# Optimized Economic Loading of Distribution Transformers Using Minimum Energy Loss Computing

Authors: Ayman Agha; Hani Attar; Ashish Kr. Luhach
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

This paper presents an approach to achieve the Economic Loading of Distribution Transformer (ELDT) based on minimizing the active energy loss. The effects of the transformer on-load and no-load losses, load factor (LF), and load loss factor (LSF) on the total energy losses are considered. The adopted technique in this paper consists of two phases, where ELDT is determined analytically in the first phase, and the load profile adequated (LSF) is statistically obtained in the second one. The results declare that the proposed technique is suitable for the shifts system mode of operation such as in industrial plants. Moreover, this paper investigates the effect of the total active and reactive power and energy losses on ELDT. Finally, numerical examples with software analyses are performed as a valuable tool, which supports the decision-makers to decide trustfully the size of the transformer and its capacity (kVA) during the design stage, as well as to determine the economic loading during the operation based on the effective factors, that is, total power, energy losses, and the load profile.


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## Body

## 1. Introduction

The key power system elements that affect the cost are the generators, transmission lines, and transformers; consequently, the maximum power system benefit is achieved when these elements operate at their maximum efficiency and minimum operational cost. Indeed, the appropriate design, selection, and operational conditions for the power system elements play a crucial factor in reducing the overall operational cost and hence improve the overall system efficiency and availability. The proposed work in this paper focuses on the power and energy losses of the transformer and its effect on the ELDT when waving the other system elements (generators and transmission lines).The transformer is regarded as static equipment where most of the generated power passes through it at different stages. Moreover, the transformer is usually designed for a lifetime of 20 years or more. Apart from small-size transformers, the power and distribution transformer efficiencies usually vary from 95% to 98.5%, taking into consideration the fact that these percentages are even higher for large-size transformers. Moreover, the vast number of transformers installed in a single power system makes the transformer design and its load be significant issues on the overall system cost, efficiency, and availability. Hence, the improper selection of transformers results in additional expenses added to the total investment value, extra cost to the power in terms of power and energy losses, and the extra availability/reliability of the overall system.Based on the above, it is essential to optimize the transformer’s loading versus the technical losses, which might have been placed during the power system operation period.Different approaches were proposed to assess the technical and economic transformers’ cost and design issues, such as in [1], where the reduction of overall losses in the distribution transformer was discussed. The impact of the transformer overexcitation on the loading conditions was investigated in [2]. The effect of the transformer location on cost minimization was introduced by [3], where [4] showed how to minimize the iron losses. The research work in [5] classified the losses of transformers into three groups of losses, which are tank losses, core joints losses, and stray losses. The determination of the no-load losses magnitude by using finite element analysis was presented in [6]. The authors of [7] proposed a mathematical model to determine the uncertainty of the transformer on-load and no-load losses. The usage of load loss factor is utilized in [8] to find the amount of the transformer power and energy losses. The reduction of the core cost which resulted from newly developed materials was given in [9]. In [10], the relationship between the no-load current and the level of the total harmonic distortion was investigated, wherein, in [11], the impact of the harmonic level over the nonlinear load was investigated in good detail. The work in [12] proposed a thermal model to monitor the operation and identify the hotspot points. The oil breakdown strength based on nanofluids was discussed in [13]. The developed methods to determine the transformer’s life span considering the annual load factor were presented in [14, 15]. In addition, the economics of the transformer and the evaluation of the total owning cost have been thoroughly discussed and analyzed in [3, 16–18].In this paper, the ELDT is achieved for the transformer’s capacity, based on the effective factors, parameters, and nature of the working mode during the transformer’s lifetime operation.The rest of the paper is arranged as follows. In Section2, the formulation of the problem is presented, supported by the required mathematical equations, and the transformer losses calculations are presented in Section 3. In Section 4, numerical examples are shown for the proposed technique to simplify the understanding of implementing the technique. Finally, the results, conclusions, and future work proposals are presented in Section 5.


## 2. Problem Formulation

The optimum transformer loading is subject to all power and energy losses throughout the operational time, where one year is usually considered enough for the propose of the economic evaluation. Moreover, the objective function can be mathematically formulated as follows:Find(1)SLecon..It is subject to(2)Sn≥SD≥SLecon.And(3)ΔE=ΔENLL+ΔELL,where SL(econ.) is the economic loading of the transformer (kVA), ∆ELL is the on-load energy loss of the transformer (kWh), ENLL is the no-load energy loss of the transformer (kWh), Sn is the nominal capacity of the transformer (kVA), ∆E represents overall average active energy losses (kWh), and SD is the designed capacity of the transformer (kVA).In the other sections of this work, the authors look out to optimize the loading of the transformer by finding ELDT, which leads to finding the value ofSL(econ.) in (1) in (kVA); this obtained value shall be less than the capacity of the transformer, which was determined during the design stage SD in (2) to avoid the overloading of the transformer (i.e., SD ≥ SL (econ.)). Usually, the designed capacity (kVA) of the transformer is not matched with the standard unit capacity of the transformers; therefore, the designer should select a transformer with a unit size capacity that is nearest to greater standard capacity (Sn ≥ SD). Consider the total energy losses (∆E) (kWh) in (3), taking into account the future expansion in operation (if any) and the increase in the demand on power as illustrated in numerical examples in Section 4.


## 3. Transformer Losses Calculation

The current flows through the transformer windings, causing a power loss even if the secondary winding of the transformer is open. However, when the transformer is loaded, an additional power loss occurs. The flow of the current in the transformer windings causes active and reactive power losses. The active and reactive power losses in the transformer are explained as follows.


### 3.1. Transformer Active Power Losses

The active power lossesΔP caused by the current flow in the transformer are of two types, which are the iron losses or the no-load losses ΔPFe and the copper losses or the on-load lossesΔPCu. The first active power losses are independent of the load current, which are called the iron losses or the no-load losses ΔPFe; they are hysteresis losses and eddy current losses. The second type of active power loss is due to the flow of the load current in the transformer’s windings, which is called the copper losses or load losses (ΔPCu). The total power losses ΔP then can be calculated from the sum of the power losses caused by the flow of active and reactive power which consist of two components. The first is the active power losses due to flow of active power ΔPp and the second is active power losses due to the flow of reactive power ΔPq, expressed as follows:(4)ΔP=ΔPp+ΔPq,where ΔP represents the total active power losses (kW). ΔPp represents the active power losses due to the flow of active power (kW). ΔPq represents the active power losses due to the flow of reactive power (kW).The active power losses of the transformer with respect to the transformer loading can be expressed as follows:(5)ΔPp=ΔPFe+ΔPCu⋅SLSn2,where ΔPCu represents active power on-load losses (kW), ΔPFe represents active power no-load losses (kW), and SL represents the loading of the transformer (kVA).The values ofΔPFe and ΔPCu are usually a part of the transformer nameplate information or the technical data sheet of the transformer provided by the manufacturers.In addition, the flow of the reactive power is causing reactive power lossesΔQ related to the no-load current ΔQoand the on-load current ΔQL, respectively. When the transformer is loaded with a power of SL of its nominal capacity Sn, then ΔQ with respect to the transformer loading can be presented as follows:(6)ΔQ=ΔQo+ΔQL⋅SLSn2,where ΔQ represents reactive power losses (kVAR), ΔQL represents reactive power on-load losses (kVAR), and ΔQo represents reactive power no-load losses (kVAR).ΔQo and ΔQL can be calculated from the two following formulas, respectively:(7)ΔQo=Io%100⋅Sn,(8)ΔQL=ΔVx%100⋅Sn,where Io% is the no-load current (%) and ΔVx% is the impedance voltage of the transformer (%).Typically the value of current (Io%) is about 2.20% to 8.5% for distribution transformer and can even vary between 0.1% and 10% of total load current in some special transformers [10]. The value of ∆Vx% usually varies between 3.5% and 6.5% and is up to 10% or more in some particular case transformers [19].The flow of the reactive power causes a loss in active power which can be represented as an equivalent coefficientCe. Thus the active power losses due to the flow of reactive power are expressed as follows:(9)ΔPq=Ce⋅ΔQ,where Ce is the equivalent coefficient converting the reactive power losses to active power losses (kW/kVAR).Substituting (5) and (6) into (4) with considering (9) yields the total active power losses that are shown in the following formula:(10)ΔP=ΔPFe+ΔPCu⋅SLSn2+Ce⋅ΔQo+ΔQL⋅SLSn2.The equivalent coefficientCe can be calculated from the expression in (11) according to [25] as follows:(11)Ce=∂∂QP2+Q2V2⋅RTr=2QVn2⋅RTr,where P is the active load power (kW), Q is the reactive load power (kVAR), and RTr is the transformer windings resistance (Ω).The transformer windings resistanceRTr according to [18] can be calculated as follows:(12)RTr=ΔPCu⋅Vn2Sn2.


### 3.2. Transformer Active Energy Losses

The product of the power losses (ΔP(t)) integration over a time (T) gives the energy losses (ΔE), and if the time-varying power losses ΔPt over a time (T) are arranged in descending order from ΔP(max) to ΔP(min) in equal integrated areas, then the active energy losses can be expressed as follows:(13)ΔE=∫0TΔPtdt=ΔPt⋅T=ΔPmax.⋅τ,where τ is the equivalent load losses hours (h/year), T represents the year’s hours = 8760 h, ΔPt represents the time-varying active power losses (kW), and ΔPmax is maximum value of active power losses (kW).Substituting (10) into (13) yields the following formula:(14)ΔE=ΔPFe+Ce⋅ΔQoT+ΔPCu+Ce⋅ΔQLSLSn2⋅τ.


### 3.3. The Transformer Power Economic Loading due to the Transferred Energy

If the time-varying loading power (P(t)) over a time (T) is arranged in descending order from P(max) to P(min) in equal integrated areas, then the active energy consumed over a time period (Teq) becomes(15)E=∫0TPtdt=ΔPt⋅T=Pmax.⋅Teq,where Teq represents equivalent working hours over a year (h).Based on the triangular power, the relation between the active power, apparent power of the transformer, and power factorcosφ is as follows:(16)Pmax=SLcosφ,where cos φ is the power factor. Substituting (15) into (16) yields the following formula:(17)E=SL⋅cosφ⋅Teq,where E is the transferred energy in one year (kWh).The relative energy losses for the transformed energy over the year (T = 8760 h) can be calculated by dividing transformer energy losses in (14) by the transformed energy in (17) which yields the following formula:(18)ΔEE=ΔPFe+Ce⋅ΔQo⋅TSL⋅cosφ⋅Teq+ΔPCu+Ce⋅ΔQL⋅SL/Sn2⋅τSL⋅cosφ⋅Teq.Equation (18) represents the relative active energy losses of the transformer. When performing some mathematical manipulation, the final relative active energy loss is(19)ΔEE=ΔPFe+Ce⋅ΔQo⋅TeqSL⋅cosφ⋅Teq+ΔPCu+Ce⋅ΔQL⋅SL⋅τSL2⋅cosφ⋅Teq⋅Sn2.The minimum function is obtained when differentiating (19) with respect to (SL) and equalizing the result to zero as in (20), resulting in (21).(20)dΔE/EdSL=0,(21)−ΔPFe+Ce·ΔQo·TeqSL2⋅cosφ⋅Teq+ΔPCu+Ce·ΔQL⋅τcosφTeq·Sn2=0.In this case,SL is the optimum loading that is equal to SL(econ.); accordingly, the transformer optimum loading is obtained by performing some mathematical arrangement as shown in the following equation:(22)SLecon.=SnΔPFe+Ce⋅ΔQo⋅TeqΔPCu+Ce⋅ΔQL⋅τ.The second derivative of (19) is checked and it is found that d2ΔE/E/dSl2>0, which means that the obtained result has a local minimum.If the solution of equation (22) equals the standard unit size of the transformer, then the achieved solution is regarded as an optimum; however, if it does not equal the standard unit size (which is most probably), then the optimum solution will be obtained by selecting the nearest greater standard unit size; hence, the obtained solution is the economic loading.In equation (22), the values of Teq and τ are still unknown. However, Teq is determined in the following clause (4) of this section.


### 3.4. Equivalent Working Hours Determination

Based on the formula published in [20–22] which determines the equivalent working hours per month Tw/meq (also called maximum load utilization time), the expression of Tw/meq is shown in the following equation: (23)Tw/meq.=2⋅Dw/y⋅ns3+3−ns3⋅ApawApw+2⋅365−Dw/y⋅ApawApw,(24)Teq=N⋅Tw/meq.The proof of equations (23) and (24) will be introduced in Appendix A at the end of this study. In equations (23) and (24), Apw is the sum of the consumed active energies (day and night) during the time of operation per month (kWh/N), Apaw is the consumed active energy after the time of operation per month (kWh/N), Dw/y is the number of working days over the year (excluding holidays, shutdowns, weekends, etc.), n(h/m) is the number of hours per month (i.e., year’s hours/12 months = 730 h), ns is the number of working shifts during the time of operation in one day, Teq represents the equivalent working hours (h), Tw/meq represents the equivalent working time over the month in hours (h/month), and N is The number of months (N = 1, 2, …, 12, where N = 12 for one year).The equivalent working hours (Teq) in (24) can be calculated by multiplying Tw/meq in (23) with the number of the months, which is normally one year (12 months).


### 3.5. Maximum Loss Time Determination

Based on (13), where the energy loss is determined, and when the time-varying power losses values (ΔP(t)) are arranged in descending order from ΔP(max) to ΔP(min) as depicted in Figure 1, the areas under the curve of the dotted and straight line are equal.Figure 1
Electrical energy losses behavior [22].Consider the relation between LSF and the year’s hours (T = 8760 h) to obtain τ as in the following equation:(25)τ=LSF⋅T,where LSF is load loss factor (unit less). When multiplying the LSF factor by the peak power losses (ΔP(max)) for a period (T), the overall average energy lost (∆E) was given in [23].Substituting (25) into (13) yields the following formula:(26)ΔE=ΔPmax⋅T⋅LSF.The relationship between LF and LSF is given in [24], where the classical computational formula is presented as follows:(27)LSF=1−k⋅LF2+k⋅LF.The value ofk in (27) could be any value between zero and one. For instance, k = 0.2 in the UK, 0.3 in the USA, and 0.33 in PL. In this paper and for the proceeding calculations, the value of k is assumed as 0.33.Many authors in many literature works are usingk = 0.30 or k = 0.33, which is appropriate to the urban power grid. In fact, the value of k in (27) might have different values subject to the loads’ profile curve. Hence, it is possible to obtain different mathematical equations by using the fitting curve techniques to represent the relation between LSF and LF for the analyzed period.LF in equation (27) is defined as the ratio of the average load (P(avg)) supplied during the designated period (T) to the maximum load (P(max)) occurring in the same period (T) [25]. Consequently, LF can be expressed as follows:(28)LF=PavgPmax=EPmax⋅T=Pmax⋅TeqPmax⋅T=TeqT,where Pavg is the average load power (kW).


## 3.1. Transformer Active Power Losses

The active power lossesΔP caused by the current flow in the transformer are of two types, which are the iron losses or the no-load losses ΔPFe and the copper losses or the on-load lossesΔPCu. The first active power losses are independent of the load current, which are called the iron losses or the no-load losses ΔPFe; they are hysteresis losses and eddy current losses. The second type of active power loss is due to the flow of the load current in the transformer’s windings, which is called the copper losses or load losses (ΔPCu). The total power losses ΔP then can be calculated from the sum of the power losses caused by the flow of active and reactive power which consist of two components. The first is the active power losses due to flow of active power ΔPp and the second is active power losses due to the flow of reactive power ΔPq, expressed as follows:(4)ΔP=ΔPp+ΔPq,where ΔP represents the total active power losses (kW). ΔPp represents the active power losses due to the flow of active power (kW). ΔPq represents the active power losses due to the flow of reactive power (kW).The active power losses of the transformer with respect to the transformer loading can be expressed as follows:(5)ΔPp=ΔPFe+ΔPCu⋅SLSn2,where ΔPCu represents active power on-load losses (kW), ΔPFe represents active power no-load losses (kW), and SL represents the loading of the transformer (kVA).The values ofΔPFe and ΔPCu are usually a part of the transformer nameplate information or the technical data sheet of the transformer provided by the manufacturers.In addition, the flow of the reactive power is causing reactive power lossesΔQ related to the no-load current ΔQoand the on-load current ΔQL, respectively. When the transformer is loaded with a power of SL of its nominal capacity Sn, then ΔQ with respect to the transformer loading can be presented as follows:(6)ΔQ=ΔQo+ΔQL⋅SLSn2,where ΔQ represents reactive power losses (kVAR), ΔQL represents reactive power on-load losses (kVAR), and ΔQo represents reactive power no-load losses (kVAR).ΔQo and ΔQL can be calculated from the two following formulas, respectively:(7)ΔQo=Io%100⋅Sn,(8)ΔQL=ΔVx%100⋅Sn,where Io% is the no-load current (%) and ΔVx% is the impedance voltage of the transformer (%).Typically the value of current (Io%) is about 2.20% to 8.5% for distribution transformer and can even vary between 0.1% and 10% of total load current in some special transformers [10]. The value of ∆Vx% usually varies between 3.5% and 6.5% and is up to 10% or more in some particular case transformers [19].The flow of the reactive power causes a loss in active power which can be represented as an equivalent coefficientCe. Thus the active power losses due to the flow of reactive power are expressed as follows:(9)ΔPq=Ce⋅ΔQ,where Ce is the equivalent coefficient converting the reactive power losses to active power losses (kW/kVAR).Substituting (5) and (6) into (4) with considering (9) yields the total active power losses that are shown in the following formula:(10)ΔP=ΔPFe+ΔPCu⋅SLSn2+Ce⋅ΔQo+ΔQL⋅SLSn2.The equivalent coefficientCe can be calculated from the expression in (11) according to [25] as follows:(11)Ce=∂∂QP2+Q2V2⋅RTr=2QVn2⋅RTr,where P is the active load power (kW), Q is the reactive load power (kVAR), and RTr is the transformer windings resistance (Ω).The transformer windings resistanceRTr according to [18] can be calculated as follows:(12)RTr=ΔPCu⋅Vn2Sn2.


## 3.2. Transformer Active Energy Losses

The product of the power losses (ΔP(t)) integration over a time (T) gives the energy losses (ΔE), and if the time-varying power losses ΔPt over a time (T) are arranged in descending order from ΔP(max) to ΔP(min) in equal integrated areas, then the active energy losses can be expressed as follows:(13)ΔE=∫0TΔPtdt=ΔPt⋅T=ΔPmax.⋅τ,where τ is the equivalent load losses hours (h/year), T represents the year’s hours = 8760 h, ΔPt represents the time-varying active power losses (kW), and ΔPmax is maximum value of active power losses (kW).Substituting (10) into (13) yields the following formula:(14)ΔE=ΔPFe+Ce⋅ΔQoT+ΔPCu+Ce⋅ΔQLSLSn2⋅τ.


## 3.3. The Transformer Power Economic Loading due to the Transferred Energy

If the time-varying loading power (P(t)) over a time (T) is arranged in descending order from P(max) to P(min) in equal integrated areas, then the active energy consumed over a time period (Teq) becomes(15)E=∫0TPtdt=ΔPt⋅T=Pmax.⋅Teq,where Teq represents equivalent working hours over a year (h).Based on the triangular power, the relation between the active power, apparent power of the transformer, and power factorcosφ is as follows:(16)Pmax=SLcosφ,where cos φ is the power factor. Substituting (15) into (16) yields the following formula:(17)E=SL⋅cosφ⋅Teq,where E is the transferred energy in one year (kWh).The relative energy losses for the transformed energy over the year (T = 8760 h) can be calculated by dividing transformer energy losses in (14) by the transformed energy in (17) which yields the following formula:(18)ΔEE=ΔPFe+Ce⋅ΔQo⋅TSL⋅cosφ⋅Teq+ΔPCu+Ce⋅ΔQL⋅SL/Sn2⋅τSL⋅cosφ⋅Teq.Equation (18) represents the relative active energy losses of the transformer. When performing some mathematical manipulation, the final relative active energy loss is(19)ΔEE=ΔPFe+Ce⋅ΔQo⋅TeqSL⋅cosφ⋅Teq+ΔPCu+Ce⋅ΔQL⋅SL⋅τSL2⋅cosφ⋅Teq⋅Sn2.The minimum function is obtained when differentiating (19) with respect to (SL) and equalizing the result to zero as in (20), resulting in (21).(20)dΔE/EdSL=0,(21)−ΔPFe+Ce·ΔQo·TeqSL2⋅cosφ⋅Teq+ΔPCu+Ce·ΔQL⋅τcosφTeq·Sn2=0.In this case,SL is the optimum loading that is equal to SL(econ.); accordingly, the transformer optimum loading is obtained by performing some mathematical arrangement as shown in the following equation:(22)SLecon.=SnΔPFe+Ce⋅ΔQo⋅TeqΔPCu+Ce⋅ΔQL⋅τ.The second derivative of (19) is checked and it is found that d2ΔE/E/dSl2>0, which means that the obtained result has a local minimum.If the solution of equation (22) equals the standard unit size of the transformer, then the achieved solution is regarded as an optimum; however, if it does not equal the standard unit size (which is most probably), then the optimum solution will be obtained by selecting the nearest greater standard unit size; hence, the obtained solution is the economic loading.In equation (22), the values of Teq and τ are still unknown. However, Teq is determined in the following clause (4) of this section.


## 3.4. Equivalent Working Hours Determination

Based on the formula published in [20–22] which determines the equivalent working hours per month Tw/meq (also called maximum load utilization time), the expression of Tw/meq is shown in the following equation: (23)Tw/meq.=2⋅Dw/y⋅ns3+3−ns3⋅ApawApw+2⋅365−Dw/y⋅ApawApw,(24)Teq=N⋅Tw/meq.The proof of equations (23) and (24) will be introduced in Appendix A at the end of this study. In equations (23) and (24), Apw is the sum of the consumed active energies (day and night) during the time of operation per month (kWh/N), Apaw is the consumed active energy after the time of operation per month (kWh/N), Dw/y is the number of working days over the year (excluding holidays, shutdowns, weekends, etc.), n(h/m) is the number of hours per month (i.e., year’s hours/12 months = 730 h), ns is the number of working shifts during the time of operation in one day, Teq represents the equivalent working hours (h), Tw/meq represents the equivalent working time over the month in hours (h/month), and N is The number of months (N = 1, 2, …, 12, where N = 12 for one year).The equivalent working hours (Teq) in (24) can be calculated by multiplying Tw/meq in (23) with the number of the months, which is normally one year (12 months).


## 3.5. Maximum Loss Time Determination

Based on (13), where the energy loss is determined, and when the time-varying power losses values (ΔP(t)) are arranged in descending order from ΔP(max) to ΔP(min) as depicted in Figure 1, the areas under the curve of the dotted and straight line are equal.Figure 1
Electrical energy losses behavior [22].Consider the relation between LSF and the year’s hours (T = 8760 h) to obtain τ as in the following equation:(25)τ=LSF⋅T,where LSF is load loss factor (unit less). When multiplying the LSF factor by the peak power losses (ΔP(max)) for a period (T), the overall average energy lost (∆E) was given in [23].Substituting (25) into (13) yields the following formula:(26)ΔE=ΔPmax⋅T⋅LSF.The relationship between LF and LSF is given in [24], where the classical computational formula is presented as follows:(27)LSF=1−k⋅LF2+k⋅LF.The value ofk in (27) could be any value between zero and one. For instance, k = 0.2 in the UK, 0.3 in the USA, and 0.33 in PL. In this paper and for the proceeding calculations, the value of k is assumed as 0.33.Many authors in many literature works are usingk = 0.30 or k = 0.33, which is appropriate to the urban power grid. In fact, the value of k in (27) might have different values subject to the loads’ profile curve. Hence, it is possible to obtain different mathematical equations by using the fitting curve techniques to represent the relation between LSF and LF for the analyzed period.LF in equation (27) is defined as the ratio of the average load (P(avg)) supplied during the designated period (T) to the maximum load (P(max)) occurring in the same period (T) [25]. Consequently, LF can be expressed as follows:(28)LF=PavgPmax=EPmax⋅T=Pmax⋅TeqPmax⋅T=TeqT,where Pavg is the average load power (kW).


## 4. Numerical Example

The proposed method is implemented in a numerical example, where the calculations are as follows.


### 4.1. Load Input Data

It is intended to build an industrial plant with an expected maximum power of 584 kW to supply the processes requirement and associated buildings, with an expected power factor equaling 0.80. The load is expected to increase in the future by 20%.The plant could initially work five days a week in one shift or two shifts or work continuously in three shifts and one shift off.To find the ELDT, the calculation in this example is conducted to one shift of work: 8 hours a day, five days a week, excluding holidays. To maintain the process outside the working hours’ time some of equipment and lighting remain ON (essential load). The amount of energy required to maintain the process outside the time of operation is equal to 20% of the energy consumed during the normal time of operation.However, at the end of the example, the results of two- and three-shift modes of work are presented.Solution:Calculate:(i)
Teq represents the equivalent working hours (h).The consumed energy (E) over the year is estimated by considering the essential load to maintain the plant in the running state after the working hours. It is calculated using the following formula:Apaw/Apw=20%.Teq is calculated for one shift (8 hours a day, five days a week, excluding holidays, etc.), as in (23).(29)Tw/meq.=2×24413+3−13×20100+2365−244×20100=278.6h/m.And, based on equation (24),(30)Teq=12×278.6=3343.2h.(ii)
E is the active energy consumed over the year.From equation (15), E is obtained:(31)E=584×3343.2=1952.43MWh/yr.(iii)
SL is the apparent power of the load.From equation (16), we obtain(32)SL=Pmaxcosφ=5840.8=730kVA.(iv)
LF is the load factor.Equation (28) is used to calculate the load factor:(33)LF=EPmax⋅T=1952.43MWh584kW×8760h=0.382.(v)
LSF is load loss factor.From equation (27), LSF can be calculated (for k = 0.333) as(34)LSF=1−0.333×0.3822+0.333×0.382=0.224086h.(vi)
τ represents the equivalent load losses’ hours (h/year).By using equation (25), we obtain(35)τ=0.224086×8760=1963h.(v)
Q is reactive power load.From the triangular power, for the relation between active, reactive, and apparent power, the reactive power equals(36)Q=SLTr.⋅sinφ=SLTr⋅sincosφ−1=730×0.6=438.51kVAR.(vi)
Sn is the nominal capacity of the transformer (kVA).Considering the constraints in (2), the selection of the transformer capacity shall be equal to or more than the designed value; also, taking into account the future expansion in the load by 20%, the capacity of the transformer (Sn) is(37)Sn=730×1.2=876kVA.Round to the nearest standard available size, select a transformer with a capacity equal toSn = 1000 kVA.Based on (24), (25), (27), and (28), the load data for the one-shift mode of work is calculated. The result of the calculations is summarized in Table 1.Table 1
Load character data (calculated for one shift).
E (MWh/yr)1952.43SL (kVA)730.0Q (kVAR)438.52LF (/)0.382Teq (h/yr)3343τ (h)1963


### 4.2. Transformers Technical Data

The technical data for the transformer (data belonging to transformer type (THUE23/1000) manufactured by ALSTHOM) which is provided in the manufacturer’s technical data sheet and the calculated data are given in Table2.Table 2
Transformer 1000 kVA technical data1.
Given dataCalculated dataItemValueItemValueSn (kVA)1000∆Qo (kVAR)21.20Vn (kV)6.6/0.4∆QL (kVAR)50.00∆PCu (kW)12.1RTr (Ω)0.527∆PFe (kW)2.3Ce kW/kVAR0.1061Io (%)2.12——∆Vsc (%)4——1Data belonging to transformer type (THUE23/1000) manufactured by ALSTHOM.The given data of the transformer and the calculated data are summarized in Table2, where ΔQo and ∆QL are calculated based on (7) and (8), respectively, as follows:(38)ΔQo=2.12100×1000=21.20kVAR,ΔQL=4100×1000=50.00kVAR.Moreover,RTr and Ce can be calculated based on (11) and (12), respectively, as follows:(39)RTr=12.1×6.6210002×1000=0.527Ω,Ce=2×438.526.62×1000×0.527=0.1061kW/kVAR.


### 4.3. Transformer Power Economic Loading

Based on the given data, the ELDT can be calculated from equation (22) for one working condition for one shift.(40)SLecon.=1000×2.3+0.1061×21.2×334312.1+0.1061×50×1963=667.21kVA.However, the ELDT for two shifts and that for three shifts are similar; the difference is only in calculating the time ofTeq and τ. The results for the work in shifts system are presented in Table 3.Table 3
Transformer economic loading and working shifts.
Shift numberOne shiftTwo shiftsThree shiftsSLecon (kVA)667.21607.81552.76However, when neglecting the effect ofTeq time and τ in (22), the ELDT reduces to 511 [kVA] for this transformer.So, when selecting a transformer with less on-load losses by 25%, 50%, and 78%, where the no-load losses remain constant, the ELDT can be, respectively, increased to 72%, 82%, and 100%.A lot of exercises can be performed by the recently devolved software.As a future application of the proposed approach, it can be used for future smart cities where high-performance energy systems in cooperation with many other technologies must be accessible, for example, IT infrastructure [26–28], wireless sensor systems [29], intelligent transport systems [30–32], industrial developments and cyber-physical systems [33–38], and monitoring systems [39–42].


## 4.1. Load Input Data

It is intended to build an industrial plant with an expected maximum power of 584 kW to supply the processes requirement and associated buildings, with an expected power factor equaling 0.80. The load is expected to increase in the future by 20%.The plant could initially work five days a week in one shift or two shifts or work continuously in three shifts and one shift off.To find the ELDT, the calculation in this example is conducted to one shift of work: 8 hours a day, five days a week, excluding holidays. To maintain the process outside the working hours’ time some of equipment and lighting remain ON (essential load). The amount of energy required to maintain the process outside the time of operation is equal to 20% of the energy consumed during the normal time of operation.However, at the end of the example, the results of two- and three-shift modes of work are presented.Solution:Calculate:(i)
Teq represents the equivalent working hours (h).The consumed energy (E) over the year is estimated by considering the essential load to maintain the plant in the running state after the working hours. It is calculated using the following formula:Apaw/Apw=20%.Teq is calculated for one shift (8 hours a day, five days a week, excluding holidays, etc.), as in (23).(29)Tw/meq.=2×24413+3−13×20100+2365−244×20100=278.6h/m.And, based on equation (24),(30)Teq=12×278.6=3343.2h.(ii)
E is the active energy consumed over the year.From equation (15), E is obtained:(31)E=584×3343.2=1952.43MWh/yr.(iii)
SL is the apparent power of the load.From equation (16), we obtain(32)SL=Pmaxcosφ=5840.8=730kVA.(iv)
LF is the load factor.Equation (28) is used to calculate the load factor:(33)LF=EPmax⋅T=1952.43MWh584kW×8760h=0.382.(v)
LSF is load loss factor.From equation (27), LSF can be calculated (for k = 0.333) as(34)LSF=1−0.333×0.3822+0.333×0.382=0.224086h.(vi)
τ represents the equivalent load losses’ hours (h/year).By using equation (25), we obtain(35)τ=0.224086×8760=1963h.(v)
Q is reactive power load.From the triangular power, for the relation between active, reactive, and apparent power, the reactive power equals(36)Q=SLTr.⋅sinφ=SLTr⋅sincosφ−1=730×0.6=438.51kVAR.(vi)
Sn is the nominal capacity of the transformer (kVA).Considering the constraints in (2), the selection of the transformer capacity shall be equal to or more than the designed value; also, taking into account the future expansion in the load by 20%, the capacity of the transformer (Sn) is(37)Sn=730×1.2=876kVA.Round to the nearest standard available size, select a transformer with a capacity equal toSn = 1000 kVA.Based on (24), (25), (27), and (28), the load data for the one-shift mode of work is calculated. The result of the calculations is summarized in Table 1.Table 1
Load character data (calculated for one shift).
E (MWh/yr)1952.43SL (kVA)730.0Q (kVAR)438.52LF (/)0.382Teq (h/yr)3343τ (h)1963


## 4.2. Transformers Technical Data

The technical data for the transformer (data belonging to transformer type (THUE23/1000) manufactured by ALSTHOM) which is provided in the manufacturer’s technical data sheet and the calculated data are given in Table2.Table 2
Transformer 1000 kVA technical data1.
Given dataCalculated dataItemValueItemValueSn (kVA)1000∆Qo (kVAR)21.20Vn (kV)6.6/0.4∆QL (kVAR)50.00∆PCu (kW)12.1RTr (Ω)0.527∆PFe (kW)2.3Ce kW/kVAR0.1061Io (%)2.12——∆Vsc (%)4——1Data belonging to transformer type (THUE23/1000) manufactured by ALSTHOM.The given data of the transformer and the calculated data are summarized in Table2, where ΔQo and ∆QL are calculated based on (7) and (8), respectively, as follows:(38)ΔQo=2.12100×1000=21.20kVAR,ΔQL=4100×1000=50.00kVAR.Moreover,RTr and Ce can be calculated based on (11) and (12), respectively, as follows:(39)RTr=12.1×6.6210002×1000=0.527Ω,Ce=2×438.526.62×1000×0.527=0.1061kW/kVAR.


## 4.3. Transformer Power Economic Loading

Based on the given data, the ELDT can be calculated from equation (22) for one working condition for one shift.(40)SLecon.=1000×2.3+0.1061×21.2×334312.1+0.1061×50×1963=667.21kVA.However, the ELDT for two shifts and that for three shifts are similar; the difference is only in calculating the time ofTeq and τ. The results for the work in shifts system are presented in Table 3.Table 3
Transformer economic loading and working shifts.
Shift numberOne shiftTwo shiftsThree shiftsSLecon (kVA)667.21607.81552.76However, when neglecting the effect ofTeq time and τ in (22), the ELDT reduces to 511 [kVA] for this transformer.So, when selecting a transformer with less on-load losses by 25%, 50%, and 78%, where the no-load losses remain constant, the ELDT can be, respectively, increased to 72%, 82%, and 100%.A lot of exercises can be performed by the recently devolved software.As a future application of the proposed approach, it can be used for future smart cities where high-performance energy systems in cooperation with many other technologies must be accessible, for example, IT infrastructure [26–28], wireless sensor systems [29], intelligent transport systems [30–32], industrial developments and cyber-physical systems [33–38], and monitoring systems [39–42].


## 5. Results, Conclusions, and Future Work

This paper presents a detailed approach for studying and calculating the ELDT. Different schedules of working hours for different transformers input technical data were taken into account. The evaluation shows that the ELDT is accruing when the objective is to minimize the active energy losses rather than active power losses, which was justified by the product of the equivalent working hours to the equivalent load losses, which is always more than or equal to one; the results of calculations are presented in Table3.The submitted formula for the ELDT shows the effect of the on-load losses (the copper losses) and the no-load losses (the core losses) on the result of the transformer economic loading. The optimality is achieved when the load losses are equal to no-load losses; however, this solution is very costly and sometimes practically is impossible; hence, a compromise is needed.The explained numerical examples show the impact of the technical parameters (in terms of load and no-load losses) and operational time on the ELDT, where the ELDT could reach a value equal to or less than the transformer rated capacity. The more the relative value of the on-load losses to the no-load losses is reduced, the more enhancement of the ELDT is achieved, the utilization end is enhanced, and the operational cost of the transformer is reduced.In addition, the software has been devolved to find the ELDT faster and easier, which is very beneficial, especially when having two or more transformers with different capacity and technical data working in parallel to feed a variable load.This paper, along with the numerical examples, could be a base (platform) for other researches or future work, such as the effect of energy loss cost in terms of money on ELDT and the effect of load changing and power factor on the ELDT.



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*Source: 8081212-2021-12-15.xml*