Principles of Hydrostatic Pressure
Principles of Hydrostatic Pressure
Principles of Hydrostatic Pressure
dF
p
dA
Note: If the pressure is uniform over the area A, then
F
p
A
Otherwise this formula gives only the average pressure
Units:
lb 2 psf
lb 2 psi
English System:
or
ft
in
Metric System:
gr
cm
S.I. :
m2
2. Pascals Law: states that at any point in a fluid at rest the pressure is
the same in all directions.
p h
h
p = .h
3.1
Fatm
F
= W + Fatm
p x A = (h x A) + patm x A
p
= h + patm
= gh + patm
p h patm
p gh patm
F
p h patm
p gh patm
equation (2)
N/m2
kg/m3
N/m3
m/s2
N
N
N/m2
3.2
Metal tube
spring
Pinion gear
joint
Dial scale
Axial
3.2.2 Piezometer
h
p+
Piezometers
This is a very simple instrument used for measuring liquid pressure only. The Piezometer is a
glass tube with an end open to the atmosphere. The lower end of the glass tube is connected to
the liquid source to be measured causing the liquid inside the source of container to move
freely within the tube as shown in Fig. above. The height of the liquid in the tube varies directly
with the magnitude of that liquid pressure by the following relation:
p gh h
where:
N/m3
kg/m3
N/m3
m/s2
m
3.2.3 Manometer
(a) A Simple manometer or open-end manometer.
p+
h
Differential manometer
The differential manometer consists of a U-shape glass tube for measuring the
pressure difference of the fluids in two different containers as shown in Fig.
above. The level difference of liquid in the U tube can be used to determine the
pressure difference of the fluids in the two
containers.
An inclined manometer
In the case the pressure in the container is very low which causes the level difference in the
liquid in the U tube to be very small. This makes it very difficult to read the value of h. Thus if
the tube on the right hand side is inclined the linear length of the liquid in the right hand
tube will increase resulting in a more accurate reading of the length H as shown in Fig.
above. The ratio of h /H is sin or
h = H.sin
Where: sin can be 1/10, 1/5 or 1/2 or any other fractions.
Thereby the value of H obtained from the length reading of the inclined tube can then be
used to determine the fluid pressure in the container.
patm
Barometer
This type of pressure measuring equipment is essentially a manometer for measuring
atmospheric pressure. It consists of a glass tube closed at one end and open at the other end.
The tube is initially filled with a liquid (normally mercury). It is then turned over into an upside
down position with the open end submerged in a small container of the same liquid as show in
Fig. above.
The height of the liquid column h rising above the surface of the liquid in the container can be
used to determine the atmospheric pressure.
po
dp dh
0
b) If is considered to vary,
p
n 1 gh
1
po
n RTo
n
n 1
p
e
po
g h
RTo
po K
To
p
po To K h
po = absolute pressure at sea level
p = absolute pressure at h above sea level
R = gas constant
To = absolute temperature at sea level
K = -0.00356 R/ft = -0.00650 K/m
Case c
Case d
pabs patm pg
Atmospheric Level
z
B
Atmospheric Pressure
v
Absolute Zero Level
p1 p2 h
F1
W1
Liquid,
W2
A1
A2
PRESSURE
14.69 psi = 101.26 kPa
14.17 psi = 97.67 kPa
13.66 psi = 94.16 kPa
13.16 psi = 90.71 kPa
12.68 psi = 87.40 kPa
12.21 psi = 84.16 kPa
12.08 psi = 83.26 kPa
11.76 psi = 81.06 kPa
11.32 psi = 78.03 kPa
10.89 psi = 75.06 kPa
10.48 psi = 72.24 kPa
10.09 psi = 69.55 kPa
69.55
3049
F
0
20
40
50
Pv
F
80
100
120
140
0.05
0.13
0.28
0.41
1.16
2.17
3.87
6.63
Pv
F
160
180
200
212
10.9
17.28
26.52
33.84
Temperature
C
0.06
60
2.03
10
0.12
70
3.20
20
0.25
80
4.96
30
0.44
90
7.18
40
0.76
100
10.33
50
1.26
ps po or
s hs o ho
where:
N
N
ss w 2.95 9810 m3 28,939.5 m3
N
N
so w 0.84 9810 m3 8,240.4 m3
N
N
28,939.5 3 hs 8,240.4 3 4.57m
m
m
hs 1.3m
pw pL or
w hw L hL
w hw sL w hL
hw sL hL
Example 3.
A water barometer reads 10 m. If the temperature is 20
C, what is the pressure of the atmosphere in kPa?
Solution:
From table, the pressure head of water at 20 C is
pv
0.25m
w
The height of water in the barometer is expressed by the
difference of atmospheric pressure head and the vapor pressure
head.
p a pv
N
h
pa 9810 3 10.25m
w w
m
N
pa
pa 100,553 2 or pa 100.553kPa
10
0.25
m
pa
10 0.25
Example 4.
head in:
(a) m of oil;
(b) m of water
ho
(a)
o so w 0.80 9810 N
3
m
3 N
414
x
10
2
p
m
hw
42.202m
(b)
N
w
9810 3
m
N
414 x103 2
p
p
m
3.103m 3103mm
hm
( c)
m sm w 13.6 9810 N
3
m
Example 5.
The pressure in a gas tank is 2.75 atmosphere. Compute
the absolute pressure in psi , the gage pressure in psi and the pressure
head in ft. of water assuming the specific weight of water is 62.4 lb/ft3.
Solution:
The absolute pressure in psi is,
14.7 psi
pabs 2.75atm
40.425 psi
atm
The gage pressure in psi is,
p g pabs patm 40.425 14.7 25.725 psi
lb
w
62.4 3
62.4
ft
144in 2
x
ft 2
59.365 ft
lb
ft 3
Example 6.
A gage on the suction side of a pump shows a vacuum of
250 mm of mercury. Compute (a) the pressure head in m of water;
(b) The pressure in kPa; ( c) absolute pressure in kPa if the barometer
reads 737 mm of mercury. The specific weight of water is 9810 N/m3.
Solution:
(a) The pressure head in m of water is,
3 N
101.325 x10 2
m
250mmHg
760mmHg
p 250mmHg
hw
N
w 9810 N
9810 3
m
m3
(b) The pressure in kPa
101.325kPa
33.33kPa
p 250mmHg
760mmHg
3.4m
101.325kPa
pabs 64.928kPa
po
dp dh
N
p dp 12 m3
o
305m
dh
0
N
p po p 12 3 305m 0
m
N
p 3,660Pa 3,660 2
m
po K
To
p
po To K h
N
, abs
2
m
hm 765mmHg
p 73x103
To 293K
po m hm sm w hm
K 0.00649 K / m
N
o 12.01 3
m
N
N
1m
102
,
514
13.66 9810 3 765mm
2
m
m
1000
mm
continuation
o To
po K
To
p
po To K h
73x10
293
102,514 293 0.00649 h
3
12.01 293
102, 514 0.00649
73x103
12.01293
293
ln
ln
102,514 102,514 0.00649 293 0.00649h
ln
12.01293
293 0.00649h
102,514
293
e
293 0.00649h
12.01 293
102, 514
continuation
293
e
293 0.00649h
12.01 293
102, 514
293
293 0.00649 h
e
12.01 293
102, 514
293
0.00649 h 293
e
12.01 293
102, 514
1
293
h
293
73x103
102, 514 0.00649
0.00649
ln
102, 514
12.01 293
h 2807.15m
Example 9.
Given the figure below, determine the pressure at m if
x = 760 mm and y = 760 mm.
1
Air
y
0
CaCl4 ( s = 1.60)
Oil ( s = 0.856)
x
2
Solution:
Using addition and subtraction of pressure heads, expressed
in head of water. The addition and subtraction maybe started at
atmospheric level (level 0) and proceeding from level to level up to
to obtain the algebraic sum of the pressure heads at the level in
consideration ( In this case, it is point m).
1
Air
y = 760 mm
0
CaCl4 ( s = 1.60)
m
Oil ( s = 0.856)
x = 760 mm
Starting from level 0, then to level 1, then to level 2 and lastly to level m
pm
N
pm 18,311 2
0 y 1.6 x0.856
m
w
kN
pm
18.311 2
0 0.760m 1.6 0.760m 0.856
m
N
or
9810 3
m
pm 18.311kPa
s = 2.95
x
1
125 mm dia.
4 mm dia. tube
s = 0.915
Solution:
(a) Starting at the 0 level and proceeding up to the level m.
0 x2.95 y 0.915
pm
12 mm dia. tube
s = 2.95
X = 0.30 m
125 mm dia.
4 mm dia. tube
y = 0.50 m
s = 0.915
0 x2.95 y0.915
pm
pm
N
9810 3
m
or 13.170kPa
1252 d 122 r
4
2
2
12
125
d
r
r
d or
Note when the pressure at m is
125
12
increased, the fluid inside tends
to push outward.
0
New level
Original level
12 mm dia. tube
s = 2.95
X = 0.30 m
125 mm dia.
New level
4 mm dia. tube
y = 0.50 m
s = 0.915
pm
New level
Original level
12 mm dia. tube
s = 2.95
X = 0.30 m
125 mm dia.
New level
4 mm dia. tube
y = 0.50 m
s = 0.915
13.17 7
9.81
2
2
20.17
12
12
0 r 0.30
r
2
.
95
0
.
50
r
0
.
915
125
125
9.81
r 0.24m
y = 0.75 m
Level n
Level 2
x = 1.50 m
Level m
Fluid A
w
Fluid A
Note : x + z = y + w
or x y = w z
1.50 0.75 = w z
w z = 0.75
Fluid B
Level 1
y = 0.75 m
Level n
Level 2
Fluid A
x = 1.50 m
Level m
n
Fluid A
Note: w z = 0.75
N
p m pn 13,611 2
m
or p m pn 13.611kPa
Exercise Problems:
1. If the pressure 4 m below the free surface is 150 kPa, calculate
its specific weight and specific gravity.
2. If the pressure at a point in the ocean is 1500 kPa, what is the
pressure 20 m below this point? The specific gravity of salt
water is 1.03.
3. An open vessel contains carbon tetrachloride ( s = 1.50 ) to a
depth of 2.5 m and water above this liquid to a depth of 1.5 m.
What is the pressure at the bottom?
4. How many meters of water are equivalent to a pressure of
120 kPa? How many cm of mercury?
5. What is the equivalent pressure in kPa corresponding to one
meter of air at 15 C under standard atmospheric conditions?
6. At sea level a mercury barometer reads 750 mm and at the
same time on the top of the mountain another mercury
barometer reads 745 mm. The temperature of the air is
assumed constant at 15 C and its specific weight assumed
uniform at 12 N/m3. Determine the height of the mountain.