Pressure scales (leacture 2)

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28-Oct-20
LECTURE 2
Pressure Scales
Pressure Scales
In this lecture we will be covering the following topics:
- Pressure; Fluid Pressure and Hydrostatic head.
- Atmospheric Pressure, Absolute Pressure and Gauge Pressure.
- Fluid Pressure Measurement.
2
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28-Oct-20
Pressure
Pressure
ratio of Force
to the Area on
which the
force acts
Units: Force
units divided by
Area
SI Units:
𝑵/𝒎𝟐,
𝒅𝒚𝒏𝒆𝒔/𝒄𝒎𝟐,
𝒍𝒃/𝒊𝒏𝟐 𝐨𝐫 𝐩𝐬𝐢 h(m) 𝑷(𝑵/𝒎𝟐)
𝟑
𝑭𝒍𝒖𝒊𝒅 𝒅𝒆𝒏𝒔𝒊𝒕𝒚 𝝆(𝒌𝒈/𝒎 )
𝟐
𝑷(𝑵/𝒎 )
𝑨(𝒎𝟐)
F(𝑵/𝒎𝟐)
Fluid pressure in a tank and pipe
𝑷𝒐(𝑵/𝒎𝟐)
𝑨(𝒎𝟐)
𝑷 = 𝑷𝒐 + 𝝆𝒈𝒉
Fluid pressure at the base of a fluid column
F(𝑵/𝒎𝟐)
𝑨(𝒎𝟐)
𝑷(𝑵/𝒎𝟐)
Pressure
Pressure
Force/Area
head of a particular fluid
Height of a hypothetical column
of this fluid would exert the
given pressure at its base if the
pressure at the top were zero
𝑷𝒐 = 𝟎
𝑷
So now we can speak of a pressure 14.7 psi or
equivalently of a pressure(or head) of
𝟑𝟑. 𝟗 𝒇𝒕 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓 (𝟑𝟑. 𝟗 𝒇𝒕 𝑯𝟐𝑶) or
𝟕𝟔 𝒄𝒎 𝒐𝒈 𝒎𝒆𝒓𝒄𝒖𝒓𝒚 (𝟕𝟔 𝒄𝒎 𝑯𝒈)
The equivalence between a pressure P
(force/area) and corresponding head 𝑷𝒉
(head of fluid is given by: (with 𝑷𝒐 = 𝟎)
𝑷
𝒇𝒐𝒓𝒄𝒆
𝑨𝒓𝒆𝒂
= 𝝆𝒇𝒍𝒖𝒊𝒅 𝒈𝑷𝒉
3
2

28-Oct-20
Example 3.4-1
Express a pressure of 2 𝑥 105 Pa in terms of mmHg.
Solution:
Using, 𝑷
𝑷𝒉 =
𝝆𝑯𝒈𝒈
=
2𝑥 105 𝑁
𝑚2
𝜌𝐻𝑔 = 𝑆. 𝐺. 𝑥𝜌𝑤𝑎𝑡𝑒𝑟
𝜌𝐻𝑔 = 13.6𝑥1000 𝑘𝑔/𝑚3
𝜌𝐻𝑔 = 13600 𝑘𝑔/𝑚3
𝝆𝒇𝒍𝒖𝒊𝒅
𝑺. 𝑮. =
𝝆𝒘𝒂𝒕𝒆𝒓
13600 𝑘𝑔
𝑚3
9.8 𝑚
𝑠2
𝐹 = 𝑚 𝑥 𝑎
𝑁 = 𝑘𝑔 𝑥 𝑚/𝑠2
1 𝑁 1 𝑚
𝑘𝑔. 𝑚/𝑠2 1000 𝑚𝑚
= 𝟏. 𝟓𝟎 𝒙 𝟏𝟎𝟑 𝒎𝒎𝑯𝒈
Pressure Scale
• Absolute Pressure: The actual pressure at a given position is called the absolute
pressure, and it is measured relative to absolute vacuum (i.e., absolute zero
pressure)
• Gage Pressure: Gage pressure is the pressure relative to the atmospheric
pressure. In other words, how much above or below is the pressure with respect to
the atmospheric pressure.
• Vacuum Pressure: Pressures below atmospheric pressure are called vacuum
pressures and are measured by vacuum gages that indicate the difference
between the atmospheric pressure and the absolute pressure.
• Atmospheric Pressure: The atmospheric pressure is the pressure that an area
experiences due to the force exerted by the atmosphere
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2

28-Oct-20
Pressure Scale
0 pressure
𝑷𝒂𝒃𝒔 = 𝟎
𝑷𝒂𝒕𝒎
𝑷𝒂𝒕𝒎
𝑷𝒈𝒂𝒈𝒆
𝑷𝒂𝒃𝒔
𝑷𝒂𝒃𝒔 = 𝑷𝒈𝒂𝒈𝒆 + 𝑷𝒂𝒕𝒎
𝑷𝒂𝒕𝒎
𝑷𝒗𝒂𝒄
𝑷𝒂𝒃𝒔
P vac = P atm − P abs
• A typical value of the atmospheric pressure at sea level is 760 mmHg which
has been designated as a standard pressure of 1 atmosphere.
• Many pressure measuring devices give the gauge pressure of the fluid or the
pressure related to atmospheric pressure.
• Gage Calibration:
𝑷𝒈𝒂𝒈𝒆 = 𝟎 ⇒ 𝑷𝒂𝒕𝒎 = 𝟏 𝒂𝒕𝒎
• The abbreviations 𝑝𝑠𝑖𝑎 and 𝑝𝑠𝑖𝑔 are commonly used to denote absolute and
gauge pressure in lb/𝑖𝑛2
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2

28-Oct-20
Conversion factors of pressure
1 atm = 14.7 psi
1 atm = 101325 Pa = 101.325 kPa
1 atm = 1.01325 bar
1 atm = 760 mmHg
1 atm = 33.9 ft 𝐻2𝑂
1 atm = 1.013x106 dyne/𝑐𝑚2
1 m 𝐻2𝑂 = 9806.65 Pa
Example 3.4-2
What is the pressure 30 m below the surface of a lake? Atmospheric pressure
(the pressure at the surface) is 10.4 𝑚 𝐻2𝑂 and the density of water is
1000 𝑘𝑔/𝑚3. Assume that 𝑔 = 9.8 𝑚/𝑠2.
Solution:
𝑃 = 𝑃𝑜+ 𝜌𝑔𝑕
=
10.4 𝑚 𝐻2𝑂 101325 𝑃𝑎
10.33 𝑚 𝐻2𝑂
+
1000 𝑘𝑔 9.8 𝑚
𝑚3 𝑠2
30 𝑚
6
= 𝟑. 𝟗𝟓𝟗 𝒙 𝟏𝟎𝟓 𝑵/𝒎𝟐
2

28-Oct-20
Problem 3.32 (Felder)
Perform the following pressure conversions, assuming when necessary that
atmospheric pressure is 1 atm. Unless otherwise stated, the given pressures
are absolute.
a) 2600 mmHg to psi
b) 275 ft 𝐻2𝑂 to kPa
c) 3 atm to 𝑁/𝑐𝑚2
d) 280 cm Hg to 𝑑𝑦𝑛𝑒/𝑚2
e) 20 cm Hg of vacuum to atm (abs)
f) 25 psig to mm Hg (gage)
g) 25 psig to mm Hg (abs)
Solution:
a. 2600 mmHg to psi
=
b. 2.75 ft 𝐻2𝑂 to kPa
=
760 𝑚𝑚𝐻𝑔
2600 𝑚𝑚𝐻𝑔 1 𝑎𝑡𝑚 = 𝟓𝟎. 𝟐𝟖 𝒎𝒎 𝑯𝒈
2.75 𝑓𝑡 𝐻2𝑂 1 𝑎𝑡𝑚
33.9 𝑓𝑡 𝐻2𝑂 1 𝑎𝑡𝑚
101.325 𝑘 𝑃𝑎 = 𝟖. 𝟐𝟐 𝒌𝑷𝒂
1 𝑎𝑡𝑚
7
14.7 𝑝𝑠𝑖
2

28-Oct-20
Solution
c. 3 atm to 𝑁/𝑐𝑚2
=
d. 280 cm Hg to 𝑑𝑦𝑛𝑒/𝑚2
=
3 𝑎𝑡𝑚
1 𝑎𝑡𝑚
101325 𝑃𝑎 303975 𝑁
𝑚2
1 𝑚2
(100)2 𝑐𝑚2
=
𝑵
= 𝟑𝟎. 𝟒
𝒄𝒎𝟐
1 𝑎𝑡𝑚
280 𝑐𝑚 𝐻𝑔 1.013x106 dyne/𝑐𝑚2
1 𝑚2
1 𝑎𝑡𝑚
76 𝑐𝑚 𝐻𝑔
(100)2 𝑐𝑚2 =
𝑵
𝟑𝟕, 𝟑𝟐𝟏
𝒄𝒎𝟐
Solution
e. 20 cm Hg of vacuum to atm (abs)
= =
8
10 𝑚𝑚
20 𝑐𝑚 𝐻𝑔 1 𝑎𝑡𝑚
1 𝑐𝑚 760 𝑚𝑚 𝐻𝑔
⇒ P abs = P g + P atm
P abs = 0 + 0.263 = 0.263 atm
0.263 𝑎𝑡𝑚
2

28-Oct-20
Solution
f. 25 psig to mm Hg (gage)
=
g. 25 psig to mm Hg (abs)
⇒ 𝑷𝒂𝒃𝒔 = 𝑷𝒈 + 𝑷𝒂𝒕𝒎
⇒ 𝑷𝒂𝒃𝒔 = 1293 𝑚𝑚 𝐻𝑔 + 760 𝑚𝑚 𝐻𝑔
⇒ 𝑷𝒂𝒃𝒔 = 𝟐𝟎𝟓𝟑 𝒎𝒎𝑯𝒈
= 𝟏𝟐𝟗𝟑 𝒎𝒎𝑯𝒈
25 𝑝𝑠𝑖𝑔
1 𝑎𝑡𝑚
760 𝑚𝑚𝐻𝑔 1 𝑎𝑡𝑚
14.7 𝑝𝑠𝑖
Fluid Pressure Measurement
• They can be categorized as:
 elastic-element methods – Bourdon tubes, bellows or diaphragms
Diaphragm Gauge
Bourdon Tubes
9
Bellows Pressure Gauge
2

28-Oct-20
Fluid Pressure Measurement
• liquid-column methods – manometers
Manometers
General Manometer Equation:
𝑷𝟏 + 𝝆𝟏𝒈𝒅𝟏 = 𝑷𝟏 + 𝝆𝟐𝒈𝒉𝟐 + 𝝆𝒇𝒈𝒉
10
2

28-Oct-20
Differential Manometer
𝑷𝟏 − 𝑷𝟏 = (𝝆𝒇 − 𝝆) 𝒈𝒉
Differential Manometer Equation:
Example 3.4-3
A differential manometer is used to measure the drop in pressure between
two points in a process line containing water. The specific gravity of the
manometer fluid is 1.05. The measured levels in each arm are shown below.
𝑐𝑚2
Calculate the pressure drop between points 1 and 2 in 𝑑𝑦𝑛𝑒𝑠
.
11
2

28-Oct-20
Solution
⇒ 𝑃𝑎= 𝑃𝑏  (A)
⇒ 𝑃𝑎= 𝑃1 + 𝜌𝑔𝑕
1
⇒ 𝑃𝑏 = 𝑃2 + 𝜌𝑔𝑕
2 + 𝜌𝑓𝑔𝑕
3
Equation (A):
⇒ 𝑃1 + 𝜌𝑔𝑕
1 = 𝑃2 + 𝜌𝑔𝑕
2 + 𝜌𝑓𝑔𝑕
3
𝑷𝟏 − 𝑷𝟐 = 𝝆𝒈𝒉𝟐 + 𝝆𝒇𝒈𝒉𝟑 − 𝝆𝒈𝒉𝟏
𝑃1− 𝑃2 = 𝜌𝑔𝑥 + 𝜌𝑓𝑔(8) − 𝜌𝑔(8 + 𝑥)
𝑃1 − 𝑃2 = 𝜌𝑔 𝑥 − 8 + 𝑥 + 𝜌𝑓𝑔 8
𝑃1 − 𝑃2 = −𝜌𝑔 8 + 𝜌𝑓𝑔 8
𝑃1− 𝑃2 = (𝜌𝑓−𝜌)𝑔 8
𝟑𝟖𝟐 − 𝟑𝟕𝟒
= 𝟖𝒎𝒎
Solution
𝑃1− 𝑃2 = (𝜌𝑓−𝜌)𝑔 8
= 1.05 − 1 𝑔
𝑐𝑚3
9.8 𝑚
𝑠2
1 𝑐𝑚
10 𝑚𝑚
8 𝑚𝑚 100 𝑐𝑚
1 𝑚
= 39.2 𝑔 1 𝑑𝑦𝑛𝑒
𝑐𝑚 . 𝑠2 1 g. cm/𝑠2
= 𝟑𝟗. 𝟐 𝒅𝒚𝒏𝒆/𝒄𝒎𝟐
12
2

Pressure scales (leacture 2)

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