4.4.

2 Ventilation of switchgear and transformer rooms

Design criteria for room ventilation

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

4
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. If HVAC (air-conditioning) installations and air ducts
are installed, the required space and the configuration must be included at an early
stage of planning.
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.
Figs. 4-27 and 4-28 show frequently used examples of room ventilation.

Fig. 4-27
Compartment ventilation: a) Simple compartment ventilation, b) compartment
ventilation with exhaust hood above the switchboard, c) ventilation with false floor,
d) ventilation with recirculating cooling system

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Fig. 4-28
Cross section through transformer cells:
a) incoming air is channelled over ground, exhaust air is extracted through a chimney.
b) as in a), but without chimney. c) incoming air is channelled below ground, exhaust
air is removed through an opening in the wall of the transformer compartment.
d) transformer compartment with fan. 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 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.
In ventilated and air-conditioned compartments occupied by personnel for extended
periods the quality regulations for room air specified by DIN 1946 must be observed.
The resistance of the air path is generally:
R = R1 + m2 R2.
Here: 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. Fig. 4-28 shows
common configurations.
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:
acceleration 1 slow change of direction 0…0.6
right-angle bend 1.5 wire screen 0.5…1
rounded bend 1 slats 2.5…3.5
a bend of 135 ° 0.6 cross section widening 0.25…0.91)
1) 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.

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Calculation of the quantity of cooling air:
· QL
V0 = ———— ; ∆ϑ = T2 – T1
cpL · ∆ϑ
With temperature and height correction1) the following applies for the incoming air flow:
g·H
· · T1 g·H
– R———
V1 = V0 · — · e —— L · T0
T0 RL · T0
V0 = standard air volume flow at sea level, p0 = 1013 mbar, T0 = 273 K = 0 °C,

4
T1 = cooling air temperature (in K),
T2 = exhaust air temperature (in K),
m
g = gravitational acceleration, g = 9.81 —2 ,
s
H0 = height above sea level,
kJ
RL = gas constant of the air, RL = 0.287 ——— ,
kg · K
kJ
cpL = specific heat capacity of the air, cpL = 1.298 ———— ,
m3 · K
QL = total quantity of heat exhausted by ventilation: QL = PV + ΣQ,
PV = device power loss,
ΣQ = heat exchange with the environment.
1)
May be neglected at up to medium installation height and in moderate climates

At high power dissipation and high temperatures, solar radiation and thermal
conduction through the walls can be neglected. Then QL = PV.
Example:
At given incoming air and exhaust air temperature, the power dissipation PV should be
exhausted by natural ventilation. The volume of air required should be calculated:
T2 = 40 °C = 313 K, T1 = 30°C = 303 K, PV = 30 kW = 30 kJ/s, height above sea
level = 500 m
g·H
· PV T1 – ———
g·H m3 m3
L · T0
V1 = —————— · — · e R——— = 2,4 — = 8640 —
cpL (T2 – T1) T0 RL · T0 s h

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 Fig. 4-28a, 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

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If the exhaust air duct is 10 % larger than the incoming air duct, then
A 1
m = —–1 = —– = 0.91 and m2 = 0.83,
A2 1.1
then R = 2.9 + 0.83 · 5.5 = 7.5.
The ventilation ratios can be calculated with the formula
P2
(∆ ϑ)3 · H = 13.2 —2V– (R1 + m2 R2).
A1
numerical value equation with ∆ ϑ in K, H in m, PV in kW and A1 in m2.
Example:
transformer losses PV = 10 kW, ∆ ϑ = 12 K, R = 7.5 and H = 6 m yield:
A1 ≈ 1 m2.
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. For the design of transformer substations and fire-prevention
measures, see Section 4.7.5 to 4.7.6.

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:
.
V · ∆p
PL = ——–—, η = efficiency
η
Example:
For the cooling air. requirement of the transformer in the example above, where
Pv = 30 kW, with V = 2.4 m3/s, η = 0.2, ∆p = 0.35 mbar = 35 Ws/m3
the fan capacity is calculated as:
2.4 · 0.35
PL = ————— = 0.42 kW.
0.2
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.

4.4.3 Forced ventilation and air-conditioning of switchgear installations

Overview and selection

When planning switchgear installations, thermal loads resulting from heat dissipation
from the installation and environmental conditions (local climate) must be taken into
account. This is generally done by:

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