Calculation between wire size and power consumption

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Calculation of the wire size based on power consumption: First, estimate the load current. 1. Purpose: This is a formula for calculating the current in amperes based on the power of the electrical equipment (in kilowatts or kVA). The magnitude of the current is directly related to power, as well as to voltage, phase, and power factor (also known as the power factor). Generally, there are formulas available for calculation. Since factories typically use a 380/220 volt three-phase four-wire system, the current can be calculated directly based on the power level. 2. Mnemonic: For low-voltage 380/220 volt systems, the current in amperes per kilowatt. Kilowatts, current – how to calculate them? Electric power doubles, electric heating increases by half. ① Single-phase kilowatt, 4.5 amps. ② Single-phase 380, current two and a half amps. ③ 3. Note: The mnemonic is based on three-phase equipment in a 380/220 volt three-phase four-wire system, to calculate the amperage per kilowatt. For certain single-phase or single-phase devices with different voltages, a separate mnemonic is provided for the amperes per kilowatt. ① In these two mnemonics, \"electric power\" specifically refers to electric motors. At 380 volts three-phase (with a power factor of around 0.8), the current per kilowatt for the motor is approximately 2 amps. In other words, multiply the kilowatt rating by two to get the current in amps. This current is also known as the motor’s rated current. 【Example 1】 For a 5.5-kilowatt motor, the current calculated using the \"doubled power\" method is 11 amps. 【Example 2】 For a 40-kilowatt water pump motor, the current is calculated as 80 amps using the \"doubling of power\" method. Electric heating refers to resistive furnaces and similar devices that use electrical resistance for heating. For three-phase 380-volt electric heating equipment, the current per kilowatt is 1.5 amps. Simply \"multiply the kilowatts by half\" (i.e., multiply by 1.5) to get the current, in amps. 【Example 1】 For a 3-kilowatt electric heater, the current is calculated as 4.5 amps based on the formula \"electrical heating plus half\". 【Example 2】 For a 15-kilowatt resistance furnace, the current is calculated to be 23 amps based on the formula \"electrical heating plus half\". This mnemonic is not specific to electric heating; it also applies to lighting. Although the lighting bulb is single-phase rather than three-phase, the three-phase four-wire main supply for the lighting remains three-phase. Calculations can be done this way as long as the three phases are roughly balanced. In addition, electrical devices measured in kVA (such as transformers or rectifiers) and phase-shifting capacitors measured in kvar (used to improve power factor) are also applicable. To put it immediately, although the second part of this sentence refers to electric heating, it includes all electrical devices measured in kilovatts-ampere and kvar, as well as electric heating and lighting devices measured in kilowatts. 【Example 1】 For a 12-kilowatt three-phase (balanced) lighting main circuit, the current is calculated as 18 amps using the \"electrical heating plus half\" method. 【Example 2】 For a 30 kVA rectifier, the current is calculated as 45 amps based on the \"electrical heating plus half\" method (referring to the 380-volt three-phase AC side). 【Example 3】 For a 320 kVA distribution transformer, the current is calculated as 480 amps using the \"electrical heating plus half\" method (referring to the 380/220 V low-voltage side). 【Example 4】 For a 100 kVAR phase-shifting capacitor (380 volts, three-phase), the current is calculated as 150 amps using the \"electrical heating plus half\" method. ②In a 380/220 volt three-phase four-wire system, single-phase devices that have one wire connected to the phase line and the other connected to the neutral line (such as lighting fixtures) are single-phase 220 volt devices. The power factor of such devices is usually 1; therefore, the mnemonic directly states “4.5 amps per single-phase kilowatt”. For calculations, simply \"multiply the kilowatts by 4.5\" to get the current in amps. As mentioned above, it is applicable to all single-phase 220-volt electrical devices measured in kVA, as well as electric heating and lighting devices measured in kilowatts; it is also suitable for 220-volt direct current. 【Example 1】 For a floor lamp transformer with a capacity of 500 volt-amps (0.5 kVA) (on the 220-volt side), the current is calculated as 2.3 amps based on the formula of \"1 kilowatt per phase, 4.5 amps\". 【Example 2】 For a 1000-watt spotlight, calculated as \"single-phase kilowatt, 4.5 amps,\" the current is 4.5 amps. For single-phase systems with lower voltages, it is not mentioned in the mnemonic. 220 volts can be taken as a standard; the more the voltage drops, the more the current increases accordingly. For example, with a voltage of 36 volts, and using 220 volts as the standard, since this voltage is 1/6 of the standard value, the current should increase by a factor of 6; thus, the current per kilowatt would be 6*4.5=27 amps. For example, for a 36-volt, 60-watt work light, the current per unit is 0.06*27=1.6 amps; therefore, 5 such lights would require a total of 8 amps. ③In a 380/220 volt three-phase four-wire system, the two wires of single-phase equipment are connected to the phase wires; it is conventionally referred to as single-phase 380 volt equipment (when in fact it is connected to two phases). For such equipment, when expressed in kilowatts, the power factor is usually 1; the mnemonic also states directly: “Single-phase 380, current two and a half amps.” It also includes 380-volt single-phase equipment measured in kVA. For calculations, simply \"multiply kilowatts or kVA by 2.5\" to get the current in amps. 【Example 1】 A 32-kilowatt molybdenum wire resistance furnace connected to single-phase 380 volts has a current of 80 amps, calculated based on a current of two and a half amps. 【Example 2】 For a 2 kVA portable lamp transformer with a single-phase 380V input, the current is calculated to be 5 amps based on a current value of 2.5 amps. 【Example 3】 For an AC welding transformer with a capacity of 21 kVA, connected to a single-phase 380-volt supply, the current is calculated to be 53 amps based on the rule of \"2.5 amps per kilovaolt\". After estimating the current required by the load, the cross-sectional area of the appropriate wire is selected based on this current. Several factors need to be taken into account when choosing the wire’s cross-sectional area: first, the mechanical strength of the wire; second, the current density of the wire (i.e., its safe current-carrying capacity); and third, the allowable voltage drop. Estimation of voltage drop: 1. Purpose – To estimate the voltage loss in the power supply line based on the load on that line, thereby assessing the quality of power supply in that line. 2. Mnemonic: A baseline value for estimating voltage loss is provided, and through some simple calculations, the voltage loss in the power supply circuit can be estimated. The pressure loss is expressed in “kilowatts·meter”; for 2.5 aluminum wire, it ranges from 20 to 1. As the cross-section increases, the bending moment increases, while the voltage decreases in proportion to the square of that increase. ① For three-phase four-wire systems, use a factor of 6; for copper wires, multiply by 1.7. ② The inductive load causes high pressure loss; the impact on the cross-section at 10 is minimal. If a power factor of 0.8 is used, it increases by 0.2 to 1 at 10. ③ 3. Note: The calculation of voltage loss is related to numerous factors, making it a complex process. During estimation, the conductors and cross-sections for the line have already been selected based on the load conditions; in other words, the relevant requirements are basically met. Voltage loss is measured as “what percentage of the rated voltage it represents”. The mnemonic mainly lists the most basic data for estimating voltage loss – at what level of \"load torque\" will the voltage loss be 1%? When the load torque is high, the voltage loss increases accordingly. Therefore, the load torque of this circuit should be calculated first. The so-called load torque is the load (in kilowatts) multiplied by the length of the circuit (the circuit length refers to the length of the wire run in meters, that is, the path covered by the wire, regardless of the number of wires in the circuit). ), and the unit is “kilowatt·meter”. For radial circuits, the calculation of the load torque is very simple. As shown in Figure 1, the load torque is 20*30=600 kilowatts·meter. But with the trunk-style circuit in Figure 2, it’s more troublesome. For the 5-kilowatt equipment, the load torque at its installation location should be calculated as follows: starting from the power supply point of the circuit, it is divided into three sections depending on the branching of the circuit. In each section of the circuit, all three loads (10, 8, and 5 kilowatts) are present; therefore, the load torque is as follows: First section: 10 * (10 + 8 + 5) = 230 kilowatt-meters. Second section: 5 * (8 + 5) = 65 kilowatt-meters. Third section: 10 * 5 = 50 kilowatt-meters. The total load torque up to the 5-kilowatt device is: 230 + 65 + 50 = 345 kilowatt-meters. Below is an explanation of the formula used: ① First, it is stated that the most basic basis for calculating voltage loss is the load torque, measured in kilowatt-meters. Then, a reference value is given: for 2.5 square millimeter aluminum wire, at 220 volts single-phase, with resistive loads (power factor of 1), the voltage loss per 20 kilowatt-meters of load torque is 1%. This is the “2.5 aluminum wire 20—1” mentioned in the mnemonic. Based on a 1% voltage loss, a larger cross-section allows for a greater load torque, with the relationship being proportional. For example, with an aluminum wire of 10 square millimeters in cross-section, which is 4 times larger than 2.5 square millimeters, the value is 20*4=80 kilowatts·meter; thus, the load torque for this type of wire is 80 kilowatts·meter, with only a 1% voltage loss. Analogous reasoning applies to the remaining cross-sections. When the voltage is not 220 volts but some other value, such as 36 volts, first determine what 1/6 of 220 volts corresponds to. At this point, the load torque corresponding to a 1% voltage drop in such a circuit is not 20 kilowatt-meters; rather, it should be reduced by a factor of 1/6 squared, that is, 1/36. Therefore, the value is 20 * (1/36) = 0.55 kilowatt-meters. In other words, at 36 volts, the voltage drop decreases by 1% for every 0.55 kilowatt-meters (i.e., every 550 watts-meters). ““The voltage decreases as the square of the reduction” applies not only to cases with lower rated voltages but also to those with higher rated voltages. At this point, it has to be increased by squaring. For example, in the case of single-phase 380 volts, since 380 volts is 1.7 times 220 volts, the load torque resulting from a 1% voltage loss should be 20*1.7 squared = 58 kilowatts-per-meter. It can be seen from the above that the mnemonic is: “A larger cross-section results in a greater bending moment, and the voltage decreases in proportion to the square of that value.” All are referenced to the baseline data “2.5 aluminum wire 20—1”. 【Example 1】 A 220-volt lighting circuit using 2.5 mm² aluminum wire, with a load torque of 76 kilowatts·meter. Since 76 is 3.8 times 20 (76/20=3.8), the voltage loss is 3.8%. 【Example 2】 For a 40-meter long circuit equipped with an aluminum wire of 4 square millimeters, supplying two 1-kilowatt single-phase electric stoves operating at 220 volts, the estimated voltage loss is as follows: First, calculate the load torque as 2*40 = 80 kilowatts·meter. Calculating the load torque for a 4 mm² aluminum wire with 1% voltage loss, and based on the principle that an increased cross-sectional area leads to a greater load torque, comparing 4 with 2.5 shows that the cross-sectional area increases by a factor of 1.6 (4/2.5=1.6); therefore, the load torque increases to 20*1.6=32 kilowatts·meter (this is the value corresponding to 1% voltage loss). Finally, 80/32 = 2.5, which means the voltage loss for this line is 2.5%. ②When the circuit is three-phase four-wire rather than single-phase, (it is generally required that the three-phase load be fairly balanced in such a three-phase four-wire system.) Its voltage corresponds to that of single-phase. If the single-phase voltage is 220 volts, the corresponding three-phase voltage is 380 volts, that is, 380/220 volts. ) For aluminum wires with the same cross-sectional area of 2.5 square millimeters, the load torque resulting from a 1% voltage loss is 6 times that of the baseline value in ①, that is, 20*6=120 kilowatts·meter. As for changes in cross-section or voltage, the value of this load torque also changes accordingly. When the wire is copper instead of aluminum, the load torque value for the aluminum wire must be multiplied by 1.7; for example, if \"2.5 aluminum wire 20–1\" is replaced with a copper wire of the same cross-section, the load torque becomes 20*1.7=34 kilowatt-meters, with only a 1% voltage loss. 【Example 3】 For the lighting circuit illustrated earlier, if copper wires are used, then 76/34 = 2.2, meaning the voltage loss is 2.2%. For the circuit that supplies power to the electric furnace, if it is a copper wire, then 80/(32*1.7) = 1.5, meaning the voltage loss is 1.5%. 【Example 4】 A 380-volt three-phase circuit with 50 square millimeter aluminum wires, 30 meters in length, is used to supply a 60-kilowatt three-phase electric furnace. The voltage loss estimation is as follows: first calculate the load torque: 60*30=1800 kilowatt-meters. Next, calculate the load torque when there is a 1% voltage loss in a 380-volt three-phase system with 50 square millimeter aluminum wires: According to the principle that an increase in cross-sectional area leads to an increase in load torque, since 50 is 20 times greater than 2.5, a factor of 20 must be applied. Additionally, since it is a three-phase four-wire system, a factor of 6 must also be applied. Thus, the load torque increases to 20*20*6 = 2400 kilowatts-meter. Finally, 1800/2400=0.75, meaning the voltage loss is 0.75%. ③All of the above applies to resistive loads. For inductive loads (such as motors), the calculation method is more complex than the one mentioned above. But the mnemonic first states that for the same load torque – in kilowatts·meter – the voltage drop in inductive loads is slightly higher than that in resistive loads. It is related to the cross-sectional size and the distance between the wires. For wires of 10 square millimeters or less, the impact is minimal, and it is not necessary to increase it. For circuits with a cross-sectional area of 10 square millimeters or more, the estimation can be done as follows: first calculate the voltage loss using methods ① or ②, and then \"increase it by 0.2 to 1\", which means multiplying the value by 1.2 to 2. This can be determined based on the size of the cross-section; a larger cross-section requires a higher value. For example, 70 square millimeters can be multiplied by 1.6, and 150 square millimeters can be multiplied by 2. The above refers to the situation where the lines are overhead or laid openly on supports. For cable or conduit-mounted circuits, since the distance between the circuits is very small and the impact is minimal, estimation can still be carried out according to rules ① and ②; there is no need to increase the values, or only a slight increase is required for wires with large cross-sections (within 0.2). 【Example 5】 In Figure 1, if the 20-kilowatt motor is a 380-volt three-phase motor with 3*16 aluminum wires installed in an exposed manner, the voltage loss can be estimated as follows: the known load torque is 600 kilowatts·meter. To calculate the load torque at 1% voltage loss for a 16 mm² aluminum wire at 380 volts in a three-phase system: since 16 is 6.4 times 2.5, and the three-phase load torque is 6 times that of a single phase, the load torque increases to: 20*6.4*6 = 768 kilowatts·meter. 600/768 = 0.8; thus, the estimated voltage loss is 0.8%. But now it is a motor load, and the wire cross-section is over 10, so it should be increased a bit. Based on the cross-sectional conditions and considering 1.2, the estimate is 0.8*1.2=0.96; it can be assumed that the voltage loss is approximately 1%. The above is the method for estimating voltage loss. Finally, a few points on issues related to this matter: First, at what level of voltage loss on the line does the quality become poor? Generally, 7~8% is the standard. (More strictly speaking: the voltage loss is based on the rated voltage of the electrical equipment (such as 380/220 volts), with a tolerance of up to 5% below this rated voltage (2.5% for lighting). However, the voltage at the low-voltage bus terminal of the distribution transformer is specified to be 5% higher than the rated voltage (400/230 volts). Therefore, in the entire circuit from the transformer to the electrical equipment, a loss of 5% + 5% = 10% is theoretically possible, but usually only 7–8% is allowed. This is because the voltage losses inside the transformer, as well as the effects of a low power factor of the transformer, must also be taken into account. ) However, this 7–8% refers to the entire circuit from the low-voltage side of the distribution transformer up to the electrical device being calculated. It usually includes sections such as outdoor overhead lines, indoor main lines, and branch lines. It should be the sum of the results for each segment, totaling around 7~8%. II. Estimating voltage loss is a task involved in the design process, aimed primarily at preventing poor voltage quality during future use. Since there are many factors that affect the calculations (the main ones being the accuracy of calculating the main load and the stability of the voltage on the transformer’s supply side, etc.), it is not necessary to achieve high precision in these calculations; it is sufficient to have a general idea of the results. For example, the ratio of cross-sections can also be simplified to 1.5 times for 4 to 2.5, 2.5 times for 6 to 2.5, and 6 times for 16 to 2.5. It will be more convenient to calculate this way. III. In estimating the voltage loss in the motor circuit, there is another scenario involving the estimation of the voltage loss during motor startup. This is to prevent the motor from starting directly if the loss is too large. Due to the high current and low power factor during startup, it is generally specified that the voltage drop during startup can reach 15%. The calculation of voltage loss during startup is more complex, but it can be assessed using the results obtained from the formula mentioned above. Generally, aluminum wires with a cross-sectional area of 25 square millimeters or less can meet the 5% requirement and thus be suitable for direct startup; aluminum wires with a cross-sectional area of 35 or 50 square millimeters can also satisfy the requirements if the voltage loss is within 3.5% ； Aluminum wires with a cross-sectional area of 70 or 95 square millimeters can also meet the requirements as long as the voltage loss is within 2.5% ； And for an aluminum wire of 120 square millimeters, the voltage loss should be within 1.5. Only then can it be satisfied. These 3.5%, 2.5%, and 1.5% correspond exactly to 70%, 50%, and 30% of 5%, so it can be simply remembered as: “Above 35, 70%, 50%, 30%.” IV. If it is indeed found during use that the voltage drop is too large and affects the quality of power supply, the load can be reduced (by transferring part of the load to other, less burdened circuits, or by adding an additional circuit), or the cross-sectional area of some sections of the wires can be increased (it is best to increase the cross-sectional area of the main wires first) to resolve the issue. For motor circuits, cables can also be used to reduce voltage loss. When the motor cannot start directly, in addition to the solutions mentioned above, voltage-reduction starting devices (such as star-delta starters or autotransformer starters) can also be used to address this issue. The cross-sectional area of the wires should be selected based on the current. 1. Purpose: The current-carrying capacity of various wires (for safe electricity use) can usually be found in manuals. But by using a mnemonic together with some simple mental arithmetic, the calculation can be done directly without needing to consult a table. The current-carrying capacity of a wire is related to its cross-sectional area, as well as factors such as the material of the wire (aluminum or copper), its type (insulated wire or bare wire, etc.), the method of installation (exposed or in conduits, etc.), and the ambient temperature (around 25°C or higher). There are many influencing factors, making the calculations complex. 2. Mnemonic: The relationship between the current-carrying capacity of aluminum-insulated wires and their cross-sectional area: S (cross-sectional area) = 0.785 * D (diameter) squared. For a diameter of 10, the value is 5; for 100, it’s 2. For diameters of 25 and 35, the values are 4 and 3 respectively. For diameters of 70 and 95, the value is 2.5 times the corresponding diameter. ① Tube insertion, temperature: 20-30% discount. ② Add half of the bare wire. ③ Copper wire upgrade counts. ④ 3. Note: The mnemonic is based on aluminum-core insulated wires installed exposed, under the condition of an ambient temperature of 25°C. If the conditions are different, there are separate instructions for the mnemonic. Insulated wires include various types of rubber-insulated or plastic-insulated wires. The mnemonic does not specify the current flow (current, in amps) for various cross-sections directly; instead, it is expressed as \"the cross-section multiplied by a certain factor\". To do this, one should first become familiar with the sequence of wire cross-sections in square millimeters: 1, 1.5, 2.5, 4, 6, 10, 16, 25, 35, 50, 70, 95, 120, 150, 185…… The cross-sections of aluminum-core insulated wires manufactured by producers usually start at 2.5, while those of copper-core insulated wires start at 1 ； It starts at 16 for bare aluminum wires, and at 10 for bare copper wires. ①This mnemonic states that the current-carrying capacity of aluminum-core insulated wires, in amps, can be calculated as \"a multiple of the cross-sectional area\". In the mnemonic, Arabic numerals represent the wire cross-section (in square millimeters), while Chinese numerals indicate the multiplier. Arranging the \"section area and multiple relationship\" from the mnemonic gives the following: . . . 10*5, 16; 25*4, 35; 45*3, 70; 95*2.5, 120*2. . . . . . ?? Now, by comparing it with the mnemonic, it becomes clearer – \"10 times five\" means that when the section area is 10 or less, the flow rate is always five times the value of the section area. “\"100 times two\" means that for a cross-sectional area of over 100, the flow rate is twice the value corresponding to that cross-sectional area. Sections 25 and 35 are the boundaries for four times and three times, respectively. This is the mnemonic phrase “25, 35: four and three realms”. While sections 70 and 95 are two and a half times. As can be seen from the above arrangement: except for values below 10 and above 100, the cross-sectional area of the wires in the middle is always the same multiple for every two specifications. The following is an example using exposed aluminum-core insulated wire at an ambient temperature of 25°C: [Example 1] For a 6 square millimeter wire, the current-carrying capacity is 30 amps according to the \"10 down 5\" formula. 【Example 2】For an area of 150 square millimeters, the current-carrying capacity is 300 amps according to the rule of \"100 plus 20\». 【Example 3】For 70 square millimeters, the current-carrying capacity is calculated as 70 and 95 times 1.5, resulting in 175 amps. It can also be seen from the above arrangement that the multiples decrease as the cross-section increases. At the boundary of the multiple transitions, the error is slightly larger. For example, sections 25 and 35 represent the boundary between four times and three times the value. Section 25 falls within the four-times range, but it is closer to the side where the value shifts to three times; according to the formula, it should be 100 amps, but in reality it is less than four times that amount (97 amps according to the manual). On the other hand, section 35, per the formula, should correspond to three times the value, which is 105 amps, but in actuality it is 117 amps. However, this has little impact on practical use. Of course, if one has a clear idea of the requirements, it is more accurate to ensure that for a 25 mm² wire the current rating does not reach 100 amps, while for a 35 mm² wire the current can slightly exceed 105 amps. Similarly, the wire with a cross-sectional area of 2.5 square millimeters is located at five times the initial (left) end; in practice, the current can be more than five times that value (up to over 20 amps). However, to reduce power losses within the wire, such high currents are generally not used, and the manuals usually specify only 12 amps. ②From here on, the mnemonic is for handling changes in conditions. The actual name “Cable installation, temperature; 20–30% discount” means that in the case of cable installation (including installation using trays, etc., i.e., when the wires are covered with a protective layer and not exposed), the cost is calculated according to method ① and then a 20% discount is applied (multiplied by 0.8). If the ambient temperature exceeds 25°C, calculate according to ① and then apply a 10% discount (multiply by 0.9). Regarding the ambient temperature, it is defined as the average maximum temperature of the hottest month in summer. In fact, temperature varies, and under normal circumstances, its impact on the current-carrying capacity of a conductor is not significant. Therefore, a discount is only considered in certain high-temperature workshops or hot areas where the temperature exceeds 25°C significantly. In another case, when both conditions change (tubing is used and the temperature is high), the calculation is done according to①, followed by a 20% discount and then a 10% discount. Or simply calculate it at a 30% discount at once (that is, 0.8*0.9=0.72, approximately 0.7). This can also be interpreted as meaning “pipe installation, temperature – 20-30% discount”. For example: For 10 square millimeter aluminum-core insulated wires, when used in tubes (with a 20% discount), the current-carrying capacity is 40 amps (10*5*0.8=40). In high-temperature conditions (with a 10% discount), it is 45 amps (10*5*0.9=45). When used in tubes under high-temperature conditions (with a 30% discount), the capacity is 35 amps (10*5*0.7=35 amps). For 95 square millimeter wires, when used in tubes (with a 20% discount), the current-carrying capacity is 190 amps (95*2.5*0.8=190). In high-temperature conditions (with a 10% discount), it is 214 amps (95*2.5*0.9=213.8). When used in tubes under high-temperature conditions (with a 30% discount), the capacity is 166 amps (95*2.5*0.7=166.3). ③ Regarding the current-carrying capacity of bare aluminum wires, the rule states that “add half to the value obtained from method ①”, that is, multiply the result from method ① by 1.5. This means that, compared to bare aluminum wire, an aluminum-core insulated core with the same cross-section can handle twice as much current. 【Example 1】 16 mm² bare aluminum wire: 96 amps (16*4*1.5=96); at high temperatures, 86 amps (16*4*1.5*0.9=86.4). 【Example 2】 35 mm² bare aluminum wire: 158 amps (35*3*1.5=157.5). 【Example 3】 120 mm² bare aluminum wire: 360 amps (120*2*1.5=360). ④ For the current-carrying capacity of copper wires, a rule states that \"the copper wire should be upgraded in grade\"; that is, the cross-sectional area of the copper wire is increased by one level according to the grading sequence, and then the calculation is performed using the corresponding parameters for aluminum wires. 【Example 1】 35 mm² bare copper wire at 25°C. Upgrade to 50 square millimeters; for bare aluminum wire at 25°C, the current is 225 amps (50*3*1.5). 【Example 2】 16 mm² copper insulated wire at 25°C. Under the same conditions as 25 square millimeter aluminum insulated wire, it is calculated to be 100 amps (25*4). 【Example 3】 95 mm² copper insulated wire at 25°C, in a conduit. Under the same conditions as 120 square millimeter aluminum insulated wire, it is calculated to be 192 amps (120*2*0.8). By the way: regarding cables, nothing is mentioned in the mnemonic. For high-voltage cables that are buried directly in the ground, it is generally possible to use the relevant multiples listed in ① for direct calculation; for example, the current-carrying capacity of a 35 square millimeter high-voltage armored aluminum-core cable buried in the ground is approximately 105 amps (35*3). 95 square millimeters is approximately 238 amps (95*2.5). The estimation formula below is similar in principle to the one above: multiply 2.5 by 9, then subtract 1 and move forward sequentially. Thirty-five times 3.5, pair them up and subtract 0.5. If conditions change, a discount is applied; at high temperatures, copper upgrades at a 10% discount. Two, three, four roots in the tube; 80%, 70%, 60% capacity for current conduction. 2.5 square mm * 9, 4 square mm * 8, 6 square mm * 7, 10 square mm * 6, 16 square mm * 5, 25 square mm * 4, 35 square mm * 3.5, 50 and 70 square mm * 3, 95 and 120 square mm * 2.5 ..................... Finally, it should be noted that using current to estimate the cross-sectional area is suitable for situations where the load is close to the power source, while voltage drop considerations are relevant for long-distance connections. Here’s a mnemonic to help you remember this: “Ten or less, multiply by five; over a hundred, multiply by two; 25, 35, 45, 35 – multiply by four or three respectively; 70 and 95, multiply by 2.5; account for a 10-20% reduction due to temperature effects; for copper wires, upgrade to the next higher grade; for bare wires, add half to the calculated value.” Explanation: “Ten or less, multiply by five” means that values below 10 are multiplied by 5; “over a hundred, multiply by two” means that values above 100 are multiplied by 2; “25, 35, 45, 35 – multiply by four or three respectively” refers to multiplying 25 by 4 and 35 by 3; “70 and 95, multiply by 2.5” means that both 70 and 95 are multiplied by 2.5; “account for a 10-20% reduction due to temperature effects” means that the calculated safe current value is adjusted by multiplying it by 0.8 or 0.9 depending on the temperature; “for copper wires, upgrade to the next higher grade” means that if using an aluminum wire with the same cross-sectional area, it should be upgraded to the next higher grade, for example, a 2.5 square millimeter copper wire would be treated as a 4 square millimeter aluminum wire; “for bare wires, add half to the calculated value” means that half is added to the already calculated safe current value. How is the relationship between wire thickness and electrical appliance power calculated? Hello! Calculation of wire cross-sectional area and current-carrying capacity I. Current-carrying capacity of general copper wires The safe current-carrying capacity of a wire is determined based on the maximum allowable temperature of the wire core, cooling conditions, and installation conditions. The safe current-carrying capacity of copper wires is generally 5~8 A/mm2, while that of aluminum wires is 3~5 A/mm2. The safe current-carrying capacity of copper wires is generally 5~8 A/mm2, while that of aluminum wires is 3~5 A/mm2. For example: the recommended safe current-carrying capacity for a 2.5 mm2 BVV copper wire is 2.5×8A/mm2 = 20A; the recommended safe current-carrying capacity for a 4 mm2 BVV copper wire is 4×8A/mm2 = 32A. II. Calculating the cross-sectional area of the copper wire: Using the recommended safe current-carrying capacity of 5–8A/mm2, the upper and lower limits for the cross-sectional area S of the chosen copper wire can be calculated as follows: S = I / (5–8) = 0.125I – 0.2I (mm2). Where S represents the cross-sectional area of the copper wire in mm2, and I represents the load current in amps. III. Power calculation: Ordinary loads (which can also be referred to as electrical appliances such as lights, refrigerators, etc.) are divided into two types: resistive loads and inductive loads. For resistive loads, the calculation formula is: P=UI. For fluorescent lamp loads, the formula is: P=UIcosφ, where the power factor cosφ for fluorescent lamp loads is 0.5. Different inductive loads have different power factors; when calculating household appliances collectively, a power factor of cosφ = 0.8 can be used. In other words, if the total power of all the electrical appliances in a household is 6000 watts, then the maximum current is I=P/U*cosφ=6000/220*0.8=34(A). However, in practice, the appliances in a household are not used simultaneously; therefore, a correction factor is applied, with a value of around 0.5. Therefore, the calculation above should be rewritten as I=P*common coefficient/Ucosφ=6000*0.5/220*0.8=17(A). In other words, the total current consumption for this household is 17A. In that case, the main circuit air switch cannot use a 16A rating; one with a rating higher than 17A should be used. Estimation formula: Multiply 2.5 by 9, then subtract 1 and move forward in sequence. Thirty-five times 3.5, pair them up and subtract 0.5. If conditions change, a discount is applied; at high temperatures, copper upgrades at a 10% discount. Two, three, four roots in the tube; 80%, 70%, 60% capacity for current conduction. Note: (1) This section’s mnemonic does not specify the current-carrying capacity (safe current) for various insulated wires (rubber- and plastic-insulated wires) directly; instead, it is expressed as \"the cross-sectional area multiplied by a certain factor,\" which can be calculated mentally. As can be seen from Table 5-3, the multiple decreases as the cross-section increases. “\"2.5 down multiplied by 9, then subtract one and move forward in sequence\" refers to aluminum-core insulated wires with cross-sections of 2.5 mm² or less; their current-carrying capacity is approximately 9 times the value of their cross-sectional area. For example, a 2.5mm’ wire has a current-carrying capacity of 2.5×9=22.5(A). The multiple relationship between the current-carrying capacity and the number of cross-sections for wires of 4 mm² and above is arranged in ascending order of wire size, with the multiple decreasing by 1 each time; that is, 4×8, 6×7, 10×6, 16×5, 25×4. “\"35 multiplied by 3.5; group them together and subtract 0.5,\" which means that the current-carrying capacity of a 35mm” wire is 3.5 times the area of its cross-section, that is, 35×3.5=122.5(A). For wires of 50 mm² and above, the multiplicative relationship between the current-carrying capacity and the cross-sectional area changes such that wire sizes are grouped in pairs, with the multiplier decreasing by 0.5 for each subsequent pair. That is, the current-carrying capacity of 50 and 70 mm’ wires is 3 times the cross-sectional area ； The current-carrying capacity of 95, 120mm” wires is 2.5 times their cross-sectional area, and so on. “If conditions change, a discount is applied; at high temperatures, a 10% discount applies and the copper grade is upgraded. The above formula is based on aluminum-core insulated wires installed exposed, under an ambient temperature of 25°C. If aluminum-core insulated wires are installed exposed in areas where the ambient temperature remains above 25°C for extended periods, the current-carrying capacity of the wires can be calculated using the method outlined above, and then that value should be reduced by 10% ； When insulated copper wires are used instead of aluminum wires, their current-carrying capacity is slightly higher than that of aluminum wires of the same specification; the current-carrying capacity can be calculated by using the method outlined above, by increasing the wire gauge by one step compared to aluminum wires. For example, the current-carrying capacity of a 16 mm² copper wire can be estimated based on that of a 25 mm² aluminum wire. How is the relationship between wire thickness and electrical appliance power calculated? The relationship between the wire diameter and the power load also depends on the working environment of the wire – whether it is buried or run exposed, whether it is a cable or a plastic wire, and whether it is made of copper or aluminum. You can buy an \"Electrical Engineer’s Handbook\" from a bookstore, or download it from shared resources. As a general estimate, the maximum current can be 10 amperes per square millimeter for overhead wiring, and 5–6 amperes for buried wiring. If we know the electrical power that a wire can handle, and also its cross-sectional area, how can we calculate the maximum electrical power that wire can sustain? Or, if the required electrical power is known, how can one determine what size of mm2 wire should be used? Answer: We can consult electrical engineering manuals to find out the maximum current-carrying capacity of the wire. Using the formula Power P = Voltage U × Current I, the power calculated in this way represents the maximum electrical power that the wire can handle. For example: With a 220-volt power supply and a 1.5 square millimeter wire, what is the maximum power rating of an appliance that can be connected? Solution: According to the manual, the current-carrying capacity of a 1.5 square millimeter wire is 22 A. Using the formula Power P = Voltage U × Current I, we get 220 volts × 22 amps = 4840 watts. Therefore, the maximum electrical appliance that can be connected is one with a power consumption of 4840 watts. Conversely, if the required power is known, we can calculate the current using the above formula: Current = Power ÷ Voltage. Once the current value is known, the manual can be consulted to determine what size of wire is needed. For example: What size wire is needed for a 10-kilowatt appliance connected to a 220-volt power supply? Solution: According to the formula Current = Power ÷ Voltage, we have 10,000 watts ÷ 220 volts = 45.5 amps. Referring to the relevant manual, a wire with a cross-sectional area of 6 square millimeters is required. Answer: A wire with a 6 square millimeter cross-sectional area is needed. For copper-core plastic-insulated wires rated at 500V or less, installed in air at a working temperature of 30°C, the current-carrying capacity under continuous 100% load is as follows: 1.5 square millimeters – 22A; 2.5 square millimeters – 30A; 4 square millimeters – 39A; 6 square millimeters – 51A; 10 square millimeters – 70A; 16 square millimeters – 98A. For copper-core wires: Cross-sectional area of the wire… Allowable continuous current… 2.5 square millimeters (16A–25A); 4 square millimeters (25A–32A); 6 square millimeters (32A–40A). For aluminum-core wires: Cross-sectional area of the wire… Allowable continuous current… 2.5 square millimeters (13A–20A); 4 square millimeters (20A–25A); 6 square millimeters (25A–32A). // Example: // 1. Each computer consumes approximately 200–300 watts (about 1–1.5A). Therefore, 10 computers would require a copper-core wire with a 2.5 square millimeter cross-sectional area to supply power; otherwise, a fire might occur. 2. Three large air conditioners consume approximately 3000W of power (about 14A), so each air conditioner requires its own copper wire with a cross-sectional area of 2.5 square millimeters to supply power. 3. The wiring used for power supply to homes these days is generally 4 square millimeter copper wire; therefore, the total power consumption of household appliances connected simultaneously should not exceed 25A (that is, 5500 watts). It is pointless for some people to replace the wires in their homes with 6 square millimeter copper wire, since the wire leading to the electricity meter is already 4 square millimeters in diameter. 4. For early-era housing (15 years ago), the incoming wire was usually 2.5 square millimeters of aluminum wire; therefore, no more than 13A (i.e., 2800 watts) of power could be used by household appliances turned on at the same time. 5. Home appliances with high power consumption include: air conditioners at 5A (1.2 horsepower), electric water heaters at 10A, microwaves at 4A, rice cookers at 4A, dishwashers at 8A, washing machines with drying function at 10A, and electric kettles at 4A. In fires caused by electrical issues, 90% are resulting from overheating of connections; therefore, all connections must be welded. Contact components that cannot be welded need to be replaced every 5–10 years (such as sockets and circuit breakers). . . . . . . . . . . . . . . . The long-term current permitted by national standards is 25–32A for 4 square millimeters, and 32–40A for 6 square millimeters. . . . . . . . . . . . . . In fact, these are all theoretical safety values; the actual limit values are higher than these. For a copper wire with a cross-sectional area of 2.5 square millimeters, the maximum power that can be used is 5,500 watts. For a wire with a cross-sectional area of 4 square millimeters, it’s 8,000 watts; for one with a cross-sectional area of 6 square millimeters, 9,000 watts is no problem at all. A digital meter rated for 40 amps can handle 9,000 watts without any issues, and even mechanical meters can handle 12,000 watts without being damaged.\