Hydraulics Exercise 1 A few useful constants: g = 9.80665 m/s 2 = 32.174 ft/s 2
ID: 1842539 • Letter: H
Question
Hydraulics Exercise 1
A few useful constants:
g = 9.80665 m/s2 = 32.174 ft/s2 use 9.8 m/s2 and 32 ft/s2
rwater = 1000 kg/m3
gwater = 62.4 lb/ft3=9800 N/m3
goil 55.7 lb/ft3 8760 N/m3 (generic oil SG .893)
1 atmosphere=14.7 psi
Give the two equivalent definitions of specific gravity, SG.
Amazon offers an “Accugage standard tire press gauge 3-60 psi.” It is shown in the picture. In the picture it is exposed to the atmosphere and reads 0 psi. Does it read gage or absolute pressure? Explain.
When the pressure gage in problem 2 is used to measure the pressure in a car tire it reads 32 psi.
What is the gage pressure?
What is the absolute pressure?
Explanation / Answer
Specific Gravity- Density of any fluid with respect to water is known as specific gravity of that fluid.
Specific gravity=Denstiy of given fluid / Density of water.
Example- Suppose density of mercury is13600 kg/m^3. so it's specific gravity will be 13.6.
It read gage pressure .
Gage Pressure- Pressure with respect to atmosphere is known as gage pressure.
Gage pressure may be positive ,nagative or zero.
Absolute Pressure- Pressure with respect to vacuum is known as absolute pressure.
Absolute pressure is always positive.
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