2. A compressed air tank is connected to one end of a mercury manometer and the other is open to the atmosphere. The mercury level in the arm open to the atmosphere is 10.0 in lower (h) than the level in the arm attached to the tank. The density of mercury is 847 Ibm/ft³ and atmospheric pressure is 14.7 psia. The pressure in the tank is: Patm= 14-7 psia ap-pgh=? 2 7 fto up Patm = P + pgh /9: 71m1/5? P=Patm-pgh = ? 9=32.2 ft/s? SHg = 8471bm/ft ³ 3 N → unit conversion to fi, lbn to lbf, ps1 ILD

Refrigeration and Air Conditioning Technology (MindTap Course List)
8th Edition
ISBN:9781305578296
Author:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Publisher:John Tomczyk, Eugene Silberstein, Bill Whitman, Bill Johnson
Chapter1: Heat, Temperature, And Pressure
Section: Chapter Questions
Problem 14RQ: Atmospheric pressure at sea level under standard conditions is ______inches of mercury (Hg) or_____...
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2. A compressed air tank is connected to one end of a mercury manometer and the other is open
to the atmosphere. The mercury level in the arm open to the atmosphere is 10.0 in
lower (h) than the level in the arm attached to the tank. The density of mercury is 847
Ibm/ft³ and atmospheric pressure is 14.7 psia. The pressure in the tank is:
ap=Pgh=?
2
T
fto
LIP
Patm = P + pgh
/7.2/11/6²
P= Patm-pgh=
= ?
Patm= 14-7 psia
9=32.2 ft/s ²
SHg = 8471bm/ft
→ unit conversion to fi, Ibn to
H
I ft
12 in
1bf
32.2 lbm Sz
X 1
3
164, p51
1 psl
x
* (194 the
→P=14-7-4.9
=9.8751
Transcribed Image Text:2. A compressed air tank is connected to one end of a mercury manometer and the other is open to the atmosphere. The mercury level in the arm open to the atmosphere is 10.0 in lower (h) than the level in the arm attached to the tank. The density of mercury is 847 Ibm/ft³ and atmospheric pressure is 14.7 psia. The pressure in the tank is: ap=Pgh=? 2 T fto LIP Patm = P + pgh /7.2/11/6² P= Patm-pgh= = ? Patm= 14-7 psia 9=32.2 ft/s ² SHg = 8471bm/ft → unit conversion to fi, Ibn to H I ft 12 in 1bf 32.2 lbm Sz X 1 3 164, p51 1 psl x * (194 the →P=14-7-4.9 =9.8751
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