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Equipment utilization rate

If the machine was supposed to work 160 hours in a given month, and practically due to downtime not provided for by the plan of loss of working time, it worked 150 hours, then the equipment utilization rate in time (extensive load factor) is 93.8% (6.2% - loss of machine time). It is important to ensure the operation of the equipment not only without downtime, but also with the installed capacity and productivity.

If the machine, according to the standards, must process six parts of the same type per hour, but in fact only five are processed, then the equipment utilization factor in terms of power (intensive load factor) is 83.3%. (5: 6 = 0.833). The use of the power of the equipment depends on its condition, timely and high-quality care, on the qualifications and diligence of workers.

The equipment utilization factor in terms of the volume of work (integral load factor) reflects both the time and the degree of utilization of its power and is equal to the ratio of the volume of products actually produced on it to the planned volume that should be obtained during operation without downtime and with the installed capacity. If, according to the plan, it is planned to process 960 parts per month on the machine, but actually processed 750, then the generalizing integral equipment utilization factor is 78.1% (the product of the equipment utilization factors in terms of time and power: 0.938X0.833). Increasing the utilization rate of equipment is the most important prerequisite for intensifying production and increasing output at existing facilities.

At the 27th Party Congress it was noted: “Planning and economic bodies, collectives of enterprises must do everything possible to ensure that the created capacities operate at the design level. Only in heavy industry would it be possible to almost double the rate of increase in production ”(Materials of the XXVII Congress of the CPSU, p. 41). The increase in the utilization rate of equipment is achieved by eliminating downtime, increasing the shift rate, improving preventive repair and maintenance of equipment, strengthening labor discipline, and improving the qualifications of workers. Decommissioning and sale of low-productivity, unloaded equipment based on certification of workplaces also contributes to an increase in the utilization rate of equipment.

Productive capacity- the maximum possible annual output of products with the optimal use of production equipment, is determined for the entire range of products.

The capacity for this type of product is determined by the minimum capacity of the leading shop, the capacity of the leading shop is determined by the minimum capacity of the department or section, the capacity of the section is determined by the capacity of the leading equipment. The leading workshops and departments include those in which the main technological processes and operations are carried out. Leading equipment power:

M =ngT ef

where n is the number of pieces of equipment;

g - hourly productivity of each piece of equipment;

Teff - effect, equipment operating time

Where K n is the calendar number of days in a year;

B - the number of days off and holidays in the planned period;

C is the number of shifts per day;

D is the duration of the shift in hours. If necessary, losses for equipment overhaul are taken into account.

P p - the percentage of planned current downtime

During the planned year, production capacities can be introduced and retired, therefore, in order to determine the volume of production for the planned year, it is necessary to calculate the average annual capacity:

M srg = M n + M Vv - M select 

where M n - capacity at the beginning of the year;

М Вв - newly commissioned power;

M select - retiring power;

k - the number of months of work during the year.

23. Indicators of the use of production capacity

Generalizing indicators of the use of production capacity are:

Power utilization factor (K them), as the ratio of the actual volume of output (gross, marketable) to the average annual production capacity (PM).

To them = Vproduction / PM. (one)

2. The equipment load factor (Кз), as the ratio of the labor intensity of the production program (∑ Т) to the planned fund of the operating time of all equipment (Фп * К).

Кз = ∑ Т / Фп * К. (2)

3. The coefficient of equipment replacement (Ks), as the ratio of the labor intensity of the production program (∑ T) to the planned fund of the equipment operation time per shift (F 1s K).

Ks = ∑ T / F 1s K. (3)

4. An integral indicator of the utilization of production capacities (Ki), as the product of the equipment utilization rates in terms of time and capacity.

Ki = Kw * Km. (4)

5. The coefficient of proportionality of capacities, which is calculated as the ratio of the production capacity of the shop to the production capacity of the plant (the capacity of the shop and the site).

The analysis of the use of production capacity is carried out using the named indicators, which are calculated according to planned and actual data. The object of analysis should be all units, production sites, workshops and the plant as a whole.

24. The concept and structure of OPF

Fixed assets - these are material values ​​(means of labor) that repeatedly participate in the production process, do not change their natural material form and transfer their value to finished products in parts as they wear out. By functional purpose, the fixed assets of the enterprise are divided into production and non-production.

Production assets directly or indirectly related to the production of products. Non-productive assets serve to meet the cultural and everyday needs of workers

Composition and classification of fixed assets:

About 70% of all electricity generated in our country is consumed by receivers of industrial enterprises. Receivers of electrical energy are devices, units, mechanisms designed to convert electrical energy into another type of energy. The power that the load receives is the product of voltage and current, corrected for capacity utilization. The latter, one way or another, is related to the number of phases.

For information. An AC electrical system has a characteristic line-to-line or phase-to-neutral voltage. In office buildings, the phase voltage is 220 V. In factory shops, the line voltage (for example, to start a pump motor) is usually 460 V. Some production capacity is “single-phase”, some is “three-phase”.

Currently, industrial enterprises are supplied with three-phase alternating voltage. Line and phase voltages are usually different from each other in any case.

The central axiom of circuit theory is that power is proportional to the product of voltage and current. The higher the load current, the more electrical power it receives. In the case of a pump, the more current it consumes, the more liquid it can pump, thereby increasing technical indicators, including production capacity.

The problem, however, arises from the fact that electricity is transmitted to consumers by alternating, not direct current. This brings some important advantages to several types of electrical machines, but it also has some disadvantages.

One disadvantage is that the current must remain in phase with the voltage. If it lags behind the phase, then the power to the load will be less than it should be. In theory, the current could alternate with the phase with similar inefficiency, but the lagging case is more typical, so the lagging case is considered more often.

In an AC voltage system, the current follows in a wave-like manner, as the voltage changes over a period of time. But if the current does not peak at the same time as the voltage, then the power will be provided to a lesser extent than it should. The picture shows an example of a graph of current (red sine wave) and voltage (blue sine wave) for an inductive load.

Indeed, if the current lags behind the voltage by a quarter of a cycle (only 1/240 seconds), it gives no real power at all. It will take a fairly intensive review of trigonometry to explain this issue in fine analytical detail, but in general it is not so difficult to understand it based on the relationship formulas and the ratios of physical quantities.

Interconnection of circuit parameters

The power that is actually consumed in the circuit is called active or real. It is denoted by P. Wattmeters indicate the active power of the circuit. The current in phase with the voltage generates true (active) power. Therefore, the formula for the calculation looks like this:

P = U * I * cos φ.

Active power produces heat in heaters, torque in motors, light in lamps and is expressed in watts or kilowatts. The reactive component of the current (i.e., I * sin φ), when multiplied by the circuit voltage, results in reactive power, which is denoted Q. Therefore, this physical quantity is equal to:

Q = U * I * sin φ

and it is expressed in VAR (reactive volt amperes) or KVAR (reactive kilovolt amperes). Reactive power does no useful work in the circuit: it is supplied by the source during the first half cycle and returned to the source during the next half cycle. It is this parameter that determines the cos φ.

The product of the rms current and voltage is called the total power S, which is measured in VA (volt-amperes) or KVA (kilovolt-amperes) and is calculated by the formula:

Power utilization factor

This parameter of the AC circuit is defined only as the cosine of the angular displacement between voltage and current. Namely:

1. In the case of a pure resistive circuit, the alternating current is in phase with the applied voltage, i.e. φ = 0. Therefore, cos φ of a pure resistive circuit is equal to 1;
2. In the case of a pure capacitive or pure inductive circuit, the 90o current is out of phase with the circuit voltage, i.e. φ = 90o. Therefore, the cos φ of the circuit is zero.

In the case of inductive loads (such as motors, transformers ... whatever has windings), the current will lag behind the applied voltage. For capacitive loads (capacitors), the current will be ahead of the applied voltage.

Important! The power factor of the RLC circuit is between 0 and 1 and can never be greater than unity. In practice, cos φ always appears, because most of the loads used are inductive in nature. In the AC voltage circuits of the power system, cos φ plays a rather significant role.

Since the power of the chain is determined by the ratio:

P = U * I * cos φ or I = P / (U * cos φ),

then at a fixed power at a constant voltage, the current increases with decreasing cos φ.

Important! Cos φ is an important factor in power generation, distribution and transmission. This is the fraction of the maximum possible power that the current provides due to the voltage delay.

Low cos φ problems

The cos φ parameter is very important for every power system or company as it helps to maintain an inductive load. At cos φ less than unity, the “missing” power, known as reactive power, increases. The latter is necessary to provide the magnetizing field required for motors and other inductive loads that perform their functions.

Poor cos φ is usually the result of a significant phase difference between voltage and current at the load terminals, or it may be due to high harmonic content or distorted current waveforms.

Power factor:

• 100% is ideal and occurs when the current is not lagging behind the voltage;
• 90% is generally considered acceptable;
• 80% applies depending on the application;
• less than 80% is usually difficult.

Cos φ is 80%, which means that 80% of the power is actually delivered. What happens to the other 20%? The remaining 20% ​​are not lost, they remain in the system. This is small but can damage the bearings of the electric motor and generator. If cos φ = 100% is needed, then to correct the coefficient, 125% of the required current is gained to make up the difference.

The main disadvantages of low cos φ in an alternating voltage circuit can be noted:

• conductors must carry more current at the same power, so they require a larger cross-sectional area;
• conductors must carry more current for the same power, which increases losses and leads to low system efficiency;
• voltage drop increases resulting in poor system regulation.

The problem with low cos φ is that it forces the load to pull additional current. The latter requires heavier wires that are expensive. The total capacity increases, which means that the utility has to provide more capacity. Therefore, the utility is billing industrial customers with poor cos φ.

A cable line with poor cos φ has a bad effect on conductors that become hot and heat dissipation is high. This forces the utility company to generate more electricity to offset consumer demand. The cost of electricity will increase, the cost of equipment will also increase. If it is possible to increase the cos φ, then only the fine and all these problems can be avoided.

Important! An uncorrected power factor results in power system losses in the distribution system. As losses increase, you may encounter a voltage drop. Excessive voltage drops can cause overheating and premature failure of motors or other inductive equipment. Thus, by increasing the cos φ, voltage drops are minimized. This allows the motors to run more efficiently with a slight increase in power and starting torque.

Low cos φ solution

Understanding power factor is very simple when you understand the nature of inductance and capacitor. Power factor is only observed in inductive or capacitive circuits. As far as production is concerned, cos φ is usually corrected for it by adding capacitors.

In the interest of reducing losses in the distribution system, power factor correction is added to neutralize some of the magnetizing current of the motor. Typically, the corrected power factor will be 0.92-0.95.

For information. An inductive load requires a magnetic field to operate, and when such a magnetic field is created, the current will be out of phase with the voltage. Power factor correction is the process of compensating for a lagging current by creating a drive current by connecting capacitors to a power source.

Electrical equipment and machines connected to the power system, such as transformers, switching mechanisms, alternators, usually have lower cos φ values. To increase this indicator of the AC circuit, a capacitor is connected in parallel with the circuit. In the case of a DC circuit, cos φ is zero, since the inductive and capacitive reactance are zero due to zero frequency.

It is preferable to use a switched capacitor unit in the system. Thus, a switched capacitor unit is usually installed in the primary network of a power substation, which also helps to improve the capacity of the entire system. The capacitor bank can automatically turn on and off based on the status of various system parameters.

When the power factor of the system is below the set value, the bank will automatically turn on to improve the power factor. The function of the capacitor bank is to compensate or neutralize the reactive power of the system.

The installed capacity utilization factor is the most important characteristic of the efficiency of the electric power industry enterprises. Any system with a cos φ close to 1 is considered a good or excellent system, while any system with a cos φ close to 0 (e.g. 0.2, 0.3, 0.4, 0.5, 0.6) is considered a bad system, for which the organization has to pay something as a penalty to the utility company, because this imposes serious costs on the power supply side.

Video

Equipment statistics

In the structure of fixed assets, a large place belongs to machines and equipment, as an active part of fixed assets.

Equipment classification:

1. By types:

a. Energy equipment is machines and devices for the production of various types of energy from natural resources and for the conversion of some types of energy into others.

Primary

Secondary

b. Production equipment is an instrument of labor, with the help of which a direct impact on the object of labor is carried out, with the aim of transforming it into a product necessary for society.

Mechanical equipment

Thermal

Chemical

Equipment included in each of these groups can be subdivided according to the following criteria:

1) By the nature of specialization:

Universal

Specialized

2) By scope:

For use in specific industries

Diversified application

3) By the degree of automation:

Hand-operated (foot-operated) machines

Machines without a rigid worker connection

Body with the subject of labor

4) By the type of material processing:

Metalworking

Woodworking

5) By the degree of technical improvement:

Technically perfect

Not perfect enough

Obsolete, requiring modernization

6) By technical condition:

Serviceable, fit for work

Requiring overhaul

Write-off, unusable

7) By affiliation:

Domestic

Imported

8) By age:

10 and more years

The availability of equipment is characterized by an indicator of its number by category of equipment.

1. Use indicators, by number:

- Utilization rate of available equipment- share of working equipment in the total number of available equipment

- Installed equipment utilization rate- share of operating equipment in the total number of installed equipment

2. Equipment utilization rate over time:

- Shift ratio- shows how many shifts each piece of equipment worked on average during the day. It is calculated for both installed and actually operating equipment.

Kcm. = (total number of shifts worked by all pieces of equipment for the period) / (number of machine-days)

Number of machine days = average number of pieces of equipment * number of days of work of the enterprise in this period

In addition to the shift factor, the shift mode utilization rate- the ratio of the shift ratio to the number of shifts of the enterprise according to the established regime

To isp. see dir. = (shift ratio) / (number of shifts) * 100%

- Extensive utilization rate- is calculated as the ratio of the time actually worked by the equipment to one of the time funds (calendar, routine or scheduled)

To ex. = (actually worked time) / (working time fund (calendar, routine, planned)) * 100%

Calendar fund of time (KFV)- the number of calendar hours in the period attributable to all units of installed equipment,

e.g. KF per year = calendar days (365) * 24 hours * number of pieces of equipment

Regime fund of time (RFV)= Calendar time fund reduced by off-shift times, holidays and weekends, and

Regime fund of time = duration of a shift * number of shifts * number of working days * number of pieces of equipment

Planned time fund (PFV)= RFV - scheduled repair time - standby time

Kekst. = Тfact. / Тmax * 100%,

Tfact - actually worked time

T max - the maximum fund of time (calendar, regime or planned)

The Extensive Load Ratio shows the proportion of hours actually worked in the total Time Fund.

Difference (100% -Kext.) reflects the proportion of unused time due to downtime, repairs and other reasons.

3.Equipment utilization rate by capacity:

- Equipment load intensity factor- shows the degree of use of the technical capabilities of the equipment per unit of time.

To int. = (average actual capacity of the equipment) / (potential capacity (i.e., nameplate or planned))

Kint. = Msr / M max

Difference ( 100% -Kint.) reflects the reserves for the growth of output or energy production per unit of time.

4. Indicator of equipment utilization by volume of work:

1) Integral load factor- gives a comprehensive description of the use of equipment both in time and in power. It is calculated as the ratio of the actually performed amount of work to the maximum possible amount of work for the billing period.

To integra. Load = Qact / Qmax,

where Q is the volume of products produced, or processed raw materials, or produced energy.

To integra. = To ext. * To int.

5. The indicator of production capacity is used as a generalizing indicator of the production potential of an enterprise.

Production capacity of the enterprise- this is the maximum possible volume of annual output of products or processed raw materials with the full use of production equipment under the conditions of a certain range of products and the operating mode of the enterprise. It is determined both in kind and in value terms.

Production capacity utilization indicator = (volume of actually manufactured products per year or processed raw materials) / to the average annual production capacity.

Average annual production capacity determined by the formula:

where Мвв - capacities commissioned during the year

Мvyb - capacities retired during the year,

Т1, Т2 - the number of months, respectively, from the moment of commissioning and disposal of capacity until the end of the year.