Thermal calculation (using the example of a high school). Project of a heat supply system for the Koskovo school in the village. Koskovo Kichmengsko-Gorodetsky district General information about the building

CALCULATION of the annual need for heat and fuel using the example of a boiler house high school with 800 students, Central Federal District.

Appendix No. 1 to the letter of the Ministry of Economy of Russia dated November 27, 1992 No. BE-261 / 25-510

LIST of data that must be submitted along with the application to establish the type of fuel for enterprises (associations) and fuel consuming installations.

1.General questions

Questions	Answers
Ministry (department)	MO
The enterprise and its location (republic, region, locality)	Central Federal District
Object distance to: A) railway station B) gas pipeline (its name) B) petroleum product bases D) the nearest source of heat supply (CHP boiler house), indicating its power, load and ownership	B) 0.850 km
The readiness of the enterprise to use fuel and energy resources (operating, reconstructed, under construction, projected), indicating its category	Current
Documents, approvals (date, number, name of organization) A) on the use of natural gas, coal and other types of fuel B) on the construction of an individual or expansion of an existing boiler house (CHP) On the basis of what document is the enterprise designed, built, expanded, or reconstructed?	MO task
Type and quantity (thousands, here) of fuel currently used and on the basis of what document (date, number) the consumption is established (for solid fuel indicate its origin and brand) Type of fuel requested, total annual consumption (thousands here) and year of start of consumption Year the enterprise reached its design capacity, total annual consumption (thousands here) this year	Natural gas; 0.536; 2012 2012; 0.536

2. Boiler plants and thermal power plants
A) Heat energy demand

For what needs	Attached max. heat load (Gcal/h)		Hours of work per year	Annual heat demand (thousand Gcal)		Covering heat demand thousand Gcal/year
For what needs	Noun	Etc. incl. noun	Hours of work per year	Noun	Etc. incl. noun	Boiler house (CHP)	Secondary energy resources	Parties
1	2	3	4	5	6	7	8	9
Heating		1,210	5160		2,895	2,895
Ventilation		0,000	0,000		0,000	0,000
		0,172	2800		0,483	0,483
Technological needs		0,000			0,000	0,000
Own needs of the boiler house (CHP)		0,000			0,000	0,000
Losses in heating networks		0,000			0,000	0,000
		1,382			3,378	3,378

B) Composition and characteristics of boiler house equipment, type and annual fuel consumption

Type of boilers by group	Qty	Total power Gcal/h	Fuel used			Requested fuel
Type of boilers by group	Qty	Total power Gcal/h	Type of main (backup)	Specific consumption kg.e.t/Gcal	Annual consumption thousand t.e.t.	Type of main (backup)	Specific consumption kg.e.t/Gcal	Annual consumption thousand t.e.
1	2	3	4	5	6	7	8	9
Active
Dismantled
Installable Buderus boilers Logano SK745-820 VAXI (820 kW)	2	1,410				Natural gas (none)	158.667	0,536
Reserve

Note:

1. Indicate the total annual fuel consumption for groups of boilers.

2. Specify specific fuel consumption taking into account the own needs of the boiler house (CHP)

3. In columns 4 and 7, indicate the method of fuel combustion (layer, chamber, fluidized bed).

4. For thermal power plants, indicate the type and brand of turbine units, their electrical power in thousand kW, annual production and supply of electricity in thousand kWh,

annual heat supply in Gcal., specific fuel consumption for electricity and heat supply (kg/Gcal), annual fuel consumption for electricity and heat production in general at the CHP plant.

5. When consuming more than 100 thousand tons of equivalent fuel per year, the fuel and energy balance of the enterprise (association) must be presented.

2.1 General part

Calculation of the annual fuel requirement for a modular boiler house (heating and hot water supply) of a secondary school was carried out according to the instructions of the Moscow Region. The maximum winter hourly heat consumption for heating a building is determined based on aggregated indicators. Heat consumption for hot water supply is determined in accordance with the instructions of clause 3.13 of SNiP 2.04.01-85 "Internal water supply and sewerage of buildings." Climatological data are accepted according to SNiP 23-01-99 "Construction climatology and geophysics". Estimated average temperatures internal air taken from " Guidelines to determine the consumption of fuel, electricity and water for heat production by heating boiler houses of municipal heat and power enterprises." Moscow 1994.

2.2 Heat source

For heat supply (heating, hot water supply) to the school, it is planned to install two Buderus Logano SK745 boilers (Germany) with a capacity of 820 kW each in a specially equipped boiler room. The total capacity of the installed equipment is 1,410 Gcal/h. Natural gas is requested as the main fuel. No backup required.

2.3 Initial data and calculation

No.	Indicators	Formula and calculation
1	2	3
1	Design outdoor temperature for heating design	T(R.O)= -26
2	Estimated outside air temperature for ventilation design	T(R.V)= -26
3	Average outside air temperature during the heating period	T(SR.O)= -2.4
4	Estimated average temperature of the internal air of heated buildings	T(VN.)=20.0
5	Duration of the heating season	P(O)=215 days.
6	Number of operating hours of heating systems per year	Z(O)=5160 h
7	Number of operating hours of ventilation systems per year	Z(V)=0 h
8	Number of operating hours of hot water supply systems per year	Z(G.V)=2800 h
9	Number of working hours technological equipment per year	Z(V)=0 h
10	Coeff. simultaneity of action and use. Maksim. technological loads	K(T)=0.0 h
11	Coeff. working days	KRD=5.0
12	Average hourly heat consumption for heating	Q(O.SR)= Q(O)[T(VN)-T(CP.O)]/ [T(BH)-T(R.O))= 1.210 [(18.0)-( -2.4)]/ [(18.0)-(-26.0)]= 0.561 Gcal/h
13	Average hourly heat consumption for ventilation	Q(B.CP)= Q(B)[T(BH)-T(CP.O)]/ [T(BH)-T(P.B))= 0.000 [(18.0)-( -2.4)]/ [(18.0)-(-26.0)]= 0.000 Gcal/h
14	Average hourly heat consumption for hot water supply for heating. period	Q(G.W.SR)= Q(G.W)/2.2=0.172/2.2=0.078 Gcal/h
15	Average hourly heat consumption for hot water supply in summer period	Q(G.V.SR.L)= (G.V.SR)[(55-1 5)/(55-5)]0.8= 0.078[(55-15)/(55-5) ]0.8=0.0499 Gcal/h
16	Average hourly heat consumption per technology per year	Q(TECH.SR)= Q(T)* K(T)=0.000*0.0=0.000 Gcal/h
17	Annual heat demand for heating	Q(O.YEAR)=24* P(O)* Q(O.SR)=242150.561=2894.76 Gcal
18	Annual heat requirement for ventilation	Q(V.YEAR)= Z(V)* Q(V.SR)=0.0*0.0=0.00 Gcal
19	Annual heat demand for water supply	Q(G.V.YEAR)(24* P(O)* Q(G.V.SR)+24* Q(G.V.SR.L)) KRD= (24* 2150.078 +24 0.0499 (350-215)) 6/7=483.57 Gcal
20	Annual heat demand for technology	Q(T.YEAR)= Q(TECH.CP)* Z(T)=0.000*0=0.000 Gcal
21	Total annual heat demand	Q(YEAR)= Q(O.YEAR)+ Q(V.YEAR)+ Q(YEARYYEAR)+ Q(T.YEAR)= 2894.76 + 0.000+483.57+0.000=3378.33 Gcal
	TOTAL for existing buildings:
	Annual heat demand for Heating Ventilation Hot water supply Technology Losses in t/s Own needs of the boiler room	Q(O.YEAR)= 2894.76 Gcal Q(V.YEAR)= 0.000 Gcal Q(G.V.YEAR)= 483.57 Gcal Q(T.YEAR)= 0.000 Gcal ROTER= 0.000 Gcal SOBS= 0.000 Gcal
	TOTAL:	Q(YEAR)=3378.33 Gcal
	Specific consumption of equivalent fuel	В= 142.8*100/90=158.667 KG.U.T./Gcal
	Annual consumption of equivalent fuel for heat supply of existing buildings	B=536.029 T.U.T

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Heating system for a school, kindergarten, college, university: a range of services from our company

project development internal system heating of educational institutions;
thermal and hydraulic calculation school boiler room, kindergarten, university;
reconstruction and modernization of the heating system;
installation of internal networks and heating equipment;
selection and boiler installation heating systems for children's and educational institutions;
calculation, selection and installation water heated floor systems;
maintenance and repair heating and boiler equipment;
coordination with supervisory authorities.

For educational institutions in areas with an estimated outside air temperature of -40°C and below, it is allowed to use water with additives that prevent it from freezing (harmful substances of the 1st and 2nd hazard classes according to GOST 12.1.005 should not be used as additives), and in buildings of preschool institutions it is not allowed to use coolant with additives of harmful substances of hazard classes 1–4.

Design and installation of autonomous boiler houses and heating systems in schools, preschool and educational institutions

Heating system for schools, kindergartens and other children's and educational institutions (universities, vocational schools, colleges) in cities is connected to central system heating and hot water, which is powered from the city thermal power plant or its own boiler house. IN rural areas use stand-alone circuit, having its own boiler room in a special room. In the case of gasified areas, the boiler runs on natural gas; in small schools and preschool institutions, boilers are used low power working on solid or liquid fuel or electricity.

When designing an internal heating system, microclimatic standards for air temperature in classrooms, school classrooms, canteens, gyms, swimming pools and other premises should be taken into account. Various by technical purpose building areas must have their own heating networks with water and heat meters.

For heating gyms, along with a water system, it is used air system heating combined with forced ventilation and operating from the same boiler room. A water floor heating device may be present in locker rooms, bathrooms, showers, swimming pools and other premises, if available. On entrance groups In large educational institutions, thermal curtains are installed.

Heating system of a kindergarten, school, educational institution - list of works on the organization and reconstruction of the heating system:

identifying needs when creating a project or sketch diagram heat supply;
choice way and place installation of pipelines;
selection equipment and materials appropriate quality;
thermal and hydraulic calculation of the boiler room, determination of technology and testing it against the requirements of SNiP;
possibility of increasing productivity, connection additional equipment (if needed);
load calculation and the performance of the heating system as a whole and by area of heated premises;
during the reconstruction of the facility – site preparation, foundation and walls for subsequent installation;
defective sections of the building heating system;
calculation of terms and costs works and equipment, coordination of estimates;
supply of equipment and execution of work on time at a pre-agreed cost estimate.

For heating devices and pipelines in children's preschool premises, stairwells and vestibules, it is necessary to provide protective barriers and thermal insulation pipelines.

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Introduction

Calculation of heating, ventilation and hot water supply for 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 heating surface area and selection heating devices central heating systems

4 Calculation of school air exchange

5 Selection of heaters

6 Calculation of heat consumption for hot water supply to 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 enlarged standards for residential and public buildings

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

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

1 Construction of an annual heat load graph

3.2 Selection of coolant

3 Selection of boilers

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 sector of the national economy. A large amount of energy is spent on heating industrial, residential and public buildings, creating an artificial microclimate in livestock buildings and protective soil structures, drying agricultural products, manufacturing products, obtaining artificial cold and for many other purposes. Therefore, energy supply to agricultural enterprises includes a wide range of tasks related to the production, transmission and use of thermal and electrical energy, using traditional and non-traditional energy sources.

This course project proposes an option for integrated energy supply to a populated area:

· 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 heating, ventilation and hot water supply loads is carried out;

· the required power of the boiler house is determined, which could meet the household’s heat needs;

· selection of boilers is carried out.

· calculate gas consumption,

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

1.1 Brief description of the school

Dimensions 43.350x12x2.7.

Room volume V = 1709.34 m 3.

External longitudinal walls are load-bearing, made of facing and finishing, thickened brick of grade KP-U100/25 in accordance with GOST 530-95 on cement - sand solution M 50, thickness 250 and 120 mm and 140 mm of insulation - polystyrene foam between them.

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

Partitions are made of brick KP-U75/15 in accordance with GOST 530-95, with M 50 mortar.

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

Floors - concrete M300 and soil compacted with crushed stone.

Double windows with paired wooden frames, window sizes 2940x3000 (22 pieces) and 1800x1760 (4 pieces).

External wooden single doors 1770x2300 (6 pcs)

Design parameters of outside air tн = - 25 0 С.

Estimated winter ventilation temperature of outside air tn.v. = - 16 0 C.

Estimated internal air temperature tв = 16 0 С.

The area's humidity zone is normal dry.

Barometric pressure 99.3 kPa.

1.2 Calculation of school air exchange

The learning process takes place at school. Characterized by long-term presence of a large number of students. There are no harmful emissions. The air change coefficient for a school will be 0.95...2.

K ∙ Vп,

where Q is air exchange, m³/h; Vп - room volume, m³; K - air exchange rate is taken = 1.

Fig.1. Room dimensions.

Room volume: = 1709.34 m 3 . = 1∙1709.34 = 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 using the formula: F = Q / (3600 ∙ ν k.in). , having previously determined the air speed in the exhaust shaft with a height of h = 2.7 m

ν k.in. =

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

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

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

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

We do not take into account heat loss through the internal enclosures of the room, because the temperature difference in the separated rooms does not exceed 5 0 C. We determine the heat transfer resistance of the enclosing structures. Heat transfer resistance outer wall(Fig. 1) will be found using the formula using the data in the table. 1, knowing that thermal resistance to heat perception inner surface fence Rв=0.115 m 2 ∙ 0 С/W

where Rв is the 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 t - layer fencing with thickness δi (m), made of materials with thermal conductivity λi, W / (m·ºС), values of λ are given in Table 1; Rн - thermal resistance to heat transfer of the outer surface of the fence Rн=0.043 m 2 ∙ 0 C/W (for external walls and attic floors).

Fig.1 Structure of wall materials.

Table 1 Thermal conductivity and width of wall materials.

Heat transfer resistance of the outer wall:

R 01 = m²·ºС/W.

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

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

) Resistance to heat transfer of the ceiling for a roofless ceiling (Rв=0.115 m 2 ∙ 0 С/W, Rн=0.043 m 2 ∙ 0 С/W).

Calculation of heat losses through ceilings:

Fig.2 ceiling structure.

Table 2 Thermal conductivity and width of floor materials

Ceiling heat transfer resistance

m 2 ∙ 0 C/W.

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

Area of floor zones minus basement area: = 43 ∙ 2 + 28 ∙ 2 = 142 m 2

F1=12 ∙ 2 + 12 ∙ 2 = 48 m 2 ,= 43 ∙ 2 + 28 ∙ 2=148 m 2

F2=12 ∙ 2 + 12∙ 2 = 48 m 2 ,= 43 ∙ 2 + 28 ∙ 2=142 m 2

F3=6 ∙ 0.5 + 12 ∙ 2 = 27 m 2

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

F1=6 ∙ 2 + 6 ∙ 2 = 24 m 2 ,= 15 ∙ 2 + 15 ∙ 2=60 m 2

F2=6 ∙ 2 = 12 m 2

F1 = 15 ∙ 2 + 15 ∙ 2=60 m 2

Floors located directly on the ground are considered uninsulated 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 the 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=Rн.п. for the first zone it is 2.15, for the second - 4.3, for the third - 8.6, the rest - 14.2 m 2 ∙ 0 C/W.

) Total area of window openings: approx = 2.94∙3∙22+1.8∙1.76∙6 = 213 m2.

Total area of external doorways: dv = 1.77 ∙ 2.3 ∙ 6 = 34.43 m2.

External wall area 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 = 62 m2.

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

) Ceiling area: pot = 42.85 ∙ 12+3∙ 8.5 = 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 enclosing structures are determined with an accuracy of 0.1 m, following the measurement rules); tв and tн - calculated temperatures of internal and external air, ºС (add. 1…3); R 0 - total heat transfer resistance, m 2 ∙ 0 C / W; n is a 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=1 (for external walls, roofless roofs, attic floors with a steel, tiled or asbestos-cement roof over sparse lathing, floors on the ground)

Heat losses through external walls:

FNS = 601.1 W.

Heat losses through the external walls of the basement:

Fn.s.p = 130.1 W.

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

Heat loss through windows:

Fock = 25685 W.

Heat losses 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 losses through the basement floor:

Fpol.p = 100 W.

∑F floor =F floor. +F half p. =6240.5+100=6340.5 W.

Additional heat losses through external vertical and inclined (vertical projection) walls, doors and windows depend on various factors. Fdob values 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%, and to the southeast and west - 5%.

Additional losses for infiltration of outside air for industrial buildings are assumed to be 30% of the main losses through all fences:

Finf = 0.3 · (Fn.s. + Fok. + Fpot. + Fdv + Fpol.) = 0.3 · (731.2 + 25685 + 13093.3 + 6565.72 + 6340.5) = 15724, 7 W

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

1.4 Calculation of heating surface area and selection of heating devices for central heating systems

The most common and universally used heating devices 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 heating system pipelines. The heat flow given to the room by openly laid non-insulated pipelines is determined by formula 3:

Ftr = Ftr ∙ ktr · (ttr - tv) ∙ η,

where Ftr = π ∙ d l - 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 - pipe heat transfer coefficient W/(m 2 ∙ 0 C) is determined according to Table 4 depending on the temperature pressure and type of coolant in the pipeline, ºC; η - coefficient equal to 0.25 for the supply line located under the ceiling, for vertical risers - 0.5, for the return line located above the floor - 0.75, for connections to the heating device - 1.0

Supply pipe:

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

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 = 67.5 ºС is taken equal to 9.2 W/(m²∙ºС). in accordance with the data in Table 4.

Direct heat conduction:

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

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

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

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

Return heat pipe:

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

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

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

Total heat flow from all pipelines:

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

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

where Fogr-Ftr is the heat transfer of heating devices, W; Ftr - 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 C). for water heating tpr = (tg+tо)/2; tg and tо - calculated temperature of hot and chilled water in the device; for steam heating low pressure take tpr=100 ºС, in high-pressure systems tpr is equal to the steam temperature in front of the device at its corresponding pressure; tв - estimated air temperature in the room, ºС; β 1 - correction factor taking into account the installation method of the heating device. When freely installed against a wall or in a niche 130 mm deep, 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 covered with a board in the form of a shelf with a distance between the board and the heating device of 40...100 mm 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 heating device of 40...100 mm, coefficient β 1 = 1.11...1.06; c) the device is installed in a wall without a niche and covered 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 heating device equal to 150, 180, 220 and 260 mm, coefficient β 1 is respectively equal to 1.25; 1.19; 1.13 and 1.12; β 1 - correction factor β 2 - correction factor taking into account the cooling of water in the pipelines. With open installation of water heating pipelines and with steam heating β 2 =1. for a hidden pipeline, with pump circulation β 2 = 1.04 (single-pipe systems) and β 2 = 1.05 (two-pipe systems with overhead distribution); during natural circulation, due to the increase in cooling of water in pipelines, the values of β 2 should be multiplied by a coefficient of 1.04.pr= 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 n value is approximate. If necessary, it is divided into several devices and, by introducing a correction factor β 3, taking 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 = 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

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

The selection of air heaters is determined in the following order:

We determine the heat flow (W) used to heat the air:

Фв = 0.278 ∙ Q ∙ ρ ∙ c ∙ (tв - tн), (10)

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

Air Density:

ρ = 346/(273+18) 99.3/99.3 = 1.19;

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

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

m².

3. Then, according to Table 7, we select the model and number of the heater with the open air cross-sectional area close to the calculated one. When installing several heaters in parallel (along the air flow), their total open cross-sectional area is taken into account. We select 1 K4PP No. 2 with a clear air cross-sectional 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 this, 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 graph, heat transfer coefficient k = 16 W/(m 2 0 C)

We determine the actual heat flow (W) transferred by the heating unit to the heated air:

Фк = k ∙ F ∙ (t´ср - tср),

where k is the heat transfer coefficient, W/(m 2 ∙ 0 C); F - heater heating surface area, m²; t´av - average coolant temperature, ºС, for coolant - steam - t´av = 95 ºС; tср - average temperature of heated air t´ср = (tк + tн) /2

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

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

The heat transfer margin is =6%.

1.6 Calculation of heat consumption for hot water supply to a school

At school, hot water is needed for sanitary and domestic needs. A school with 90 seats consumes 5 liters per day hot water per day. Total: 50 liters. Therefore, we place 2 risers with a water flow rate of 60 l/h each (that is, only 120 l/h). Considering that on average hot water is used for sanitary needs for about 7 hours during the day, we find the amount of hot water is 840 l/day. School consumption per hour is 0.35 m³/h

Then the heat flow to the water supply will be

Fgv. = 0.278 · 0.35 · 983 · 4.19 · (55 - 5) = 20038 W

The number of shower cabins for the school is 2. The hourly consumption of hot water per cabin is Q = 250 l/h, let us assume that on average the shower operates 2 hours a day.

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

Fgv. =0.278 · 1 · 983 · 4.19 · (55 - 5) = 57250 W.

∑F g.v. =20038+57250=77288 W.

2. Calculation of heat load for centralized heating

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

Photo = φ ∙ F,

Photo.j.=0.25∙Phot.j., (19)

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

1. For thirteen 16-apartment buildings with an area of 720 m2, we obtain:

Photo = 13 ∙ 170 ∙ 720 = 1591200 W.

For eleven 8-apartment buildings with an area of 360 m2 we get:

Photo = 8 ∙ 170 ∙ 360 = 489600 W.

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

Photototal=0.25∙170∙6∙6=1530 W;

For an office with dimensions 6x12 m:

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

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

Ph = qot Vn (tv - tn) a,

Фв = qв · Vн · (tв - tн.в.),

where q from and q in are the 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 external measurement without the basement, m 3, is taken according to standard designs or determined by multiplying its length by its width and height from the planning level of the ground to the top of the cornice; t in = average design air temperature, typical for most rooms of the building, 0 C; t n = calculated winter outside air temperature, - 25 0 C; t n.v. - estimated winter ventilation temperature of outside air, - 16 0 C; a - correction factor taking into account the influence of local climatic conditions on the specific thermal characteristics at tn = 25 0 C a = 1.05

Ph = 0.7 ∙ 18∙36∙4.2 ∙ (10 - (- 25)) ∙ 1.05 = 5000.91 W,

Fv.tot.=0.4∙5000.91=2000 W.

Brigade house:

Ph = 0.5∙ 1944 ∙ (18 - (- 25)) ∙ 1.05 = 5511.2 W,

School workshop:

Ph = 0.6 ∙ 1814.4 ∙ (15 - (- 25)) 1.05 = 47981.8 W,

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

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

The average heat flow (W) spent during the heating period on hot water supply to buildings is found by the formula:

F g.v. = q g.v. n f,

Depending on the rate of water consumption at a temperature of 55 0 C, the aggregated indicator of the average heat flow (W) spent on hot water supply for one person will be equal to: At water consumption - 115 l/day q g.w. is 407 W.

For 16 apartment buildings with 60 residents, the heat flow for hot water supply will be: F g.w. = 407 60 = 24420 W,

for thirteen such houses - F g.v. = 24420 · 13 = 317460 W.

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

F g.v.l. = 0.65 · F g.v. = 0.65 317460 = 206349 W

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

F g.v. = 407 · 30 = 12210 W,

for eleven such houses - F g.v. = 12210 · 11 = 97680 W.

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

F g.v.l. = 0.65 · F g.v. = 0.65 · 97680 = 63492 W.

Then the heat flow to the office water supply 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.v.l. = 0.65 ∙ F g.v. = 0.65 ∙ 47690 = 31000 W

Heat flow to medical water supply. point will be:

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

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

F g.v.l. = 0.65 ∙ F g.v. = 0.65 ∙ 13167 = 8559 W

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

The workshop contains 2 risers with a water flow rate of 30 l/h each (that is, 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. = 0.278 · 0.68 · 983 · 4.19 · (55 - 5) = 38930 W

Heat flow consumed for hot water supply to a school workshop in the summer:

Fgv.l = 38930 · 0.65 = 25304.5 W

Summary table of heat flows

Calculated heat flows, W
Name	Heating	Ventilation	Technical needs

School for 90 students
16 sq.m. house
Honey. paragraph
8 apartment building
School workshop

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

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

.1 Construction of an annual heat load graph

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 room throughout the year. Such a graph is constructed depending on the duration of various temperatures in a given area, which is determined according to Appendix 3.

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

First, a graph is drawn for changes in 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 ordinate axis, and the found point is connected by a straight line to the point corresponding to the outside air temperature equal to the average design temperature of residential buildings; public and industrial buildings tв = 18 °С. 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 tн is an inclined straight line 3 from tв = 18 °С to the calculated ventilation temperature tн.в. for a given climatic region. At lower temperatures, room air is mixed with the supply outside air, i.e. recirculation occurs, and the heat consumption remains unchanged (the graph is parallel to the abscissa axis). In a similar way, graphs of heat consumption for heating and ventilation of various industrial buildings are constructed. The average temperature of industrial buildings tв = 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 as straight line 5.

The total graph of heat consumption depending on the outside air temperature is shown by broken line 6 (the break point corresponds to tn.v.), cutting off on the ordinate axis a segment equal to the maximum heat flow spent on all types of consumption (∑Phot + ∑Fv + ∑Fg. V. + ∑Ft) at the calculated external temperature tн.

Adding up the total loads I got 2.9W.

To the right of the abscissa axis, for each external temperature, the number of hours of the heating season (cumulatively) during which the temperature remained equal to or lower than that for which the construction was made was kept (Appendix 3). And vertical lines are drawn through these points. Next, ordinates corresponding to the maximum heat consumption at the same external temperatures are projected onto these lines from the total heat consumption graph. The resulting points are connected by a smooth curve 7, which represents a graph of the heat load during 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 graph, mm²; m Q and m n are the scale of heat consumption and operating time of the boiler room, respectively W/mm and h/mm.year = 3.6 ∙ 10 -6 ∙ 9871.74 ∙ 23548 ∙ 47.8 = 40001.67 J/year

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

3.2 Selection of coolant

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

The use of superheated water as a coolant provides greater savings in pipe metal by reducing their diameter, and reduces the energy consumption of network pumps, since the total amount of water circulating in the system is reduced.

Since some consumers require steam for technical purposes, consumers need to install additional heat exchangers.

3.3 Selection of boilers

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

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

Estimated power Фр = 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 a.a. = 0.8

When operating in steam mode, excess steam pressure is 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 outside air temperature, the operating mode of the ventilation and air conditioning system, water consumption for hot water supply and technological needs, economical modes of thermal energy generation in the boiler room must be ensured 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 represented by tп = f (tн, ºС), tо = f (tн, ºС). Having constructed a graph using the method given in the work for tн = 95 ºС; tо = 70 ºС for heating (it is taken into account that the temperature of the coolant in the hot water supply network should not fall below 70 ºС), tпв = 90 ºС; tov = 55 ºС - for ventilation, we determine the ranges of temperature changes of the coolant in the heating and ventilation networks. The values of the external temperature are plotted along the abscissa axis, and the temperature of the supply water is plotted along the ordinate axis. The origin 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 tп = 95 ºС, tн = -25 ºС, point A is found, and by drawing a horizontal line from the return water temperature of 70 ºС, point B is found. Connecting points A and B with the beginning coordinates, we obtain a graph of changes in the temperature of forward and return water in the heating network depending on the outside air temperature. If there is 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 supply water has an inflection point C, to the left of which τ p =const. The supply of heat to heating at a constant temperature is controlled by changing the coolant flow rate. The minimum return water temperature is determined by drawing a vertical line through point C until it intersects with the return water graph. The projection of point D onto the ordinate axis shows the smallest value of τto. The perpendicular, restored from the point corresponding to the calculated outside temperature (-16 ºС), intersects straight lines AC and BD at points E and F, showing the maximum temperatures of forward and return water for ventilation systems. That is, temperatures are 91 ºС and 47 ºС, respectively, which remain unchanged in the range from tн.в and tн (lines EK and FL). In this range of outside air 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 presented in Fig. 4.

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

Bibliography

1. Efendiev A.M. Design of energy supply for agricultural enterprises. Toolkit. Saratov 2009.

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

Zakharov A.A. Application 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.

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Introduction

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

1.1 Brief description of the school

1.2 Determination of heat loss through the external fences of the garage

1.3 Calculation of heating surface area and selection of heating devices for central heating systems

1.4 Calculation of school air exchange

1.5 Selection of heaters

1.6 Calculation of heat consumption for hot water supply to a school

2. 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 enlarged standards for residential and public buildings

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

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

3.1 Construction of an annual heat load graph

3.2 Selection of coolant

3.3 Selection of boilers

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

Bibliography

Introduction

This course project proposes an option for integrated energy supply to a populated area:

· 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 heating, ventilation and hot water supply loads is carried out;

· the required power of the boiler house is determined, which could meet the household’s heat needs;

· selection of boilers is carried out.

· calculate gas consumption,

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

1 . 1 Brief haschool characteristics

Dimensions 43.350x12x2.7.

Room volume V = 1709.34 m 3.

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

Internal walls are made of hollow, thickened ceramic bricks of grade KP-U100/15 in accordance with GOST 530-95, with M50 mortar.

Partitions are made of brick KP-U75/15 in accordance with GOST 530-95, with M 50 mortar.

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

Floors - concrete M300 and soil compacted with crushed stone.

Double windows with paired wooden frames, window sizes 2940x3000 (22 pieces) and 1800x1760 (4 pieces).

External wooden single doors 1770x2300 (6 pcs)

Design parameters of outside air tн = - 25 0 С.

Estimated winter ventilation temperature of outside air tn.v. = - 16 0 C.

Estimated internal air temperature tв = 16 0 С.

The area's humidity zone is normal dry.

Barometric pressure 99.3 kPa.

1.2 School air exchange calculation

where Q is air exchange, m?/h; Vп - volume of the room, m?; K - air exchange rate is taken = 1.

Fig.1. Room dimensions.

Room volume:

V=1709.34 m3.

Q = 1 1709.34 = 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 ? n k.vn). , having previously determined the air speed in the exhaust shaft with a height of h = 2.7 m

n k.in. = = 1.23 m/s

F = 1709.34 / (3600 1.23) = 0.38 m?

Number of exhaust shafts

n wsh = F / 0.04 = 0.38 / 0.04 = 9.5? 10

We accept 10 exhaust shafts 2 m high with a live cross-section of 0.04 m? (with dimensions 200 x 200 mm).

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

We do not take into account heat loss through the internal enclosures of the room, because the temperature difference in the separated rooms does not exceed 5 0 C. We determine the heat transfer resistance of the enclosing structures. We will find the heat transfer resistance of the outer wall (Fig. 1) using the formula, using the data in table. 1, knowing that the thermal resistance to heat absorption of the inner surface of the fence Rв = 0.115 m 2 0 C/W

where Rв is the thermal resistance to heat absorption of the inner surface of the fence, m?·?С/W; - the sum of thermal resistances to thermal conductivity of individual layers m - layer fencing with thickness di (m), made of materials with thermal conductivity li, W / (m·? C), values \u200b\u200bare given in Table 1; Rн - thermal resistance to heat transfer of the outer surface of the fence Rн = 0.043 m 2 0 C/W (for external walls and attic floors).

Fig.1 Structure of wall materials.

Table 1 Thermal conductivity and width of wall materials.

Heat transfer resistance of the outer wall:

R 01 = m?·?С/W.

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

Heat transfer resistance of external doors and gates is 0.215 m 2 0 C/W (we find it from the table on page 8)

3) Heat transfer resistance of the ceiling for a roofless ceiling (Rв=0.115 m 2 0 С/W, Rн=0.043 m 2 0 С/W).

Calculation of heat losses through ceilings:

Fig.2 ceiling structure.

Table 2 Thermal conductivity and width of floor materials

Ceiling heat transfer resistance

m 2 0 C/W.

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

Area of floor zones minus basement area:

F1 = 43 2 + 28 2 = 142 m 2

F1=12 2 + 12 2 = 48 m 2,

F2 = 43 2 + 28 2 = 148 m 2

F2=12 2 + 12 2 = 48 m 2,

F3 = 43 2 + 28 2 = 142 m 2

F3=6 0.5 + 12 2 = 27 m 2

Areas of basement floor areas:

F1 = 15 2 + 15 2 = 60 m 2

F1=6 2 + 6 2 = 24 m 2,

F2 = 15 2 + 15 2 = 60 m 2

F2=6 2 = 12 m 2

F1 = 15 2 + 15 2 = 60 m 2

Floors located directly on the ground are considered uninsulated if they consist of several layers of materials, the thermal conductivity of each of which is l? 1.16 W/(m 2 0 C). Floors are considered insulated if the insulating layer has l<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 l? 1.16 W/m 2 0 C. So, heat transfer resistance Ro = Rn.p. for the first zone it is 2.15, for the second - 4.3, for the third - 8.6, the rest - 14.2 m 2 0 C/W.

5) Total area of window openings:

Fok = 2.94 3 22 + 1.8 1.76 6 = 213 m 2.

Total area of external doorways:

Fdv = 1.77 2.3 6 = 34.43 m2.

External wall area minus window and door openings:

Fn.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 = 62 m2 .

Basement wall area:

Fn.s.p =14.5 2.7+5.5 2.7-4.1=50

6) Ceiling area:

Fpot = 42.85 12+3 8.5 = 539.7 m2,

where F is the area of the fence (m?), which is calculated with an accuracy of 0.1 m? (the linear dimensions of enclosing structures are determined with an accuracy of 0.1 m, observing the measurement rules); tв and tн - calculated temperatures of internal and external air, ? C (add. 1…3); R 0 - total heat transfer resistance, m 2 0 C / W; n is a 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=1 (for external walls, roofless roofs, attic floors with a steel, tiled or asbestos-cement roof over sparse lathing, floors on the ground)

Heat losses through external walls:

Fns = 601.1 W.

Heat losses through the external walls of the basement:

Fn.s.p = 130.1 W.

F n.s. =F n.s. +F n.s.p. =601.1+130.1=731.2 W.

Heat loss through windows:

Focal = 25685 W.

Heat losses 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 losses through the basement floor:

Fpol.p = 100 W.

F floor = F floor. +F half p. =6240.5+100=6340.5 W.

Additional losses for infiltration of outside air for industrial buildings are assumed to be 30% of the main losses through all fences:

Finf = 0.3 · (Fn.s. + Fok. + Fpot. + Fdv + Fpol.) = 0.3 · (731.2 + 25685 + 13093.3 + 6565.72 + 6340.5) = 15724, 7 W

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

Fogr=78698.3 W.

1.4 Calculation of heating surface area and selectionheating devices for central heating systems

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

Ftr = Ftr ktr · (ttr - tv) z,

where Ftr = p? d l - area of the outer surface of the pipe, m?; 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 C) is determined according to Table 4 depending on the temperature pressure and the type of coolant in the pipeline, ? C; z - coefficient equal to 0.25 for the supply line located under the ceiling, for vertical risers - 0.5, for the return line located above the floor - 0.75, for connections to the heating device - 1.0

Supply pipe:

Diameter-50mm:

F1 50mm =3.14 73.4 0.05=11.52 m?;

Diameter 32mm:

F1 32mm =3.14 35.4 0.032=3.56 m?;

Diameter - 25 mm:

F1 25mm =3.14 14.45 0.025=1.45 m?;

Diameter-20:

F1 20mm =3.14 32.1 0.02=2.02 m?;

Return pipeline:

Diameter-25mm:

F2 25mm =3.14 73.4 0.025=5.76 m?;

Diameter-40mm:

F2 40mm =3.14 35.4 0.04=4.45 m?;

Diameter-50mm:

F2 50mm =3.14 46.55 0.05=7.31 m?;

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 = 67.5 °C is taken equal to 9.2 W/(m? °C). in accordance with the data in Table 4.

Direct heat conduction:

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

Ф p1.32mm =3.56 9.2 · (95 - 16) 1=2620.16 W;

Ф p1.25mm =1.45 9.2 · (95 - 16) 1=1067.2 W;

Ф p1.20mm =2.02 9.2 · (95 - 16) 1=1486.72 W;

Return heat pipe:

Ф p2.25mm =5.76 9.2 · (70 - 16) 1=2914.56 W;

Ф p2.40mm =4.45 9.2 · (70 - 16) 1=2251.7 W;

Ф p2.50mm =7.31 9.2 · (70 - 16) 1=3698.86 W;

Total heat flow from all pipelines:

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

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

where Fogr-Ftr is the heat transfer of heating devices, W; Ftr - heat transfer of open pipelines located in the same room with heating devices, W;

kpr - heat transfer coefficient of the device, W/(m 2 0 C). for water heating tpr = (tg+tо)/2; tg and tо - calculated temperature of hot and chilled water in the device; for low-pressure steam heating, tpr = 100 °C is taken; in high-pressure systems, tpr is equal to the steam temperature in front of the device at its corresponding pressure; tв - design air temperature in the room, ?С; in 1 - correction factor taking into account the method of installation of the heating device. When freely installed against a wall or in a niche 130 mm deep, 1 = 1; in other cases, values of 1 are taken based on the following data: a) the device is installed against a wall without a niche and covered with a board in the form of a shelf with a distance between the board and the heating device of 40...100 mm, coefficient of 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 heating device of 40...100 mm, coefficient 1 = 1.11...1.06; c) the device is installed in a wall without a niche and 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 heating device equal to 150, 180, 220 and 260 mm, the coefficient of 1 is 1.25, respectively; 1.19; 1.13 and 1.12; in 1 - correction factor; in 2 - correction factor taking into account the cooling of water in the pipelines. With open installation of water heating pipelines and with steam heating in 2 =1. for a hidden pipeline, with pump circulation at 2 = 1.04 (single-pipe systems) and at 2 = 1.05 (two-pipe systems with overhead distribution); with natural circulation, due to increased cooling of water in pipelines, values of 2 should be multiplied by a factor of 1.04.

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

n = Fpr / fsection,

where fsection is the heating surface area of one section, m? (Table 2).

n = 96 / 0.31 = 309.

The resulting n value is approximate. If necessary, it is divided into several devices and, by introducing a correction factor of 3, taking 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:

nset = n · in 3;

nset = 309 · 1.05 = 325.

We install 27 radiators in 12 sections.

heating water supply school ventilation

1.5 Selection of heaters

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

The selection of air heaters is determined in the following order:

1. Determine the heat flow (W) used to heat the air:

Fv = 0.278 Q ? With? c (tв - tн), (10)

where Q is the volumetric air flow rate, m?/h; с - air density at temperature tк, kg/m?; ap = 1 kJ/ (kg ? C) - specific isobaric heat capacity of air; tk - air temperature after the heater, ? C; tn - initial temperature of air entering the heater, ? C

Air Density:

c = 346/(273+18) 99.3/99.3 = 1.19;

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

Estimated mass air speed is 4-12 kg/s m?.

3. Then, according to Table 7, we select the model and number of the heater with the open air cross-sectional area close to the calculated one. When installing several heaters in parallel (along the air flow), their total open cross-sectional area is taken into account. We choose 1 K4PP No. 2 with a clear air cross-sectional area of 0.115 m? and a heating surface area of 12.7 m?

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

5. After this, 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 ns value. According to the graph, heat transfer coefficient k = 16 W/(m 2 0 C)

6. Determine the actual heat flow (W) transferred by the heating unit to the heated air:

Фк = k F (t?ср - tср),

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

Fk = 16 12.7 (95 -(16-16)/2) = 46451 2 = 92902 W.

2 plate heaters KZPP No. 7 provide a heat flow of 92902 W, and the required one is 83789.85 W. Consequently, heat transfer is fully ensured.

The heat transfer margin is = 6%.

1.6 Calculation of heat consumption for hot water supply to a school

At school, hot water is needed for sanitary and domestic needs. A school with 90 seats consumes 5 liters of hot water per day. Total: 50 liters. Therefore, we place 2 risers with a water flow rate of 60 l/h each (that is, only 120 l/h). Considering that on average hot water is used for sanitary needs for about 7 hours during the day, we find the amount of hot water is 840 l/day. School consumption per hour is 0.35 m?/h

Then the heat flow to the water supply will be

Fgv. = 0.278 · 0.35 · 983 · 4.19 · (55 - 5) = 20038 W

The number of shower cabins for the school is 2. The hourly consumption of hot water per cabin is Q = 250 l/h, let us assume that on average the shower operates 2 hours a day.

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

Fgv. =0.278 · 1 · 983 · 4.19 · (55 - 5) = 57250 W.

F g.v. =20038+57250=77288 W.

2. Calculation of heat load for centralized heating

2.1 RCalculation of heat consumption for heating and ventilation according toenlarged standards

Photo = c? F,

Photo.j.=0.25 Photo.j., (19)

where c is an aggregated indicator of the maximum specific heat flow spent on heating 1 m2? living area, W/m?. The values of c are determined depending on the calculated winter outside air temperature according to the schedule (Fig. 62); F - living area, m?.

1. For thirteen 16-apartment buildings with an area of 720 m2, we obtain:

Photo = 13,170,720 = 1591200 W.

2. For eleven 8-apartment buildings with an area of 360 m2 we obtain:

Photo = 8,170,360 = 489600 W.

3. For honey. point with dimensions 6x6x2.4 we get:

Photototal=0.25 170 6 6=1530 W;

4.For an office with dimensions 6x12 m:

Photo general = 0.25 170 6 12 = 3060 W,

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

Ph = qot Vn (tv - tn) a,

Фв = qв · Vн · (tв - tн.в.),

Ph = 0.7 18 36 4.2 (10 - (- 25)) 1.05 = 5000.91 W,

Fv.tot.=0.4 5000.91=2000 W.

Brigade house:

Ph = 0.5 1944 (18 - (- 25)) 1.05 = 5511.2 W,

School workshop:

Ph = 0.6 1814.4 (15 - (- 25)) 1.05 = 47981.8 W,

Fv = 0.2 1814.4 (15 - (- 16)) = 11249.28 W,

2.2 RCalculation of heat consumption for hot water supply forresidential and public buildings

The average heat flow (W) spent during the heating period on hot water supply to buildings is found by the formula:

F g.v. = q g.v. n f,

For 16 apartment buildings with 60 residents, the heat flow for hot water supply will be: F g.w. = 407 60 = 24420 W,

for thirteen such houses - F g.v. = 24420 · 13 = 317460 W.

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

F g.v.l. = 0.65 · F g.v. = 0.65 317460 = 206349 W

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

F g.v. = 407 · 30 = 12210 W,

for eleven such houses - F g.v. = 12210 · 11 = 97680 W.

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

F g.v.l. = 0.65 · F g.v. = 0.65 · 97680 = 63492 W.

Then the heat flow to the office water supply 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.v.l. = 0.65 F g.v. = 0.65 47690 = 31000 W

Heat flow to medical water supply. point will be:

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

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

F g.v.l. = 0.65 F g.v. = 0.65 13167 = 8559 W

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

Fgv. = 0.278 · 0.68 · 983 · 4.19 · (55 - 5) = 38930 W

Heat flow consumed for hot water supply to a school workshop in the summer:

Fgv.l = 38930 · 0.65 = 25304.5 W

Summary table of heat flows

Calculated heat flows, W
	Name	Heating	Ventilation	Technical needs

	School for 90 students
	16 sq.m. house
	Honey. paragraph
	8 apartment building
	School workshop

F total =F from +F to +F g.v. =2147318+13243+737078=2897638 W.

3. Construction of an annual scheduleheating load and selection of boilers

3.1 Construction of an annual heat load graph

The total graph of heat consumption depending on the outside air temperature is shown by broken line 6 (the break point corresponds to tn.v.), cutting off on the ordinate axis a segment equal to the maximum heat flow spent on all types of consumption (? Ph + ? Fv + ? Fg. V. + ?Ft) at the calculated external temperature tн.

Adding up the total loads I got 2.9W.

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):

Qyear = 3.6 10 -6 F m Q m n,

where F is the area of the annual heat load graph, mm?; m Q and m n are the scale of heat consumption and operating time of the boiler room, W/mm and h/mm, respectively.

Qyear = 3.6 10 -6 9871.74 23548 47.8 = 40001.67 J/year

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

3.2 Selection of coolant

We use water as a coolant. So what is the thermal design load Fr? 2.9 MW, which is less than the condition (Fr? 5.8 MW), it is allowed to use water with a temperature of 105 °C in the supply line, and in the return pipeline the water temperature is assumed to be 70 °C. At the same time, we take into account that the temperature drop in the consumer network can reach 10%.

Since some consumers require steam for technical purposes, consumers need to install additional heat exchangers.

3.3 Selection of boilers

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

Estimated power Фр = 2128 kW

Installed power Fu = 2337 kW

Heating surface area - 40.6 m?

Number of sections - 26

Dimensions 2249? 2300? 2361 mm

Maximum water heating temperature - 115? C

Efficiency when operating on gas zk.a. = 0.8

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

When operating in steam mode, power is reduced by 4 - 7%

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

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 represented by tп = f (tн, ?С), tо = f (tн, ?С). Having constructed a graph using the method given in the work for tn = 95? C; tо = 70?С for heating (it is taken into account that the temperature of the coolant in the hot water supply network should not fall below 70?С), tпв = 90?С; tov = 55? C - for ventilation, we determine the ranges of temperature changes of the coolant in the heating and ventilation networks. The values of the external temperature are plotted along the abscissa axis, and the temperature of the supply water is plotted along the ordinate axis. The origin coincides with the calculated internal temperature for residential and public buildings (18? C) and the coolant temperature, also equal to 18? C. At the intersection of perpendiculars restored to the coordinate axes at points corresponding to temperatures tп = 95 °С, tн = -25 °С, point A is found, and by drawing a horizontal line from the return water temperature of 70 °С, point B is found. By connecting points A and With the origin of coordinates, we obtain a graph of changes in the temperature of forward and return water in the heating network depending on the outside air temperature. If there is 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 supply water has an inflection point C, to the left of which f p =const. The supply of heat to heating at a constant temperature is controlled by changing the coolant flow rate. The minimum return water temperature is determined by drawing a vertical line through point C until it intersects with the return water graph. The projection of point D onto the ordinate axis shows the smallest value of pho. The perpendicular, restored from the point corresponding to the calculated outside temperature (-16? C), intersects straight lines AC and BD at points E and F, showing the maximum temperatures of forward and return water for ventilation systems. That is, the temperatures are 91 °C and 47 °C, respectively, which remain unchanged in the range from tн.в and tн (lines EK and FL). In this range of outside air 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 presented in Fig. 4.

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

Bibliography

1. Efendiev A.M. Design of energy supply for agricultural enterprises. Toolkit. Saratov 2009.

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

3. Zakharov A.A. Application of heat in agriculture. Moscow Kolos 1980.

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

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

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Appendix No. 1 to the letter of the Ministry of Economy of Russia dated November 27, 1992 No. BE-261 / 25-510

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