Calculation of heat and humidity load of computer room:
According to Article 5.2.2 of GB50174-2008 "Design Code for Electronic Computer Room", the heat and humidity load of the air conditioning of electronic computer room should include the following:
A. Heat dissipation of computers and other equipments;
B. Heat transmission of the building envelope;
C. Solar radiation;
B. Heat transfer of building envelope;
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C. solar radiation heat;
D. human body heat and moisture dissipation;
E. lighting fixtures heat dissipation;
F. fresh air load.
1) Heat and humidity load analysis:
By analyzing the above six elements that constitute the heat and humidity load of computer room air conditioning, it can be learned that the heat load of computer room air conditioning consists of heat dissipation by computers and other equipment, heat transfer from the building maintenance structure, solar heat, human body heat dissipation, heat dissipation by the lighting fixtures and the fresh air heat load; the humidity load of computer room air conditioning consists of humidity dissipation by the human body and the fresh air humidity load; the humidity load of computer room air conditioning consists of humidity dissipation by the human body and the fresh air heat load. Dissipation of moisture as well as the new air humidity load consists of.
2) Calculation and analysis of heat load:
A. Heat dissipation of computers and other equipment; Q1=860×P×η1η2η3 Kcal/h
Where:
Q1: heat dissipation load of the computers and other equipment;
P: total power consumption of the computers and other equipment;
η1: the coefficient of simultaneous use;
The heat load is calculated by the following factors:
The heat load is analyzed by the calculation of the heat load of the computer room. coefficient;
η2: utilization coefficient;
η3: load work uniformity coefficient;
usually, η1η2η3 is taken to be between 0.6-0.8
B. Heat transfer of the building envelope; Q2==K×F×(t1-t2) Kcal/h
where:
Q2: heat transfer load of the building envelope;
K: heat transfer coefficient of the building maintenance structure, ordinary concrete is 1.4-1.5;
F: area of the building maintenance structure;
t1: outdoor calculated temperature;
t2: indoor calculated temperature of the computer room;
In addition. Roof and floor heat transfer should be considered correction factor calculation.
C. Solar radiation heat; Q3=K×F×q Kcal/h
Where:
Q3: solar radiation heat load;
K: solar radiation heat penetration coefficient, usually taken as 0.36-0.4;
F: the area of the glass window;
q: intensity of solar radiation heat through the window;
Thread: the area of the window;
q: the area of the window. intensity of solar radiation heat;
D. Human body heat;
Heat emitted by the human body varies with the working condition.
Staff in the machine room may be treated as light work. When the room temperature is 24 ℃, its sensible heat load is 56cal, latent heat load is 46cal; when the room temperature is 21 ℃, its sensible heat load is 65cal, latent heat load is 37ca1. in both cases, its total heat load is 102cal
E. Lighting fixtures to dissipate heat; Q4 = C × P Kcal/h
Where:
C : heat per lW output (kcal/hW), usually 0.86 for self-incandescent lamps and 1.0 for fluorescent lamps;
P: nominal rated output power of lighting fixtures;
F. Fresh air load.
In order to continuously replenish fresh air for the staff in the computer room, as well as to maintain the positive pressure of the computer room with air exchange, it is necessary to send fresh air from outside to the computer room through the fresh air outlet of the air-conditioning equipment, and this fresh air will also become the heat load.
The amount of outdoor air intruded through the door, window gaps and switches, the degree of sealing of the random room, the number of people in and out of the number of times and the outdoor wind speed and change. This heat load is usually very small, if necessary, it can be split into room air volume to determine the heat load.
1.2.3, area calculation method Qt = S * P
-Qt total cooling capacity (KW)
-S area of the machine room (㎡)
-P cold estimation index
For example:
Basic machine room Situation UPS capacity 6KVA * 1, the area of 17 square meters;
Electric heat load
Machine room load according to the UPS full load, then the electric power of 6 * 0.8 = 4.8KW, 90% of which is converted into heat
Q1 = 4.8 * 0.9 = 4.32KW
Machine room environment heat Q2 = 17 * 0.18 = 3.06KW
Electric heat Q1 = 4.8* 0.9 = 3.06KW
Machine room environment heat Q2 = 17* 0.18 = 3.06KW. 3.06KW
Total heat generation of machine room
Q=Q1+Q2=4.32+3.06=7.38KW
Therefore: the minimum cooling capacity provided by machine room air-conditioner is 7.38KW
Because of the large area of the machine room, according to the actual situation on site,
Recommended one Emerson DME07WC machine room air-conditioner with a single unit with a total cooling capacity of 7.5KW. DME07WC special air conditioner for equipment room fully meets the cooling requirements.