Abstract: The selection and calculation of rectifier transformers are technical problems that generator users often encounter. Due to the special environmental conditions of small hydropower stations, the results of classical theoretical calculation formulas are often unrealistic. This article makes some discussion on this and proposes a set of concise calculation methods. The discussion in this paper is based on the following conditions: 400V low-voltage synchronous generator set, grid-connected operation, three-phase thyristor rectification, and grass-roots user use (customers provide ordering data to the manufacturer, involving no parameter design for transformer manufacturing)
Keywords: dry-rectifier transformer, thyristor rectification, excitation, synchronous generator, small hydropower
1. Overview In the work of selection and maintenance of small hydropower excitation equipment, users of power stations often encounter problems in the calculation of rectifier transformer parameters. Many electrician design manuals provide the design formulas of rectifier transformers. However, these formulae apply to the standard application conditions and are different from the actual operating environment of small hydropower. The transformers designed accordingly may not be practical. At the same time, there is also a shortage of professional technicians at the grassroots level of small hydropower stations. Users often find it difficult to select and calculate rectifier transformers. Therefore, it is very necessary to propose a simple calculation method for grassroots users.
1.1 Selection of rectification mode: At present, low-voltage units are basically self-excited static thyristor excitation. The rectification method generally has three-phase full-wave semi-controlled rectification and three-phase half-wave rectification (Figure 1). Full-wave rectification of the transformer is relatively high efficiency (95%), the waveform is better. Less half-wave rectifier silicon components, but the transformer secondary winding DC current through, the efficiency is relatively low (74%), waveform distortion, used in less than 10kW rectifier circuit, but some of the larger units of the early design is also a half-wave Rectification. The calculation formulas of the two types of rectifier transformers are different.
1.2 Rectifier Transformer: Epoxy dry type transformer. Capacity is generally within 10-100kVA, nominal primary voltage (network) 400V, secondary voltage (valve) within 100V, current 100-300A. Due to the relatively small capacity, a power distribution board is located in the same position as the rectifier. The rectifier transformer cooling method is self-cooling, and the cooling conditions are better when no closing plate is installed on the side of the disk.
1.3 Insulation grade and heat dissipation method: The insulation grade of the dry epoxy transformer used by small hydropower is generally Class B, and the maximum temperature of the insulation system is 130°C. Therefore, it is normal that the transformer face temperature is hot when the transformer is working at full load. If the transformer is effectively forced air cooled, its output power can be increased by 10% to 30%. Conversely, if the transformer is operating in a sealed distribution box, the cooling conditions are poor and its current capacity must be reduced by 10% or more.
1.4 Impedance voltage: In the excitation system of the generator, there may be factors such as rectifier tube breakdown or short circuit of the DC circuit, so the short-circuit impedance voltage of the rectifier transformer is higher than the ordinary transformer to limit the excessive short-circuit current. . The parameters of the short-circuit impedance voltage are designed by the transformer manufacturer. We will not discuss it, but the user must specify the thyristor rectifier transformer when ordering from the manufacturer.
2 Wiring group The wiring group of the rectifier transformer must match the phase required by the thyristor rectification control. If it is a new design, it can be considered according to the following principles.
Generally, the method of D, y11 is adopted, that is, the network side (one time) adopts the â–³ connection method, and the valve side (secondary) adopts the y connection method. The secondary phase voltage of this connection is behind 300 in phase by one phase voltage.
The connection of D and y11 adapts to both three-phase half-wave and full-wave rectification simultaneously. If the existing rectifying transformer wiring group is Y, d11, it can be used, but it cannot be used for three-phase half-wave rectification.
As for Y, y wiring group is not recommended. We know that the third harmonic voltage generated by the three-phase controllable rectification is very high, which can reach more than 50% of the fundamental value. The D-connection method of the transformer can cancel the third harmonic flux and reduce the influence to the minimum. However, if the Y, y wiring groups are used, the closed loop of the magnetic flux generated by the third harmonic of the rectifier circuit cannot be cancelled out. Excessive third-order harmonics can cause excessive distortion of the electrical waveform, affecting the normal operation of transformers and generators and other instrumentation electrical equipment.
When the power station submits ordering data to the manufacturer, it should clearly explain the transformer connection group, primary and secondary voltages (at the same time, the phase or line voltage must be indicated).
3 Selection of Primary Line Voltage U1 The nominal line voltage of the small unit is 400V, but the small hydropower station is generally located at the far end of the power grid, and has a large impedance from the substation line. The voltage at the end of the network is too high, especially during peak periods in the flood season. The network voltage (end) is often as high as 460V. If the voltage is still designed in accordance with 400V at this time, the transformer will withstand overvoltage, so that the loss increases and the heat is exceeded.
The relationship between the iron loss of the rectifier transformer and its withstand voltage ratio is 4th power. For example, a rectifier transformer designed for 400V increases its core loss to (480/400) = 2.07 times when operating at 1.2 times (480V). . These losses are eventually converted to heat in the transformer, which increases the temperature rise of the transformer.
What is more, when the power supply voltage reaches a certain level, the magnetic flux density of the core of the transformer will enter the saturation region, causing the surge in the primary current to cause the coil to burn. For some rectifier transformers, due to cost considerations, the Bm value of the magnetic flux density of the core is selected to be high, and the voltage value of the primary winding is still selected to be 400V. Therefore, it is not uncommon to burn transformers in areas where the voltage is too high.
In this regard, the voltage value of the primary winding should be appropriately increased so that the net voltage +20 % transformer can cope with the work. General transformers still have 5% voltage overload capability, so we can use empirical formulas to select primary winding rated line voltage values.
U1 = 0.95 U1(MAX) ,
In the formula, U1 (MAX) is the maximum voltage value of the network power (converted to the machine). If the calculation result is less than 400V, it is selected by 400V.
After the primary voltage selection value is increased, the secondary voltage should also increase by the same ratio, keep the transformation ratio unchanged, and maintain the excitation voltage increase and decrease in the same proportion with the terminal voltage, because the higher the generator voltage, the required excitation power. The bigger it is.
Increasing the primary voltage is equivalent to increasing the number of volts per volt, all to reduce the magnetic flux density of the transformer core. Prevent entry into the saturation section of the flux density curve. The benefits are also reduced transformer no-load current and iron loss.
Of course, this also has some negative effects. As the number of winding turns increases, the internal resistance of the transformer increases, and the current loss (copper loss) increases slightly, but it does not affect the normal operation of the transformer. Voltage adjustment factor is n=U1/400
For simple calculations, U1 = 440V can be accessed, which can meet the requirements of most grid conditions (400V-470V).
4 Calculation of secondary voltage U2 The secondary voltage selection value relates to the top value (strong excitation) voltage of the excitation system, the maximum excitation current, the thyristor conduction angle and harmonic distortion, and the power factor of the rectifier circuit.
According to the relevant specifications, the excitation circuit must provide a 1.6-1.8 times stronger excitation voltage, that is, the secondary voltage of the transformer needs to be 1.6-1.8 times the rated value. However, in fact, China's small hydropower generating units rarely operate isolated power grids, and most of them are integrated into the operation of large power grids for power sale, and there is no need for and the ability to provide power to the power grid. The impact of a large, single hydropower unit on it is negligible.
If the secondary voltage is selected by increasing the value of 1.6 to 1.8 times, the rectified voltage is relatively high. The thyristor rectifier system is in a state of being controlled for a long time during excitation, the conduction angle of the thyristor is small, the waveform distortion is increased, and the power factor is deteriorated. The faulted short-circuit current increases, and these factors are detrimental to the operation of the transformer and the unit equipment. At the same time, under the same transformer power capacity, the high voltage will inevitably lead to the current reduction, the cross-sectional area of ​​the winding of the coil winding will decrease, and the current loss will also increase.
According to our experience, choosing the maximum rectified voltage to be 1.3 times the rated excitation voltage is relatively moderate. In addition to dealing with the disadvantage of avoiding the above-mentioned high voltage, it also retains a certain amount of rectified power margin and adapts to changes in operating conditions.
Calculation of phase voltage U2 Three-phase full-wave rectification U2 = 1.3*1.06 (n UE +2.5)/ 2.34 = 0.59 n UE + 1.47
Three-phase half-wave rectification U2 = 1.3*1.06 (n UE+1.7)/ 1.17 =1.18 n UE + 2.0
Explain the above formula:
UE-generator rated excitation voltage (V);
The coefficient 1.3 - as previously mentioned, is the margin value of the excitation voltage;
Coefficient 1.06—The voltage drop compensation caused by the internal resistance of the transformer when the current is full. Here a simple fixed value is used instead of a complex calculation, and the error is not too great;
N—voltage adjustment factor, as described in the previous section;
Coefficient 2.34 (or 1.17)—the ratio of the output DC voltage to the input AC phase voltage when the three-phase full-wave (or half-wave) rectifying element is fully on, ie, UE /U2 = 2.34 (or UE / U2 = 1.17);
The figure 2.5 (or 1.7)—the sum of the voltage drops of the excitation circuit, including the forward voltage drop of the rectifying element (1.5V or 0.75V) and the voltage drop (1.0 V) of the feed conductor and the carbon brush slip ring.
Transformer ratio K = U2 / U1.
For simple calculations, the first and second terms of the above equation can be combined with U2 = 0.71 UE (full-wave), or U2 = 1.4 UE (half-wave)
5. Current calculation Three-phase full-wave rectification Primary current I2 = 0.816 K IE, secondary phase current I1 = 0.816 IE
Three-phase half-wave rectifier primary current I2 = 0.472 K IE, secondary phase current I2 = 0.577 IE
The I1 value here has not considered the efficiency of the transformer.
6 Power Calculation Excitation Power: PE = UE IE(W)
Transformer secondary side power:
Full-wave rectification P2 =3 U2 I2 =3(0.59 n UE + 1.47)* 0.816 IE = 1.45 n PE +3.60 IE (W)
Half-wave rectification P2 =3 U2 I2 =3(1.18 n UE + 2.0)* 0.577 IE = 2.04 n PE +3.46 IE (W)
Transformer capacity calculation:
Full-wave rectification S1 = P2/(0.8 * 97%) = 1.29 P2 = 1.29 (1.45 n PE + 3.60 IE) = 1.87 n PE + 4.64 IE (VA)
Half-wave rectification S1 = P2 / (0.8 * 97%) = 1.29 P2 = 1.29 (2.04 n PE +3.46 IE) = 2.63 n PE + 4.46 IE (VA)
In the formula, 0.8 is the rated power factor of the transformer, and 97% is the primary side efficiency of the transformer.
For simple calculations, the first and second terms of the above formula can be merged.
S1 = 2.2 PE (full wave), or S1 = 3 PE (half wave)
In actual orders, we have sometimes discovered that some manufacturers are trying to reduce the cost of transformer manufacturing materials are tight, so that the transformer operating temperature rise is high. In this case, for insurance reasons, it is best to increase the order of the transformer capacity 10% to deal with, at this time the voltage value of the transformer does not change, the primary and secondary currents must be increased proportionally.
7 Calculation example A small hydropower station needs to order an excitation rectifier transformer. The generator parameters are:
Generator power 400kW, rated voltage 400V, generator excitation voltage UE = 49.4V, excitation current IE = 153A.
Working conditions: The exciter is a thyristor three-phase full-wave semi-controlled rectification, and the unit is operated in a large grid. The maximum voltage of the grid (machine terminal) is 460V.
The solution steps are as follows
1) Connection group D, y11
2) Excitation power PE = UE IE = 49.4×153 = 7558.2 (W)
2) The primary line voltage U1 = 460 × 0.95 = 437 (V), take 440V
Voltage adjustment factor n = U1/400 = 1.1
3) The secondary phase voltage U2 = 0.59 n UE + 1.07 = 0.59 x 1.1 x 49.4 + 1.07 = 33.13V, optional 34V;
4) Transformer ratio K = U2/U1 = 34/440 = 0.0773
4) Secondary current I2 = 0.816 IE = 0.816 × 153 = 124.85 (A), optional 125A.
5) Primary current I1 = 0.816 KIE = 0.816 × 153 × 0.0773 = 9.65 (A), optional 9.7A
6) Power capacity S1 = 1.87 n PE + 4.64 IE =1.87 x 1.1 x 7558.2 + 4.64 x 153
= 16257(VA), optional 17 kVA
Finally, fill in the order list:
Three-phase dry rectifier transformer Ordering data Transformer capacity: 17 kVA Connection group: D,y11
Voltage ratio: 440 V (line) / 34 V (phase) current (primary current may not be reported): 9.7 A / 125 A
Power factor: 0.80 Work system: Long-term continuous operation According to the calculation results in this paper, the load ratio of the transformer is 0.77 at rated condition and it is in an economical operation state. When the rated field current is output, the control angle of the rectifier thyristor is about 650 Ω.
We used the ms excel (spreadsheet) software to write the calculation program. The calculation only needs three parameters: excitation voltage UE, current IE, and maximum network voltage U1 (MAX). The program will automatically calculate all the data of the rectifier transformer. Give the order list. This program can be downloaded from the website of Shenzhen Puwei Electric Company (szpwr.com), or the website can be calculated on behalf of the
8. Simple calculation method Excitation power: P = 40.3×161.4 = 7558.24 (W)
Take a line voltage: U1 = 440V
Phase voltage U2 = 1.4 UE = 1.4 × 49.4 = 69.2 V for three-phase half-wave rectification, optional 70 V
Transformer capacity S1 = 3×7558. = 22674(VA), optional 23 kVA
Phase voltage U2 = 0.71 for three-phase full-wave rectification UE = 0.71 × 49.4 = 35.1 (V), optional 35V
Transformer capacity S1 = 2.2×7558.2 = 1662(VA), select 17 kVA
Compared with the calculation results in the previous section, the error in the simple calculation method is not large and can be applied. When the excitation voltage UE is large, the U2 value of the simplified method will be slightly larger, but it will not affect the actual operation.