Heat load calculation

   

To determine the total cooling capacity required for a building in HVAC design, you need to perform a comprehensive heat load calculation, considering both sensible and latent heat loads. Here’s a detailed step-by-step process with formulas and a complete example.

1. Sensible and Latent Heat Load Calculation

Sensible heat affects the temperature of a space, while latent heat affects the humidity.

A. Sensible Heat Load Calculation

1. Heat Gain through Building Envelope

Walls, Roofs, and Floors

Formula:

$$Q=U\times A\times \mathrm{\Delta }T$$

• Q = Sensible heat gain/loss (BTU/hr or W)
• U = Heat transfer coefficient (BTU/hr·ft²·°F or W/m²·K)
• A = Area of the surface (ft² or m²)
• ΔT = Temperature difference (°F or K)

Example:

• Wall area: 1,000 ft²
• U-value: 0.35 BTU/hr·ft²·°F
• Indoor temperature: 75°F
• Outdoor temperature: 95°F
• ΔT = 95 - 75 = 20°F

$$Q=0.35\times 1,000\times 20=7,000\text{ BTU/hr}$$

2. Heat Gain through Windows

Formula:

$$Q=SHGC\times A\times I$$

• SHGC = Solar Heat Gain Coefficient
• A = Area of the window (ft² or m²)
• I = Solar radiation intensity (W/m²)

Example:

• Window area: 200 ft²
• SHGC: 0.3
• Solar intensity: 400 W/m²

$$Q=0.3\times 200\times 400\times 3.414=8,500\text{ BTU/hr}$$

3. Internal Heat Gains

a. People

Formula:

$$Q=\text{Number of occupants}\times \text{Heat per person}$$

Example:

• Number of occupants: 20
• Heat per person: 400 BTU/hr

$$Q=20\times 400=8,000\text{ BTU/hr}$$

b. Lighting

Formula:

$$Q=\text{Wattage of lights}\times \text{Heat gain factor}$$

(Heat gain factor is typically 3.414 BTU/hr per watt)

Example:

• Lighting wattage: 1,000 watts

$$Q=1,000\times 3.414=3,414\text{ BTU/hr}$$

c. Equipment

Formula:

$$Q=\text{Heat output of equipment}$$

Example:

• Equipment heat output: 2,500 BTU/hr

$$Q=2,500\text{ BTU/hr}$$

B. Latent Heat Load Calculation

1. Latent Heat Gain from People

Formula:

$$Q=\text{Number of occupants}\times \text{Moisture per person}\times \text{Latent heat factor}$$

(Latent heat factor is 0.68 BTU/hr per pint of moisture)

Example:

• Number of occupants: 20
• Moisture per person: 0.5 pints/hr

$$Q=20\times 0.5\times 0.68=6.8\text{ BTU/hr}$$

2. Latent Heat Gain from Equipment

Formula:

$$Q=\text{Moisture output of equipment}\times \text{Latent heat factor}$$

Example:

• Moisture output: 300 grams/hr
• Latent heat factor: 2260 kJ/kg = 970 BTU/kg

$$Q=0.3\times 970=291\text{ BTU/hr}$$

3. Latent Heat Gain from Cooking

Formula:

$$Q=\text{Moisture generated}\times \text{Latent heat factor}$$

Example:

• Moisture generated: 2 liters/hr
• Latent heat factor: 2260 kJ/kg (2260 BTU/kg)

$$Q=2\times 2260=4,520\text{ BTU/hr}$$

2. Total Cooling Load Calculation

Combine the sensible and latent heat loads to determine the total cooling load.

Example Calculation for a Building

Assumptions:

• Building Area: 10,000 ft²
• Wall U-value: 0.35 BTU/hr·ft²·°F
• Window Area: 800 ft²
• SHGC: 0.25
• Lighting Wattage: 2,000 watts
• Occupants: 50
• Internal Equipment: 10,000 BTU/hr

Sensible Heat Load:

1. Walls: $$Q=0.35\times 10,000\times 20=70,000\text{ BTU/hr}$$
2. Windows: $$Q=0.25\times 800\times 400\times 3.414=32,740\text{ BTU/hr}$$
3. People: $$Q=50\times 400=20,000\text{ BTU/hr}$$
4. Lighting: $$Q=2,000\times 3.414=6,828\text{ BTU/hr}$$
5. Equipment: $$Q=10,000\text{ BTU/hr}$$

Total Sensible Heat Load:

$$Q_{sensible}=70,000+32,740+20,000+6,828+10,000=139,568\text{ BTU/hr}$$

Latent Heat Load:

1. People: $$Q=50\times 0.5\times 0.68=17\text{ BTU/hr}$$
2. Equipment: $$Q=0.3\times 970=291\text{ BTU/hr}$$
3. Cooking: $$Q=2\times 2260=4,520\text{ BTU/hr}$$

Total Latent Heat Load:

$$Q_{latent}=17+291+4,520=4,828\text{ BTU/hr}$$

Total Cooling Load:

$$Q_{total}=Q_{sensible}+Q_{latent}=139,568+4,828=144,396\text{ BTU/hr}$$

Convert to Tons of Refrigeration (TR): 1 Ton of refrigeration = 12,000 BTU/hr

$$\text{TR}=\frac{144,396}{12,000}=12.03\text{ TR}$$