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SWITCHROOM HEAT LOAD CALCULATION

This report presents a comprehensive analysis of the heat load for a switchroom or control building, determining the required HVAC cooling capacity based on equipment, building envelope, solar gains, occupancy, and infiltration.

Room Geometry

Given floor area; used for floor conduction, lighting, room volume and the default roof area

Formula: $$ A_{floor} = A_{floor,given} $$
Result: $$ = 77.00 \; \text{m}^2 $$

Total perimeter wall area

Formula: $$ A_{wall} = 2 \times (L + W) \times H $$
Substitution: $$ = 2 \times (11.00 + 7.00) \times 3.00 $$
Result: $$ = 108.00 \; \text{m}^2 $$

Internal room volume

Formula: $$ V_{room} = A_{floor} \times H $$
Substitution: $$ = 77.00 \times 3.00 $$
Result: $$ = 231.00 \; \text{m}^3 $$

Temperature Difference

Outside design temperature 35.0 °C, inside target 22.0 °C

Formula: $$ \Delta T = T_{outside} - T_{inside} $$
Substitution: $$ = 35.0 - 22.0 $$
Result: $$ = 13.0 \; \text{K} $$

Humidity Ratio Calculation

Tetens' equation for saturation vapor pressure at outside temperature

Formula: $$ P_{sat} = 611.2 \times e^{\frac{17.67 \times T}{T + 243.5}} $$
Substitution: $$ = 611.2 \times e^{\frac{17.67 \times 35.0}{35.0 + 243.5}} $$
Result: $$ = 5631.2 \; \text{Pa} $$

Where P_vapor = P_sat x RH/100 = 5631.2 x 60/100 = 3378.7 Pa

Formula: $$ W_{out} = \frac{0.622 \times P_{vapor}}{P_{atm} - P_{vapor}} $$
Substitution: $$ = \frac{0.622 \times 3378.7}{101325 - 3378.7} $$
Result: $$ = 0.021456 \; \text{kg/kg} $$

Tetens' equation at inside design temperature

Formula: $$ P_{sat} = 611.2 \times e^{\frac{17.67 \times T}{T + 243.5}} $$
Substitution: $$ = 611.2 \times e^{\frac{17.67 \times 22.0}{22.0 + 243.5}} $$
Result: $$ = 2642.8 \; \text{Pa} $$

Where P_vapor = P_sat x RH/100 = 2642.8 x 50/100 = 1321.4 Pa

Formula: $$ W_{in} = \frac{0.622 \times P_{vapor}}{P_{atm} - P_{vapor}} $$
Substitution: $$ = \frac{0.622 \times 1321.4}{101325 - 1321.4} $$
Result: $$ = 0.008219 \; \text{kg/kg} $$

Difference in moisture content used for latent infiltration load

Formula: $$ \Delta W = W_{out} - W_{in} $$
Substitution: $$ = 0.021456 - 0.008219 $$
Result: $$ = 0.013237 \; \text{kg/kg} $$

Equipment Heat Load

Equipment heat dissipation based on nameplate ratings and load factors: * VSD: 132.0 kW x 1 x 0.03 = 3.96 kW

Equipment Heat Load Total

Sum of 1 equipment item(s)

Formula: $$ Q_{equip} = \sum (P_{rated} \times n \times LF) $$
Result: $$ = 3.96 \; \text{kW} $$

Conduction Heat Transfer

U = 0.60 W/m²K, A = 108.0 m²

Formula: $$ Q = \frac{U \times A \times \Delta T}{1000} $$
Substitution: $$ = \frac{0.60 \times 108.0 \times 13.0}{1000} $$
Result: $$ = 0.842 \; \text{kW} $$

U = 0.40 W/m²K, A = 77.0 m²

Formula: $$ Q = \frac{U \times A \times \Delta T}{1000} $$
Substitution: $$ = \frac{0.40 \times 77.0 \times 13.0}{1000} $$
Result: $$ = 0.400 \; \text{kW} $$

U = 0.30 W/m²K, A = 77.0 m²

Formula: $$ Q = \frac{U \times A \times \Delta T}{1000} $$
Substitution: $$ = \frac{0.30 \times 77.0 \times 13.0}{1000} $$
Result: $$ = 0.300 \; \text{kW} $$

Sum of 3 surface(s)

Formula: $$ Q_{cond} = \sum (U \times A \times \Delta T) / 1000 $$
Result: $$ = 1.543 \; \text{kW} $$

Solar Heat Gain

Sun-exposed wall area = 40.0 m², factor = 150 W/m²

Formula: $$ Q_{wall} = \frac{A_{wall} \times F_{wall}}{1000} $$
Substitution: $$ = \frac{40.0 \times 150}{1000} $$
Result: $$ = 6.00 \; \text{kW} $$

Sun-exposed roof area = 77.0 m², factor = 300 W/m²

Formula: $$ Q_{roof} = \frac{A_{roof} \times F_{roof}}{1000} $$
Substitution: $$ = \frac{77.0 \times 300}{1000} $$
Result: $$ = 23.10 \; \text{kW} $$

Solar radiation through exposed surfaces and windows

Formula: $$ Q_{solar} = Q_{wall} + Q_{roof} + Q_{window} $$
Substitution: $$ = 6.00 + 23.10 + 0.00 $$
Result: $$ = 29.10 \; \text{kW} $$

Lighting Heat Load

Lighting density = 10.0 W/m²

Formula: $$ Q_{lighting} = \frac{A_{floor} \times D_{lighting}}{1000} $$
Substitution: $$ = \frac{77.00 \times 10.0}{1000} $$
Result: $$ = 0.770 \; \text{kW} $$

Personnel Heat Load

ASHRAE standard: 75 W sensible heat per person

Formula: $$ Q_{pers,sensible} = n \times 0.075 $$
Substitution: $$ = 1 \times 0.075 $$
Result: $$ = 0.075 \; \text{kW} $$

ASHRAE standard: 55 W latent heat per person

Formula: $$ Q_{pers,latent} = n \times 0.055 $$
Substitution: $$ = 1 \times 0.055 $$
Result: $$ = 0.055 \; \text{kW} $$

Infiltration Heat Load

Air changes per hour = 0.50

Formula: $$ \dot{V} = \frac{V_{room} \times ACH}{3600} $$
Substitution: $$ = \frac{231.00 \times 0.50}{3600} $$
Result: $$ = 0.0321 \; \text{m}^3/\text{s} $$

Coefficient 1.23 = air density x specific heat (kW per m³/s per K)

Formula: $$ Q_{infil,s} = \dot{V} \times \Delta T \times 1.23 $$
Substitution: $$ = 0.0321 \times 13.0 \times 1.23 $$
Result: $$ = 0.513 \; \text{kW} $$

Coefficient 3010 = air density x latent heat of vaporization

Formula: $$ Q_{infil,l} = \dot{V} \times \Delta W \times 3010 $$
Substitution: $$ = 0.0321 \times 0.013237 \times 3010 $$
Result: $$ = 1.278 \; \text{kW} $$

Total Heat Load

Sum of all sensible heat sources

Formula: $$ Q_{sensible} = Q_{equip} + Q_{cond} + Q_{solar} + Q_{lighting} + Q_{pers,s} + Q_{infil,s} $$
Substitution: $$ = 3.96 + 1.54 + 29.10 + 0.77 + 0.07 + 0.51 $$
Result: $$ = 35.96 \; \text{kW} $$

Sum of all latent heat sources

Formula: $$ Q_{latent} = Q_{pers,l} + Q_{infil,l} $$
Substitution: $$ = 0.06 + 1.28 $$
Result: $$ = 1.33 \; \text{kW} $$

Total cooling load to be removed

Formula: $$ Q_{total} = Q_{sensible} + Q_{latent} $$
Substitution: $$ = 35.96 + 1.33 $$
Result: $$ = 37.29 \; \text{kW} $$

Ratio of sensible to total load; higher SHR means less dehumidification needed

Formula: $$ SHR = \frac{Q_{sensible}}{Q_{total}} \times 100 $$
Substitution: $$ = \frac{35.96}{37.29} \times 100 $$
Result: $$ = 96.4 \% $$

Required Cooling Capacity

Safety factor of 15% applied for uncertainties and future growth

Formula: $$ Q_{required} = \max\left(0,\; Q_{total} \times \left(1 + \frac{SF}{100}\right)\right) $$
Substitution: $$ = \max\left(0,\; 37.29 \times \left(1 + \frac{15}{100}\right)\right) $$
Result: $$ = 42.89 \; \text{kW} $$

Air Conditioning System Sizing

Rounded up to next whole unit; each unit rated at 10.00 kW

Formula: $$ N_{active} = \left\lceil \frac{Q_{required}}{Q_{unit}} \right\rceil $$
Substitution: $$ = \left\lceil \frac{42.89}{10.00} \right\rceil $$
Result: $$ = 5 \; \text{units} $$

0 redundant unit(s)

Formula: $$ N_{total} = N_{active} + N_{redundant} $$
Substitution: $$ = 5 + 0 $$
Result: $$ = 5 \; \text{units} $$

Install 5 air conditioning unit(s) with total installed capacity of 50.00 kW

Formula: $$ Q_{installed} = N_{total} \times Q_{unit} $$
Substitution: $$ = 5 \times 10.00 $$
Result: $$ = 50.00 \; \text{kW} $$

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