(3.36)

In the above equation, i stands for an assayed metal and j for a process unit.

The calculation of these best estimates requires the calculation of the derivatives of S for each of the unknowns, that is, the non-measured solid mass flowrates. The best estimates are those unknown variable values for which the derivatives are all together equal to zero:

$\frac{\delta S}{\delta W_K}=0$ (3.37)

(3.37)

It is worth noting that with the obtained best estimates the mass conservation equations are not rigorously verified; essentially, the node imbalances are only minimized. By experience, the estimates obtained by this method are as good as the measured values used for the calculation are themselves (Example 3.10).

Example 3.10

For the example data set in Table ex 3.9, apply the Node Imbalance Minimization method to estimate the solid split value and compare it with the 2-product formula estimates.

Solution

In this example there is one process unit and three metals assayed; hence, there are four mass conservation equations and hence four imbalances. The criterion S can be written as:

$\begin{array}{ll}S\hfill & =W_F^2+W_C^2+W_T^2-2W_F{W}_C-2W_F{W}_T+2W_C{W}_T\hfill \\ \hfill & +W_F^2\sum _{i=1}^3{x}_{Fi}^2+W_C^2\sum _{i=1}^3{x}_{Ci}^2+W_T^2\sum _{i=1}^3{x}_{Ti}^2\hfill \\ \hfill & -2W_F{W}_C\sum _{i=1}^3{x}_{Fi}{x}_{Ci}-2W_F{W}_T\sum _{i=1}^3{x}_{Fi}{x}_{Ti}+2W_C{W}_T\sum _{i=1}^3{x}_{Ci}{x}_{Ti}\hfill \end{array}$ (3.38)

(3.38)

where i stands for the assayed metal.

The derivatives of S for W_C and W_T are then:

$\frac{\delta S}{\delta W_C}=2\left(W_C-W_F+W_T+W_C\sum _{i=1}^3{x}_{Ci}^2-W_F\sum _{i=1}^3{x}_{Fi}{x}_{Ci}+W_T\sum _{i=1}^3{x}_{Ci}{x}_{Ti}\right)=0$ (3.39)

(3.39)

$\frac{\delta S}{\delta W_T}=2\left(W_T-W_F+W_C+W_T\sum _{i=1}^3{x}_{Ti}^2-W_F\sum _{i=1}^3{x}_{Fi}{x}_{Ti}+W_C\sum _{i=1}^3{x}_{Ci}{x}_{Ti}\right)=0$ (3.40)

(3.40)

Now, the solid split value can be obtained from the above 2 equations:

$\frac{W_T}{W_F}=\frac{\left(1+\sum _{i=1}^3{x}_{Fi}{x}_{Ti}\right)\left(1+\sum _{i=1}^3{x}_{Ci}^2\right)-\left(1+\sum _{i=1}^3{x}_{Fi}{x}_{Ci}\right)\left(1+\sum _{i=1}^3{x}_{Ci}{x}_{Ti}\right)}{\left(1+\sum _{i=1}^3{x}_{Ti}^2\right)\left(1+\sum _{i=1}^3{x}_{Ci}^2\right)-{\left(1+\sum _{i=1}^3{x}_{Ci}{x}_{Ti}\right)}^2}$ (3.41)

(3.41)

and $\frac{W_C}{W_F}=1-\frac{W_T}{W_F}$

Using the data set of Table ex 3.9, Table ex 3.10 is derived.

Table ex 3.10

Node Imbalance Example

	$\frac{W_T}{W_F}(\%)$	$\frac{W_C}{W_F}(\%)$
Using Zn, Cu and Fe	80.8	19.1
Using Zn and Cu	93.1	6.87
Using Zn only	93.3	6.67

The results of Table ex 3.10 show that the method is sensitive to the accuracy of the metal assays. Indeed, the Fe assay is bad as demonstrated by the application of the 2-product formula (see Table 3.2).

3.6.3 Two-step Least Squares Minimization

The 2-step least squares minimization method proposes a way to adjust and correct metal assays for the measurement errors that affect them (Wiegel, 1972; Mular, 1979). It is a compromise between the limitations of the node imbalance minimization method (Section 3.6.2) and the complexity of the generalized least squares minimization method (Section 3.6.4).

In the node imbalance method, the deviations to the mass conservation of solids and of metals are considered all together. The obtained estimated flowrate values minimize the deviations to all the mass conservation equations.

In the 2-step least squares method, flowrate values that rigorously verify the mass conservation equations of solids and, at the time, minimize the deviations to the mass conservation equations of metals, are first estimated. In a second step, corrected values of metal assays that rigorously verify the mass conservation equations of metals are estimated.

To achieve, taking Eqs. (3.23) and (3.24) as an example, the mass conservation equations are re-written as:

$\frac{W_T}{W_F}=1-\frac{W_C}{W_F}$ (3.42)

(3.42)

$x_F=\frac{W_C}{W_F}{x}_C+\left(1-\frac{W_C}{W_F}\right)x_T$ (3.43)

(3.43)

and the node imbalance equations for metals become:

$x_F-\frac{W_C}{W_F}{x}_C-\left(1-\frac{W_C}{W_F}\right)x_T=I_x$ (3.44)

(3.44)

or:

$(x_F-x_T)-\frac{W_C}{W_F}(x_C-x_T)=I_x$ (3.45)

(3.45)

The general problem can now be stated as: find the best estimate of $\frac{W_C}{W_F}$ which minimizes the sum of squared imbalances I_x for all nodes and assayed metals:

$S={\sum _j\sum _i(i_{x_j}^i)}^2$ (3.46)

(3.46)

By comparison with the node imbalance minimization method, the criterion only contains the node imbalances for metals and the search variables are a set of independent relative solid flowrates. Similar to the node imbalance minimization method, the calculation of the best estimates of the independent relative solid flowrates requires the calculation of the derivatives of S for each of the unknowns, that is, the independent relative solid flowrates. The best estimates are those values for which the derivatives are all together equal to zero. Due to the way the node imbalances for metals are written (Eq. (3.45)), the mass conservation of relative solid flowrates is rigorously verified for the estimated relative solid flowrate values.

Having determined $\frac{W_C}{W_F}$, assuming the feed solid flowrate $W_F$ is measured, it is possible to estimate W_C and W_T. All these flowrate values rigorously verify their mass conservation equations.

In a second step, it is possible to adjust the metal assay measured values ($x$) to values $(\hat{x})$ that verify their own mass conservation equations as well. Let us call r_x the required adjustment for each metal assay value, then it follows that:

$r_x=x-\hat{x}$ (3.47)

(3.47)

and since:

$({\hat{x}}_F-{\hat{x}}_T)-\frac{W_C}{W_F}({\hat{x}}_C-{\hat{x}}_T)=0$ (3.48)

(3.48)

it follows by difference with Eq. (3.45) that:

$I_x-(r_{x_F}-r_{x_T})-\frac{W_C}{W_F}(r_{x_C}-r_{x_T})=0$ (3.49)

(3.49)

The problem is now to find the values of r_x which minimize the following least squares criterion:

$S=\sum _i\sum _j{(r_{x_j}^i)}^2$ (3.50)

(3.50)

under the equality constraint K of Eq. (3.49). In Eq. (3.50), i stands for the number of streams (or samples) and j for the number of assayed metals. The problem is best solved using the Lagrange technique where the criterion becomes:

$L=\sum _i\sum _j{(r_{x_j}^i)}^2+\sum _k{\lambda }_k{K}_k$ (3.51)

(3.51)

where k is the number of nodes (or mass conservation equations), and λ_k is the Lagrange coefficient of equality constraint K_k.

The problem is solved by calculating the derivatives of L for each of the unknown, that is, $r_{x_j}^i$ and λ_i. Criterion L is minimal when all its derivatives are equal to zero, which leads to a set of equations to be solved. Solving the set of equations provides the best estimates of the $r_{x_j}^i$ values, that is, the adjustments to the measured assay values which will make the adjusted assay values coherent from a mass conservation point of view (Example 3.11).

Example 3.11

For the example data set of Table ex 3.9 apply the 2-step least squares minimization method to estimate the solid split value and compare it with the 2-product formula and node imbalance estimates. Calculate the adjusted metal assay values and compare with the measured values.

Solution

In the present case, Eq. (3.46) is:

$S=\sum _j\sum _i{(I_{x_j}^i)}^2={\sum _{j=1}^3\left[(x_{F_j}-x_{T_j})-\frac{W_C}{W_F}(x_{C_j}-x_{T_j})\right]}^2$

where j stands for Zn, Cu and Fe. Deriving S for the unknown $w_C=\frac{W_C}{W_F}$ gives:

$\frac{\delta S}{\delta w_C}=2\sum _{j=1}^3\left[(x_{F_j}-x_{T_j})-w_C(x_{C_j}-x_{T_j})\right]\left(x_{C_j}-x_{T_j}\right)=0$

The solution of which is:

$w_C=\frac{W_C}{W_F}=\frac{\sum _{j=1}^3(x_{F_j}-x_{T_j})(x_{C_j}-x_{T_j})}{\sum _{j=1}^3{(x_{C_j}-x_{T_j})}^2}$

Using the data set of Table ex 3.9, Table ex 3.11a is derived.

Table ex 3.11a

Two-step Least Squares Minimization Example

	$\frac{W_T}{W_F}(\%)$	$\frac{W_C}{W_F}(\%)$
Using Zn, Cu and Fe	93.4	6.57
Using Zn and Cu	93.3	6.67
Using Zn only	93.3	6.67

The results of Table ex 3.11a show that the method is much less sensitive to the accuracy of the metal assays than the results of the node imbalance method shown in Table ex 3.10. Having determined the solid split, the next step is to adjust the metal assay values to make them coherent with the calculated solid split and coherent from a mass conservation point of view.

In the present case, Eq. (3.51) is:

$L=\sum _{i=1}^3{(r_{x_F}^i)}^2+{(r_{x_C}^i)}^2+{(r_{x_T}^i)}^2+\sum _{i=1}^3{\lambda }_i\left[I_x^i-(r_{x_F}^i-r_{x_T}^i)-w_C(r_{x_C}^i-r_{x_T}^i)\right]$

and the derivatives for each variable are:

$\frac{\delta L}{d{r}_{x_F}^i}=2r_{x_F}^i-{\lambda }_i=0$

$\frac{\delta L}{d{r}_{x_F}^i}=2r_{x_F}^i+{\lambda }_i{w}_C=0$

$\frac{\delta L}{d{r}_{x_T}^i}=2r_{x_F}^i+{\lambda }_i(1-w_C)=0$

$\frac{\delta L}{d{\lambda }_i}=I_x^i-(r_{x_F}^i-r_{x_T}^i)-w_C(r_{x_C}^i-r_{x_T}^i)=0$

Solving the system of equations leads to:

$r_{x_F}^i=\frac{I_x^i}{1+w_C^2+{(1-w_C)}^2}$

$r_{x_C}^i=\frac{-I_x^i{w}_C}{1+w_C^2+{(1-w_C)}^2}$

$r_{x_T}^i=\frac{-I_x^i(1-w_C)}{1+w_C^2+{(1-w_C)}^2}$

and the numerical solution is:

	$I_x^i$	Adjustments $r_{x_j}^i$			Adjusted values ${\hat{x}}_j^i$
		Feed	Concentrate	Tailings	Feed	Concentrate	Tailings
Zn	0.049885	0.049885	0.049885	0.049885	3.903426	52.0717466	0.514828
Cu	−0.01418	−0.01418	−0.01418	−0.01418	0.167552	0.65950362	0.132944
Fe	−1.62385	−1.62385	−1.62385	−1.62385	12.43504	14.6131448	12.28181

Table ex 3.11b shows a comparison of the measured assays (Table ex 3.9) with the adjusted ones where adjustment=(measured – adjusted)/measured as a percent. Note the concentrate stream assays are barely adjusted by comparison with the other 2 stream assays.

Table ex 3.11b

Adjustment Distribution (%)

	Feed	Concentrate	Tailings
Zn	0.68	0.00	−5.07
Cu	−4.72	0.08	5.04
Fe	−7.48	0.39	6.17

3.6.4 Generalized Least Squares Minimization

The generalized least squares minimization method proposes a way to estimate flowrate values and adjust metal assay values in a single step, all together at the same time (Smith and Ichiyen, 1973; Hodouin and Everell, 1980). There is also no limitation on the number of mass conservation equations.

The mass conservation equations are written for the theoretical values of the process variables. The process variable measured values carry measurement errors and hence do not verify the mass conservation equations. It is assumed that the measured values are unbiased, uncorrelated to the other measured values and belong to Normal distributions, N(µ,σ). The balances can be expressed as follows:

$W_F^{*}=W_C^{*}+W_T^{*}$ (3.52)

(3.52)

$W_F^{*}{x}_F^{*}=W_C^{*}{x}_C^{*}+W_T^{*}{x}_T^{*}$ (3.53)

(3.53)

where * denotes the theoretical value of the variable. Since the theoretical values are not known, the objective is to find the best estimates of the theoretical values. With the statistical assumptions made, the maximum likelihood estimates are those which minimize the following generalized least square criterion:

$S=\sum _i\frac{{(W_j-{\hat{W}}_j)}^2}{{\sigma }_{W_j}^2}+\sum _j\sum _i\frac{{(x_{ij}-{\hat{x}}_{ij})}^2}{{\sigma }_{x_{ij}}^2}$ (3.54)

(3.54)

where the ^ denotes the best estimate value and σ² is the variance of the measured value.

It is evident that the best estimates must obey the mass conservation equations while minimizing the generalized least squares criterion S. Therefore, the problem consists in minimizing a least squares criterion under a set of equality constraints, the mass conservation equations. Such a problem is solved by minimizing a Lagrangian of the form:

$L=\sum _i\frac{{(W_j-{\hat{W}}_j)}^2}{{\sigma }_{W_j}^2}+\sum _j\sum _i\frac{{(x_{ij}-{\hat{x}}_{ij})}^2}{{\sigma }_{x_{ij}}^2}+\sum _k{\lambda }_k{K}_k$ (3.55)

(3.55)

where λ_k are the Lagrange coefficients, and K_k the mass conservation equations.

As seen previously, the problem is solved by calculating the derivatives of L for each of the unknowns, that is, the $\hat{W}$, $\hat{x}$ and the λ. Criterion L is minimal when all its derivatives are equal to zero, which leads to a set of equations to be solved. Solving the set of equations provides the best estimate values $\hat{W}$ and $\hat{x}$, which verify the mass conservation equations. It is important to note that even non-measured variables can be estimated, since they appear in the mass conservation equations of the Lagrangian, as long as there is the necessary redundancy.

In Eq. (3.55), the variance of the measured values, σ², are weighting factors. An accurate measured value being associated with low variance cannot be markedly adjusted without significantly impacting the whole criterion value. In contrast, a bad measured value associated to a high standard deviation can be significantly adjusted without impacting the whole criterion value. Hence, adjustments, small or large, will be made in accordance to the confidence we have in the measured value whenever possible considering the mass conservation constraints that must always be verified by the adjusted values (Example 3.12).

Example 3.12

For the example data set of Table ex 3.9 apply the generalized least squares minimization method to estimate the solid split value and compare it with the 2-step least squares minimization estimate. Calculate the adjusted metal assay values and compare with the measured values.

Solution

In the present case, the Lagrangian is:

$L=\sum _{j=1}^3\sum _{i=1}^3\frac{{(x_{ij}-{\hat{x}}_{ij})}^2}{{\sigma }_{x_{ij}}^2}+\sum _{i=1}^3{\lambda }_i{K}_i$

There is no measured mass flowrate, there are 3 stream samples with 3 metal assays each and there are 3 mass conservation equations:

${\hat{x}}_{F_i}=\frac{{\hat{W}}_C}{{\hat{W}}_F}{\hat{x}}_{C_i}+\left(1-\frac{{\hat{W}}_C}{{\hat{W}}_F}\right){\hat{x}}_{T_i}$

where i stands for Zn, Cu, or Fe, and the problem variables are: λ_i, ${\hat{x}}_{ij}$, and ${\hat{w}}_C=\frac{{\hat{W}}_C}{{\hat{W}}_F}$.

The Lagrangian derivatives are:

$\frac{\delta L}{d{\hat{w}}_C}=-\sum _{i=1}^3({\hat{x}}_{C_i}-{\hat{x}}_{T_i})=0$

$\frac{\delta L}{d{\lambda }_i}={\hat{x}}_{F_i}-{\hat{w}}_C{\hat{x}}_{C_i}-(1-{\hat{w}}_C){\hat{x}}_{T_i}=0$

$\frac{\delta L}{d{\hat{x}}_{F_i}}=-2\frac{(x_{F_i}-{\hat{x}}_{F_i})}{{\sigma }_{x_{F_i}}^2}+{\lambda }_i$

$\frac{\delta L}{d{\hat{x}}_{C_i}}=-2\frac{(x_{C_i}-{\hat{x}}_{C_i})}{{\sigma }_{x_{C_i}}^2}-{\lambda }_i{w}_C=0$

$\frac{\delta L}{d{\hat{x}}_{T_i}}=-2\frac{(x_{T_i}-{\hat{x}}_{T_i})}{{\sigma }_{x_{T_i}}^2}-{\lambda }_i(1-w_C)=0$

where i stands for Zn, Cu or Fe, for a total of 13 equations and 13 unknowns. It can be shown that once the value of ${\hat{w}}_C$ is known then there is an analytical solution for ${\hat{x}}_{ij}$:

${\hat{x}}_{F_i}=x_{F_i}-\frac{x_{F_i}-x_{C_i}{\hat{w}}_C-x_{T_i}(1-{\hat{w}}_C)}{\frac{1}{{\sigma }_{x_{F_i}}^2}+\frac{{\hat{w}}_C^2}{{\sigma }_{x_{C_i}}^2}+\frac{{(1-{\hat{w}}_C)}^2}{{\sigma }_{x_{T_i}}^2}}\times \frac{1}{{\sigma }_{x_{F_i}}^2}$

${\hat{x}}_{C_i}=x_{C_i}+\frac{x_{F_i}-x_{C_i}{\hat{w}}_C-x_{T_i}(1-{\hat{w}}_C)}{\frac{1}{{\sigma }_{x_{F_i}}^2}+\frac{{\hat{w}}_C^2}{{\sigma }_{x_{C_i}}^2}+\frac{{(1-{\hat{w}}_C)}^2}{{\sigma }_{x_{T_i}}^2}}\times \frac{{\hat{w}}_C}{{\sigma }_{x_{C_i}}^2}$

${\hat{x}}_{T_i}=x_{T_i}+\frac{x_{F_i}-x_{C_i}{\hat{w}}_C-x_{T_i}(1-{\hat{w}}_C)}{\frac{1}{{\sigma }_{x_{F_i}}^2}+\frac{{\hat{w}}_C^2}{{\sigma }_{x_{C_i}}^2}+\frac{{(1-{\hat{w}}_C)}^2}{{\sigma }_{x_{T_i}}^2}}\times \frac{1-{\hat{w}}_C}{{\sigma }_{x_{T_i}}^2}$

where i stands for Zn, Cu or Fe.

Therefore, it is only required to find the best value of ${\hat{w}}_C$ and the algorithm consists in:

1. Assuming an initial value for ${\hat{w}}_C$

2. Calculating the ${\hat{x}}_{ij}$ values

3. Calculating the criterion S of Eq. (3.54)

4. Iterating on the value of ${\hat{w}}_C$ to minimize the value of S

The solution is given in Table ex 3.12a for variance values all equal to 1 for simplicity. The initial solid split ${\hat{w}}_C$ has been estimated at 6.67% (based on result in Example 3.11).

Table ex 3.12a

Generalized Least Squares Minimization Example

	Feed	Concentrate	Tailings
Zn	3.89	52.1	0.527
Cu	0.167	0.660	0.133
Fe	12.4	14.6	12.3

Table ex 3.12b shows a comparison of the measured assays with the adjusted ones. The obtained results are similar to the ones obtained with the previous method.

Table ex 3.12b

Adjustment Distribution (%)

	Feed	Concentrate	Tailings
Zn (%)	1.00	0.00	−7.47
Cu (%)	−4.64	0.07	4.96
Fe (%)	−7.47	0.38	6.17

While Example 3.12 presents the calculations in detail for a simple case, there exists a more elegant and generic mathematical solution that can be programmed. The solution uses matrices and requires knowledge in matrix algebra.

3.6.5 Mass Balance Models

The formulation of the generalized least squares minimization method enables resolving complex mass balance problems. The Lagrange criterion consists of 2 types of terms: the weighted adjustments and the mass conservation constraints. The criterion can easily be extended to include various types of measurements and mass conservation equations.

The whole set of mass conservation equations that apply to a given data set and mass balance problem is called the mass balance model. Depending on the performed measurements and sample analyses, the following types of mass conservation equations may typically apply in mineral processing plants:

There are also additional mass conservation constraints such as:

All these constraint types can be handled and processed by the Lagrangian of the generalized least squares minimization algorithm. As mentioned, all the necessary mass conservation equations for a given data set constitute the mass balance model.

It is convenient to define the mass balance model using the concept of networks. Indeed, the structure and the number of required mass conservation equations depend on the process flow diagram and the measurement types. Therefore, it is convenient to define a network type by process variable type, knowing that the structure of mass conservation equations for solid flowrates (Eq. (3.23)) is different from that of metal assays (Eq. (3.24)), for example.

Not only does a network type mean a mass conservation equation type, but it also expresses the flow of the mass of interest (solids, liquids, metals…).

For the 2-product process unit of Figure 3.33, assuming 3 metals (Zn, Cu, and Fe) have been assayed on each stream, then Figure 3.35 shows the 2 networks that can be developed, one for each equation type (Eqs. (3.23) and (3.24)).

For the more complex flow diagram of Figure 3.34, assuming again 3 metals (Zn, Cu, and Fe) have been assayed on each of the 6 streams, then Figure 3.36 shows the 2 networks that can be developed.

Assuming a fourth metal, Au for example, has been assayed on the main Feed, main Tailings and main Concentrate streams only, then a third network should be developed for Au as shown in Figure 3.37.

Networks are conveniently represented using a matrix (Cutting, 1976). In a network matrix, an incoming stream is represented with a+1, an outgoing stream with a–1 and any other stream with a 0. For the networks of Figure 3.35 and Figure 3.36, the network matrices are, respectively:

$M=\left[\begin{array}{ccc}1& -1& -1\end{array}\right]$ (3.56)

(3.56)

and

$M=\left[\begin{array}{cccccc}1& -1& -1& 0& 0& 0\\ 0& 1& 0& -1& -1& 0\\ 0& 0& 1& 0& 1& -1\end{array}\right]\begin{array}{c}\mathrm{for}\mathrm{node}\mathrm{A}\\ \mathrm{for}\mathrm{node}\mathrm{B}\\ \mathrm{for}\mathrm{node}\mathrm{C}\end{array}$ (3.57)

(3.57)

Each row represents a node; each column stands for a stream. Equations (3.56) and (3.57) expressed in a matrix form become:

$M_{W}W=0$ (3.58)

(3.58)

$M_x{\overline{W}}_x{X}_i=0$ (3.59)

(3.59)

where M_W and M_x are the network matrices for the solid network and the metal assay network, W is the column matrix of solid mass flowrates, ${\overline{W}}_x$ the diagonal matrix of the solid mass flowrates used in the metal assay network, and X_i the column matrix of metal assay i on each stream of the metal assay network (i stands for Zn, Cu and Fe) (Example 3.13).

Example 3.13

A survey campaign is being designed around the grinding circuit in Figure ex 3.13. For the location of flow meters and samplers, and for the analysis types performed on the samples, develop the mass conservation networks for each data type and mass conservation equation type that apply.

Solution

With the slurry feed flowrate to the grinding circuit being measured, water addition stream flowrates being measured and % solids in slurry being measured, the slurry and water mass balances can be calculated and associated measured values adjusted. The mass conservation equations that apply obey to the Slurry and Water Mass Conservation Networks are given in Figure ex 3.13-S.

The mass conservation equations for the dry solid obey the same network except water addition streams are removed.

Mass size fractions are not conserved through grinding devices (SAG mill and ball mills). Therefore there is no mass conservation equation for size fractions around the grinding devices and consequently there is no node for grinding devices in mass size fraction networks. The same reasoning applies to Cu-by-size fractions.

The (overall) Cu on stream samples is conserved through grinding. Since Cu has been assayed only on the SAG mill feed and the cyclone overflows feeding the flotation circuit, the mass conservation network for Cu on stream samples consists of only one node.

Assuming eleven sieves have been used for measuring the size distributions, then the total system of mass conservation equations consist in:

• Seven mass conservation equations for slurry flowrate variables

• Seven mass conservation equations for % solids variables

• Seven mass conservation equations for water flowrate variables

• Seven mass conservation equations for solid flowrate variables

• Forty eight (4 nodes×12 mass fractions) mass conservation equations for size fractions

• Forty eight mass conservation equations for Cu-by-size fractions

• One mass conservation equation for Cu

for a total of 135 equations and 316 variables.

3.6.6 Error Models

In the generalized least square minimization method, weighting factors prevent large adjustments of trusted measurements and, on the contrary, facilitate large adjustments of poorly measured process variables. Setting the weighting factors is the main challenge of the method (Almasy and Mah, 1984; Chen et al., 1997; Darouach et al., 1989; Keller et al., 1992; Narasimhan and Jordache, 2000; Blanchette et al., 2011).

Assuming a measure obeys to a Normal distribution, the distribution variance is a measure of the confidence we have in the measure itself. Hence, a measure we trust exhibits a small variance, and the variance increases when the confidence decreases. The variance should not be confused with the process variable variation, which also includes the process variation: the variance is the representation of the measurement error only.

Measurement errors come from various sources and are of different kinds. Theories and practical guidelines have been developed to understand the phenomenon and hence render possible the minimization of such errors as discussed in Section 3.2. A brief recap of measurement errors as they pertain to mass balancing will help drive the message home.

Errors fall under two main categories: systematic errors or biases, and random errors. Systematic errors are difficult to detect and identify. First, they can only be suspected over time by nature and definition. Second, a source of comparison must be available. That is achievable to some extent through data reconciliation (Berton and Hodouin, 2000). A bias can be suspected if a measure over time is systematically adjusted to a lower (or higher) value. Once a bias is detected and identified, the bias source and the bias itself should be eliminated. Remember, the generalized least square method is a statistical method for data reconciliation by mass balancing and bias is a deterministic error, not a statistical variable, and is therefore a non-desired disturbance in the statistical data processing.

Measurement errors can also be characterized by their amplitude, small and gross errors, and by the frequency distribution of the measure signal over time: low frequency, high frequency or cyclical. Although there is no definition for a gross error, gross errors are statistically barely probable. As such, if a gross error is suspected it should be corrected before the statistical adjustment is performed. Typically, gross errors have a human source or result from a malfunctioning instrument.

Measurement errors originate from two main sources: sampling errors, and analysis errors. Sampling errors mainly result from the heterogeneity of the material to sample and the difficulty in collecting a representative sample; analysis errors result from the difficulty to analyze sample compositions. The lack of efficiency of the instruments used for measuring process variables also contributes to measurement errors. Analysis errors can be measured in the laboratory and therefore the variance of the analysis error is estimable. Sampling errors are tricky to estimate and one can only typically classify collected samples from the easiest to the more difficult to collect. Assuming systematic and gross errors have been eliminated, then purely random measurement errors can be modeled using 2 functions: one that represents the sampling error and the other one that represents the analysis error. The easiest model is the so-called multiplicative error model where the analysis error variance $({\sigma }_{AE}^2)$ is multiplied by a factor (k_SE) representing the sample representativeness or sampling error: the higher the factor value, the less representative the sample.

${\sigma }_{ME}^2=k_{SE}{\sigma }_{AE}^2$ (3.60)

(3.60)

The multiplicative error model can serve as a basis to develop more advanced error models.

3.6.7 Sensitivity Analysis

The remaining unknown, once a mass balance is obtained for a given set of measured process variables and measurement errors, is: how trustable are the mass balance results or is it possible to determine a confidence interval for each variable estimated by mass balance? That is the objective of the sensitivity analysis.

Assuming systematic and gross errors have been eliminated, and measurement errors are random and obey to Normal distributions, there are two main ways to determine the variance of the estimated variable values (stream flowrates and their composition): Monte Carlo simulation, and error propagation calculation. Each method has its own advantages and disadvantages; each method provides an estimate of the variance of the adjusted or estimated variables.

In the Monte Carlo simulation approach (Laguitton, 1985), data sets are generated by disturbing the mass balance results according to the assumed error model. Each new data set is statistically reconciled and the statistical properties of the reconciled data sets determined.

In the error propagation approach (Flament et al., 1986; Hodouin et al., 1989), the equations through which measurement errors have to be propagated are complex and therefore it is preferable to linearize the equations around a stationary point. Obviously, the obtained mass balance is a good and valid stationary point. The set of linearized equations is then solved to determine the variable variances. This method enables the calculation of covariance values, a valuable feature for calculating confidence intervals around key performance indicators, for instance.

From a set of measured process variables and their variances, therefore, it is possible to determine a set of reconciled values and their variances:

$\left\{\begin{array}{c}\left(W_j,{\sigma }_{W_j}^2\right)\\ \left(x_{ij},{\sigma }_{x_{ij}}^2\right)\end{array}\right\}\Rightarrow \left\{\begin{array}{c}\left({\hat{W}}_j,{\sigma }_{{\hat{W}}_j}^2\right)\\ \left({\hat{x}}_{ij},{\sigma }_{{\hat{x}}_{ij}}^2\right)\end{array}\right\}$ (3.61)

(3.61)

It can be demonstrated that the variance of the adjusted values is less than the variance of the measured values. While the variance of the measured values contributes the most to the variance of the adjusted values, the reduction in the variance values results from the statistical information provided. The data redundancy and the topology of the mass conservation networks are the main contributors to the statistical content of the information provided for the mass balance calculation.

3.6.8 Estimability and Redundancy Analysis

We have seen with the n-product formula that n–1 components need be analyzed on the n+1 streams around the process unit to estimate the n product stream flowrates, assuming the feed stream flowrate is known (or taken as unity to give relative flowrates). The n-product formula is the solution to a set of n equations with n unknowns.

In such a case, there are just enough data to calculate the unknowns: no excess of data, that is, no data redundancy, and the unknowns are mathematically estimable. A definition of redundancy is therefore:

It follows from the definition that the measured value is itself an estimate of the variable value. Other estimates can be obtained by calculation using other measured variables and mass conservation equations.

With the n-product formula, determining data redundancy and estimability is easy. In complex mass balances, determining data redundancy and estimability is quite tricky (Frew, 1983; Lachance and Flament, 2011). Estimability and redundancy cannot be determined just by the number of equations for the number of unknowns. It is not unusual to observe global redundancy with local lack of estimability (Example 3.14).

Example 3.14

In the following mass balance problem (Figure ex 3.14), determine the estimability and redundancy of each variable entering the mass balance considering the Feed stream mass flowrate is measured and:

1. One metal has been analyzed on each sample;

2. Two metals have been analyzed on each sample.

Solution

When one metal is analyzed on each of the 6 streams:

There are 4 equations (2 for the mass flowrates and 2 for the metal assay) and 5 unknowns. The system is globally underdetermined since there are not enough equations for the number of unknowns. Around node A, the 2-product formula can be applied and therefore the mass flowrates of unit A products can be estimated. However, around node B, a 3-product formula cannot be applied since only one metal assay has been performed. Unit B product stream flowrates are not estimable. Metal assays are not redundant and cannot be adjusted by statistical data reconciliation.

When two metals are analyzed on each of the 6 streams:

There are a total of 6 equations and 5 unknowns. The system is therefore globally redundant and estimable a priori. Around node A, the 2-product formula can be applied with each metal assay and therefore the two product stream mass flowrates are estimable and the metal analyses on the 3 streams around unit A are redundant and adjustable. Around node B, the 3-product formula can be applied and therefore the 3 product stream mass flowrates are estimable now that the mass flowrate of the feed stream to unit B is known. However, the metal analyses are not redundant and not adjustable.

What are the factors influencing estimability and redundancy? Obviously, from Example 3.14, the number of measured variables is a strong factor, but it is not the only one. The network topology is also a factor. In the case of Example 3.14, if one of the product streams of unit B was recycled to unit A, then, with two metals analyzed on each stream, all the metal analyses would be redundant and adjustable.

It is worth noting that analyzing additional stream components does not necessarily increase redundancy and enable estimability. A component that is not separated in the process unit has the same or almost same concentration in each stream sample. The associated mass conservation equation is then similar (collinear) to the mass conservation of solids preventing the whole set of equations to be solved.

Estimability and redundancy analyses are complex analyses that are better performed with mathematical algorithm. However, the mathematics is too complex to be presented here.

3.6.9 Mass Balancing Computer Programs

It might be tempting to use spreadsheets to compute mass balances. Indeed, spreadsheets offer most of the features required to easily develop and solve a mass balance problem. However, spreadsheets are error prone, expensive to troubleshoot and maintain over time, and become very quickly limited to fulfill the needs and requirements of complex mass balances and their statistics (Panko, 2008).

There exist several providers of computer programs for mass balance calculations for a variety of prices. While all the features of an advanced solution may not be required for a given type of application (research, process survey, modeling and simulation project, on-line mass balancing for automatic control needs, rigorous production and metallurgical accounting), a good computer program should offer:

Furthermore, depending on the mass balance types and application, the program should provide:

Additional information can be found in Crowe (1996), Romagnoli and Sánchez (1999) and Morrison (2008).

3.6.10 Metallurgical Balance Statement

One principal output of the sampling, assaying, mass balancing/data reconciliation exercise is the metallurgical balance, a statement of performance over a given period: a shift, a day, a week, etc. Assuming that the reconciled data in Example 3.12 correspond to a day when throughput was 25,650 t (measured by weightometer and corrected for moisture), the metallurgical balance would look something like in Table 3.3.

The calculations in the Wt (weight) column is from the solid (mass) split using Eq. (3.25). It should be evident using the reconciled data in Table ex 3.12a that regardless of which metal assay is selected to perform the calculation that the result for solid split is the same (including using Fe), giving W_C/W_F=0.065. (In checking you will note that it is important to retain a large number of significant figures to avoid “rounding errors”, rounding at the finish to create the values given in Table 3.3.) From the solid split the tonnage to concentrate is calculated by multiplying by the feed solids flowrate (i.e., 25,650×0.065=1,674.1), which, in turn, gives the tonnage to tailings. The recovery to concentrate is given using Eq. (3.27), and thus the recovery (loss) to tails is known; together recovery and loss are referred to as “distribution” in the table. The information in the table would be used, for instance, to compare performance between time periods, and to calculate Net Smelter Return (Chapter 1).

3.7.1 Use of Particle Size

Many units, such as hydrocyclones and gravity separators, produce a degree of size separation and the particle size data can be used for mass balancing (Example 3.15).

Example 3.15

The streams around the hydrocyclone in the circuit (Figure ex 3.15) were sampled and dried for particle size analysis with the results in Table ex 3.15. Determine the solids split to underflow and the circulating load.

Table ex 3.15

Size Distribution Data Obtained on Cyclone Feed, Overflow and Underflow

Size* (μm)	CF	COF	CUF	f-o	u-o	(f–o)/(u–o)
+592	7.69	0.01	12.84	7.68	12.83	0.599
−592+419	4.69	0.34	7.18	4.35	6.84	0.636
−419+296	6.68	0.7	9.96	5.98	9.26	0.646
−296+209	7.03	2.61	9.32	4.42	6.71	0.659
−209+148	11.29	7.63	13.6	3.66	5.97	0.613
−148+105	13.62	13.55	14.91	0.07	1.36	0.051
−105+74	11.39	11.52	10.42	−0.13	−1.1	0.118
−74+53	9.81	15	7.19	−5.19	−7.81	0.665
−53+37	5.65	9.37	3.32	−3.72	−6.05	0.615
−37	22.15	39.27	11.26	−17.12	−28.01	0.611
	100	100	100

Solution

Solids split: We can set up the calculation in the following way, first recognizing the two balances:

Solids: $F=O+U$

Particle size: $Ff=Oo+Uu$

where F, O, U refer to solids (dry) flowrate of feed, overflow and underflow, and f, o, u refer to weight fraction (or %) in each size class. Since the question asks for solids split to underflow, U/F, we can use the solids balance to substitute for O (=F – U) in the particle size balance. After gathering terms and re-arranging we arrive at:

$\frac{U}{F}=\frac{(f-o)}{(u-o)}$

While this may seem that we could simply apply the two-product formula, it is always advisable to set up the solution starting from the basic balances. Attempting to remember formulae is an invitation to error.

The estimates of U/F derived from each size class are included in the table. This is another example where errors in the data produce uncertainty in the mass balance. Note that the two estimates that are far from the others, for the 105 μm and 74 μm size fractions, correspond to situations where there is little difference in data between the three streams, making the calculation numerically unstable (if all three were equal no solution can be found).

By re-arranging we can write:

$(f-o)=\frac{U}{F}(u-o)$

This has the form of a linear equation passing through the origin (y=mx) where the slope is solids split U/F. Figure ex 3.15-S shows the resulting plot and includes the solids split resulting from linear regression using the function in Excel.

The plot agrees with expectation and the solids split is therefore:

$\frac{U}{F}=0.616$

Circulating load: While this is encountered and defined later (e.g., Chapter 7), it is sufficient here to note it is given by the flowrate returning the ball mill (U) divided by the fresh feed rate to the circuit (N); that is, and observing that N=O (at steady state), we can write:

$CL=\frac{U}{O}$

We could approach the problem in the same way as we did to solve for U/F (by substituting for F in this case). More directly CL is given by noting that:

$U=0.616F\mathrm{and}\mathrm{thus}O=0.384F$

giving:

$\frac{U}{O}=\frac{0.616}{0.384}=1.60$

Thus the circulating load is 1.60, often quoted as a percentage, 160%. We will see that this is typical value for a closed ball mill-cyclone circuit (Chapters 7 and 9).

Example 3.15 is an example of node imbalance minimization; it provides, for example, the initial value for the generalized least squares minimization. This graphical approach can be used whenever there is “excess” component data; in Example 3.9 it could have been used.

Example 3.15 uses the cyclone as the node. A second node is the sump: this is an example of 2 inputs (fresh feed and ball mill discharge) and one output (cyclone feed). This gives another mass balance (Example 3.16).

Example 3.16

At the same time as the streams around the cyclone were sampled, the streams around the sump were also sampled, and sized. The measurement results for two size classes, +592 μm and −37 μm are given in the Table ex 3.16.

Table ex 3.16

Raw (unadjusted) Size Distribution Data for New Feed, Ball Mill Discharge, and Cyclone Feed

	n	b	f	n-f	f-b	(n–f)/(f–b)
592	19.86	2.16	7.69	12.17	5.53	2.201
−37	19.84	20.27	22.15	−2.31	1.88	−1.229

Determine the circulating load.

Solution

The mass balance equations for the sump are:

Solids: $N+B=F$

Particle size: $Nn+Bb=Ff$

and $CL=\frac{B}{N}$

Combining equations to eliminate F we deduce that:

$CL=\frac{n-f}{f-b}$

The estimates of CL are included in the table. Clearly, there is error, the result for −37 μm even being negative. After reconciliation the adjusted data and mass balance are given below:

592	19.42	1.44	8.27	11.15	6.83	1.633
−37	20.66	21.61	21.25	−0.59	−0.36	1.639

Clearly now the data do agree more closely with the result in Example 3.15. By experience, mass balancing around the sump is less reliable than around the cyclone.

In Chapter 9 we return to this grinding circuit example using adjusted data to determine the cyclone partition curve.

3.7.2 Use of Percent Solids

A unit that gives a large difference in %solids between streams is a thickener (Examples 3.17); another is the hydrocyclone (Example 3.18).

Example 3.17

Feed to a thickener is 30% solids by weight and the underflow is 65% by weight. Calculate the recovery of water to the overflow R_w/o assuming no solids report to the overflow.

Solution

We can approach the problem in two ways: 1, starting with a balance on the slurry, and 2, using dilution ratios. Using F, O, U to represent feed, overflow and underflow flowrates, f, o, u to represent % solids (as a fraction), and subscripts sl, s and w to represent slurry, solids, and water we can write the appropriate balances.

Balances based on slurry: We can write three balances:

Slurry: $F_{sl}=O_{sl}+U_{sl}$

Solids: $F_{sl}{f}_s=O_{sl}{o}_s+U_{sl}{u}_s$

Water: $F_{sl}(1-f_s)=O_{sl}(1-o_s)+U_{sl}(1-u_s)$

and thus water recovery will be given by:

$R_{w/o}=\frac{O_{sl}}{F_{sl}}\frac{1-o_s}{1-f_s}\to \frac{O_{sl}}{F_{sl}}\times \frac{1}{1-f_s}$

Since U_sl does not appear in the recovery equation it can be substituted by (F_sl – O_sl) from the slurry balance into the water balance (this is effectively combining the two equations). In this manner the ratio O_sl/F_sl is obtained from the water balance equation:

$\frac{O_{sl}}{F_{sl}}=\frac{u_s-f_s}{u_s}$

and thus R_w/o is determined:

$R_{w/o}=\frac{u_s-f_s}{u_s(1-f_s)}$

Balances based on dilution ratios: Using dilution ratios (Section 3.4) the water mass flowrate can be obtained, as illustrated using the feed stream as an example:

Feed solids: $F_s=F_{sl}{f}_s$

Feed water: $F_w=F_{sl}(1-f_s)$

From the ratio: $F_w=F_s\frac{1-f_s}{f_s}$

Analogous expressions for water flowrate in the other streams can be written. It will be noticed that the expression for O_w in this example is not determinable (as O_s and o_s are zero). Thus the R_w/o is determined from:

$R_{w/o}=1-\frac{U_w}{F_w}$

After substituting by the dilution ratios and re-arranging we obtain:

$R_{w/o}=\frac{u_s-f_s}{u_s(1-f_s)}$

Thus the two approaches give the same result, as must be the case.

Solving we obtain:

$\begin{array}{l}{R}_{w/o}=\frac{0.65-0.30}{0.65\times (1-0.3)}\\ R_{w/o}=0.769\mathrm{or}77\%\end{array}$

Example 3.18

Referring to Example 3.15, at the same time % solids data were determined on each stream: feed, 55%; overflow, 41%; and underflow, 70%. Determine the split of solids to underflow and the split of water to the underflow.

Solution

The problem could be set up starting with the slurry balance, but it lends itself to the dilution ratio approach. Using the same symbolism as in Example 3.17 the dilution ratios are:

$\begin{array}{l}{F}_w=F_s\frac{100-f_s}{f_s}=F_s{f}_s^{\text{'}}=F_s(0.818)\\ O_w=O_s\frac{100-o_s}{o_s}=O_s{o}_s^{\text{'}}=O_s(1.439)\\ U_w=U_s\frac{100-u_s}{u_s}=U_s{u}_s^{\text{'}}=U_s(0.429)\end{array}$

Solids split: The two balances are:

Solids: $F_s=O_s+U_s$

Water: $F_s{f}_s^{\text{'}}=O_s{o}_s^{\text{'}}+U_s{u}_s^{\text{'}}$

giving:

Solids split: $R_{s/u}=\frac{U_s}{F_s}$

By combining equations to eliminate O_s we derive:

$\begin{array}{l}{R}_{s/u}=\frac{f_s^{\text{'}}-o_s^{\text{'}}}{u_s^{\text{'}}-o_s^{\text{'}}}=\frac{0.818-1.439}{0.429-1.439}\\ R_{s/u}=0.614(\mathrm{or}61.4\%)\end{array}$

This compares well with the estimate in Example 3.15, which is often not the case using raw, that is, unadjusted data; the % solids data could be included in mass balance/data reconciliation, and must be if use is to be made of stream values derived from those data.

Water split: The water split is:

$\begin{array}{l}{R}_{w/u}=\frac{U_w}{F_w}=\frac{U_s{u}_s^{\text{'}}}{F_s{f}_s^{\text{'}}}\\ R_{w/u}=0.614\times \frac{0.429}{0.818}\\ R_{w/u}=0.322(\mathrm{or}32.2\%)\end{array}$

The water split to underflow, also referred to as water recovery to underflow, is an important parameter in modeling the performance of a cyclone (Chapter 9).

3.7.3 Illustration of Sensitivity of Recovery Calculation

See Example 3.19.

Example 3.19

A concentrator treats a feed of 2.0% metal. Compare the 95% confidence interval on the recovery for the following two conditions:

a. Producing concentrate grading 40% metal and a tailings of 0.3% metal.

b. Producing concentrate grading 2.2% metal and a tailings of 1.3% metal.

Take that the relative standard deviation on the metal assays is 5% in both cases.

Solution a)

Recovery:

From Eq. (3.27): R=85.6%

Standard deviation on R:

Solving for V_R (Eq. (3.33)) gives:

$V_R=0.14492(394.023V_f+0.001806V_c+16044.44V_t)$

To insert the values of V_f, V_c, and V_t the relative standard deviations must be changed to absolute values, σ (remembering the Eq. (3.33) is written in terms of fractions):

σ_f=5% of 2.0%=0.1% (or 0.001);

σ_c=5% of 40%=2% (or 0.02);

and σ_t=5% of 0.3%=0.015% (or 0.00015)

Substituting gives (recalling that V=σ²):

V_R=0.14492 (0.000394+7.225E–10+0.000361)

V_R=0.00011

That is: σ_R=0.01046 (or 1.05%)

The standard deviation on the recovery is ±1.05% or an approximate 95% confidence interval of 2×1.05% or 2.1%. The recovery is best reported, therefore, as:

$R=85.6\pm 2.1\%$

Solution b)

Recovery:

From Eq. (3.27): R=85.6% (i.e., the same)

Standard deviation of R:

Solving for V_R (Eq. (3.33)):

$V_R=3.11675(0.2025V_f+0.10124V_c+0.02367V_t)$

To insert the values of V_f, V_c, and V_t the relative standard deviations must be changed to absolute values, σ:

σ_f=5% of 2.0%=0.1% (or 0.001);

σ_c=5% of 2.2%=0.11% (or 0.0011);

and σ_t=5% of 1.3%=0.065% (or 0.00065)

Substituting gives (recalling that V=σ²):

$\begin{array}{l}{V}_R=3.11675(2.025\mathrm{E}-07+1.225\mathrm{E}-07+1\mathrm{E}-08)\hfill \\ V_R=0.01044\hfill \end{array}$

That is:

${\sigma }_R=0.102(\mathrm{or}10.2\%)$

And the 95% confidence interval is:

$R=85.6\pm 20.4\%$

The much higher uncertainty in the Example 3.19 (b) compared to a) again illustrates the greater uncertainty in the value of the recovery when the metal (or any component) is not well separated.

Example 3.19 raises a question as to how to estimate the variance. In this case a relative standard deviation (standard deviation divided by the mean) was taken and assumed equal for all assays. The standard deviation could be broken down into that associated with sampling and that associated with assaying, or some other “error model”. Regardless, there is some questioning of the common use of relative standard deviation as the absolute standard deviation decreases as the assay value decreases and some would argue this is not realistic. Alternative error models which avoid this have been suggested (Morrison, 2010), but none appear to be universally accepted.