Density and Pressure Variation in a Real Fluid

(a) Expression for Density $\rho$ as a function of depth $h$

We start with the definition of Bulk Modulus ($B$) and Compressibility ($\beta$). Compressibility is the reciprocal of Bulk Modulus.

$$ B = \frac{\Delta P}{-\frac{\Delta V}{V}} = \frac{dP}{\frac{d\rho}{\rho}} $$ $$ \beta = \frac{1}{B} = \frac{1}{\rho} \frac{d\rho}{dP} $$

From the hydrostatic condition, the pressure increase due to a fluid column of height $dh$ is given by:

$$ dP = \rho g dh \quad \text{— (ii)} $$

Equating (i) and (ii):

$$ \frac{1}{\beta} \frac{d\rho}{\rho} = \rho g dh $$ $$ \frac{d\rho}{\rho^2} = \beta g dh $$

Now, we integrate both sides. At the surface ($h=0$), the density is $\rho_0$. At depth $h$, the density is $\rho$.

$$ \int_{\rho_0}^{\rho} \frac{d\rho}{\rho^2} = \int_{0}^{h} \beta g dh $$ $$ \left[ -\frac{1}{\rho} \right]_{\rho_0}^{\rho} = \beta g h $$ $$ -\frac{1}{\rho} – \left( -\frac{1}{\rho_0} \right) = \beta g h $$ $$ \frac{1}{\rho_0} – \frac{1}{\rho} = \beta g h $$

Rearranging to solve for $\rho$:

$$ \frac{1}{\rho} = \frac{1}{\rho_0} – \beta g h = \frac{1 – \rho_0 \beta g h}{\rho_0} $$ $$ \rho = \frac{\rho_0}{1 – \rho_0 \beta g h} $$ $$ \rho = \rho_0 (1 – \rho_0 \beta g h)^{-1} $$

Since $\beta$ is typically very small for fluids, the term $\rho_0 \beta g h$ is much less than 1. We can use the Binomial approximation $(1-x)^{-n} \approx 1+nx$.

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(b) Expression for Pressure $p$ as a function of depth $h$

To find the pressure at depth $h$, we integrate the hydrostatic equation $dp = \rho g dh$ using our new expression for density.

$$ p = p_0 + \int_{0}^{h} \rho g dh $$ Substitute $\rho \approx \rho_0 (1 + \rho_0 \beta g h)$: $$ p = p_0 + \int_{0}^{h} \rho_0 (1 + \rho_0 \beta g h) g dh $$ $$ p = p_0 + \rho_0 g \int_{0}^{h} (1 + \rho_0 \beta g h) dh $$ $$ p = p_0 + \rho_0 g \left[ h + \frac{\rho_0 \beta g h^2}{2} \right]_{0}^{h} $$