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Density is
mass per unit
volume;
specific volume
is volume per
unit mass.
Density
Specific weight: The
weight of a unit volume
of a substance.
Specific volume
1
Additional Information
2
Absolute pressure: The actual pressure at a given position. It is measured
relative to absolute vacuum (i.e., absolute zero pressure).
Gage pressure: The difference between the absolute pressure and the local
atmospheric pressure. Most pressure-measuring devices are calibrated to
read zero in the atmosphere, and so they indicate gage pressure.
Vacuum pressures: Pressures below atmospheric pressure.
Throughout this text, the pressure P will denote absolute pressure
unless specified otherwise.
3
40 kPa
100 kPa.
100  40 60 kPa
Question: A wooden log with a density of 500 kg/m3 and volume
of 5m3 is immersed in the water. Determine if the log will float or
sink?
Solution: Weight of the log:
W= ρlog g V
W = (500 kg/m3) (9.81 m/s2) (5 m3) W = 24525 N
Buoyant Force:
FB = ρ g V
FB = (1000 kg/m3) (9.81 m/s2) (5 m3) FB = 49050 N
FB > W, hence the log will float.

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Miscellaneous Formulas-just to try-1.pptx

  • 1. Density is mass per unit volume; specific volume is volume per unit mass. Density Specific weight: The weight of a unit volume of a substance. Specific volume 1 Additional Information
  • 2. 2 Absolute pressure: The actual pressure at a given position. It is measured relative to absolute vacuum (i.e., absolute zero pressure). Gage pressure: The difference between the absolute pressure and the local atmospheric pressure. Most pressure-measuring devices are calibrated to read zero in the atmosphere, and so they indicate gage pressure. Vacuum pressures: Pressures below atmospheric pressure. Throughout this text, the pressure P will denote absolute pressure unless specified otherwise.
  • 3. 3 40 kPa 100 kPa. 100  40 60 kPa Question: A wooden log with a density of 500 kg/m3 and volume of 5m3 is immersed in the water. Determine if the log will float or sink? Solution: Weight of the log: W= ρlog g V W = (500 kg/m3) (9.81 m/s2) (5 m3) W = 24525 N Buoyant Force: FB = ρ g V FB = (1000 kg/m3) (9.81 m/s2) (5 m3) FB = 49050 N FB > W, hence the log will float.