Transcript Basic Hydraulics Pressure and Force Math for Water Technology MTH 082

Basic Hydraulics Pressure and
Force
Math for Water Technology
MTH 082
Lecture 5
Hydraulics Chapter 2
(pgs. 213-224)
Pressure
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)
In the water industry we deal with pressure exerted by water
or the height of water
Pressure =Water Height
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/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)
2
=
0.433
lb/in
0.433 lb
1 ft
of water
=
0.433 psi
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. *****
The pressure exerted by a column
of water one inch square when at
rest, is the _________ pressure. It
is usually measured in psi.
100%
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Static
Dynamic
Theoretical
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A pound of water weighs
______lbs.
43%
62
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8.
34
0%
7.
48
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7.48
8.34
62.4
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A column of water 12" high and
1 square inch in surface area will
produce a pressure of _______
lbs.
1.0 lb
2.31 lbs
0.433 lbs
62.4 lbs
13%
13%
62
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Pressure
=
Water
Height
Can use a ratio to determine feet of water are equivalent to
psi
1 ft
( x) ft

0.433 psi 1 psi
1 ft
0.433 lb
of water
(1 ft )(1 ft )
x
0.433 psi
2.31 ft = x
1 psi is equivalent to the pressure created by
a column of water 2.31 ft high
Pressure 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
Same hydrostatic pressure in a circle or a square
pool, 15 feet below water surface
Same Pressures/Different Containers
7 ft
3 psi
3 psi
3 psi
14 ft
6 psi
7 ft
7 ft
14 ft
14 ft
6 psi
6 psi
Water Tank Pressure
Same pressure at bottom
Different pressure at bottom
50,000 gallons of water
25,000 gallons of water
140 ft
140 ft
130 ft
70 ft
61 psi
61 psi
50 psi
30.5 psi
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 of
each column, read?
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
82%
9%
9%
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.8
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.8
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i
ps
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;8
8.
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..
.
0%
Pressure 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.
Pressure Types
Absolute Pressure
64.7 psia
Gauge Pressure
50 psig
14.7 psia
0 psig
12.7 psia
-2 psig
0 psia
-14.7 psig
Condition
116 ft of head
Empty line
Atmosph. press
Partial line
vacuum
Total line
vacuum
Pressure Conversions
1 psig =
1 ft head =
1kPa
=
2.31 ft head
0.433 psig
0.0109 m of head
Pressure Problems
Example 1. Convert gauge pressure of 14 ft to pounds per
square inch gauge.
1 psig
2.31 ft of head
14 ft 1 psig
(
)  6.06 psig
1 2.31 ft
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
250 ft 1 psig
(
)  108.23 psig
1
2.31 ft
Pressure Problems
Example 3. A pressure of 210 kPa (gauge) is equivalent to
how many meters of head?
1 kPa
0.1019 m
of head
210kPa 0.1019m
(
)  21.4m
1
1kPa
Pressure Problems
Example 3. What would be the psi gauge readings at point A
and B?
200 ft
80 ft
A ? psi
Po int A
200 ft 1 psig
(
)  86.58 psig
1
2.31 ft
B? psi
80 ft 1 psig
(
)  34.63 psig
1 2.31 ft
Pressure Problems
Example 4. Psi gauges are used in this water system, What is
the pressure in feet at each point?
h1
h2
h3
56 psig
A
h4
32 psig
20 psig
44 psig
B
C
D
Po int D
56 psig 2.31 ft
Po int A
(
)  129.36 ft of head
1
1 psig
Po int B
Po int C
20 psig 2.31 ft
(
)  46.2 ft of head
1
1 psig
32 psig 2.31 ft
(
)  73.92 ft of head
1
1 psig
44 psig 2.31 ft
(
)  101.64 ft of head
1
1 psig
Force
Force = Pressure X Area
F= P X A
Example 5. If a pressure of 5 psig is exerted on a surface 2
in by 3 in, what is its force?
F= P X A
F= (5 psig) (2in) (3 in)
F= 5 psig (6 in2)
F= 30 lb of force
5 lb
5 lb
5 lb
5 lb
5 lb
5 lb
2 in
3 in
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
100%
56
ps
i
0
10
0%
ps
i
0%
ps
i
0%
2
ps
i
14
ps
i
0%
.2
24.2 psi
14 psi
2 psi
100 psi
56 psi
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5.
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?
80%
11.8 ft
25.0 ft
34.6 ft
38.9 ft
10%
10%
ft
38
.9
ft
34
.6
ft
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25
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ft
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Force
Example 6. The pressure on a surface is 12 psig. If the
surface is 120 in2. What is the force?
F= P X A
F= (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?
F= P X A
F= (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
Force
The jack has an operating piston with a surface area of 5 in and a lifting
2
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?
Force = Pressure X Area
Force on operating system= Pressure on jack X Area of
operating piston
Operating System
Operating Cylinder
F=P X A
F= P X A
150 lb= (x psig) (5in2)
F= (30 psig)(100 in2)
150lb/5in2 = x
F= 3000 lb
30 psig =x
Today’s objective: to become proficient with the
concept of pressure and force has been met.
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