Heating calculation

1. Method for calculating the air permeability resistance of the wall enclosing structure

1.
Determine the specific gravity of the outer and
internal air, N/m2

,
(6.1)

.
(6.2)

2.
Determine the difference in air pressure
on outer and inner surfaces
building envelope, Pa

(6.3)

where Vhall–

maximum from average wind speeds rumbam for January, m/s, , (see Table 1.1).

3. Calculate
required air permeation resistance,
m2hPa/kg

, (6.4)

where Gn–

normative air permeability of enclosing structures, m2hPa/kg, .

4.
Find the total actual resistance
breathability of the outer
fences, m2hPa/kg

,
(6.5)

where Rtheir–

resistance breathability of individual layers building envelope, m2hPa/kg .

If
the condition
,
then the enclosing structure responds
air permeability requirements, if
condition is not met, then
take steps to increase
breathability.

Example
10

Payment
breathability resistance

wall enclosing structure

Average calculation and exact

Given the factors described, the average calculation is carried out according to the following scheme. If for 1 sq. m requires 100 W of heat flow, then a room of 20 sq. m should receive 2,000 watts. A radiator (popular bimetallic or aluminum) of eight sections emits about 150 watts. We divide 2,000 by 150, we get 13 sections. But this is a rather enlarged calculation of the thermal load.

The exact one looks a little intimidating. Actually, nothing complicated. Here is the formula:

• q₁ – type of glazing (ordinary = 1.27, double = 1.0, triple = 0.85);
• q₂ – wall insulation (weak or absent = 1.27, 2-brick wall = 1.0, modern, high = 0.85);
• q₃ - the ratio of the total area of ​​window openings to the floor area (40% = 1.2, 30% = 1.1, 20% - 0.9, 10% = 0.8);
• q₄ - outdoor temperature (the minimum value is taken: -35 o C = 1.5, -25 o C = 1.3, -20 o C = 1.1, -15 o C = 0.9, -10 o 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).

Using any of the methods described, it is possible to calculate the heat load of an apartment building.

3. Method for calculating the effect of infiltration on the temperature of the inner surface and the heat transfer coefficient of the building envelope

1.
Calculate the amount of air entering
through the outer fence, kg/(m2h)

.
(6.7)

2.
Calculate the internal temperature
the surface of the fence during infiltration,
С

,
(6.8)

where Cv–	specific heat capacity of air, kJ/(kgС);
e –	base natural logarithm;
RXi –	thermal resistance to heat transfer of the enclosing structures, starting from the outer air up to a given section in the thickness fences, m2С/W:

.
(6.9)

3.
Calculate the internal temperature
the surface of the fence in the absence
condensation, С

.
(6.10)

4. Determine
heat transfer coefficient of the fence
taking into account infiltration, W/(m2С)

.
(6.11)

5.
Calculate the heat transfer coefficient
fencing in the absence
infiltration according to equation (2.6), W/(m2С)

.
(6.12)

Example
12

Payment
influence of infiltration on temperature
inner surface
and coefficient
building envelope heat transfer

Initial
data

Values
quantities required for calculation:
Δp= 27.54 Pa;t_n = -27 С;
t_v = 20 С;
V_hall= 4.4 m/s;
= 3.28 m2С/W;
e= 2,718;
= 4088.7m2hPa/kg;
R_v = 0.115 m2С/W;
WITH_V = 1.01 kJ/(kgС).

Order
calculation

Calculate
the amount of air passing through
external fence, according to equation (6.7),
kg/(m2h)

G_and = 27,54/4088,7 = 0,007
g/(m2h).

Calculate
inner surface temperature
fencing during infiltration, С,
and thermal resistance to heat transfer
enclosing structure, starting from
outside air up to a given section
in the thickness of the fence according to equations (6.8) and
(6.9).

m2С
/W;

C.

Counting
inner surface temperature
guards in the absence of condensation,
С

C.

From
calculations it follows that the temperature
inner surface during filtration
lower than without infiltration ()
by 0.1С.

Determine
heat transfer coefficient of the fence
taking into account infiltration according to the equation
(6.11), W/(m2С)

W/(m2С).

Calculate
heat transfer coefficient of the fence
in the absence of infiltration
equation (2.6), W/(m2S)

W/(m2С).

So
Thus, it was found that the coefficient
heat transfer taking into account infiltration
k_andmore
corresponding coefficient without
infiltrationk(0,308 > 0,305).

Control
questions for section 6:

1.
What is the main purpose of calculating the air
outdoor mode
fences?

2.
How does infiltration affect temperature?
inner surface
and coefficient
heat transfer of the building envelope?

7.
Requirements
to the consumption of thermal energy for heating
and building ventilation

Infiltration volume calculation

Calculation of the volume of infiltration.

In order for the effect of acid on carbonate inclusions to be noticeable, in precipitation seeping through the aeration zone, the pH must be less than 4, which is very rare (mainly in industrial areas and not always). In this case, acidic solutions are completely neutralized in the rocks of the aeration zone. At the same time, according to calculations, 6 g 3042″ will flow to the surface of the aquifer with an area of ​​\u200b\u200b1 m2, and the increase in concentration in groundwater will be only 4 mg / l. Consequently, the pollution of groundwater with sulfur compounds due to the ingress of polluted precipitation from the atmosphere is insignificant. In terms of the volumes of runoff entering the groundwater and the area of ​​their distribution during infiltration, the leakage of conditionally clean industrial waters in the territory of the ESR and ZLO and the leakage of fresh industrial waters in the ASZ territory are of the greatest importance. Wastewater, infiltrating through the aeration zone, interacts with rocks. Filtration losses from the ESR are approximately 120-130 thousand m3/year (or -0.23 ad/year, or 6.33 m3/day). The value of infiltration on EDT without taking into account evaporation and transpiration is 2.2.10-3m/day (or 0.77 ad/year). Filtering through the aeration zone, these solutions change their composition. Due to the leaching of gypsum from the rocks, the ionic strength of the solution increases. In addition, the dissolution of calcite occurs first, which is contained in rocks in a small amount. Then, according to the simulation data, due to the violation of the ratio of Ca2+ ions in the solution, dolomite precipitation will be observed during the dissolution of gypsum. Also, when the solution interacts with rocks, migratory forms of aluminum (A102 and A1(0H)4 mainly) will pass into it.

In the general case, the protection of groundwater is assessed on the basis of four indicators: the depth of groundwater or the thickness of the aeration zone, the structure and lithological composition of the constituent rocks of this zone, the thickness and prevalence of low-permeable deposits above groundwater, and the filtration properties of rocks above the groundwater level. The last two signs have the greatest influence on the speed and volume of infiltrating polluted waters, and the depth of groundwater is of subordinate importance. Therefore, in preliminary assessments of protection categories, the aeration zone thickness parameter and calculations of the depths and rates of polluted water infiltration are used. In more detailed assessments, such parameters as absorbing and sorption properties of rocks and ratios of aquifer levels are introduced into calculations or predictive models in order to assess horizontal directions and the volume of lateral migration of polluted waters. At the same stage, along with natural ones, it is necessary to take into account technogenic physical and chemical processes (liquid properties).

The estimated hourly heat load of heating should be taken according to standard or individual building projects.

If the value of the calculated outdoor air temperature adopted in the project for designing heating differs from the current standard value for a particular area, it is necessary to recalculate the estimated hourly heat load of the heated building given in the project according to the formula:

Q_op = Q_{o pr}

where: Q_op — calculated hourly heat load of the building heating, Gcal/h (GJ/h);

t_v is the design air temperature in the heated building, C; taken in accordance with the head of SNiP 2.04.05-91 and according to Table. one;

t_nro - design outdoor air temperature for designing heating in the area where the building is located, according to SNiP 2.04.05-91, C;

Table 1 CALCULATED AIR TEMPERATURE IN HEATED BUILDINGS

Building name	Estimated air temperature in the building t C
Residential building	18
Hotel, hostel, administrative	18 — 20
Kindergarten, nursery, polyclinic, outpatient clinic, dispensary, hospital	20
Higher, secondary specialized educational institution, school, boarding school public catering enterprise, club	16
Theater, shop, fire station	15
Garage	10
Bath	25

In areas with an estimated outdoor air temperature for heating design of 31 C and below, the design air temperature inside heated residential buildings should be taken in accordance with chapter SNiP 2.08.01-85 20 C.

Easy Ways to Calculate Heat Load

Any calculation of the heat load is needed to optimize the parameters of the heating system or improve the thermal insulation characteristics of the house. After its implementation, certain methods of regulating the heating load of heating are selected. Consider non-labor-intensive methods for calculating this parameter of the heating system.

The dependence of heating power on the area

For a house with standard room sizes, ceiling heights and good thermal insulation, a known ratio of room area to required heat output can be applied. In this case, 1 kW of heat will be required per 10 m². To the result obtained, it is necessary to apply a correction factor depending on the climatic zone.

Let's assume that the house is located in the Moscow region. Its total area is 150 m². In this case, the hourly heat load on heating will be equal to:

15*1=15 kWh

The main disadvantage of this method is the large error. The calculation does not take into account changes in weather factors, as well as building features - heat transfer resistance of walls and windows. Therefore, it is not recommended to use it in practice.

Enlarged calculation of the thermal load of the building

The enlarged calculation of the heating load is characterized by more accurate results. Initially, it was used to pre-calculate this parameter when it was impossible to determine the exact characteristics of the building. The general formula for determining the heat load on heating is presented below:

Where q°
- specific thermal characteristic of the structure. The values ​​must be taken from the corresponding table, a
- correction factor, which was mentioned above, Vn
- external volume of the building, m³, Tvn
and Tnro
– temperature values ​​inside the house and outside.

Suppose that it is necessary to calculate the maximum hourly heating load in a house with an external volume of 480 m³ (area 160 m², two-story house). In this case, the thermal characteristic will be equal to 0.49 W / m³ * C. Correction factor a = 1 (for the Moscow region). The optimum temperature inside the dwelling (Tvn) should be + 22 ° С. The outside temperature will be -15°C. We use the formula to calculate the hourly heating load:

Q=0.49*1*480(22+15)= 9.408 kW

Compared to the previous calculation, the resulting value is less. However, it takes into account important factors - the temperature inside the room, on the street, the total volume of the building. Similar calculations can be made for each room.The method of calculating the heating load according to aggregated indicators makes it possible to determine the optimal power for each radiator in a particular room. For a more accurate calculation, you need to know the average temperature values ​​\u200b\u200bfor a particular region.