Question

5.6. A slurry contains crystals of copper sulfate pentahydrate [CuSO4 5H20(s), specific gravity 2.3 suspended in an aqueous copper sulfate solution (liquid SG 1.2). A sensitive transducer is used to measure the pressure difference, AP(Pa), between two points in the sample container separated by a vertical distance of h meters. The reading is in turn used to determine the mass fraction of crystals in the slurry, xe(kg crystals/kg slurry). Liquid solution SG 1.2 Crystals SG 2.3 Solution-crystal slurry (a) Derive an expression for thetransducer reading, ΔΡ(Pa), in terms of the overall slurry density. Psl(kg/m3), assuming that the equation used to calculate the pressure head in Chapter 3 (P Po +pgh) is valid for this two-phase system.

Answer 1

Consider a point at the top of the container and a point at the bottom of the container, between which the pressure difference is measured.

Let P₁ be the pressure at the bottom of the container, at a height, h₁. Let P₂ be the pressure at the top of the container, at a height, h₂.

Pressure at point 1, P₁, is

$P_{1}=P_{0}+g\rho_{sl}h_{1}$

where P₀ is the Atmospheric Pressure

$\rho_{sl}$ is the Overall density of slurry

g is the Acceleration due to the gravity

Pressure at point 2, P₂, is

$P_{2}=P_{0}+g\rho_{sl}h_{2}$

Pressure difference between the two points, $\Delta$ P, is

$\Delta P=P_{1}-P_{2}$

$\Rightarrow \Delta P=(P_{0}-g\rho_{sl}h_{1})-(P_{0}-g\rho_{sl}h_{2})$

$\Rightarrow \Delta P=g\rho_{sl}(h_{1}-h_{2})$

The difference between the two points is h=h₁-h₂

$\Rightarrow \Delta P=g\rho_{sl}h$

Considering the units of the expression

$\Rightarrow \Delta P=g(m/s^{2})*\rho_{sl}(kg/m^{3})*h(m)$

$\Rightarrow \Delta P=\rho_{sl}gh(kg/ms^{2})$

(Unit: kg/m s² is Pa)

$\mathbf{\Delta P(Pa)=\boldsymbol{\rho}_{sl}gh}$

The expression for the transducer reading is $\mathbf{\Delta P=\boldsymbol{\rho}_{sl}gh}$