By A, GRADENER When attempting to achieve the most economical
conditions of power supply in a sugar factory, it is not only the power generation part of the system which requires attention, but also the components of the distribution network. The electrical supply system may be said to consist of the following components:
Generators, Main switchgear, Transformers,
Motor-control-centres, Drives and cable network.
The total power requirement of consumers in a factory determines the choice of generation and distribution voltages. When, for reasons to be dealt with in the latter part of this paper, high tension is chosen, the location of the power house in relation to the factory and also the positions of transformer stations and low tension main switchgear, have a considerable influence on the final layout of the system.
The transformer stations may be arranged either:
a) —at load centres as separate stations where two or three units are installed together with their LT- switchboard.
b) —at a central location where all units and the LT-main switchboard are installed.
For operation of the factory on an LT supply system, case b) would apply.
Let us first discuss case a), where the stations are placed in centres of power. It depends on the size of the transformers whether their LT sides should be connected in parallel or to separate busbars. In view of rupturing capacity considerations, more than 2,500 to 3,000 KVA should not be connected in parallel. Coupling facilities are, however, advisable, so that the power supply of any section, where a transformer fails, may be maintained. This however, necessitates that transformers of load centres are dimensioned accordingly. If, for instance, three transformers are installed, any two of them must be capable of carrying the total load connected to the three. If there is a possibility of temporarily discon- necting non-essential consumers however, the trans- formers need not be so amply dimensioned. If it is preferred to minimise the size of transformers another method to ensure constant supply, even in the event of a transformer failing, may be employed. This involves an LT-ring cable system with isolators in the stations, where all switchboards of power centres are interconnected. It depends on the distances between stations and the power to be transferred, which solution is superior and more economical. An HT ring system of course is also possible. This however, necessitates HT switchgear in each load centre and consequently represents the more expensive way.
Normally, transformers of load centres are radially
fed by separate cables, which are run from the HT switchboard of the power house.
In case b), where all transformers are located in one station, a costly ring main is unnecessary as the LT system has common busbars. In order to avoid heavy short circuit currents, these busbais .must be divided, into sections, with the possibility of coupling in the event of a failure of any transformer.
Whether case a), or case b), is chosen depends on the physical layout of the factory. The primary aim is to have minimum lengths of cables between the LT main switchboard and the motor control centres.
This is of the utmost importance as losses incurred in an LT cable system are a multiple of those of an HT system. The reason for this is simple to explain as losses go into account by the square of the current but only linear to the resistance—i.e. of the cross section—of cables. For this reason, electrical losses being economical losses, HT should be brought as closely as possible to consumers. If the power house is not situated close to the factory, it would be very uneconomical to have the transformers there, neces- sitating long runs of LT cables of considerable sizes to motor control centres. In this case the power centre system as described under a), lends itself.
If the power house is located in the centre of the factory, the centralized transformer station as des- cribed in b), is mostly the advisable solution.
Considerations of power are decisive for the design of electrical equipment and plant. It is not only the power requirement that is of interest, but also the power factor involved. This defines the KVA rating to which the electrical plant must be designed.
In Fig. 1, the conditions which prevail when electri- city flows are shown. There is hardly an electric consumer which does not possess resistance, induc- tance and capacitance simultaneously and it is only
162
Proceedings of The South African Sugar Technologists' Association—March 1966 163
the predominance of one or the other which defines the final behaviour.
The electric current will either lag or lead the voltage by an angle / between 0 and TT/2. This can be illus- trated either by drawing the sine curves of voltage and current in relationship to one another or by drawing a vector diagram which shows voltage and current vectors. The two illustrations of Fig. I, both show the same conditions, i.e. the voltage V and the current I at an angle / to one another.
The lengths of the vectors are in direct proportion to the r.m.s. values. It may now be imagined, that the actual current (I) is made up of two components, one of which is "in phase", with the voltage (active current), while the other is out of phase by 90° (reactive current). As, according to the rules of geometry:
the active current la -.- Ix cos / a n d the reactive current Ir -••-: ix sin / the
resulting power may be arrived at by multiplying both sides of the equations by the voltage across phases (three phase system) and %/J"
active power Pa --•• V >; v T > ' la ": V ;•: •v/J'x 1 :-: c o s /
reactive power Pr V .: vT • If V vT : 1 x sin f
The maximum electrical active power of a turboset is determined by the mechanical output of the turbine.
The required maximum KVA rating of the alternator results from the turbine output and the power factor at which the alternator supplies this power. The con- sumers of a factory draw both reactive and active power. Each consumer has a certain power factor which indicates the ratio of Watts/VA.
The mean value of power factors of all consumers is the power factor at which the alternator must supply power. The power factors of individual motors (being the main consumers in a sugar factory), depend on their HP rating, their speed and their loading. Tacle I shows the influence of these factors.
TABLE I
Influence of H.P. rating, speed and loading on the power factor of an electrical motor
Influence of rating:
(all at same speed)
Power 10 kW 100 kW 1,000 kW p.f. 0.85 0.88 0.89 Speed influence:
Speed 500 rpm 750 ipm l.OOOrpm 1,500
(at same HP) rpm pf 0.80 0.86 0.87 0.88
Loading influence:
(at same
HP) 4/4 3/4 1/2 Loading Loading Loading at 500 rpm p.f. 0.80 0.76 0.65 at 1500 rpm p.f. 0.88 0.85 0.77
As may be observed, the influence of the rating is not great. This also applies to the speed, not taking very slow speeds into account. The loading however, influences the power factor considerably. Conse- quently overdesigned motors, which in effect are not properly loaded, reduce the power factor to a con- siderable extent. It is always good practice to design equipment with a margin of safety but this does not apply to the ratings of electrical motors where an overdimensioning results in a poor power factor.
In turn the reactive current required must be generated in expensive machinery. Apart from that, excessive HP adds to the cost of the motors involved.
As a rule, the power factor of a sugar factory is about 0.7, if power factor correction capacitors are not installed. It is therefore advisable to choose a KVA rating of the alternator, which is large enough to allow full utilization of the given power of the turbine. (It must be born in mind, that alternators are normally designed for a power factor of 0.8). In order to demonstrate the foregoing clearly, conditions of a 3,000 kW alternator are investigated in the following and demonstrated in Fig. 2.
Power Limitation Diagram of Synchronous Alternators
The kW rating is divided by 0.8, to arrive at the KVA rating of the alternator. Therefore, this rating is ' = 3,750 KVA. It is represented by the line
0.8
A — C, of the diagram in Fig. 2.
When ascertaining the power which the alternator is capable of supplying at a power factor of 0.7, it must be realised that the permissible power, in KVA, of the alternator operating between a power factor of 0.8 to 0, decreases along the curve C-E-F. The reason for this is that an increase of reactive power requires additional excitation. The latter mentioned however, has its limits as the permissible temperature of the alternator must not be surpassed. At a power factor of 0.7 the KVA would be represented by the line A-E, corresponding to 3,500 KVA, i.e. to 93 % of the original rating. As however, at this poor power The lengths of the vectors are in direct proportion
to the r.m.s. values. It may now be imagined, that the actual current (I) is made up of two components, one of which is "in phase", with the voltage (active current), while the other is out of phase by 90° (reactive current). As, according to the rules of geometry:
The maximum electrical active power of a turboset is determined by the mechanical output of the turbine.
The required maximum KVA rating of the alternator results from the turbine output and the power factor at which the alternator supplies this power. The con- sumers of a factory draw both reactive and active power. Each consumer has a certain power factor which indicates the ratio of Watts/VA.
The mean value of power factors of all consumers is the power factor at which the alternator must supply power. The power factors of individual motors (being the main consumers in a sugar factory), depend on their HP rating, their speed and their loading. Tacle I shows the influence of these factors.
TABLE I
Influence of H.P. rating, speed and loading on the power factor of an electrical motor
Influence of rating:
(all at same speed) Power 10 kW p.f. 0.85
100 kW 0.88 Speed influence:
Speed 500 rpm 750 ipm (at same HP)
pf 0.80 Loading influence:
(at same HP)
at 500 rpm p.f.
at 1500 rpm p.f.
0.86
1,000 kW 0.89
1,0001 0.87 4/4 3/4 Loading Loading
0.80 0.88
0.76 0.85
rpm 1,500 rpm 0.88
1/2 Loading
0.65 0.77
164 Proceedings of The South African Sugar Technologists'' Association- -March 1966
factor the reactive power has been increased, as per line G-E, the active power available is also reduced to 2,400 kW (line A-G), i.e. to 80% of its original.
Consequently, the turbine is not being fully utilized under these conditions as part of the available power is idle.
In order to comply with the required conditions a larger alternator must be chosen, say 4,500 KVA at a power factor of 0.8, corresponding to the line A - C At a power factor of 0.7 this alternator will only be capable of supplying 4,230 KVA, corresponding to line A-E'. If full turbine power of 3,000 kW is utilized at a power factor of 0.7, 4,280 KVA would result, being represented by the line A-E." As the alternator is not quite large enough to cope, the only other possibilities which remain, other than employing a larger alternator, are to utilize either the full power of 3,000 kW but at a power factor of 0.71 or to operate at only 2,940 kW at a power factor of 0.7.
As may be seen from the foregoing, the alternator KVA must be adapted to the power factor as given by the consumers, to ensure that it can generate the active power corresponding to the turbine power and at the same time supply the reactive power required.
It is of course possible to load the alternator with a greater active power if the reactive power is generated by another source, such as by capacitors. In the following, a price comparison is given, in order to ascertain which solution is more economical. Fig. 3 compares the situation between the alternators as mentioned before, both working at 400 Volts.
In example 1, a 4,230 KVA alternator works at a power factor of 0.71 and. in example 2, a 3,750 KVA alternator works at a power factor of 0.8. In both cases the load comprises 7 consumers each of which, for the sake of simplicity, is 605 KVA. In the 2nd example however, the reactive power is compensated by 7 capacitors, one of 100 KVA r, at each consumer.
In this case therefore, the power which the alternator must supply, is reduced to 3,750 KVA and conse- quently a smaller machine is required. In both cases, the alternator is connected to the main switchboard by means of a busbar with a length of 60 feet. The lengths of cables from this switchboard to the con- sumers is 300 feet. The required cross section is smaller in. example No. 2, as the reactive power is generated at the consumers and therefore need not pass through the conductors. The busbars of the switchboard and the circuit breakers have been neglected, as in both cases a 6,000 Amp alternator breaker and 1,000 Amp circuit breakers are required for the feeders. If the price of the smaller alternator is represented by 100%, the following result is obtained:
The price of the larger alternator is 106%. The busbars in example I, are 14% and in example 2, 11 %.
The cable price is 36 % in example 1 and in example 2, 28%. In the 2nd example an additional 17% is required for capacitors and power factor control equipment. The total price for example 1 is then 156%
and for example 2, also 156 %. Both solutions are equal in price and it is adviseable to chose the techni- cally simpler method, i.e. to use a larger alternator.
As a rule, it is more economical, in the case of low power, to generate the reactive power in the alter- nator. This is clearly shown in examples 3 and 4.
Both alternators of examples 4 and 5 generate 1,000 kW, the one of example 4, being for 1,250 KVA at a power factor of 0.8 and the one of example 3, being for 1,490 KVA, at a power factor of 0.68. Again, 7 equal consumers are supplied, each of 210 KVA. In example 4, the reactive power is compensated in 7 capacitors of 50 KVA r each.
If the price of the smaller alternator is represented by 100% the following result is obtained:
The price of the larger alternator is 113 %, the prices of the busbars are 10% and 12%, while the cable prices are 12% and 15%,. However, the price of the power factor equipment is 35% in example 4, as compared with 17% in example 2. The conclusion is, that the plant with the larger alternator, totals up to 140% while the plant with the smaller alternator in combination with the power factor connection equipment, totals 157%. The latter mentioned is therefore 12% more expensive than the former.
Generally speaking therefore, it is more economical and technically simpler to have the alternators laid out for a power factor of 0.7. Wherever new alter- nators are required, this should be taken into account.
With existing generating plant, the only solution is to install capacitors which should be connected very closely to consumers, in order to relieve circuit breakers and cables. Motors of about 50 kW and
Proceedings of The South African Sugar Technologists' Association —
more, would be individually compensated, especially when continuously in operation. Smaller motors however, would be compensated in groups or all together. In both cases however, it is essential to control automatically the switching of capacitors in order to adapt the capacitances to the reactive power requirements and thereby avoid over-compensation.
Over-compensation bears the danger of self excitation and voltage increases which in turn endanger the insulation of electrical equipment and machinery.
With individual compensation of motors, the capacitor is generally laid out to compensate about 90 % of the no-load reactive power. A power factor of approximately 0.9 at full load and 0.95 at partial load is thereby achieved. The capacitor is connected in parallel to the motor terminals and switched on and off together with the motor. With larger capacitors damping reactances or resistances are required to limit the current resp. the steepness of the current increase.
The question whether or not to generate power at high, or low voltage is of great importance for the electrical supply system. The limit to which 400 Volt alternators can be manufactured is 5,000 K.VA. The rated current of this alternator is 7,250 amps for which no circuit breakers are available, the maximum size of low voltage switchgear being 6,000 amps.
It would of course be possible to cope with 12,000 amps by connecting two circuit breakers in parallel but the rupturing capacity cannot be doubled by this measure. In the case of a fault, one of the circuit breakers will always open a fraction of a second earlier than the other, leaving only one breaker to deal with the full rupturing power. The rupturing capacity, as a matter of fact, is the primary consider- ation with regard to the choice of high or low voltage for a system and before going on, a few basic details on this subject are given.
The severest conditions prevail in the case of a 3- phase short circuit with the alternator fully excited, if occurring at the moment of the voltage passing through zero. These conditions are illustrated in Fig. 4, where the behaviour of a short circuit current over a period of time is shown.
— March 1966 165
The short circuit current reaches its peak after the first half wave, i.e. at 50 cycles, a period of 10 milli- seconds (Length b, in Fig. 4). From cycle to cycle the amplitudes become smaller and the unsymmetrical behaviour becomes more and more symmetrical.
Within about 250 milliseconds the short circuit current oscillates symmetrically. A steady state, the
"continuous short circuit current", is attained after a number of seconds. As may be observed from Fig. 4, a fading DC component occurs in the first instance of a short circuit and displaces the amplitudes of the AC current asymmetrically. The AC component, as indicated with broken lines, attains its peak at the moment when the short circuit is initiated (length a).
The amplitude of the AC component decreases rapidly at first and then decreases slower until it finally attains a constant value. This decrease of current is caused by the armature reaction of the short circuit current which weakens the alternator field and therefore reduces the voltage at the terminals of the alternator.
It must be realized that the peak value of the short circuit current is responsible for the dynamic stress of electrical equipment. The dynamic power developed is in direct relationship to the square of the current.
As the time constant of circuit breakers is generally between iOO and 250 milliseconds and the peak value of the short circuit current is attained within 10 milliseconds, there is practically no possibility of avoiding its full dynamic effect on electrical equipment in circuit, other than by using HRC fuses. These have the ability to interrupt the circuit before the peak value is reached (this solution is however, only possible with Amperages within values of fuses on the market). The peak value of the asymmetrical short circuit current is therefore decisive foi the con- struction of switchgear, busbars, current trans- formers, etc. and consequently for the plant costs involved.
As may be seen from the foregoing, it is not the asymmetrical peak short circuit value, but the current value, at the moment of interruption, which defines the layout of circuit breakers. As the DC component has faded away within 100 to 250 milli-seconds, i.e. 10 to 25 half cycles, only the AC component need be taken into account. The latter however, has also decreased within this period and the greater the time constant of the circuit breaker, the smaller the current is. Consequently the actual current at the time of separation, of contacts generally is smaller than the maximum symmetrical short circuit current.
It can be calculated by multiplying the maximum symmetrical value, with a correction factor. The circuit breaker is then laid out to meet this value. By multiplying the current which prevails at the moment of separation of contacts, with the voltage across phases and \ / 3 , the rupturing capacity is arrived at.
This is only a theoretical figure however, as the two values do not occur at the same time.
In the same way as the actual current at the time of switching may be calculated, by multiplying the maximum symmetrical short circuit current by a correction factor (which is dependent on the conditions
Proceedings of The South African Sugar Technologists' Association — March 1966 165