Given The following figure illustrates a \"closed\" steel storage tank with air
ID: 1712061 • Letter: G
Question
Given The following figure illustrates a "closed" steel storage tank with air and three difterent fluids of different heights. The air at the top of the tank is pressurized and there are two pressure gages and a manometer. The water which has a unit weight - 62.4 lb/T. The manometer is open to atmosphere at the top and uses mercury. Use the diagram below to for all the dimensions and information. (All the pressure readings are "gage." ne "absolute") roblem Total ions: is a tota Air pressure = 30 psi t of: (a) Sp.gr. 0.70 nly refe Gage A ote she 14 itions dit for u per Mercuryh S.G. 13.6 Water Gage B 1.5 ou mi 3e 25' CHF ANI Manometer Sp.gr.# 1.6 Gage C Find: Using the diagram and information above answer the following: a. What is the pressure reading of Gage A in psi? b. What is the pressure reading of Gage B in psi?Explanation / Answer
given that a closed steel storage contains air and three different fluids of different heights
Air Pressure P1= 30 psi = 0.206N/mm2
Height of fluid below air = 8' = 2.138 m
Specific gravity of the fluid below air S1= 0.7
Density of fluid = S1 x 1000 = 700 kg/m3
Height of water = 14' = 4.267
density of water =62.4 lb/ft3 = 1000 kg/m3
Height of fluid below water =25' =7.62 m
Specific gravity of fluid below water S2= 1.6
Density of fluid = S2 x 1000 = 1600 kg/m3
Pressure due to fluid below air P2= density x gravity x height = 700 x 9.81 x 2.138 = 14681.646 N/m2
= 0.0146N/mm2
Pressure due to water P3 = density x gravity x height = 1000 x 9.81 x4.267 = 41859.27 N/m2 = 0.0415 N/mm2
Pressure due to fluid below water P3 = density x gravity x height = 1600 x 9.81 x 7.62 = 119603.52 N/m2
= 0.119 N/mm2
In th question given that pressure gages are used to calculate pressures at point A and B but gages are just an instrument, the values should be equal if we calculate the same pressure by using any instrument.
So, calculating the pressures at A and B by general method
Pressure at A = Pressure due to air + Pressure due to fluid below air = 0.206+ 0.0146 = 0.2206 N/mm2
Pressure at B = Pressure due to air + Pressure due to fluid below air + Pressure due to water
= 0.206 + 0.0146 + 0.0415 = 0.2621 N/mm2
Pressure at C = Pressure due to air + Pressure due to fluid below air + Pressure due to water + Pressure due to fluid below water = 0.2621 + 0.119 = 0.3811 N/mm2
so the required pressure are
Pressure reading of Gage A = 0.2206 N/mm2 = 32 psi
Pressure reading of Gage B = 0.2621 N/mm2 = 38.01 psi
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