21-Mat-A3 Structure and Characterization of Materials · December 2016
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 2016 — 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, ironmaking, magnesium and aluminum production, hydrometallurgy and electrometallurgy — and is answered as such.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
F. Habashi, Textbook of Pyrometallurgy — roasting, smelting, mass and heat balances.
T. Rosenqvist, Principles of Extractive Metallurgy, 2nd ed. — blast-furnace ironmaking, thermal and electrolytic reduction, electrometallurgy.
B. A. Wills and J. Finch, Wills' Mineral Processing Technology, 8th ed. — pulp density, gravity and electrostatic separation, froth flotation.
D. R. Gaskell, Introduction to the Thermodynamics of Materials — heat capacities, transformation enthalpies, electrochemistry.
ASM Handbook, Vol. 15 (Casting) and the CIM / Canadian Institute of Mining 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
15 m3/h
20 m3/h
30 m3/h
Pulp density
ρp
1500 kg/m3
to be found
to be found
Solids by weight
x
to be found
20 %
30 %
Ore density
ρs
3000 kg/m3
3000 kg/m3
3000 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 15 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 20 % and 30 %.
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 $x$,
$$x=\frac{\dfrac{1}{\rho_p}-\dfrac{1}{\rho_w}}{\dfrac{1}{\rho_s}-\dfrac{1}{\rho_w}}
=\frac{\dfrac{1}{1500}-\dfrac{1}{1000}}{\dfrac{1}{3000}-\dfrac{1}{1000}}
=\frac{-3.3333\times10^{-4}}{-6.6667\times10^{-4}}=0.5000$$
$$\boxed{x=50.00\ \%\ \text{solids by weight}}$$
This particular pulp density and ore density combine to a clean half-and-half split, which is a useful check that the arithmetic has not been inverted.
Part (a)(ii) — convert to a solids mass flow. The total pulp mass flow is
$$\dot m_p=Q\,\rho_p=15\ \text{m}^3\text{/h}\times1500\ \text{kg/m}^3=22\,500\ \text{kg/h}$$
and the ore is the fraction $x$ of that:
$$\dot m_s=x\,\dot m_p=0.5000\times22\,500=11\,250\ \text{kg/h}$$
$$\boxed{\dot m_s=11\,250\ \text{kg/h}=11.25\ \text{t/h of dry ore}}$$
Check the answer by rebuilding the volume. The ore occupies $11\,250/3000=3.750$ m3/h and the water $(22\,500-11\,250)/1000=11.250$ m3/h. Those sum to 15.00 m3/h, the stated feed rate, so the split is internally consistent.
Part (b)(i) — get the density of each feed. Here the concentrations are known and the densities are not, so the mixing rule is used forward. For stream 1 at $x_1=0.20$,
$$\rho_1=\left(\frac{0.20}{3000}+\frac{0.80}{1000}\right)^{-1}=(6.6667\times10^{-5}+8.0000\times10^{-4})^{-1}=1153.8\ \text{kg/m}^3$$
and for stream 2 at $x_2=0.30$,
$$\rho_2=\left(\frac{0.30}{3000}+\frac{0.70}{1000}\right)^{-1}=(1.0000\times10^{-4}+7.0000\times10^{-4})^{-1}=1250.0\ \text{kg/m}^3$$
The denser stream is again the more concentrated one.
Convert each feed to mass flows. Multiplying by the stated volumetric rates,
$$\dot m_1=20\times1153.8=23\,076.9\ \text{kg/h},\qquad \dot m_2=30\times1250.0=37\,500.0\ \text{kg/h}$$
so the ore carried is $\dot m_{s1}=0.20\times23\,076.9=4615.4$ kg/h and $\dot m_{s2}=0.30\times37\,500.0=11\,250.0$ kg/h.
Close the balance over the pump. Nothing is added or removed at the junction, so ore and water are each conserved:
$$\dot m_{s,\text{tot}}=4615.4+11\,250.0=15\,865.4\ \text{kg/h},\qquad
\dot m_{\text{tot}}=23\,076.9+37\,500.0=60\,576.9\ \text{kg/h}$$
The combined concentration is therefore
$$x_{\text{mix}}=\frac{15\,865.4}{60\,576.9}=0.2619$$
$$\boxed{x_{\text{mix}}=26.19\ \%\ \text{solids by weight}}$$
This sits well above the naive volumetric average of 20 % and 30 %, 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:
$$\boxed{\dot m_{s,\text{tot}}=15\,865.4\ \text{kg/h}=15.87\ \text{t/h of dry solids}}$$
As a closure check, the combined pulp density is $60\,576.9/50=1211.5$ kg/m3, and feeding $x_{\text{mix}}=0.2619$ back through the mixing rule returns the same 1211.5 kg/m3.
