RU2560318C2 - Clean room air conditioning method - Google Patents

Abstract

FIELD: heating.

SUBSTANCE: invention relates to clean room air conditioning methods. A clean room air conditioning method using a straight-through conditioning circuit is characterised by the fact that air heating is performed by a refrigerating machine working in a heat pump mode during a warm season due to cooling of ambient air to the temperature below dew point for the purpose of drying to the required moisture content; with that, excess heat is used in a heat supply system, and during a cold season due to heat of the air removed from the room.

EFFECT: improvement of energy efficiency of straight-through air conditioning systems of clean rooms.

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Description

The invention relates to methods for air conditioning in cleanrooms.

A known method of air conditioning of the premises [Ananyev V.A., Balueva L.N., Halperin A.D. and other ventilation and air conditioning systems. Theory and practice. - M .: "Euroclimate" publishing house "Arina", 2000. - 416 p.] Using a channel air conditioner with air recirculation, which allows to reduce energy consumption by mixing recirculated air from the room.

The disadvantage of this method is that for cooling air, energy is expended on the refrigeration machine of the indoor unit and energy is expended on heating the air with an electric air heater. And also contaminated recirculated air from the room is mixed in, which is unacceptable for some types of clean rooms.

There is also a method of air conditioning [Ananiev V.A., Balueva L.N., Halperin A.D. and other ventilation and air conditioning systems. Theory and practice. - M .: "Euroclimate" publishing house "Arina", 2000 - 416 p.] Using a roof air conditioner.

The disadvantage of this method is that for cooling the air, energy is expended on the refrigeration machine and energy is expended on heating the air with an electric air heater or hot water.

Closest to the proposed, is a direct-flow air conditioning method of the premises [Shtokman EA, Shilov VA, Novgorodsky E.E. and others. Ventilation, air conditioning and air purification at food industry enterprises. DIA, 2001. - 687 p.]. In the warm period, the outdoor air is cooled and dried at the expense of energy costs for the operation of the chiller, and then heated in the heater due to the energy consumption of electric heating or hot water to a temperature determined by sanitary standards, for supply to an air-conditioned room. In the cold period, the air is heated to a temperature determined by sanitary standards for supply to an air-conditioned room and moistened.

However, to cool the air, energy is expended on the refrigeration machine and energy is expended on heating the air with an electric air heater or hot water.

The technical result of the claimed invention is to increase the energy efficiency of direct-flow air conditioning systems in clean rooms.

The specified technical result is achieved due to the fact that in the method of air conditioning in cleanrooms using a direct-flow conditioning scheme according to the invention, the air is heated by a refrigeration machine operating in the heat pump mode in the warm period by cooling the outside air, while excess heat is used in the system heat supply during the cold period due to the heat of the air removed from the room.

In FIG. Figures 1 and 2 show a scheme for air conditioning in clean rooms in the warm period and a scheme for air conditioning in clean rooms in the cold period, respectively.

Schemes include the following elements: 1 - compressor of the refrigeration machine; 2 - throttle control valve; 3 and 4 - heat exchangers; 5 - humidifier; 6 - fan; 7 - air-conditioned room; 8 - heat supply system; 9 - recuperative heat exchanger.

The method is as follows.

a) The warm period. Outside air with a flow rate of L _n , an enthalpy of I _n and a moisture content of d _H enters the heat exchanger 3, where it is cooled to a temperature below the dew point temperature and drained thereby to the desired moisture content of d _o . The heat is removed by the refrigeration machine due to the operation of the compressor 1, with the energy spent on the drive of the compressor N. The refrigerant vapor from the compressor is directed to the heat exchanger 4 and the regenerative heat exchanger 9, in which they give off heat, heating the air to the temperature required by sanitary standards and the coolant in the heat supply system 8. The condensate of the refrigerant is throttled to 2 and sent to the heat exchanger 3. And the heated air with enthalpy I _{B by the} fan 6 is supplied to the air-conditioned room 7, and then it is vented to the atmosphere.

b) The cold period. The heat exchanger 4 in the cold period is disconnected from the circuit of the refrigeration machine. Outside air with a flow rate of L _n , enthalpy I _n and moisture content d _n enters the heat exchanger 3, where it is heated to the temperature required by sanitary standards, and then adiabatically moistened with water with enthalpy I _w = I _p to the required moisture content d _p. Heat is supplied by refrigeration by the machine due to the operation of compressor 1, with energy costs for compressor N. The refrigerant condensate is throttled at 2 and sent to a recuperative heat exchanger 9. A heated air with enthalpy I in a fan 6 is supplied to an air-conditioned room 7, and then to a recuperative heat exchanger 9, after which the air is removed to the atmosphere or used in technological processes for cooling. Heat from the air is transferred to the refrigerant, the vapor of which is sucked out of 9 by compressor 1.

The schemes provide for the installation of filters for air purification, which for the operation of the air conditioner in this case do not matter and are not conditionally shown.

An example implementation of the method.

Equations of heat balances of an air conditioner:

a) The warm period.

Heat exchanger 3 Q _TO3 = L _n (I _n -I _o ), kW;

Heat exchanger 4 Q _TO4 = L _n (I _p -I _o ), kW;

Since the amount of heat taken from the air in the heat exchanger 3 is greater than the amount of heat given to the air in the heat exchanger 4, for a recuperative heat exchanger Q _PT = Q _TO3 -Q _TO4 , kW, where these values are determined by the equations of the heat balances of the air conditioner:

a) The warm period.

Heat exchanger 3 Q _TO3 = L _n (I _n -I _o ), kW;

Heat exchanger 4 Q _TO4 = L _n (I _p -I _o ), kW,

where I _H - enthalpy of the outdoor air, determined by its design temperature. For example, for the city of Stavropol - I _n = 58 kJ / kg (at a temperature of 25 ° C and a relative humidity of 65%; SNiP 23-01-99 (2003) Construction climatology);

I _o - enthalpy of air at a temperature below the temperature of the "dew point". I _about = 29 kJ / kg (at a temperature of 10 ° C; relative humidity φ = 100% - the process of heating the air in the heat exchanger 4 to the supply air temperature, point "P");

I _p - enthalpy of supply air, is taken at a temperature in the working area: + 22 ° C (± 1 ° C) and relative humidity: 45% (± 5%). I _p = 41 kJ / kg.

Thus, the amount of heat released in the heat exchanger 3 (I _n -I _o = 58-29 = 29 kJ / kg) is approximately 2.4 times the amount of heat absorbed in the heat exchanger 4 (I _p -I _o = 41-29 = 12 kJ / kg). In this case, excess heat is generated in the heat pump cycle, and not due to the recovery of heat from the air removed from the room.

Since in the proposed scheme in the warm period, energy is spent on ensuring the operation of the compressor with power N, as a result of which, in addition to the main task (cooling the supply air in the heat exchanger 3), heat energy Q _{TO4 is generated} in the heat exchanger 4, while the heat exchanger 9 acts as a cooler Q _PT performance.

Saving thermal energy Q _e = Q _TO4 + Q _PT , kW,

b) The cold period.

Heat exchanger 3 Q _TO3 = L _n (I _n -I _n ), kW;

Recuperative heat exchanger Q _PT = L _n (I _in -I _y ), kW;

Because the scheme proposed in the cold period of the energy expended to provide the capacity of the compressor N, whereby, in addition to the main task (heating the supply air in a heat exchanger 3) is made Q _PT thermal energy in an indirect heat exchanger 9.

Saving thermal energy Q _e = Q _PT , kW.

For example, we take a complex as an object, in the clean rooms of which work is carried out with substances that are dangerous to human health (representing a danger of bacteriological or other type of infection).

All main parameters are accepted as for operating rooms (including relative humidity, indoor temperature, etc.)

The parameters are adopted according to SanPiN 2.1.3.1375-03, the air purity of the working area must comply with ISO5 class, according to GOST R ISO 14644-4-2002. For clean rooms of this class, only unidirectional air flow is used in the working area at a speed of 0.2-0.5 m / s.

In premises of this type, only direct-flow ventilation and air conditioning systems are allowed. Air recirculation is not allowed. The air inside the cleanroom (within the boundaries of the working area) must meet the following requirements:

Temperature in the working area: + 22 ° C (± 1 ° C);

Relative humidity: 45% (± 5%).

The applied air distribution system in accordance with the required cleanliness class: unidirectional air flow.

Unidirectional airflow velocity at the outlet of high-performance ceiling filters: 0.25-0.27 m / s.

Area of the working area: 10.5 (3 × 3.5 m).

The density of the supply air (at a temperature of 22 ° C and a relative humidity of 45%) is 1.12 kg / m ³ .

Heat inflows from solar radiation and other external sources (i.e. located outside a clean room) are absent. Since the cleanroom is located within the boundaries of the "external" room and has a high degree of isolation.

Number of people in a clean room: 4 people

The nature of the work performed: hard work.

Heat dissipation from one person: Q _l = 130 W / person.

Heat influx (explicit), including all the heat gain indoors, with the exception of heat gain from people: Q _T = 6 kW.

The air performance of the supply and exhaust ventilation and air conditioning systems is determined by the conditions for removing harmful substances from the area where people are located, affecting the formation of determining air parameters. The hazards may include: heat and moisture.

When calculating the performance of air conditioning and ventilation systems, it is necessary to calculate the required amount of air for:

- removal of heat surplus (typical for residential and public buildings);

- removal of harmful substances present in the air (CO ₂ , etc.);

- removal of moisture surplus;

Clean rooms with unidirectional air flow are characterized by high supply air flow. In this regard, the determining factor in the calculation of the supply air is the calculation of the required amount of air to ensure a given class of cleanliness.

The required volumetric flow rate L to ensure a given purity class can be calculated as follows:

where F is the area of ceiling high-performance filters through which air is supplied to the working area of a clean room, m ² ;

ν - normalized air velocity at the outlet of high-performance filters, m / s.

Thus, the required volumetric flow rate for the case under consideration will be equal to:

(for the convenience of calculations, we take 10000 m ³ / h)

Determine the required mass air flow G to ensure a given purity class:

where L is the volumetric air flow;

ρ is the density of air in a clean room.

Mass air flow equal to:

G = 10000-1.12 = 11200, kg / h.

Let us determine the temperature difference of air at the inlet and outlet of a clean room. To do this, we express the temperature difference from the formula for finding the required air flow for the assimilation of heat surplus:

Express the temperature difference:

where (t _y -t _p ) is the temperature difference at the outlet and inlet of air from the room, ° C;

G - mass air flow, kg / h;

Q _i - obvious heat surpluses inside a clean room, W;

with _in - the heat capacity of the air, taken equal to 1.005 kJ / (kg · K).

where Q _{l - I} - explicit heat gain from people, W;

Q _T - total heat inflow from lighting, equipment, technological processes, etc., W.

In the working area, a temperature of 22 ° C must be maintained. At the same time, the working area itself is, as a rule, at a distance of 1-1.5 m from the floor level of a clean room. As a result, we assume that the air temperature at the entrance to the clean room (at the level of ceiling filters) is 21 ° С, and at the exit (at the floor level) - 23 ° С. This ensures a difference of 2 ° C.

Heat and moisture surplus in a clean room depend only on internal conditions, which do not change depending on the season of the year.

Thus, the total angular beam of the process can be defined as:

Thermal energy savings of 144 kW, while the cost of electricity to ensure the operation of the compressor is 30.4 kW.

Moreover, the cooling capacity of the cooler Q _TO3 = 113.2 kW is approximately 2.7 times higher than the heat capacity of the heat exchanger Q _TO4 = 40.9 kW (excess heat used in the heat supply system).

Saving thermal energy of 93 kW at the cost of electricity to ensure the operation of the compressor 24.9 kW.

Claims (1)

The method of air conditioning in clean rooms using a direct-flow conditioning scheme, characterized in that the air is heated by a refrigeration machine operating in the heat pump mode in the warm period by cooling the outside air to a temperature below the dew point temperature, for drying, due to this, to the required moisture content , while the excess heat is used in the heat supply system, and in the cold period - due to the heat of the air removed from the room.

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