HVAC Assessment

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Disclaimer: The views and opinions expressed herein do not necessarily reflect those of the ITER Organization
HVAC Assessment

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results
• Results for the room concerning the ICL, just two
rooms may run above 25C:
– 14-B2-20
– 14-L1-21
Room
excess cooling
power (HVAC
+ LAC - heat
loads) at 35C
(kW)
excess
cooling
power (HVAC
+ LAC - heat
loads) at 25C
(kW)
14-B1-23 15.8 9.3
14-B1-22A 11.1 8.5
14-B1-22 26.6 22.9
14-B1-24 14.2 7.4
14-B2-20 9.9 -3.6
14-B2-21 13.6 11.4
14-L1-22 38.8 23.3
14-L1-21 -0.7 -4.2

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How I got the results
• Calculated how many kW the HVAC can remove based
on Air Handling Unit (AHU) specs and the following
assumption, :
– Air supply temp (C) to HVAC: 18C (as per spec)
– Humidity: 8.2 g water/kg dry air (as per spec)
– 1-Assumed outlet temperature of HVAC air toward exhaust
fan:35C thermal equilibrium with the room at 35C
– 2-Assumed outlet temperature of HVAC air toward exhaust
fan:25C thermal equilibrium with the room at 25C
– Calculated the delta enthalpy between inlet and outlet of each
room assuming constant humidity, using HVAC air flow from
P&IDs
– Results in last column of the table

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How I got the results
• Using several online tools for domestic application, I
have estimated which would be the required A/C power
to keep a certain temperature difference VS
environment with three different level of insulation:
– Temperature difference VS external environment of:
• 5C
• 10C
• 15C
– Insulation level:
• Good
• Normal
• Poor
• Results in the table

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How I got the results
• Assessed what normal, good and poor insulation mean
and how they compare with the rooms in the tritium
building:
– Domestic (0.25m wall thickness)
• 5C of delta T (assuming 40C outside and 35C inside the room)
– Good insulation: 0.15 W/(m*K) 300W every 100m2 of
exchange surface
– Normal insulation: 1 W/(m*K) 2000W every 100m2 of
exchange surface
– B14 (0.7m wall thickness)
• 5C of delta T (assuming 40C outside and 35C inside the room)
– Solid concrete: 1.8 W/(m*K) 1286W every 100m2
– Solid concrete: 1 W/(m*K)714 W every 100m2

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How I got the results
• Based on the thermal conductivity numbers and the
different wall thicknesses between a typical domestic
situation and the rooms in B14, we can assume that:
– We can consider around 60-65% of the estimated A/C
consumption by the online tools for the situation in B14 (e.g.):
• For domestic application, the expected cooling power required is
2.6kW to keep a 5C difference with the external environment
(assuming 40C outside 35C inside)
• We can consider around 65% of 2.6kW for a room inside
B141.7 kW

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Uncertainty
• The required cooling power, for each B14 interior room,
is over estimated since they do not share walls with the
exterior.
• This way of estimate heat load is pretty simple so has a
significant level of uncertainty