Design criteria for room ventilation
In order to design a good ventilation of switchgear and transformer rooms, the air in the room must meet various requirements. The most important is not to exceed the permissible maximum temperature. Limit values for humidity and air quality, e.g. dust content, may also be set.
Switchboards and gas-insulated switchgear have a short-term maximum temperature of 40 °C and a maximum value of 35°C for the 24h average. The installation requirements of the manufacturers must be observed for auxiliary transformers, power transformers and secondary installations. The spatial options for ventilation must also be considered.
Ventilation cross sections may be restricted by auxiliary compartments and buildings. If necessary, the loss heat can be vented through a chimney.
Ultimately, economic aspects such as procurement and operating expenses must be taken into account as well as the reliability (emergency power supply and redundancy) of the ventilation. At outside air temperatures of up to 30 °C, natural ventilation is generally sufficient. At higher temperatures there is danger that the permissible temperature for the equipment may be exceeded.
Figures 1 and 2 show frequently used examples of room ventilation.
The ventilation efficiency is influenced by the configuration and size of the incoming air and exhaust air vents, the rise height of the air (centre of incoming air opening to centre of exhaust air opening), the resistance in the path of the air and the temperature difference between incoming air and outgoing air. The incoming air vent and the exhaust air vent should be positioned diagonally opposite to each other to prevent ventilation short circuits.
If the calculated ventilation cross section or the chimney opening cannot be dimensioned to ensure sufficient air exchange, a fan will have to be installed. It must be designed for the required quantity of air and the pressure head.
If the permissible room temperature is only slightly above or even below the maximum outside temperature, refrigeration equipment or air-conditioning is used to control the temperature.
The resistance of the air path is generally: R = R_{1} +m^{2} R_{2}
Where:
- R1 resistance and acceleration figures in the incoming air duct,
- R2 resistance and acceleration figures in the exhaust air duct,
- m ratio of the cross section A1 of the incoming air duct to the cross section A2 of the exhaust air duct.
Figure 2 shows common configurations.
Where //
- A1 = incoming air cross section,
- A2 = exhaust air cross section,
- H = “chimney” height,
- 1 = fan,
- 2 = exhaust air slats,
- 3 = inlet air grating or slats,
- 4 = skirting,
- 5 = ceiling.
The total resistance consists of the components together. The following values for the individual resistance and acceleration figures can be used for an initial approximation:
Figures | Value |
Acceleration | 1 |
Right-angle bend | 1.5 |
Rounded bend | 1 |
Bend of 135° | 0.6 |
Slow change of direction | 0…0.6 |
Wire screen | 0.5…1 |
Slats | 2.5…3.5 |
Cross section widening | 0.25…0.9** |
** The smaller value applies for a ratio of fresh air cross section to compartment cross section of 1:2, the greater value for 1:10.
Calculation of the quantity of cooling air:
With temperature and height correction the following applies for the incoming air flow:
Where:
- V_{0} = standard air volume flow at sea level, p_{0} = 1013 mbar, T_{0} = 273 K = 0 °C,
- T_{1} = cooling air temperature (in K),
- T_{2} = exhaust air temperature (in K),
- g = gravitational acceleration, g = 9.81 m/s^{2}
- H_{0} = height above sea level,
- R_{L} = gas constant of the air, RL = 0.287 kJ/kg·K
- c_{pL} = specific heat capacity of the air, c_{pL} = 1.298 kJ/m^{3}·K
- Q_{L} = total quantity of heat exhausted by ventilation: Q_{L} = P_{V} + ΣQ,
- P_{V} = device power loss,
- ΣQ = heat exchange with the environment.
Example //
At given incoming air and exhaust air temperature, the power dissipation P_{v} should be exhausted by natural ventilation. The volume of air required should be calculated:
- T_{2} = 40°C = 313 K,
- T_{1} = 30°C = 303 K,
- P_{V} = 30kW = 30 kJ/s,
- Height above sea level = 500 m
If the warm air is exhausted directly over the heat source, this will increase the effective temperature difference ∆θ to the difference between the temperature of the outside air and the equipment exhaust air temperature. This will allow the required volume of cooling air to be reduced.
Calculation of the resistances in the air duct and the ventilation cross section: Based on the example in Figure 2a, the following applies:
for incoming air // | Acceleration | 1 |
Screen | 0.75 | |
Widening in cross section | 0.55 | |
Gradual change of direction | 0.6 | |
R1 = | 2.9 | |
for exhaust air // | Acceleration | 1 |
Right-angle bend | 1.5 | |
Slats | 3 | |
R2 = | 5.5 |
If the exhaust air duct is 10 % larger than the incoming air duct, then:
m = A1/A2 = 1/1.1 = 0.91 and m^{2 }= 0.83
then R = 2.9 + 0.83·5.5 = 7.5
The ventilation ratios can be calculated with the formula:
Numerical value equation with ∆υ in K, H in m, P_{V} in kW and A_{1} in m^{2}.
Example #2
- Transformer losses P_{V} = 10kW,
- ∆θ = 12K,
- R = 7.5 and
- H = 6m yield:
- A1 ≈ 1 m^{2}.
Practical experience has shown that the ventilation cross sections can be reduced if the transformer is not continuously operated at full load, the compartment is on the north side or there are other suitable intervals for cooling. A small part of the heat is also dissipated through the walls of the compartment.
The accurate calculation can be done as per DIN 4701.
Fans for switchgear and transformer rooms
Ventilation fans, in addition to their capacity, must compensate for the pressure losses in the air path and provide blow-out or dynamic pressure for the cooling air flow. This static and dynamic pressure can be applied with ∆p ≈ 0.2…0.4 mbar.
Then the propulsion power of the fan is:
Example #3
For the cooling air requirement of the transformer in the example above, where //
- P_{v} = 30 kW, with
- V = 2.4 m^{3}/s,
- η = 0.2,
- ∆p = 0.35 mbar = 35 Ws/m^{3}
the fan capacity is calculated as:
Resistances in the ventilation ducts and supplementary system components, such as dust filters, must be considered separately in consultation with the supplier. For sufficient air circulation, a minimum clearance between the equipment and the wall is required, depending on the heat output. For auxiliary transformers, this is about 0.4 m, for power transformers about 1 m.
Reference // Switchgear Manual by ABB (Order yourself a copy)
Euan Taylor
When I try the example number 1, Using the V1 formula I’m unable to achieve the answer of 2.4m3/s using the values provided, the only way for me to make it work is to convert the 0.287 kJ/kg.K to 287 J/kg.K this way I end up with a result of 2.4m3/s.
Is this correct?
Using the basic V0 formula I get 2.31 m3/s.
Thanks.
EHSAN
in the name of god
dear csanyi
thank you foe information. I need to accurate information for dimensions & calculation SKIRTING AND CEILING in the ventilation transformer room.
Manolis
I think there is an inconsistency in using H and H0.
At the height correction formula H is used (height difference between inlet and outlet) but it is explained as H0. In the example if we use H0 the result for V1 is very small.
James Bergman
The only exhaust systems I have seen in switch rooms have been done with a false floor. It means that the switch room ceilings have to be higher, or the switch boxes shorter, but it seems like a really efficient cooling method to me. Mostly for the switch boxes in the middle.
DSRMURTHY
Nice article.
Freddy
Edvard, your article comments on the need to consider criteria for ventilation in switchgear, which is right.
Taking advantage photograph switchgear GIS which starts your article , I think that in the case of GIS also be considered the criterion for the extraction of SF6 (gas ) in case of leaks , I understand the SF6 is heavier than air and has the tendency to concentrate near the floor , in this case extraction equipment should also be provided
Freddy Quijaite
fquijaite@gmail.com; freddy.quijaite@applus.com