How to independently calculate the heat loss of a building. Calculation of heat losses to the surrounding space

Calculation of the heating system, hot water supply and ventilation

Explanatory note to the course work on the discipline

"Heating, ventilation and air conditioning"

Completed:

student of group 31 E

Zakharets A.V.

Supervisor

Art. teacher of the department T

Koksharov M.V.

According to the option you need:

1) Calculate the heat loss of the building.

3) Calculate the hot water supply system.

4) Draw an isometric diagram of the hot water supply system, indicate the diameters of the pipelines

5) Calculate the ventilation system, determine the amount of heat for heating the ventilated air.


UDC 621.313.333

The course work contains 28 pages, 7 figures, 4 tables, 5 sources, 2 applications.

Heat losses, enclosing structures, heating system, radiator, coolant, infiltration, hot water supply, riser, plank bed, pipeline, ventilation.

The object of study is a two-story residential building.

The purpose of the work is the development and consolidation of methods for calculating the heat losses of a building, heating systems, hot water supply, ventilation.

Research methods - calculation and graphic.

Coursework done in a text editor Microsoft Word 2007


Introduction. 5

1 Initial data. 6

2 Calculation of heat losses of the building. 7

2.1 Filling in the table.. 7

2.2 Calculation of the diameters of pipelines of the heating system. 20

3 Calculation DHW systems.. 23

3.1 Determination of the estimated water consumption in DHW systems .. 23

3.2 Determining the diameters of the DHW pipeline .. 23

4 Calculation of the ventilation system. 26

4.1 Consumption supply air. 26

4.2 Determination of heat consumption for heating ventilated air. 26

Conclusion. 28

Bibliographic list. 29

Annex A

Annex B


Introduction

The heat loss calculation is milestone design of heating, hot water and ventilation systems.



To determine the thermal power covering maximum load on the heating system, it is necessary to know the heat loss of the building in the most severe settlement part of the cold period of the year. To address the issue of compliance of the level of heat consumption by the heating system of the building with modern requirements, especially considering the problem of energy saving, it is necessary to determine the heat loss of the building for the entire heating period.

There are various approaches to choosing the calculated values ​​of the thermal conductivity coefficients building materials. At the same time, care in choosing the value given coefficient extremely important. It is also necessary to correctly evaluate the values ​​of the heat transfer coefficients on the surfaces of the fences, especially the heat transfer coefficient on inner surface, because if its value is too high, the calculated temperature on the inner surface, for example, of a window, will also be too high. When determining the heat loss of a building, it is important to correctly assess the heat transfer coefficients of the building envelope.

The paper presents calculations of the heat loss of the building and the need for heat for heating infiltration air, calculated and designed heating, hot water and ventilation systems.

The purpose of this work is to gain knowledge, skills in the calculation and design of heating, hot water and ventilation systems.

Initial data

Figure 1.1 - Plan of the first (second) floor of the building

Table 1.1 - Initial data


Calculation of building heat losses

With a careful approach to the design of a home heating system, it is necessary to start by calculating the heat loss of the building. Heat loss in the house occurs through walls, windows, entrance doors, roof and floor of the first floor. Heat also escapes with air when it infiltrates through gaps in structures, windows and doors.

For the convenience of calculating and presenting information, the result of the second section of this term paper the table will be filled. For each room, 25 parameters will be determined or calculated. The calculation is made in accordance with SNiP 23-02-2003 " Thermal protection buildings."

Filling in the table

2.1.1 Name of the room

This column indicates the room number according to the building plan. Usually the numbering of rooms starts from the entrance and goes clockwise. The first digit is the floor number, the rest are the room number.

Figure 2.1 - Plan of the first floor of the task

Figure 2.2 - Plan of the second floor of the task.

2.1.2 Outside air temperature.

In this column, in accordance with SNiP 23-01-99 "Construction climatology", the air temperature of the coldest five-day period with a security of 0.92 tn, ° C for the desired city or region is indicated.

For St. Petersburg t n \u003d -26 ° С

2.1.3 Design indoor air temperature

In this column, in accordance with GOST 30494-2011 "Residential and public buildings" indicates optimum temperature indoor air t in, ° C, depending on its type. Yes, for living rooms

t in \u003d 18 - 20 ° С, for bathrooms t in \u003d 24 - 26 ° С, for kitchens t in \u003d 19 - 21 ° С.

In calculations for bathrooms, we take t in = 25 ° C, for all other rooms t in = 20 ° C

2.1.4 Name of the surface.

The following abbreviations are introduced to designate enclosing structures:

HC - outer wall

TO - window

DN - outer door

2.1.5 Surface orientation

The orientation of the vertical enclosing structures to the cardinal points is indicated:

B - east

2.1.6 Surface length

The length or, in the case of a vertical surface, the height of the building envelope is indicated in meters.

2.1.7 Surface width

Specifies the width of the surface in meters.

2.1.8 Surface area

The surface area is defined as the product of the length (height) and the width of the surface according to the formula:

, (2.1)

a – length (height), m

b - width, m

When calculating heat losses, the area of ​​individual fences A, m2, is determined in compliance with the following rules measurements:

1. The area of ​​windows, doors and lanterns is measured by the smallest building opening.

2. The area of ​​the ceiling and floor is measured between the axes of the inner walls and the inner surface of the outer wall. The area of ​​walls and floors located on the ground, including on logs, is determined with a conditional breakdown of them into zones.

3. The area of ​​the outer walls is measured

In plan - along the outer perimeter between the axes of the inner walls and the outer corner of the wall;

In height - on all floors except the lower one: from the level of the finished floor to the floor of the next floor. On the last floor the top of the outer wall coincides with the top of the cladding, or attic floor. On the lower floor, depending on the floor design: a) from the inner surface of the floor on the ground; b) from the preparation surface for the floor structure on the logs; c) from the lower edge of the ceiling over an unheated underground or basement.

4. When determining heat loss through internal walls their areas are measured along the inner perimeter. Heat loss through the internal enclosures of the premises can be ignored if the air temperature difference in these premises is 3°C or less.

The transfer of heat from the room through the structure of the floor or wall and the thickness of the soil with which they come into contact is subject to complex laws. To calculate the resistance to heat transfer of structures located on the ground, a simplified method is used. The floor surface on the ground is divided into strips 2 m wide, parallel to the junction of the outer wall and the ground surface. The counting of zones starts along the wall from the ground level, and if there are no walls along the ground, then zone I is the floor strip closest to the outer wall. The next two strips will be numbered II and III, and the rest of the floor will be zone IV. (see figure 2.3)

Thus, total area the floor is divided into zones and the area is entered in a column for each floor zone, and for the first zone, the area in the corners of the building is counted twice.

Figure 2.3 - The principle of dividing the floor of the building into zones

Figure 2.4 - Splitting the floor of the 1st floor into zones

2.1.9 Calculated temperature difference

,ºС is determined as the difference between the temperatures of the indoor air in the room and the temperature of the outdoor air of the coldest five-day period according to the formula:

(2.2)

2.1.10 Coefficient n

We choose the coefficient n, which takes into account the position of the building envelope in relation to the outside air:

n = 1. External walls and coverings (including those ventilated by outside air), attic floors (with a roof made of piece materials) and over driveways; ceilings over cold (without enclosing walls) undergrounds in the Northern building-climatic zone.

n = 0.9. Ceilings over cold cellars communicating with outside air; attic ceilings (with a roof made of roll materials); ceilings over cold (with enclosing walls) undergrounds and cold floors in the Northern building-climatic zone.

n=0.75. Ceilings over unheated basements with skylights in the walls.

n = 0.6. Ceilings over unheated basements without skylights in the walls, located above ground level.

n = 0.4. Ceilings over unheated technical undergrounds located below ground level

2.1.11 Heat transfer coefficient of the building envelope

The heat transfer coefficient of the enclosing structure k, W / (m 2 ∙ ° С) - a value numerically equal to surface density heat flow passing through the building envelope with a difference between the internal and outdoor temperature air is calculated by the formula:

where R i is the normative value of the heat transfer resistance of the i-th zone of the floor.

For each zone of an uninsulated floor, standard values heat transfer resistance:

zone I - R I \u003d 2.1 m 2 ° C / W;

zone II - R II \u003d 4.3 m 2 ° C / W;

zone III - R III \u003d 8.6 m 2 ° C / W;

zone IV - R IV \u003d 14.2 m 2 ° C / W.

2.1.12 Main heat losses

The formula for calculating the main heat loss Q main, W of the room through the building envelope:

(2.5)

where k is the heat transfer coefficient of the building envelope, W / (m 2 ∙ ° С);

A - surface area, m 2

2.1.13 Additional loss factor β 1

The addition to the orientation of the fence along the cardinal points is accepted for all external vertical fences or vertical projections of external inclined fences:

· for northern, northeastern, northwestern, eastern orientation ß 1 = 0.1;

southeastern and western ß 1 = 0.05;

southern and southwestern ß 1 = 0.

Figure 2.5 - The value of the coefficient ß 1

2.1.14 Additional loss factor β 2

The addition for a corner room with two or more external walls takes into account that the radiation temperature in such a room is lower than in a standard room. Therefore, in corner room in a residential building, the internal air temperature is taken 2 ° C higher than in an ordinary room, and in buildings for other purposes, increased heat losses are taken into account by adding ß 2 \u003d 0.05 to the main heat losses of vertical external fences.

2.1.15 Additional loss factor β 3

The addition to the penetration of cold air through external doors into a building that is not equipped with an air-thermal curtain, when they are opened for a short time, is taken into account as the main heat loss of the doors. So, in a building with a height H for triple doors with two vestibules , for double doors with tambour , for double doors without vestibule , for single doors . For external gates in the absence of a vestibule and air curtain heat losses are calculated with an additive, and if there is a vestibule at the gate - with an additive. The specified additives do not apply to summer and spare exterior doors and gates.

2.1.16 Total additional loss factor

The total coefficient of additional losses is determined by the formula:

(2.6)

2.1.17 Heat losses taking into account additional losses Q β

To find heat losses, taking into account additional losses, it is necessary to multiply the values ​​of the twelfth and sixteenth columns, i.e. the influence of additional coefficients on the main heat losses is taken into account.

2.1.18 Rated air permeability

Rated air permeability G n is the maximum permitted air permeability of the structure for any weather conditions, taken in accordance with SNiP 23-02-2003, the values ​​​​of which are given in table. 2.1

Table 2.1 - Values ​​G n

fencing Breathability G n, kg / (m 2 h)
1. outer wall, overlapping and covering a residential, public, administrative and domestic building or premises 0,5
2. Exterior wall, slab and cover production building or premises 1,0
3. The joint between the panels of the outer walls of the building: residential industrial 0,5* 1,0*
4. Entrance door to the apartment 1,5
5. Entrance door to a residential, public, household building 7,0
6. Window and balcony door of a residential, public, domestic building or premises in wooden binding; window, lantern of industrial building with air conditioning 6,0
7. Window and balcony door of a residential, public, domestic building or premises in plastic or aluminum binding 5,0
8. Window, door, gate of the production building 8,0
9. Industrial building lantern 10,0

2.1.19 Air pressure difference

The flow rate of outdoor air entering the premises as a result of infiltration under design conditions depends on the space-planning solution of the building, as well as the density of windows, balcony doors, and stained-glass windows. The task of engineering calculation is reduced to determining the flow rate of infiltration air G inf, kg / h, through separate fences of each room. Infiltration through walls and coverings is small, so it is usually neglected and calculated only through the filling of skylights, as well as through closed doors and gates, including those that, under normal operating mode do not open. The heat costs for air rushing in through opening doors and gates in the design mode are taken into account as additions to the main heat losses through the entrance doors and gates.

The calculation reveals the maximum possible infiltration, so each window or door is considered to be on the windward side of the building.

The calculated pressure difference Δp, Pa for a window or door of each floor is determined by the formula:

For doors:

(2.9)

R inf.ok R inf.dv - the required resistance to air penetration of the window and door, respectively, m 2 ∙ h / kg;

Δр – calculated pressure difference, Pa;

Δр 0 – 10 Pa.

2.1.21 Infiltration heat transfer coefficient

Coefficient taking into account the influence of transmission heat flow:

k = 0.7. For wall butt panels and for triple-glazed windows;

k = 0.8. For windows and balcony doors with separate bindings;

k = 1. For windows and balcony doors with paired or adjacent sashes.

2.1.22 Heat consumption for infiltration

Heat consumption for infiltration Q inf, W is calculated by the formula:

2.1.24 Power of the heater unit

As heater cast-iron radiator M-140, which is widely known in the CIS, was chosen. Cast iron sectional radiators are traditional appliances for our country.

Their main advantage is the ability to use open systems. Unlike other radiators, cast iron ones are practically insensitive to system emptying, that is, they allow you to drain water from it as often as you like. When cast iron is poured on its surface, a particularly durable layer with high content silicon, therefore, in the raw form, cast iron is quite resistant to corrosion, including from the effects of solid particles present in the coolant. Speaking of performance cast iron radiators, it should be noted their high thermal conductivity and large thermal inertia.

Radiator sections are cast from gray cast iron, they can be combined into devices of various sizes. Sections are connected on nipples with gaskets made of cardboard, rubber or paronite.

Let's take the power of one section of the M-140 radiator equal to 140 watts.

There is no heating in the bathroom. The room is heated by installing a heated towel rail on the DHW pipeline. Let's take the power of the heated towel rail equal to 260 watts.

2.1.25 Number of heating appliances

In order to find the number of sections of the M-140 radiator for one room, you need to divide the total heat loss of this room by the power of one section of the M-140 radiator.

General thermal load the first floor of the building is 25.152 kW, the second floor is 23.514 kW.

All calculations of the previous paragraphs are performed for each floor of the building and are tabulated in Appendix A (for the first floor) and Appendix B (for the second floor)

HEAT LOSS CALCULATION

UNINSULATED PIPING

FOR ABOVE-GROUND LAYING

METHODOLOGICAL INSTRUCTIONS

Introduction

This document discusses the features of calculating heat losses by uninsulated pipelines of heat networks during above-ground laying and proposes a practical method for performing the calculation.

The calculation of heat losses by insulated pipelines must be carried out in accordance with the methods set forth in the current normative documents/12/. Characteristic for this situation is that the heat flow is mainly determined by the thermal resistance of the thermal insulation. In this case, the heat transfer coefficient on the outer surface of the cover layer has little effect on the amount of heat loss and, therefore, can be taken according to average values.

The operation of a heating network pipeline without thermal insulation is an atypical situation, since, according to the norms, all heat pipelines must have thermal insulation in order to avoid significant heat losses. That is why no regulatory documents provide methods for calculating the heat loss of pipelines for this case.

However, during the operation of heat networks, situations can and do arise when separate sections pipelines are deprived of thermal insulation. To ensure the possibility of calculating heat losses by such pipelines, this method has been developed. It is based on the most general theoretical dependencies for the heat transfer of a pipeline under conditions of forced convection, which are given in educational and reference literature.

According to customer's requirement, all formulas and calculated values are given not in the international system of units, but in relation to the measurement of heat loss in kcal/hour.

1. Theoretical basis heat loss calculation

uninsulated pipelines

with above-ground laying

The pipeline of the heating network is a horizontally located heated pipe, blown by the wind or located in calm air. Therefore, the heat transfer of such a pipeline can be determined from known dependencies using the heat transfer coefficient through the pipe wall:

Q = Fp · (Tp - Tv) / K, (1.1)

K = 1 / (1/αp + δm/λm + 1/αw), (1.2)

Q

αp

Fp

Tp

TV

To

αp

δm

λm

aw

Tp

heat transfer of the pipeline, kcal/hour;

area of ​​the outer surface of the pipeline, m2;

outside air temperature, °С.

heat transfer coefficient through the wall of the considered pipeline, kcal/(h m2 °С);

heat transfer coefficient on the outer surface of the pipeline, kcal/(h m2 °C);

thickness of the metal wall of the pipe, m;

thermal conductivity of the pipe wall material, kcal/(h m °C);

heat transfer coefficient on the inner surface of the pipeline, kcal/(h m2 °C);

temperature of the outer surface of the pipeline, °C;

As the calculated temperatures, the average temperatures for the period under consideration should be taken. At the same time, the pipeline surface temperature can be taken equal to the water temperature in the pipeline, since the thermal resistance of the pipe wall δm/λm and resistance to heat transfer on the inner surface 1/αw for a clean pipe, many times less than the resistance to heat transfer on the outer surface 1/αp . This assumption makes it possible to significantly simplify the calculation and reduce the number of required initial data, since then it is not required to know the water velocity in the pipe, the pipe wall thickness, and the degree of wall contamination on the inner surface. The calculation error associated with such a simplification is small and much less than the errors associated with the uncertainty of other calculated values.

The area of ​​the outer surface of the pipeline is determined by its length and diameter:

Fp = π Dp L, (1.3)

Taking into account the above, expression (1) can be converted to the form:

Q = αp π Dp L (Tp - Tv), (1.4)

The most important thing in calculating heat losses is the correct determination of the heat transfer coefficients on the outer surface of the pipeline. The issue of heat transfer from a single pipe is well studied, and the calculated dependences are given in textbooks and reference books on heat transfer. According to the theory, the total heat transfer coefficient is defined as the sum of the convective and radiant heat transfer coefficients:

αp = αk + αl (1.5)

The coefficient of convective heat transfer depends on the air velocity and flow direction with respect to the axis of the pipeline, the diameter of the pipeline, and the thermophysical characteristics of the air. In the general case, the expression for determining the heat transfer coefficient on the outer surface of the pipeline with transverse airflow will be:

in the laminar mode of air movement (Reynolds criterion Re less than 1000)

αc = 0.43 βφ Re0.5 λv / Dn (1.6)

In the transitional and turbulent regime of air movement (Reynolds criterion Re equal to or greater than 1000)

αc = 0.216 βφ Re0.6 λv / Dn , (1.7)

Re = U β u Dn/v in , (1.8)

U

βu

estimated air speed;

correction factor that takes into account the height of the pipeline above the ground and the nature of the terrain.

7. Determine the radiant heat transfer coefficient:

αl \u003d 4.97 εp (((Tp + 273) / 100)4 - ((Tv + 273) / 100) 4) / (Tp - Tv) (3.4)

8. Determine the total heat transfer coefficient:

αp = αk + αl (3.5)

9. Determine the sentries heat loss pipeline:

Q = αp π Dp L (Tp - Tv) / 1000 (3.6)

10. We determine the heat loss, for billing period time, Gcal/hour:

QN = 24 QN / 1000000, (3.7)

where N - the number of days in the billing period.

Further actions should be taken if there is concern that the temperature drop in the area is large and the calculation should be performed according to a non-linear relationship. For further calculation, the flow rate of the coolant in the area must be known.

11. Determine the module of the exponent BUT L :

BUT L = αп π Dп L / (106 gw ) (3.8)

If the obtained value slightly differs from 0, then the error in calculating the heat loss is approximately half of the calculated value. So, if the obtained value is equal to 0.05, then we can assume that the heat losses were determined with an accuracy of about 2.5%. If the obtained calculation accuracy suits, then go to step 13. If necessary, you can correct the heat loss value in accordance with a certain error:

Q = Q (1 - AL / 2) (3.9)

12. If the value of the module of the exponent BUT L greater than 0.05, or if a higher accuracy of calculation is required, we calculate the decrease in the temperature of the coolant in the area due to heat losses according to the exponential dependence:

Tw = ( Tw - T in ) (1 - e--A L )

13. We determine the final temperature of the coolant to make sure that the pipeline does not freeze:

Twk = Tw - ∆Tw (3.10)

13. Determine the corrected value of heat loss:

Q = 1000 Gw ∆Tw (3.11)

14. We determine the adjusted heat loss for the estimated period of time in accordance with clause 10.

4. An example of calculating the heat losses of the pipeline

Initial data:

It is required to determine the heat loss by the supply pipeline for February with the following initial data:

Dp = 426 mm L= 750 m Tw = 78°С, T in = -21 °С, Uv = 6.4 m/s,

gw = 460 t/h, N = 28 days, rugged terrain.

Calculation:

1. We determine according to the tables of Appendix A at T in = -21 °С: λv = 1,953

vv = 11,69

2. According to table 1, we determine for rough terrain: βu = 0,707

3. We take the average value: βφ , = 0,821

4. We calculate: Re= 1000 6.4 0.707 426 / 11.69 = 164890

5. We calculate: αk = 2.16 0.821 1625670.6 1.953 / 420 = 10.975

6. We take the average value: εп = 0,9

7. We calculate:

αl = 4.97 0.9 (((78+273)/100)4 – ((-21+273)/100)4) / (78+21) = 4.348

8. We calculate: αp = 10,975 + 4,348 = 15,323

9. We calculate:

Q= 16.08 3.14 420 750 (78+21) / 1000 = 1522392 kcal/hour

11. We calculate: BUT L = 16.08 3.14 420 750 / (106 460) = 0.03343

Therefore, heat losses were determined with an error of about 0.03343 / 2 100 = 1.7%. Non-linear dependency calculations are not required. To correct the value of heat loss, we calculate:

Q= 1522392 (1 - 0.03343 / 2) = 1496945 kcal/hour

12. We calculate: Tw = 1496945 /(103 460) = 3.254 °С

13. Calculate: Q N = 24 1496945 28 / 1000000 = 1005.95 Gcal

When calculated by exponential dependence, we would get the following results:

Tw = (78 + 21) (1 - EXP(0.03343)) = 3.255 °С

Q= 1000 460 3.255 = 1497300 kcal/hour

Q N = 24 1497300 28 / 1000000 = 1006.2 Gcal

Annex A

Thermophysical characteristics of air

Table A1 - Air thermal conductivity coefficients λw 102

TV, °C

TV< 0

TV > 0

Table A2 - Air kinematic viscosity coefficients 106

TV, °С

TV< 0

TV > 0


Literature

1. Nashchokin VV Technical thermodynamics and heat transfer. Tutorial for non-energy specialties of universities - M.: Higher School, 1975 - 496 p. ill.

2. Internal sanitary devices. At 3 pm Part I. Heating / V. N. Bogoslovsky, B. A. Krupnov, A. N. Skanavi and others: Ed. I. G. Staroverov and Yu. I. Schiller. - 4th ed., revised. and additional -M.: Stroyizdat, 1990 - 344 p.: ill. - (Designer's Handbook).

3. Nesterenko A. V. Fundamentals of thermodynamic calculations of ventilation and air conditioning - 3rd ed., revised. and additional -M.: Higher School, 1971 - 460 p. ill.

Choose a city Choose a city Brest Vitebsk Volgograd Dnepropetrovsk Yekaterinburg Zaporozhye Kazan Kyiv Lugansk Lvov Minsk Moscow Nizhny Novgorod Novosibirsk Odessa Omsk Perm Riga Rostov-on-Don Samara St. Petersburg Simferopol Ufa Kharkiv Chelyabinsk Chernigov t out = - o C

Enter the air temperature in the room; t ext = + o C

Heat loss through walls expand collapse

Front view Default Without ventilated air gap With ventilated air gap α =

External wall area, sq.m.

Thickness of the first layer, m.

Thickness of the second layer, m.

Thickness of the third layer, m.

Heat loss through walls, W

Heat loss through windows expand collapse

Select glazing

Default Single glazing Double glazing Single glazing with selective coating Double glazing with argon filling Double glazing in separate sashes Two single-chamber double-glazed windows in twin bindings k =

Enter the area of ​​windows, sq.m.

Heat loss through windows

Heat loss through ceilings expand collapse

Choose the type of ceiling

Default is attic. Between ceiling and roof air layer Attic. Roof close to ceiling Ceiling under unheated attic α =

Enter the ceiling area, sq.m.

First layer material Select material Concrete Reinforced concrete Foam concrete 1000 kg/cu.m. Foam concrete 800 kg/cu.m. Foam concrete 600 kg/cu.m. Gas block D400 Aeroc glued Slag concrete Cement-sand mortar Porotherm P+W on thermoiz. mortar Masonry from hollow ceramics. brick masonry silicate brick Solid ceramic masonry. bricks Wood Plywood Fibreboard Chipboard Mineral wool Styrofoam Styrofoam Drywall λ =

Thickness of the first layer, m.

Second layer material Select material Concrete Reinforced concrete Foam concrete 1000 kg/cu.m. Foam concrete 800 kg/cu.m. Foam concrete 600 kg/cu.m. Gas-block D400 Aeroc on glue Slag concrete Cement-sand mortar Porotherm P + W on thermoiz. mortar Masonry from hollow ceramics. brick Masonry of silicate brick Masonry of solid ceramics. bricks Wood Plywood Fibreboard Chipboard Mineral wool Styrofoam Styrofoam Drywall λ =

Thickness of the second layer, m.

Third layer material Select material Concrete Reinforced concrete Foam concrete 1000 kg/cu.m. Foam concrete 800 kg/cu.m. Foam concrete 600 kg/cu.m. Gas-block D400 Aeroc on glue Slag concrete Cement-sand mortar Porotherm P + W on thermoiz. mortar Masonry from hollow ceramics. brick Masonry of silicate brick Masonry of solid ceramics. bricks Wood Plywood Fibreboard Chipboard Mineral wool Styrofoam Styrofoam Drywall λ =

Thickness of the third layer, m.

Heat loss through the ceiling

Heat loss through the floor expand collapse

Select type of floor

By default Above a cold basement communicating with outside air Above an unheated basement with skylights in the walls Above an unheated basement without skylights in the walls Above technical underground below ground level Floor on the ground α =

Enter floor area, sq.m.

First layer material Select material Concrete Reinforced concrete Foam concrete 1000 kg/cu.m. Foam concrete 800 kg/cu.m. Foam concrete 600 kg/cu.m. Gas-block D400 Aeroc on glue Slag concrete Cement-sand mortar Porotherm P + W on thermoiz. mortar Masonry from hollow ceramics. brick Masonry of silicate brick Masonry of solid ceramics. bricks Wood Plywood Fibreboard Chipboard Mineral wool Styrofoam Styrofoam Drywall λ =

Thickness of the first layer, m.

Second layer material Select material Concrete Reinforced concrete Foam concrete 1000 kg/cu.m. Foam concrete 800 kg/cu.m. Foam concrete 600 kg/cu.m. Gas-block D400 Aeroc on glue Slag concrete Cement-sand mortar Porotherm P + W on thermoiz. mortar Masonry from hollow ceramics. brick Masonry of silicate brick Masonry of solid ceramics. bricks Wood Plywood Fibreboard Chipboard Mineral wool Styrofoam Styrofoam Drywall λ =

Thickness of the second layer, m.

Third layer material Select material Concrete Reinforced concrete Foam concrete 1000 kg/cu.m. Foam concrete 800 kg/cu.m. Foam concrete 600 kg/cu.m. Gas-block D400 Aeroc on glue Slag concrete Cement-sand mortar Porotherm P + W on thermoiz. mortar Masonry from hollow ceramics. brick Masonry of silicate brick Masonry of solid ceramics. bricks Wood Plywood Fibreboard Chipboard Mineral wool Styrofoam Styrofoam Drywall λ =

Thickness of the third layer, m.

Heat loss through the floor

First layer material Select material Concrete Reinforced concrete Foam concrete 1000 kg/cu.m. Foam concrete 800 kg/cu.m. Foam concrete 600 kg/cu.m. Gas-block D400 Aeroc on glue Slag concrete Cement-sand mortar Porotherm P + W on thermoiz. mortar Masonry from hollow ceramics. brick Masonry of silicate brick Masonry of solid ceramics. bricks Wood Plywood Fibreboard Chipboard Mineral wool Styrofoam Styrofoam Drywall λ =

Thickness of the first layer, m.

Second layer material Select material Concrete Reinforced concrete Foam concrete 1000 kg/cu.m. Foam concrete 800 kg/cu.m. Foam concrete 600 kg/cu.m. Gas-block D400 Aeroc on glue Slag concrete Cement-sand mortar Porotherm P + W on thermoiz. mortar Masonry from hollow ceramics. brick Masonry of silicate brick Masonry of solid ceramics. bricks Wood Plywood Fibreboard Chipboard Mineral wool Styrofoam Styrofoam Drywall λ =

Thickness of the second layer, m.

Third layer material Select material Concrete Reinforced concrete Foam concrete 1000 kg/cu.m. Foam concrete 800 kg/cu.m. Foam concrete 600 kg/cu.m. Gas-block D400 Aeroc on glue Slag concrete Cement-sand mortar Porotherm P + W on thermoiz. mortar Masonry from hollow ceramics. brick Masonry of silicate brick Masonry of solid ceramics. bricks Wood Plywood Fibreboard Chipboard Mineral wool Styrofoam Styrofoam Drywall λ =

Thickness of the third layer, m.

Zone 1 area, sq.m. expand (opens in a new window)

Very often, in practice, heat losses at home are taken at the average of about 100 W / sq.m. For those who count money and plan to equip a house without unnecessary investments and with low fuel consumption, such calculations will not work. It will be enough to say that the heat loss of a well-insulated house and an uninsulated house can differ by 2 times. Accurate calculations according to SNiP require a lot of time and special knowledge, but the effect of accuracy will not be felt properly on the efficiency of the heating system.

This program has been designed to offer best result price/quality, i.e. (elapsed time)/(sufficient accuracy).

The thermal conductivity coefficients of building materials are taken from Appendix 3 for the normal humidity regime of the normal humidity zone.

12/03/2017 - the formula for calculating heat loss for infiltration has been corrected. Now there are no discrepancies with the professional calculations of the designers (in terms of heat loss for infiltration).

01/10/2015 - added the ability to change the air temperature inside the premises.

FAQ expand collapse

How to calculate heat loss to neighboring unheated rooms?

According to the norms of heat loss to neighboring rooms, it must be taken into account if the temperature difference between them exceeds 3 o C. This can be, for example, a garage. How to calculate these heat losses using an online calculator?

Example. In the room we should have +20, and in the garage we plan to +5. Decision. In the field t out we put the temperature cold room, in our case a garage, with a "-" sign. -(-5) = +5 . Select the front view as default. Then we count as usual.

Attention! After calculating the heat loss from room to room, do not forget to set the temperature back.

For reduction of heat consumption strict accounting for heat losses in technological equipment and heating networks. Heat losses depend on the type of equipment and pipelines, their proper operation and the type of insulation.

Heat loss (W) is calculated by the formula

Depending on the type of equipment and pipeline, the total thermal resistance is:

for an insulated pipeline with one layer of insulation:

for an insulated pipeline with two layers of insulation:

for technological apparatuses with multilayer flat or cylindrical walls with a diameter of more than 2 m:

for technological apparatuses with multilayer flat or cylindrical walls with a diameter of less than 2 m:

carrier to the inner wall of the pipeline or apparatus and from the outer surface of the wall into the environment, W / (m 2 - K); X tr, ?. st, Xj - thermal conductivity, respectively, of the material of the pipeline, insulation, walls of the apparatus, /-th layer of the wall, W / (m. K); 5 ST. — apparatus wall thickness, m.

The heat transfer coefficient is determined by the formula

or according to the empirical equation

The transfer of heat from the walls of the pipeline or apparatus to the environment is characterized by the coefficient a n [W / (m 2 K)], which is determined by criterion or empirical equations:

according to criteria equations:

The heat transfer coefficients a b and a n are calculated according to criterion or empirical equations. If the hot coolant is hot water or condensing steam, then a b > a n, i.e. R B< R H , и величиной R B можно пренебречь. Если горячим теплоносителем является воздух или перегретый пар, то а в [Вт/(м 2 - К)] рассчитывают по критериальным уравнениям:

by empirical equations:

Thermal insulation of devices and pipelines is made of materials with low thermal conductivity. well chosen thermal insulation allows to reduce heat loss to the surrounding space by 70% or more. In addition, it increases the productivity of thermal installations, improves working conditions.

The thermal insulation of the pipeline consists mainly of a single layer, covered on top for strength with a layer of sheet metal(roofing steel, aluminum, etc.), dry plaster from cement mortars etc. In the case of using a cover layer made of metal, its thermal resistance can be neglected. If the cover layer is plaster, then its thermal conductivity differs slightly from the thermal conductivity of thermal insulation. In this case, the thickness of the cover layer is, mm: for pipes with a diameter of less than 100 mm - 10; for pipes with a diameter of 100-1000 mm - 15; for pipes with a large diameter - 20.

The thickness of the thermal insulation and the cover layer should not exceed the limiting thickness, depending on the mass loads on the pipeline and its overall dimensions. In table. 23 shows the values ​​of the maximum thickness of the insulation of steam pipelines, recommended by the standards for the design of thermal insulation.

Thermal insulation of technological devices can be single layer or multilayer. Heat loss through thermal

insulation depends on the type of material. Heat losses in pipelines are calculated for 1 and 100 m of pipeline length, in process equipment - for 1 m 2 of the apparatus surface.

A layer of contaminants on the inner walls of pipelines creates additional thermal resistance to the transfer of heat into the surrounding space. Thermal resistances R (m. K / W) during the movement of some coolants have the following values:

Pipelines supplying technological solutions to apparatuses and hot heat carriers to heat exchangers have fittings in which part of the flow heat is lost. Local heat loss (W / m) is determined by the formula

The coefficients of local resistance of fittings of pipelines have the following values:

When compiling the table. 24 calculation of specific heat losses was carried out for steel seamless pipelines (pressure< 3,93 МПа). При расчете тепловых потерь исходили из следующих данных: тем-

the air temperature in the room was taken equal to 20 °C; its speed during free convection is 0.2 m/s; steam pressure - 1x10 5 Pa; water temperature - 50 and 70 ° C; thermal insulation is made in one layer of asbestos cord, = 0.15 W / (m. K); heat transfer coefficient а„ \u003d 15 W / (m 2 - K).

Example 1. Calculation of specific heat losses in a steam pipeline.

Example 2. Calculation of specific heat losses in an uninsulated pipeline.

Given conditions

Pipeline steel diameter 108 mm. Nominal diameter d y = 100 mm. Steam temperature 110°С, environment 18 °C. Thermal conductivity of steel X = 45 W / (m. K).

The data obtained indicate that the use of thermal insulation reduces heat losses per 1 m of pipeline length by 2.2 times.

Specific heat losses, W/m 2 , in technological apparatuses of leather and felting production are:

Example 3. Calculation of specific heat losses in technological devices.

1. The Giant drum is made of larch.

2. Dryer firm "Hirako Kinzoku".

3. Longboat for dyeing berets. Made from of stainless steel[k \u003d 17.5 W / (m-K)]; there is no thermal insulation. dimensions longboat 1.5 x 1.4 x 1.4 m. Wall thickness 8 ST = 4 mm. Process temperature t = = 90 °C; air in the workshop / av = 20 °С. Air velocity in the workshop v = 0.2 m/s.

Heat transfer coefficient a can be calculated in the following way: a = 9.74 + 0.07 At. At / cp \u003d 20 ° C, a is 10-17 W / (m 2. K).

If the surface of the coolant of the apparatus is open, the specific heat losses from this surface (W / m 2) are calculated by the formula

The industrial service "Capricorn" (Great Britain) proposes to use the "Alplas" system to reduce heat losses from open surfaces of coolants. The system is based on the use of hollow polypropylene floating balls that almost completely cover the surface of the liquid. Experiments have shown that at a water temperature in an open tank of 90 ° C, heat losses when using a layer of balls are reduced by 69.5%, two layers - by 75.5%.

Example 4. Calculation of specific heat losses through the walls of the drying plant.

The walls of the dryer can be made from various materials. Consider the following wall structures:

1. Two layers of steel with a thickness of 5 ST = 3 mm with insulation located between them in the form of an asbestos plate with a thickness of 5 And = 3 cm and thermal conductivity X and = 0.08 W / (m. K).

Many, building Vacation home, forget about the approach of winter cold, which is why the calculation of the heat loss of the building is done in a hurry, and as a result, heating does not create a comfortable microclimate in the premises. But making a house warm is not difficult, you just need to take into account a number of nuances.

What is the basis for calculating the heat loss of a building

Any material has such a property as thermal conductivity, only the level differs. thermal resistance, i.e throughput. From any house, even with thermal insulation arranged according to all the rules, heat leaves through windows, doors, walls, floors, ceilings (roofs), as well as through ventilation. With the difference between the external and internal temperatures, the so-called “dew point” necessarily arises, with an average value. And only on the microclimate in the premises, the material and thickness of the walls, as well as the characteristics of thermal insulation, it depends where this point will be: inside, outside or directly in the wall, as well as what temperature it will be.

If you approach the task responsibly and carry out the calculation of the heat loss of the building in accordance with all the rules, it will take you many hours and you will have to make many formulas, the calculations will take up an entire notebook. Therefore, we will determine the indicators of interest to us using a simplified method, or by contacting SNiP and GOSTs for help. And, since it was decided to make the calculations not too in-depth, let's leave aside the determination of the average annual temperature and humidity for the coldest five-day period for several years, as required by SNiP 23-01-99. Let's just note the most frosty day for the last winter season, for example, it will be -30 o C. We will also not take into account the average seasonal wind speed, humidity in the region and the duration heating period.

Building heat loss calculator

Specify the dimensions and types of walls.
On the street
average temperature per day
Select value -40°C -30°C -20°C -15°C -10°C -5°C 0°C +5C +10C
Inside
average temperature per day
Walls
Outgoing only
outside the wall!

Add walls facing the street and specify what layers the wall consists of

Rooms

Add all used rooms, even corridors, and specify what layers the floors are made of

Heat loss:
Through walls: - kW Through windows: - kW Through the top: - kW Through the bottom: - kW Through ventilation: - kW Total: -kW Click on the button to calculate

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However, what makes up the microclimate in the living room? Comfortable conditions for residents depend on air temperature t in, its humidity φ in and movement v in the presence of ventilation. And another factor affects the level of heat - radiation of heat or cold t p, characteristic of naturally heated (cooled) objects and surfaces in the environment. It determines the resulting temperature t n, using the formula [ t n = ( t p+ t in 2]. All these indicators for different rooms can be seen in the table below.

Optimal parameters of the microclimate of residential buildings in accordance with GOST 30494-96

Period of the yearroom

Indoor air temperature t c, °С

Resulting temperature t p, °С

Relates indoor air humidity φ in, %

Air speed v in, m/s

ColdLiving room
The same, in areas with t 5 from -31 °C
Kitchen
Toilet
Bathroom, combined bathroom
Rest and study area
Inter-apartment corridor
lobby, stairwell
Pantry
WarmLiving room

The letters HH denote non-normalized parameters.

We make a thermal calculation of the wall, taking into account all layers

As already said, each material has inherent resistance to heat transfer, and the thicker the walls or floors, the higher this value. However, do not forget about thermal insulation, in the presence of which the surfaces enclosing the room become multi-layered and prevent heat leakage much better. Each layer has its own resistance to the passage of heat, and the sum of all these quantities is indicated in the formulas as Σ R i (here the letter i defines the layer number).

Since the materials that make up the fencing of the premises with different properties have some resentment temperature regime in its structure, the total resistance to heat transfer is calculated. Its formula is as follows: where R in and R n correspond to the resistance on the inner and outer surfaces of the fence, whether it be a wall or a ceiling. However, heaters make adjustments to the heat engineering calculation of the wall, which are based on the coefficient of heat engineering uniformity r, defined by the formula .

Indicators with digital indices are, respectively, the coefficients of internal fasteners and the connection of the design fence with any other. The first one, that is r 1, is responsible just for fixing the heaters. If the coefficient of thermal conductivity of the latter λ = 0.08 W / (m ° C), the value r 1 will be large, but if the thermal conductivity of thermal insulation is estimated as λ = 0.03 W / (m ° C), then less.

The value of the coefficient of internal fasteners decreases as the thickness of the insulation layer increases.

In general, the picture is as follows. Suppose thermal insulation is mounted by direct anchoring on a three-layer cellular concrete wall, lined with bricks on the outside. Then with a layer of insulation of 100 millimeters r 1 corresponds to 0.78-0.91, a thickness of 150 millimeters gives an internal fastener coefficient of 0.77-0.90, the same indicator, but at 200 mm, determines r 1 as 0.75-0.88. If a the inner layer also made of bricks r 1 \u003d 0.78-0.92, and if the walls of the room are reinforced concrete, then the coefficient is shifted to 0.79-0.93. And here window slopes and ventilation make a difference r 2 = 0.90-0.95. All these data should be taken into account in the future.

Some information on how to calculate the thickness of the insulation

In order to proceed with the calculation of thermal insulation, we need, first of all, to calculate R o , then find out the required thermal resistance R req according to the following table (abbreviated version).

Required values ​​of resistance to heat transfer of enclosing structures

Building/ room

Degree-days of the heating period D d , °С day

Reduced resistance to heat transfer of fences R req, m 2 °C / W

walls

coatings

attic floor and ceiling over cold basements

windows and balcony door, showcases and stained glass windows

1. Residential, medical and preventive and children's institution, school, boarding school
a
b
2. Public, administrative, household and other premises with damp or wet conditions
a
b

Odds a and b necessary for cases where D d , °C day differs from that given in the table, then R req, m 2 ° С / W is calculated by the formula R req \u003d a D d+ b. For column 6 of the first group of buildings, there are amendments: if the degree-day value is less than 6000 ° С day, a= 0.000075, and b\u003d 0.15, if the same indicator is in the range of 6000-8000 ° С day, then a = 0,00005, b= 0.3, if more than 8000 °C day, then a= 0.000025, and b= 0.5. When all the data is collected, we proceed to the calculation of thermal insulation.

Now let's find out how to calculate the thickness of the insulation. Here you will have to turn to mathematics, so be prepared to work with formulas. Here is the first of them, according to it we determine the required conditional resistance to heat transfer R o conv. tr = R req / r. We need this parameter to determine the required heat transfer resistance of the insulation R ut tr = R o conv. tr - (R in + Σ R t. izv + R n), here Σ R t. izv is the sum of the thermal resistance of the layers of the fence without taking into account thermal insulation. We find the thickness of the insulation δ ut \u003d R ut tr λ ut (m), and λ ut is taken from Table D.1 SP 23-101-2004, and round the result obtained up to the constructive value, taking into account the manufacturer's nomenclature.

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