Heating project for a three-story school. The heating system of a school, kindergarten, educational institutions - organization and reconstruction with all approvals. Technology of installation of elements of the heat supply system

Ñîäåðæàíèå

Introduction

Calculation of heating, ventilation and hot water supply of a school for 90 students

1.1 a brief description of schools

2 Determination of heat loss through the outer fences of the garage

3 Calculation of the heating surface area and selection of heating devices of central heating systems

4 Calculation of school air exchange

5 Selection of heaters

6 Calculation of heat consumption for hot water supply of a school

Calculation of heating and ventilation of other objects according to the given scheme No. 1 with centralized and local heat supply

2.1 Calculation of heat consumption for heating and ventilation according to the aggregated standards for residential and public facilities

2.2 Calculation of heat consumption for hot water supply for residential and public buildings

3.Construction of the annual schedule of heat load and selection of boilers

1 Building an annual heat load graph

3.2 Choice of heat transfer medium

3 Boiler selection

3.4 Construction of an annual schedule for regulating the supply of a thermal boiler house

Bibliography

Introduction

The agro-industrial complex is an energy-intensive branch of the national economy. A large number of energy is spent on heating industrial, residential and public buildings, creating an artificial microclimate in livestock buildings and structures of protective ground, drying of agricultural products, production of products, obtaining artificial cold and for many other purposes. Therefore, the energy supply of agricultural enterprises includes a wide range of tasks associated with the production, transmission and use of thermal and electrical energy using traditional and non-traditional energy sources.

In this course project, a variant of the integrated energy supply of the settlement is proposed:

· for a given scheme of agro-industrial complex objects, an analysis of the need for thermal energy, electricity, gas and cold water is carried out;

Calculation of loads of heating, ventilation and hot water supply;

· the necessary power of the boiler house is determined, which could meet the needs of the economy in heat;

Boilers are selected.

calculation of gas consumption,

1. Calculation of heating, ventilation and hot water supply of a school for 90 students

1.1 Brief description of the school

Dimensions 43.350x12x2.7.

The volume of the room V = 1709.34 m 3.

External longitudinal walls - load-bearing, are made of facing and finishing, thickened bricks of the KP-U100 / 25 brand in accordance with GOST 530-95 on cement - sandy mortar M 50, 250 and 120 mm thick and 140 mm of insulation - expanded polystyrene between them.

Internal walls - are made of hollow, thickened ceramic bricks of grade KP-U100/15 according to GOST 530-95, on mortar M50.

Partitions - are made of brick KP-U75/15 according to GOST 530-95, on mortar M 50.

Roofing - roofing felt (3 layers), cement-sand screed 20mm, expanded polystyrene 40mm, roofing felt in 1 layer, cement-sand screed 20mm and reinforced concrete slab;

Floors - concrete M300 and soil compacted with crushed stone.

The windows are double with paired wooden binding, the size of the windows is 2940x3000 (22 pcs) and 1800x1760 (4 pcs).

Exterior wooden single doors 1770x2300 (6 pcs)

Design parameters of outdoor air tn = - 25 0 С.

Estimated winter outdoor air temperature tn.a. = - 16 0 С.

Estimated temperature of the internal air tv = 16 0 С.

The humidity zone of the area is normal dry.

Barometric pressure 99.3 kPa.

1.2 Calculation of air exchange school

The learning process takes place in the school. It is characterized by a long stay of a large number of students. harmful emissions no. The air shift coefficient for the school will be 0.95…2.

K ∙ Vp,

where Q - air exchange, m³/h; Vp - room volume, m³; K - the frequency of air exchange is accepted = 1.

Fig.1. Room dimensions.

Room volume: \u003d 1709.34 m 3 .= 1 ∙ 1709.34 \u003d 1709.34 m 3 / h.

In the room we arrange general ventilation combined with heating. We arrange natural exhaust ventilation in the form of exhaust shafts, the cross-sectional area F of the exhaust shafts is found by the formula: F = Q / (3600 ∙ ν k.in) . , having previously determined the air speed in the exhaust shaft with a height h = 2.7 m

ν k.in. =

ν k.in. = \u003d 1.23 m / s \u003d 1709.34 ∙ / (3600 ∙ 1.23) \u003d 0.38 m²

Number of exhaust shafts vsh \u003d F / 0.04 \u003d 0.38 / 0.04 \u003d 9.5≈ 10

We accept 10 exhaust shafts 2 m high with a living section of 0.04 m² (with dimensions of 200 x 200 mm).

1.3 Determination of heat losses through the external enclosures of the room

Heat losses through the internal enclosures of the premises are not taken into account, because the temperature difference in the shared rooms does not exceed 5 0 C. We determine the resistance to heat transfer of enclosing structures. Heat transfer resistance outer wall(Fig. 1) we find by the formula, using the data in Table. 1, knowing that the thermal resistance to heat absorption of the inner surface of the fence Rv \u003d 0.115 m 2 ∙ 0 C / W

,

where Rv - thermal resistance to heat absorption of the inner surface of the fence, m² ºС / W; - the sum of thermal resistances of thermal conductivity of individual layers of m - layered fence with a thickness of δi (m), made of materials with thermal conductivity λi, W / (m ºС), the values ​​of λ are given in Table 1; Rn - thermal resistance to heat transfer of the outer surface of the fence Rn = 0.043 m 2 ∙ 0 C / W (for external walls and bare floors).

Fig.1 Structure of wall materials.

Table 1 Thermal conductivity and width of wall materials.

Heat transfer resistance of outer wall:

R 01 \u003d m² ºС / W.

) Heat transfer resistance of windows Ro.ok \u003d 0.34 m 2 ∙ 0 C / W (we find from the table on p. 8)

Heat transfer resistance of external doors and gates 0.215 m 2 ∙ 0 C / W (find from the table on p. 8)

) Heat transfer resistance of the ceiling for a non-attic floor (Rv \u003d 0.115 m 2 ∙ 0 C / W, Rn \u003d 0.043 m 2 ∙ 0 C / W).

Calculation of heat losses through floors:

Fig.2 ceiling structure.

Table 2 Thermal conductivity and width of floor materials

Ceiling heat transfer resistance

m 2 ∙ 0 C / W.

) Heat losses through the floors are calculated by zones - strips 2 m wide, parallel to the outer walls (Fig. 3).

The areas of the floor zones minus the basement area: \u003d 43 ∙ 2 + 28 ∙ 2 \u003d 142 m 2

F1 \u003d 12 ∙ 2 + 12 ∙ 2 \u003d 48 m 2, \u003d 43 ∙ 2 + 28 ∙ 2 \u003d 148 m 2

F2 \u003d 12 ∙ 2 + 12 ∙ 2 \u003d 48 m 2, \u003d 43 ∙ 2 + 28 ∙ 2 \u003d 142 m 2

F3 \u003d 6 ∙ 0.5 + 12 ∙ 2 \u003d 27 m 2

Areas of basement floor zones: = 15 ∙ 2 + 15 ∙ 2 \u003d 60 m 2

F1 \u003d 6 ∙ 2 + 6 ∙ 2 \u003d 24 m 2, \u003d 15 ∙ 2 + 15 ∙ 2 \u003d 60 m 2

F2 \u003d 6 ∙ 2 \u003d 12 m 2

F1 \u003d 15 ∙ 2 + 15 ∙ 2 \u003d 60 m 2

Floors located directly on the ground are considered non-insulated if they consist of several layers of materials, the thermal conductivity of each of which is λ≥1.16 W / (m 2 ∙ 0 C). Floors are considered insulated if their insulating layer has λ<1,16 Вт/м 2 ∙ 0 С.

Heat transfer resistance (m 2 ∙ 0 C / W) for each zone is determined as for non-insulated floors, because thermal conductivity of each layer λ≥1.16 W / m 2 ∙ 0 C. So, heat transfer resistance Ro \u003d Rn.p. for the first zone is 2.15, for the second - 4.3, for the third - 8.6, the rest - 14.2 m 2 ∙ 0 C / W.

) The total area of ​​​​window openings: ok \u003d 2.94 ∙ 3 ∙ 22 + 1.8 ∙ 1.76 ∙ 6 \u003d 213 m 2.

The total area of ​​​​external doorways: dv \u003d 1.77 ∙ 2.3 ∙ 6 \u003d 34.43 m 2.

The area of ​​the outer wall minus window and door openings: n.s. = 42.85 ∙ 2.7 + 29.5 ∙ 2.7 + 11.5 ∙ 2.7 + 14.5 ∙ 2.7 + 3 ∙ 2.7 + 8.5 ∙ 2.7 - 213-34 ,43 \u003d 62 m 2.

Basement wall area: n.s.p = 14.5∙2.7+5.5∙2.7-4.1=50

) Ceiling area: sweat \u003d 42.85 ∙ 12 + 3 ∙ 8.5 \u003d 539.7 m 2,

,

where F is the area of ​​​​the fence (m²), which is calculated with an accuracy of 0.1 m² (the linear dimensions of the enclosing structures are determined with an accuracy of 0.1 m, observing the measurement rules); tv and tn - design temperatures of internal and external air, ºС (app. 1 ... 3); R 0 - total resistance to heat transfer, m 2 ∙ 0 C / W; n - coefficient depending on the position of the outer surface of the fence in relation to the outside air, we will take the values ​​​​of the coefficient n \u003d 1 (for external walls, non-attic coverings, attic floors with steel, tiled or asbestos-cement roofing along a sparse crate, floors on the ground)

Heat loss through external walls:

Fns = 601.1 W.

Heat loss through the outer walls of the basement:

Fn.s.p = 130.1W.

∑F n.s. =F n.s. + F n.s.p. \u003d 601.1 + 130.1 \u003d 731.2 W.

Heat loss through windows:

fok = 25685 W.

Heat loss through doorways:

Fdv = 6565.72 W.

Heat loss through the ceiling:

Fpot = = 13093.3 W.

Heat loss through the floor:

Fpol = 6240.5 W.

Heat loss through the basement floor:

Fpol.p = 100 W.

∑F floor \u003d F floor. + Ф pol.p. \u003d 6240.5 + 100 \u003d 6340.5 W.

Additional heat losses through external vertical and inclined (vertical projection) walls, doors and windows depend on various factors. The values ​​of Fdob are calculated as a percentage of the main heat losses. Additional heat loss through the outer wall and windows facing north, east, northwest and northeast is 10%, southeast and west - 5%.

Additional losses for infiltration of outdoor air for industrial buildings are taken in the amount of 30% of the main losses through all fences:

Finf \u003d 0.3 (Fn.s. + Focal. + Fpot. + Fdv + Fpol.) \u003d 0.3 (731.2 + 25685 + 13093.3 + 6565.72 + 6340.5) \u003d 15724, 7 W

Thus, the total heat loss is determined by the formula:

1.4 Calculation of the heating surface area and selection of heaters for central heating systems

The most common and versatile heating devices in use are cast-iron radiators. They are installed in residential, public and various industrial buildings. We use steel pipes as heating devices in industrial premises.

Let us first determine the heat flow from the pipelines of the heating system. The heat flux given off to the room by openly laid non-insulated pipelines is determined by formula 3:

Фfr = Ftr ∙ ktr (tfr - tv) ∙ η,

where Ftr \u003d π ∙ d l is the area of ​​the outer surface of the pipe, m²; d and l - outer diameter and length of the pipeline, m (diameters of main pipelines are usually 25 ... 50 mm, risers 20 ... 32 mm, connections to heating devices 15 ... 20 mm); ktr - heat transfer coefficient of the pipe W / (m 2 ∙ 0 С) is determined according to table 4 depending on the temperature difference and the type of coolant in the pipeline, ºС; η - coefficient equal to the supply line located under the ceiling, 0.25, for vertical risers - 0.5, for the return line located above the floor - 0.75, for connections to the heating device - 1.0

Supply pipeline:

Diameter-50mm:50mm =3.14∙73.4∙0.05=11.52 m²;

Diameter 32mm:32mm =3.14∙35.4∙0.032=3.56 m²;

Diameter-25mm:25mm =3.14∙14.45∙0.025=1.45m²;

Diameter-20:20mm = 3.14∙32.1∙0.02=2.02 m²;

Return pipeline:

Diameter-25mm:25mm =3.14∙73.4∙0.025=5.76 m²;

Diameter-40mm:40mm =3.14∙35.4∙0.04=4.45 m²;

Diameter-50mm:50mm =3.14∙46.55∙0.05=7.31 m²;

The heat transfer coefficient of pipes for the average difference between the water temperature in the device and the air temperature in the room (95 + 70) / 2 - 15 \u003d 67.5 ºС is taken to be 9.2 W / (m² ∙ ºС). in accordance with the data in table 4 .

Direct heat pipe:

Ф p1.50mm = 11.52 ∙ 9.2 (95 - 16) ∙ 1 = 8478.72 W;

Ф p1.32mm \u003d 3.56 ∙ 9.2 (95 - 16) ∙ 1 \u003d 2620.16 W;

Ф p1.25mm \u003d 1.45 ∙ 9.2 (95 - 16) ∙ 1 \u003d 1067.2 W;

Ф p1.20mm \u003d 2.02 ∙ 9.2 (95 - 16) ∙ 1 \u003d 1486.72 W;

Return heat pipe:

Ф p2.25mm \u003d 5.76 ∙ 9.2 (70 - 16) ∙ 1 \u003d 2914.56 W;

Ф p2.40mm \u003d 4.45 ∙ 9.2 (70 - 16) ∙ 1 \u003d 2251.7 W;

Ф p2.50mm \u003d 7.31 ∙ 9.2 (70 - 16) ∙ 1 \u003d 3698.86 W;

Total heat flow from all pipelines:

F tr \u003d 8478.72 + 2620.16 + 1067.16 + 1486.72 + 2914.56 + 2251.17 + 3698.86 \u003d 22517.65 W

The required heating surface area (m²) of devices is approximately determined by formula 4:

,

where Fogr-Ftr - heat transfer of heating devices, W; Фfr - heat transfer of open pipelines located in the same room with heating devices, W; pr - heat transfer coefficient of the device, W / (m 2 ∙ 0 С). for water heating tpr \u003d (tg + tо) / 2; tg and tо - design temperature of hot and chilled water in the device; for low-pressure steam heating, tpr \u003d 100 ºС is taken; in high-pressure systems, tpr is equal to the temperature of the steam in front of the device at its corresponding pressure; tv - design air temperature in the room, ºС; β 1 - correction factor, taking into account the installation method of the heater. With free installation against a wall or in a niche with a depth of 130 mm, β 1 = 1; in other cases, the values ​​of β 1 are taken based on the following data: a) the device is installed against a wall without a niche and is covered with a board in the form of a shelf with a distance between the board and the heater of 40 ... 100 mm, the coefficient β 1 = 1.05 ... 1.02; b) the device is installed in a wall niche with a depth of more than 130 mm with a distance between the board and the heater of 40 ... 100 mm, the coefficient β 1 = 1.11 ... 1.06; c) the device is installed in a wall without a niche and is closed with a wooden cabinet with slots in the top board and in the front wall near the floor with a distance between the board and the heater equal to 150, 180, 220 and 260 mm, the coefficient β 1, respectively, is equal to 1.25; 1.19; 1.13 and 1.12; β 1 - correction factor β 2 - correction factor that takes into account the cooling of water in pipelines. With open laying of water heating pipelines and with steam heating, β 2 =1. for a hidden laying pipeline, with pump circulation β 2 \u003d 1.04 (single-pipe systems) and β 2 \u003d 1.05 (two-pipe systems with top wiring); in natural circulation, due to the increase in cooling of water in pipelines, the values ​​\u200b\u200bof β 2 should be multiplied by a factor of 1.04.pr \u003d 96 m²;

The required number of sections of cast-iron radiators for the calculated room is determined by the formula:

Fpr / fsection,

where fsection is the heating surface area of ​​one section, m² (Table 2).= 96 / 0.31 = 309.

The resulting value of n is approximate. If necessary, it is divided into several devices and, by introducing the correction factor β 3, which takes into account the change in the average heat transfer coefficient of the device depending on the number of sections in it, the number of sections accepted for installation in each heating device is found:

mouth \u003d n β 3;

mouth = 309 1.05 = 325.

We install 27 radiators in 12 sections.

heating water supply school ventilation

1.5 Selection of heaters

Heaters are used as heating devices to increase the temperature of the air supplied to the room.

The selection of heaters is determined in the following order:

We determine the heat flux (W) going to heat the air:

Phv = 0.278 ∙ Q ∙ ρ ∙ c ∙ (tv - tn), (10)

where Q is the volumetric air flow, m³/h; ρ - air density at temperature tk, kg/m³; ср = 1 kJ/ (kg ∙ ºС) - specific isobaric heat capacity of air; tk - air temperature after the heater, ºС; tn - initial temperature of the air entering the heater, ºС

Air Density:

ρ = 346/(273+18) 99.3/99.3 = 1.19;

Fw = 0.278 ∙ 1709.34 ∙ 1.19 ∙ 1 ∙ (16- (-16)) = 18095.48 W.

,

Estimated mass air velocity is 4-12 kg/s∙m².

m².

3. Then, according to Table 7, we select the model and number of the air heater with an open air area close to the calculated one. With a parallel (along the air) installation of several heaters, their total area of ​​\u200b\u200bthe live section is taken into account. We choose 1 K4PP No. 2 with a free air area of ​​0.115 m² and a heating surface area of ​​12.7 m²

4. For the selected heater, calculate the actual mass air velocity

= 4.12 m/s.

After that, according to the graph (Fig. 10) for the adopted heater model, we find the heat transfer coefficient k depending on the type of coolant, its speed, and the value of νρ. According to the schedule, the heat transfer coefficient k \u003d 16 W / (m 2 0 C)

We determine the actual heat flux (W) transferred by the calorific unit to the heated air:

Фк = k ∙ F ∙ (t´av - tav),

where k is the heat transfer coefficient, W / (m 2 ∙ 0 С); F - heating surface area of ​​the air heater, m²; t´av - average temperature of the coolant, ºС, for the coolant - steam - t´av = 95 ºС; tav - average temperature of the heated air t´av = (tk + tn) /2

Fk \u003d 16 ∙ 12.7 ∙ (95 - (16-16) / 2) \u003d 46451 ∙ 2 \u003d 92902 W.

plate heater KZPP No. 7 provide a heat flow of 92902 W, and the required is 83789.85 W. Therefore, heat transfer is fully ensured.

The heat transfer margin is =6%.

1.6 Calculation of heat consumption for hot water supply of the school

The school needs hot water for sanitary needs. The school with 90 seats consumes 5 liters of hot water per day. Total: 50 liters. Therefore, we place 2 risers with a water flow of 60 l / h each (that is, a total of 120 l / h). Taking into account the fact that on average hot water for sanitary needs is used for about 7 hours during the day, we find the amount of hot water - 840 l / day. The school consumes 0.35 m³/h per hour

Then the heat flow to the water supply will be

FGV. \u003d 0.278 0.35 983 4.19 (55 - 5) \u003d 20038 W

The number of shower cabins for the school is 2. The hourly consumption of hot water by one cabin is Q = 250 l / h, we assume that on average the shower works 2 hours a day.

Then the total consumption of hot water: Q \u003d 3 2 250 10 -3 \u003d 1m 3

FGV. \u003d 0.278 1 983 4.19 (55 - 5) \u003d 57250 W.

∑ F year \u003d 20038 + 57250 \u003d 77288 W.

2. Calculation of heat load for district heating

The maximum heat flow (W) consumed for heating residential and public buildings of the village, included in the district heating system, can be determined by aggregated indicators depending on the living area using the following formulas:

Photograph = φ ∙ F,

Photo.l.=0.25∙Photo.l., (19)

where φ is an aggregated indicator of the maximum specific heat flux consumed for heating 1 m² of living space, W / m². The values ​​of φ are determined depending on the calculated winter temperature of the outside air according to the schedule (Fig. 62); F - living area, m².

1. For thirteen 16 apartment buildings with an area of ​​720 m 2 we get:

Photograph \u003d 13 170 720 \u003d 1591200 W.

For eleven 8-apartment buildings with an area of ​​​​360 m 2 we get:

Photograph = 8 ∙ 170 ∙ 360 = 489600 W.

For honey. points with dimensions 6x6x2.4 we get:

Photototal=0.25∙170∙6∙6=1530 W;

For an office with dimensions of 6x12 m:

Photo common = 0.25 ∙ 170 ∙ 6 12 = 3060 W,

For individual residential, public and industrial buildings, the maximum heat flows (W) consumed for heating and air heating in the supply ventilation system are approximately determined by the formulas:

Phot \u003d qot Vn (tv - tn) a,

Fv \u003d qv Vn (tv - tn.v.),

where q from and q in - specific heating and ventilation characteristics of the building, W / (m 3 0 C), taken according to Table 20; V n - the volume of the building according to the outer measurement without the basement, m 3, is taken according to standard designs or is determined by multiplying its length by its width and height from the planning mark of the earth to the top of the eaves; t in = average design air temperature, typical for most rooms of the building, 0 С; t n \u003d calculated winter temperature of the outside air, - 25 0 С; t N.V. - calculated winter ventilation temperature of the outside air, - 16 0 С; a is a correction factor that takes into account the impact on the specific thermal characteristic of local climatic conditions at tn=25 0 С a = 1.05

Phot \u003d 0.7 ∙ 18 ∙ 36 ∙ 4.2 ∙ (10 - (- 25)) ∙ 1.05 \u003d 5000.91W,

Fv.tot.=0.4∙5000.91=2000 W.

Brigade House:

Phot \u003d 0.5 ∙ 1944 ∙ (18 - (- 25)) ∙ 1.05 \u003d 5511.2 W,

School workshop:

Phot \u003d 0.6 ∙ 1814.4 ∙ (15 - (- 25)) 1.05 \u003d 47981.8 W,

Fv \u003d 0.2 ∙ 1814.4 ∙ (15 - (- 16)) ∙ \u003d 11249.28 W,

2.2 Calculation of heat consumption for hot water supply for residential and public buildings

The average heat flow (W) consumed during the heating period for hot water supply of buildings is found by the formula:

F = q yr. · n f,

Depending on the rate of water consumption at a temperature of 55 0 C, the aggregated indicator of the average heat flux (W) spent on hot water supply of one person will be equal to: is 407 watts.

For 16 apartment buildings with 60 residents, the heat flow for hot water supply will be: \u003d 407 60 \u003d 24420 W,

for thirteen such houses - F g.v. \u003d 24420 13 \u003d 317460 W.

Heat consumption for hot water supply of eight 16-apartment buildings with 60 inhabitants in summer

F g.w.l. = 0.65 F g.w. = 0.65 317460 = 206349 W

For 8 apartment buildings with 30 residents, the heat flow for hot water supply will be:

F \u003d 407 30 \u003d 12210 W,

for eleven such houses - F g.v. \u003d 12210 11 \u003d 97680 W.

Heat consumption for hot water supply of eleven 8-apartment buildings with 30 inhabitants in summer

F g.w.l. = 0.65 F g.w. \u003d 0.65 97680 \u003d 63492 W.

Then the heat flow to the water supply of the office will be:

FGV. = 0.278 ∙ 0.833 ∙ 983 ∙ 4.19 ∙ (55 - 5) = 47690 W

Heat consumption for office hot water supply in summer:

F g.w.l. = 0.65 ∙ F g.c. = 0.65 ∙ 47690 = 31000 W

Heat flow for water supply honey. point will be:

FGV. = 0.278 ∙ 0.23 ∙ 983 ∙ 4.19 ∙ (55 - 5) = 13167 W

Heat consumption for hot water supply honey. points in summer:

F g.w.l. = 0.65 ∙ F g.c. = 0.65 ∙ 13167 = 8559 W

In the workshops, hot water is also needed for sanitary needs.

The workshop accommodates 2 risers with a water flow of 30 l/h each (i.e. a total of 60 l/h). Considering that, on average, hot water for sanitary needs is used for about 3 hours during the day, we find the amount of hot water - 180 l / day

FGV. \u003d 0.278 0.68 983 4.19 (55 - 5) \u003d 38930 W

The flow of heat consumed for hot water supply of the school workshop in the summer:

Fgw.l \u003d 38930 0.65 \u003d 25304.5 W

Summary table of heat flows

∑Ф total =Ф from +Ф to +Ф g.v. \u003d 2147318 + 13243 + 737078 \u003d 2897638 W.

3. Construction of an annual heat load schedule and selection of boilers

.1 Building an annual heat load curve

The annual consumption for all types of heat consumption can be calculated using analytical formulas, but it is more convenient to determine it graphically from the annual heat load schedule, which is also necessary to establish the operating modes of the boiler house throughout the year. Such a schedule is built depending on the duration of various temperatures in a given area, which is determined by Appendix 3.

On fig. 3 shows the annual load schedule of the boiler house serving the residential area of ​​the village and a group of industrial buildings. The graph is built as follows. On the right side, along the abscissa axis, the duration of the boiler house operation in hours is plotted, on the left side - the outside air temperature; heat consumption is plotted along the y-axis.

First, a graph is plotted for changing the heat consumption for heating residential and public buildings, depending on the outside temperature. To do this, the total maximum heat flow spent on heating these buildings is plotted on the y-axis, and the found point is connected by a straight line to the point corresponding to the outdoor air temperature, which is equal to the average design temperature of residential buildings; public and industrial buildings tv = 18 °C. Since the beginning of the heating season is taken at a temperature of 8 °C, line 1 of the graph up to this temperature is shown as a dotted line.

The heat consumption for heating and ventilation of public buildings in the function tn is an inclined straight line 3 from tv = 18 °C to the calculated ventilation temperature tn.v. for this climatic region. At lower temperatures, room air is mixed with the supply air, i.e. recirculation occurs, and the heat consumption remains unchanged (the graph runs parallel to the x-axis). In a similar way, graphs of heat consumption for heating and ventilation of various industrial buildings are built. The average temperature of industrial buildings tv = 16 °С. The figure shows the total heat consumption for heating and ventilation for this group of objects (lines 2 and 4 starting from a temperature of 16 °C). Heat consumption for hot water supply and technological needs does not depend on tn. The general graph for these heat losses is shown by straight line 5.

The total graph of heat consumption depending on the outdoor air temperature is shown by a broken line 6 (the break point corresponds to tn.a.), cutting off on the y-axis a segment equal to the maximum heat flow consumed for all types of consumption (∑Fot + ∑Fv + ∑Fg. in. + ∑Ft) at the design outside temperature tn.

Adding the total load received 2.9W.

To the right of the abscissa axis, for each outdoor temperature, the number of hours of the heating season (on a cumulative total) is plotted, during which the temperature was kept equal to or lower than that for which the construction is being made (Appendix 3). And through these points draw vertical lines. Further, ordinates are projected onto these lines from the total heat consumption graph, corresponding to the maximum heat consumption at the same outdoor temperatures. The obtained points are connected by a smooth curve 7, which is a graph of the heat load for the heating period.

The area bounded by the coordinate axes, curve 7 and horizontal line 8, showing the total summer load, expresses the annual heat consumption (GJ / year):

year = 3.6 ∙ 10 -6 ∙ F ∙ m Q ∙ m n ,

where F is the area of ​​the annual heat load schedule, mm²; m Q and m n - scales of heat consumption and operating time of the boiler house, respectively W/mm and h/mm.year = 3.6 ∙ 10 -6 ∙ 9871.74 ∙ 23548 ∙ 47.8 = 40001.67J/year

Of which the share of the heating period is 31681.32 J / year, which is 79.2%, for the summer 6589.72 J / year, which is 20.8%.

3.2 Choice of heat transfer medium

We use water as a heat carrier. Since the thermal design load Fr is ≈ 2.9 MW, which is less than the condition (Fr ≤ 5.8 MW), it is allowed to use water with a temperature of 105 ºС in the supply line, and the water temperature in the return pipeline is assumed to be 70 ºС. At the same time, we take into account that the temperature drop in the consumer's network can reach up to 10%.

The use of superheated water as a heat carrier gives greater savings in pipe metal due to a decrease in their diameter, reduces the energy consumption of network pumps, since the total amount of water circulating in the system is reduced.

Since for some consumers steam is required for technical purposes, additional heat exchangers must be installed at consumers.

3.3 Boiler selection

Heating and industrial boilers, depending on the type of boilers installed in them, can be water-heating, steam or combined - with steam and hot-water boilers.

The choice of conventional cast-iron boilers with a low-temperature coolant simplifies and reduces the cost of local energy supply. For heat supply, we accept three Tula-3 cast-iron water boilers with a thermal power of 779 kW each with gas fuel with the following characteristics:

Estimated power Fr = 2128 kW

Installed power Fu = 2337 kW

Heating surface area - 40.6 m²

Number of sections - 26

Dimensions 2249×2300×2361 mm

Maximum water heating temperature - 115 ºС

Efficiency when operating on gas η k.a. = 0.8

When operating in steam mode, excess steam pressure - 68.7 kPa

.4 Construction of an annual schedule for regulating the supply of a thermal boiler house

Due to the fact that the heat load of consumers varies depending on the outdoor temperature, the mode of operation of the ventilation and air conditioning system, the flow of water for hot water supply and technological needs, economical modes of heat generation in the boiler house should be provided by central regulation of heat supply.

In water heating networks, high-quality regulation of heat supply is used, carried out by changing the temperature of the coolant at a constant flow rate.

The graphs of water temperatures in the heating network are tp = f (tn, ºС), tо = f (tн, ºС). Having built a graph according to the method given in the work for tн = 95 ºС; to = 70 ºС for heating (it is taken into account that the temperature of the heat carrier in the hot water supply network should not fall below 70 ºС), tpv = 90 ºС; tov = 55 ºС - for ventilation, we determine the ranges of change in the temperature of the coolant in the heating and ventilation networks. On the abscissa axis, the values ​​of the outside temperature are plotted, on the ordinate axis - the temperature of the network water. The origin of coordinates coincides with the calculated internal temperature for residential and public buildings (18 ºС) and the coolant temperature, also equal to 18 ºС. At the intersection of perpendiculars restored to the coordinate axes at points corresponding to temperatures tp = 95 ºС, tн = -25 ºС, point A is found, and by drawing a horizontal straight line from the return water temperature of 70 ºС, point B. Connecting points A and B with the beginning coordinates, we get a graph of the change in the temperature of the direct and return water in the heating network, depending on the outdoor temperature. In the presence of a hot water supply load, the temperature of the coolant in the supply line of an open type network should not fall below 70 ° C, therefore the temperature graph for the supply water has a break point C, to the left of which τ p = const. The supply of heat for heating at a constant temperature is regulated by changing the flow rate of the coolant. The minimum return water temperature is determined by drawing a vertical line through point C until it intersects with the return water curve. The projection of the point D on the y-axis shows the smallest value of τо. The perpendicular, reconstructed from the point corresponding to the calculated outdoor temperature (-16 ºС), intersects straight lines AC and BD at points E and F, showing the maximum supply and return water temperatures for ventilation systems. That is, the temperatures are 91 ºС and 47 ºС, respectively, which remain unchanged in the range from tn.v and tn (lines EK and FL). In this range of outdoor temperatures, ventilation units operate with recirculation, the degree of which is regulated so that the temperature of the air entering the heaters remains constant.

The graph of water temperatures in the heating network is shown in Fig.4.

Fig.4. Graph of water temperatures in the heating network.

Bibliography

1. Efendiev A.M. Design of energy supply for agro-industrial complex enterprises. Toolkit. Saratov 2009.

Zakharov A.A. Workshop on the use of heat in agriculture. Second edition, revised and enlarged. Moscow Agropromizdat 1985.

Zakharov A.A. The use of heat in agriculture. Moscow Kolos 1980.

Kiryushatov A.I. Thermal power plants for agricultural production. Saratov 1989.

SNiP 2.10.02-84 Buildings and premises for storage and processing of agricultural products.