Absolute Pressure vs Gauge Pressure

Concepts of Pressure

  • Absolute Pressure: The total pressure exerted on an object, including atmospheric pressure.

  • Gauge Pressure: The pressure measured relative to atmospheric pressure.

  • Atmospheric Pressure (P_a): The pressure exerted by the weight of air in the atmosphere; at sea level, it is approximately 1 atm or 101.3 kPa.

Problem 1: Gauge Pressure in Tank

  • Given: Total pressure inside tank = 4.2 atm

  • Atmospheric pressure (P_a) at sea level = 1 atm

  • Formula for Gauge Pressure:[ P_{gauge} = P_{total} - P_a ]

  • Calculation:[ P_{gauge} = 4.2 \text{ atm} - 1 \text{ atm} = 3.2 \text{ atm} ]

  • Result: Gauge pressure is 3.2 atm.

Problem 2: Gauge Pressure in Storage Tank

  • Given: Total pressure = 0.9 atm

  • Atmospheric pressure (P_a) = 1 atm

  • Calculation:[ P_{gauge} = 0.9 \text{ atm} - 1 \text{ atm} = -0.1 \text{ atm} ]

  • Interpretation:

    • A negative gauge pressure indicates that the absolute pressure inside the tank is less than atmospheric pressure.

Problem 3: Absolute Pressure in Tire

  • Given: Gauge pressure = 325 kPa

  • Atmospheric pressure (P_a) = 101.3 kPa at sea level

  • Formula for Absolute Pressure:[ P_{absolute} = P_{gauge} + P_a ]

  • Calculation:[ P_{absolute} = 101.3 \text{ kPa} + 325 \text{ kPa} = 426.3 \text{ kPa} ]

  • Result: Absolute pressure inside the tire is 426.3 kPa.

Problem 4: Diver at 50 Meters Depth

  • Understanding:

    • Gauge Pressure: Pressure due to water above the diver.

    • Absolute Pressure: Total pressure due to water weight and atmospheric pressure.

  • Gauge Pressure Calculation:

    • Formula:[ P_{gauge} = \rho g h ]

    • Given values:

      • Density of seawater (( \rho )) = 1025 kg/m³

      • Gravitational acceleration (g) = 9.8 m/s²

      • Depth (h) = 50 m

    • Calculation:[ P_{gauge} = 1025 \times 9.8 \times 50 = 502250 \text{ Pa} ]

    • Conversion:[ 502250 , ext{Pa} = 502.25 \text{ kPa} ]

  • Absolute Pressure Calculation:

    • ( P_{absolute} = P_{gauge} + P_a )

    • Calculation:[ P_{absolute} = 502.25 \text{ kPa} + 101.3 \text{ kPa} = 603.55 \text{ kPa} ]

    • Result: Gauge pressure = 502 kPa, Absolute pressure = 603.55 kPa.

Problem 5: Pressure at Oil-Water Interface

  • Given Heights:

    • Height of water = 15 m

    • Height of oil = 8 m

  • Gauge Pressure at Oil-Water Interface:

    • Formula:[ P_{gauge} = Pgh ]

    • Given values:

      • Density of oil (( \rho )) = 750 kg/m³

      • g = 9.8 m/s²

      • Height (h) of oil = 8 m

    • Calculation:[ P_{gauge} = 750 \times 9.8 \times 8 = 58800 \text{ Pa} ]

    • Conversion:[ 58800 \text{ Pa} = 58.8 \text{ kPa} ]

  • Total Pressure (Absolute Pressure) at Interface:

    • Calculation:[ P_{total} = P_{gauge} + P_a = 58.8 \text{ kPa} + 101.3 \text{ kPa} = 160.1 \text{ kPa} ]

  • Gauge Pressure at Bottom of Container:

    • Combine pressures from oil and water:

    • ( P_{gauge} = P_{gauge, oil} + P_{gauge, water} )

    • Oil: ( P_{gauge, oil} = 750 \times 9.8 \times 8 = 58800 \text{ Pa} = 58.8 \text{ kPa} )

    • Water: ( P_{gauge, water} = 1000 \times 9.8 \times 15 = 147000 \text{ Pa} = 147 \text{ kPa} )

    • Total Gauge Pressure:[ P_{gauge} = 58.8 \text{ kPa} + 147 \text{ kPa} = 205.8 \text{ kPa} ]

  • Absolute Pressure at Bottom:

    • Calculation:[ P_{absolute} = P_{gauge} + P_a = 205.8 \text{ kPa} + 101.3 \text{ kPa} = 307.1 \text{ kPa} ]

  • Results:

    • Gauge Pressure at Oil-Water Interface: 58.8 kPa

    • Absolute Pressure at Oil-Water Interface: 160.1 kPa

    • Gauge Pressure at Bottom of Container: 205.8 kPa

    • Absolute Pressure at Bottom of Container: 307.1 kPa.