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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: 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

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

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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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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