SNiP 41-01-2003 Heating, ventilation and air conditioningSNiP 41-01-2003 Heating, ventilation and air conditioning

Normative references

1. GOST
30494-96. Buildings residential and public. The parameters of the microclimate in the premises.

2. GOST
31168-2003. Buildings are residential. Method for determining the specific consumption of heat
energy for heating.

3. MGSN 3.01-01. Residential buildings.

4. SNiP
23-01-99*. Building climatology.

5. SNiP 23-02-2003. Thermal
building protection.

6. SNiP
2.04.05-91*. Heating, ventilation and air conditioning.

7. SNiP
2.04.01-85*. Internal plumbing and sewerage of buildings.

8. SP 23-101-2004.
Design of thermal protection of buildings.

9. Standard ABOK-1-2004.
Buildings residential and public. Air exchange standards.

Power in sports

It is possible to evaluate work using power not only for machines, but also for people and animals. For example, the power with which a basketball player throws a ball is calculated by measuring the force she applies to the ball, the distance the ball has traveled, and the time that force has been applied. There are websites that allow you to calculate work and power during exercise. The user selects the type of exercise, enters the height, weight, duration of the exercise, after which the program calculates the power. For example, according to one of these calculators, the power of a person with a height of 170 centimeters and a weight of 70 kilograms, who did 50 push-ups in 10 minutes, is 39.5 watts. Athletes sometimes use devices to measure the amount of power a muscle is working during exercise. This information helps determine how effective their chosen exercise program is.

Dynamometers

To measure power, special devices are used - dynamometers. They can also measure torque and force. Dynamometers are used in various industries, from engineering to medicine. For example, they can be used to determine the power of a car engine. To measure the power of cars, several main types of dynamometers are used. In order to determine the power of the engine using dynamometers alone, it is necessary to remove the engine from the car and attach it to the dynamometer. In other dynamometers, the force for measurement is transmitted directly from the wheel of the car. In this case, the car's engine through the transmission drives the wheels, which, in turn, rotate the rollers of the dynamometer, which measures the power of the engine under various road conditions.

This dynamometer measures the torque as well as the power of the car's powertrain.

Dynamometers are also used in sports and medicine. The most common type of dynamometer for this purpose is isokinetic. Usually this is a sports simulator with sensors connected to a computer. These sensors measure the strength and power of the entire body or individual muscle groups. The dynamometer can be programmed to give signals and warnings if the power exceeds a certain value

This is especially important for people with injuries during the rehabilitation period, when it is necessary not to overload the body.

According to some provisions of the theory of sports, the greatest sports development occurs under a certain load, individual for each athlete. If the load is not heavy enough, the athlete gets used to it and does not develop his abilities. If, on the contrary, it is too heavy, then the results deteriorate due to overload of the body. Physical activity during some activities, such as cycling or swimming, depends on many environmental factors, such as road conditions or wind. Such a load is difficult to measure, but you can find out with what power the body counteracts this load, and then change the exercise scheme, depending on the desired load.

Article author: Kateryna Yuri

Heat loss through building envelopes

1) We calculate the resistance to heat transfer of the wall by dividing the thickness of the material by its coefficient of thermal conductivity. For example, if the wall is built of warm ceramics 0.5 m thick with a thermal conductivity of 0.16 W / (m × ° C), then we divide 0.5 by 0.16: 0.5 m / 0.16 W/(m×°C) = 3.125 m2×°C/W Thermal conductivity coefficients of building materials can be found here.

2) Calculate the total area of ​​the outer walls. Here is a simplified example of a square house: (10 m width × 7 m height × 4 sides) - (16 windows × 2.5 m2) = 280 m2 - 40 m2 = 240 m2

3) We divide the unit by the resistance to heat transfer, thereby obtaining heat loss from one square meter of the wall per one degree temperature difference. 1 / 3.125 m2×°C/W = 0.32 W/m2×°C

4) Calculate the heat loss of the walls. We multiply the heat loss from one square meter of the wall by the area of ​​​​the walls and by the temperature difference inside the house and outside. For example, if +25°C inside and -15°C outside, then the difference is 40°C. 0.32 W / m2×°C × 240 m2 × 40 °C = 3072 W This number is the heat loss of the walls. Heat loss is measured in watts, i.e. is the heat dissipation power.

5) In kilowatt-hours it is more convenient to understand the meaning of heat loss. For 1 hour through our walls with a temperature difference of 40 ° C, thermal energy is lost: 3072 W × 1 h = 3.072 kWh Energy spent in 24 hours: 3072 W × 24 h = 73.728 kWh

22Pro GSOP hereInsulating glass unit heat transfer resistance

Power units

Power is measured in joules per second, or watts. Along with watts, horsepower is also used. Before the invention of the steam engine, the power of engines was not measured, and, accordingly, there were no generally accepted units of power. When the steam engine began to be used in mines, engineer and inventor James Watt began to improve it. In order to prove that his improvements made the steam engine more productive, he compared its power to the working capacity of horses, since horses have been used by people for many years, and many could easily imagine how much work a horse can do in a certain amount of time. In addition, not all mines used steam engines. On those where they were used, Watt compared the power of the old and new models of the steam engine with the power of one horse, that is, with one horsepower. Watt determined this value experimentally, observing the work of draft horses at the mill. According to his measurements, one horsepower is 746 watts. Now it is believed that this figure is exaggerated, and the horse cannot work in this mode for a long time, but they did not change the unit. Power can be used as a measure of productivity, as increasing power increases the amount of work done per unit of time. Many people realized that it was convenient to have a standardized unit of power, so horsepower became very popular. It began to be used in measuring the power of other devices, especially vehicles. Even though watts have been around for almost as long as horsepower, horsepower is more commonly used in the automotive industry, and it's clearer to many buyers when a car's engine power is listed in those units.

60 watt incandescent lamp

Factors

What affects the annual heat consumption for heating?

Duration of the heating season ().
It, in turn, is determined by the dates when the average daily temperature in the street for the last five days falls below (and rises above) 8 degrees Celsius.

• The degree of thermal insulation of the building
  greatly affects what will be the rate of thermal power for him. An insulated facade can reduce the need for heat by half compared to a wall made of concrete slabs or bricks.
• building glazing factor.
  Even when using multi-chamber double-glazed windows and energy-saving spraying, noticeably more heat is lost through windows than through walls. The greater part of the facade is glazed, the greater the need for heat.
• The degree of illumination of the building.
  On a sunny day, a surface oriented perpendicular to the sun's rays can absorb up to a kilowatt of heat per square meter.

Power of household electrical appliances

On household electrical appliances, the power is usually indicated. Some lamps limit the power of the bulbs that can be used in them, for example, no more than 60 watts. This is because higher wattage bulbs generate a lot of heat and the bulb holder can be damaged. And the lamp itself at a high temperature in the lamp will not last long. This is mainly a problem with incandescent lamps. LED, fluorescent and other lamps generally operate at lower wattage at the same brightness and if used in luminaires designed for incandescent lamps there are no wattage problems.

The greater the power of the electrical appliance, the higher the energy consumption and the cost of using the appliance. Therefore, manufacturers are constantly improving electrical appliances and lamps. The luminous flux of lamps, measured in lumens, depends on the power, but also on the type of lamps. The greater the luminous flux of the lamp, the brighter its light looks. For people, it is high brightness that is important, and not the power consumed by the llama, so recently alternatives to incandescent lamps have become increasingly popular. Below are examples of types of lamps, their power and the luminous flux they create.

Calculations

Theory is theory, but how are the heating costs of a country house calculated in practice? Is it possible to estimate the estimated costs without plunging into the abyss of complex heat engineering formulas?

Consumption of the required amount of thermal energy

The instruction for calculating the approximate amount of heat required is relatively simple. The key phrase is an approximate amount: for the sake of simplifying calculations, we sacrifice accuracy, ignoring a number of factors.

• The base value of the amount of thermal energy is 40 watts per cubic meter of cottage volume.
• To the base value is added 100 watts for each window and 200 watts for each door in the exterior walls.

Further, the obtained value is multiplied by a coefficient, which is determined by the average amount of heat loss through the outer contour of the building. For apartments in the center of an apartment building, a coefficient equal to one is taken: only losses through the facade are noticeable. Three of the four walls of the contour of the apartment border on warm rooms.

For corner and end apartments, a coefficient of 1.2 - 1.3 is taken, depending on the material of the walls. The reasons are obvious: two or even three walls become external.

Finally, in a private house, the street is not only along the perimeter, but also from below and above. In this case, a coefficient of 1.5 is applied.

In a cold climate zone, there are special requirements for heating.

Let's calculate how much heat is needed for a cottage measuring 10x10x3 meters in the city of Komsomolsk-on-Amur, Khabarovsk Territory.

The volume of the building is 10*10*3=300 m3.

Multiplying the volume by 40 watts/cube will give 300*40=12000 watts.

Six windows and one door is another 6*100+200=800 watts. 1200+800=12800.

Private house. Coefficient 1.5. 12800*1.5=19200.

Khabarovsk region. We multiply the need for heat by another one and a half times: 19200 * 1.5 = 28800. In total - at the peak of frost, we need about a 30-kilowatt boiler.

Calculation of heating costs

The easiest way is to calculate the consumption of electricity for heating: when using an electric boiler, it is exactly equal to the cost of thermal power. With continuous consumption of 30 kilowatts per hour, we will spend 30 * 4 rubles (approximate current price of a kilowatt-hour of electricity) = 120 rubles.

Fortunately, the reality is not so nightmarish: as practice shows, the average heat demand is about half the calculated one.

• Firewood - 0.4 kg / kW / h.
  Thus, the approximate norms for the consumption of firewood for heating in our case will be equal to 30/2 (the rated power, as we remember, can be divided in half) * 0.4 \u003d 6 kilograms per hour.
• The consumption of brown coal in terms of a kilowatt of heat is 0.2 kg.
  The consumption rates of coal for heating are calculated in our case as 30/2*0.2=3 kg/h.

Brown coal is a relatively inexpensive heat source.

• For firewood - 3 rubles (the cost of a kilogram) * 720 (hours in a month) * 6 (hourly consumption) \u003d 12960 rubles.
• For coal - 2 rubles * 720 * 3 = 4320 rubles (read others).

Determination of the flow of infiltrating air in existing residential buildings under construction until 2000

Residential building construction up
2000 are characterized by low tightness of window openings, as a result of which
the flow of infiltrating air through these openings under the action of gravitational
and wind pressure often exceeds that required for ventilation. Consumption
infiltrating air G_inf, kg/h, in the building
is found according to the following empirical dependence*:

(4.1)

where G._inf.kv - medium (according to
building) the amount of infiltration through the windows of one apartment, kg/h;

TO_sq. - the number of apartments in the building;

- the same as in
formula ();

G_inf.LLU - value
infiltration at t_n = -25 °С through
windows and external doors of the premises of the stair-elevator unit, attributable to one
floor, kg/h For residential buildings that do not have staircases, separated
outdoor transitions, G_inf.LLU accepted in
depending on the area of ​​the windows of the staircase and elevator units F_LLU, m2, one floor (Table 4.1). For residential buildings with
stairwells, separated by external passages, G_inf.LLU accepted in
depending on the height of the building Nand resistance characteristics
doors of external passages S_dvin the ranges (0.5-2)ּ10-3 Paּh/kg2
(first value for unsealed closed doors) (Table 4.2);

* This method for determining air infiltration into
residential building was developed at MNIITEP based on the generalization of a series of calculations of air
mode on the computer. It allows you to determine the total flow rate of the infiltrating
air in all apartments of the building, taking into account the depressurization of the windows of the upper floors
to ensure the sanitary norm of inflow into living rooms and taking into account the peculiarities
air infiltration through windows and doors in the staircase and elevator assembly. Method
published in the journal Water Supply and Sanitary Engineering, 1987, No. 9.

Table 4.2

N	9	12	16	22
Ginf.LLU, kg/h -at heated staircase	348-270	380-286	419-314	457-344
-at unheated stairwell	249-195	264-200	286-214	303-226

N- the number of floors in the building, multiplied by the number of sections.

Average infiltration
through the windows of one apartment G_inf.kv determined by
formula

G_inf.kv = G_{close quarters}β_fiβ_n,(4.2)

where G_{close quarter} - the average value of infiltration with closed windows for
one apartment with F_ca.rmsR_and\u003d 74.6 kg / h (see calculation example in). Values G_{close quarter} shown in
tab. 4.3;

F_ca.rms - average for
building area of ​​windows and balcony doors of one apartment, m2;

R_and — resistance to air penetration of windows according to field tests,
m2ּh/kg, at ΔР = 10Pa;

β_fi- coefficient depending on the actual for a given building
values ​​F_ca.rmsR_and, defined
according to the formula

(4.3)

R_n - coefficient,
taking into account the increase in infiltration to the ventilation rate of air due to
opening vents, transoms, etc. Determined by the table. 4.4.

Table 4.3

number of storeys	Speed wind, m/s	Gclose quarter, kg/h, at tn °С
-40	-30	-25	-15	-10	-5	5
5	126	110	102	86	78	69	60	51
3	168	149	143	124	115	108	98	91
5	198	185	176	160	152	145	137	129
7	246	231	222	207	203	196	189	183
9	157	137	127	108	97	86	75	64
3	198	180	170	150	141	130	121	111
5	227	209	199	183	174	165	156	147
7	262	248	240	224	216	208	200	192
12	167	148	138	115	104	94	80	69
3	214	194	185	165	154	143	132	121
5	240	221	213	193	183	174	165	155
7	274	259	251	236	226	216	207	199
16	180	159	150	125	113	102	88	74
3	232	210	197	176	165	157	146	136
5	253	235	227	206	198	183	178	169
7	290	278	270	249	242	233	224	215
22	192	168	158	134	122	108	95	79
3	249	228	216	194	181	169	156	143
5	267	247	238	216	208	198	187	178
7	298	283	276	256	248	239	229	219

Wind speed, m/s	βn at
0,5	0,7	0,9	1,1	1,3	1,5	2
1,02	1,05	1,11	1,22	1,35	1,5	2
More 3	1	1	1,05	1,15	1,3	1,5	2

Notes:

1) for > 2 take β_n = 2;

2) when regulating with correction according to
indoor air temperature value G_{close quarter}accept
in the absence of wind

Minimum required amount of infiltration
in apartments, including the sanitary norm of supply air for living rooms and
the amount of air entering through the closed windows in the kitchen, kg / h, is determined by the formula:

(4.4)

where F_w.sr. - average for
building living area of ​​one apartment, m2;

G_{close quarter}, β_fi, F_ca.rms, is the same as in
formula ();

F_{ok.av.kitchen}- average for
building window area in one kitchen, m2.

Coefficient To_v,
taking into account additional air infiltration in apartments compared to
the required air exchange in them, is calculated by the formula (4.5) and is substituted into the formula ():

(4.5)