Computer room precision air conditioning, heat load calculation
1: room main heat source of (1) equipment load
Computer and cabinet heat load)
(2) computer room lighting load;
(3) building maintenance structure load;
(4) add new wind load;
(5) personnel of heat load, etc.
6. Other 2: heat load analysis: (
Computer equipment heat load: Q1 = 860 XPX & eta;
3 Kcal/hQ1: computer equipment heat load P: engine rooms all sorts of equipment total power consumption & eta;
1: use coefficient & eta; at the same time
2: the utilization coefficient, & eta;
3: load uniformity coefficient, usually work & eta;
3 take 0.
Between 8, this design considering capacitance change is small, the values to 0.
Lighting equipment heat load: Q2 = CxPKcal/hP: lighting equipment calibration output power C: 1 w heat quantity per output Kcal/hw (
Incandescent lamp 0.
86 light 1)
According to national standard 'computing station site technical requirements' requirements, the room illumination should be greater than 2001 x, its power consumption is about 20 W/M2 after calculation, lighting power consumption will be calculated based on 20 W/M2.
Body heat load Q3 = PxNKcal/hN: room often personnel number P: human body heat, light manual workers heat load is the sum of sensible heat and latent heat, at room temperature is 21 ℃ and 24 ℃ 102 kcal.
Palisade structure heat conduction Q4 = KxFx (
Kcal/hK: 1 protection structure system of thermal conductivity of normal concrete.
5 f: protecting structure area of t1: room temperature ℃ in t2: calculate ℃ temperature outside the room in the later calculation, t1 -
T2 at 10 ℃ is calculated.
The roof and floor according to the correction coefficient of 0.
Fresh air heat load calculation is relatively complicated, in this scenario, we have to air conditioning equipment allowance to balance itself, not the other.
Other heat load in addition to the heat load, used in the work of the oscilloscope, soldering iron, vacuum cleaner, etc will be heat load, because these devices to small power consumption, only roughly according to the input power and the products of the joule equivalent calculation.
Q5 = 860 xp2, rooms total heat Q = Q1 plus Q2, Q3 and Q4 + Q5