Question

Convert a barometric pressure of 28.75 in. Hg at 32 F (0°C) to (a) psia (b) inches of water, (e) meters of liquid whose specific gravity is 0.8, and (d) kPa

Answer 1

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:

$28.75inHg*\frac{1psi}{2.03602inHg}=14.12psi$

and the same procedure is used for parts b and d:

b) 1inHg=13.595inH2O

so:

$28.75inHg*\frac{13.595inH2O}{1inHg}=390.86inH2O$

d) 1inHg=3.386kPa

so:

$28.75inHg*\frac{3.386kPa}{1inHg}=97.36kPa$

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.

$Sg=\frac{\gamma_{substance}}{\gamma_{H_{2}O}}$

so when solving for the specific gravity of the substance, we get that it is:

$\gamma _{substance}=(SG)\gamma _{H_{2}O}$

which can be used in the pressure equation:

$P=\gamma h$

when solving for h we get that:

$h=\frac{P}{\gamma _{substance}}$

when substituting equations we get that:

$h=\frac{P}{(SG)\gamma _{H_{2}O}}$

we know that:

$\gamma _{H_{2}O}=9.8kN/m^{3}$

and from the previous part of the problem we know the pressure in kPa, so when using this data we get that:

$h=\frac{97.36kPa}{(0.8)(9.8kN/m^{3})}$

so:

h=12.42m