1.SUMMARY

This paper is concerned with the development of an economic cost model to aid the decision-making process in belt conveyor design. The model involves the establishment of cost or objective functions which integrate the performance characteristics with the various cost factors involved. The cost factors include energy costs and annual equivalent cost of the conveyor components. With respect to the latter. consideration is given to such factors as equipment or component life, salvage value, taxation rates and rates of return. The effects of inflation and variations in the annual differential escalation in the energy component are included in the model. The application of the model to the design of single and multi-belt conveyor systems is described by reference to design examples.

2. INTRODUCTION

The handling of materials in bulk form is of major concern to a vast number and variety of industries throughout the world. Such industries rely heavily on the need to transport bulk solids over widely varying distances, the costs associated with these operations being very substantial. It is, therefore, most important that both high overall efficiency in terms of energy requirements, together with optimum least cost performance over the life of the installations is obtained. For this reason, it is necessary, at the design and planning stage, to undertake detailed appraisals of various alternative handling schemes that may be employed for particular installations.

When comparisons are made between continuous modes of transport such as slurry pipelines, belt conveyors, screw conveyors and pneumatic systems and discontinuous modes such as ship or barge, road and rail, the variations in costs may differ by several orders of magnitude. Even when one mode of conveying, such as belt conveying, is examined for a particular installation, within the range of possible combinations of conveyor size, speed and plant layout, there can be considerable variations in the overall costs. For these reasons it is important that the conditions for optimum performance of particular types of conveyors and handling equipment be established.

Of the various modes of conveying of bulk solids, belt conveyors are of considerable importance in view of their widespread use and proven reliability. Although their use is mainly confined to shorter distances, they also offer advantages for long distance transportation. While their application to long distance transportation is increasing, their potential in this respect has yet to be fully realised. Following the work of Jonkers [1], when comparisons are made on an energy loss per unit distance basis, belt conveyors rank second lowest of the continuous modes, with slurry or hydraulic transport being the lowest. However, in the case of slurry pipelines, since the energy losses are a function of the transportation velocity, there is a need to specify a maximum economical velocity. Again, on an energy loss per unit basis, belt conveyors rank approximately equal to rail transport.

As demonstrated by Roberts et al [2-6], energy costs cannot be considered in isolation and design and performance evaluations of conveyor systems are only meaningful when complete economic studies, based on the total life of the plant, are made. In the work of Roberts et al an optimum design methodology has been developed in which cost functions are derived which take into account the energy costs and annual equivalent cost computed on a life cycle basis. The latter requires consideration of such factors as equipment or component life, salvage value, taxation rates and rates of return. The effects of inflation and variations in the annual differential escalation in the energy and component costs are included in the model. Optimum, minimum cost solutions to the conveyor design problem are sought taking into account constraints imposed by design, manufacturing and operational limitations. The optimum solutions may be implemented directly or used as a 'yardstick' against which the actual conveyor performance can be measured.

The purpose of this paper is to describe the application of the afore-mentioned economic analysis to the optimum design of belt conveyors. Both single and multiple conveyor systems are discussed.

3. THE GENERALISED CONVEYOR DESIGN PROBLEM

3.1 Introductory Remarks

The problem of conveying a bulk solid by belt conveyor from one point to another is depicted, schematically, in Figure 1. Over short distances a single conveyor will be used as in Figure 1(a) whereas for long distance transportation, a multiple conveyor system as in Figure 1(b) will need to be considered. In the latter case it becomes necessary to determine the most desirable number of individual conveyors and their corresponding lengths.

It is convenient, in the first instance, to consider the optimum design of belt conveyors in terms of the procedures applicable to the general class of continuous conveyors for bulk solids handling. The problem is formulated in generalised form taking into account the various steps necessary in the design process.

3.2 Performance Characteristics

The design of a conveyor or system of conveyors depends on a knowledge of the relevant performance characteristics. Of particular importance are the relationships for throughput and power. While the design and operating features of the various types of mechanical conveyors differ widely, the basic performance characteristics, in functional form, are quite similar.
For a given bulk material, the general relationships for throughput and power are:

      Throughput (kg/s or t/h)

Q_m = f₂(x₁, x₂, ...., x_n, ρ_m, s, α)          {Equation 1}

      Power (kW)

P_M = f₁(x₁, x₂, ...., x_n, ρ_m, L, s, α)        {Equation 2}

In the above relationships (x₁, x₂, .... , x_n) are geometrical design variables applicable to the particular conveyor, p_m is the bulk density of the material being conveyed, L is the overall conveyor length, s is the conveyor speed and α is the angle of elevation. It is to be noted that for most continuous type conveyors the throughout Q_m is independent of the conveyor length, as indicated by equation (1).

The geometrical variables are those that express the carrying capacity (and power) in terms of a unit conveyor length. For example, in the case of a belt conveyor, they include the belt width, number of plies, idler configuration (number of rollers and troughing angles) and idler troughing angles) and idler a screw conveyor they include the screw diameter, pitch, core diameter, choke length and casing clearance. For a bucket elevator they include the belt width, bucket capacity and bucket spacing.

In the case of multiple length conveyers the overall length will be given by

	N	ℓi             {Equation 3}
L =	Σ
	i = l

where ℓ_i = length of i th conveyor

and N = number of conveyors involved

For conveyors used to elevate bulk materials. the overall efficiency is of importance in the performance assessment. The overall efficiency relates the theoretical power to elevate a bulk material in the absence of friction to the actual power. That is

η0 =	Qm g L sinα	{Equation 4}
	3600 PM

where QM = t/h

In addition to the performance equations (1), (2) and (4), it is necessary to establish a relationship which takes into account the over-riding geometrical requirements governing the conveying distance and height of lift. That is, a relationship is needed which expresses the effective height of lift H(m) as a function of conveyor length and angle of elevation α. Knowing the required height of lift, an additional height allowance must be made to permit the gravity flow of the bulk material from the outlet of the conveyor, through some discharge device such as a transfer chute.

In general terms

H = f₃(L, α)            {Equation 5}

By way of example, an appropriate function for the conveyor Figure 1(a) is

H =	L sin α	for 0 < α <	π	{Equation 6}
	1+CHsinα		2

Where C_H = coefficient based on conveyor geometry

3.3 Design Constraints

In addition to the need to satisfy the performance and geometrical requirements, the design analysis must take into account certain additional constraints. These fall into two main groups:

3.3.1 Functional Constraints

These govern the need for the conveyor components to be designed for strength, safety and reliability. The actual magnitude of the conveyor geometrical variables may often be dictated by the strength and life characteristics of the component materials. For example, the number of plies p in a conveyor belt, as well as depending on the belt width B and maximum belt tension F_max, also depends on the safe working stress of the belt material σ_b

That is

p = f₄(σ_b, B, F_max)            {Equation 7}

3.3.2 Constraints on System Variables

Practical considerations usually create the need to constrain the conveyor variables so that they lie within fixed limits. For instance the upper limit on belt conveyor speed is dictated by the need to ensure efficient tracking and to minimise component wear. Normally belt speeds are limited to a maximum value of 6 m/s but in view of the current improved technology, belt speeds greater than this are being used.*^{(* The Rheinische Braunkolinwerke in West Germany operate 2. 5 m wide belt conveyors running at 7.5 m/s.)} Belt widths are limited by manufacturing capabilities. Hence in general terms we can write

x_1ℓ < x₁ < x_1u}
------------------}
x_nℓ < x_n < x_nu}
s_ℓ < s < s_u      }
α_l < α < α_u     }                {Equation 8}

3.4 Design Solution

The design of a conveyor to satisfy the specified performance requirements will involve the consideration of a number of alternative solutions. In theory, computations based on the relationships of (1) and (2) together with the constraints of the type given by (5), (7) and (8) will yield a large number (in fact an infinite number) of solutions in the 'solution space', all of which meet the required performance conditions. A decision needs to be made as to which of the possible solutions is the most appropriate. For this reason additional criteria need to be laid down to aid the decision-making process. Such criteria, inevitably, must take account of the need for efficient and economic operation.

By establishing appropriate cost or objective functions it is possible to obtain optimal solutions to conveyor design problem. Such objective functions need to be derived on the basis of detailed economic considerations as outlined in the next sections of this paper. Objective functions obtained in this way will have the functional form

I = I(x₁, x₂, .... , x_n, ρ_m, s, L, α)          {Equation 9}

The objective is to determine the system variables (x_l, x₂, .... X_n, s, L, α) that minimise I subject to the required performance condition or throughput expressed by (2). The solution must take account of the various system constraints such as those given by equations (3), (5), (7) and (8).

In general terms the design analysis is transformed into a constrained, non-linear optimisation problem. While several known computing algorithms for this class of problem exist [7], two have been found to be effective for the solutions of conveyor optimisation problems. These are the constrained Fletcher-Powell (Conmin) algorithm which has been adapted by Khaw [8] for the solution of screw conveyor design problems and the Box (complex) algorithm which was also examined by Khaw and more recently by Lim [10] for belt conveyor problems. A modification of the complex algorithm based on reference [9] has also been examined by Lim for the case of discrete valued variables optimisation problems which are more representative of actual belt conveyor design problems.

While optimum solutions may be found in this way, it is important to note that the optimisation algorithm is not going to be the final answer in all cases. At least perturbations of the solution about the optimum will show how sensitive the operating costs are to variations in the performance variables. If the independent variables are few in number and the general range of these variables for a feasible solution is known, then it is often easier to determine the optimum solution by repeated computation of the cost function I for the selected range of variables. This procedure may be preferable where, as in many cases, the variables need to be discrete values or integers such as belt width and the number of plies comprising the belt.

4. ECONOMIC CONSIDERATIONS

The costs incurred in a conveyor system may be divided into two categories

1. Capital costs

2. Operating costs.

Any form of economic evaluation must express both types of cost in some common measure. By specifying a rate of return required on the capital funds employed, costs may be expressed as present equivalent costs or as annual equivalent costs over the life of the system.

4.1 Annual Equivalent of Capital Cost

For the belt conveyor the capital cost items include

• Drive system, motor, speed reducers

• Belt

• Idlers

• Structure

• Transfer stations (in multi-conveyor systems)

The costs need to be determined on a life cycle basis. Depending on the accuracy required, the analysis may be on the basis of cash flows

• Before tax

• After tax without considering inflation

• After tax considering inflation

Only the last of these will be described as the other two are readily derived from it.

If all costs are expressed as the inflated dollar expenses expected to be incurred at the time they occur, then for capital items not needing replacement during the life of the system, n, the present equivalent of these capital costs, PEC, is given by [14]

 

PEC =	A - V (	P	) if - t(PED)	{Equation 10}
		fn
	________________
	1 - t

where

• A = First cost of the item
  

• V = Salvage value
  

• t = Tax rate
  

• PED = Present equivalent of depreciation
  

• i_f - inflation modified rate of return

• (	P	)	if	=	1	= Present equivalent factor.
  	f		n		(1 + if)n

  
  

For a company maintaining a constant proportion of debt capital, r_d then if is given by

i_f = (1-t)r_d i_d + (1-r_d)[(1+r) (1+i_e) - 1]           {Equation 11}

where i_d = interest rate on debt
i_e = after tax return required on equity funds with zero inflation rate.

The annual equivalent cost, in year zero dollars, may be obtained by multiplying the present equivalent by the capital recovery factor,

(	a	)	ifo	=	ifo(1+ifo) n
	p		n		(1+ifo)n - 1

                                 that is

AEC = PEC (	a	)	ifo	{Equation 12}
	p		n

The factor i_f^o which expresses the time value of money when all cash flows are expressed in constant year zero dollars rather than inflated dollars is given by

ifo =	(1-t)rd id - r rd	+ (1-rd)ie	{Equation 13}
	(1+r)

For any item which may require replacement during the life of the conveyor system, the values of A, V and PED in equation (10) are modified to include the present equivalents of the original and all replacement items required.

By expressing salvage values and depreciations as a fraction of first cost A, the annual equivalent cost of any item can be expressed as a fraction of its first cost and hence as a function of the operating and geometrical variables of the system. By expressing all costs in this form and summing, the total annual equivalent cost is obtained.

4.3 Operating Costs

For a given conveyor or handling system the operating costs include

• Energy

• Repairs and maintenance

• Labor

4.3.1 Energy Costs

The annual energy cost may be calculated from the annual energy consumption. That is

I₁ = k₁ e_c p       {Equation 14}

where e_c = annual hours of operation times the unit cost of energy

            k₁ = annual equivalent energy cost coefficient taking into account inflation and annual escalation rate of             energy costs

It is worth noting that a considerable amount of energy is expended in overcoming frictional resistance. This will vary with the speed and design features of the particular conveyor and the bulk density of the material. A typical belt conveyor could require as much as 30% of the total power requirement to move the empty belt.

In some studies it may be desirable to allow for the possibility of energy costs rising more rapidly than the general inflation rate. If

    r = general inflation rate
    r_e = annual escalation rate of energy costs
    e_co = energy cost at time zero
    e_cz = energy cost in year z

then

e_cz = e_co (1 + r_e)^z           {Equation 15}

The present equivalent of energy costs over the life of the system n is given by

 

PEC =	n Σ z=1	eco (	1+re	)z     {Equation 16}
			1 + if

The annual equivalent in year zero dollars is obtained by multiplying by the capital recovery factor

(	a	)	ifo	=	ifo(1+ifo) n
	p		n		(1+ifo)n - 1

That is

 

AEC (energy) = eco (	a	) ifo	n Σ z=1	[	1 + re	]z         {Equation 17}
	pn				1 + if

 

Thus k₁ in equation (14) becomes

 

k1 = (	a	) ifo	n Σ z=1	[	1 + rez	]z         {Equation 18}
	pn				1 + if

 

4.3.2 Repairs and Maintenance

Although this is an important item, little is known of the relation with the operating parameters such as speed and belt width. For overall estimates it is often taken as some percentage of the overall cost of capital items plus some percentage of the belt cost [12]. This gives no insight into the variation with operating parameters. Intuitively on general grounds it may be expected that maintenance cost and costs associated with overall reliability would increase with the speed of the conveyor. This may be seen, for example, when the life of the idlers is considered. The life of the idler bearings decreases as both the load and rotational speed (and hence belt speed) increase. However, due to ignorance of the form of the repairs and maintenance cost functions, these costs have not been included in the model. The assumption of these costs being a fixed percentage of the overall cost is made; as a consequence the, costs based on this assumption will not influence the operating parameters which justifies the exclusion of these cost items.

4.3.3 Labor

In comparing the belt conveyor with an alternative for a particular application, labor costs for operation may be quite significant. In optimising a belt conveyor for a particular application, the labor is unlikely to change with a different choice of operating parameters. For this reason no labor costs have been included.

5. ECONOMIC ANALYSIS APPLIED TO BELT CONVEYOR DESIGN

5.1 Problem Specification

Consider the problem of designing a typical belt conveyor installation as shown in Figure 1(a) or 1(b). The conveyor is required to transport a bulk solid of density ρm (kg/m³) at a rate of Q_m (kg/s) over a distance L (m) and height of lift H (m).

5.1.1 Design Assumptions

The design requires some assumptions to be made, such as

• The proposed idler configuration

• The proposed belt type

• The proposed belt and drive configuration
  to suit the particular application.

5.1.2 Conveyor Variables

The relevant geometrical design variables are

    x₁ = B = Belt width
    x₂ = p = Number of plies (note: p must be an integer)
    x₃ = ß = idler troughing angle for two roller or three roller configuration
    x₄ = λ = idler contact perimeter ratio for nominated idler configuration
    x₅ = a_o = idler spacing on carrying side (m)
    x₆ = a_u = idler spacer on return side (m)
    x₇ = ℓ = length of individual conveyors in multiple conveyor system (m)

Also v = s = belt velocity (m/s)

5.2 Performance Characteristics

While the general design procedures for belt conveyors are well documented, for the purpose of the present discussion the essential equations given in references [5] and [6] are summarised here.

5.2.1 Throughput Q_m

This is given by

Q_m = ρ_m A v cos α       {Equation 19}

Normally the angles of inclination α are low enough for cos α = 1. Hence

Q_m = ρ_mA v        {Equation 20}

Where A = U b²       {Equation 21}

U = Non-dimensional cross-sectional area shape factor

b = contact or 'wetted' perimeter (m)

Shape factors can be determined for different idler configurations:

• Single Idler - Flat Belt

  U1 =	tan δ	{Equation 22}
  	6

  Where δ = surcharge angle

• Two Idler System

  
  Figure 2 Two Idler System

  U2 =	sin 2 ß	+	tan δ	(cos 2 ß + 1)         {Equation 23}
  	8		12

  For Maximum U₂

  tan 2 ß* =	3	{Equation 24}
  	2tan δ

• Three idler system
  
  Figure 3 Three Idler System

  U3 =	1	{λsinß +	λ²	sin 2 ß +	tan δ	[1+4 λ cos ß + 2 λ²(1+cos 2ß)]}      {25}
  	1+2λ)²		2		6

  Where λ = y/x

  For given values of δ and λ, the angle ß* for maximum U₃ is given by

  cos ß* + λ cos 2ß* -	2	tan δ (sin ß* + λ sin 2ß) = 0         {Equation 26}
  	3

  Alternatively, for given values of δ and ß, the ratio λ* for maximum U₃ is given by

   

  λ* =	2/3tan δ (1 - cos ß) - sin ß	{Equation 27}
  	sin2ß + 2/3tan δ (1+cos 2 ß) - 2 sin ß - 4/3 tan δ cos ß

  The belt width B is expressed in terms of the contact perimeter b allowing for edge effects

  B = 1.11b + 0.056 (m)           {Equation 28}

5.2.2 Conveyor Power

In simplified terms the total resistance F_U of a conveyor belt is

F_U = C[F_H1 + F_H2} + F_st + F_s1 + F_s2 (kN)          {Equation 29}

Where

• F_H1 = Empty belt frictional resistance

• F_H2 = Load frictional resistance

• F_st = Slope resistance

• F_s1, F_s2 = Special resistances

• C = Factor to allow for secondary resistances such as those due to accelerating the material onto the belt.

The factor C is given in reference [11]. Alternatively it may be expressed as

C = 0.85 + 13.31 L^-0.576 for 10 < L < 1500m
                           C = 1.025 for 1500 < L < 5000m                       {Equation 30}

The required motor power is

PM =	FUv	{Equation 31}
	η

where η = drive efficiency

5.3 Belt Design

Based on the simplified drive analysis the wrap factor C_W for a given angle of wrap ø (radians) on the driving pulley(s) may be approximated by

C_W = 1/e^µø - 1      {Equation 32}

Where µ = Friction coefficient between the belt and pulley

The slack side tension is

F₂ = C_W F_U        {Equation 33}

and tight side tension is

F₁ = F_U + F₂        {Equation 34}

The maximum belt tension will be the larger of the values of F₁ and that computed from the conveyor layout where it is necessary to limit the belt sag between the idlers.

For an assumed belt type, the allowable stress is given by o_b (kN/m, ply). Thus for a given belt width B the required number of plies is given by

p = F_max/B o_b        {Equation 35}

where p = integer value

5.4 Economic Considerations

The cost components are expressed as an annual equivalent cost based on the life of the plant. The actual cost values are in dollars Australian but are readily convertible to other currencies. While the relative differences between the energy and component costs may vary from one country to another, the objective of this study is to present the general principles by which economic design may be achieved. Individual cost variations can be readily catered for via the use of the various coefficients incorporated in the design model.

5.4.1 Energy Costs

The annual equivalent cost of the energy is given by equation (14)

I₁ = k₁ e_c p        {Equation 14}

5.4.2 Capital Cost

For all these items it has been found possible, over restricted ranges, to express the first cost as a linear function of the operating and design variables [5,6]. The annual equivalent capital cost is obtained by multiplying the first cost by a coefficient determined on the basis of the economic analysis previously described. The annual equivalent cost relationships may be summarised as follows:

              Motor

I₂ = k₂(c₁ + c₂ P_M)       {Equation 36}

              Gear Reducer

I₃ = k₃(c₃ + c₄ T_R)       {Equation 37}

              Conveyor Belting

I₄ = k₄(c₅ + c₆ B)KL       {Equation 38}

              Idler Pulleys

I₅ = k₅(c₇ + c₈ B) L/a_o       {Equation 39}

              for carrying side

I₆ = k₆(c₉ + c₁₀ B)L/a_u       {Equation 40}

              for return side

In the above equations c₁,c₂ ....c₁₀ are first cost coefficients; k₂, k₃ .... k₆ are annual equivalent cost coefficients; P_M is the power; T_R is the transmitted torque; B is the belt width; a_o and a_u are spacing of idlers on carrying and return side respectively; L is the conveyor length. The factor K in equation (33) allows for the total belt length, taking into account such factors as take-up pulleys and trippers; (K > 2). It should be noted that the coefficients c₅ and c₆ for the conveyor belting are a fraction of the belt type and number of plies.

As shown by Heaney [13], the conveyor structure, which is a function of the belt width, is a major component of the overall cost and should be taken into consideration. The supporting structure is assumed to consist of two steel main beams, braced at intervals, with a walkway at each side. A typical cross-section is shown in Figure 4. The structure is assumed to be supported at 5 m intervals.

 

Figure 4 Conveyor Structure

Other capital cost items include the belt tensioning arrangements, discharge arrangements and drive couplings. Although the cost of these items is dependent, to some extent, on the conveyor width, capacity and operating parameters, as far as the overall cost is concerned their contribution may be assumed constant. For this reason they need not be included in the cost or objective function.

The overall cost or objective function is the summation of all the component costs, as indicated by equation (9).

5.4.3 General Comment

It should be noted that the component costs, such as, for example, those associated with the gear reducer, are not really continuous functions as assumed in the preceding equations. The costs are really stepwise functions which are dependent on the available standard size range of the components. This is illustrated in Figure 5.

 

Figure 5 Cost Functions for Conveyor Components

6. BELT CONVEYOR EXAMPLE

6.1 Design of Single Conveyor

A belt conveyor, 500 m long, is required to convey a bulk material of bulk density ρ_m = 850 kg/m³ up an incline of 1 in 10 and discharge it at a rate of Q_m = 600 t/h.

6.1.1 Principal Design Assumptions

• Idlers - 3 roll system with λ = 1 and ß = 35°

• Surcharge angle δ = 20°

• belt type - Kuralon/Nylon (Type KN 150)
  Allowable stress o_b = 15.8 kN/m, ply

• Gear reducer - helical type

• Operation - 12 hours/day over 300 days per year

6.1.2 Design Constraints

Belt width 0.65 < B < 1.6 (m)

Plies 2 < p < 8

Belt speed 0.5 < v < 6 (m/s)

6.1.3 Economic Considerations

The basic assumptions are

• General inflation rate is 10%.

• Energy costs - unit cost of energy is $0.06 per kW/hr.- annual cost escalation rate is 15%.

• Installation - life of the conveyor and drive components is 12 years. Salvage value is zero. Cost escalation
  rate per year for drive and structure is 10%.

• Conveyor belting and idler pulleys - life of 7 years is assumed with zero salvage value. The belt and idlers
  are replaced after the seventh year and are then depreciated and written off at the end of the twelfth year. The price escalation rate per year for these components is 10%.

• After tax rate of return on equity capital is 5%.

• Taxation rate is 0.46.

• Depreciation by straight line method.

With these assumptions the annual equivalent cost coefficients are:

Energy k₁ = 1.317

Motor k₂ = 0.166

Gear reducer k₃ = 0.166

Conveyor belting k₄ = 0.260

Carrying idlers k₅ = 0.260

Return idlers k₆ = 0.260

Structure k₇ = 0.166

The method of calculating the above coefficients is illustrated in the Appendix.

6.1.4 Design Solutions

To gain some appreciation for the cost variations involved by considering alternative design solutions, cost functions have been computed for a range of belt widths. The results are shown in Figure 6. Figure 6(a) shows the variation of velocity, power and number of plies as a function of belt width; the corresponding annual equivalent cost curves are presented in Figure 6(b). In Figure 6(b) the lower curve is the cost excluding the structure while the upper curve is the total cost including the structure.

In the solution of the conveyor design problem consideration needs to be given to the constraints which have been previously indicated, as well as to the need for some of the design variables to be discrete values. A possible solution in this case is

Belt width B = 0.65 m (a preferred size)

Number of plies = 4

Belt speed = 4.21 m/s

Power = 146 kW

Total annual cost I_c = $61,300, excluding structure
                              I_T = $113,200, including structure

It is useful to examine the contributions of each item in the overall cost. This information is presented in Figure 7 and, as can be seen, the over-riding contribution is that due to the structure. With respect to the actual conveyor components, the substantial cost is that due to the belt. The energy cost is a major component in this case, the energy to overcome the slope resistance being a major factor.

The example clearly demonstrates the advantages of using narrower, faster running belts. For instance. doubling the belt width reduces the required belt speed and the power requirements but results in an increase of 28% in the overall cost.

Figure 8 shows the break-up of the various components of the power as a function of belt width. For comparison purposes the belt velocity is also shown.

PERFORMANCE AND OVERALL COST
CHARACTERISTICS FOR BELT CONVEYOR
Q= 600 T/HR, L= 500M, H= 50 M

Figure 6

BELT CONVEYOR COMPONENT COSTS

Figure 7

 

CONVEYOR POWER COMPONENTS

Q = 600T/HR, L = 500M, H = 50 M

Figure 8

6.2 Influence of Variable Inflation Rates

To illustrate the effects of variations due to cost escalation, the same problem as in 6.1 was examined for the case where the annual cost escalation rate is 20% for the energy costs and 15% for the belt, all other parameters being the same as before. In this case

k_l = 1.751

and k₄ = 0.298 (See Appendix for computation of this coefficient)

The comparable annual costs for the conveyor alone and for the conveyor structure are, respectively,

I_c = $76,415

I_T = $128,364

6.3 Comparison between Two and Three Idler System

Consider the same problem as in 6.1 but in this case a two idler system with ß = 35° is to be used. With the same conditions applicable as in 6.1 the suggested solution is indicated in Table 1. For comparison purposes the solution for the three idler system is also included.

TABLE 1 Comparison between Two and Three Idler Systems
Q_m = 600 t/h, ρ = 850 kg/m³, L = 500 m, H = 50 m

	Two Idler System ß = 35°	Three Idler System ß = 35° λ = 1,0
Belt width B (m)	0.65	.65
Number of plies	4	4
Belt Speed m/s	5.70	4.21
Power kW	138	146
Annual cost Ic	$63,420	$61,300
Total Annual Cost IT	$114,700	$113,200

7. MULTIPLE CONVEYOR SYSTEMS

Where materials are to be conveyed by belts over long distances, multiple conveyors, as illustrated in Figure 1(b), need to be employed. The choice of the number of conveyors and the individual length of each component conveyor will often be dictated by the design constraints such as the need to limit the maximum belt tension to suit the maximum number of plies available. However, it is also evident that in view of the cost variations per unit length, it is important that economic factors are taken into account.

By way of illustration, the annual equivalent costs per unit length of conveyor have been determined for the case where the throughput is Q_m = 600 t/h bulk density ρ_m = 850 kg/m³ and the slopes are zero in one case and 1 in 10 in another. All other relevant parameters are the same as in the previous example of Section 6.1. Figures 9 and 10 show, for the two cases, the annual equivalent cost per metre length as a function of belt length for a range of belt widths. The range of belt lengths considered is from zero to one kilometre.

In both cases the costs per unit length for very short length conveyors are very high, as would be expected. For the zero slope case (Figure 9) and for the range of lengths beyond 200m, the costs per unit length are substantially constant. This trend is also shown to occur for the slope of 1 in 10 (Figure 10) for the narrower belts, but as the belt widths increase, the cost per unit length starts to increase with belt length. Furthermore, the belt length becomes restricted due to the limitations in plies. A stronger belt would then be necessary. For wider belts the indications favor the use of several short conveyors rather than one long belt. However, the cost advantage in employing several conveyors would be either partially or totally offset by the additional costs due to the transfer stations.

As shown by Lim [10], the optimum number and length of component conveyors may be determined as part of the solution to the general belt conveyor design problem. Solutions to such problems have been obtained using the Box (complex) algorithm. One example considered by Lim involved the requirement to convey a bulk material of density ρ_m = 800 kg/m³ over a distance of 5 km up a slope of 1 in 100. It was assumed that the component conveyors were all of equal length and constraints were imposed to limit the number of plies to 8, the belt speed to 6 m/s and the minimum belt width.

BELT CONVEYOR - TOTAL COST PER METER -
600 T/HR. SLOPE 0.0

Figure 9

BELT CONVEYOR - TOTAL COST PER METER -
Q = 600 T/HR, SLOPE 0.1

Figure 10

to 0.75 m. Considering the overall costs, including the structure but neglecting the transfer stations, gave the optimum solution

Number of belt sections N = 11

Belt Width B = 0.75 m

Number of plies p = 2

Belt speed v = 3.27 m/s

If the cost of transfer stations is included, the number of belt sections decreased to N = 3, and number of plies increased to p = 6. The other parameters remained unchanged.

8. CONCLUSIONS

This paper has drawn attention to the costs of bulk handling operations and the consequent need to design more efficient and economical systems. To achieve this objective a design methodology has been developed which integrates the underlying principles of engineering economic analysis with the concepts of optimisation theory.

The paper has dealt specifically with the economic analysis and optimum design of belt conveyors for bulk solids handling. The various conveyor component costs as functions of the overall costs have been examined and the conclusions drawn favor the use of narrower, faster-running belts. Analysis of annual equivalent costs per unit length has provided guidelines for the selection of optimum lengths of conveyors in multi-conveyor system.

On the basis of the design and analysis procedure presented, comparisons between various types of conveyors can readily be made. It is clear, from the results presented, that the global problem of optimisation applied to large, integrated handling systems can only be meaningful when the best operating conditions of individual conveyors and other items of handling equipment are understood.

ACKNOWLEDGEMENT

The author wishes to express his appreciation of the valuable contribution rendered by his colleague Air. J.W. Hayes, in the development of the economic models. The assistance rendered by Dr. F. Huang in the development of the computer programs and associated plotting routines is gratefully acknowledged.

9. REFERENCES

1. JONKERS, C.O., "Loss Factor in Transport". Fördern und Heben, 1981.

2. ROBERTS, A. W. and CHARLTON, W.H., "Optimum Design and Control of Grain Conveyor Systems". Proceedings of the International Grain and Forage Harvesting Conference, American Society of Agricultural Engineers (ASAE) and Commission Internatimale eu Genie Rural (CIGR) held at Iowa State University, U.S.A., Sept. 1977.

3. ROBERTS, A.W., SCOTT, 0.J. and HAYES, J.W., "Design Criteria for Optimum Performance of Conveyor Systems used in Agricultural Production". Proceedings of Conference on Agricultural Engineering, The Institution of Engineers, Australia, Toowoomba, Qld., Aug. 1978.

4. ROBERTS, A.W., HAYES, J.W. and SCOTT, 0.J., "Economic Considerations in the Optimum Design of Conveyors for Bulk Solids Handling". Proceedings of International Powder and Bulk Solids Handling and Processing Conference, Philadelphia, U.S.A., May 15-17, 1979, pp. 101-116.

5. ROBERTS, A.W. and HAYES, J.W., "Economic Analysis in the Optimum Design of Conveyors", TUNRA, The University of Newcastle, N.S.W., Australia (ISBN 0 7259 340 6) 2nd Edition, 1980.

6. ROBERTS, A.W., HAYES, J.W. and SCOTT, 0.J., "Optimal Design of Continuous Conveyors". Bulk Solids Handling, Vol. 1, No. 2, 1981.

7. KUESTER, J. L. and MIZE, J.H., "Optimisation Techniques with Fortran" McGraw Hill, 1973

8. KHAW, B.W., "Evaluation of Computing Algorithms for Solving Non-Linear Multi-variable Optimisation Problems". Unpublished B.E. Thesis, Dept. of Mech. Eng., The University of Newcastle, N.S.W., Australia, 1977

9. BEVERIDGE, G.S.G. and SCHECHTER, "optimisation Theory and Practice", McGraw Hill.

10. LIM, B.C., "Optimisation of Bulk Solids Handling Systems". Unpublished B.E. thesis, Dept. of Mech. Eng., The University of Newcastle, N.S.W., Australia, 1979.

11. ISO/DISS048 Draft International Standard: "Continuous Mechanical
    Handling Equipment - Belt Conveyors with Carrying Idlers Calculation of Operating Power and Tensile Forces".

12. CUMMlNS, A.B. and GIVEN, I.A., "SME Mining Engineering Handbook", Vol. 2, The American Institute of Mining, Metallurgical and Petroleum Engineers Inc., 1973.

13. HEANEY, S.J., "An Investigation into the Design of Belt Conveyors". Unpublished M. Eng.5c. Thesis, Dept. of Mech. Eng., The University of Newcastle, N.S.W., Australia, 1979.

14. SMITH, G.W., "Engineering Economy: Analysis of Capital Expenditures, 2nd Ed., Iowa State University Press, 1973.

APPENDIX

Calculation of Annual Equivalent Cost Coefficients. Example - coefficient for Belt Cost, k₄.

Assumptions

Capital is all equity capital, that is r_d = 0

Required return on equity, i_e = 5%

Income tax rate, t = 46%

Conveyor system life, n = 12 years

Estimated belt life = 7 years

General inflation rate r = 10%

Annual cost escalation rate of belt r_b = 15%

Salvage value at any time, V = 0

Depreciation by straight line based on 7 year life with remaining value written off at the end of year 12

i_f = (l-t) r_di_d + (1-r_d)[(1+r)(1+i_e) - 1]
   = 0 + (1+.10)(1+.05) - 1
   = 0.155

ifo	=	(1-t) rdid - rrd	+ (1-rd) ie
		1+r
	=	0 + 1(.05)
	=	.05

Price of replacement belt = A(1.15)⁷

Present equivalent of first cost of belts	=	A[1 + (1.15)7]
		(1.155)
	=	1.9701 A

Depreciation of first belt = A/7 each year for seven years

Depreciation of second belt = A/7(1.15) each year plus an additional 2A/7(1.15)⁷ in year 12

 

PED =

A

(

p

7if +

A

(1.15)7

(

f

)5if (

p

) 12if +

2A

(1.15)7

(

p

)12if

7

a

7

a

f

7

f

= A[	(1.155)7 - 1	+	(1.15)7	(	1.1555-1	)	1	+	2	(1.15)7	1	]
	(7)(.155)(1.155)7		7		.155		(1.155)12		7		(1.155)12

=	A[.5855 + .4591 + .1348]
=	1.1794A

 

PEC =	A - V(p/f)nif - t(PED)
	1 - t
=	1.9701A - .46(1.1794A)
	.54
=	2.6436A

 

AEC =	PEC (	a	)125
		p
=	(2.6436)(.11283)A
=	.2983A

Thus the annual equivalent cost coefficient

 

k =	AEC	= 0.2983
	A