CN105373707A - Method for calculating heating loads of solar heating system and building - Google Patents

Abstract

The invention discloses a method for calculating heating loads of a solar heating system and a building. The calculation method comprises the following steps: calculating a building heating load q required by a building winter heating ventilation system; calculating a unit power Q<2> required in a building heating area; calculating an area S of a heat collector in the solar heating system; and calculating a volume V of a water tank in the solar heating system. Through adoption of the calculation method, a building heating load numerical value of the building can be calculated accurately, so that the area of the heat collector and the volume of the water tank in the solar heating system can be configured reasonably, and the aim of lowering the building energy consumption while enhancing the indoor thermal environment quality is fulfilled. The calculation method is suitable for calculating the heating load of the solar heating system.

Description

The computing method of solar heating system and building heat supplying load

Technical field

The invention belongs to solar heating technical field, be specifically related to the computing method of a kind of solar heating system and building heat supplying load.

Background technology

Sun power is reproducible clean energy resource, solar heating application can be saved fossil energy under construction, avoid environmental pollution, alleviate greenhouse effect, in urban architecture, effectively utilize solar heating, not only can reduce heating system initial outlay and maintenance management expense, ensure that living environment is comfortable, good energy-saving effect can be obtained simultaneously, therefore the use of large solar heating system is the trend of cities and towns heating in the future, and application prospect is very good.

Solar heating system is made up of solar thermal collector, water tank, connecting tube, control system etc.Solar heating system refers to and the sun power of dispersion is passed through heat collector, sun power is converted to hot water, by hot water storage in water tank, then by delivery to heating terminal (such as: floor panel heating, radiator heating), the demand of building heat supplying is provided.

The configuration of solar thermal collector and water tank is the power decision needed for building and heating area, and the power needed for building and heating area is relevant to heating load.Heating load is not only relevant to floor area of building, also relevant with building external envelope structure.And in prior art, also not relevant easy computing method clearly can determine the numerical value of the heating load of single buildings, the numerical value of heating load is freely chosen according to certain numerical range often, if the heating load numerical value chosen is higher, make the configuration requirement of each equipment in solar heating system high, costly; If the heating load numerical value chosen is lower, do not reach again building heating demand.

Summary of the invention

For solving the above deficiency existed in prior art, the present invention aims to provide the computing method of a kind of solar heating system and building heat supplying load, accurately can calculate the heating load numerical value of buildings, thus the area of solar thermal collector in reasonable disposition solar heating system and the volume of water tank, with reach improve indoor thermal environment quality while reduce the object of building energy consumption.

For achieving the above object, the technical solution adopted in the present invention is as follows:

Computing method for solar heating system and building heat supplying load, the method, for the volume V of the area S and water tank that calculate the heat collector in solar heating system, specifically comprises the following steps:

Steps A.

The energy Qa that through type (1) calculating building external envelope structure every a part of unit interval in the winter time consumes

$Qa=\frac{\mathrm{\&Delta;}T\&CenterDot;A}{\underset{i=1}{\overset{n}{\&Sigma;}}\frac{b_i}{{\mathrm{\&lambda;}}_i}+\frac{1}{{\mathrm{\&alpha;}}_1}+\frac{1}{{\mathrm{\&alpha;}}_2}}---\left(1\right)$

Parameters in formula is the basic architecture information using conventional method to gather, wherein:

Δ T-architectural exterior-protecting construction temperature difference DEG C;

A-architectural exterior-protecting construction area m ²;

B _i-building structural materials thickness m;

λ _i-building structural materials imports coefficient W/ (mk);

α ₁-building inside surface Air Heat Transfer Coefficient W/ (mk);

α ₂-outer surface of building Air Heat Transfer Coefficient W/ (mk);

The heat summation consumed by each several part, draws the energy Q of building wastage in bulk or weight in winter

$Q=\underset{a=1}{\overset{n}{\&Sigma;}}Qa---\left(2\right)$

The a kind peripheral protective structure of a-building;

Step B.

The Q calculated in steps A is brought into formula (3), calculate the heating load q of building needed for winter heating

$q=\frac{Q}{A_o}---\left(3\right)$

In formula, A ₀-floor area of building m ²;

Step C.

The q calculated in step B is brought into formula (4), calculate the power Q needed for building and heating area ₂

Q ₂＝q·A ₁·H(4)

In formula, A ₁-building and heating area m ²;

The H-heating duration h of a day;

The Q that step D. will calculate in step C ₂bring formula (5) into, calculate the area S of the heat collector in solar heating system

$S=\frac{Q_2}{q_i\&CenterDot;H_1}\&CenterDot;f---\left(5\right)$

In formula, q _i-heat collector heat absorption capacity W/ (m ²h), be the performance parameter of heat collector;

H ₁the effective light application time h of-heating season sun;

F-correction factor, f=1.2.

Step e. by the Q calculated in step C ₂bring formula (6) into, calculate the volume V of the water tank in solar heating system

$V=\frac{Q_2}{{\mathrm{c\&Delta;T}}_1\mathrm{\&rho;}}---\left(6\right)$

In formula, c-specific heat of water J/ (kg DEG C);

Δ T ₁-tank hot water can supply the temperature difference DEG C;

The density kg/m of ρ-water ³.

Owing to have employed technique scheme, compared with prior art, acquired technical progress is in the present invention:

Computing method of the present invention accurately can calculate the heating load q needed for building winter heating ventilating system and the power Q needed for building winter heating area ₂if, Q ₂≤ 40w, then this building can adopt solar energy heating system heat supply, thus recycling Q ₂the area of the solar thermal collector in reasonable disposition solar heating system and the volume of water tank, with reach improve indoor thermal environment quality while reduce the object of building energy consumption.

The present invention is applicable to the heating demand calculating solar heating system.

Embodiment

Embodiment 1

Computing method for solar heating system and building heat supplying load, the method, for the volume V of the area S and water tank that calculate the heat collector in solar heating system, specifically comprises the following steps:

Steps A.

The energy Qa that through type (1) calculating building external envelope structure every a part of unit interval in the winter time consumes

$Qa=\frac{\mathrm{\&Delta;}T\&CenterDot;A}{\underset{i=1}{\overset{n}{\&Sigma;}}\frac{b_i}{{\mathrm{\&lambda;}}_i}+\frac{1}{{\mathrm{\&alpha;}}_1}+\frac{1}{{\mathrm{\&alpha;}}_2}}---\left(1\right)$

Parameters in formula is the basic architecture information using conventional method to gather, wherein:

Δ T-architectural exterior-protecting construction temperature difference DEG C;

A-architectural exterior-protecting construction area m ²;

B _i-building structural materials thickness m;

λ _i-building structural materials imports coefficient W/ (mk);

α ₁-building inside surface Air Heat Transfer Coefficient W/ (mk);

α ₂-outer surface of building Air Heat Transfer Coefficient W/ (mk);

The heat summation consumed by each several part, draws the energy Q of building wastage in bulk or weight in winter

$Q=\underset{a=1}{\overset{n}{\&Sigma;}}Qa---\left(2\right)$

The a kind peripheral protective structure of a-building;

Step B.

The Q calculated in steps A is brought into formula (3), calculate the heating load q of building needed for winter heating

$q=\frac{Q}{A_o}---\left(3\right)$

In formula, A ₀-floor area of building m ²;

Step C. brings the q calculated in step B into formula (4), calculates the power Q needed for building and heating area ₂

Q ₂＝q·A ₁·H(4)

In formula, A ₁-building and heating area m ²;

The H-heating duration h of a day;

If the Q calculated in step D. step C ₂≤ 40w, then by Q ₂bring formula (5) into, calculate the area S of the heat collector in solar heating system

$S=\frac{Q_2}{q_i\&CenterDot;H_1}\&CenterDot;f---\left(5\right)$

In formula, q _i-heat collector heat absorption capacity W/ (m ²h), be the performance parameter of heat collector;

H ₁the effective light application time h of-heating season sun;

F-correction factor, f=1.2.

Step e. by the Q calculated in step C ₂bring formula (6) into, calculate the volume V of the water tank in solar heating system

$V=\frac{Q_2}{{\mathrm{c\&Delta;T}}_1\mathrm{\&rho;}}---\left(6\right)$

In formula, c-specific heat of water J/ (kg DEG C);

Δ T ₁-tank hot water can supply the temperature difference DEG C;

The density kg/m of ρ-water ³.

The solar heating system of embodiment 2XX residential quarter and the calculating of building heat supplying load

The indoor temperature setting this building winter heating is 18 DEG C, and public non-heating portion temperature is 0 DEG C, and outdoor temperature is-5 DEG C.

This building is totally 5 kinds of building enclosures, is respectively exterior wall, the public parting wall of not heating, outer door and window, enclosed balcony breast board and roof.

Steps A. calculate the hot merit that five kinds of building enclosures consume respectively

A1. exterior wall

Exterior wall is that shale porous brick wall all pastes extruded polystyrene board, the coefficient of heat conductivity λ of shale porous brick wall ₁=0.58w/mk, the thickness b of shale porous brick wall ₁=0.37m, the coefficient of heat conductivity λ of extruded polystyrene board ₂=0.033w/mk, the thickness b of extruded polystyrene board ₂=0.05m, outer surface of wall amasss A=347m ², due to building inside surface Air Heat Transfer Coefficient α ₁, outer surface of building Air Heat Transfer Coefficient α ₂very low on the impact of hot merit Q1 numerical value, therefore ignore, draw according to formula (1) hot merit that exterior wall consumes

$Q1=\frac{\mathrm{\&Delta;}T\&CenterDot;A}{\frac{b_1}{{\mathrm{\&lambda;}}_1}+\frac{b_2}{{\mathrm{\&lambda;}}_2}}=\frac{\&lsqb;18-(-5)\&rsqb;\&times;347}{\frac{0.37}{0.58}+\frac{0.05}{0.033}}\&ap;3707w$

A2. the public parting wall of not heating

The public parting wall of not heating is that shale porous brick pastes rock cotton board, the coefficient of heat conductivity λ of shale porous brick wall ₁=0.58w/mk, the thickness b of shale porous brick wall ₁=0.24m, the coefficient of heat conductivity λ of rock cotton board ₂=0.04w/mk, the thickness b of rock cotton board ₂=0.015m, the public parting wall wall area A=199.68m of not heating ², due to building inside surface Air Heat Transfer Coefficient α ₁, outer surface of building Air Heat Transfer Coefficient α ₂very low on the impact of hot merit Q2 numerical value, therefore ignore, draw according to formula (1) hot merit that the public parting wall of not heating consumes

$Q2=\frac{\mathrm{\&Delta;}T\&CenterDot;A}{\frac{b_1}{{\mathrm{\&lambda;}}_1}+\frac{b_2}{{\mathrm{\&lambda;}}_2}}=\frac{\&lsqb;18-0\&rsqb;\&times;199.68}{\frac{0.24}{0.58}+\frac{0.015}{0.04}}\&ap;4555w$

A3. outer door and window

Outer door and window adopts double glazing plastic-steel window, heat transfer coefficient k=2.8w/mk, external door window ara A=298.456m ², the hot merit that outer door and window consumes

Q3＝k·A·ΔT＝2.8×298.456×(18-(-5))≈19221w

A4. enclosed balcony breast board

Enclosed balcony breast board is that air-entrained concrete building block pastes extrusion molding benzene plate, the coefficient of heat conductivity λ of air-entrained concrete building block ₁=0.16w/mk, the thickness b of air-entrained concrete building block ₁=0.15m, the coefficient of heat conductivity λ of extrusion molding benzene plate ₂=0.033w/mk, the thickness b of extrusion molding benzene plate ₂=0.03m, balcony fence area A=152.496m ², due to building inside surface Air Heat Transfer Coefficient α ₁, outer surface of building Air Heat Transfer Coefficient α ₂very low on the impact of hot merit Q4 numerical value, therefore ignore, draw according to formula (1) hot merit that enclosed balcony breast board consumes

$Q4=\frac{\mathrm{\&Delta;}T\&CenterDot;A}{\frac{b_1}{{\mathrm{\&lambda;}}_1}+\frac{b_2}{{\mathrm{\&lambda;}}_2}}=\frac{\&lsqb;18-(-5)\&rsqb;\&times;152.496}{\frac{0.15}{0.16}+\frac{0.03}{0.033}}\&ap;1899w$

A5. roof

Roof is that armoured concrete slab pastes extrusion molding polystyrene board heat-insulatation layer, the coefficient of heat conductivity λ of armoured concrete slab ₁=1.74w/mk, the thickness b of armoured concrete slab ₁=0.12m, the coefficient of heat conductivity λ of extrusion molding benzene plate ₂=0.033w/mk, the thickness b of extrusion molding benzene plate ₂=0.1m, roof area A=525.7m ², due to building inside surface Air Heat Transfer Coefficient α ₁, outer surface of building Air Heat Transfer Coefficient α ₂very low on the impact of hot merit Q5 numerical value, therefore ignore, draw according to formula (1) hot merit that roof consumes

$Q5=\frac{\mathrm{\&Delta;}T\&CenterDot;A}{\frac{b_1}{{\mathrm{\&lambda;}}_1}+\frac{b_2}{{\mathrm{\&lambda;}}_2}}=\frac{\&lsqb;18-(-5)\&rsqb;\&times;525.7}{\frac{0.12}{1.74}+\frac{0.1}{0.033}}\&ap;3900w$

According to formula (2), total hot merit that this building consumes

Q＝Q1+Q2+Q3+Q4+Q5＝3707+4555+19221+1899+3900＝33282w

Step B. calculates this heating load q needed for building winter heating ventilating system

The area A of this building ₀=900m ²

$q=\frac{Q}{A_o}=\frac{33282}{900}\&ap;36.98w/m^2$

Round, q=37w/m ²

Step C. calculates this power Q needed for building and heating area ₂

This building is totally 66 families, each household heating area 109.1m ², the total heating area A of this building ₁=7200m ².One day heating time was that (wherein 23:00-5:00 was heat supply mode at night, setting heating 1 hour, stopped for 1 hour in 19 hours; Daytime, noon, 11:00-13:00 did not heat), draw the power Q needed for this building and heating area according to formula (4) ₂

Q ₂＝q·A ₁·H＝37×7200×19＝5061600w(4)

Step D. calculates the area S of the heat collector in solar heating system

Flat plate collector is greater than 550w/m at photo-irradiation intensity ²time, its heat absorption capacity can reach q _i=420w/m ², accept solar energy impinges time H every day ₁=5.5h, draws the area of heat collector according to formula (5)

$S=\frac{Q_2}{q_i\&CenterDot;H_1}\&CenterDot;f=\frac{5061600}{420\&times;5.5}\&times;1.2\&ap;2630m^2$

The heat collector that this engineering is chosen is flat plate collector, and adopt blue film absorber technology, instantaneous heat absorption efficiency >=80%, absorptivity >=95%, emissivity≤5%, size 2000x1000, endothermic tube is copper pipe.Every block collecting plate area is 2m ², then need 2630/2=1315 altogether, according to the actual conditions of the roofing of community, place collecting plate quantity to greatest extent, this engineering finally chooses 1350 pieces.

According to the power of this building and heating area that the quantity of collecting plate calculates

$Q_2=\frac{S\&CenterDot;q_i\&CenterDot;H_1}{f}=\frac{1350\&times;2\&times;420\&times;5.5}{1.2}=5197500w\&ap;5198kw$

Step e. calculate the volume V of the water tank in solar heating system

The minimum heating temperature t1=32 DEG C of heat storage water tank, rule of thumb water tank flat plate collector is heated to 62-65 DEG C water, and its heat transfer efficiency is higher, therefore makes t2=64 DEG C, Δ T ₁=t2-t1=32 DEG C, the density p=1000kg/m of water ³, specific heat of water c=1kCal/kg DEG C, the unit conversion 1kwh=859.9931kcal of mechanical equivalent of heat, draw according to formula (6)

$V=\frac{Q_3}{\mathrm{\&rho;}\&CenterDot;c(t2-t1)}=\frac{5198\&times;859.99}{1000\&times;1\&times;(64-32)}\&ap;140m^3$

It is 150m that this engineering chooses water tank capacity ³.

Claims (1)

1. computing method for solar heating system and building heat supplying load, is characterized in that, the method, for the volume V of the area S and water tank that calculate the heat collector in solar heating system, specifically comprises the following steps:

Steps A.

The energy Qa that through type (1) calculating building external envelope structure every a part of unit interval in the winter time consumes

$Qa=\frac{\mathrm{\&Delta;}T\&CenterDot;A}{\underset{i=1}{\overset{n}{\&Sigma;}}\frac{b_i}{{\mathrm{\&lambda;}}_i}+\frac{1}{{\mathrm{\&alpha;}}_1}+\frac{1}{{\mathrm{\&alpha;}}_2}}---\left(1\right)$

Parameters in formula is the basic architecture information using conventional method to gather, wherein:

Δ T-architectural exterior-protecting construction temperature difference DEG C;

A-architectural exterior-protecting construction area m ²;

B _i-building structural materials thickness m;

λ _i-building structural materials imports coefficient W/ (mk);

α ₁-building inside surface Air Heat Transfer Coefficient W/ (mk);

α ₂-outer surface of building Air Heat Transfer Coefficient W/ (mk);

The heat summation consumed by each several part, draws the energy Q of building wastage in bulk or weight in winter

$Q=\underset{a=1}{\overset{n}{\&Sigma;}}Qa---\left(2\right)$

The a kind peripheral protective structure of a-building;

Step B.

The Q calculated in steps A is brought into formula (3), calculate the heating load q of building needed for winter heating

$q=\frac{Q}{A_o}---\left(3\right)$

In formula, A ₀-floor area of building m ²;

Step C.

The q calculated in step B is brought into formula (4), calculate the power Q needed for building and heating area ₂

Q ₂＝q·A ₁·H(4)

In formula, A ₁-building and heating area m ²;

The H-heating duration h of a day;

The Q that step D. will calculate in step C ₂bring formula (5) into, calculate the area S of the heat collector in solar heating system

$S=\frac{Q_2}{q_i\&CenterDot;H_1}\&CenterDot;f---\left(5\right)$

In formula, q _i-heat collector heat absorption capacity W/ (m ²h), be the performance parameter of heat collector;

H ₁the effective light application time h of-heating season sun;

F-correction factor, f=1.2.

Step e. by the Q calculated in step C ₂bring formula (6) into, calculate the volume V of the water tank in solar heating system

$V=\frac{Q_2}{{\mathrm{c\&Delta;T}}_1\mathrm{\&rho;}}---\left(6\right)$

In formula, c-specific heat of water J/ (kg DEG C);

Δ T ₁-tank hot water can supply the temperature difference DEG C;

The density kg/m of ρ-water ³.