3-10 Determine the atmospheric pressure at a location where the barometric reading is 750 mmHg. Take the density of mercury to be 13.600 kg/m.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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3-7 A vacuum gage connected to a chamber reads 24 kPa
at a location where the atmospheric pressure is 92 kPa.
Determine the absolute pressure in the chamber.
3-7
Solution
absolute pressure in the chamber is to be determined.
The pressure in a vacuum chamber is measured by a vacuum gage. The
24 kPa
Pas
Analysis
The absolute pressure in the chamber is determined from
PaPan- P -92-24- 68 kPa
P 92 kPa
Điscussion We must remember that "vacuum pressure" is the negative of gage pressure - hence the negative sign.
3-9
The water in a tank is pressurized by air, and the pres-
sure is measured by a multifluid manometer as shown in Fig.
P3-9. Determine the gage pressure of air in the tank if h,
= 0.2 m, h, = 0.3 m, and h, = 0,46 m. Take the densities of
water, oil, and mercury to be 1000 kg/m, 850 kg/m, and
13,600 kg/m', respectively.
- Oil
AIR
WATER
Mercury
FIGURE P3-9
3-9
Solution
The pressure in a pressurized water tank is measured by a multi-fluid manometer. The gage pressure of air
in the tank is to be determined.
Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density).
and thus we can determine the pressure at the air-water interface.
Properties
The densities of mercury, water, and oil are given to be 13,600, 1000, and 850 kgim', respectively.
Analysis
go down) or subtracting (as we go up) the pgh terms until we reach point 2, and setting the result equal to Pa since the
tube is open to the atmosphere gives
Starting with the pressure at point I at the air-water interface, and moving akong the tube by adding (as we
P+ Pa sh + Paish, - Pamay gh, = Pa
Air
Solving for P.
P- Pan Pean h, -Pagh, + Pan gh,
or,
-P.
Noting that P= P,- Pa and substituting.
Water
Roy = (9.81 mis (13,600 kp'm j(0.46 m) - (1000 kgim x0.2 m)
IN
IkPa
- (850kgim'x0.3 m
Ikg-m's A1000 Nim
-56.9 kPa
Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the
same fluid simplifies the analysis greatly.
3-10 Determine the atmospheric pressure at a location
where the barometric reading is 750 mmHg. Take the density
of mercury to be 13.600 kg/m.
ーミー
ーミー
Transcribed Image Text:3-7 A vacuum gage connected to a chamber reads 24 kPa at a location where the atmospheric pressure is 92 kPa. Determine the absolute pressure in the chamber. 3-7 Solution absolute pressure in the chamber is to be determined. The pressure in a vacuum chamber is measured by a vacuum gage. The 24 kPa Pas Analysis The absolute pressure in the chamber is determined from PaPan- P -92-24- 68 kPa P 92 kPa Điscussion We must remember that "vacuum pressure" is the negative of gage pressure - hence the negative sign. 3-9 The water in a tank is pressurized by air, and the pres- sure is measured by a multifluid manometer as shown in Fig. P3-9. Determine the gage pressure of air in the tank if h, = 0.2 m, h, = 0.3 m, and h, = 0,46 m. Take the densities of water, oil, and mercury to be 1000 kg/m, 850 kg/m, and 13,600 kg/m', respectively. - Oil AIR WATER Mercury FIGURE P3-9 3-9 Solution The pressure in a pressurized water tank is measured by a multi-fluid manometer. The gage pressure of air in the tank is to be determined. Assumptions The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density). and thus we can determine the pressure at the air-water interface. Properties The densities of mercury, water, and oil are given to be 13,600, 1000, and 850 kgim', respectively. Analysis go down) or subtracting (as we go up) the pgh terms until we reach point 2, and setting the result equal to Pa since the tube is open to the atmosphere gives Starting with the pressure at point I at the air-water interface, and moving akong the tube by adding (as we P+ Pa sh + Paish, - Pamay gh, = Pa Air Solving for P. P- Pan Pean h, -Pagh, + Pan gh, or, -P. Noting that P= P,- Pa and substituting. Water Roy = (9.81 mis (13,600 kp'm j(0.46 m) - (1000 kgim x0.2 m) IN IkPa - (850kgim'x0.3 m Ikg-m's A1000 Nim -56.9 kPa Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly. 3-10 Determine the atmospheric pressure at a location where the barometric reading is 750 mmHg. Take the density of mercury to be 13.600 kg/m. ーミー ーミー
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