21-Mat-A3 Structure and Characterization of Materials · December 2019
Question 2 of 7: Mass Balance (20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2019 — 10-Met-A3, Metal Extraction Processes. Three hours, closed book, one approved calculator (Casio or Sharp). Seven problems of 20 marks each; the rubric asks for any five, and only the first five in the answer book are marked. All seven are solved here, because this set is a study resource rather than an exam script.
Note on the exam title. The printed exam header reads 10-Met-A3, Metal Extraction Processes. The content is extractive metallurgy — mineral processing, mass and heat balances, pyrometallurgical roasting, refining, magnesium production and electrometallurgy — and is answered as such.
Note on the data. The Question 6 iron heat-balance data set (Cp expressions and transformation enthalpies for α/β/γ/δ-Fe) uses a mass of 55.85 kg (chosen so it equals exactly 1000 mol) and temperature endpoints of 160–1735 °C, crossing the 1535 °C melting point into the liquid.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
F. Habashi, Textbook of Pyrometallurgy — roasting, refining, mass and heat balances.
T. Rosenqvist, Principles of Extractive Metallurgy, 2nd ed. — thermal and electrolytic reduction, electrometallurgy, refining.
B. A. Wills and J. Finch, Wills' Mineral Processing Technology, 8th ed. — pulp density, gravity/electrostatic separation, froth flotation.
D. R. Gaskell, Introduction to the Thermodynamics of Materials — heat capacities, transformation enthalpies, electrochemistry.
ASM Handbook, Vol. 15 (Casting) and CIM (Canadian Institute of Mining, Metallurgy and Petroleum) practice literature for Canadian smelter and refinery practice.
Given. Two independent slurry problems share one physical model: a two-phase pulp of ore and water whose bulk density follows from the volumes the two phases occupy.
Given data
Quantity
Symbol
Part (a)
Part (b), stream 1
Part (b), stream 2
Volumetric flow
Q
10 m3/h
15 m3/h
25 m3/h
Pulp density
ρp
1250 kg/m3
to be found
to be found
Solids by weight
x
to be found
15 %
25 %
Ore density
ρs
2850 kg/m3
2700 kg/m3
2700 kg/m3
Water density
ρw
1000 kg/m3 (process water, taken as pure)
Find. In (a), the solids mass fraction and the dry-solids mass flow carried by the 10 m3/h stream; in (b), the solids concentration of the two streams after they merge in the pump, and the dry tonnage the pump handles each hour.
Figure 2.1 — Part (b): two slurry feeds combine in one pump. Solids and water are each conserved across the junction, so the combined concentration is a mass-weighted average, never the arithmetic mean of 15 % and 25 %.
Approach. Write the reciprocal (volume-additive) mixing rule that links pulp density to solids mass fraction, invert it in part (a) to get the concentration and hence the solids flow, and in part (b) use it forward on each feed to get the two pulp densities, then close a solids balance and a total-mass balance over the pump.
State the two-phase mixing rule. One kilogram of pulp is made of $x$ kg of ore occupying $x/\rho_s$ and $(1-x)$ kg of water occupying $(1-x)/\rho_w$. Volumes add, so
$$\frac{1}{\rho_p}=\frac{x}{\rho_s}+\frac{1-x}{\rho_w}$$
where $\rho_p$ is the pulp (slurry) density, $\rho_s$ the ore density and $\rho_w$ the water density.
Part (a)(i) — invert the rule for the solids mass fraction. Solving for $x$,
$$x=\frac{\dfrac{1}{\rho_p}-\dfrac{1}{\rho_w}}{\dfrac{1}{\rho_s}-\dfrac{1}{\rho_w}}
=\frac{\dfrac{1}{1250}-\dfrac{1}{1000}}{\dfrac{1}{2850}-\dfrac{1}{1000}}
=\frac{-2.0000\times10^{-4}}{-6.4912\times10^{-4}}=0.30811$$
$$\boxed{x=30.81\ \%\ \text{solids by weight}}$$
Part (a)(ii) — convert to a solids mass flow. The total pulp mass flow is
$$\dot m_p=Q\,\rho_p=10\ \text{m}^3\text{/h}\times1250\ \text{kg/m}^3=12\,500\ \text{kg/h}$$
and the ore is the fraction $x$ of that:
$$\dot m_s=x\,\dot m_p=0.30811\times12\,500=3851.4\ \text{kg/h}$$
$$\boxed{\dot m_s=3851\ \text{kg/h}=3.851\ \text{t/h of dry ore}}$$
Check the answer by rebuilding the volume. The ore occupies $3851.4/2850=1.351$ m3/h and the water $(12\,500-3851.4)/1000=8.649$ m3/h. Those sum to 10.00 m3/h, the stated feed rate, so the split is internally consistent.
Part (b)(i) — get the density of each feed. The concentrations are known and the densities are not, so the mixing rule is used forward. For stream 1 at $x_1=0.15$,
$$\rho_1=\left(\frac{0.15}{2700}+\frac{0.85}{1000}\right)^{-1}=(5.5556\times10^{-5}+8.5000\times10^{-4})^{-1}=1104.3\ \text{kg/m}^3$$
and for stream 2 at $x_2=0.25$,
$$\rho_2=\left(\frac{0.25}{2700}+\frac{0.75}{1000}\right)^{-1}=(9.2593\times10^{-5}+7.5000\times10^{-4})^{-1}=1186.8\ \text{kg/m}^3$$
The denser stream is again the more concentrated one.
Convert each feed to mass flows. Multiplying by the stated volumetric rates,
$$\dot m_1=15\times1104.3=16\,564.4\ \text{kg/h},\qquad \dot m_2=25\times1186.8=29\,670.3\ \text{kg/h}$$
so the ore carried is $\dot m_{s1}=0.15\times16\,564.4=2484.7$ kg/h and $\dot m_{s2}=0.25\times29\,670.3=7417.6$ kg/h.
Close the balance over the pump. Nothing is added or removed at the junction, so ore and water are each conserved:
$$\dot m_{s,\text{tot}}=2484.7+7417.6=9902.2\ \text{kg/h},\qquad
\dot m_{\text{tot}}=16\,564.4+29\,670.3=46\,234.7\ \text{kg/h}$$
The combined concentration is therefore
$$x_{\text{mix}}=\frac{9902.2}{46\,234.7}=0.21418$$
$$\boxed{x_{\text{mix}}=21.42\ \%\ \text{solids by weight}}$$
This sits above the naive volumetric average of 15 % and 25 %, because the higher-flow, higher-concentration second stream dominates the weighting.
Part (b)(ii) — report the dry tonnage. The solids balance already gives it directly:
$$\boxed{\dot m_{s,\text{tot}}=9902.2\ \text{kg/h}=9.902\ \text{t/h of dry solids}}$$
As a closure check, the combined pulp density is $46\,234.7/40=1155.9$ kg/m3, and feeding $x_{\text{mix}}=0.21418$ back through the mixing rule returns the same value.
