A slurry contains crystals of copper sulfate pentahydrate [CuSO4•5H2O(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, ΔP(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, xc(kg crystals/kg slurry). 

(a) Derive an expression for the transducer reading, ΔP(Pa), in terms of the overall slurry density, ρ_sl(kg/m³), assuming that the equation used to calculate the pressure head in Chapter 3 (P = P₀ + ρgh) is valid for this two-phase system. 

(b) Validate the following expression relating the overall slurry density to the liquid and solid crystal densities (ρ_l and ρ_c) and the mass fraction of crystals in the slurry: 

(c) Suppose 175 kg of the slurry is placed in the sample container with h = 0:200 m and a transducer reading ΔP = 2775 Pa is obtained. Calculate (i) ρ_sl, (ii) x_c, (iii) the total slurry volume, (iv) the mass of crystals in the slurry, (v) the mass of anhydrous copper sulfate (CuSO₄ without the water of hydration) in the crystals, (vi) the mass of liquid solution, and (vii) the volume of liquid solution. 

(d) Prepare a spreadsheet to generate a calibration curve of x_c versus ΔP for this device. Take as inputs ρc (kg/m₃), ρl(kg/m₃), and h(m), and calculate ΔP(Pa) for x_c = 0.0; 0.05; 0.10; ... ; 0.60. Run the program for the parameter values in this problem (ρ_c = 2300, ρ_l = 1200, and h = 0:200). Then plot x_c versus ΔP (have the spreadsheet program do it, if possible), and verify that the value of x_c corresponding to ΔP = 2775 Pa on the calibration curve corresponds to the value calculated in Part (c). 

(e) Derive the expression in Part (b). Take a basis of 1 kg of slurry [x^c (kg), V_c (m³) crystals, (1 - xc)(kg), V_l (m³) liquid], and use the fact that the volumes of the crystals and liquid are additive.

Transcribed Image Text:

Liquid solution SG = 1.2 Crystals SG = 2.3 h Solution-crystal - slurry AP