负荷计算公式（Load calculation formula）

负荷计算公式（Load calculation formula） 负荷计算公式(Load calculation formula) Calculation formula of air conditioning cooling load Calculation formula for heat transfer of external wall and roof heat transfer Cooling load Q tau (W) is calculated at the calculated time of the formation of external wall or roof heat transfer. Q tau tau = k. f. Δ t - factor (1.1) Type: F - computing area, ?; Tao -- computing time, o 'clock; The time when the temperature wave ACTS on the outer wall or the outside of the roof. Δ t tau - function factor - moment, through the wall or roofing cooling load calculation of the temperature difference, referred to as "load temperature difference, ?. Note: for the outer walls of the delay time is 5 hours, for example, in room 16 points to determine the cooling load of the heat transfer, the computing time to tau = 16, time delay factor = 5, part time for tau factor = 16-5 = 11. This is because the cooling load on the surface of the 16-hour external wall due to temperature fluctuation is the result of the temperature fluctuation outside the external wall at 5 hours. When the attenuation coefficient of the outer wall or the roof is beta < 0.2, the daily average cold load Qpj can be used instead of the cold load Q tau of each calculation moment: Qpj = k. f. Δ TPJ (1.2) Type: Δ TPJ - load average daily temperature difference, ?. The temperature difference of the outside window is hot and cold The calculation of cold load Q tau is calculated according to the calculation of the calculation of heat transfer in the external window. Q tau tau = a. k. f. Δ t (2.1) Type: Δ t tau - calculating moment under the temperature load, ?; K - heat transfer coefficient; A - window frame correction factor. External window solar radiation cooling load The calculation of the cooling load Q tau at the time of solar radiation from the outer window shall be calculated according to the following categories according to different conditions: When the outer window has no shading facilities Q Jw tau tau = F, Xg, (3.1) Type: Xg-window construction correction coefficient; Through the Jw tau - calculating moment not shading glass solar radiation intensity of cooling load, W / ?. When the outer window only has inner shading facilities Q tau = F Xg, Xz, Jacqueline Nottingham tau (3.2) Type: Xz - inner shading coefficient; Jacqueline Nottingham tau - computing time, through a glass shading facilities in solar radiation intensity of cooling load, W / ?. When the outer window is only outside the visor Q tau = [the Jw F1 Jw tau tau + F1 (F -) ? 0], Xg (3.3) Type: F1 - window under the sun point-blank area, ?. Through the Jw tau 0 - calculating moment not shading glass scattering solar radiation intensity of cooling load, W / ?. When the window has inner shading facilities and external visor Q tau = [F1, Jacqueline Nottingham tau + F1 (F -), Jacqueline Nottingham tau 0], Xg, Xz (3.4) Type: Jacqueline Nottingham tau 0 - computing time, through a window shading facilities in scattering solar radiation intensity of cooling load, W / ?. Heat transfer and cooling load of inner envelope The adjacent space is well ventilated When the adjacent space is well ventilated, the cooling load of the inner wall or laminates can be calculated according to the following equation: Q = k. f. (TWP - tn) (4.1) Type: TWP - summer outdoor calculated daily average temperature of air conditioning, ?; Adjacent space has heat The temperature difference heat transfer load of the inner window, partition wall, floor or inner door of the air conditioning room shall be calculated according to the following formula: Q = k. f. (TWP + Δ TLS - tn) (4.2) Type: Q - steady state cold load, same as W; Tn - summer air conditioning indoor temperature calculation, ?; Δ TLS, adjacent room temperature rise, according to the adjacent room cooling intensity of use, ?. Body cooling load The calculation of heat dissipation of human body is calculated by the cooling load Q tau. Q tau tau = phi, n. q1, X - tau (5.1) Type: Phi - group coefficients; N - calculate the total number of air conditioning rooms in the air conditioning room; Q1 -- an adult male with heat and heat, W; Tao -- computing time, h; Tau - time of entering the air conditioning zone, h; Tao - tau - from the time when the person enters the air conditioning area to calculate the duration of the computation time, h; The cooling load coefficient of heat dissipation in the human body at the time of X - tau - tau - tau - tau. Light load The calculation of heat dissipation of the lighting equipment is calculated according to the type and installation of the lamp. The cooling load formed by the heat dissipation of incandescent bulbs Q tau tau = n1, N, X - tau (6.1) The ballast is in the fluorescent lamp outside the air conditioning area Q tau tau = n1, N, X - tau (6.2) A fluorescent lamp installed in an air-conditioned area Q tau tau = 1.2, n1, n. X - tau (6.3) The fluorescent lamp in the glass cover of the air conditioning room Q tau tau = n0, n1, n. X - tau (6.4) Type: N -- installation power of lighting equipment, W; N0, reflection, glass ceiling inside ventilation condition coefficient, when the fluorescent lamp shade have holes, use natural ventilation in the ceiling, cooling off for 0.5-0.6, fluorescent lamp shade, no vent, according to the condition of ceiling ventilation inside take 0.6 0.8; N1 -- the use coefficient is generally 0.5-0.8; Tao -- computing time, h; Tau -- the moment of light, h; Tau tau tau -- it's time to compute the moment from the turn on the light, h; The cooling load coefficient of heat dissipation of X - tau - tau - tau - tau. Equipment cooling load Heat equipment and thermal surface heat dissipation formation calculation time cold load Q tau, according to the calculation: Q tau tau = qs, X - tau (7.1) Type: Tao -- the moment when the heat source is put into use, h; Tau tau tau -- the time that we're going to use from the heat source to compute the duration of the computation, h; X - tau - tau - tau - tau time equipment, cooling load coefficient of heat dissipation; Qs -- actual heat loss of heat source, W. Heat transfer process equipment Qs = n1, n2, n3, n4 interchange, N (7.2) The distribution of motor and process equipment in air conditioning room Qs = n1, n2, n3 n. / eta (7.3) Only the motor is in the air conditioning room Qs = n1, n2, n3, n. (1 - eta)/eta (7.4) Only the heat of the process equipment in the air conditioning room Qs = n1, n2, n3 n. (7.5) Type: N - total installation power of equipment, W; Eta - motor efficiency; N1 -- the use coefficient is generally recommended to be 0.5 to 1.0; N2 - installation coefficient, generally recommended 0.7-0.9; N3 - load factor, i.e., the ratio of the average real consumption power to the maximum power of the design, generally is about 0.4 to 0.5; N4 - ventilation insulation coefficient; The permeability of air is hot and cold The visible cooling load of the air is calculated according to the following equation: Q = 0.28 ? G ? (tw - tn) (8.1) Type: G - the total amount of air infiltrated into the indoor air, kg/h; Tw - summer outdoor dry bulb temperature of air conditioning, ?; Tn - indoor temperature calculation, ?. The obvious heat and cooling load of food The heat loss of food should be considered when calculating the cooling load of the restaurant. The cooling load formed by the heat dissipation of food can be considered by every dining guest. Bulk moisture and latent heat cooling load Body dispersion and latent heat and cooling load The body dispersion is calculated according to the calculation D tau tau = 0.001, phi, n, g (10.1.1) Type: D -- bulk humidity, kg/h; Phi - group coefficients; N tau - calculating the total number of air conditioning units; G - an adult man's hour of dehumidification, g/h. The latent heat and cold load Q tau (W) formed by the body dispersion is calculated by the following formula: Q tau = phi, n tau, q2 (10.1.2) Type: Q2 - an adult male with an hour of latent heat, W. Infiltration of air dispersion and latent heat and cold The moisture in the air is brought into the indoor moisture content D (kg/h), and the calculation is: D = 0.001 ? G ? (dw - dn) (10.2.1) The latent heat and cooling load (W) formed by the infiltration of air is calculated by the following formula: Q = 0.28 ? G ? (hw - hn) (10.2.2) Type: Dw - moisture content of outdoor air, g/Kg; Dn - moisture content of indoor air, g/Kg; Hw -- enthalpy of outdoor air, KJ/Kg; Hn -- enthalpy of indoor air, KJ/Kg. Food dispersion and latent heat and cooling load The food dispersion D tau (kg/h) in the restaurant is calculated by the following formula: D tau = 0.012, n tau, phi (10.3.1) Type: N tau -- total number of meals. The latent heat and cooling load of food dispersion is calculated by the following formula: Q tau tau = 700 ? D (10.3.2) Evaporation of surface water and latent heat and cooling load The evaporative dehumidification of the open water is D (kg/h). D = (a + 0.00013 v), (Pqb - Pq), a. B/B1 (10.4.1) Type: A - evaporating surface, ?; A - diffusion coefficient at different water temperatures; V - air flow velocity on evaporating surface; Pqb - water vapor pressure that corresponds to saturated air under water surface temperature; Pq - water vapor pressure of indoor air; B - standard atmosphere, 101325Pa; B1 -- local atmospheric pressure (Pa). Q (W) of latent heat and cold load formed by evaporation of the surface of water. Q = (2500-2.35 t) 1000 (10.4.2) d. Type: T - water surface temperature, ?. Evaporation of water vapor and latent heat and cooling load The surface of water flowing on the surface should be calculated by the following formula: D = G c. (t1, t2)/gamma (10.5.1) Type: G -- the flow of water, kg/h; C - specific heat of water, 4.1868 kJ/(kg. K); The initial temperature of t1 - water, ?; T2 - the terminal temperature of the water discharged into water pipeline network of water temperature, ?; Gamma - water vaporization latent heat, averaging 2450kJ/kg. Q (W) of latent heat and cold load formed by evaporation of the surface of water. Q = (2500-2.35 (t1 + t2) / 2), d. 1000 (10.5.2) The heat and dispersion of chemical reactions Q = n1 n2 g. Q / 3600 (10.6.1) W = n1, n2, g. w. (10.6.2) Qq = 628 w. (10.6.2) Type: Q - total heat and heat loss of chemical reaction, W; N1 -- consider the coefficient of incomplete combustion, which is preferable to 0. 95; N2 - load factor, i.e., the ratio of actual fuel consumption per combustion point to its maximum fuel consumption, according to the process usage; G - maximum fuel consumption per hour, m3 / h; Q - heat value of fuel, kJ/m3; W - unit bulk moisture of fuel, kg/m3; W - the dispersion of chemical reaction, kg/h; Qq - the latent heat of chemical reaction, W. Heat load calculation formula The calculation of the basic heat loss of the enclosure The heat transfer heat (also known as the basic heat loss of the enclosure) is calculated by using the heat transfer heat of one side of the heating room (also known as the basic heat loss of the enclosure). Qj = k. f. a. (tn - tw) (1.1) Type: - the encroachment of the heat transfer coefficient k, W/(? ?); F - the encroachment of the surface heat dissipation area, ?; Tn - indoor air temperature calculation, ?; Tw - outdoor calculated temperature of heating, ?; A - temperature correction coefficient. External wall, roof heat bridge calculation The heat transfer coefficient of the outer wall and roof should be taken into account with the average heat transfer coefficient of the outer wall when considering the influence of the heat Bridges of beams, floors and columns. According to the regulation, take the weighted average of each component area. Surface heat transfer calculation When the enclosure is non-insulated ground with soil, the heat transfer heat transfer Qj. D (W) is calculated in the following formula: Qj. D = KPJ. D., Fd (tn - tw), (1.2) Type: KPJ. D - the insulation the average heat transfer coefficient of the ground, W/(? ?); Fd - room area on the ground, ?. Additional heat consumption The additional consumption of heat is calculated as a percentage of the basic calorie consumption. After consideration of the various additional, heat consumption of heat consumption of the surrounding enclosure is Q1 (W) : Q1 = Qj ? (1 + beta ch + beta f + beta lang + beta m) (1 + beta fg) (1 + beta jian) (2.1) Type: Qj - the basic heat loss of the enclosure, W; Beta ch - orientation correction; Beta f - wind correction; Beta lang - two-sided external wall correction; The area ratio of beta m - window wall is larger than that. Beta fg - room height correction; Beta jian - intermittent attachment. Heat dissipation of heat 2 (W) by cold air permeability through door and window gaps Q2 = 0.28 ? Cp ? V ? rho w ? (tn-tw) (3.1) Type: Cp - dry air quality of constant pressure specific heat, Cp = 1.0 Kj/(Kg ? ?); V - infiltration volume of air flow rate, m ^ 3 / h. Rho w - outdoor temperature of air density, Kg/m ^ 3; Tn - indoor air temperature calculation, ?; Tw - outdoor heating temperature calculation, ?. Aperture method Ignoring the increasing of hot pressure and outdoor wind speed up the room, only the calculation method of V when wind pressure is included: V = sigma (l? L? N) (3.1.1) Type: L -- the length of open doors and window gaps in the orientation of the room, m; L - volume of air permeability of doors and doors in doors and Windows, m3 / (m? H); N - the orientation correction factor for air permeability. Considering the combined effect of heat pressure and wind pressure, and the calculation method of outdoor wind speed with height increasing (the method of heating and air conditioning design specifications) : V = l1 ? L0 ? pow (m, b) (3.1.2) Type: L1 - door and window slit length, m; L0 - door window is the benchmark of aperture infiltration air volume per meter, m ^ 3 / h.m. M -- the comprehensive correction factor of the air permeability of the crack of doors and Windows; B - door and window crack permeability index, b = 0.56 ~ 0.78. When no actual data is measured, b = 0.67. Determination of L0: L = a1 ? pow ((v10 ? v10 ? rho w / 2), b) (3.1.3) A1 - door crack seepage coefficient, m ^ 3 / (m * h * Pab), note: Pab representative: Pa (mpa) b; V10 - the average wind speed of the most wind direction in winter outdoor, m/s. Determination of m: M = Cr ? Cf ? (pow (n, 1 / b) + C) ? Ch (3.1.4) Type: Cr - coefficient of heat pressure; Cf - the coefficient of wind pressure difference, m/s, when there is no measured data, it is desirable 0.7; C - the ratio of effective thermal pressure difference and effective air pressure difference on both sides of the door and window analysis; Ch - height correction coefficient can be calculated according to the calculation. Ch = 0.3 ? pow (h, 0.4) (3.1.5) H - the elevation of the center line of window and window. The determination of C C = 70 {(hz-h) / [Cf ? v10 ? v10 ? pow (h, 0.4)]} ? [(tn '-tw)/(273 + tn')] (3.1.6) Type: Hz - thermal pressure alone, the elevation of the building neutralization, m; Tn '- the temperature of the shaft in which the thermal pressure is formed in the building. Ventilation frequency method V = k. Vf (3.2.1) Type: V -- cold air permeability of room, m3 / h; K -- the number of ventilation, 1 / h; Vf -- the net area of the room, m3. The heat and heat loss of the door and window of the single-storey industrial workshop can be determined by the value of the second edition of the second edition of the design manual of practical heating and air conditioning Layers of industrial workshop outside door crack seepage hot air loss, when there is no other artificial ventilation system work in workshop and no skylight, generating a large amount of waste heat, per meter aperture infiltration air volume can be calculated on civil multi-storey building seepage volume, with proper gap method, calculated after infiltration volume, then calculating the heat. The external door opens into the cold air heat consumption Q3 (W) Please refer to P314, the second edition of the design manual of practical heating and air conditioning. The gate of the single-storey plant opens to the cold wind energy consumption Q3 (W) The opening time of each shift is equal to or less than 15min, and the additional rate method is adopted to determine the heat consumption of the gate Additional on the basic consumption heat of the gate, the additional rate is 200% ~ 500% The opening time of each shift is greater than 15min. According to the following experience formula, the gate will be opened and the cold wind amount of G (kg/s) will be turned on. G = A + (A + N, vw), F (5.1) Type: G -- wind in the cold wind, kg/s A - constant N - constant, when the gate size is 3.0 m x 3.0 m, N = 0.25 When the gate size is 4.0 m x 4.0 m, N = 0.2 When the gate size is 4.7 m x 5.6 m, N = 0.15 Vw - winter outdoor average wind speed, m/s F - the area of vent window or vent that may be opened at the top of the workshop, m2 Door opening into the air cooling heat consumption of multi-story factory buildings according to civil multi-storey building outside door open into air cooling heat consumption calculation, on the condition that no mechanical ventilation in the workshop of pressure (or positive or negative), without the skylight, without a lot of heat.