Green Energy - House Construction

With Minimum Meterological data, how i can build model for Green Energy new construction

WIth Minimum Meterological data, how i can build model for Green Energy new 

With minimal meteorological data, you can still build a basic model for green energy construction using these steps:

1. Identify Key Data: Focus on crucial weather parameters:

   • Solar: Solar radiation or sunlight hours (for solar panels).
   • Wind: Wind speed and direction (for wind turbines).
   • Temperature: Helps estimate energy efficiency.
   • Precipitation: Affects solar efficiency or rainwater harvesting.

2. Collect Data: Use historical or open-source meteorological data from sources like:

   • NASA POWER dataset
   • Government weather portals
   • Local weather stations

3. Simplify Energy Models:

   • Solar Energy: Use models like PVWatts to estimate solar power output based on sunlight.
   • Wind Energy: Apply the Power Law to estimate wind turbine energy output with basic wind speed data.

4. Site-Specific Adjustments:

   • Use elevation, terrain, and shading data from tools like Google Earth or GIS tools.

5. Build Predictive Model:

   • Use simple algorithms like Linear Regression to predict energy output.
   • For software, use Python with libraries like Pandas, Scikit-learn, and NumPy.

6. Validate & Optimize: Compare model results with small-scale installations for accuracy.

This approach works as a basic starting point.

You said:

i have min, max tempearture in a particular place. How i can build model. Explain in step by step with use case for a typical class room for 60 students

ChatGPT

with minimum and maximum temperature data, you can build a model to estimate energy needs for maintaining a comfortable classroom for 60 students. Here's a step-by-step approach:

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Objective:

Model the cooling or heating energy required to maintain a comfortable temperature (e.g., 24°C) in a classroom with 60 students.

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Step 1: Define the Use Case

• Classroom dimensions: Assume typical size, e.g., 10m x 8m x 3m = 240 cubic meters.
• Occupancy: 60 students.
• Comfort Temperature: 24°C (standard indoor comfort).
• Given Data: Minimum and maximum temperatures for the location.

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Step 2: Estimate Heat Load

The heat load includes factors that increase room temperature:

1. External Temperature Influence:

   • Use the average temperature: $$T_{\text{avg}} = \frac{T_{\text{min}} + T_{\text{max}}}{2}$$

2. Heat Gain from Occupants:

   • Each person generates around 100W of heat.
     For 60 students: $$Q_{\text{occupants}} = 60 \times 100 = 6000 \, \text{W} \, (6 \, \text{kW})$$

3. External Heat Transfer (Walls and Windows):

   • Simplify using the U-value (thermal transmittance) of walls/windows.
   • Heat transfer: $$Q_{\text{walls}} = U \times A \times \Delta T$$
     • U: Thermal transmittance (e.g., 1.5 W/m²K for walls).
     • A: Area (walls, windows, roof).
     • ΔT: Difference between outside and desired indoor temperature.

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Step 3: Total Heat Load Calculation

Add all sources of heat load:

$$Q_{\text{total}} = Q_{\text{occupants}} + Q_{\text{walls}}$$

This gives the energy required to cool or heat the classroom.

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Step 4: Model the Energy Requirement

• If average temperature $T_{\text{avg}}$ > 24°C → Cooling needed.
• If $T_{\text{avg}}$ < 24°C → Heating needed.

Energy Consumption can be calculated using:

$$\text{Energy (kWh)} = \frac{Q_{\text{total}} \times \text{Operating Hours}}{1000}$$

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Step 5: Use Case with Assumptions

Example Scenario:

• Min Temp = 28°C, Max Temp = 36°C → $T_{\text{avg}} = 32°C$ (Cooling needed).
• Classroom: 10m x 8m x 3m.
• Occupants: 60 students.
• Comfort Temp: 24°C.
• Operating hours: 6 hours/day.
• U-value = 1.5 W/m²K for walls.
• Total surface area (walls + roof): 150m².

1. Heat from students:

$$Q_{\text{occupants}} = 6000 \, \text{W}$$

2. Heat transfer through walls:

$$Q_{\text{walls}} = U \times A \times \Delta T = 1.5 \times 150 \times (32 - 24) = 1800 \, \text{W}$$

3. Total Heat Load:

$$Q_{\text{total}} = 6000 + 1800 = 7800 \, \text{W} \, (7.8 \, \text{kW})$$

4. Energy Consumption:
For 6 hours:

$$\text{Energy} = \frac{7800 \times 6}{1000} = 46.8 \, \text{kWh/day}$$

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Step 6: Build the Model in Python

Use Python to automate this calculation. Here’s a simple script:

# Inputs


# Inputs
T_min = 28  # Minimum temperature
T_max = 36  # Maximum temperature
T_comfort = 24  # Desired temperature
U_value = 1.5  # Thermal transmittance (W/m²K)
area = 150  # Surface area of walls+roof (m²)
occupants = 60
heat_per_person = 100  # Heat generated per person (W)
hours = 6  # Operating hours
# Average temperature
T_avg = (T_min + T_max) / 2
# Heat load calculations
delta_T = T_avg - T_comfort
Q_occupants = occupants * heat_per_person  # Heat from students
Q_walls = U_value * area * delta_T  # Heat transfer through walls
Q_total = Q_occupants + Q_walls  # Total heat load
# Energy consumption
energy_kWh = (Q_total * hours) / 1000
# Results
print(f"Total Heat Load: {Q_total:.2f} W")
print(f"Daily Energy Consumption: {energy_kWh:.2f} kWh")

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Results

Total Heat Load: 7800.00 W

Daily Energy Consumption: 46.80 kWh

Step 7: Insights

• You can use this model to size AC systems or heating units.
• Compare energy requirements for different locations (changing temperatures).
• Use this to assess energy costs or design energy-efficient classrooms.

Let me know if you need further tweaks, bro! 🚀

How much AC i have to provide. Please advise for some use cases in hot climate regions like Ramnad in Tamilnadu

ChatGPT said:

In hot climate regions like Ramanathapuram (Ramnad) in Tamil Nadu, where temperatures often exceed 35-40°C, you need to size the AC units properly. Here's a step-by-step guide with use cases for a typical classroom:

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Step 1: Cooling Load Estimation

Cooling load is the amount of heat that needs to be removed to maintain a comfortable 24°C.

1. Key Heat Sources:

   • External heat gain: Through walls, roof, windows.
   • Internal heat gain: Students, lights, and other equipment.

2. Simplified Formula:
   For a classroom, use 1 ton of AC for every 12,000 BTU/hr cooling load.
   1 Ton = 3.517 kW (energy).

   Cooling load can be calculated as:

   $$\text{Total Cooling Load (BTU/hr)} = Q_{\text{walls}} + Q_{\text{occupants}} + Q_{\text{lights}}$$

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Step 2: Typical Classroom in Ramnad

Let’s assume:

• Classroom Size = 10m x 8m x 3m (standard dimensions).
• 60 students (100W per student).
• Lights: Assume 10 lights at 50W each.
• High external temperatures: Min = 30°C, Max = 40°C.

Comfort Temp = 24°C.

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Step 3: Cooling Load Calculation

1. Heat Gain from Students:

   $$Q_{\text{occupants}} = 60 \times 100 = 6000 \, \text{W} \, (6 kW)$$

2. Heat Gain from Lights:

   $$Q_{\text{lights}} = 10 \times 50 = 500 \, \text{W} \, (0.5 kW)$$

3. External Heat Gain (Walls + Roof):
   Assume:

   • Total Area (walls + roof) = 150m².
   • U-value = 1.5 W/m²K (standard wall insulation).
   • Temperature Difference (ΔT): Max Temp - Comfort Temp = $40 - 24 = 16°C$.
   $$Q_{\text{walls}} = U \times A \times \Delta T = 1.5 \times 150 \times 16 = 3600 \, \text{W} \, (3.6 kW)$$

4. Total Cooling Load:

   $$Q_{\text{total}} = Q_{\text{occupants}} + Q_{\text{lights}} + Q_{\text{walls}}$$ $$Q_{\text{total}} = 6000 + 500 + 3600 = 10,100 \, \text{W} \, (10.1 \, \text{kW})$$

Convert to BTU/hr:

$$\text{BTU/hr} = 10.1 \, \text{kW} \times 3412 = 34,500 \, \text{BTU/hr}$$

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Step 4: AC Sizing

1 Ton of AC = 12,000 BTU/hr.
For 34,500 BTU/hr:

$$\text{Required AC Capacity} = \frac{34,500}{12,000} \approx 3 \, \text{Tons}$$

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Step 5: Use Cases for Hot Climate

Use Case 1: Basic Classroom in Ramnad

• AC Capacity: 3 Tons (Split AC or Cassette AC).
• Configuration: Use 2 x 1.5 Ton ACs for better air distribution.

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Use Case 2: Large Classroom or Higher Load

• If room size increases to 12m x 10m x 3m or more students (e.g., 80 students):
  • Cooling load increases to ~4-5 Tons.
  • Use 2 x 2 Ton ACs or 1 centralized 5 Ton unit.

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Use Case 3: Energy-Efficient Classroom

To reduce AC load:

• Use insulation on walls/roof.
• Install cool roofing to reduce heat gain.
• Use blinds or reflective glass for windows.
• Opt for ceiling fans + AC to reduce AC usage.

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Final Recommendation for Ramnad

For a 60-student classroom:

• Install 3 Tons of AC (2 x 1.5 Ton units).
• Choose Inverter ACs for energy efficiency.

This ensures comfort even in peak summer temperatures, bro! 🌞 Let me know if you need cost estimates or further tweaks. 🚀