Question 2.1: What is the intensity of pressure in the ocean at a depth of...
What is the intensity of pressure in the ocean at a depth of 1500 m, assuming (a) salt water is incompressible with a specific weight of 10050 N/m^{3} and (b) salt water is compressible and weighs 10050 N/m^{3} at the free surface? E (bulk modulus of elasticity of salt water) = 2070 MN/m^{2} (constant).
(a) For an incompressible f luid, the intensity of pressure at a depth, according to Eq. (2.16), is p (pressure in gauge) = \rho g h = 10050 (1500) N/m^{2} = 15.08 MN/m^{2} gauge
p-p_{0}=\rho g\left(z_{0}-z_{1}\right)=\rho g h (2.16)
(b) The change in pressure with the depth of liquid h from free surface can be written according to Eq. (2.14) as
\frac{ d p}{ d z}=-\rho g (2.14)
\frac{ d p}{ d h}=\rho g (2.68)
Again from the definition of bulk modulus of elasticity E (Eq. (1.5)),
E=\lim _{\Delta \rho \Rightarrow 0} \frac{\Delta p}{(\Delta \rho / \rho)}=\rho \frac{ d p}{ d \rho} (1.5)
d p=E \frac{ d \rho}{\rho} (2.69)
Integrating equation (2.69), for a constant value of E, we get
p=E \ln \rho+C (2.70)
The integration constant C can be found out by considering p=p_{0} \text { and } \rho=\rho_{0} at the free surface.
Therefore Eq. (2.70) becomes
p-p_{0}=E \ln \left(\frac{\rho}{\rho_{0}}\right) (2.71)
Substitution of dp from Eq. (2.68) into Eq. (2.69) yields
d h=\frac{E d \rho}{g \rho^{2}}
After integration
h=-\frac{E}{g \rho}+C_{1}
The constant C_{1} is found out from the condition that, \rho=\rho_{0} \text { at } h = 0 (free surface)
Hence, h=\frac{E}{g}\left(\frac{1}{\rho_{0}}-\frac{1}{\rho}\right)
from which \frac{\rho}{\rho_{0}}=\frac{E}{E-h \rho_{0} g}
Substituting this value \text { of } \rho / \rho_{0} \text { in } Eq. (2.71), we have
p-p_{0}=E \ln \left(\frac{E}{E-h \rho_{0} g}\right)
Therefore,
p(\text { in gauge })=2.07 \times 10^{9} \ln \left[\frac{2.07 \times 10^{9}}{2.07 \times 10^{9}-(10050)(1500)}\right] N / m ^{2} \text { gauge }
= 15.13 MN / m ^{2} \text { gauge }