = (500 lb)x cos180o + 225, 465.84 ft-lb

225, 465.84 ft-lb = (500 lb)x cos180o

x= 225, 465.84 ft-lb

(500 lb)(cos 180o)

x=225, 465.84 ft-lb

500 lb

x= 450.93 ft total braking distance

A s 19


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Answer 19

The truck can stop in a shorter distance than the sports car (about half the stopping distance). Thereason for this is due to how much kinetic energy each vehicle possesses as it travels down the highway. Masshas a linear effect on kinetic energy, but velocity has a square effect on kinetic energy. Thus, while doublinga vehicles mass will double its Ek at any given velocity, doubling a vehicles velocity will quadruple its Ekfor any given mass.

A helpful thought experiment to try here is to imagine a vehicle traveling at the slow speed of the pickuptruck, and with the light mass of the sports car, and then use this vehicle as the standard of comparison.The truck will have twice the kinetic energy of this third vehicle. The sports car will have four times asmuch kinetic energy as this third vehicle. Therefore, the sports car will have twice the kinetic energy of thetruck.

Answer 20

As the wrecking ball is pulled sideways until its cable forms a 20 degree angle from vertical, it will be

lifted up 1.809 feet from the height it started (a right triangle with a hypotenuse of 30 feet and an adjacentangle of 20 degrees will have a side length of 28.19 feet, or 1.809 feet less than 30 feet).

From this known amount of vertical lift, we may calculate the wrecking balls maximum velocity byassuming all that potential energy gets converted into kinetic energy:

1

2mv2 =mgh

1

2v2 =gh

v2 = 2gh

v=

2gh

v= (2)(32.2 ft/s

2)(1.809 ft)

v= 10.79 ft/s

Answer 21


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Answer 21

When the bullet is fired upward at ground level, its kinetic energy is at a maximum but its potentialenergy is near zero. In rising against gravity, that kinetic energy will be translated into potential energy untilthe bullets apogee, where there will be zero kinetic energy and maximum potential energy. As it falls, thereverse will happen: potential energy will translate into kinetic energy again, resulting in the bullet travelingat 1100 feet per second when it returns to ground level.

The time for the bullet to rise to its apogee will be equal to the time required for it to fall back toground level from that maximum altitude. So, if we were able to calculate that rise time (or that fall time),all we would have to do is multiply by two to obtain the total time spent in the air.

The attraction of Earths gravity results in a downward acceleration of 32.2 feet per second squared.This means that any object dropped from altitude will accelerate at this rate, increasing in velocity by 32.2ft/s every second. It also means that any object rising by its own inertia (no propulsive force upward) willslow down by 32.2 ft/s each and every second until its vertical velocity is zero and it begins to fall back

down. This rate of acceleration holds true for anyobject, aerodynamic resistance notwithstanding.

For this rate of acceleration to increase a falling objects velocity from 0 ft/s to 1100 ft/s will requirethis much time:

v= at

t= v

a

t= 1100 ft/s

32.2 ft/s2

= 34.16 seconds

This is the time required to either accelerate the bullet from zero ft/s (at apogee) to 1100 ft/s downward,or to decelerate it from 1100 ft/s upward to 0 ft/s (at apogee). Doubling this figure gives us 68.32 secondstotal time. In other words, the stupid person has a little over a minute to duck and cover before the bulletfinds its way back to terra firma.

Answer 22


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Answer 22

The torque () in this case is obviously 550 lb-ft, since the 550 pound weight is acting on a momentarm 1 foot long (the drums radius). All we need to do is translate the vertical velocity of 1 foot per secondinto drum rotation in units of RPM, and well have the data we need to calculate k :

Circumference of drum =D= 2r = 6.283 ftThis is the amount of cable that travels in one revolution of the drum (1 rev = 6.283 ft), and this

equality constitutes a conversion factor which we may use to convert the linear velocity of 1 ft/s into arotational velocity:

1 ft

sec

1 rev

6.283 ft

60 sec

1 min

= 9.5493 RPM

Therefore,

P =k S

1 hp = (k)(550 ft-lb)(9.5493 RPM)

k= 0.0001904

P = 0.0001904 S

. . . or . . .

P= S

5252

Where,P= Shaft power in horsepower= Shaft torque in lb-ftS= Shaft speed in revolutions per minute (RPM)

By coincidence, the factor of 5252 happens to be close to the number of feet in a mile (5280 feet = 1mile). This might come in handy as an approximation!

Answer 23

One gallon of gasoline burned over a period of 2 hours = 16.8 kW

The battery charger outputs a total of 875.16 kJ over the 2.5 hour time period, at a rate of 97.24 watts.

The power plant must burn coal at a rate of 3.061 metric tons per minute.

Data for the heat value of various fuels taken from the following web document:http://bioenergy.ornl.gov/papers/misc/energy conv.html

Simple machines


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Metric prefixes and conversion constants


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Metric prefixes Yotta = 1024 Symbol: Y Zeta = 1021 Symbol: Z Exa = 1018 Symbol: E Peta = 1015 Symbol: P

Tera = 1012

Symbol: T Giga = 109 Symbol: G Mega = 106 Symbol: M Kilo = 103 Symbol: k Hecto = 102 Symbol: h Deca = 101 Symbol: da Deci = 101 Symbol: d Centi = 102 Symbol: c Milli = 103 Symbol: m

Micro = 106 Symbol: Nano = 109 Symbol: n Pico = 1012 Symbol: p Femto = 1015 Symbol: f Atto = 1018 Symbol: a Zepto = 1021 Symbol: z Yocto = 1024 Symbol: y

100

103

106

109

1012

10-3

10-6

10-9

10-12

(none)kilomegagigatera milli micro nano picokMGT m n p

10-2

10-1

101

102

deci centidecahectoh da d c

METRIC PREFIX SCALE

Conversion formulae for temperature

oF = (oC)(9/5) + 32

oC = (oF - 32)(5/9)

oR = oF + 459.67

K = oC + 273.15

Conversion equivalencies for distance

1 inch (in) = 2.540000 centimeter (cm)

1 foot (ft) = 12 inches (in)

Conversion equivalencies for volume

1 gallon (gal) = 231 0 cubic inches (in3) = 4 quarts (qt) = 8 pints (pt) = 128 fluid ounces (fl oz )


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1 gallon (gal) = 231.0 cubic inches (in ) = 4 quarts (qt) = 8 pints (pt) = 128 fluid ounces (fl. oz.)= 3.7854 liters (l)

1 milliliter (ml) = 1 cubic centimeter (cm3)

Conversion equivalencies for velocity1 mile per hour (mi/h) = 88 feet per minute (ft/m) = 1.46667 feet per second (ft/s) = 1.60934kilometer per hour (km/h) = 0.44704 meter per second (m/s) = 0.868976 knot (knot international)

Conversion equivalencies for mass

1 pound (lbm) = 0.45359 kilogram (kg) = 0.031081 slugs

Conversion equivalencies for force

1 pound-force (lbf) = 4.44822 newton (N)

Conversion equivalencies for area

1 acre = 43560 square feet (ft2) = 4840 square yards (yd2) = 4046.86 square meters (m2)

Conversion equivalencies for common pressure units (either all gauge or all absolute)

1 pound per square inch (PSI) = 2.03602 inches of mercury (in. Hg) = 27.6799 inches of water (in.W.C.) = 6.894757 kilo-pascals (kPa) = 0.06894757 bar

1 bar = 100 kilo-pascals (kPa) = 14.504 pounds per square inch (PSI)

Conversion equivalencies for absolute pressure units (only)

1 atmosphere (Atm) = 14.7 pounds per square inch absolute (PSIA) = 101.325 kilo-pascals absolute(kPaA) = 1.01325 bar (bar) = 760 millimeters of mercury absolute (mmHgA) = 760 torr (torr)

Conversion equivalencies for energy or work

1 british thermal unit (Btu International Table) = 251.996 calories (cal International Table)= 1055.06 joules (J) = 1055.06 watt-seconds (W-s) = 0.293071 watt-hour (W-hr) = 1.05506 x 1010

ergs (erg) = 778.169 foot-pound-force (ft-lbf)

Conversion equivalencies for power

1 horsepower (hp 550 ft-lbf/s) = 745.7 watts (W) = 2544.43 british thermal units per hour(Btu/hr) = 0.0760181 boiler horsepower (hp boiler)

( )

Physical constants

Speed of light in a vacuum (c) = 2 9979 108 meters per second (m/s) = 186 281 miles per second


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Speed of light in a vacuum (c) = 2.9979 10 meters per second (m/s) = 186,281 miles per second(mi/s)

Avogadros number (NA) = 6.022 1023 per mole (mol1)

Electronic charge (e) = 1.602 1019 Coulomb (C)

Boltzmanns constant (k) = 1.38 1023 Joules per Kelvin (J/K)

Stefan-Boltzmann constant () = 5.67 108 Watts per square meter-Kelvin4 (W/m2K4)

Molar gas constant (R) = 8.314 Joules per mole-Kelvin (J/mol-K)

Properties of Water

Freezing point at sea level = 32oF = 0oC

Boiling point at sea level = 212oF = 100oC

Density of water at 4oC = 1000 kg/m3 = 1 g/cm3 = 1 kg/liter = 62.428 lb/ft3 = 1.94 slugs/ft3

Specific heat of water at 14oC = 1.00002 calories/goC = 1 BTU/lboF = 4.1869 Joules/goC

Specific heat of ice 0.5 calories/goC

Specific heat of steam 0.48 calories/goC

Absolute viscosity of water at 20oC = 1.0019 centipoise (cp) = 0.0010019 Pascal-seconds (Pas)

Surface tension of water (in contact with air) at 18oC = 73.05 dynes/cm

pH of pure water at 25o C = 7.0 (pH scale = 0 to 14)

Properties of Dry Air at sea level

Density of dry air at 20oC and 760 torr = 1.204 mg/cm3 = 1.204 kg/m3 = 0.075 lb/ft3 = 0.00235

slugs/ft3

Absolute viscosity of dry air at 20oC and 760 torr = 0.018 centipoise (cp) = 1.8 105 Pascal-seconds (Pas)

Questions

Question 1


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Question 1

Identify what classeach of these levers is:

Cable cutter

Laboratoryscale

Crane

Hydrauliccylinder

file i02617

Question 2

Calculate the mechanical advantageof this lever:

Lever

4 ft12 ft

Mass

file i02618

Question 3

Calculate the mechanical advantage for this lever system:

F

3 ft 4 ft

3 ft 5 ft

Question 4

/


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The piston on a hydraulic bottle jack lifts up 1/8 of an inch for every stroke of the lever. In eachlever stroke, the handle end moves 14 inches. Calculate the mechanical advantageof this hydraulic jack:

14 inches

1/8inch

file i02625

Question 5A screw jack has a thread pitch of 4 threads per inch, and is turned by a handle 1.5 feet long (measured

from the screw center to the handles end). Calculate the mechanical advantageof this screw jack:

1.5 ft

4 t.p.i.

file i02626

Question 6

H h t i i th i th h it tt h t th ili ?


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How much tension is there in the rope where it attaches to the ceiling?

F

Scale(registers 8 pounds

of pulling force)

Mass

Pulley

Rope

Also, calculate how much this suspended object weighs in units of pounds, and how much mass it hasin units of slugs.

Finally, calculate the mechanical advantage of this pulley system.file i02627

Question 7

If the pulling force exerted on the end of the rope in this pulley system is 8 pounds how much upward


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If the pulling force exerted on the end of the rope in this pulley system is 8 pounds, how much upwardpulling force is exerted on the mass by the lower pulley?

Scale(registers 8 pounds

of pulling force)

Mass

F

Calculate the amount of work done in lifting the mass 3 feet while pulling on the rope with 8 pounds offorce.

file i02628

Question 8

If the free end of the rope is pulled a distance of 10 cm how far will the mass be lifted?


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If the free end of the rope is pulled a distance of 10 cm, how far will the mass be lifted?

Mass

Assuming the mass is 55 kg, how much work is done while pulling the rope 10 cm?file i02629

Question 9

Determine a way to arrange pulleys to provide a 3:1 mechanical advantage.file i02630

Question 10

Calculate the mechanical advantage (MA) for each of these pulley systems:


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Calculate the mechanical advantage (MA) for each of these pulley systems:

Mass

Mass

file i02824

Question 11

Suppose a mechanic pulls perpendicularly at the end of a wrench 0.6 meters in length with a steadyforce of 190 newtons for two complete revolutions. Calculate the amount of work done by the mechanic (innewton-meters or joules), and also calculate his power output in watts if those two turns were completed in6 seconds.

BoltWrench

F = 190 Newtons

length=0.6meters

file i03777

Question 12

How much linear force will the cars tire exert on the ground if the axle exerts a torque of 1500 lb-ft on


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g qthe wheel, and the tires radius is 11 inches?

11 inches

= 1500 lb-ft

file i01402

Question 13

Calculate the net torque applied to the drum from the two forces shown. The drums outside radius is


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6 feet, and the radius of the smaller pulley (attached to the drum) is 2 feet:

Drum

Axis ofrotation

r2= 6 feet

r1= 2 feet

F1= 4 lb

F2= 3.2 lb

Also, calculate the mechanical advantage of this system, ifF1 is considered the inputforce.file i01428

Question 14

Suppose an electric actuator is used to lift a large concrete gate in an irrigation water flow control


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facility. The gate effectively acts as a control valve for water flowing through an open irrigation channel,and a powerful winch is necessary to control its position:

WaterGate

Channel wall

Channel wall

Winch drum

Motor/gearbox

The winch drum measures 20 inches in diameter, and the concrete gate weighs 12,740 pounds. Calculatethe torque required at the drum to lift the gate, and also the torque required by the electric motor given agearbox speed-reduction ratio of 1200:1.

Assuming the electric motor powering this speed-reducing gearbox spins at 1720 RPM (at full load),calculate the vertical lifting speed of the gate in feet per minute. Finally, calculate the horsepower outputof the electric motor lifting this much weight (12,740 pounds) at this vertical speed.


Like all story problems involving mathematical calculation, the most important aspect of your answeris howyou arrived at it, not the numerical value(s) of your answer. Explain how you were able to setup the proper equations to solve for drum torque, motor torque, lifting speed, and motor output power.

A useful problem-solving technique is to sketch a simple diagram of the system you are asked to analyze.This is useful even when you already have some graphical representation of the problem given to you, asa simple sketch often reduces the complexity of the problem so that you can solve it more easily. Drawyour own sketch showing how the given information in this problem inter-relates, and use this sketch toexplain your solution.

file i00584

Question 15

The General Electric Frame 6 gas turbine engine is a popular choice for natural gas powered electricali h U i d S i h f i l 4 Th bi i lf


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generators in the United States, with an output power of approximately 42 megawatts. The turbine itselfspins at 5163 RPM, but the 2-pole AC generator must spin at a different speed in order to generate 60 Hzelectrical power.

Calculate the gear ratio necessary to couple the turbines shaft to the generators shaft, and identifywhether this gear set needs to reducespeed or increasespeed.

file i04790

Question 16

The Siemens model SGT-800 gas turbine engine has an output power of approximately 47 megawatts.The turbine itself spins at 6608 RPM, but the 2-pole AC generator must spin at a different speed in orderto generate 50 Hz electrical power for European power markets.

Calculate the gear ratio necessary to couple the turbines shaft to the generators shaft, and identifywhether this gear set needs to reducespeed or increasespeed.

file i04789

Question 17

Roy has the meanest pulling tractor in his county: its engine outputs a maximum torque of 1200 lb-ft,and the total geartrain (transmission combined with rear axle differential gearing) has a 12:1 reduction ratioin the lowest gear. With 5.5 foot tall tires, how much horizontal pulling force can this tractor (theoretically)exert?

If Roys tractor drags a weight 300 feet along the ground while pulling at maximum engine torque, howmuch work was done by the tractor?

Rate the horsepower of Roys tractor if it took exactly 1 minute to drag that weight 300 feet along theground.

When Roy goes to the county fair to compete in the tractor-pull contest, he notices that the front endof the tractor tends to raise up off the ground when pulling a heavy load. Explain to Roy why this happens.

file i01429

Question 18

Calculate the RPM of the engine in a truck as it drives down the highway at 70 MPH with 30-inch tallwheels and an axle gear speed reduction ratio of 3.55:1, in top gear (a transmission ratio of 1:1).

Next, calculate the engines RPM at the same highway speed if the truck is equipped with an overdrivegear in the transmission (turning the axle driveshaft faster than the engine is turning), having an effectiveratio of 0.7:1.

file i04791

Question 19

Electric motors usually rotate at too high of speed to be used directly as valve actuators. Nearlyll l t i l t t h i t d th d f th l t i t ( d lti l


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all electric valve actuators use gear mechanisms to reduce the speed of the electric motor (and multiplyits torque). One of the more popular gear mechanisms for achieving great speed reduction (and torquemultiplication) is called the worm gear:

Worm gear mechanism

Worm screw

Worm wheel

The worm wheels teeth match the pitch of the threads on the worm screw, allowing the two pieces tomesh like gears. It should be evident from inspection that it takes many, many turns of the work screw toobtain one revolution of the work wheel. In electric valve actuators, the motor couples to the worm screwand the wheel turns the valve mechanism.

What might not be so evident is how torque on the worm wheel directly translates to linear thruston the worm screw. In other words, the more twisting force output by the worm wheel, the greater thestraight-line force experienced by the screw:

Torque

Thrust

If we can find a way to measure this linear thrust on the worm screw, we may infer the torque outputby the wheel. Explain how this could be done in an electric valve actuator mechanism.

file i01390

Answers

Answer 1

1st class = laboratory scale


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1st class = laboratory scale 2nd class = cable cutter 3rd class = crane

Answer 2For any lever regardless of class the mechanical advantage (MA) is always calculated as the ratio

of radii between the input and output points on the lever (i.e. the ratio of lengths also called momentarms between each point and the fulcrum).

In this case, since the ratio of moment arms lengths is 12:4 (3:1), the mechanical advantage will be 3:1as well. The person only needs to exert 1/3 the weight of the mass on the right-hand end of the lever inorder to lift the mass, but the motionat that end of the lever will be three times as much as the motion ofthe mass.

Answer 3The first lever (where force F works on the right-hand end) is a first-class, with a mechanical advantage of

4:3. It connects to a second lever (third-class), with an mechanical advantage ratio (actually, a disadvantageratio) of 5:8. The overall mechanical advantage is the product of these two advantage ratios:

MA=

4

3

5

8

=

5

6= 0.83333

In other words, for every pound of force applied at F, there will be 0.8333 pounds of force available tomove the mass.

Answer 4

The mechanical advantage of any machine may be empirically determined by dividing input displacementby output displacement:

MA= sin

sout

In this case, an input displacement of 14 inches yields an output displacement of 0.125 inches, so:

MA= 14 in0.125 in

= 112

This means the handle tip moves 112 times farther than the jacks lifting piston, but the lifting pistonexerts 112 times more force than it takes to move the handle.

Answer 5

For each turn of the screw (or nut), the jack will lift 1/4 of an inch. We know this because there arefour threads per inch which means four complete turns are required to lift one inch Now all we need to do


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four threads per inch, which means four complete turns are required to lift one inch. Now all we need to dois calculate how far the tip of the handle travels in one turn, and we have the necessary data to calculatemechanical advantage (MA=

sin

sout).

Given a radius of 1.5 feet (18 inches), the circumference of the circle described by one full rotation ofthe handle will be 113.1 inches according to the formula C=D= 2r. Thus, with an input displacementof 113.1 inches and an output displacement of 0.25 inch, the mechanica advantage must be:

MA=113.1 in

0.25 in = 452.4

This means the handle tip moves 452.4 times farther than the jacks lifting screw, but the lifting screwexerts 452.4 times more force than it takes to move the handle.

Answer 6

Since the scale on the right-hand end of the rope registers a tension of 8 pounds, the tension at theother end of the rope must be 8 pounds as well, not counting any friction in the pulley. All the pulley doesis redirectthe force pulling on the rope.

Since the object is being supported by the tension in two rope lengths, its weight must be twice thetension:

W= (2)(8 lb) = 16 lb

The relationship between mass and weight is the basic F =ma formula where force Fis the weight ofthe object, m is its mass, and a is the acceleration of gravity:

m=F

a =

16 lb

32.2 ft/s2

= 0.4969 slugs

The mechanical advantage of any machine is (ideally) the ratio between output force and input force.Since in this case the output force is the 16 lb weight and the input force is the 8 lb rope tension, the

calculation looks like this:

MA=Fout

Fin

MA=16 lb

8 lb = 2

In other words, for every pound of force applied on the rope, there will be 2 pounds of force availableto move the mass.

Answer 7

With 8 pounds of tension in the cable, and fourcables pulling upward on the lower pulley assembly, thetotal upward force exerted on the mass will be four times the tension, or 32 pounds.


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total upward force exerted on the mass will be four times the tension, or 32 pounds.

With a mechanical advantage of 4:1, the mass moves 14

the distance that the rope is pulled. So, if themass moves 3 feet, the rope must be pulled 12 feet. This gives the following values for work done (calculatedeither at the mass or at the ropes end, yields the same result):

W=F x= (8 lb)(12 ft) = 96 ft-lb (calculated at ropes end)

W =F x= (32 lb)(3 ft) = 96 ft-lb (calculated at mass)

Answer 8

Seeing that the lower pulley assembly is supported by fourlengths of rope, the mechanical advantage inthis system must be 4:1. Thus, the output displacement will be four times less than the input displacement:

MA= xin

xout

xout = xin

MA

xout =10 cm

4 = 2.5 cm

Expressing the same result in meters instead of centimeters:

xout=0.1 m

4 = 0.025 m

A mass of 55 kg weighs 539.99 newtons in Earths gravity (F = ma, where Fis the force that gravityexerts on the mass, m is the amount of mass, and a is the acceleration of Earth gravity: 9.81 meters persecond squared). This is the amount of force exerted upward on the mass by the pulley system. Given our

mechanical advantage of 4:1, it means the ropes tension at the pulled end must be

1

4 this value, or 134.89newtons.

We may calculate the amount of work done at the mass or at the ropes end. Either way, we will getthe exact same result:

W =F x= (134.89 N)(0.1 m) = 13.489 N-m (calculated at ropes end)

W =F x= (539.55 N)(0.025 m) = 13.489 N-m (calculated at mass)

Answer 9


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Mass

Answer 10

MA = 2:1 (on left) MA = 3:1 (on right)

Answer 11

The mechanics work is 1432 Newton-meters, or 1432 Joules. The mechanics average power outputduring the 6 seconds is 238.7 watts.

Answer 12

F= 1636.36 lb

( )


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To solve for force, we simply need to manipulate the torque equation so that force (F) is by itself onone side of the equality sign:

=r F

F =

r

Since we happen to know in this problem that all three vectors are orthogonal (perpendicular) to eachother, we may re-write the equation in simpler terms of scalar quantities instead of vector quantities:

F =

r

Before we may insert the given values for torque and moment arm length, we need to convert units oflength for the moment arm:

(11 inches)(1 foot / 12 inches) = 0.916667 feet

Now, solving for force:

F= 1500 lb-ft

0.916667 ft

F= 1636.36 lb

Answer 13

net = 11.2 lb-ft, clockwise

MA=Fout

Fin=

3.2 lb

4 lb = 0.8

Answer 14

Partial answer:

motor = 8.8472 lb-ft

Motor output = 2.897 horsepower

Answer 15

A 60 Hz, 2-pole generator must spin at 3600 RPM exactly (60 revolutions per second). We thereforerequire a reductiongear set between the turbine and the generator with a gear ratio of 1.434:1 ( 5163

3600).

Answer 16

A 50 Hz, 2-pole generator must spin at 3000 RPM exactly (50 revolutions per second). We therefore

Answer 17

Maximum pulling force = 5236.36 pounds


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W =F x= (5236.36 lb)(300 ft) = 1570909.1 ft-lb

1 horsepower is 550 ft-lb of work done per second. If Roys tractor did 1,570,909.1 ft-lb of work in 60seconds, it is equivalent to 26,181.8 ft-lb/s of power, which is 47.6 horsepower.

As the tractor mechanism exerts torque on the wheels, and the weight of the load opposes the wheelsturning, the tractor experiences this torque about the axis of rotation: the axles. As the wheels tryto rotatein a forward direction, but are impeded by the resistance of the load, the reaction torque tries to rotate thetractor backward about the same axis. This manifests itself in the form of the front tires of the tractor liftingoff the ground.

Answer 18

Each revolution of the trucks 30-inch diameter wheels is equivalent to 30 inches of linear motion alongthe highway. Converting 70 miles per hour into inches per minute, and then from linear inches per minuteinto revolutions per minute, gives us the rotational speed of the wheels:

70 miles

hour

5280 feet

1 mile

12 inches

1 foot

1 rev

30 inches

1 hour

60 minutes

= 784.3 RPM

Since the axles gear ratio is a speed reducerfrom driveshaft to wheels, it acts as a speed increaserfromwheels to driveshaft. Thus, with a ratio of 3.55:1, the driveshaft will spin at a speed of 2784.3 RPM.

With a transmission ratio of 1:1, this means the engine will also spin at 2784.3 RPM.

With an overdrive transmission ratio of 0.7:1, this means the engine will also spin at 1949.0 RPM.

Answer 19

One way to measure worm screw thrust force is with a load cell. Another way is to spring-load the screwshaft and use an LVDT or other motion-sensing device to measure displacement.

Static fluids

This worksheet and all related files are licensed under the Creative Commons Attribution License,version 1.0. To view a copy of this license, visit http://creativecommons.org/licenses/by/1.0/, or send a


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py , p // g/ / y/ /,letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA. The terms andconditions of this license allow for free copying, distribution, and/or modification of all licensed works bythe general public.

This worksheet introduces the basic concepts of gases and liquids.

Questions

Question 1

Suppose we were to steadily pour a liquid into the leftmost vertical tube until it reaches a mark fourinches from the bottom. Given the diameters of the other tubes, how high will the liquid level settle in each


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inches from the bottom. Given the diameters of the other tubes, how high will the liquid level settle in eachwhen all columns are in a condition of equilibrium (no liquid flowingthrough any part of the system)?

1 inch

2inches

1/2inch

inches

2-1/2

inches

1-1/2

inches1-

1/

2

4

inches inches1-

1/

2

Now consider the same set of vertical tubes (same diameters, same step heights) connected at the

bottom by an inclined pipe. If we were to pour a liquid into the leftmost vertical tube until it reaches amark two inches from its bottom, how high will the liquid level settle in each column when all columns arein a condition of equilibrium?

inches2

5

inches

Question 2

Which of these tubes will generate the most hydrostatic pressure, assuming they all contain the sametype of liquid at precisely the same (vertical) height?


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? ? ?

file i02953

Question 3

Toluene has a density of 0.8669 g/cm3 at 20o C. Calculate its density in units of pounds per cubic feetand its specific gravity (unitless).

file i00233

Question 4

A liquid has a density of 865 kilograms per cubic meter. Calculate its specific gravity.

A liquid has a density of 59 pounds per cubic foot. Calculate its specific gravity.

file i00237

Question 5

If force is exerted on the piston of this hydraulic cylinder, in what direction(s) will this force betransmitted to the cylinder walls? In other words, how does a fluid under pressure push against itssurrounding container?


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surrounding container?

Piston

Rod

Fluid

Force

Hydraulic cylinder

Steel cylinderwall

Steel cylinderwall

file i00142

Question 6

Suppose a small rubber ball is floating inside the fluid of a hydraulic cylinder as shown below. Whatwill happen to the ball when a pushing force is exerted on the cylinders rod? What will happen to the ballwhen a pulling force is exerted on the rod?

Rubber ball

file i00143

Question 7

A surface-mounted water pump pulls water out of a well by creating a vacuum, though it might be moretechnically accurate to say that the pump works by reducing pressure in the inlet pipe to a level less thanatmospheric pressure, allowing atmospheric pressure to then push water from the well up the pumps inlet


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atmospheric pressure, allowing atmospheric pressure to then push water from the well up the pump s inletpipe:

Pump

Water

Atmospheric

pressure

Based on this description of pump operation, what is the theoretical maximum height that any pumpcan lift water out of a well, assuming the well is located at sea level?

Water wells located at altitudes other than sea level will have different theoretical maximum lifting

heights (i.e. the farthest distance a surface-mounted pump may suck water out of the well). Research theaverage barometric pressure in Denver, Colorado (the mile-high city) and determine how far up a surfacepump may draw water from a well in Denver.

Domestic water wells may be hundreds of feet deep. How can water be pumped out of wells this deep,given the height limitation of vacuum pumping?


If the liquid in question was something other than water, would the maximum lift depth be different?Why or why not?

file i00147

Question 8

Water pressure available at a fire hydrant is 80 PSI. If a fire hose is connected to the hydrant and thehydrant valve opened, how high can the end of the hose be raised and still have water flow out the end?


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80 PSI

How high???

Now, suppose that a spray nozzle attached to the end of the hose requires at least 30 PSI of pressureat the coupling in order to create a proper spray of water. How high can the hose be raised then, and stillhave enough water pressure at the nozzle to allow for the fighting of a fire?

80 PSI

How high???

required hereAt least 30 PSI

file i00148

Question 9

Explain how a vertical height of liquid is able to create pressure, such as in this example:


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Swimming pool

Pressure registered

by the gauge

20 feet

6 feet

No pressureregistered by

the gauge

The deeper you descend into the water, the more pressure there is.

Recall that pressure is defined as force divided by area:

P =F

A

Calculate the total weight of the water contained in this swimming pool (assuming the pool is circular

in shape, the 20-foot dimension being its diameter), and use this figure to calculate pressure at the bottomof the pool, knowing that pressure is defined as force exerted over an area. Remember that the density ofwater is 62.428 lb/ft3.

Weight of water = lbs

Pressure at bottom of pool = PSI

file i00749

Question 10

Specific gravityis defined as the ratio of densities between a particular fluid and a reference fluid. Forliquids, the reference fluid is water; for gases, the reference fluid is air.

For example, the density of olive oil is 57.3 lb/ft3 and the density of water is 62.4 lb/ft3. Calculating


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the ratio of these two densities yields the specific gravity of olive oil: 0.918. That is to say, the density ofolive oil is 91.8% that of water.

A useful definition of specific gravity when performing hydrostatic pressure calculations for variousliquids is the ratio of equivalent water column height to the height of a particular liquid. Using the specificgravity of olive oil (0.918) as an example, we could say that 0.918 units of water column height will generatethe same hydrostatic pressure as 1 unit of olive oil height. The unit could be inches, centimeters,millimeters, cubits, or anything else:

0.918 unit W.C. pressure = 1 unit olive oil pressure

We may make a unity fraction from this equality, since we are dealing with two physically equal

quantities: the amount of hydrostatic pressure generated by two vertical columns of different liquids.0.918 unit W.C.

1 unit olive oil = unity

Apply this unity fraction to the calculation of hydrostatic pressure at the bottom of a 20 foot tallstorage tank filled to the top with olive oil, expressing that pressure in units of kPa. Show how the unitscancel in your calculation(s), beginning with feet of olive oil and ending in kilo-Pascals (kPa):

20 ft

Olive oil

P = ??? kPa

Gf= 0.918

(vent)

file i02956

Question 11

A vessel contains three different liquids of different specific gravities: glycerine, water, and olive oil.These three liquids settle at different levels in the vessel, so that there is a 3 foot deep layer of glycerine, a2 foot deep layer of water, and a 4.5 foot deep layer of olive oil:


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Glycerine

Water

Olive oil

Pressure = ???

3 feet

2 feet

4.5 feet

Calculate the total hydrostatic pressure at the bottom of the vessel, in units of PSI and kPa.file i00235

Question 12

A device called a manometer is a very simple and yet very precise pressure measuring instrument. Itworks on the principle of a differential pressure displacing a vertical liquid column. The distance betweenthe tops of the two liquid columns is proportional to the difference in pressure applied to tops of thetwo vertical tubes. This is where we get pressure units of inches/centimeters of water column and

inches/centimeters/millimeters of mercury from the operation of a manometer:

Appliedpressure

(greater)

Appliedpressure

(lesser)

HeadTransparenttube allows

liquid columnsto be seen

Manometer

Explain how this instrument may serve as a standard for pressure measurement just as a deadweight

Question 13

How much pressure is being applied to this U-tube water manometer, in units of inches of watercolumn and pounds per square inch?

Applied


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(vented)Appliedpressure

4.5"

4.5"

Water

Water levelat zero pressure

What would happen to the liquid levels if the water were replaced by an oil with a lesser density? Giventhe same applied pressure, would the distance between the two liquid columns be greater, less, or the sameas shown in the above illustration?

file i00161

Question 14

How much pressure, in units of inches of water column, is being applied to the right-hand tube of thisU-tube water manometer?

Appliedpressure

Water


= 20 "W.C.

Appliedpressure= ???

3.25"

3.25"

Also, convert this pressure into units of Pascals.


How much differential pressure is registered by this manometer?

Question 15

How much pressure is being applied to this U-tube water manometer, in units of inches of watercolumn and pounds per square inch?

( t d)Applied


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(vented)Appliedpressure

Water


1" diameter

4"diameter

3"0.1875"

file i00162

Question 16

Calculate the height of glycerine (= 78.6 lb/ft3) in a vertical tube if there is 21 PSI of hydrostaticpressure at the bottom of the tube:

Tube


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Tube

Glycerine

P = 21 PSI

Pressure gauge

Height = ?

Glycerine height = ft

Also, calculate the height of castor oil (= 60.5 lb/ft3) necessary to generate the exact same amountof pressure:

Castor oil height = ftfile i02951

Question 17

The following illustration shows a rather strange manometer, one with two different liquids inside,coupled by a hand valve:


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GlycerineWater

If the two columns of liquid are just right, they will remain at their respective (different) heights whenthe valve is opened. Unlike a normal manometer where the two liquid columns always equalize to the sameheight when vented, this manometer is content to rest at different heights. Explain why.

Also, calculate two possible heights that will balance each other, given the liquids of water and glycerine.file i02952

Question 18

How much pressure, in inches of water column, is being applied to this inclined water manometer todisplace water 5 inches along the length of the tube, inclined at an angle of 30 o from horizontal? Assume anegligible change in liquid level inside the well throughout the measurement range of the instrument:

water

Appliedpressure

(vented)

5"

30oWell

Question 19

A tube containing a 10 foot long column of water is angled 40o from horizontal. Calculate the hydrostaticpressure at the bottom of this tube in units of inches water column (W.C.) and also in units of atmospheres.


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10 feet

40o

Pressure = ???

Water

file i00234

Question 20

A simple way to make a micromanometer(an extremely sensitive manometer) is to connect two large-diameter vertical tubes by a small-diameter, transparent tube with an air bubble in it. The air bubblebecomes the marker for reading pressure along a scale:

bubble

air

Scale

A simple micromanometer

Water

If both of the large vertical tubes are 2.5 inches in diameter, and the transparent, horizontal tube is0.25 inches in diameter, how much differential pressure will be indicated by 1 inch of horizontal bubble

displacement? Assume the use of water for the manometer liquid.file i00169

Question 21

Determine what will happen at the following steps in the sequence (when prompted for a response) inthis pressure transmitter calibration setup:

Manometer 12


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Handpump

Pressurevessel

under test

Vent

3

Vent

4HL

Instrument

Step 1: Open valves 1 and 2

Step 2: Close valves 3 and 4

Step 3: Operate hand pump until manometer registers maximum pressure

Step 4: (4 points) Quickly open and close valve 4 does the manometer indication drop greatly,slightly, or not at all?

Step 5: Close valve 2


Step 7: Close valve 1


file i00463

Question 22

Complete the following table of equivalent pressures:

PSIG PSIA inches Hg (G) inches W.C. (G)18


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18400

3360

45212

1-5

There is a technique for converting between different units of measurement called unity fractionswhich is imperative for students of Instrumentation to master. For more information on the unity fractionmethod of unit conversion, refer to the Unity Fractions subsection of the Unit Conversions and PhysicalConstants section of the Physics chapter in your Lessons In Industrial Instrumentationtextbook.

file i02938

Question 23

Complete the following table of equivalent pressures:

Atm PSIG inches W.C. (G) PSIA

3.581

88340

7.12368

2100

There is a technique for converting between different units of measurement called unity fractionswhich is imperative for students of Instrumentation to master. For more information on the unity fractionmethod of unit conversion, refer to the Unity Fractions subsection of the Unit Conversions and PhysicalConstants section of the Physics chapter in your Lessons In Industrial Instrumentationtextbook.

file i02939

Question 24

Explain what is wrong with this attempt to convert a gauge pressure of 65 PSI into units of atmospheres(atm):

65 PSI

1

1 atm

14 7 PSI

= 4.422 atm


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1

14.7 PSI


The mistake made here is common for new students to make as they learn to do pressure unit conversions.Identify a sure-fire way to identify and avoid this mistake.

file i02940

Question 25

2.036 inches of mercury (Hg) is an equivalent pressure to 27.68 inches of water (W.C. or H2O). Thisfact allows us to create a unity fraction from these two quantities for use in converting pressure units frominches mercury to inches water or vice-versa. Two examples are shown here:

310 Hg

1

27.68 W.C.

2.036 Hg

= 4215 W.C.

45 W.C.

1

2.036 Hg

27.68 W.C.

= 3.31 Hg

But what if we are performing a unit conversion where the initial pressure is given in inches of mercuryor inches of water absolute? Can we properly make a unity fraction with the quantities 2.036 HgA and27.68 W.C.A as in the following examples?

310 HgA

1

27.68 W.C.A

2.036 HgA

= 4215 W.C.A

45 W.C.A

1

2.036 HgA

27.68 W.C.A

= 3.31 HgA

Explain why or why not.file i02942

Question 26

A pressure transmitter with a remote seal measures the pressure of a gas inside a process vessel. Apressure gauge directly attached to the vessel registers 19.3 PSI. The transmitter is located 22 feet 5 inchesbelow this point, with a capillary tube filled with fluid having a specific gravity of 0.94:


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PG

19.3 PSI

PT

Capilla

rytube

22 5"

Fill fluid S.G. = 0.94

Vessel

How much pressure will the transmitter register?file i00240

Question 27

A business owns a large storage tank which was used to hold water for fire protection. This tank isequipped with a pressure gauge at the bottom to infer water level. The face of the gauge reads out in feetof water rather than PSI or some other common pressure unit:


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"Empty"(0% level)

"Full"(100% level)

30 feet

Pressure(level)gauge

(vent)

Water

0 ft 10 ft

20 ft30 ft

The operation of this level-indicating pressure gauge is quite simple: as the water level changes in thetank, the amount of hydrostatic pressure generated at the bottom changes proportionally.

When the local municipality upgrades the size of the water supply line to the company property, thereis no longer a need for the fire-water storage tank. Not wanting to abandon the tank, a manager at thecompany decides to use it for gasoline fuel storage instead.

After emptying the water and re-filling the tank with gasoline, however, they notice a problem with thelevel-indicating gauge: it no longer reads correctly. With gasoline in the tank instead of water, the gaugesreading no longer correlates with tape-measure readings of liquid level like it used to. Instead, the gauge

consistently registers low: there is always more gasoline in the tank than the gauge indicates.Someone at this company asks you to explain what the problem is, because you have studied

instrumentation technology. Describe the nature of the problem in your own words, and propose a solutionto this problem that does not involve purchasing any new equipment.


If there is actually 10 feet of gasoline in the tank, how many feet with the water-calibrated gauge read?

file i02949


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Question 30

Two pressure-actuated lifts are used to raise a heavy weight off the ground. One lift uses oil underpressure (from a hydraulic pump) while the other lift uses air under pressure (from an air compressor). Eachlift is equipped with a shut-off valve on the line feeding fluid to the cylinder, so that the pistons motion maybe halted:


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Shut-offvalve

Oil

Weight

. . . fromhydraulic

pump

Shut-offvalve

Weight

. . . fromair compressor

Air

What will happen if the weight were to fall off the lift platform after it had been raised up from groundlevel, in each case? Assume that the shut-off valve is closed (no fluid flow from pump or compressor into thecylinder) when this happens.


What general lessons may we draw from this example regarding pressurized fluid safety? Does the calculation of piston force based on pressure (F=PA) change at all if the fluid in question is

a gasrather than a liquid?

file i00750


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Question 33

A double-acting hydraulic cylinder has 500 PSI of pressure applied to the side without the rod and 750PSI of pressure applied to the rod-side. Calculate the resultant force generated at the piston and transmittedthrough to the rod, and also determine this forces direction. The piston is 5 inches in diameter, and therod is 1 inch in diameter.

F ???


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piston

rod

5"

1"

750 PSI

500 PSI

Force ???


Identify which fundamental principles of science, technology, and/or math apply to each step of yoursolution to this problem. In other words, be prepared to explain the reason(s) why for every step ofyour solution, rather than merely describing those steps.

Would the piston experience a resultant force if both ports were connected together with a length oftubing (made common to each other) and then pressurized with the exact same amount of fluidpressure? Why or why not?

file i00156

Question 34

The following hydraulic system is made up of three cylinders connected together by the same tube:


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tubing

cylinder#1

cylinder#2

cylinder#3

Assuming that all three pistons are the same size, calculate the force generated by the pistons of cylinders#2 and #3 if the piston of cylinder #1 is pushed with 500 pounds of force.

file i00152


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Question 38

A solid metal cube measuring exactly 1 inch on a side is submerged in an open container filled withwater. The bottom of the cube is 24 inches down from the waters surface:


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Water

24"

1"Cube

23"

Based on your calculations of hydrostatic pressure (in PSI), determine the force applied to each sideof the cube (in units of pounds), and the net, or resultantof these six forces (one force for each side of thecube).

Based on the figure for water density of 62.428 lb/ft3, how much does one cubic inch of water happento weigh?

file i00265


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Answers

Answer 1


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4

inches

inches2

5

inches

Answer 2

This is a trick question: they all generate the exact same amount of hydrostatic pressure.

The principle at work here is the relationship between vertical height and hydrostatic pressure. Cross-sectional area of the liquid column is irrelevant! Only column height, liquid density, and the gravitationalpull of the Earth matter when calculating hydrostatic pressure. Since all three of these variables are preciselythe same in this scenario, the hydrostatic pressures must likewise be precisely the same.


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Answer 6

A pushing force on the rod will compress the rubber ball to a smaller diameter. A pulling force willexpand it to a larger diameter.


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Rubber ball Rubber ballcompresses expands

Answer 7

406.9 inches, which is a little bit less than 34 feet. For this amount of lift height, the pump wouldhave to create a near-perfect vacuum in the inlet pipe. To calculate this figure, convert 14.7 PSIA into inches

of water column absolute (14.7 PSIA)(27.68 W.C. / PSI).Since this kind of water pump works by creating a vacuum (reducing the inlet pressure to somethingless than 14.7 PSIA), it is inherently limited in lift height. Since atmospheric pressure is always 14.7 PSIA(on Earth, anyway), this kind of pump simply cannot suck water any higher than this amount of pressureexpressed in inches or feet of water.

The average barometric pressure in Denver is 24.63 inches of mercury absolute (12.097 PSIA). Thisequates to a water-lifting height of 334.9 inches, or 27.9 feet.

Submersible pumps overcome this limit by creating a positive pressure rather than a vacuum. Thepumping action is therefore not limited by the relatively low pressure of Earths atmosphere, but only bythe capacity and design of the pump itself:

Pump


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Answer 22

PSIG PSIA inches Hg (G) inches W.C. (G)18 32.7 36.65 498.25

385.3 400 784.5 1066516.21 30.91 33 448.6


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2.168 16.87 4.413 60222.0 236.7 452 6145.10.4335 15.13 0.8826 12-13.7 1 -27.89 -379.2

-5 9.7 -10.18 -138.4

Answer 23

Atm PSIG inches W.C. (G) PS

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