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What is the solar gain heat inside the substation? The room size is 30 ft in length, 20 ft in width, and 14 ft in height, the window size is 54" x 54" and one door size is 72” by 96” (6ftx8ft). The total number of windows is 2. Environment condition: The location is Bangladesh, with high temperatures and humidity, using weather data from Dhaka city. Now calculate the total solar gain heat in kW.

How to calculate the hot and humid conditions of weather substation ventilation in industrial areas? the following condition will consider:

-Substation volume(V): 30’x20’x14’

-Transformer capacity: 1250 KVA, where heat dissipated 5% of capacity?

-HT Panel heat dissipated rate is 2% of capacity total capacity

-HT Panel heat dissipated rate is 1.5% of capacity total capacity

-PFI Panel heat dissipated rate is 1.0% of capacity total capacity

-Window size is 60" x 60" and total windows is 2 [solar gain area]

-Main door size is 72" x 96" [most of the time is close]

-Air Changing per Hour (ACH) is 25 [as per ASHRAE 62.1]

-Ambient Temperature: 40 deg C

-Sub-station insider Room temperature required: 30 deg C?

-Static Pressure of the Ventilation Fan consider is 200 Pa

(1) Calculate the total heat load inside the sub-station room?

(2) Calculate the total solar gain load inside the substation room?

(3) Calculate the Total CFM required for the substation room?

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We'll calculate the heat load from each component as follows:

Transformer Heat Load = 1250 kVA * 0.05 = 62.5 kW

HT Panel Heat Load = 1250 kVA * 0.02 = 25 kW

LT Panel Heat Load = 1250 kVA * 0.015 = 18.75 kW

PFI Panel Heat Load = 1250 kVA * 0.01 = 12.5 kW

Summing these up gives:

Total Heat Load = 62.5 kW + 25 kW + 18.75 kW + 12.5 kW = 118.75 kW

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First, convert window and door dimensions from inches to meters:

Window area = 2 * (60 in * 60 in * (0.0254 m/in)^2) = 5.57 m2

Door area = 72 in * 96 in * (0.0254 m/in)^2 = 4.27 m2

Assuming worst-case conditions (SHGC = 0.86, I = 1000 W/m2), calculate the solar gain from the windows and door:

Window solar gain = 5.57 m2 * 0.86 * 1000 W/m2 = 4792 W = 4.792 kW

Door solar gain (assuming the door is often open, there will be some solar gain) = 4.27 m2 * 0.86 * 1000 W/m2 = 3672 W = 3.672 kW

Total Solar Gain Load = Window solar gain + Door solar gain = 4.792 kW + 3.672 kW = 8.464 kW

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To maintain the desired operating temperature, the ventilation system must remove the total heat load (including solar gain) from the room. First, convert the total heat load to BTU/hr:

Total Heat Load = (118.75 kW + 8.464 kW) * 3412 BTU/hr/kW = 434,132 BTU/hr

Then, using the formula for CFM:

CFM = Total Heat Load / (1.08 * Temp_difference)

where Temp_difference = (Ambient temp - Required room temp) in °F = (104°F - 86°F) = 18°F

CFM = 434,132 BTU/hr / (1.08 * 18°F) = 22,207 CFM

Considering the desired ACH:

Volume = 30 ft * 20 ft * 14 ft = 8,400 ft3

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The greater of these two values should be used to ensure adequate ventilation. Therefore, you should design for approximately 22,207 CFM to adequately cool the substation.

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[The actual CFM that each fan can deliver will depend on the fan's performance curve at the given static]