Let's solve this step-by-step. --- # Step 1: Understand the problem We need to calculate the **total heat load** for the office, considering: - **Heat gains** from people, lights, and equipment. - **Heat transfer** through the wall and window, including conduction, convection, and solar heat gain. --- # Step 2: Gather given data | Parameter | Value | | --- | --- | | Office size | 3 m x 4 m | | Wall area | 10 m² | | Latitude | 26° S | | U-factor (wall) | .348 W/m²·K | | Wall area | 10 m² | | Inside temperature | 25°C | | Outdoor temperature | 32°C | | Max outdoor temperature | 32°C | | Average outdoor temperature | 28°C | | Glass area | 2 m² | | U-factor (glass) | 6 W/m²·K | | SHGF | 350 W/m² | | Number of people | 2 | | Heat generated per person | 150 W | | Lights | 20 W/m² | | Office equipment | 15 W/m² | | Shading coefficient (SC) | .65 | | Cooling load factor (CLF) | .83 | --- # Step 3: Calculate internal heat gains ### 3.1 Heat from people \[ Q_{people} = 2 \times 150\,W = 300\,W \] ### 3.2 Heat from lighting \[ Q_{lights} = 20\,W/m^2 \times \text{area} \] \[ = 20\,W/m^2 \times (3\,m \times 4\,m) = 20 \times 12 = 240\,W \] ### 3.3 Heat from office equipment \[ Q_{equip} = 15\,W/m^2 \times 12\,m^2 = 15 \times 12 = 180\,W \] --- # Step 4: Calculate heat transfer through the wall ### 4.1 Temperature difference \[ \Delta T_{wall} = T_{inside} - T_{outside} = 25°C - 32°C = -7°C \] (Heat flow will be outward because outside is hotter) ### 4.2 Heat conduction through wall \[ Q_{wall} = U_{wall} \times A_{wall} \times |\Delta T_{wall}| \] \[ = .348\,W/m^2·K \times 10\,m^2 \times 7\,K = .348 \times 10 \times 7 = 24.36\,W \] (Note: Since the temperature difference is negative, heat flows outward, so this is heat gain/loss from the room. For cooling load, we consider heat entering the room, which is zero here because heat flows outward.) --- ### 4.3 Solar heat gain through the window \[ Q_{solar} = SHGF \times A_{window} \times \text{SC} \] \[ = 350\,W/m^2 \times 2\,m^2 \times .65 = 350 \times 2 \times .65 = 455\,W \] --- # Step 5: Calculate heat transfer through the glass ### 5.1 Temperature difference across glass \[ \Delta T_{glass} = T_{inside} - T_{outside} = -7°C \] ### 5.2 Heat transfer via glass (conduction) \[ Q_{glass} = U_{glass} \times A_{glass} \times |\Delta T_{glass}| \] \[ = 6\,W/m^2·K \times 2\,m^2 \times 7\,K = 6 \times 2 \times 7 = 84\,W \] --- # Step 6: Adjust for shading and cooling load factors Since the heat gain through the glass is affected by shading coefficient: \[ Q_{glass, adjusted} = Q_{glass} \times SC = 84 \times .65 = 54.6\,W \] --- # Step 7: Sum all heat gains | Heat source | Power (W) | | --- | --- | | People | 300 | | Lights | 240 | | Equipment | 180 | | Solar gain through glass (adjusted) | 54.6 | | Heat gain through wall | (outward flow, so no gain) | | Solar heat gain (SHGF) | 455 | **Total internal + solar heat gains:** \[ Q_{total} = 300 + 240 + 180 + 54.6 + 455 = 1,229.6\,W \] --- # **Final Step: Apply the cooling load factor** The total heat load considering the cooling load factor: \[ Q_{cooling} = Q_{total} \times CLF = 1,229.6 \times .83 \approx 1,020\,W \] --- # **Final Answer:** \[ \boxed{ \text{Total heat load} \approx \mathbf{102\,W} } \] This is the approximate cooling power required to maintain the indoor conditions considering all internal and external heat gains and losses. --- Let me know if you'd like further clarification!