Answer:
[tex] P_{gauge}= 195020 Pa= 195.02 Kpa[/tex]
And the absolute pressure at the bottom would be:
[tex] P_{abs}= P_{atm} +P_{gauge}= 101325 Pa +195020 Pa= 296345 Pa= 296.345 Kpa[/tex]
Step-by-step explanation:
From the figure attached we know the following two conditions:
[tex] Z=0 , T=40C[/tex]
[tex] Z=20 , T=4[/tex]
And we want a linear model like this one:
[tex] T = mZ +b[/tex]
Where m abd b represent the slope and intercept.
We can find the slope with the following expression:
[tex] m =\frac{4-40}{20-0}= -\frac{9}{5}[/tex]
And then we can find the intercept using the first point:
[tex] 40 = -\frac{9}{5} *0 +b , b= 40[/tex]
So then the linear model is:
[tex] T = -\frac{9}{5} Z +40[/tex]
Assuming the following formula for the density:
[tex] \rho_{H2O} = 1 +\frac{T-40}{3600}[/tex]
If we replace the expression for T into the density we got:
[tex] \rho_{H2O} = 1+ \frac{-\frac{9}{5} Z +40 -40}{3600}[/tex]
[tex] \rho_{H2O} = 1 -\frac{z}{2000}[/tex]
Since the units for [tex] \rho_{H2O}[/tex] are g/ml we can convert this into kg/m^3 multiplying by 1000 and we got:
[tex] \rho_{H2O} =1000- \frac{z}{2}[/tex]
Then we have the following differential equation:
[tex] \frac{dP}{dz} = -\rho g[/tex]
If we replace the expression for the density in terms of z we got:
[tex] \frac{dP}{dz} = (\frac{z}{2} -1000) g[/tex]
And we can separate like this:
[tex] dP = (\frac{z}{2} -1000) g dz[/tex]
Now w can integrate both sides and we got:
[tex] \Delta P = g[\frac{(\Delta z)^2}{4} -1000\Delta z][/tex]
And that would be the expression for the change of pressure:
For our case we have 20m of difference between the surface and the bottom so then the change of pressure between these two levels is:
[tex] \Delta P= 9.8 \frac{m}{s^2} [\frac{(20m)^2}{4} -1000(20m)]= -195020 Pa[/tex]
We know that the pressure increase with the depth. So then the gage pressure at the bottom would be:
[tex] P_{gauge}= 195020 Pa= 195.02 Kpa[/tex]
And the absolute pressure at the bottom would be:
[tex] P_{abs}= P_{atm} +P_{gauge}= 101325 Pa +195020 Pa= 296345 Pa= 296.345 Kpa[/tex]
