Question
Asked Sep 10, 2019
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The closed tank below is filled with water and is 5 ft long. The pressure gage on the tank reads 7 psi. Determine: (a) the height, h, in the open water column; (b) the gage pressure acting on the bottom tank surface AB, and; (c) the absolute pressure of the air in the top of the tank if the local atmospheric pressure is 14.7 psia. [Answer in units of ft. and psi]

Open
7 psi
Air
T
2 ft
Water
2 ft
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Open 7 psi Air T 2 ft Water 2 ft

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Expert Answer

Step 1

Given: height of tank=5ft

            Pressure Pair =7psi

            Local atmospheric pressure  Po= 14.7 psi

 

Step 2

a)   Pressure at atmospheric level, Patm=0 as it is open to atmosphere

  • To convert psi to lb/ft2 multiply the value 144
  • Specific weight of water = 62.4 pounds per cubic foot, lbs/ft3.
  • From the given figure pressure at air is:
Pap P (-2ft)
atm
Р. — Р.
h=
+2 ft
atm
w
Par7psi144
1008/b/ ft2
YM=62.4lb/ ft
(1008-0)
h
+2
62.4
18.15 ft
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Pap P (-2ft) atm Р. — Р. h= +2 ft atm w Par7psi144 1008/b/ ft2 YM=62.4lb/ ft (1008-0) h +2 62.4 18.15 ft

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Step 3

b)Pressure at surface ...

Рав 3 Ра, + 7, X4 ft
1
=7psi +62.4lb / f x4 ftx-
-— *
ft / lbf
144
=7+1.73
8.73psi
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Рав 3 Ра, + 7, X4 ft 1 =7psi +62.4lb / f x4 ftx- -— * ft / lbf 144 =7+1.73 8.73psi

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