For pressure above atmospheric for pressure below

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For pressure above atmospheric: 𝑃 𝑔?𝑔𝑒 = 𝑃 ??? − 𝑃 ??𝑚 For pressure below atmospheric: 𝑃 𝑣?? = 𝑃 ??𝑚 − 𝑃 ??? In this course, P will represent absolute pressure i.e. P = P abs 20 P gage = 0 bar
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21 P vac P abs P atm P abs = 0 Absolute vacuum P atm P gage P atm P abs Absolute vacuum P gage = 0 Atmospheric pressure (P atm ) Absolute pressure (P abs ) Gage pressure (P gage ) Vacuum pressure (P vac )
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Example A vacuum gage connected to a chamber reads 40 kPa. The atmospheric pressure is 100 kPa. Determine the absolute pressure in the chamber Solution: From, P vac = P atm P abs P abs = P atm P vac = 100 40 = 60 kPa 22
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Pressure Variation with Depth Pressure reading is constant at horizontal position and vary in depth 𝑃 = 𝑃 ??𝑚 + 𝜌?ℎ where; ρ = density g = gravity h = depth from the free surface Gage pressure 𝑃 = 𝜌?ℎ 23 P Pressure distribution Free surface h P gage = ρ gh
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Pascal’s Principle The pressure of fluid is constant in horizontal direction Therefore “the pressure applied to a confined fluid increases the pressure throughout by the same amount the Pascal’s Principle” We also know that the pressure force on a surface is proportional to the surface area 𝑃 1 = 𝑃 2 ? 1 𝐴 1 = ? 2 𝐴 2 ? 2 ? 1 = 𝐴 2 𝐴 1 24 A B C D M N P A = P B = P c = P D P M = P N h P atm F 2 = P 2 A 2 F 1 = P 1 A 1 P 1 A 1 P 2 A 2 1 2
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Pressure Measurement Devices Manometer From figure, ρ s is the specific gravity Density of the fluid in the column ρ = ρ s x ρ H20 The pressure of fluid in the tank: 𝑃 = 𝑃 ??𝑚 + 𝜌?ℎ Example If P atm = 95kPa, ρ s = 0.85 and h = 55cm, determine pressure in the chamber Solution 𝜌 = 𝜌 ? × 𝜌 𝐻 2 ? = 0.85 × 1000 = 850 𝑘? 𝑚 3 𝑃 = 𝑃 ??𝑚 + 𝜌?ℎ = 95 + 850 9.81 0.55 = 100.6𝑘𝑃? 25 P = ? h ρ s P atm Manometer
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