How to calculate the mining machine load

1.

The normal working hours are 20 hours a day. The peak time is 24 hours a day, calculated as 21 hours a day, 26.5 days a month, and 26.5 days a month × 21 = 556.5 (hours), 15000 / 556.5 = 27 (pieces / hour) the load of the equipment is 27 per hour. When the demand coefficient method and binomial method are used to calculate the load, they should be converted to the load ration rate ε Is the active power at 25%

formula: 225.0rnp εε== KW (5-2-1)

in load calculation, the electrical equipment should be divided into different groups according to its nature, and then the equipment power should be determined. The rated power R P or rated capacity r s of electrical equipment refers to the data on the nameplate. The rated power or rated capacity under different load rations should be converted to the active power under unified load rations, i.e. the equipment power NP

the equipment power of short-term or periodic working motor (such as crane motor) refers to the conversion of rated power into active power under uniform load ration rate

extended data:

equipment (some common equipment) load formula:

1, gantry hoist (one)

demand coefficient KX = 0.7 cos Φ= 75, TG Φ= 0.88

Pj=2Kx × Pe=2 × zero point seven × twelve ×= 10.6(KW)

Qj=Pj × tg Φ= ten point six × 0.88 = 9.35 (kV)

2, winch (one)

demand coefficient KX = 0.7 cos Φ= 75, TG Φ= 0.88

Pj=Kx × Pe=0.7 × 7.5=5.25(KW)

Qj=Pj × tg Φ= five point two five × 0.88 = 4.62 (kV)

3, lighting

demand coefficient KX = 1 cos Φ= 1 TG Φ= 0

Pj=Kx × Pe=1 × 10 = 10 (kw) (incandescent lamp, iodine tungsten lamp)

PJ = 1.2kx × Pe=1.2 × one × 3.5 = 4.2 (kw) (dysprosium lamp)

QJ = PJ × tg Φ=( 10+4.2) × 0 = 0 (kV) (all lighting)

the calculated load is a hypothetical continuous load, and its thermal effect is equal to the maximum thermal effect proced by the actual variable load at the same time. In power distribution design, the maximum average value of 30 minutes is usually used as the basis for selecting electrical appliances or conctors according to heating conditions

2. The main calculation formula is: Active Power: P30 = PE · KD, reactive power: q30 = P30 · tan φ
apparent power: s3o = P30 / cos φ
calculation current: i30 = S30 / √ 3un
where: PE is the equipment capacity and KD is the demand coefficient, that is, the demand coefficient of the electrical equipment group, which is the ratio of the half-hour maximum load of the electrical equipment group to its equipment capacity. cos φ Is the average power factor of the consumer group, and UN is the rated voltage of the consumer group.

3.

1. Need coefficient method. The calculated load can be directly calculated by multiplying the equipment power by the demand factor and the simultaneous factor. This method is simple and widely used, especially for load calculation of distribution and substation

2. The average load of the maximum load shift is calculated by using the utilization factor, and then the calculated load is obtained by multiplying the maximum factor related to the number of effective units with the influence of the number of equipment and power difference. The theoretical basis of this method is probability theory and mathematical statistics, so the calculation results are close to the reality. It is suitable for all kinds of load calculation, but the calculation process is a little complicated

Power per unit area, index per unit and power consumption per unit proct. The former two are mostly used in civil buildings, while the latter is used in some instrial buildings. When the power and number of electrical equipment cannot be determined, or in the early stage of design, these methods are the main methods to determine the equipment load

In addition to the above methods, there are binomial method, ABC method and variable demand coefficient method. Some of these methods have been replaced by other methods, some are simplified by coefficient method, and some are not popularized because of few practical data

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extended data:

the main calculation formula is: Active Power: P30 = PE · KD, reactive power: q30 = P30 · tan φ

apparent power: s3o = P30 / cos φ

calculation current: i30 = S30 / √ 3un

where: PE is the equipment capacity, KD is the demand coefficient, that is, the demand coefficient of the electrical equipment group, which is the ratio of the half-hour maximum load of the electrical equipment group to its equipment capacity. cos φ Is the average power factor of the consumer group, and UN is the rated voltage of the consumer group

4. 1. Load rate is defined as the percentage of the average load in a period of time to the maximum load in that period
2. Daily load rate can be calculated according to the following formula:
load rate = (average value of daily load curve / maximum value of daily load curve) × 100%
load rate refers to the percentage of the ratio of the average load to the maximum load in the statistical period (day, month and year)
vehicle load rate: the ratio of the maximum effective power of the internal combustion engine at the same speed is called the load rate, expressed as a percentage. Load rate is usually referred to as load.

5.

Load calculation formula

the main calculation formulas are: Active Power: P30 = PE · KD, reactive power: q30 = P30 · tan φ

apparent power: s3o = P30 / cos φ

calculation current: i30 = S30 / √ 3un

where: PE is the equipment capacity, KD is the demand coefficient, that is, the demand coefficient of the electrical equipment group, which is the ratio of the half-hour maximum load of the electrical equipment group to its equipment capacity. cos φ Is the average power factor of the consumer group, and UN is the rated voltage of the consumer group

6.

Actual capacity / rated capacity of transformer * 100% = load rate

for servo motor, load rate refers to the ratio of working time / (working time + non working time) of motor in each working cycle

if the ty cycle is low, the motor is allowed to run at 3 times of continuous current for a short time, so that it can obtain greater force than the rated continuous operation

extended data:

Load refers to the electronic components connected at both ends of the power supply in the circuit. Circuit should not be no load and directly connected to the two poles of the power supply, this connection is called short circuit

commonly used loads are resistors, engines, bulbs and other power consuming components. A device that converts electrical energy into other forms of energy is called a load

electric motor can convert electric energy into mechanical energy, resistance can convert electric energy into heat energy, light bulb can convert electric energy into heat energy and light energy, loudspeaker can convert electric energy into sound energy. Motors, resistors, light bulbs, loudspeakers, etc. are called loads. Crystal triode for the front of the signal source, can also be regarded as a load. The most basic requirement of load is impedance matching and the power it can bear

for the communication power supply, the load is the communication equipment. For example, a multimedia server needs communication power to provide it with power, and the multimedia server is the load of the communication power. Our common communication equipment is the load of communication power supply, such as optical transmission equipment, switching equipment, microwave equipment, core network equipment, communication base station and so on

7. For the power grid, the capacity to load ratio is used to analyze the substation capacity
the estimation formula of variable capacitance load ratio is:

rs = K1 · K4 / K2 · K3

where:
RS is the capacitance load ratio (kVA / kW), & 57347
K1 is the load dispersion coefficient, 57347
K2 is the average power factor
K3 is the operation rate of transformer, 57347
K4 is the reserve coefficient

the above parameters can be retrieved according to the actual situation, but there are many related factors< The capacitance load ratio of urban power grid transformer is generally:
220kV power grid 1.6-1.9,
35-110kv power grid 1.8-2.1.

8. The total power is 107.55kw, and the working current is about 204a. However, all electrical appliances may not be used at the same time. According to the 50% utilization rate, the current is about 102a. Therefore, the meter reads 3x20 (80A), 3x220 / 380, which has been overloaded Generally speaking, it should be calculated according to the actual utilization rate of electrical appliances. The general sum is not accurate.)

9. The combination of all outgoing line loads (power factor, demand factor, etc., plus reserve capacity, etc.) is the load applied to the power supply department.

10. 1. Considering the principle of three-phase load balance, the air conditioner and water heater should be connected to three phases a, B and C respectively, that is, a and B are respectively equipped with 17 air conditioners, and C is only equipped with 16 air conditioners
2. Similarly, a only has 16 water heaters, B and C all have 17 water heaters
3. According to the above connection method, the installation capacity of phase a = 17 * 2000 + 16 * 1500 = 58000w; Phase B installation capacity = 17 * 2000 + 17 * 1500 = 59500w; C phase installation capacity = 16 * 2000 + 17 * 1500 = 57500w
4. Calculate the line according to phase B with large load
5. Consider [demand coefficient] 0.85 and [simultaneous coefficient] 0.85 (select according to the use situation)
6. Consider the power factor of 0.8
7. Calculated current IJs = (59500 * 0.85 * 0.85) / (220 * 0.8) = 244.3a
8. Select (4 + 1) cable with phase line section of 120, that is yjv-4 * 120 + 1 * 70< The current transformer ratio is 300 / 5.
supplement:
this shows that the actual demand coefficient and simultaneous coefficient are relatively low, about 0.5 ~ 0.6
it is assumed that 0.85 is taken in the previous calculation. In fact, the power distribution of civil buildings is usually between 0.45 and 0.75. If there are many similar equipment, the corresponding coefficient is relatively small.