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

a) 14.12psi, b) 390.86inH2O, d) 97.36kPa, d)12.42m

Step-by-step explanation:

In order to do the conversions we need to use conversion rates:

a) 1psi=2.03602inHg

so:

`28.75inHg*(1psi)/(2.03602inHg)=14.12psi`

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

b) 1inHg=13.595inH2O

so:

`28.75inHg*(13.595inH2O)/(1inHg)=390.86inH2O`

d) 1inHg=3.386kPa

so:

`28.75inHg*(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=(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=(97.36kPa)/((0.8)(9.8kN/m^(3)))`

so:

h=12.42m