How to calculate the heat load on a building. Thermal calculation of the heating system: formulas, reference data and a specific example

Calculation of heat loss in the house

Boiler power determination

Features of the selection of radiators

Hydraulic calculation of water supply

Thermal calculation example

Conclusions and useful video on the topic

Moscow 2005

Moscow 2005

In what cases is the calculation of the heat load

Method of calculation

An example of calculating the thermal loads of a commercial facility

Calculation formula in Gcal

Calculation for heating radiators per area

Initial data for designing a heating system

Formulas for calculations and reference data

Analysis of calculations on a specific example

Program of the Kurgan region "Regional energy program of the Kurgan region for the period up to 2010" Basis for development

Explanatory note Rationale for the draft master plan Director General

On what basis can the heat load on the heating of the building be recalculated?

Collection of initial data on the heat load object

Energy audit of the building

Technical report

Initial data for the object

Enlarged calculation

Accurate calculation

General part and initial data

Hello dear readers! Today a small post about the calculation of the amount of heat for heating according to aggregated indicators. In general, the heating load is taken according to the project, that is, the data that the designer calculated are entered into the heat supply contract.

But often there is simply no such data, especially if the building is small, such as a garage, or some kind of utility room. In this case, the heating load in Gcal / h is calculated according to the so-called aggregated indicators. I wrote about this. And already this figure is included in the contract as the estimated heating load. How is this number calculated? And it is calculated according to the formula:

Qot \u003d α * qo * V * (tv-tn.r) * (1 + Kn.r) * 0.000001; where

α is a correction factor that takes into account climatic conditions district, it is used in cases where the calculated outdoor air temperature differs from -30 ° C;

qo is the specific heating characteristic of the building at tn.r = -30 °С, kcal/m3*С;

V - the volume of the building according to the external measurement, m³;

tv is the design temperature inside the heated building, °С;

tn.r - design outdoor air temperature for heating design, °C;

Kn.r is the infiltration coefficient, which is due to thermal and wind pressure, that is, the ratio of heat losses from the building with infiltration and heat transfer through external fences at the outdoor air temperature, which is calculated for heating design.

So, in one formula, you can calculate the heat load on the heating of any building. Of course, this calculation is largely approximate, but it is recommended in the technical literature on heat supply. Heat supply organizations also contribute this figure heating load Qot, in Gcal/h, to heat supply contracts. So the calculation is correct. This calculation is well presented in the book - V.I. Manyuk, Ya.I. Kaplinsky, E.B. Khizh and others. This book is one of my desktop books, a very good book.

Also, this calculation of the heat load on the heating of the building can be done according to the "Methodology for determining the amount of thermal energy and heat carrier in public water supply systems" of RAO Roskommunenergo of the Gosstroy of Russia. True, there is an inaccuracy in the calculation in this method (in formula 2 in Appendix No. 1, 10 to the minus third power is indicated, but it should be 10 to the minus sixth power, this must be taken into account in the calculations), you can read more about this in the comments to this article.

I fully automated this calculation, added reference tables, including the table climatic parameters all regions former USSR(from SNiP 23.01.99 "Construction climatology"). You can buy a calculation in the form of a program for 100 rubles by writing to me at e-mail [email protected]

I will be glad to comments on the article.

Designing and thermal calculation of the heating system is an obligatory stage in the arrangement of home heating. The main task of the computational measures is to determine the optimal parameters of the boiler and the radiator system.

Agree, at first glance it may seem that holding thermotechnical calculation only an engineer can do it. However, not everything is so difficult. Knowing the algorithm of actions, it will be possible to independently perform the necessary calculations.

The article details the calculation procedure and provides all the necessary formulas. For a better understanding, we have prepared an example of a thermal calculation for a private house.

The classical thermal calculation of the heating system is a summary white paper, which includes mandatory step-by-step standard calculation methods.

But before studying these calculations of the main parameters, you need to decide on the concept of the heating system itself.

Image gallery

The heating system is characterized by forced supply and involuntary removal of heat in the room.

The main tasks of calculating and designing a heating system:

most reliably determine heat losses;
determine the amount and conditions for the use of the coolant;
select the elements of generation, movement and heat transfer as accurately as possible.

And here room temperature air in winter period provided by the heating system. Therefore, we are interested in temperature ranges and their deviation tolerances for the winter season.

Most regulatory documents stipulate the following temperature ranges that allow a person to be comfortable in a room.

For non-residential premises office type up to 100 m2:

22-24°C — optimum temperature air;
1°C- allowable fluctuation.

For office-type premises with an area of more than 100 m 2, the temperature is 21-23 ° C. For non-residential premises of an industrial type, the temperature ranges vary greatly depending on the purpose of the premises and established norms labor protection.

Comfortable room temperature for each person "own". Someone likes to be very warm in the room, someone is comfortable when the room is cool - it's all quite individual

As for residential premises: apartments, private houses, estates, etc., there are certain temperature ranges that can be adjusted depending on the wishes of the residents.

And yet, for specific premises of an apartment and a house, we have:

20-22°C- residential, including children's, room, tolerance ± 2 ° С -
19-21°C- kitchen, toilet, tolerance ± 2 ° С;
24-26°C- bath, shower, swimming pool, tolerance ± 1 ° С;
16-18°C— corridors, hallways, stairwells, storerooms, tolerance +3°С

It is important to note that there are several more basic parameters that affect the temperature in the room and which you need to focus on when calculating the heating system: humidity (40-60%), the concentration of oxygen and carbon dioxide in the air (250: 1), the speed of movement of air masses (0.13-0.25 m/s), etc.

According to the second law of thermodynamics (school physics), there is no spontaneous transfer of energy from less heated to more heated mini or macro objects. A special case of this law is the "desire" to create a temperature equilibrium between two thermodynamic systems.

For example, the first system is an environment with a temperature of -20°C, the second system is a building with an internal temperature of +20°C. According to the above law, these two systems will tend to balance through the exchange of energy. This will happen with the help of heat losses from the second system and cooling in the first.

We can definitely say that the ambient temperature depends on the latitude at which the private house is located. And the temperature difference affects the amount of heat leakage from the building (+)

By heat loss is meant an involuntary release of heat (energy) from some object (house, apartment). For ordinary apartment this process is not so “noticeable” in comparison with a private house, since the apartment is located inside the building and “adjacent” to other apartments.

In a private house, heat “leaves” to one degree or another through the external walls, floor, roof, windows and doors.

Knowing the magnitude of heat loss for the most unfavorable weather conditions and the characteristics of these conditions, it is possible to calculate the power of the heating system with high accuracy.

So, the volume of heat leakage from the building is calculated by the following formula:

Q=Q floor +Q wall +Q window +Q roof +Q door +…+Q i, where

qi- the volume of heat loss from a homogeneous type of building envelope.

Each component of the formula is calculated by the formula:

Q=S*∆T/R, where

Q– thermal leakage, V;
S- the area of a particular type of structure, sq. m;
∆T– temperature difference between the ambient air and indoors, °C;
R- thermal resistance of a certain type of construction, m 2 * ° C / W.

The very value of thermal resistance for real existing materials it is recommended to take from auxiliary tables.

In addition, thermal resistance can be obtained using the following relationship:

R=d/k, where

R- thermal resistance, (m 2 * K) / W;
k- coefficient of thermal conductivity of the material, W / (m 2 * K);
d is the thickness of this material, m.

In old houses with a damp roof structure, heat leakage occurs through the upper part of the building, namely through the roof and attic. Carrying out activities on or solve the problem.

If you insulate the attic space and the roof, then the total heat loss from the house can be significantly reduced.

There are several more types of heat losses in the house through cracks in the structures, the ventilation system, kitchen hood, opening windows and doors. But it makes no sense to take into account their volume, since they make up no more than 5% of total number major heat leaks.

To maintain the temperature difference between environment and temperature inside the house, an autonomous heating system is needed that maintains desired temperature in every room of a private house.

The basis of the heating system is different: liquid or solid fuel, electric or gas.

The boiler is the central node of the heating system that generates heat. The main characteristic of the boiler is its power, namely the rate of conversion of the amount of heat per unit of time.

Having calculated the heat load for heating, we obtain the required nominal power of the boiler.

For an ordinary multi-room apartment, the boiler power is calculated through the area and specific power:

P boiler \u003d (S rooms * P specific) / 10, where

S rooms- the total area of the heated room;
R specific- specific power relative to climatic conditions.

But this formula does not take into account heat losses, which are sufficient in a private house.

There is another ratio that takes this parameter into account:

P boiler \u003d (Q losses * S) / 100, where

Boiler P- boiler power;
Q loss— heat loss;
S- heated area.

The rated power of the boiler must be increased. The reserve is necessary if it is planned to use the boiler for heating water for the bathroom and kitchen.

In most heating systems of private houses, it is recommended to use an expansion tank, in which the supply of coolant will be stored. Every private house needs hot water supply

In order to provide for a boiler power reserve, the safety factor K must be added to the last formula:

P boiler \u003d (Q losses * S * K) / 100, where

TO- will be equal to 1.25, that is, the calculated power of the boiler will be increased by 25%.

Thus, the power of the boiler makes it possible to maintain standard temperature air in the rooms of the building, as well as have an initial and additional volume hot water in the House.

Radiators, panels, underfloor heating systems, convectors, etc. are standard components for providing heat in a room. The most common parts of a heating system are radiators.

The heat sink is a special hollow, modular type alloy structure with high heat dissipation. It is made of steel, aluminum, cast iron, ceramics and other alloys. The principle of operation of the heating radiator is reduced to the radiation of energy from the coolant into the space of the room through the "petals".

aluminum and bimetal radiator heating replaced massive cast-iron batteries. Ease of production, high heat dissipation, good construction and design have made this product a popular and widespread tool for radiating heat in a room.

There are several methods in the room. The following list of methods is sorted in order of increasing accuracy of calculations.

Calculation options:

By area. N \u003d (S * 100) / C, where N is the number of sections, S is the area of \u200b\u200bthe room (m 2), C is the heat transfer of one section of the radiator (W, taken from those passports or certificates for the product), 100 W is the amount of heat flow , which is necessary for heating 1 m 2 (empirical value). The question arises: how to take into account the height of the ceiling of the room?
By volume. N=(S*H*41)/C, where N, S, C are similar. H is the height of the room, 41 W is the amount of heat flow that is necessary to heat 1 m 3 (empirical value).
By odds. N=(100*S*k1*k2*k3*k4*k5*k6*k7)/C, where N, S, C and 100 are similar. k1 - taking into account the number of cameras in the double-glazed window of the room window, k2 - thermal insulation of the walls, k3 - the ratio of the window area to the area of the room, k4 - average subzero temperature in the coldest week of winter, k5 is the number of external walls of the room (which "facing" to the street), k6 is the type of room from above, k7 is the height of the ceiling.

This is the most accurate option for calculating the number of sections. Naturally, fractional calculation results are always rounded to the next integer.

Of course, the “picture” of calculating heat for heating cannot be complete without calculating such characteristics as the volume and speed of the coolant. In most cases, the coolant is ordinary water in a liquid or gaseous state of aggregation.

The actual volume of the coolant is recommended to be calculated by summing up all the cavities in the heating system. When using a single-circuit boiler, this is best option. When using double-circuit boilers in the heating system, it is necessary to take into account the consumption of hot water for hygienic and other domestic purposes

Calculation of the volume of water heated double-circuit boiler to provide residents hot water and heating of the coolant, is made by summing up the internal volume of the heating circuit and the real needs of users in heated water.

The volume of hot water in heating system calculated by the formula:

W=k*P, where

W is the volume of the heat carrier;
P- power of the heating boiler;
k- power factor (number of liters per unit of power, equal to 13.5, range - 10-15 liters).

As a result, the final formula looks like this:

W=13.5*P

The coolant velocity is the final dynamic assessment of the heating system, which characterizes the rate of fluid circulation in the system.

This value helps to evaluate the type and diameter of the pipeline:

V=(0.86*P*μ)/∆T, where

P- boiler power;
μ — boiler efficiency;
∆T is the temperature difference between the supply water and the return water.

Using the above methods, it will be possible to obtain real parameters that are the "foundation" of the future heating system.

As an example of a thermal calculation, there is an ordinary 1-storey house with four living rooms, a kitchen, a bathroom, a "winter garden" and utility rooms.

Foundation made of monolithic reinforced concrete slab(20 cm), external walls - concrete (25 cm) with plaster, roof - ceilings made of wooden beams, roof - metal tile and mineral wool(10 cm)

Let us designate the initial parameters of the house necessary for the calculations.

Building dimensions:

floor height - 3 m;
small window of the front and back of the building 1470 * 1420 mm;
large facade window 2080*1420 mm;
entrance doors 2000*900 mm;
rear doors (exit to the terrace) 2000*1400 (700 + 700) mm.

The total width of the building is 9.5 m 2 , length 16 m 2 . Only living rooms (4 units), a bathroom and a kitchen will be heated.

For accurate calculation of heat loss on the walls from the area external walls you need to subtract the area of \u200b\u200ball windows and doors - this is a completely different type of material with its own thermal resistance

We start by calculating the areas of homogeneous materials:

floor area - 152 m 2;
roof area - 180 m 2, given the height of the attic 1.3 m and the width of the run - 4 m;
window area - 3 * 1.47 * 1.42 + 2.08 * 1.42 \u003d 9.22 m 2;
door area - 2 * 0.9 + 2 * 2 * 1.4 \u003d 7.4 m 2.

The area of the outer walls will be equal to 51*3-9.22-7.4=136.38 m2.

We turn to the calculation of heat loss on each material:

Q floor \u003d S * ∆T * k / d \u003d 152 * 20 * 0.2 / 1.7 \u003d 357.65 W;
Q roof \u003d 180 * 40 * 0.1 / 0.05 \u003d 14400 W;
Q window \u003d 9.22 * 40 * 0.36 / 0.5 \u003d 265.54 W;
Q door =7.4*40*0.15/0.75=59.2W;

And also Q wall is equivalent to 136.38*40*0.25/0.3=4546. The sum of all heat losses will be 19628.4 W.

As a result, we calculate the boiler power: P boiler \u003d Q losses * S heating_rooms * K / 100 \u003d 19628.4 * (10.4 + 10.4 + 13.5 + 27.9 + 14.1 + 7.4) * 1.25 / 100 \u003d 19628.4 * 83.7 * 1.25 / 100 \u003d 20536.2 \u003d 21 kW.

Let's calculate the number of radiator sections for one of the rooms. For all others, the calculations are similar. For example, a corner room (on the left, lower corner of the diagram) has an area of 10.4 m2.

So N=(100*k1*k2*k3*k4*k5*k6*k7)/C=(100*10.4*1.0*1.0*0.9*1.3*1.2*1.0*1.05)/180=8.5176=9.

This room requires 9 sections of a heating radiator with a heat output of 180 watts.

We proceed to the calculation of the amount of coolant in the system - W=13.5*P=13.5*21=283.5 l. This means that the coolant velocity will be: V=(0.86*P*μ)/∆T=(0.86*21000*0.9)/20=812.7 l.

As a result, the full turnover of the entire volume of the coolant in the system will be equivalent to 2.87 times per hour.

A selection of articles on thermal calculation will help determine the exact parameters of the elements of the heating system:

A simple calculation of the heating system for a private house is presented in the following overview:

All the subtleties and generally accepted methods for calculating the heat loss of a building are shown below:

Another option for calculating heat leakage in a typical private house:

This video talks about the features of the circulation of an energy carrier for heating a home:

The thermal calculation of the heating system is individual in nature, it must be carried out competently and accurately. The more accurate the calculations are made, the less the owners will have to overpay country house during operation.

Do you have experience in performing thermal calculation of the heating system? Or have questions about the topic? Please share your opinion and leave comments. The feedback block is located below.

Home > Document

PAYMENT

thermal loads and annual

heat and fuel for the boiler house

individual residential building

OOO OVK Engineering

This calculation is made to determine the annual consumption of heat and fuel required for a boiler house intended for heating and hot water supply of an individual residential building. The calculation of thermal loads is carried out in accordance with the following regulatory documents:

electrical energy

Building characteristics:

total area

Living space

Klimatol logical data of the construction area:

Place of construction: Russian Federation, Moscow region, Domodedovo

Design temperatures

air:

For designing a heating system: t = -28 ºС For designing a ventilation system: t = -28 ºС In heated rooms: t = +18 C

Correction factor α (at -28 С) – 1.032

Specific heating characteristic of the building - q = 0.57 [Kcal / mh С]

Heating period:

Duration: 214 days Average temperature of the heating period: t = -3.1 ºС Average of the coldest month = -10.2 ºС Boiler efficiency - 90%

Initial data for the calculation of hot water supply:

DHW operation

summer period

tap water

Humidity zone - "normal"

The maximum hourly loads of consumers are as follows:

For heating - 0.039 Gcal/hour For hot water supply - 0.0025 Gcal/hour For ventilation - no

The total maximum hourly heat consumption, taking into account heat losses in networks and for own needs - 0.0415 Gcal / h

gas boiler

Heating boiler power - 0.0413 Gcal / h

Boiler capacity – 0.0087 Gcal/h

Fuel - natural gas; the total annual consumption of natural fuel (gas) will be 0.0155 million Nm³ per year or 0.0177 thousand tce. per year of reference fuel. The calculation was made by: L.A. Altshuler

SCROLL

Data submitted by the regional main departments, enterprises (associations) to the Administration of the Moscow Region along with a request to establish the type of fuel for enterprises (associations) and heat-consuming installations.

General issues

Questions	Answers
Ministry (department)	Burlakov V.V.
The enterprise and its location (region, district, locality, the outside)	Individual residential building located at: Moscow region, Domodedovo st. Solovinaya, 1
The distance of the object to: - the railway station - the gas pipeline - the base of oil products - the nearest source of heat supply (CHP, boiler house) with an indication of its capacity, workload and ownership
The readiness of the enterprise to use fuel and energy resources (operating, designed, under construction) with an indication of the category	under construction, residential
Documents, approvals (conclusions), date, number, name of the organization: - on the use of natural gas, coal; - on the transportation of liquid fuel; - on the construction of an individual or expanded boiler house.	PO Mosoblgaz permission No. ______ from ___________ Permission from the Ministry of Housing and Public Utilities, Fuel and Energy of the Moscow Region No. ______ from ___________
Based on what document is the enterprise designed, built, expanded, reconstructed
The type and quantity (toe) of fuel currently used and on the basis of which document (date, number, established consumption), for solid fuel indicate its deposit, and for Donetsk coal - its brand	not used
Type of fuel requested, total annual consumption (toe) and year of start of consumption	natural gas; 0.0155 thousand tce in year; 2005 year
The year the enterprise reached its design capacity, the total annual fuel consumption (thousand tce) this year	2005 year; 0.0177 thousand tce

Boiler plants

a) the need for heat

For what needs	Attached maximum heat load (Gcal/h)		Number of hours of work per year	Annual heat demand (Gcal)		Heat demand coverage (Gcal/year)
For what needs	Existing	ruable, including	Number of hours of work per year		Design-may, including	Boiler room	energy go re-sources	Due to others



hot water supply
what needs
consumption
stven-nye boiler room
Heat loss

Note: 1. In column 4, indicate in brackets the number of hours of operation per year of technological equipment with maximum loads. 2. In columns 5 and 6 show heat supply to third-party consumers.

b) the composition and characteristics of boiler room equipment, type and annual

fuel consumption

Boiler type by groups		Fuel used			Requested fuel
Boiler type by groups		Type of bases leg (reserve-	flow rate	howling expense	Type of bases leg (reserve-	flow rate	howling expense

Operating of them: dismantled
"Ishma-50" "Ariston SGA 200"	0,050						thousand tce in year;

Note: 1. Indicate the total annual fuel consumption by groups of boilers. 2. Specify the specific fuel consumption taking into account the own needs of the boiler house. 3. In columns 4 and 7, indicate the method of fuel combustion (stratified, chamber, fluidized bed).

Heat consumers

Heat demand for production needs

Heat consumers

Name of product

products

Specific heat consumption per unit

products

Annual heat consumption

Technological fuel-consuming installations

a) the capacity of the enterprise for the production of main types of products

Product type	Annual output (specify unit of measurement)		Specific fuel consumption (kg c.f./unit. Product)
	existing	projected	actual	estimated

b) composition and characteristics of technological equipment,

type and annual fuel consumption

Note: 1. In addition to the requested fuel, indicate other types of fuel on which technological installations can operate.

Use of fuel and heat secondary resources

Fuel secondary resources			Thermal secondary resources
View, source	thousand tce	Amount of fuel used (thousand t.o.e.)	View, source	thousand tce	The amount of heat used (thousand Gcal/hour)
		Existing			Being-

PAYMENT

hourly and annual costs of heat and fuel

Maximum hourly heat consumption per

consumer heating is calculated by the formula:

Qot. = Vsp. x qot. x (Tvn. - Tr.ot.) x α [Kcal / h]

Where: Vzd. (m³) - the volume of the building; qfrom. (kcal/h*m³*ºС) - specific thermal characteristic of the building; α is a correction factor for a change in the value heating characteristic buildings at temperatures other than -30ºС.

Maximum hourly flow

The heat input for ventilation is calculated by the formula:

Qvent = Vн. x qvent. x (Tvn. - Tr.v.) [Kcal / h]

Where: qvent. (kcal/h*m³*ºС) – specific ventilation characteristic of the building;

The average heat consumption for the heating period for the needs of heating and ventilation is calculated by the formula:

for heating:

Qo.p. = Qot. x (Tvn. - Ts.r.ot.) / (Tvn. - Tr.ot.) [Kcal / h]

For ventilation:

Qo.p. = Qvent. x (Tvn. - Ts.r.ot.) / (Tvn. - Tr.ot.) [Kcal / h]

The annual heat consumption of the building is determined by the formula:

Qfrom.year = 24 x Qav. x P [Gcal/year]

For ventilation:

Qfrom.year = 16 x Qav. x P [Gcal/year]

Average hourly heat consumption for the heating period

for hot water supply residential buildings is determined by the formula:

Q \u003d 1.2 m x a x (55 - Tkh.z.) / 24 [Gcal / year]

Where: 1.2 - coefficient taking into account the heat transfer in the room from the pipeline of hot water supply systems (1 + 0.2); a - the rate of water consumption in liters at a temperature of 55ºС for residential buildings per person per day, should be taken in accordance with the chapter of SNiP on the design of hot water supply; Тх.з. - temperature cold water(plumbing) during the heating period, taken equal to 5ºС.

The average hourly heat consumption for hot water supply in the summer period is determined by the formula:

Qav.op.g.c. \u003d Q x (55 - Tkh.l.) / (55 - Tkh.z.) x V [Gcal / year]

Where: B - coefficient taking into account the decrease in the average hourly water consumption for hot water supply of residential and public buildings in the summer in relation to the heating period, is taken equal to 0.8; Tc.l. - the temperature of cold water (tap) in the summer, taken equal to 15ºС.

The average hourly heat consumption for hot water supply is determined by the formula:

Qyear of year \u003d 24Qo.p.g.vPo + 24Qav.p.g.v * (350 - Po) * V =

24Qavg.vp + 24Qavg.gv (55 – Tkh.l.)/ (55 – Tkh.z.) х V [Gcal/year]

Total annual heat consumption:

Qyear = Qyear from. + Qyear vent. + Qyear of year + Qyear wtz. + Qyear tech. [Gcal/year]

Calculation of annual fuel consumption is determined by the formula:

Wu.t. \u003d Qyear x 10ˉ 6 / Qr.n. x η

Where: qr.n. - lower calorific value reference fuel, equal to 7000 kcal/kg of reference fuel; η – boiler efficiency; Qyear is the total annual heat consumption for all types of consumers.

PAYMENT

heat loads and annual fuel quantity

Calculation of the maximum hourly heating loads:

1.1. House: Maximum hourly heating consumption:

Qmax. \u003d 0.57 x 1460 x (18 - (-28)) x 1.032 \u003d 0.039 [Gcal / h]

Total for residential building: Q max. = 0.039 Gcal/h Total, taking into account the own needs of the boiler house: Q max. = 0.040 Gcal/h

Calculation of average hourly and annual heat consumption for heating:

2.1. House:

Qmax. = 0.039 Gcal/h

Qav.ot. \u003d 0.039 x (18 - (-3.1)) / (18 - (-28)) \u003d 0.0179 [Gcal / h]

Qyear from. \u003d 0.0179 x 24 x 214 \u003d 91.93 [Gcal / year]

Taking into account the own needs of the boiler house (2%) Qyear from. = 93.77 [Gcal/year]

Total for residential building:

Average hourly heat consumption for heating Q cf. = 0.0179 Gcal/h

Total annual heat consumption for heating Q year from. = 91.93 Gcal/year

Total annual heat consumption for heating, taking into account the own needs of the boiler house Q year from. = 93.77 Gcal/year

Calculation of the maximum hourly loads on DHW:

1.1. House:

Qmax.gws \u003d 1.2 x 4 x 10.5 x (55 - 5) x 10 ^ (-6) \u003d 0.0025 [Gcal / h]

Total for residential building: Q max.gws = 0.0025 Gcal/h

Calculation of hourly averages and year new heat consumption for hot water supply:

2.1. House: Average hourly heat consumption for hot water supply:

Qav.d.h.w. \u003d 1.2 x 4 x 190 x (55 - 5) x 10 ^ (-6) / 24 \u003d 0.0019 [Gcal / hour]

Qav.dw.l. \u003d 0.0019 x 0.8 x (55-15) / (55-5) / 24 \u003d 0.0012 [Gcal / h]

Godothowl heat consumption for hot water supply: Qyear from. \u003d 0.0019 x 24 x 214 + 0.0012 x 24 x 136 \u003d 13.67 [Gcal / year] Total for DHW:

Average hourly heat consumption during the heating period Q sr.gvs = 0.0019 Gcal/h

Average hourly heat consumption during the summer Q sr.gvs = 0.0012 Gcal/h

Total annual heat consumption Q DHW year = 13.67 Gcal/year

Calculation of the annual amount of natural gas

and reference fuel :

∑ Qyear = ∑Qyear from. +QDHW year = 107.44 Gcal/year

The annual fuel consumption will be:

Vgod \u003d ∑Q year x 10ˉ 6 / Qr.n. x η

Annual natural fuel consumption

(natural gas) for the boiler house will be:

Boiler (efficiency=86%) : Vgod nat. = 93.77 x 10ˉ 6 /8000 x 0.86 = 0.0136 mln.m³ per year Boiler (efficiency=90%): per year nat. = 13.67 x 10ˉ 6 /8000 x 0.9 = 0.0019 mln.m³ per year Total : 0.0155 million nm  in year

The annual consumption of reference fuel for the boiler house will be:

Boiler (efficiency=86%) : Vgod c.t. = 93.77 x 10ˉ 6 /7000 x 0.86 = 0.0155 mln.m³ per yearBulletin

Production index of electrical, electronic and optical equipment in November 2009 compared to the corresponding period of the previous year amounted to 84.6%, in January-November 2009.

Program

In accordance with paragraph 8 of article 5 of the Law of the Kurgan region "On forecasts, concepts, programs of socio-economic development and target programs of the Kurgan region",

Explanatory note

Development of urban planning documentation for territorial planning and Rules for land use and development municipality urban settlement Nikel, Pechenga district, Murmansk region

In the cold season in our country, the heating of buildings and structures is one of the main cost items of any enterprise. And here it does not matter whether it is a residential, industrial or warehouse space. Everywhere you need to maintain a constant positive temperature so that people do not freeze, equipment does not fail, or products or materials do not deteriorate. In some cases, it is required to calculate the heat load for heating a particular building or the entire enterprise as a whole.

to optimize heating costs;
to reduce the calculated heat load;
in the event that the composition of heat-consuming equipment has changed (heaters, ventilation systems, etc.);
to confirm the calculated limit on consumed heat energy;
in the case of designing your own heating system or heat supply point;
if there are sub-subscribers consuming thermal energy, for its correct distribution;
In case of connection to the heating system of new buildings, structures, industrial complexes;
to revise or conclude a new contract with an organization supplying heat energy;
if the organization has received a notification requiring clarification of heat loads in non-residential premises;
if the organization has the opportunity to install heat meters;
in the event of an increase in heat consumption for unknown reasons.

Order of the Ministry of Regional Development No. 610 dated December 28, 2009 "On approval of the rules for establishing and changing (revising) heat loads"() establishes the right of heat consumers to calculate and recalculate heat loads. Also, such a clause is usually present in every contract with a heat supply organization. If there is no such clause, discuss with your lawyers the issue of including it in the contract.

However, in order to revise the contractual amounts of consumed heat energy, a technical report must be submitted with the calculation of new heat loads for heating the building, in which justifications for reducing heat consumption must be given. In addition, the recalculation of thermal loads is carried out after such events as:

overhaul of the building;
reconstruction of internal engineering networks;
increasing the thermal protection of the facility;
other energy saving measures.

To calculate or recalculate the heat load on the heating of buildings already in operation or newly connected to the heating system, the following work is carried out:

Collection of initial data about the object.
Conducting an energy audit of the building.
Based on the information obtained after the survey, the heat load for heating, hot water and ventilation is calculated.
Drawing up a technical report.
Coordination of the report in the organization providing heat energy.
Signing a new contract or changing the terms of an old one.

What data needs to be collected or received:

Agreement (copy) for heat supply with all annexes.
Certificate issued on company letterhead on the actual number of employees (in the case of industrial buildings) or residents (in the case of a residential building).
BTI plan (copy).
Data on the heating system: one-pipe or two-pipe.
Top or bottom filling of the heat carrier.

All these data are required, because. based on them, the heat load will be calculated, as well as all the information will be included in the final report. The initial data, in addition, will help determine the timing and volume of work. The cost of the calculation is always individual and may depend on factors such as:

area of heated premises;
type of heating system;
availability of hot water supply and ventilation.

Energy audit involves the departure of specialists directly to the facility. This is necessary in order to conduct a complete inspection of the heating system, to check the quality of its insulation. Also, during the departure, the missing data about the object are collected, which cannot be obtained except by means of a visual inspection. The types of heating radiators used, their location and number are determined. A diagram is drawn and photographs are attached. Be sure to inspect the supply pipes, measure their diameter, determine the material from which they are made, how these pipes are connected, where the risers are located, etc.

As a result of such an energy audit (energy audit), the customer will receive a detailed technical report, and on the basis of this report, the calculation of the heat loads for heating the building will already be carried out.

The technical report on the heat load calculation should consist of the following sections:

Initial data about the object.
Scheme of the location of heating radiators.
DHW outlet points.
The calculation itself.
Conclusion based on the results of the energy audit, which should include comparison table maximum current thermal loads and contractual.
Applications.
1. Certificate of membership in the SRO energy auditor.
2. Floor plan of the building.
3. Explication.
4. All appendices to the contract for energy supply.

After drawing up, the technical report must be agreed with the heat supply organization, after which changes are made to the current contract or a new one is concluded.

This room is on the first floor of a 4-storey building. Location - Moscow.

Address of the object	Moscow city
Floors of the building	4 floors
The floor on which the surveyed premises are located	first
The area of the surveyed premises	112.9 sq.m.
Floor height	3.0 m
Heating system	Single pipe
temperature graph	95-70 deg. FROM
Estimated temperature chart for the floor on which the room is located	75-70 deg. FROM
Bottling type	Upper
Estimated indoor air temperature	+ 20 degrees C
Heating radiators, type, quantity	Cast iron radiators M-140-AO - 6 pcs. Radiator bimetallic Global (Global) - 1 pc.
Diameter of pipes of the heating system	Du-25 mm
Heating supply line length	L = 28.0 m.
DHW	missing
Ventilation	missing
0.02/47.67 Gcal

Estimated heat transfer installed radiators heating, taking into account all losses, amounted to 0.007457 Gcal/h.

The maximum heat energy consumption for space heating was 0.001501 Gcal/h.

The final maximum consumption is 0.008958 Gcal/hour or 23 Gcal/year.

As a result, we calculate the annual savings for heating this room: 47.67-23 = 24.67 Gcal / year. Thus, it is possible to reduce the cost of heat energy by almost half. And if we take into account that the current average cost of Gcal in Moscow is 1.7 thousand rubles, then the annual savings in monetary terms will be 42 thousand rubles.

The calculation of the heat load on the heating of the building in the absence of heat meters is carried out according to the formula Q \u003d V * (T1 - T2) / 1000, where:

V- the volume of water consumed by the heating system is measured in tons or cubic meters,
T1- hot water temperature. It is measured in C (degrees Celsius) and the temperature corresponding to a certain pressure in the system is taken for calculations. This indicator has its own name - enthalpy. If it is impossible to accurately determine the temperature, then average values \u200b\u200bof 60-65 C are used.
T2- temperature of cold water. Often it is almost impossible to measure it, and in this case constant indicators are used, which depend on the region. For example, in one of the regions, in the cold season, the indicator will be 5, in the warm season - 15.
1 000 - coefficient for obtaining the result of the calculation in Gcal.

For a heating system with a closed circuit, the heat load (Gcal / h) is calculated in a different way: Qot \u003d α * qo * V * (tin - tn.r) * (1 + Kn.r) * 0.000001, where:

α - a coefficient designed to correct climatic conditions. It is taken into account if the street temperature differs from -30 C;
V- the volume of the building according to external measurements;
qo- specific heating index of the building at a given tn.r = -30 C, measured in Kcal / m3 * C;
tv is the calculated internal temperature in the building;
tn.r- estimated street temperature for drafting a heating system;
Kn.r is the infiltration coefficient. It is due to the ratio of heat losses of the calculated building with infiltration and heat transfer through external structural elements at the street temperature, which is set within the framework of the project being drawn up.

If for 1 sq.m. area requires 100 W of thermal energy, then a room of 20 sq.m. should receive 2,000 watts. A typical eight-section radiator puts out about 150 watts of heat. We divide 2,000 by 150, we get 13 sections. But this is a rather enlarged calculation of the thermal load.

The exact calculation is carried out according to the following formula: Qt = 100 W/sq.m. × S(rooms) sq.m. × q1 × q2 × q3 × q4 × q5 × q6 × q7, where:

q1- type of glazing: ordinary = 1.27; double = 1.0; triple = 0.85;
q2– wall insulation: weak or absent = 1.27; wall laid out in 2 bricks = 1.0, modern, high = 0.85;
q3- the ratio of the total area window openings to floor area: 40% = 1.2; 30% = 1.1; 20% - 0.9; 10% = 0.8;
q4- minimum outdoor temperature: -35 C = 1.5; -25 C \u003d 1.3; -20 C = 1.1; -15 C \u003d 0.9; -10 C = 0.7;
q5- the number of external walls in the room: all four = 1.4, three = 1.3, corner room = 1.2, one = 1.2;
q6- type of design room above the design room: cold attic = 1.0, warm attic = 0.9, residential heated room = 0.8;
q7- ceiling height: 4.5 m = 1.2; 4.0 m = 1.15; 3.5 m = 1.1; 3.0 m = 1.05; 2.5 m = 1.3.

The thermal calculation of the heating system seems to most to be easy and does not require special attention occupation. A huge number of people believe that the same radiators should be chosen based only on the area of \u200b\u200bthe room: 100 W per 1 sq. m. Everything is simple. But this is the biggest misconception. You cannot limit yourself to such a formula. What matters is the thickness of the walls, their height, material and much more. Of course, you need to set aside an hour or two to get the numbers you need, but everyone can do it.

To calculate the heat consumption for heating, you need, firstly, a house project.

The plan of the house allows you to get almost all the initial data that is needed to determine the heat loss and the load on the heating system

Secondly, data on the location of the house in relation to the cardinal points and the construction area will be needed - the climatic conditions in each region are different, and what is suitable for Sochi cannot be applied to Anadyr.

Thirdly, we collect information about the composition and height of the outer walls and the materials from which the floor (from the room to the ground) and the ceiling (from the rooms and outward) are made.

After collecting all the data, you can get to work. Calculation of heat for heating can be performed using formulas in one to two hours. You can, of course, use a special program from Valtec.

To calculate the heat loss of heated rooms, the load on the heating system and heat transfer from heating devices, it is enough to enter only the initial data into the program. A huge number of functions make it an indispensable assistant for both the foreman and the private developer.

It greatly simplifies everything and allows you to get all the data on heat losses and hydraulic calculation heating systems.

The calculation of the heat load for heating involves the determination of heat losses (Tp) and boiler power (Mk). The latter is calculated by the formula:

Mk \u003d 1.2 * Tp, where:

Mk - thermal performance of the heating system, kW;
Tp - heat loss at home;
1.2 - safety factor (20%).

A 20% safety factor makes it possible to take into account the possible pressure drop in the gas pipeline during the cold season and unforeseen heat losses (for example, broken window, low-quality thermal insulation entrance doors or extreme cold). It allows you to insure against a number of troubles, and also makes it possible to widely regulate the temperature regime.

As can be seen from this formula, the power of the boiler directly depends on the heat loss. They are not evenly distributed throughout the house: the outer walls account for about 40% of the total value, the windows - 20%, the floor gives 10%, the roof 10%. The remaining 20% disappear through the doors, ventilation.

Poorly insulated walls and floors, a cold attic, ordinary glazing on windows - all this leads to large heat losses, and, consequently, to an increase in the load on the heating system. When building a house, it is important to pay attention to all the elements, because even ill-conceived ventilation in the house will release heat into the street.

The materials from which the house is built have the most direct impact on the amount of heat lost. Therefore, when calculating, you need to analyze what the walls, and the floor, and everything else consist of.

In the calculations, to take into account the influence of each of these factors, the appropriate coefficients are used:

K1 - type of windows;
K2 - wall insulation;
K3 - the ratio of floor area and windows;
K4 - minimum temperature on the street;
K5 - the number of external walls of the house;
K6 - number of storeys;
K7 - the height of the room.

For windows, the heat loss coefficient is:

ordinary glazing - 1.27;
double-glazed window - 1;
three-chamber double-glazed window - 0.85.

Naturally, the last option will keep the heat in the house much better than the previous two.

Properly executed wall insulation is the key not only to a long life of the house, but also comfortable temperature in the rooms. Depending on the material, the value of the coefficient also changes:

concrete panels, blocks - 1.25-1.5;
logs, timber - 1.25;
brick (1.5 bricks) - 1.5;
brick (2.5 bricks) - 1.1;
foam concrete with increased thermal insulation - 1.

How more area windows relative to the floor, the more heat the house loses:

The temperature outside the window also makes its own adjustments. At low rates of heat loss increase:

Up to -10С - 0.7;
-10C - 0.8;
-15C - 0.90;
-20C - 1.00;
-25C - 1.10;
-30C - 1.20;
-35C - 1.30.

Heat loss also depends on how many external walls the house has:

four walls - 1.33;%
three walls - 1.22;
two walls - 1.2;
one wall - 1.

It’s good if a garage, a bathhouse or something else is attached to it. But if it is blown from all sides by winds, then you will have to buy a more powerful boiler.

The number of floors or the type of room that is located above the room determine the K6 coefficient as follows: if the house has two or more floors above, then for calculations we take the value 0.82, but if it is an attic, then for warm - 0.91 and 1 for cold .

As for the height of the walls, the values \u200b\u200bwill be as follows:

4.5 m - 1.2;
4.0 m - 1.15;
3.5 m - 1.1;
3.0 m - 1.05;
2.5 m - 1.

In addition to the above coefficients, the area of \u200b\u200bthe room (Pl) and the specific value of heat loss (UDtp) are also taken into account.

The final formula for calculating the heat loss coefficient:

Tp \u003d UDtp * Pl * K1 * K2 * K3 * K4 * K5 * K6 * K7.

The UDtp coefficient is 100 W/m2.

Heat consumers

Maximum heat loads (Gcal/h)

Technology

Heat consumers

Hot water supply

Technology

Total for residential building

The house for which we will determine the load on the heating system has double-glazed windows (K1 \u003d 1), foam concrete walls with increased thermal insulation (K2 \u003d 1), three of which go outside (K5 \u003d 1.22). The area of windows is 23% of the floor area (K3=1.1), on the street about 15C frost (K4=0.9). The attic of the house is cold (K6=1), the height of the premises is 3 meters (K7=1.05). The total area is 135m2.

Fri \u003d 135 * 100 * 1 * 1 * 1.1 * 0.9 * 1.22 * 1 * 1.05 \u003d 17120.565 (Watts) or Fri \u003d 17.1206 kW

Mk \u003d 1.2 * 17.1206 \u003d 20.54472 (kW).

Calculation of load and heat loss can be done independently and quickly enough. You just need to spend a couple of hours putting the source data in order, and then just substitute the values into the formulas. The numbers that you will receive as a result will help you decide on the choice of a boiler and radiators.