Calcular a pressão no oceano a uma profundidade de 1500 m, considerando a água salgada como um líquido compressível com ε = 21.000 kg/cm2 e densidade absoluta de 1025 kg/m3 na superfície.
ρ0
1025 kg/m3
hf
1500 m
pf
a calcular
ε = 21.000 kg/cm2
Da definição de módulo de elasticidade volumétrica
ϵ
=
−
d
p
(
d
V
V
)
⇒
d
V
V
=
−
1
ϵ
d
p
{\displaystyle \epsilon \;=\;-{\frac {dp}{({\frac {dV}{V}})}}\Rightarrow \;\;\;{\frac {dV}{V}}\;=\;-{\frac {1}{\epsilon }}\;dp}
d
ρ
ρ
=
−
d
V
V
=
1
ϵ
d
p
{\displaystyle {\frac {d\rho }{\rho }}\;=\;-{\frac {dV}{V}}\;=\;{\frac {1}{\epsilon }}\;dp}
∫
d
ρ
ρ
=
1
ϵ
∫
d
p
⇒
p
f
=
p
0
+
ϵ
ln
(
ρ
f
ρ
0
)
{\displaystyle \int {\frac {d\rho }{\rho }}\;=\;{\frac {1}{\epsilon }}\int dp\Rightarrow \;\;\;p_{f}\;=\;p_{0}+\epsilon \ln \left({\frac {\rho _{f}}{\rho _{0}}}\right)}
d
p
d
z
=
−
ρ
g
{\displaystyle {\frac {dp}{dz}}\;=\;-\rho g}
d
ρ
ρ
=
1
ϵ
d
p
=
1
ϵ
ρ
d
h
⇒
d
ρ
ρ
2
=
1
ϵ
d
h
{\displaystyle {\frac {d\rho }{\rho }}\;=\;{\frac {1}{\epsilon }}\;dp\;=\;{\frac {1}{\epsilon }}\rho dh\Rightarrow \;\;\;{\frac {d\rho }{\rho ^{2}}}\;=\;{\frac {1}{\epsilon }}\;dh}
∫
d
ρ
ρ
2
=
1
ϵ
∫
d
h
⇒
(
1
ρ
0
−
1
ρ
f
)
=
1
ϵ
(
h
f
−
h
0
)
{\displaystyle \int {\frac {d\rho }{\rho ^{2}}}\;=\;{\frac {1}{\epsilon }}\int dh\Rightarrow \;\;\;\left({\frac {1}{\rho _{0}}}\;-\;{\frac {1}{\rho _{f}}}\right)\;=\;{\frac {1}{\epsilon }}(h_{f}\;-\;h_{0})}
1
ρ
f
=
1
ρ
0
−
1
ϵ
(
h
f
−
h
0
)
⇒
ρ
f
=
1
1
ρ
0
−
1
ϵ
(
h
f
−
h
0
)
{\displaystyle {\frac {1}{\rho _{f}}}\;=\;{\frac {1}{\rho _{0}}}\;-\;{\frac {1}{\epsilon }}(h_{f}\;-\;h_{0})\Rightarrow \;\;\;\ \rho _{f}\;=\;{\frac {1}{{\frac {1}{\rho _{0}}}\;-\;{\frac {1}{\epsilon }}(h_{f}\;-\;h_{0})}}}
ρ
f
ρ
0
=
1
1
−
ρ
0
ϵ
(
h
f
−
h
0
)
{\displaystyle {\frac {\rho _{f}}{\rho _{0}}}\;=\;{\frac {1}{1\;-\;{\frac {\rho _{0}}{\epsilon }}(h_{f}\;-\;h_{0})}}}
p
f
=
p
0
+
ϵ
ln
(
ρ
f
ρ
0
)
=
p
0
+
ϵ
ln
(
1
1
−
ρ
0
ϵ
(
h
f
−
h
0
)
)
{\displaystyle p_{f}\;=\;p_{0}+\epsilon \ln \left({\frac {\rho _{f}}{\rho _{0}}}\right)\;=\;p_{0}\;+\;\epsilon \ln \left({\frac {1}{1\;-\;{\frac {\rho _{0}}{\epsilon }}(h_{f}\;-\;h_{0})}}\right)}
p
f
=
p
0
−
ϵ
ln
(
1
−
ρ
0
ϵ
(
h
f
−
h
0
)
)
{\displaystyle p_{f}\;=\;p_{0}\;-\;\epsilon \ln \left(1\;-\;{\frac {\rho _{0}}{\epsilon }}(h_{f}\;-\;h_{0})\right)}
p
f
=
0
−
21000
k
g
/
c
m
2
ln
(
1
−
1025
k
g
/
m
3
21000
k
g
/
c
m
2
(
1500
m
−
0
)
)
=
−
210000000
k
g
/
m
2
ln
(
1
−
1025
k
g
/
m
3
210000000
k
g
/
c
m
2
⋅
1500
m
)
{\displaystyle p_{f}\;=\;0\;-\;21000\;kg/cm^{2}\ln \left(1\;-\;{\frac {1025\;kg/m^{3}}{21000\;kg/cm^{2}}}(1500\;m\;-\;0)\right)\;=\;\;-\;210000000\;kg/m^{2}\ln \left(1\;-\;{\frac {1025\;kg/m^{3}}{210000000\;kg/cm^{2}}}\cdot 1500\;m\right)}
=
1540000
k
g
/
m
2
=
154
k
g
/
c
m
2
{\displaystyle \;=\;1540000\;kg/m^{2}\;=\;154\;kg/cm^{2}}
