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Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224)

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Page 1: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Basic Hydraulics Pressure and Force

Basic Hydraulics Pressure and Force

Math for Water TechnologyMTH 082Lecture 5

Hydraulics Chapter 2(pgs. 213-224)

Math for Water TechnologyMTH 082Lecture 5

Hydraulics Chapter 2(pgs. 213-224)

Page 3: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

What is Pressure and Force?What is Pressure and Force?• Flow of water in a system is dependant on the

amount of force causing the water to move.

Force= pressure X area

• Pressure is the amount of force acting (pushing) on a unit area.

• Units of pressure = psi (pounds per square inch)• Units of pressure = kPa (kilopascals)

• Flow of water in a system is dependant on the amount of force causing the water to move.

Force= pressure X area

• Pressure is the amount of force acting (pushing) on a unit area.

• Units of pressure = psi (pounds per square inch)• Units of pressure = kPa (kilopascals)

In the water industry we deal with pressure exerted by wateror the height of water

In the water industry we deal with pressure exerted by wateror the height of water

Page 4: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Pressure =Water HeightPressure =Water HeightA container that is 1 ft by 1 ft by 1 ft (a cubic foot container) is filled with water. What is the pressure on the square foot bottom of the container.

A container that is 1 ft by 1 ft by 1 ft (a cubic foot container) is filled with water. What is the pressure on the square foot bottom of the container.

Water density = 62.4 lb/ft2Water density = 62.4 lb/ft2

Convert = 62.4 lb/ft2 = 62.4 lb/ (1 ft) (1 ft) = 62.4 lb/(12 in) (12 in)

= 62.4 lb/(144 in2) = 0.433 lb/in2

= 0.433 psi

Convert = 62.4 lb/ft2 = 62.4 lb/ (1 ft) (1 ft) = 62.4 lb/(12 in) (12 in)

= 62.4 lb/(144 in2) = 0.433 lb/in2

= 0.433 psiA foot high column of water over a square inch surface area weighs 0.433 lb which equals 0.433 psi.

***Thus, 0.433 converts pressure from feet of water to pressure in pounds per square inch. *****

A foot high column of water over a square inch surface area weighs 0.433 lb which equals 0.433 psi.

***Thus, 0.433 converts pressure from feet of water to pressure in pounds per square inch. *****

1 ft 0.433 lbof water

Page 5: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

The pressure exerted by a column of water one inch square when at rest, is the _________ pressure. It

is usually measured in psi.

Sta

tic

Dyn

amic

Theo

retic

al

Pra

ctic

al

100%

0%0%0%

1. Static

2. Dynamic

3. Theoretical

4. Practical

1. Static

2. Dynamic

3. Theoretical

4. Practical

Page 6: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

A pound of water weighs ______lbs.

17.

488.

3462

.4

57%

0%

43%

0%

1. 1

2. 7.48

3. 8.34

4. 62.4

1. 1

2. 7.48

3. 8.34

4. 62.4

Page 7: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

A column of water 12" high and 1 square inch in surface area will produce a pressure of _______

lbs.

1.0

lb

2.3

1 lb

s

0.4

33 lb

s

62.

4 lb

s

0%

13%

75%

13%

1. 1.0 lb

2. 2.31 lbs

3. 0.433 lbs

4. 62.4 lbs

1. 1.0 lb

2. 2.31 lbs

3. 0.433 lbs

4. 62.4 lbs

Page 8: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Pressure = Water HeightPressure = Water Height

psi

ftx

psi

ft

1

)(

433.0

1

Can use a ratio to determine feet of water are equivalent to psiCan use a ratio to determine feet of water are equivalent to psi

1 ft 0.433 lbof water

xpsi

ftft

433.0

)1)(1(

2.31 ft = x

1 psi is equivalent to the pressure created by a column of water 2.31 ft high

Page 9: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Pressure RulePressure Rule

• Rule 1- The height of water determines pressure over square inch area. This is termed head which is measured in feet.

• Rule 2- As long as the height of water stays the same, changing the shape of the container does not change the pressure at the bottom of an object

• Rule 1- The height of water determines pressure over square inch area. This is termed head which is measured in feet.

• Rule 2- As long as the height of water stays the same, changing the shape of the container does not change the pressure at the bottom of an object

Same hydrostatic pressure in a circle or a square pool, 15 feet below water surface

Same hydrostatic pressure in a circle or a square pool, 15 feet below water surface

Page 10: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Same Pressures/Different ContainersSame Pressures/Different Containers

7 ft7 ft

14 ft14 ft6 psi6 psi

3 psi3 psi

7 ft7 ft

14 ft14 ft6 psi6 psi

3 psi3 psi

7 ft7 ft

14 ft14 ft6 psi6 psi

3 psi3 psi

Page 11: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Water Tank PressureWater Tank Pressure

140 ft140 ft

61 psi61 psi61 psi61 psi

140 ft140 ft

50,000 gallons of water50,000 gallons of water

Same pressure at bottomSame pressure at bottom Different pressure at bottomDifferent pressure at bottom

70 ft70 ft

30.5 psi30.5 psi50 psi50 psi

130 ft130 ft

25,000 gallons of water25,000 gallons of water

Page 12: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Two columns of water are filled completely at sea level to a height of 88 feet. Column A is 0.5

inches in diameter. Column B is 5 inches in diameter. What will two pressure gauges, one

attached to the bottom ofeach column, read?

Two columns of water are filled completely at sea level to a height of 88 feet. Column A is 0.5

inches in diameter. Column B is 5 inches in diameter. What will two pressure gauges, one

attached to the bottom ofeach column, read?

A. 3

.8 p

si; 3

8...

B. 8

.8 p

si; 8

....

C. 2

0.3

psi; 2

...

D. 3

8.0

psi; 3

...

9%

82%

9%0%

A. 3.8 psi; 38.0 psi

B. 8.8 psi; 8.0 psi

C. 20.3 psi; 20.3 psi

D. 38.0 psi; 38.0 psi

A. 3.8 psi; 38.0 psi

B. 8.8 psi; 8.0 psi

C. 20.3 psi; 20.3 psi

D. 38.0 psi; 38.0 psi

Page 13: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Pressure TypesPressure Types• Atmospheric pressure is 14.7 psi at sea level.exerted everywhere so oftentimes its neglected

• Gauge pressure is water pressure in a main or container that is measured by a gauge.

• The absolute pressure pounds per square inch

absolute (psia) is obtained by adding the gauge and atmospheric pressure. height of water determines pressure over square inch area. This is termed head which is measured in feet.

• Atmospheric pressure is 14.7 psi at sea level.exerted everywhere so oftentimes its neglected

• Gauge pressure is water pressure in a main or container that is measured by a gauge.

• The absolute pressure pounds per square inch

absolute (psia) is obtained by adding the gauge and atmospheric pressure. height of water determines pressure over square inch area. This is termed head which is measured in feet.

Page 14: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Pressure TypesPressure Types

Absolute PressureAbsolute Pressure Gauge PressureGauge Pressure ConditionCondition

64.7 psia64.7 psia 50 psig50 psig 116 ft of head116 ft of head

14.7 psia14.7 psia 0 psig0 psigEmpty line Atmosph. press Empty line Atmosph. press

12.7 psia12.7 psia -2 psig-2 psig Partial linevacuumPartial linevacuum

0 psia0 psia -14.7 psig-14.7 psig Total linevacuumTotal linevacuum

Page 15: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Pressure ConversionsPressure Conversions

1 psig = 2.31 ft head1 ft head = 0.433 psig1kPa = 0.0109 m of head

Page 16: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Pressure ProblemsPressure ProblemsExample 1. Convert gauge pressure of 14 ft to pounds per

square inch gauge. Example 1. Convert gauge pressure of 14 ft to pounds per

square inch gauge.

1 psig 2.31 ft of head psigft

psigft06.6)

31.2

1(

1

14

Example 2. A head of 250 ft of water is equivalent to what pressure in pounds per square inch? Example 2. A head of 250 ft of water is equivalent to what pressure in pounds per square inch?

1 psig 2.31 ft of head psigft

psigft23.108)

31.2

1(

1

250

Page 17: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Pressure ProblemsPressure ProblemsExample 3. A pressure of 210 kPa (gauge) is equivalent to how many meters of head? Example 3. A pressure of 210 kPa (gauge) is equivalent to how many meters of head?

1 kPa0.1019 m of head

mkPa

mkPa4.21)

1

1019.0(

1

210

Page 18: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Pressure ProblemsPressure Problems

psigft

psigftAPo 58.86)

31.2

1(

1

200int

Example 3. What would be the psi gauge readings at point A and B? Example 3. What would be the psi gauge readings at point A and B?

B? psiB? psiA ? psiA ? psi

200 ft200 ft

80 ft80 ft

psigft

psigft63.34)

31.2

1(

1

80

Page 19: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Pressure ProblemsPressure Problems

headofftpsig

ftpsigDPo 2.46)

1

31.2(

1

20int

headofftpsig

ftpsigBPo 64.101)

1

31.2(

1

44int

Example 4. Psi gauges are used in this water system, What is the pressure in feet at each point?Example 4. Psi gauges are used in this water system, What is the pressure in feet at each point?

AA BB CC DD

h1h1h2h2 h3h3 h4h4

56 psig56 psig44 psig44 psig

32 psig32 psig20 psig20 psig

headofftpsig

ftpsigCPo 92.73)

1

31.2(

1

32int headofft

psig

ftpsigAPo 36.129)

1

31.2(

1

56int

Page 20: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

ForceForceForce = Pressure X Area

F= P X AForce = Pressure X Area

F= P X AExample 5. If a pressure of 5 psig is exerted on a surface 2 in by 3 in, what is its force?Example 5. If a pressure of 5 psig is exerted on a surface 2 in by 3 in, what is its force?

2 in2 in

3 in3 in

5 lb5 lb

5 lb5 lb

5 lb5 lb

5 lb5 lb5 lb5 lb

5 lb5 lb

F= P X AF= (5 psig) (2in) (3 in)F= 5 psig (6 in2)F= 30 lb of force

F= P X AF= (5 psig) (2in) (3 in)F= 5 psig (6 in2)F= 30 lb of force

Page 21: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

You have a water storage tank that is 90' tall and 45' in diameter, it

currently has 56' of water in it, what is the pressure in the bottom of the

tank

24.

2 psi

14

psi

2 p

si

100

psi

56

psi

100%

0% 0%0%0%

1. 24.2 psi

2. 14 psi

3. 2 psi

4. 100 psi

5. 56 psi

1. 24.2 psi

2. 14 psi

3. 2 psi

4. 100 psi

5. 56 psi

Page 22: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

The pressure gauge on the bottom of a water holding tank reads 15 psi. The tank is 15 ft in diameter and 40 ft high. How many feet of

water are in the tank?

11.

8 ft

25.

0 ft

34.

6 ft

38.

9 ft

10% 10%

80%

0%

1. 11.8 ft

2. 25.0 ft

3. 34.6 ft

4. 38.9 ft

1. 11.8 ft

2. 25.0 ft

3. 34.6 ft

4. 38.9 ft

Page 23: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

ForceForce

Example 6. The pressure on a surface is 12 psig. If the surface is 120 in2. What is the force? Example 6. The pressure on a surface is 12 psig. If the surface is 120 in2. What is the force?

F= P X AF= (12 psig) (120 in2)F= 1440 lb of force

F= P X AF= (12 psig) (120 in2)F= 1440 lb of force

Example 7. The pressure is 40 psig against a surface that is 1 ft by 2 ft. What is the force against the surface? Example 7. The pressure is 40 psig against a surface that is 1 ft by 2 ft. What is the force against the surface?

F= P X AF= (40 psig) (288 in2)F= 11,520 lb

F= P X AF= (40 psig) (288 in2)F= 11,520 lb

A= (1 ft) (2ft)A=(2 ft2)A =(2 ft2) ( 12 in/1ft)2

A= 288 in2

A= (1 ft) (2ft)A=(2 ft2)A =(2 ft2) ( 12 in/1ft)2

A= 288 in2

Page 24: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

ForceForce

Force = Pressure X AreaForce on operating system= Pressure on jack X Area of

operating pistonOperating System Operating CylinderF=P X A F= P X A150 lb= (x psig) (5in2) F= (30 psig)(100 in2)150lb/5in2 = x F= 3000 lb30 psig =x

Force = Pressure X AreaForce on operating system= Pressure on jack X Area of

operating pistonOperating System Operating CylinderF=P X A F= P X A150 lb= (x psig) (5in2) F= (30 psig)(100 in2)150lb/5in2 = x F= 3000 lb30 psig =x

The jack has an operating piston with a surface area of 5 in2 and a lifting piston with a surface area of 100 in2. A force of 150 lb is applied to the operating piston. What pressure is created within the hydraulic system of the jack?

The jack has an operating piston with a surface area of 5 in2 and a lifting piston with a surface area of 100 in2. A force of 150 lb is applied to the operating piston. What pressure is created within the hydraulic system of the jack?

Page 25: Basic Hydraulics Pressure and Force Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 2 (pgs. 213-224) Math for Water Technology MTH 082 Lecture

Today’s objective: to become proficient with the concept of pressure and force has been met.

Today’s objective: to become proficient with the concept of pressure and force has been met.

Stro

ngly A

gree

Agre

e

Neu

tral

Dis

agre

e

Stro

ngly D

isag

...

20% 20% 20%20%20%

1. Strongly Agree

2. Agree

3. Neutral

4. Disagree

5. Strongly Disagree

1. Strongly Agree

2. Agree

3. Neutral

4. Disagree

5. Strongly Disagree