Building name cooling load index
W/m[] Building area Building name Cooling load index
Building area (square meters)
Hotel 80-90 Gymnasium 100- 135
200-350 (according to the number of seats)
Office building 85- 100
Library Room 35-40 190-380
Hospital 80-90 Data Processing 320-400
Store 105- 125
When the business hall is air-conditioned, the theater of 200-250 is based on the business hall area 126- 160.
200-300 (according to the area of the audience hall)
Hall 180-225
Note: l, the above indicators are the cooling load indicators of the total construction area: when the total construction area is less than 5000 square meters, the upper limit is taken; More than l0000 square meters, take down the restrictions.
2, according to the above indicators to determine the cooling load, that is, the capacity of the refrigerator, do not need to add coefficient.
3. Due to regional differences, the above indicators are based on Beijing. The southern region can adopt the upper limit.
Thermal load estimation
(l) Estimate according to the thermal index of the building area.
Note: The total building area, thermal performance of large-scale external envelope structure and window area are good, so smaller indicators are adopted; Instead, use a larger indicator.
(2) the window wall ratio formula method:
q =(7a+ 1.7)W/F(TN-tw)W/m[]2；
Description: Q-heating index of the building, W/m[].
A—— the area of external windows and external walls (including the proportion of windows);
W- total external wall area (including windows), m []
F total construction area, m []
TN-- indoor heating design temperature,℃
Tw-outdoor heating design temperature,℃
3.？ Unit selection
Unit selection step
A. estimate or calculate the cooling load
The total cooling load is estimated by the estimation method in Section 3.2.2, or calculated by the relevant load calculation method.
B. estimate or calculate the heat load
The total heat load is estimated by the estimation method in Section 3.2.2, or calculated by the relevant load calculation method.
C. initial unit model
According to the total cooling load, the model and number of units are selected for the first time.
D, determine the unit model.
According to the total heat load, check the initial unit model and quantity. And determine the unit model.
4.？ Case study of unit selection
Example: Building conditions: The building area of an office building in Beijing is 1 1000 m[]2? Air conditioning area 10000 m[], including large conference room area of 500 m[]2, small conference room area of 1500 m[] and office building area of 8000 m[]2, including fresh air.
A. calculate the cooling load.
A. according to the air conditioning cold load method:
Large conference room 500 x 358 =179000 w =179 kw.
Small conference room:1500 x 235 = 352500 = 352.5kw.
Office area: 7000x151=1057000 =1057kw.
Total: 358 ten 235+1208 =1588.5kw.
Load when selecting a host:1588.5x0.70 =112kw.
B. According to the construction area method:
1 1000 x98 = 12 12000 w = 1078 kw
C. According to the calculation results of 1) and 2), the cooling load is calculated as112kw.
B. Calculating heat load
Calculated by air conditioning heat load method:
1 1000 X 60 = 660000 w = 660 kw
C. initially selected models and quantities:
1, if the scheme adopts water source heat pump.
① Determine the unit model: the total cooling load is112kw, the cooling capacity of two GSHP580 water source heat pump units is 16 ~ 18℃, and the supply and return water temperature is 7 ~ 17℃. Slightly greater than the cooling load, meet the requirements.
The total heat load is 660kw. When the water temperature is 16 ~ 18℃ and the supply and return water temperature is 55 ~ 45℃, the heating capacity of a ground source heat pump unit is 665kw. Slightly greater than the heat load, meet the requirements.
② Two GSHP580 water source units were finally determined, of which two were used for cooling in summer and one for heating in winter (two were used for heating when the outdoor temperature was low).
2. If the scheme adopts air-cooled heat pump central air conditioning unit.
① Determine the unit model:
According to the above calculation, the total cooling load is112kw, and the cooling capacity of two LSBLGRF560M modular heat pump series air-cooled (hot) pump units is1120k at the return water temperature of 7 ~ 17℃.
The total heat load is 660kw, and the heating capacity of a LSBLGRF560M unit is 588kw. When the supply and return water temperature is 55 ~ 45℃, it is slightly less than the heat load and meets the requirements.
② Two LSBLGRF560M modular heat pump series air-cooled (hot) pump units are finally determined, of which two are used for cooling in summer and 1 set is used for heating in winter (two are used for heating when the outdoor temperature is low).
3. If water-cooled central air conditioning unit is adopted.
① according to the above calculation, the total cooling load is112kw, and the cooling capacity of two LSBLG640Z water-cooled central air conditioning units is 1278kw. When the return water temperature is 7 ~ 17℃, it is slightly higher than the cooling load and meets the requirements.
② Two LSBLG640Z water-cooled central air conditioning units were finally determined, two of which were used for cooling in summer.
5.？ ancillary equipment
1)? Selection of water pump:
Cooling load q =1112kw; On this basis, the heat taken by the water loop of the air conditioning system is multiplied by 1.3, and the water flow rate is g when the utilization coefficient is 0.7.
g =(q×a× 1.3)÷( 1. 163×T)
A: Utilization rate
G: water flow
T: temperature difference between supply and return water of air conditioning water system
G = (1112× 0.7×1.3) ÷ (1.163×10) = 87m [
That is, the flow rate of the pump is 87 m[]3.
/h .
2)? Resistance calculation
The pipe diameter is about 300, and the specific friction is 200 Pa/m..
Then H 1=300×200 Pa=6mH2O.
When the local resistance is 0.5, H2=0.5×6=3mH2O.
Brake control valve H3=5mH2O
Unit pressure drop H4=50Kpa=5mH2O
Heat exchanger pressure drop H5 = 4h2o
Total lift h =1.2h = (6+3+5+5+4) = 28.8H2O.
Therefore, the circulating pump g = 87m [] h = 32m H2O n =17.5kw n =1450 rpm is selected.
3)? Selection of constant pressure pump:
The constant pressure point is the highest point, and 5m H2O is added.
H = 32+5 = 37 H2O
The water capacity of this building is 1.3L/m2.
VC = 1 1000× 1.3 = 14300 l = 14.3m[]3
The hourly flow is 10% of Vc.
Then g = 0.10×14.3 =1.43m []
Therefore, the constant pressure pump needs 2 m[ central air conditioning project budget table
]3
/h H=37m n= 1450rpm
? 6.？ Water quantity calculation of heat pump central air conditioning system
(1) Water quantity calculation of central air conditioning system in summer;
According to the laws of thermodynamics, the water quantity of source water and frozen water can be obtained by the following formula.
Gr=0.86(QL+N)/△Ty
G2=0.86QL/△TL
Description: Gr source water, m3/h;
GL: chilled water volume, m3/h;
QL: cooling capacity of central air conditioning system, kW;
N: the electric power of the central air conditioner host, kW;
△Ty: temperature difference between inlet and outlet water of central air conditioner,℃;
△TL: temperature of chilled water entering and leaving the central air conditioner.
(2) Calculation of winter water quantity of heat pump central air conditioning system:
According to the law of thermodynamics, the amount of water and hot water in winter can also be obtained by the following formula.
Gy=0.86(Qr-N)/△Ty
Gr=0.86Qr/△Tr
Description: Gy source water, m3/h;
Gr: hot water quantity, m3/h;
Qr: heat of main mechanism of water source central air conditioning system, kW;
N: electric power of main engine of water source central air conditioner, kW;
△Ty: the temperature difference between the source water and the central air-conditioning host,℃;
△TL: temperature of chilled water entering and leaving the central air conditioner,℃.
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