DETERMINATION OF THE MOST ADVANTAGEOUS TRAIN COMPOSITION FOR UNDERGROUND HAULAGE BY BATTERY ELECTRIC LOCOMOTIVES

Associate Professor A. S. Betekhtin

INTRODUCTION

In our technical literature, several works have been devoted to the calculation of underground electric-locomotive haulage: the work of Prof. F. N. Shklyarsky, of Associate Professor Volotkovsky, of Engineer Livshits, of the collective of workers of the “Dynamo” plant, of Engineer Lebedev, etc. However, of all the authors listed above, only Engineer Livshits at one time provided a method for the techno-economic calculation of underground electric-locomotive haulage by battery electric locomotives. The other authors confined themselves only to a technical solution of the problem. Since the data presented in Engineer Livshits’s work, “Techno-Economic Calculation of Electric-Locomotive Haulage,” have by now become obsolete, and since there are as yet no other methods in our literature for a techno-economic solution of this problem, we consider it necessary and timely to fill the existing gap. This is all the more necessary because underground electric-locomotive haulage has become an integral and very important area of mine operations, and because, in connection with the transition of a number of large mines to the development of lower levels, the field of application of battery electric locomotives will increase considerably. Thus, the question of the most advantageous variant of underground battery-locomotive haulage ceases to be a purely local question and grows into a question of all-Union significance. In other words, a rational method of techno-economic calculation of underground battery-locomotive haulage, when applied to reconstructed and newly designed mines, promises savings in annual operating costs amounting to tens and hundreds of thousands of rubles.

§ 1. Brief Analysis of Existing Methods for Calculating Underground Haulage by Battery Electric Locomotives

The methods currently existing for calculating underground battery-locomotive haulage (with the exception of Engineer Livshits’s method) are essentially reduced to the following:

1. The type of electric locomotive is selected depending on the haulage capacity, and in particular cases also on the already existing track.

2. The size of the train is determined, usually from the conditions of movement of the empty train upgrade, taking into account that some of the mine cars will be loaded with timber (coefficient 1.1).

3. It is checked whether the adhesive weight of the electric locomotive is sufficient, with the selected train, to carry out the most difficult shunting operations, in particular, starting a loaded train upgrade.

4. The tractive efforts, speeds, and currents are determined for the established motion of the loaded and empty trains.

5. A braking check is performed; in particular, the length of the braking distance is determined.

6. The durations of the periods of acceleration and deceleration and of motion at uniform speed are determined for the loaded and empty trains, as well as the distances covered in these periods.

7. The electric-locomotive motors are checked for heating.

8. The average discharge current of the battery is determined.

9. The duration of battery discharge at the discharge current found is determined (unfortunately, very approximately, without taking into account the influence of the magnitude of the discharge current on the ampere-hour and watt-hour capacity of the battery).

10. From the complete trip period, the number of trips per shift is determined.

11. From the trip output in tons or ton-kilometers, the required number of working electric locomotives is determined.

12. From the number of working electric locomotives and the duration of battery discharge and charging, the number of batteries being charged simultaneously and the total number of batteries, including the reserve, are determined.

13. From the number of batteries being charged simultaneously, the number of mercury rectifiers (working and reserve), the number of charging tables, the capacity and number of transformers, and the amount of high-voltage and low-voltage apparatus are determined.

14. The dimensions of the depot and lamp room are determined.

15. The capital expenditures and annual operating costs are found.

16. The cost of one ton-kilometer of haulage is determined.

These are the principal stages of the most widespread method for calculating underground haulage by battery electric locomotives. With slight changes, this method is used both in design organizations and in mining higher educational institutions when students study the course on electric-locomotive haulage.

Formally, this method is correct, since it solves the problem on a sound technical basis. In essence, however, it does not answer whether the selected haulage variant is the most economical. Indeed, it does not answer a question that is far from idle: which is more advantageous—to have 1–2 extra electric locomotives but work a shift without recharging the batteries, or, conversely, to use

make use of the limiting power of locomotive motors and, with a minimal number of locomotives, have a large accumulator-battery fleet. The total expenses per locomotive per day, including wages, amount to about 100 rubles; but analogous expenses for each battery, rectifier, share of the transformer capacity, etc., also amount to a considerable sum. Finally, the number of locomotives has a substantial influence on the size of the wagon fleet, the expense for depreciation and repair of which is one of the largest expenditure items in haulage. The above-mentioned method does not answer any of these questions.

§ 2. On the essence of the method for calculating underground haulage by accumulator locomotives proposed by Eng. I. I. Livshits.

In his works, which appeared in the journal Ugol’ No. 65 in 1931, and later in Essays on Mining Electrical Mechanics¹, Eng. I. I. Livshits approached the solution of the question under consideration in the most correct way and in fact solved it for the conditions of that time, i.e., for the types of locomotives and batteries then planned for operation at the then-existing rates for labor. Eng. I. I. Livshits’s method is basically reduced to the fact that all the variables characterizing haulage and its costs are expressed by Eng. I. I. Livshits as functions of the train composition and, ultimately, he determines the magnitude of the composition at which the annual operating expenses for haulage are minimal. To facilitate the calculation, as well as to check the results obtained, the author makes extensive use of nomograms, so that his method is essentially graphical-analytical. In addition, he makes a number of assumptions and simplifications that do not particularly affect the results of the calculation. The exception is the assumption that the capacity of batteries does not depend on the magnitude of the discharge current—an assumption that, to a certain extent, is valid for alkaline batteries, but is completely unacceptable for lead-acid batteries.

Eng. Livshits’s interesting and valuable work greatly facilitated the labor of design organizations in its time. Unfortunately, as already indicated above, Eng. Livshits’s work has by now become obsolete: while the methodology is correct, the economic and, in part, technical indicators (for example, ETT tubular batteries) are different from what they were 8 years ago.

§ 3. Basic premises and formulas for determining the most advantageous magnitude of train composition.

Let us denote:

$P$ — adhesive weight of the locomotive, in tons;

$n$ — number of mine cars in the train;

$C_{gr}$ — weight of a loaded mine car, in tons;

$C_{por}$ — weight of an empty mine car, in tons;

$\mu$ — specific resistance to uniform motion, in kg per ton;

$i$ — grade in ‰ (average);

$F_{gr}$ — force required to move the loaded train;

$F_{por}$ — the same for the empty train;

$L$ — weighted-average haulage distance, in meters;

$T_1$ — period of motion of the loaded train, in sec;

$T_2$ — the same for the empty train;

$G$ — time spent on stops and maneuvers, in sec;

$J_{gr}$ — current (per motor) for the loaded train;

$J_{por}$ — the same for the empty train;

$A$ — average shift haulage productivity, in tons;

$K$ — coefficient of nonuniformity of output;

$Q$ — battery capacity, in ampere-hours;

$\lambda$ — coefficient allowing for energy consumption for maneuvers;

$v_{gr}, v_{por}$ — steady speeds of motion of the loaded and empty train.

The forces required to move the loaded and empty trains under steady motion will be:

\[ F_{gr} = (P+nG_{gr})(\mu-i)\,\kappa z \tag{1} \]

\[ F_{por} = (P+nG_{por})(\mu+i)\,\kappa z \tag{2} \]

The dependence of the force at the motor shaft (or at the wheel rim) on the current in the armature may be considered rectilinear, i.e., it may be assumed that

\[ \frac{F_{gr}}{2}=\alpha J_{gr} \quad \text{and} \quad \frac{F_{por}}{2}=\alpha J_{por}, \]

and the dependence of the speeds $v_{gr}$ and $v_{por}$ on the armature currents $J_{gr}$ and $J_{por}$ may, without great error, be taken as hyperbolic, i.e.

\[ J_{gr}v_{gr}=J_{por}v_{por}=C, \]

where $C=\text{const}$ for the given type of motor.

Then expressions (1) and (2) may be represented in the form:

\[ 2\alpha J_{gr}=(P+nG_{gr})(\mu-i) \tag{3}; \]

\[ 2\alpha J_{por}=(P+nG_{por})(\mu+i) \tag{4} \]

Expressions (3) and (4) may be represented in a form more convenient for further analysis, namely:

\[ J_{gr}=\frac{P(\mu-i)}{2\alpha}+\frac{C_{gr}(\mu-i)}{2\alpha}, \]

or

\[ I_{gr}=i_1+I_1 n \tag{5} \]

and, by analogy,

\[ I_{por}=i_2+I_2 n \tag{6} \]

Here

\[ i_1=\frac{P(\mu-i)}{2\alpha}, \qquad i_2=\frac{(\mu+i)}{2\alpha}, \]

\[ I_1=\frac{C_{gr}(\mu-i)}{2\alpha} \quad \text{and} \quad I_2=\frac{C_{por}(\mu+i)}{2\alpha}^{\,1)} \]

—are quantities constant for the adopted type of electric locomotive and for the given haulage conditions (gradient \(i^{\circ}/_{00}\)).

Physically, \(i_1\), \(i_2\), \(I_1\), and \(I_2\) represent the current forces per motor required for motion at the established speed: of the locomotive itself down the gradient, of the locomotive itself on the normal upgrade, of a loaded car down the gradient, and of an empty car on the normal upgrade.

The periods of motion of the loaded and empty trains, allowing for the time for acceleration and for slowing at the end of the run, may be represented as:

\[ T_1=\frac{L}{b_{gr}}+30 \ \text{sec} \quad \text{or} \quad T_1=\frac{L\cdot I_{gr}}{C}+30 \ \text{sec} \tag{7}, \]

where \(L\) is in meters,

and similarly

\[ T_2=\frac{L I_{por}}{C}+30 \ \text{sec} \tag{8} \]

The complete period of a trip is \(T=T_1+T_2+\Theta\);

or

\[ T=\frac{L(I_{gr}+I_{por})+C(60+\Theta)}{C}\ \text{sec} \tag{9} \]

The number of trips during 6.5 hours of net haulage work per shift will be

\[ n_p=\frac{6.5\cdot 3600}{T}, \]

or

\[ n_p=\frac{23400\,C}{L(I_{gr}+I_{por})+C(60+\Theta)}. \tag{10} \]

The shift output of the electric locomotive, in tons:

\[ a=nq n_p=\frac{23400\,C\cdot nq}{L(I_{sr}+I_{por})+C(60+\Theta)}, \tag{11} \]

where \(q\) is the capacity of a car, in tons.

\({}^{1)}\) If a significant quantity of support material is delivered during the shift, a coefficient of 1.05–1.15 must be introduced into the expression for \(I_2\).

The required number of working electric locomotives:

\[ n_{\mathrm{e}}=\frac{kA}{a}\quad \text{or}\quad n_{\mathrm{e}}=\frac{kA\,[L(l_{\mathrm{gr}}+l_{\mathrm{por}})+C(60+\Theta)]}{23400\,Cqn} \tag{12} \]

where \(k\) is the coefficient of nonuniformity of output.

If waste rock haulage is carried out, this circumstance must be taken into account by a corresponding increase in \(A\), for example, by a coefficient of 1.05.

Substituting into expression (12), in place of \(l_{\mathrm{gr}}\) and \(l_{\mathrm{por}}\), their values as determined by expressions (5) and (6), we obtain \(n_{\mathrm{e}}\) as a function of only one variable:

\[ n_{\mathrm{e}}= \frac{kA\,[((l_1+l_2)Ln+(i_1+i_2)L+C(60+\Theta))]} {23400\,qnC} \tag{13} \]

or

\[ n_{\mathrm{e}}= \frac{kA(l_1+l_2)L}{23400\,Cq} + \frac{kA[(i_1+i_2)L+C(60+\Theta)]}{23400\,Cqn} \tag{13 bis}, \]

where the first term of the right-hand side does not depend on \(n\).

For the selected type of electric locomotive and mine car

\[ \frac{l_1+l_2}{23400\,cq}=B=\mathrm{const};\qquad B_1=\frac{i_1+i_2}{23400\,Cq} \]

is also constant, as is

\[ \frac{C(60+\Theta)}{23400\,cq}=B_2 \]

(for the selected haulage organization).

Then

\[ n_{\mathrm{e}}=kBAL+\frac{kB_1AL+kAB_2}{n} = kA\left(BL+\frac{B_1L+B_2}{n}\right) \tag{14} \]

To this number it is necessary to add 2 electric locomotives: one in reserve and the other undergoing scheduled preventive repair.

The size of the mine-car fleet, depending on the haulage organization system,

\[ n_w=n(n_{\mathrm{e}}+N), \tag{15} \]

where \(N\) is the number of trains other than those operating with electric locomotives. The value \(N\) also includes the fleet in reserve and repair, as well as mine cars in the development faces.

Under the most commonly encountered system of haulage organization, the size of the mine-car fleet may, without great error, be taken as equal to:

\[ n_w=1.35\cdot 3\,n\,n_{\mathrm{e}}\quad \text{or}\quad n_w\simeq 4\,n n_{\mathrm{e}}, \tag{16} \]

which, after substituting \(n_z\), gives:

\[ n_w = 4\, kA \left(BLn + B_1L + B_2\right) \tag{17} \]

The average discharge current of the battery can be determined from the expression

\[ I_{\mathrm{avg}} = \frac{\lambda\left(2 I_{\mathrm{gr}} T_1 + 2 I_{\mathrm{por}} T_2\right)} {T_1 + T_2 + \mu\Theta}, \tag{18} \]

where \(\mu\) is a coefficient allowing for the fact that, during pauses and maneuvers, the battery operates only for part of the time \(\Theta\). The exact value of \(\mu\) is determined if the durations of all electric-locomotive operations at the loading point and in the mine yard are known, as well as the amount of waiting time at sidings. As an approximate value, \(\mu = 0.5\).

The battery capacity in ampere-hours, as a function of the magnitude of the average discharge current, is determined by Peukert’s formula:

\[ Q_x = Q_5 \left(\frac{I_5}{I_{\mathrm{avg}}}\right)^{n-1}, \]

where \(Q_5\) and \(I_5\) are the capacity and current under the five-hour regime, and \(n\) for ETT batteries is approximately 1.5 (more precisely, 1.56).

The duration of battery operation without recharging (with continuous battery discharge) is

\[ t_{\mathrm{dis}} = \frac{Q_x}{I_{\mathrm{avg}}} \quad \text{or} \quad t_{\mathrm{dis}} = \frac{Q_5\sqrt{I_5}}{I_{\mathrm{avg}}^{3/2}}. \tag{19} \]

Expression (19) is the most correct for determining the duration of battery discharge, since it takes into account the dependence of capacity on the magnitude of the discharge current, which, unfortunately, is not reflected by other formulas generally accepted here for determining \(t_{\mathrm{dis}}\).

In reality, however, the battery is not discharged continuously during a trip: over the course of a trip, equal to \(T_1 + T_2 + \Theta\), it is discharged only for \(T_1 + T_2 + \mu\Theta\) seconds, so the true duration of discharge will not be \(t_{\mathrm{dis}}\), but

\[ \frac{T_1 + T_2 + \Theta}{T_1 + T_2 + \mu\Theta}\, t_{\mathrm{dis}}. \]

Let us denote it by \(T_{\mathrm{dis}}\).

Thus,

\[ T_{\mathrm{dis}} = \frac{Q_5\sqrt{I_5}\left(T_1 + T_2 + \Theta\right)} {I_{\mathrm{avg}}^{3/2}\left(T_1 + T_2 + \mu\Theta\right)} \tag{20} \]

For the haulage distances most frequently encountered in practice, the quantity

\[ \frac{T_1 + T_2 + \Theta}{T_1 + T_2 + \mu\Theta} \]

varies within the limits from 1.27

(for \(L = 1000\) m) to 1.11 (for \(L = 3000\) m). With a sufficient degree of accuracy this fraction may be expressed as \(1.27 - \dfrac{L}{2 \cdot 10^4}\), where \(L\) is in m. Then expression (20), after the corresponding substitutions, will take the form:

\[ T_{\mathrm{dis}} = B_3 \cdot \left\{ \frac{(I_1 + I_2)n + (i_1 + i_2) + \frac{C(60 + \mu \Theta)}{L}} {2\lambda \left[(i_1 + I_1 n)^2 + (i_2 + I_2 n)^2 + \frac{30C(I_1 + I_2)n}{L} + \frac{30C(i_1 + i_2)}{L}\right]} \right\} \]

where \(B_3 = (1.27 - 0.5 L \cdot 10^4) Q_5 \sqrt{I_5}\) is a quantity constant for the given type of battery and the average haulage distance \(L\).

The required number of working batteries is:

\[ n_b = n_e \left(1 + \frac{8}{T_{\mathrm{dis}}}\right), \tag{22} \]

where 8 is the duration (in hours) of charging a tubular battery, including time for cooling and gas evolution.

Thus,

\[ n_b = n_e \left(1 + \frac{ 8 \left\{ 2\lambda \left[(i_1 + I_1 n)^2 + (i_2 + I_2 n)^2 + \frac{30C(I_1 + I_2)n}{L} + \frac{30C(i_1 + i_2)}{L} \right]^{3/2} + \frac{C(60 + \mu \Theta)}{L} \right\}^{3/2} }{ B_3 \left[(I_1 + I_2)n + i_1 + i_2 + \right. } \right. \tag{23} \]

A rather complex expression for \(n_b\) can be simplified by denoting the coefficients of \(n^2\), \(n\), as well as the terms in the numerator free of \(n\), respectively, by

\[ a = I_1^2 + I_2^2; \qquad b = 2 i_1 I_1 + 2 C_2 I_2 + \frac{30C(I_1 + I_2)}{L} \]

and

\[ d = i_1^2 + i_2^2 + \frac{30C(i_1 + i_2)}{L}, \]

and the terms free of \(n\) in the denominator by

\[ f = i_1 + i_2 + \frac{C(60 + \mu \Theta)}{L}. \]

Then expression (23) assumes the simpler form:

\[ n_b = n_e \left[ \left(1 + \frac{8 \left(2\lambda (a n^2 + bn + d)\right)^{3/2}} {B_3 \left[(I_1 + I_2)n + f\right]^{3/2}} \right. \right. \tag{23 bis} \]

Further simplification of the expression for \(n_b\) consists in the fact that, as a preliminary analysis of the fraction in the expression shows—

... of 23 bis; for cases that may occur in practice, the expression

\[ \frac{a n^2 + bn + d}{(l_1 + l_2)n + f} \]

may be replaced (with an error not exceeding 1–1.5%, i.e., lying within the limits of calculation accuracy) by the expression \(gn + h\), where

\[ g = \frac{a}{l_1 + l_2} \tag{24} \]

and

\[ h = \frac{b(l_1 + l_2) - af}{(l_1 + l_2)^2} \tag{25} \]

The final expression for \(n_b\) will be:

\[ n_b = \frac{ KAL \left( Bn + B_1 + \dfrac{B_2}{L} \right) \left[ 1 + \dfrac{8 \cdot 2\lambda^{3/2}}{B_3}(gn + h)^{3/2} \right] }{n} \tag{26} \]

To the resulting number of batteries it is necessary to add a reserve of 1–2 batteries, on average 20%.

The number of batteries charged simultaneously is

\[ n'_b = n_b - n_e \tag{27} \]

The number of charging stands \(n_{zs}\) will be, at minimum, \(n'_b + 2\), or

\[ n_{zs} = \frac{8(2\lambda)^{3/2}KAL}{B_3 \cdot n} \left( Bn + B_1 + \frac{B_2}{L} \right) (gn + h)^{3/2} + 2. \tag{28} \]

The number of mercury rectifiers \(n_{rv}\) will obviously be equal to the number of charging stands, i.e. \(n_{rv} = n_{zs}\).

There will obviously be the same number of sets of distribution apparatus and small equipment for the mercury rectifiers and stands (distribution boxes YAG-200 or AKG-3, boxes P-60 or PBT-2, end and tee couplings, plugs, cables, measuring instruments, etc.).

The transformer capacity depends on the number of batteries charged simultaneously and on the stage of their charging (beginning, middle, end).

The power consumed by the ZVN-100 mercury rectifier during the initial stage of battery charging is about 10 kW.

The standard transformer capacity is

\[ P_{tr} = 1.46 \cdot 0.8 \cdot 10 \cdot n'_b \ \mathrm{KVA}, \tag{29} \]

where 0.8 is a coefficient taking into account the nonuniformity of the charging stages of different batteries.

According to the existing rules, in addition to the operating transformer, there must be a reserve one.

The number of high-voltage boxes does not depend on the number of mercury rectifiers and, consequently, on the number simultaneously cha...

charging batteries; therefore, the cost of their depreciation and repair will not enter into the basic equation for determining the minimum annual haulage costs.

The dimensions of the garage and of the charging chamber depend on the number of batteries being charged simultaneously or, what is the same, on the number of charging tables.

The total length of the garage and charging chamber is:

\[ L_{zk}=(l_{zs}+1000)(n_{zs}+2l_{zs}+20000\ \text{mm}^{1}) \tag{30} \]

For a standard table length \(l_{zs}=3000\ \text{mm}\),

\[ L_{zk}=4n_{zs}+26\ \text{m} \]

or

\[ L_{zk}=4n^{1}_{b}+34\ \text{m}. \tag{31} \]

With the width of the charging chamber (garage) equal to \(3.7\) and its height \(3.5\ \text{m}\), its volume will be

\[ V_{zk}=(4n^{1}_{b}+34)\cdot 3.7\cdot 3.5=52n^{1}_{b}+440\ \text{m} \tag{32} \]

The cubic capacity of the converter substation, depending on the number of mercury rectifiers, may be represented in the form:

\[ V_{np}=0.8^{\prime}(n^{\prime}_{b}+2+1)4\cdot 3^{2})=9.72n^{\prime}_{b}+29.2\ \text{m}^{3}, \tag{33} \]

where \(4\ \text{m}\) is the chamber width and \(3\ \text{m}\) its height. These figures are averages for installations with transformers from 100 to 180 kva. Finally, the size of the train will affect the cost of the second tracks at loading points, and, with a single-track line, the cost of passing sidings.

The length of the second track at a loading point, depending on the size of the train, will be:

\[ l_{n}=2nl_{w}+30\ \text{m}, \tag{34} \]

where \(l_{w}\) is the length of a mine car, including the coupling.

In particular, for standard 2-ton and 3-ton mine cars we have:

for 2-ton cars, \(l_{w}=2.810\ \text{m}\) (without coupling) and \(3\ \text{m}\) with coupling,

for 3-ton cars, \(l_{w}=3.284\ \text{m}\) (without coupling) and \(3.5\ \text{m}\) with coupling.

With the number of loading points \(n_{pn}\), the length of the second tracks at loading points, as a function of the train consist, will be:

\[ L_{n}=2l_{n}n_{pn}=2n\cdot n_{pn}\cdot l_{w}+30n_{pn}\ \text{m} \tag{35} \]

The number of passing sidings with a single-track line depends on the number of panels in the mine and on the number of operating electric locomotives.

¹ “Standard electrical-mechanical equipment of coal mines.” ONTI, 1935.

² Ibid.

For a single-wing mine the number of sidings is \(n - 1\); for a two-wing mine, \(n_e - 2\). With a siding length of \(n l_w + 30\) m, the total length of sidings for the normal type of mine, i.e., a two-wing mine, will be

\[ L_p = (n l_w + 30)(n_e - 2)\ \text{m} \tag{36} \]

The total length of the second tracks at sidings and loading points is

\[ L_p + L_n = (2 n l_w + 30)(n_e + n_{pp} - 2)\ \text{m} \tag{37} \]

Annual operating expenses.

Operating expenses, as is known, consist of depreciation charges, repair costs, and the cost of labor, energy, lubricants, and wiping materials.

The costs of depreciation and repair of equipment, chambers, and workings, which depend on the size of the train consist, are given in Table 1.

Table 1

No.	Objects of capital expenditure	Cost of one unit with transport and installation, in rubles	Depreciation, in % of cost	Repair, %	Total for depreciation and repair
1	Electric locomotives	\(A_e\)	12	5	\(0,17 A_e (n_e + 2)\)
2	ETT batteries	\(B\)	50	\(10^{1)}\)	\(0,60 B (n_b + 2)\)
3	Charging tables and their equipment	\(B_{zc}\)	10	3	\(0,13 B_{zc} (n_b + 2)\)
4	Mercury rectifiers with sets of equipment for them	\(B_b\)	10	\(5^{2)}\)	\(0,15 B_b (n_b + 2)\)
5	Transformers for rectifiers	\(1,46 \cdot 0,8 \cdot 10 \beta^{3)}\)	10	2	\(2,8 \beta n'_b\)
6	Car fleet	\(A_w\)	20	\(15^{4)}\)	\(0,35 \cdot 4 \cdot A_w n \cdot n_e\)
7	Charging and forming chamber, including the chamber for electrolyte	\(D\) rub./m\(^3\)	\(r^{5)}\)	2	\(\dfrac{r + 2}{100} D (61,8 n'_b + 46\beta)\)
8	Sidings and second tracks of loading points	\(E\) rub./linear m	\(r'^{6)}\)	5	\(\left(\dfrac{r' + 5}{100}\right) E (2 n l_p + 30)(n_e + n_{pp} - 2)\)

\(^{1)}\) Taking into account washing and replacement of electrolyte.
\(^{2)}\) Including replacement of bulbs.
\(^{3)}\) Cost of 1 kVA.
\(^{4)}\) Cleaning and lubrication.
\(^{5)}\) Depending on the life of the level.
\(^{6)}\) Depending on the life of the level.

Labor.

a) For servicing locomotives.
Drivers: 4 persons per day per locomotive, at \(m_1\) rubles per shift; in all, \(4m_1\) rubles per day.

Assistants: 4 persons per day per locomotive, at \(m_2\) rubles per shift; in all, \(4m_2\) rubles per day.

Note. The daily number of persons is determined from the calculation of 25 working days per month and one month’s leave. Hence the daily personnel complement is

\[ 3 \cdot \frac{30 \cdot 12}{25 \cdot 11} \cong 4 \text{ persons} \]

Taking payroll additions as 100%, including bonuses, we obtain the annual labor costs for the locomotive fleet as \(8(m_1+m_2)\cdot345\,n_{\mathrm{e}}\) rubles, or

\[ 2760 (m_1+m_2)n_{\mathrm{e}} \text{ rubles per year} \tag{38} \]

b) For servicing the charging station.
Usually one charger is assigned for every 3–4 batteries being charged.

We shall take one charger for 3 batteries, in order to ensure better supervision of the charging process.

The number of chargers per day is

\[ \frac{4 \cdot n^{1}_{\mathrm{b}}}{3} \]

With payment for a shift of \(p\) rubles, the annual costs of paying the chargers, including payroll additions and bonuses for good maintenance of the batteries, are

\[ \frac{2 \cdot 4 n_{\mathrm{b}}}{3}\,p \cdot 345 \text{ rubles, or } 920\,p\,n^{1}_{\mathrm{b}} \text{ rubles per year} \tag{39} \]

The costs for the depot senior man, the duty man in the transformer room, and the transport manager (where such positions exist) do not depend directly on the number of locomotives and batteries.\(^1\)

Energy.

With a charging duration of 8 hours (including one hour for settling and gas evolution) and a discharge duration \(T_{\text{dis}}\), the duration of the cycle will be \((8+T_{\text{dis}})\) hours. Each battery undergoes per day

\[ \frac{24}{8+T} \]

cycles.

\(^1\) In fact, with a large locomotive and battery fleet, there is usually also an increased staff of supervisory personnel.

Energy consumption for each charge

\[ E=\frac{1.05\, Q_{wh}}{\eta_{wh}\eta_{pT}\eta_c}\ .\, kwh \]

where \(Q_{wh}\) is the capacity of the batteries in \(kwh\) (for a 5-hour discharge regime).

\(1.05\) is a coefficient allowing for a certain overcharge.

\(\eta_{wh}\) is the watt-hour efficiency coefficient; for a lead battery, \(0.75\).

\(\eta_{pT}\) is the efficiency coefficient of the converter installation, equal to \(0.70\).

\(\eta_c\) is the efficiency coefficient of the high-voltage network, equal to \(0.95\).

The annual cost of the energy expended for charging the batteries, at a cost of 1 \(kw h\) equal to \(S\) rubles, will be:

\[ E_{\text{year}}=\frac{1.05\,Q}{\eta_{wh}\eta_{pT}\eta_c}\cdot\frac{24}{8+T_{\text{dis}}}\cdot345\,S\,n'_b \tag{40} \]

We take the expenditures for lubricating and wiping material for the electric locomotives and batteries as 5% of the cost of energy. Then the total expenditure for energy and lubrication will be (after substituting the values of \(\eta\))

\[ E^1=\frac{18240\,Q\,S\,n'_b}{8+T_{\text{dis}}}\ \text{kilowatt-hours per year} \tag{41} \]

It must be noted that, in general, the expenditures for energy and lubrication of electric locomotives constitute an insignificant percentage (usually less than 5%) of the total haulage expenditures. The fluctuations in the consumption of energy and lubrication depending on the size of the train, and therefore also in their cost, will be negligible in comparison with such cost items as trackmen and the battery or car fleet. Therefore, in what follows, when determining the minimum annual expenditures, we do not introduce the quantity \(E^1\) into the equation.

§ 4. Condition for the most advantageous train composition in battery haulage.

Let us denote by \(Z\) the total annual operating expenditures. To obtain the value of \(n\) at which the annual operating expenditures will be minimal, it is necessary to take the first derivative of \(Z\) with respect to \(n\) and, equating it to zero, solve the resulting equation with respect to \(n\). In doing so, in order not to complicate the already very complex expression for \(Z\), we shall at once discard the terms independent of \(n\), \(n_e\), \(n_b\), etc., since the derivatives of these quantities will become zero.

Under this condition we obtain:

\[ \begin{aligned} Z ={}& 0.17\,A_{\mathrm{e}} n_{\mathrm{e}} + 0.60\,B n_{\mathrm{b}} + 0.13\,B_{\mathrm{zc}} n_{\mathrm{b}}^{1} + 0.15\,B_{\mathrm{b}} n_{\mathrm{b}}^{1} + 2.8\,\beta n_{\mathrm{b}} \\ &+ 1.4\,A_{w} n n_{\mathrm{e}} + \frac{61.8}{100}(r+2)D n_{\mathrm{b}}^{1} + \frac{(r^{1}+5)E}{100}\,[2(l_{\mathrm{u}} n n_{\mathrm{e}} + l_{w} n_{\mathrm{pn}} n) \\ &\qquad - 2 l_{w} n + 30 n_{\mathrm{e}}] + 2760\,(m_{1}+m_{2}) n_{\mathrm{e}} + 920\,\rho\,n_{\mathrm{b}}^{1} \end{aligned} \tag{42} \]

After replacing \(n_{\mathrm{e}}, n_{\mathrm{b}}\), and \(n_{\mathrm{b}}^{1}\) by functions of \(n\), and combining like terms, we obtain:

\[ \begin{aligned} Z ={}& KA(B_{1}L+B_{2})\left[0.17\,A_{\mathrm{e}}+0.6\,B + \frac{30(r^{1}+5)E}{100} + 2760(m_{1}+m_{2})\right]\frac{1}{n} \\ &+ \left[1.4\,A_{w}KABL + \frac{2(r^{1}+5)E l_{w}}{100}(KABL+n_{\mathrm{pn}}-2)\right]n \\ &+ \left[0.13\,B_{\mathrm{zc}}+0.15\,B_{\mathrm{b}}+0.6\,B+2.8\,\beta + \frac{61.8(r+2)D}{100}+920\,\rho\right] \times \\ &\quad \frac{(2\lambda)^{3/2}}{B3}\cdot 8KABL(gn+h)^{3/2} + \frac{8(2\lambda)^{3/2}KA(BL_{1}+B_{2})}{B3}\cdot \\ &\quad \cdot \left(0.13\,B_{\mathrm{zc}}+0.15\,B_{\mathrm{b}}+0.6\,B+2.8\,\beta + \frac{61.8(r+2)D}{100}+920\,\rho\right) \\ &\quad \cdot \frac{1}{n}(gn+h)^{3/2} \end{aligned} \tag{43} \]

Denoting the coefficients of \(n\) respectively by \(M\), \(N\), \(R\), and \(R^{1}\), equation (43) takes the form

\[ Z=\frac{M}{n}+Nn+R(gn+h)^{3/2}+\frac{R^{1}}{n}(gn+h)^{3/2} \tag{44} \]

We take the first derivative of \(Z\) with respect to \(n\) and set it equal to zero:

\[ \frac{dz}{dn} = -\frac{M}{n^{2}}+N+1.5\,g\,R(gn+h)^{1/2} +\frac{1.5\,g\,R^{1}n(gn+h)^{1/2}}{n^{2}} -\frac{R_{1}(gn+h)^{3/2}}{n^{2}} =0 \tag{45} \]

After clearing the denominator, equation (45) may be written as:

\[ (gn+h)^{1/2}(1.5\,gR\,n^{2}+0.5\,gR^{1}n-R^{1}h)+Nn^{2}-M=0 \tag{46} \]

Solving the resulting equation by one of the generally known methods (for example, by root selection), we find the value of \(n\) of interest to us. Since \(n\) can only be an integer, it is obvious that, in solving equation (46), the nearest integer should be taken, the one that most nearly satisfies equation (46).

§ 5. Example.

Given: Shift output of the level
\(A = 1000\ \mathrm{t}\)

Coefficient of nonuniformity of extraction
\(K = 1.15\)

Weighted-average haulage distance
\(L = 2000\ \mathrm{m}\)

Electric locomotive of type ZAR-113 with coupling weight
\(P = 6.5\ \mathrm{t}\)

Capacity of the mine car
\(q = 2.0\ \mathrm{t}\)

Dead weight of the mine car
\(G_n = 1.2\ \mathrm{t}\)

Average gradient of the tracks
\(i = 3\text{‰}\)

Specific resistance in steady motion
\(W = 8\ \mathrm{kg/t}\)

Coefficient accounting for energy consumption for maneuvers
\(\lambda = 1.15\)

Capacity of the ETT-250 storage battery
\(Q = 250\ \mathrm{Ah}\)

Duration of maneuvers and pauses
\(\Theta = 600\ \mathrm{sec}\)

Service life of the level
10 years.

Number of loading points
\(n_{np} = 6\)

Course of the calculation.

1. Determine \(\alpha\); according to the rated data of the GR-10 motor, at

\[ F = 125\ \mathrm{kg}, \quad J = 35\ \mathrm{A}, \]

whence

\[ \alpha = \frac{F}{J} = 3.67\ \mathrm{kg/A}. \]

1. Determine \(C\); at \(J = 35\ \mathrm{A}\), \(V = 2.8\ \mathrm{m/sec}\), whence

\[ C = Iv = 35 \cdot 2.8 = 98 \]

1. Find \(i_1, i_2, J_1, J_2\):

\[ i_1 = \frac{6.5(8 - 3)}{2 \cdot 3.67} = 4.43\ \mathrm{A} \]

\[ i_2 = \frac{6.5(8 + 3)}{2 \cdot 3.67} = 9.73\ \mathrm{A} \]

\[ J_1 = \frac{3.2(8 - 3)}{2 \cdot 3.67} = 2.18\ \mathrm{A} \]

\[ J_2 = \frac{1.2(8 + 3)}{2 \cdot 3.67} = 1.53\ \mathrm{A} \]

1. Determine the coefficients \(B, B_1, B_2\), and \(B_3\)

\[ B = \frac{J_1 + J_2}{23400\, Cq} = \frac{2.18 + 1.53}{23400 \cdot 98 \cdot 2} \simeq 8.1 \cdot 10^{-7}, \]

\[ B_1 = \frac{i_1 + i_2}{23400\, Cq} = \frac{4.43 + 9.73}{23400 \cdot 98 \cdot 2} = 30.8 \cdot 10^{-7}, \]

\[ B_2 = \frac{60 + \Theta}{23400\, q} = \frac{60 + 600}{23400 \cdot 2} = 1.41 \cdot 10^{-2} \]

\[ B_3 = (1.27 - 0.5 \cdot L \cdot 10^{-4})\, Q_5 \sqrt{J_5} = 250 \cdot \sqrt{50}\,(1.27 - 0.5 \cdot 2000 \cdot 10^{-4}) = \]

\[ = 2068. \]

1. Find the coefficients \(a, b, d\), and \(f\).

\[ a = J_1^2 + J_2^2 = 2.18^2 + 1.53^2 = 4.75 + 2.34 \simeq 6.10 \]

\[ b = 2 i_1 J_1 + 2 i_2 J_2 + \frac{30 C (J_1 + J_2)}{L} = 2 \cdot 4.43 \cdot 2.18 + 2 \cdot 9.73 \cdot 1.53 + \]

\[ + \frac{98 \cdot 30 \cdot (2.18 + 1.53)}{2000} = 19.31 + 29.77 + 5.45 = 54.53. \]

\[ d = i_1^2 + i_2^2 + \frac{30 C (i_1 + i_2)}{L} = 4.43^2 + 9.73^2 + \]

\[ + \frac{30 \cdot 98 (4.43 + 9.73)}{2000} = 135.1^{1)} \]

\[ f = i_1 + i_2 + \frac{C(60 + \mu \Theta)}{L} = 4.43 + 9.73 + \frac{98 \cdot 360}{2000} = 17.04. \]

1. Determine the coefficients for \(n\):

a) Coefficient \(M\)

From price-reference books we find that \(A_e \simeq 25000\), \(B \simeq 10000\) rubles. We take the pay for the electric-locomotive driver \(m_1\) as 10 rubles, and for the assistant \(m_2\) as 6 rubles.

By condition \(r^1 = 10\%\). We take the cost of driving one running meter of workings for passing loops and loading points to be \(E = 150\) rubles.

Then

\[ M = K A (B_1 L + B_2) \left[ 0.17 A_e + 0.6 B + \frac{30(r^1+5)E}{100} + \right. \]

\[ \left. + 2760(m_1 + m_2) \right] = 1.15 \cdot 1000 \left(30.8 \cdot 10^{-7} \cdot 2000 + 1.41 \cdot 10^{-2}\right) \]

\[ \left[ 0.17 \cdot 25000 + 0.6 \cdot 10000 + \frac{30 \cdot 15 \cdot 150}{100} + 2760 \cdot 16 \right] = 1283480. \]

b) Coefficient \(N\)

The cost of a two-ton mine car is \(A_w = 1300\) rubles.
The length \(l_w\) with coupler is 3 m.

\[ N = 1.4 KABL A_w + 2 \frac{(r^1+5) E \cdot l_w}{100} (KABL + n_{nn} - 2) \]

\[ KABL = 1.15 \cdot 1000 \cdot 8.1 \cdot 10^{-7} \cdot 2000 = 1.863 \]

\[ 1.4\, KABL \cdot A_w = 1.4 \cdot 1.863 \cdot 1300 \simeq 3380 \]

\[ N = 3380 + \frac{2 \cdot 15 \cdot 150 \cdot 3}{100}(1.863 + 6 - 2) = 3380 + 792 = 4172 \]

\(^{1)}\) The quantity \(d\) will be required in determining the percentage error arising from simplifying the expression for \(n^6\) and \(n^{16}\).

c) Coefficient \(R\)

The quantities entering into this coefficient have in part already been found (for example, \(KABL\) or \(B_3\)). The others may be taken from reference books. Thus, \(B_{3c}=1000\) rub., \(B_b=5000\) rub., \(\beta=20\) rub., \(r=10\%\) (as also \(r^1\)) and \(D=100\) rub. per \(m^3\). The charging-station harness \(P=8\) rub. Then

\[ R=\frac{8\,KABL(2\lambda)^{3/2}}{B_3} \left[ 0.13\,B_{3c}+0.15\,B_b+2.8\,\beta+ \right. \]

\[ \left. +\frac{61.8(r+2)D}{100}+920\,P \right] = \frac{8\cdot1.863(2\cdot1.15)^{3/2}}{2068} \]

\[ [0.13\cdot1000+0.15\cdot5000+2.8\cdot20+61.8\cdot12+920\cdot8]\simeq226 \]

d) Coefficient \(R'\)

Similarly, the coefficient \(R^1\) is also found; it differs from \(R\) in that instead of \(8KABL\) it has the factor \(8KA(B_1L+B_2)\), equal under the given conditions to \(\sim 12.5\cdot8KABL\)

\[ R^1_1=2840 \]

1. We determine the coefficient

\[ g=\frac{a}{J_1+J_2} \]

and the constant term

\[ h=\frac{b(J_1+J_2)-af}{(J_1+J_2)^2} \]

\[ g=\frac{6.1}{3.71}=1.65 \]

and

\[ h=\frac{54.53\cdot3.71-6.1\cdot17.64}{3.71^2}=6.9. \]

1. We take the basic equation for the minimum of \(Z\):

\[ (gn+h)^{1/2}(1.5\,gRn^2+0.5\,gR^1n-R^1h)+Nn^2-M=0 \]

1. After substituting the values of the coefficients, we obtain:

\[ (1.65\,n+6.9)^{1/2}(560\,n^2+2350\,n-19600)+4172\,n^2-1283480=0. \]

1. Solving this equation by trial selection of roots, we obtain that the nearest integer most satisfying the equation is

\[ n=13. \]

1. We determine the percentage error possible as a result of simplifying the expression for \(n_b\) and \(n^1_b\), i.e., by comparing the values \(an^2+\)

\(bn + d\) and \([(gn + h)(J_1 + J_2)h + f]\). We find that the error is, in the present case,

\[ P = \frac{(1875 - 1868)\,100}{1875} \simeq 0.4\% \]

1. We find the principal elements of haulage.

a) Number of electric locomotives

\[ n_e = KABL + \frac{KAB_1L + KAB_2}{n} = 1.863 + \frac{1.863 \cdot 12.5}{13} \simeq 3.65. \]

In practice, it would be necessary to take 4 working electric locomotives. It is interesting to note that the next larger number of mine cars, i.e., 14 in the train, would also require 4 working electric locomotives. Only at

\[ n = \frac{1.863 \cdot 12.5}{3 - 1.863} \simeq 20.5, \]

i.e., in practice with at least 20 mine cars in the train, would it be possible to make do with 3 working electric locomotives.

b) Car fleet \(n_w = 4n_e\). Theoretically, \(n_w = 4 \cdot 13 \cdot 3.65 = 190\) mine cars; in practice, with \(n_e = 4\),

\[ n_w = 4 \cdot 13 \cdot 4 = 208 \]

mine cars.

c) Average duration of battery discharge:

\[ T_{\text{dis}} = \frac{B_z}{(2\lambda)^{3/2}(gn + h)^{1/2}} = \frac{2068}{3.48(1.65 \cdot 13 + 6.9)^{1/2}} \simeq 3.96 \text{ hours} \]

or, rounded, 4 hours.

d) Number of working batteries

\[ n_b = n_e\left(1 + \frac{8}{T_{\text{dis}}}\right) \]

\[ n_b = 4\left(1 + \frac{8}{4.00}\right) = 12. \]

e) Number of batteries being charged simultaneously

\[ n'_b = 12 - 4 = 8 \]

f) Number of charging tables and number of mercury rectifiers: \(n'_b + 2\), i.e., 10 of each.

g) Rated power of the transformer

\[ P_{\text{tr}} = 1.46 \cdot 0.8 \cdot 10 \cdot 8 = 93.44 \text{ kVA} \]

Obviously, a TMRSh 100/6 transformer may be adopted.

h) Cubic capacity of the charging chamber

\[ V_{z.k} = 52 \cdot 8 + 440 \text{ m} = 856 \text{ m}^3 \]

The cubature of the substation is

\[ V_{nn}=9.72\cdot 8+29.2=107\ \text{m}^3 \]

k) The length of sidings and passing loops

\[ L_n+L_p=(2\cdot 13\cdot 3+30)(4+6-2)=864\ \text{m}. \]

l) Annual labor costs (with wage charges)

\[ 2760\cdot 16\cdot 4+920\cdot 8\cdot 8=235520\ \text{rubles}. \]

m) Annual costs for energy and lubrication, at a cost of \(kwh\) of 5 kopecks.

\[ E'=\frac{18240\cdot 30\cdot 8\cdot 0.05}{8+4.00}\simeq 18240\ \text{rubles}. \]

Without calculating the annual deductions for depreciation and repairs by individual items, so as not to burden the article, we give the total figure for this case. It amounts to 227112 rubles. It is interesting to note that the largest amount falls on the cars (94640 rubles) and on the batteries (72000 rubles). These two items account in the present case for 74% of all expenditures under this heading.

11) Cost of a ton-kilometer

\[ \frac{235520+18240+227112}{1000\cdot 3\cdot 345\cdot 2} = \frac{480872}{2070000} = 0.232\ \text{rubles}. \]

or about 23.2 kopecks—a quite realistic figure.

1. Using the formulas given in the present article, the remaining elements of haulage can also be found. Thus, the duration of a trip, including shunting,

\[ T=\frac{L\,(t_1+l_1 n+i_2 n)+(60+\Theta)}{C}= \]

\[ =\frac{2000(14.16+3.71\cdot 13)}{98}+660 \]

\[ =1274+660=1934\ \text{sec.} \]

Number of trips per shift

\[ n_p=\frac{23400}{T}=\frac{23400}{1934}=12.1 \]

or, rounded, 12 trips.

The possible shift output of the electric locomotive is

\[ a=13\cdot 2\cdot 12=312\ t, \]

or

\[ 312\times 2=624\ \text{t-km}, \]

and taking \(k\) into account only

\[ \frac{624}{1.15}=542\ \text{t-km} \]

Actual output is

$$ \frac{1000 \cdot 2}{4} = 500 \text{ t-km}. $$

This figure is fairly high for a battery electric locomotive.

The coefficient of utilization of the electric locomotive is

$$ \frac{500}{542} = 0.92, $$

i.e., also fairly high.

Average discharge current of the batteries:

$$ I_{gr} = 4.43 + 2.18 \cdot 13 = 32.8 \, A; \qquad I_{por} = 9.73 + 1.53 \cdot 13 = 29.6 \, A $$

$$ T_1 = \frac{L I_{gr}}{C} + 30 = 669 + 30 \simeq 700 \text{ sec} $$

Similarly, $T_2 = 634 \text{ sec}$.

$$ J_{sr} = \frac{2 \cdot 1.15(32.8 \cdot 700 + 29.6 \cdot 634)} {700 + 634 + 0.5 \cdot 600} = 59 \, A $$

taking into account peaks during starts and maneuvers.

The average discharge capacity of the batteries is

$$ Q_x = 250 \cdot \left(\frac{50}{59}\right)^{0.5} \simeq 240 \, Ah $$

In conclusion, let us consider whether it is advantageous to have a train of 16 cars with the ETT—250 battery, assuming that with such a train the motors of the electric locomotive will not be overloaded.¹

It may be said a priori that this train will be less advantageous than the one we found.

Indeed, the number of working electric locomotives at $n = 16$ will remain the same, i.e., 4, since only with a train of 20 cars can one be limited to 3 working electric locomotives.² Thus there is no saving in labor. On the other hand, the cost of the car fleet will increase by 21,840 rubles, since the number of cars will now be not 208, but $4 \cdot 16 \cdot 4 = 256$.

The battery discharge time will decrease to 3.1 hours. The number of batteries will increase to 14.35, i.e., in the best case to 14.

The number of charging tables and mercury rectifiers will increase by 2. The workings will be lengthened.

Thus, from the economic point of view this variant will be completely disadvantageous, since under it the annual operating expenses will increase by 38,700 rubles, with a lower coefficient of utilization of the electric locomotive. If the GR—10 motors permitted, under the specified haulage conditions, a train of 20 cars, we would have a different picture.

In this case the expenditures on labor for servicing electric locomotives will be substantially reduced by

\[ \frac{2760 \cdot 16 \cdot 4}{4} = 44160 \]

rubles, and those for their depreciation and repair by 4,250 rubles. On the other hand, the following cost items will increase:

a) The car fleet will increase by \(4 \cdot 3 \cdot 20 - 208 = 32\) cars, and the expenditures on it by \(0.35 \cdot 1300 \cdot 32 = 14560\) rubles.

b) The unloading time will be reduced to 2.36 hours.

c) The number of working batteries will be

\[ 3\left(1 + \frac{8}{2.36}\right) = 13.2 \]

or, in the best case, 13 units.

This will give an additional increase in expenditures of another 6,000 rubles per year.

d) The number of mercury rectifiers and charging tables will increase by 2, which gives a further increase in expenditures of 1,760 rubles per year. Expenditures for depreciation and repair of the charging and forming equipment will increase by 1,483 rubles, and by 1,755 rubles for depreciation and repair of switches and second tracks at loading points. Nevertheless, in the final reckoning there would still be savings of 22,862 rubles per year.

True, alongside this there would also be certain unfavorable aspects: the need to change batteries three times during a shift, a high discharge current and, therefore, more rapid wear of the batteries, and, finally, the difficulty of braking such a train when only a hand brake is available. But with such a train the motors of the ZAR-113 electric locomotive cannot operate for long.

It is necessary to raise the question of the earliest possible introduction into the practice of underground electric-locomotive haulage, and in particular battery-locomotive haulage, of more powerful types of electric locomotives¹, with greater adhesive weight (on the order of 8–10 t), with pneumatic or electric braking. Evidently, these electric locomotives will also require batteries of greater capacity than the ETT-250. And yet, for giant mines with an output of 2 million tons per year and more, a profitable solution to the question of haulage by battery electric locomotives, as the basic prerequisite, requires more powerful electric locomotives and more powerful batteries. Both more powerful electric locomotives and higher-capacity batteries are being mastered by the manufacturing plants. The coal industry is demanding them ever more insistently, and there is every reason to hope that it will receive them in the very near future.

1. For example, with DK VOOA motors, 100/110 V, 7.6/8.5 kW, 100 A. ↩↩↩

2. According to the formula for $n_e$, at $n = 16$ 3.32 electric locomotives would be required, i.e., in practice the same 4, since it would hardly be possible to reduce $\theta$ further. In addition, it is necessary to take into account the transportation of people to the place of work. ↩