Answer:
a) 14.12psi, b) 390.86inH2O, d) 97.36kPa, d)12.42m
Explanation:
In order to do the conversions we need to use conversion rates:
a) 1psi=2.03602inHg
so:
[tex]28.75inHg*\frac{1psi}{2.03602inHg}=14.12psi[/tex]
and the same procedure is used for parts b and d:
b) 1inHg=13.595inH2O
so:
[tex]28.75inHg*\frac{13.595inH2O}{1inHg}=390.86inH2O[/tex]
d) 1inHg=3.386kPa
so:
[tex]28.75inHg*\frac{3.386kPa}{1inHg}=97.36kPa[/tex]
e)
Now part e is a little tricky since we need to review our specific gravity concept. The specific gravity is defined as the ratio between the specific weight of a substance ofver the specific weight of water.
[tex]Sg=\frac{\gamma_{substance}}{\gamma_{H_{2}O}}[/tex]
so when solving for the specific gravity of the substance, we get that it is:
[tex]\gamma _{substance}=(SG)\gamma _{H_{2}O}[/tex]
which can be used in the pressure equation:
[tex]P=\gamma h[/tex]
when solving for h we get that:
[tex]h=\frac{P}{\gamma _{substance}}[/tex]
when substituting equations we get that:
[tex]h=\frac{P}{(SG)\gamma _{H_{2}O}}[/tex]
we know that:
[tex]\gamma _{H_{2}O}=9.8kN/m^{3}[/tex]
and from the previous part of the problem we know the pressure in kPa, so when using this data we get that:
[tex]h=\frac{97.36kPa}{(0.8)(9.8kN/m^{3})}[/tex]
so:
h=12.42m
