Fluid Mechanics Exercise

At a general depth $z$ let the pressure be $P$ . Assume that the pressure increases with depth. Taking a fluid element of cross-section area $A$ and thickness $\delta z$ at a depth of $z$ , sketching out its free body diagram and balancing all vertical forces gives the following:

$PA + \rho(z) A g \ \delta z = PA + \frac{dP}{dz}\delta z$

$\implies \rho(z) g=\frac{dP}{dz}$

Substituting the density function, separating variables and integrating gives:

$\int_{P_{atm}}^{P} \frac{dP}{P+P_o} = \frac{\rho_og}{2P_o} \int_{0}^{z} dz$

$P(z) = (P_o+P_{atm}) \mathrm{e}^{\frac{\rho_o g z}{2 P_o}} - P_o$

$\implies \alpha = 1 \ ; \ \beta = 2$

Now, Having found the pressure as a function of depth, the density as a function of depth can be found by substituting the above result, which yeilds:

$\rho(z) = \frac{\rho_o (P_o+P_{atm}) }{2P_o}\mathrm{e}^{\frac{\rho_o g z}{2 P_o}}$

Finally, the mass of the fluid element specified above is:

$\delta M = \rho(z) A \delta z$

Substituting the density expression, taking the limit as $\delta z$ tends to zero and integrating throughout the height of the cylinder:

$M = \int_{0}^{h} \frac{\rho_oA (P_o+P_{atm}) }{2P_o}\mathrm{e}^{\frac{\rho_o g z}{2 P_o}} dz$ $M = \frac{A}{g} (P_o+P_{atm})\left(\mathrm{e}^{\frac{\rho_o g h}{2 P_o}} -1\right)$

$\implies \gamma = 1$

$M$ is actually $\dfrac{A}{g}(P_0+P_{atm})\left (e^{\frac{\rho_0gh}{2P_0}}-1\right )$ .