Overview
Calculates the total heat load within a switchroom or control building for proper HVAC system sizing. Determines both sensible and latent heat loads from equipment dissipation, building envelope conduction, solar radiation, lighting, personnel, and air infiltration. Supports metric (SI) and imperial unit systems. Based on ASHRAE Fundamentals methodology with Tetens' equation for psychrometric calculations.
Input Parameters
Unit System
| Parameter | Type | Default | Range | Description |
|---|---|---|---|---|
unit_system |
Enum: metric / imperial | metric | — | Select metric (SI) or imperial units; affects all dimensional inputs and outputs |
Equipment Heat Load
| Parameter | Type | Default | Range | Description |
|---|---|---|---|---|
equipment_list |
Array of objects | 1× VSD, 132 kW, LF 0.03 | — | List of electrical equipment generating heat. Each item has:
name (string),
rating (kW, min 0),
qty (integer, min 1),
load_factor (0–1) |
Room Dimensions
| Parameter | Type | Default | Range | Description |
|---|---|---|---|---|
room_length |
Number | 11.0 | min 0.1 | Internal room length (m or ft) |
room_width |
Number | 7.0 | min 0.1 | Internal room width (m or ft) |
room_height |
Number | 3.0 | min 0.1 | Internal room height (m or ft) |
Environmental Conditions
| Parameter | Type | Default | Range | Description |
|---|---|---|---|---|
temp_outside |
Number | 35.0 | — | Maximum expected ambient temperature (°C or °F) |
temp_inside |
Number | 22.0 | — | Target inside design temperature (°C or °F) |
humidity_outside |
Number | 60.0 | 0 – 100% | Outside relative humidity |
humidity_inside |
Number | 50.0 | 0 – 100% | Target inside relative humidity |
Building Envelope
| Parameter | Type | Default | Range | Description |
|---|---|---|---|---|
wall_area |
Number | 108.0 | min 0 | Total wall surface area (m² or ft²); auto-calculated from dimensions if zero |
wall_u_value |
Number | 0.6 | min 0 | Wall U-value (W/m²·K or BTU/hr·ft²·°F) |
roof_area |
Number | 77.0 | min 0 | Roof surface area (m² or ft²); auto-calculated if zero |
roof_u_value |
Number | 0.4 | min 0 | Roof U-value (W/m²·K or BTU/hr·ft²·°F) |
floor_area |
Number | 77.0 | min 0 | Floor area (m² or ft²); \(L \times W\) if zero. A given value is used for floor conduction, lighting, room volume and the default roof area |
floor_u_value |
Number | 0.3 | min 0 | Floor U-value (W/m²·K or BTU/hr·ft²·°F) |
Solar Exposure
| Parameter | Type | Default | Range | Description |
|---|---|---|---|---|
sun_exposed_wall_area |
Number | 40.0 | min 0 | Wall area receiving direct sunlight (m² or ft²) |
sun_exposed_roof_area |
Number | 77.0 | min 0 | Roof area receiving direct sunlight (m² or ft²) |
solar_wall_factor |
Number | 150.0 | min 0 | Solar heat gain factor for walls (W/m²) |
solar_roof_factor |
Number | 300.0 | min 0 | Solar heat gain factor for roof (W/m²) |
Windows
| Parameter | Type | Default | Range | Description |
|---|---|---|---|---|
window_list |
Array of objects | [] (empty) | — | List of windows. Each item has:
name (string),
area (m² or ft², min 0),
u_value (min 0) |
solar_window_factor |
Number | 250.0 | min 0 | Solar heat gain factor for windows (W/m²) |
Other Loads
| Parameter | Type | Default | Range | Description |
|---|---|---|---|---|
num_personnel |
Number | 1 | min 0 | Typical occupancy count |
lighting_density |
Number | 10.0 | min 0 | Lighting power density (W/m²) |
air_changes_per_hour |
Number | 0.5 | min 0 | Infiltration rate (air changes per hour) |
Cooling Requirements
| Parameter | Type | Default | Range | Description |
|---|---|---|---|---|
safety_factor |
Number | 15.0 | 0 – 100% | Additional capacity margin applied to total heat load |
redundant_units |
Number | 0 | min 0 | Number of backup AC units (e.g., 1 for N+1 redundancy) |
ac_unit_capacity |
Number | 10.0 | min 0.1 | Rated cooling capacity per AC unit (kW or BTU/hr) |
Output Results
| Output | Unit | Description |
|---|---|---|
room_geometry.floor_area_m2 |
m² | Floor area: the given floor_area, else \(L \times W\) |
room_geometry.wall_area_m2 |
m² | Calculated wall area \(2(L+W) \times H\) |
room_geometry.volume_m3 |
m³ | Room volume \(A_{floor} \times H\) |
environmental.temp_diff_c |
°C | Temperature difference \(T_{outside} - T_{inside}\) |
environmental.humidity_ratio_outside |
kg/kg | Outside air humidity ratio (from Tetens' equation) |
environmental.humidity_ratio_inside |
kg/kg | Inside air humidity ratio (from Tetens' equation) |
load_breakdown.equipment.total_kw |
kW | Total equipment heat dissipation |
load_breakdown.conduction.total_kw |
kW | Total conduction through building envelope |
load_breakdown.solar.total_kw |
kW | Total solar heat gain (walls + roof + windows) |
load_breakdown.lighting_kw |
kW | Lighting heat load |
load_breakdown.personnel.sensible_kw |
kW | Personnel sensible heat (75 W/person) |
load_breakdown.personnel.latent_kw |
kW | Personnel latent heat (55 W/person) |
load_breakdown.infiltration.sensible_kw |
kW | Infiltration sensible heat |
load_breakdown.infiltration.latent_kw |
kW | Infiltration latent heat |
total_heat_load.sensible |
kW | Total sensible heat load |
total_heat_load.latent |
kW | Total latent heat load |
total_heat_load.total |
kW | Total heat load (sensible + latent) |
required_capacity |
kW | Required cooling capacity including safety factor; 0 for a net heat loss |
net_heat_loss |
— | True when the total load is negative (the room loses heat): no cooling is required and heating may be required |
ac_sizing.active_units |
— | Number of active AC units required |
ac_sizing.total_units |
— | Total units including redundancy |
ac_sizing.total_installed_capacity_kw |
kW | Total installed cooling capacity |
Formulas
Equipment Heat Load
\[ Q_{equip} = \sum_{i=1}^{n} P_i \times N_i \times f_i \quad (\text{kW}) \]
Where \(P_i\) is the equipment power rating (kW), \(N_i\) is the quantity, and \(f_i\) is the load factor (fraction of rated power dissipated as heat, 0–1). For a VSD with 97% efficiency, \(f_i = 0.03\).
Conduction Heat Transfer
\[ Q_{cond} = \frac{U \times A \times \Delta T}{1000} \quad (\text{kW}) \]
Where \(U\) is the surface U-value (W/m²·K), \(A\) is the surface area (m²), and \(\Delta T = T_{outside} - T_{inside}\) (°C). Applied separately to walls, roof, floor, and each window.
Solar Heat Gain
\[ Q_{solar} = \frac{A_{wall} \times F_{wall} + A_{roof} \times F_{roof} + A_{window} \times F_{window}}{1000} \quad (\text{kW}) \]
Where \(A\) is the sun-exposed area (m²) and \(F\) is the solar heat gain factor (W/m²) for each surface type.
Lighting Heat Load
\[ Q_{light} = \frac{A_{floor} \times D_{light}}{1000} \quad (\text{kW}) \]
Where \(A_{floor}\) is the floor area (m²) and \(D_{light}\) is the lighting power density (W/m²).
Personnel Heat Load
\[ Q_{pers,sensible} = N_{pers} \times 0.075 \quad (\text{kW}) \]
\[ Q_{pers,latent} = N_{pers} \times 0.055 \quad (\text{kW}) \]
ASHRAE standard values: 75 W sensible heat and 55 W latent heat per person for light office/sedentary activity.
Saturation Vapour Pressure (Tetens' Equation)
\[ P_{sat} = 611.2 \times e^{\frac{17.67 \times T}{T + 243.5}} \quad (\text{Pa}) \]
Where \(T\) is the dry-bulb temperature in °C. Valid for −40°C to +60°C.
Humidity Ratio
\[ P_{vapor} = P_{sat} \times \frac{RH}{100} \]
\[ w = 0.622 \times \frac{P_{vapor}}{P_{atm} - P_{vapor}} \quad (\text{kg/kg}) \]
Where \(RH\) is relative humidity (%), \(P_{atm} = 101\,325\) Pa (standard atmospheric pressure), and \(w\) is the humidity ratio in kg water per kg dry air.
Infiltration Heat Load
\[ \dot{V} = \frac{V_{room} \times ACH}{3600} \quad (\text{m}^3\text{/s}) \]
\[ Q_{inf,sensible} = \dot{V} \times \Delta T \times 1.23 \quad (\text{kW}) \]
\[ Q_{inf,latent} = \dot{V} \times \Delta w \times 3010 \quad (\text{kW}) \]
Where \(V_{room}\) is the room volume (m³), \(ACH\) is air changes per hour, \(\Delta T\) is the temperature difference (°C), and \(\Delta w\) is the humidity ratio difference (kg/kg). The coefficient 1.23 combines \(\rho \times c_p\) (1.2 × 1.006), and 3010 combines \(\rho \times h_{fg}\) (1.2 × 2501).
Total Heat Load
\[ Q_{sensible} = Q_{equip} + Q_{cond} + Q_{solar} + Q_{light} + Q_{pers,sensible} + Q_{inf,sensible} \]
\[ Q_{latent} = Q_{pers,latent} + Q_{inf,latent} \]
\[ Q_{total} = Q_{sensible} + Q_{latent} \]
Required Cooling Capacity
\[ Q_{required} = \max\left(0,\; Q_{total} \times \left(1 + \frac{SF}{100}\right)\right) \quad (\text{kW}) \]
Where \(SF\) is the safety factor as a percentage (e.g., 15%). A negative \(Q_{total}\) is a net heat loss (cold outside, little equipment): no cooling is required, so \(Q_{required} = 0\) and no AC units are counted; the room may need heating instead.
AC Unit Sizing
\[ N_{active} = \left\lceil \frac{Q_{required}}{Q_{unit}} \right\rceil \]
\[ N_{total} = N_{active} + N_{redundant} \]
Where \(Q_{unit}\) is the rated capacity per AC unit (kW) and \(N_{redundant}\) is the number of backup units for redundancy (e.g., 1 for N+1).
Sensible Heat Ratio
\[ SHR = \frac{Q_{sensible}}{Q_{total}} \]
A high SHR (close to 1.0) indicates the load is predominantly temperature-driven with minimal moisture removal needed, which is typical for switchrooms.
Constants Used
| Constant | Value | Description |
|---|---|---|
| Atmospheric pressure \(P_{atm}\) | 101,325 Pa | Standard atmospheric pressure at sea level |
| Air density \(\rho\) | 1.2 kg/m³ | Dry air at standard conditions |
| Specific heat of air \(c_p\) | 1.006 kJ/kg·K | At constant pressure |
| Latent heat of vaporisation \(h_{fg}\) | 2501 kJ/kg | Water at standard conditions |
| Personnel sensible heat | 75 W/person | ASHRAE standard, light activity |
| Personnel latent heat | 55 W/person | ASHRAE standard, light activity |
Unit Conversion Factors
| From (Imperial) | To (Metric) | Factor |
|---|---|---|
| 1 ft | 0.3048 m | × 0.3048 |
| 1 ft² | 0.09290304 m² | × 0.09290304 |
| °F to °C | — | (F − 32) × 5/9 |
| 1 BTU/hr·ft²·°F | 5.678263 W/m²·K | × 5.678263 |
| 1 BTU/hr | 0.000293071 kW | × 0.000293071 |
Worked Examples
The 22 examples below are the validation cases for this calculation. Each one was produced by the calculation engine from the inputs shown, every figure was checked against the independent reference quoted at the end of the example, and the same cases are run against every engine build (see validation scorecard, tool fe_calc_heat_load; worked examples for every calculation).
| # | Case | Qtotal (kW) | Required (kW) | Units | Largest | Result |
|---|---|---|---|---|---|---|
| 1 | Harmonic filter room with a 25 % margin, N+1 | 27.64 | 34.55 | 3 + 1 | equipment | PASS |
| 2 | Control room with windows (conduction and solar) | 13.86 | 15.93 | 2 + 0 | equipment | PASS |
| 3 | Desert site: inside air moister than outside | 28.72 | 33.03 | 4 + 0 | equipment | PASS |
| 4 | Empty interior room: lighting is the largest load | 3.94 | 4.53 | 1 + 0 | lighting | PASS |
| 5 | Outside at room temperature on a humid day | 7.07 | 8.13 | 1 + 0 | equipment | PASS |
| 6 | Explicit envelope areas for a room sharing two walls | 42.74 | 49.16 | 5 + 0 | solar | PASS |
| 7 | L-shaped room: given floor area drives lighting and volume | 60.60 | 69.69 | 7 + 0 | solar | PASS |
| 8 | Indoor transformer room, no solar, N+1 units | 28.86 | 33.19 | 3 + 1 | equipment | PASS |
| 9 | Well-insulated interior UPS room | 4.61 | 5.07 | 2 + 0 | equipment | PASS |
| 10 | Large switchroom, transformers plus VSDs, N+1 | 124.71 | 149.65 | 6 + 1 | solar | PASS |
| 11 | MCC room with five equipment types, N+2 units | 77.56 | 89.20 | 4 + 2 | equipment | PASS |
| 12 | Net heat loss on a cold day: no cooling required | -2.18 | 0.00 | 0 + 1 | equipment | PASS |
| 13 | Default switchroom with a sunlit wall | 37.29 | 42.89 | 5 + 0 | solar | PASS |
| 14 | Partly shaded roof and sunlit wall | 50.48 | 58.05 | 6 + 0 | equipment | PASS |
| 15 | Site solar factors on a light-coloured roof, no margin | 35.38 | 35.38 | 3 + 0 | equipment | PASS |
| 16 | Training room: thirty occupants | 7.11 | 8.18 | 1 + 0 | people | PASS |
| 17 | Tropical site, 95 % RH outside, 2 ACH | 21.47 | 24.70 | 3 + 0 | equipment | PASS |
| 18 | Uninsulated steel shed under a canopy | 18.31 | 21.06 | 3 + 0 | conduction | PASS |
| 19 | Required capacity just over two 10 kW units | 20.01 | 20.01 | 3 + 0 | equipment | PASS |
| 20 | Required capacity just under two 10 kW units | 19.99 | 19.99 | 2 + 0 | equipment | PASS |
| 21 | Leaky shed at 10 air changes per hour | 16.30 | 18.75 | 2 + 0 | infiltration | PASS |
| 22 | Winter design day: fabric losses offset transformer heat | 21.66 | 24.91 | 3 + 0 | equipment | PASS |
Example 1 — Harmonic filter room with a 25 % margin, N+1
A 25 % safety factor on a filter-plus-VSD load, 12.5 kW units and one standby.
Given: Room 9 × 7 × 3.2 m; outside 40 °C / 50 % RH, inside 23 °C / 50 % RH. Equipment: Harmonic filter 150 kW × 1 at 4 % heat; VSD 250 kW × 2 at 3 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 15 m² wall at 150 W/m², 0 m² roof at 300 W/m². 1 occupant(s), lighting 10 W/m², 0.5 air changes/h, safety factor 25 %, AC units of 12.5 kW with 1 standby.
| Geometry | Afloor = 9 × 7 = 63.0 m²; Awall = 2 × (9 + 7) × 3.2 = 102.4 m²; V = Afloor × H = 63.0 × 3.2 = 201.6 m³; ΔT = 40 − 23 = 17.0 K |
|---|---|
| Equipment | Qequip = 150 × 1 × 0.04 + 250 × 2 × 0.03 = 21.000 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 102.4 + 0.4 × 63.0 + 0.3 × 63.0) × 17.0 / 1000 = 1.794 kW |
| Solar | Qsolar = (15 × 150 + 0 × 300 + 0 × 250) / 1000 = 2.250 kW |
| Lighting | Qlight = 63.0 m² × 10 W/m² / 1000 = 0.630 kW |
| People | Qpeople = 1 × (0.075 sensible + 0.055 latent) = 0.130 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(40 °C) = 7,395 Pa, Pv = 7,395 × 50/100 = 3,697 Pa, w = 0.622 × 3,697 / (101,325 − 3,697) = 0.02356; inside Psat(23 °C) = 2,809 Pa, Pv = 2,809 × 50/100 = 1,404 Pa, w = 0.622 × 1,404 / (101,325 − 1,404) = 0.00874; Δw = 0.02356 − 0.00874 = 0.01482 kg/kg |
| Infiltration | V̇ = 201.6 × 0.5 / 3600 = 0.0280 m³/s; sensible = 0.0280 × 17.0 × 1.23 = 0.585 kW; latent = 0.0280 × 0.01482 × 3010 = 1.249 kW |
| Total | Sensible 26.335 + latent 1.304 = 27.638 kW; largest contributor: equipment |
| Capacity | Required = 27.638 × (1 + 25/100) = 34.548 kW; units = ⌈34.548 / 12.5⌉ = 3 active + 1 standby = 4 × 12.5 kW (installed 50.0 kW, duty 37.5 kW at 92.1 %) PASS |
| Result | 1/1 checks pass → PASS; required 34.55 kW → 4 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 150 × 1 × 0.04 + 250 × 2 × 0.03 = 21.000 kW; conduction ΔT 17 K × ΣUA 105.54 W/K = 1.794 kW (walls 102.4 m² × 0.6, roof 63 × 0.4, floor 63 × 0.3); solar (15 × 150 + 0 × 300)/1000 = 2.250 kW; lighting 63 m² × 10 W/m² = 0.630 kW; people 1 × (0.075 + 0.055) = 0.130 kW; infiltration V̇ = 201.6 × 0.5/3600 = 0.02800 m³/s, sensible 0.02800 × 17 × 1.23 = 0.585 kW, Tetens P_sat(40 °C) = 7395 Pa → w_out = 0.02356, P_sat(23 °C) = 2809 Pa → w_in = 0.00874, Δw = 0.01482 kg/kg, latent 0.02800 × 0.01482 × 3010 = 1.249 kW; sensible 26.335 + latent 1.304 = total 27.638 kW; required 27.638 × 1.25 = 34.548 kW → ⌈34.548/12.5⌉ = 3 active + 1 standby = 4 × 12.5 kW (50 kW installed, duty 37.5 kW at 92.1 %); largest contributor equipment (validation case c0012-conservative-margin-25pct) |
Example 2 — Control room with windows (conduction and solar)
Two windows of different glazing add a conduction term each and a solar term at the window factor; detailed output lists the conduction by surface.
Given: Room 8 × 6 × 3 m; outside 36 °C / 60 % RH, inside 23 °C / 50 % RH. Equipment: SCADA servers 6 kW × 1 at 100 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K, windows North window 4 m² (U 5.8), Door light 1.5 m² (U 2.8). Sunlit: 20 m² wall at 150 W/m², 0 m² roof at 300 W/m², windows at 250 W/m². 3 occupant(s), lighting 10 W/m², 0.5 air changes/h, safety factor 15 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 8 × 6 = 48.0 m²; Awall = 2 × (8 + 6) × 3 = 84.0 m²; V = Afloor × H = 48.0 × 3 = 144.0 m³; ΔT = 36 − 23 = 13.0 K |
|---|---|
| Equipment | Qequip = 6 × 1 × 1 = 6.000 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 84.0 + 0.4 × 48.0 + 0.3 × 48.0 + 5.8 × 4.0 + 2.8 × 1.5) × 13.0 / 1000 = 1.448 kW |
| Solar | Qsolar = (20 × 150 + 0 × 300 + 5.5 × 250) / 1000 = 4.375 kW |
| Lighting | Qlight = 48.0 m² × 10 W/m² / 1000 = 0.480 kW |
| People | Qpeople = 3 × (0.075 sensible + 0.055 latent) = 0.390 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(36 °C) = 5,951 Pa, Pv = 5,951 × 60/100 = 3,571 Pa, w = 0.622 × 3,571 / (101,325 − 3,571) = 0.02272; inside Psat(23 °C) = 2,809 Pa, Pv = 2,809 × 50/100 = 1,404 Pa, w = 0.622 × 1,404 / (101,325 − 1,404) = 0.00874; Δw = 0.02272 − 0.00874 = 0.01398 kg/kg |
| Infiltration | V̇ = 144.0 × 0.5 / 3600 = 0.0200 m³/s; sensible = 0.0200 × 13.0 × 1.23 = 0.320 kW; latent = 0.0200 × 0.01398 × 3010 = 0.842 kW |
| Total | Sensible 12.848 + latent 1.007 = 13.855 kW; largest contributor: equipment |
| Capacity | Required = 13.855 × (1 + 15/100) = 15.933 kW; units = ⌈15.933 / 10⌉ = 2 active + 0 standby = 2 × 10 kW (installed 20.0 kW, duty 20.0 kW at 79.7 %) PASS |
| Result | 1/1 checks pass → PASS; required 15.93 kW → 2 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 6 × 1 × 1 = 6.000 kW; conduction ΔT 13 K × ΣUA 111.40 W/K = 1.448 kW (walls 84 m² × 0.6, roof 48 × 0.4, floor 48 × 0.3, windows 5.5 m²); solar (20 × 150 + 0 × 300 + 5.5 × 250)/1000 = 4.375 kW; lighting 48 m² × 10 W/m² = 0.480 kW; people 3 × (0.075 + 0.055) = 0.390 kW; infiltration V̇ = 144 × 0.5/3600 = 0.02000 m³/s, sensible 0.02000 × 13 × 1.23 = 0.320 kW, Tetens P_sat(36 °C) = 5951 Pa → w_out = 0.02272, P_sat(23 °C) = 2809 Pa → w_in = 0.00874, Δw = 0.01398 kg/kg, latent 0.02000 × 0.01398 × 3010 = 0.842 kW; sensible 12.848 + latent 1.007 = total 13.855 kW; required 13.855 × 1.15 = 15.933 kW → ⌈15.933/10⌉ = 2 active + 0 standby = 2 × 10 kW (20 kW installed, duty 20 kW at 79.7 %); largest contributor equipment; window conduction (detailed surfaces) North 5.8 × 4 × 13/1000 = 0.3016 kW, Door light 2.8 × 1.5 × 13/1000 = 0.0546 kW (validation case c0012-control-room-windows) |
Example 3 — Desert site: inside air moister than outside
At 46 °C / 8 % RH the outside humidity ratio is below the room's 24 °C / 55 % RH, so Δw and the latent infiltration term are negative (infiltration dries the room).
Given: Room 10 × 6 × 4 m; outside 46 °C / 8 % RH, inside 24 °C / 55 % RH. Equipment: Transformer 1,250 kW × 2 at 1 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 1 occupant(s), lighting 10 W/m², 0.5 air changes/h, safety factor 15 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 10 × 6 = 60.0 m²; Awall = 2 × (10 + 6) × 4 = 128.0 m²; V = Afloor × H = 60.0 × 4 = 240.0 m³; ΔT = 46 − 24 = 22.0 K |
|---|---|
| Equipment | Qequip = 1,250 × 2 × 0.01 = 25.000 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 128.0 + 0.4 × 60.0 + 0.3 × 60.0) × 22.0 / 1000 = 2.614 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 60.0 m² × 10 W/m² / 1000 = 0.600 kW |
| People | Qpeople = 1 × (0.075 sensible + 0.055 latent) = 0.130 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(46 °C) = 10,128 Pa, Pv = 10,128 × 8/100 = 810 Pa, w = 0.622 × 810 / (101,325 − 810) = 0.00501; inside Psat(24 °C) = 2,983 Pa, Pv = 2,983 × 55/100 = 1,641 Pa, w = 0.622 × 1,641 / (101,325 − 1,641) = 0.01024; Δw = 0.00501 − 0.01024 = -0.00522 kg/kg |
| Infiltration | V̇ = 240.0 × 0.5 / 3600 = 0.0333 m³/s; sensible = 0.0333 × 22.0 × 1.23 = 0.902 kW; latent = 0.0333 × -0.00522 × 3010 = -0.524 kW |
| Total | Sensible 29.191 + latent -0.469 = 28.721 kW; largest contributor: equipment |
| Capacity | Required = 28.721 × (1 + 15/100) = 33.030 kW; units = ⌈33.030 / 10⌉ = 4 active + 0 standby = 4 × 10 kW (installed 40.0 kW, duty 40.0 kW at 82.6 %) PASS |
| Result | 1/1 checks pass → PASS; required 33.03 kW → 4 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 1250 × 2 × 0.01 = 25.000 kW; conduction ΔT 22 K × ΣUA 118.80 W/K = 2.614 kW (walls 128 m² × 0.6, roof 60 × 0.4, floor 60 × 0.3); solar (0 × 150 + 0 × 300)/1000 = 0.000 kW; lighting 60 m² × 10 W/m² = 0.600 kW; people 1 × (0.075 + 0.055) = 0.130 kW; infiltration V̇ = 240 × 0.5/3600 = 0.03333 m³/s, sensible 0.03333 × 22 × 1.23 = 0.902 kW, Tetens P_sat(46 °C) = 10128 Pa → w_out = 0.00501, P_sat(24 °C) = 2983 Pa → w_in = 0.01024, Δw = -0.00522 kg/kg, latent 0.03333 × -0.00522 × 3010 = -0.524 kW; sensible 29.191 + latent -0.469 = total 28.721 kW; required 28.721 × 1.15 = 33.030 kW → ⌈33.030/10⌉ = 4 active + 0 standby = 4 × 10 kW (40 kW installed, duty 40 kW at 82.6 %); largest contributor equipment (validation case c0012-desert-dry-negative-latent) |
Example 4 — Empty interior room: lighting is the largest load
No equipment, no solar and only 2 K across the fabric: 30 W/m² of lighting outweighs every other term and one unit at 10 % utilisation covers it.
Given: Room 10 × 10 × 3 m; outside 26 °C / 60 % RH, inside 24 °C / 50 % RH. Equipment: no equipment. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 1 occupant(s), lighting 30 W/m², 0.5 air changes/h, safety factor 15 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 10 × 10 = 100.0 m²; Awall = 2 × (10 + 10) × 3 = 120.0 m²; V = Afloor × H = 100.0 × 3 = 300.0 m³; ΔT = 26 − 24 = 2.0 K |
|---|---|
| Equipment | Qequip = 0 = 0.000 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 120.0 + 0.4 × 100.0 + 0.3 × 100.0) × 2.0 / 1000 = 0.284 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 100.0 m² × 30 W/m² / 1000 = 3.000 kW |
| People | Qpeople = 1 × (0.075 sensible + 0.055 latent) = 0.130 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(26 °C) = 3,361 Pa, Pv = 3,361 × 60/100 = 2,017 Pa, w = 0.622 × 2,017 / (101,325 − 2,017) = 0.01263; inside Psat(24 °C) = 2,983 Pa, Pv = 2,983 × 50/100 = 1,492 Pa, w = 0.622 × 1,492 / (101,325 − 1,492) = 0.00929; Δw = 0.01263 − 0.00929 = 0.00334 kg/kg |
| Infiltration | V̇ = 300.0 × 0.5 / 3600 = 0.0417 m³/s; sensible = 0.0417 × 2.0 × 1.23 = 0.102 kW; latent = 0.0417 × 0.00334 × 3010 = 0.419 kW |
| Total | Sensible 3.462 + latent 0.474 = 3.935 kW; largest contributor: lighting |
| Capacity | Required = 3.935 × (1 + 15/100) = 4.526 kW; units = ⌈4.526 / 10⌉ = 1 active + 0 standby = 1 × 10 kW (installed 10.0 kW, duty 10.0 kW at 45.3 %) PASS |
| Result | 1/1 checks pass → PASS; required 4.53 kW → 1 units; governing: cooling capacity |
| Reference | Hand calculation: equipment none = 0.000 kW; conduction ΔT 2 K × ΣUA 142.00 W/K = 0.284 kW (walls 120 m² × 0.6, roof 100 × 0.4, floor 100 × 0.3); solar (0 × 150 + 0 × 300)/1000 = 0.000 kW; lighting 100 m² × 30 W/m² = 3.000 kW; people 1 × (0.075 + 0.055) = 0.130 kW; infiltration V̇ = 300 × 0.5/3600 = 0.04167 m³/s, sensible 0.04167 × 2 × 1.23 = 0.102 kW, Tetens P_sat(26 °C) = 3361 Pa → w_out = 0.01263, P_sat(24 °C) = 2983 Pa → w_in = 0.00929, Δw = 0.00334 kg/kg, latent 0.04167 × 0.00334 × 3010 = 0.419 kW; sensible 3.462 + latent 0.474 = total 3.935 kW; required 3.935 × 1.15 = 4.526 kW → ⌈4.526/10⌉ = 1 active + 0 standby = 1 × 10 kW (10 kW installed, duty 10 kW at 45.3 %); largest contributor lighting (validation case c0012-empty-room-lighting-only) |
Example 5 — Outside at room temperature on a humid day
With ΔT = 0 the conduction and sensible infiltration terms vanish; only the UPS, lighting, the occupant and the latent infiltration from 90 % RH outside air remain.
Given: Room 10 × 6 × 3.5 m; outside 24 °C / 90 % RH, inside 24 °C / 50 % RH. Equipment: UPS 100 kW × 1 at 5 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 1 occupant(s), lighting 10 W/m², 1 air changes/h, safety factor 15 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 10 × 6 = 60.0 m²; Awall = 2 × (10 + 6) × 3.5 = 112.0 m²; V = Afloor × H = 60.0 × 3.5 = 210.0 m³; ΔT = 24 − 24 = 0.0 K |
|---|---|
| Equipment | Qequip = 100 × 1 × 0.05 = 5.000 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 112.0 + 0.4 × 60.0 + 0.3 × 60.0) × 0.0 / 1000 = 0.000 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 60.0 m² × 10 W/m² / 1000 = 0.600 kW |
| People | Qpeople = 1 × (0.075 sensible + 0.055 latent) = 0.130 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(24 °C) = 2,983 Pa, Pv = 2,983 × 90/100 = 2,685 Pa, w = 0.622 × 2,685 / (101,325 − 2,685) = 0.01693; inside Psat(24 °C) = 2,983 Pa, Pv = 2,983 × 50/100 = 1,492 Pa, w = 0.622 × 1,492 / (101,325 − 1,492) = 0.00929; Δw = 0.01693 − 0.00929 = 0.00764 kg/kg |
| Infiltration | V̇ = 210.0 × 1 / 3600 = 0.0583 m³/s; sensible = 0.0583 × 0.0 × 1.23 = 0.000 kW; latent = 0.0583 × 0.00764 × 3010 = 1.341 kW |
| Total | Sensible 5.675 + latent 1.396 = 7.071 kW; largest contributor: equipment |
| Capacity | Required = 7.071 × (1 + 15/100) = 8.132 kW; units = ⌈8.132 / 10⌉ = 1 active + 0 standby = 1 × 10 kW (installed 10.0 kW, duty 10.0 kW at 81.3 %) PASS |
| Result | 1/1 checks pass → PASS; required 8.13 kW → 1 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 100 × 1 × 0.05 = 5.000 kW; conduction ΔT 0 K × ΣUA 109.20 W/K = 0.000 kW (walls 112 m² × 0.6, roof 60 × 0.4, floor 60 × 0.3); solar (0 × 150 + 0 × 300)/1000 = 0.000 kW; lighting 60 m² × 10 W/m² = 0.600 kW; people 1 × (0.075 + 0.055) = 0.130 kW; infiltration V̇ = 210 × 1/3600 = 0.05833 m³/s, sensible 0.05833 × 0 × 1.23 = 0.000 kW, Tetens P_sat(24 °C) = 2983 Pa → w_out = 0.01693, P_sat(24 °C) = 2983 Pa → w_in = 0.00929, Δw = 0.00764 kg/kg, latent 0.05833 × 0.00764 × 3010 = 1.341 kW; sensible 5.675 + latent 1.396 = total 7.071 kW; required 7.071 × 1.15 = 8.132 kW → ⌈8.132/10⌉ = 1 active + 0 standby = 1 × 10 kW (10 kW installed, duty 10 kW at 81.3 %); largest contributor equipment (validation case c0012-equal-temperatures-humid-day) |
Example 6 — Explicit envelope areas for a room sharing two walls
Wall, roof and floor areas given explicitly (two walls and part of the roof are internal) override the areas computed from the dimensions; the given floor area also sets lighting and room volume, and the sunlit roof defaults to the given roof area.
Given: Room 12 × 8 × 3.5 m (walls 70 m², roof 60 m², floor 50 m² given); outside 38 °C / 60 % RH, inside 24 °C / 50 % RH. Equipment: MCC 400 kW × 2 at 2 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 35 m² wall at 150 W/m², whole roof at 300 W/m². 2 occupant(s), lighting 10 W/m², 0.5 air changes/h, safety factor 15 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 50.0 m² as given (not 12 × 8); Awall = given = 70.0 m²; roof 60.0 m² as given; V = Afloor × H = 50.0 × 3.5 = 175.0 m³; ΔT = 38 − 24 = 14.0 K |
|---|---|
| Equipment | Qequip = 400 × 2 × 0.02 = 16.000 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 70.0 + 0.4 × 60.0 + 0.3 × 50.0) × 14.0 / 1000 = 1.134 kW |
| Solar | Qsolar = (35 × 150 + 60 × 300 + 0 × 250) / 1000 = 23.250 kW |
| Lighting | Qlight = 50.0 m² × 10 W/m² / 1000 = 0.500 kW |
| People | Qpeople = 2 × (0.075 sensible + 0.055 latent) = 0.260 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(38 °C) = 6,639 Pa, Pv = 6,639 × 60/100 = 3,983 Pa, w = 0.622 × 3,983 / (101,325 − 3,983) = 0.02545; inside Psat(24 °C) = 2,983 Pa, Pv = 2,983 × 50/100 = 1,492 Pa, w = 0.622 × 1,492 / (101,325 − 1,492) = 0.00929; Δw = 0.02545 − 0.00929 = 0.01616 kg/kg |
| Infiltration | V̇ = 175.0 × 0.5 / 3600 = 0.0243 m³/s; sensible = 0.0243 × 14.0 × 1.23 = 0.419 kW; latent = 0.0243 × 0.01616 × 3010 = 1.182 kW |
| Total | Sensible 41.453 + latent 1.292 = 42.745 kW; largest contributor: solar |
| Capacity | Required = 42.745 × (1 + 15/100) = 49.157 kW; units = ⌈49.157 / 10⌉ = 5 active + 0 standby = 5 × 10 kW (installed 50.0 kW, duty 50.0 kW at 98.3 %) PASS |
| Result | 1/1 checks pass → PASS; required 49.16 kW → 5 units; governing: cooling capacity. Solar gain (23.2 kW) dominates the load; the simplified solar model is conservative — check the sunlit areas and solar factors, or shade the roof. |
| Reference | Hand calculation: floor 50 m² as given, volume 50 × 3.5 = 175 m³; equipment = 16.000 kW; conduction ΔT 14 K × ΣUA 81.00 W/K = 1.134 kW (walls 70 m², roof 60, floor 50); solar = 23.250 kW (sunlit roof 60 m²); lighting 50 m² × 10 W/m² = 0.500 kW; people 2 × 0.130 = 0.260 kW; infiltration V̇ = 175 × 0.5/3600 = 0.02431 m³/s, sensible × 14 × 1.23 = 0.419 kW, Tetens w_out = 0.02545, w_in = 0.00929, latent × 0.01616 × 3010 = 1.182 kW; sensible 41.453 + latent 1.292 = total 42.745 kW; required 42.745 × 1.15 = 49.157 kW → ⌈49.157/10⌉ = 5 active + 0 standby (duty 50 kW at 98.3 %); largest contributor solar (validation case c0012-explicit-areas-shared-walls) |
Example 7 — L-shaped room: given floor area drives lighting and volume
A 12 × 10 m bounding box with a 4 × 5 m notch: the given 100 m² floor area (not L × W = 120 m²) sets floor conduction, the lighting load, the room volume for infiltration and the default (whole, sunlit) roof area; the wall area is given as the true perimeter × height.
Given: Room 12 × 10 × 4 m (walls 176 m², floor 100 m² given); outside 40 °C / 60 % RH, inside 25 °C / 50 % RH. Equipment: VSD 250 kW × 3 at 3 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 0 m² wall at 150 W/m², whole roof at 300 W/m². 1 occupant(s), lighting 12 W/m², 0.5 air changes/h, safety factor 15 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 100.0 m² as given (not 12 × 10); Awall = given = 176.0 m²; roof = Afloor = 100.0 m²; V = Afloor × H = 100.0 × 4 = 400.0 m³; ΔT = 40 − 25 = 15.0 K |
|---|---|
| Equipment | Qequip = 250 × 3 × 0.03 = 22.500 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 176.0 + 0.4 × 100.0 + 0.3 × 100.0) × 15.0 / 1000 = 2.634 kW |
| Solar | Qsolar = (0 × 150 + 100 × 300 + 0 × 250) / 1000 = 30.000 kW |
| Lighting | Qlight = 100.0 m² × 12 W/m² / 1000 = 1.200 kW |
| People | Qpeople = 1 × (0.075 sensible + 0.055 latent) = 0.130 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(40 °C) = 7,395 Pa, Pv = 7,395 × 60/100 = 4,437 Pa, w = 0.622 × 4,437 / (101,325 − 4,437) = 0.02848; inside Psat(25 °C) = 3,167 Pa, Pv = 3,167 × 50/100 = 1,584 Pa, w = 0.622 × 1,584 / (101,325 − 1,584) = 0.00988; Δw = 0.02848 − 0.00988 = 0.01861 kg/kg |
| Infiltration | V̇ = 400.0 × 0.5 / 3600 = 0.0556 m³/s; sensible = 0.0556 × 15.0 × 1.23 = 1.025 kW; latent = 0.0556 × 0.01861 × 3010 = 3.112 kW |
| Total | Sensible 57.434 + latent 3.167 = 60.601 kW; largest contributor: solar |
| Capacity | Required = 60.601 × (1 + 15/100) = 69.691 kW; units = ⌈69.691 / 10⌉ = 7 active + 0 standby = 7 × 10 kW (installed 70.0 kW, duty 70.0 kW at 99.6 %) PASS |
| Result | 1/1 checks pass → PASS; required 69.69 kW → 7 units; governing: cooling capacity |
| Reference | Hand calculation: floor 100 m² as given, volume 100 × 4 = 400 m³; equipment = 22.500 kW; conduction ΔT 15 K × ΣUA 175.60 W/K = 2.634 kW (walls 176 m², roof 100, floor 100); solar = 30.000 kW (sunlit roof 100 m²); lighting 100 m² × 12 W/m² = 1.200 kW; people 1 × 0.130 = 0.130 kW; infiltration V̇ = 400 × 0.5/3600 = 0.05556 m³/s, sensible × 15 × 1.23 = 1.025 kW, Tetens w_out = 0.02848, w_in = 0.00988, latent × 0.01861 × 3010 = 3.112 kW; sensible 57.434 + latent 3.167 = total 60.601 kW; required 60.601 × 1.15 = 69.691 kW → ⌈69.691/10⌉ = 7 active + 0 standby (duty 70 kW at 99.6 %); largest contributor solar (validation case c0012-explicit-floor-area-l-shaped) |
Example 8 — Indoor transformer room, no solar, N+1 units
Two 1 MVA transformers in a windowless room: equipment losses dominate; one standby unit added.
Given: Room 10 × 6 × 4 m; outside 40 °C / 60 % RH, inside 25 °C / 50 % RH. Equipment: Transformer 1,000 kW × 2 at 1.2 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 0 occupant(s), lighting 10 W/m², 0.5 air changes/h, safety factor 15 %, AC units of 15 kW with 1 standby.
| Geometry | Afloor = 10 × 6 = 60.0 m²; Awall = 2 × (10 + 6) × 4 = 128.0 m²; V = Afloor × H = 60.0 × 4 = 240.0 m³; ΔT = 40 − 25 = 15.0 K |
|---|---|
| Equipment | Qequip = 1,000 × 2 × 0.012 = 24.000 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 128.0 + 0.4 × 60.0 + 0.3 × 60.0) × 15.0 / 1000 = 1.782 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 60.0 m² × 10 W/m² / 1000 = 0.600 kW |
| People | Qpeople = 0 × (0.075 sensible + 0.055 latent) = 0.000 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(40 °C) = 7,395 Pa, Pv = 7,395 × 60/100 = 4,437 Pa, w = 0.622 × 4,437 / (101,325 − 4,437) = 0.02848; inside Psat(25 °C) = 3,167 Pa, Pv = 3,167 × 50/100 = 1,584 Pa, w = 0.622 × 1,584 / (101,325 − 1,584) = 0.00988; Δw = 0.02848 − 0.00988 = 0.01861 kg/kg |
| Infiltration | V̇ = 240.0 × 0.5 / 3600 = 0.0333 m³/s; sensible = 0.0333 × 15.0 × 1.23 = 0.615 kW; latent = 0.0333 × 0.01861 × 3010 = 1.867 kW |
| Total | Sensible 26.997 + latent 1.867 = 28.864 kW; largest contributor: equipment |
| Capacity | Required = 28.864 × (1 + 15/100) = 33.194 kW; units = ⌈33.194 / 15⌉ = 3 active + 1 standby = 4 × 15 kW (installed 60.0 kW, duty 45.0 kW at 73.8 %) PASS |
| Result | 1/1 checks pass → PASS; required 33.19 kW → 4 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 1000 × 2 × 0.012 = 24.000 kW; conduction ΔT 15 K × ΣUA 118.80 W/K = 1.782 kW (walls 128 m² × 0.6, roof 60 × 0.4, floor 60 × 0.3); solar (0 × 150 + 0 × 300)/1000 = 0.000 kW; lighting 60 m² × 10 W/m² = 0.600 kW; people 0 × (0.075 + 0.055) = 0.000 kW; infiltration V̇ = 240 × 0.5/3600 = 0.03333 m³/s, sensible 0.03333 × 15 × 1.23 = 0.615 kW, Tetens P_sat(40 °C) = 7395 Pa → w_out = 0.02848, P_sat(25 °C) = 3167 Pa → w_in = 0.00988, Δw = 0.01861 kg/kg, latent 0.03333 × 0.01861 × 3010 = 1.867 kW; sensible 26.997 + latent 1.867 = total 28.864 kW; required 28.864 × 1.15 = 33.194 kW → ⌈33.194/15⌉ = 3 active + 1 standby = 4 × 15 kW (60 kW installed, duty 45 kW at 73.8 %); largest contributor equipment (validation case c0012-indoor-transformer-room) |
Example 9 — Well-insulated interior UPS room
A small room inside a larger building (no solar, ΔT only 5 K, 0.2 ACH): the UPS losses are almost nine tenths of the load.
Given: Room 6 × 4 × 3 m; outside 27 °C / 60 % RH, inside 22 °C / 50 % RH. Equipment: UPS 50 kW × 1 at 8 % heat. U-values wall 0.3, roof 0.25, floor 0.25 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 1 occupant(s), lighting 10 W/m², 0.2 air changes/h, safety factor 10 %, AC units of 5 kW with 0 standby.
| Geometry | Afloor = 6 × 4 = 24.0 m²; Awall = 2 × (6 + 4) × 3 = 60.0 m²; V = Afloor × H = 24.0 × 3 = 72.0 m³; ΔT = 27 − 22 = 5.0 K |
|---|---|
| Equipment | Qequip = 50 × 1 × 0.08 = 4.000 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.3 × 60.0 + 0.25 × 24.0 + 0.25 × 24.0) × 5.0 / 1000 = 0.150 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 24.0 m² × 10 W/m² / 1000 = 0.240 kW |
| People | Qpeople = 1 × (0.075 sensible + 0.055 latent) = 0.130 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(27 °C) = 3,566 Pa, Pv = 3,566 × 60/100 = 2,140 Pa, w = 0.622 × 2,140 / (101,325 − 2,140) = 0.01342; inside Psat(22 °C) = 2,643 Pa, Pv = 2,643 × 50/100 = 1,321 Pa, w = 0.622 × 1,321 / (101,325 − 1,321) = 0.00822; Δw = 0.01342 − 0.00822 = 0.00520 kg/kg |
| Infiltration | V̇ = 72.0 × 0.2 / 3600 = 0.0040 m³/s; sensible = 0.0040 × 5.0 × 1.23 = 0.025 kW; latent = 0.0040 × 0.00520 × 3010 = 0.063 kW |
| Total | Sensible 4.490 + latent 0.118 = 4.607 kW; largest contributor: equipment |
| Capacity | Required = 4.607 × (1 + 10/100) = 5.068 kW; units = ⌈5.068 / 5⌉ = 2 active + 0 standby = 2 × 5 kW (installed 10.0 kW, duty 10.0 kW at 50.7 %) PASS |
| Result | 1/1 checks pass → PASS; required 5.07 kW → 2 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 50 × 1 × 0.08 = 4.000 kW; conduction ΔT 5 K × ΣUA 30.00 W/K = 0.150 kW (walls 60 m² × 0.3, roof 24 × 0.25, floor 24 × 0.25); solar (0 × 150 + 0 × 300)/1000 = 0.000 kW; lighting 24 m² × 10 W/m² = 0.240 kW; people 1 × (0.075 + 0.055) = 0.130 kW; infiltration V̇ = 72 × 0.2/3600 = 0.00400 m³/s, sensible 0.00400 × 5 × 1.23 = 0.025 kW, Tetens P_sat(27 °C) = 3566 Pa → w_out = 0.01342, P_sat(22 °C) = 2643 Pa → w_in = 0.00822, Δw = 0.00520 kg/kg, latent 0.00400 × 0.00520 × 3010 = 0.063 kW; sensible 4.490 + latent 0.118 = total 4.607 kW; required 4.607 × 1.1 = 5.068 kW → ⌈5.068/5⌉ = 2 active + 0 standby = 2 × 5 kW (10 kW installed, duty 10 kW at 50.7 %); largest contributor equipment (validation case c0012-insulated-interior-ups-room) |
Example 10 — Large switchroom, transformers plus VSDs, N+1
Two equipment types with quantities, overridden U-values and solar factors, 20 % margin and 25 kW units with one standby: the 200 m² sunlit roof alone contributes 70 kW.
Given: Room 20 × 10 × 4 m; outside 40 °C / 70 % RH, inside 22 °C / 50 % RH. Equipment: Transformer 500 kW × 2 at 2 % heat; VSD 75 kW × 4 at 3 % heat. U-values wall 0.5, roof 0.35, floor 0.3 W/m²·K. Sunlit: 60 m² wall at 200 W/m², whole roof at 350 W/m². 2 occupant(s), lighting 12 W/m², 0.3 air changes/h, safety factor 20 %, AC units of 25 kW with 1 standby.
| Geometry | Afloor = 20 × 10 = 200.0 m²; Awall = 2 × (20 + 10) × 4 = 240.0 m²; V = Afloor × H = 200.0 × 4 = 800.0 m³; ΔT = 40 − 22 = 18.0 K |
|---|---|
| Equipment | Qequip = 500 × 2 × 0.02 + 75 × 4 × 0.03 = 29.000 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.5 × 240.0 + 0.35 × 200.0 + 0.3 × 200.0) × 18.0 / 1000 = 4.500 kW |
| Solar | Qsolar = (60 × 200 + 200 × 350 + 0 × 250) / 1000 = 82.000 kW |
| Lighting | Qlight = 200.0 m² × 12 W/m² / 1000 = 2.400 kW |
| People | Qpeople = 2 × (0.075 sensible + 0.055 latent) = 0.260 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(40 °C) = 7,395 Pa, Pv = 7,395 × 70/100 = 5,176 Pa, w = 0.622 × 5,176 / (101,325 − 5,176) = 0.03349; inside Psat(22 °C) = 2,643 Pa, Pv = 2,643 × 50/100 = 1,321 Pa, w = 0.622 × 1,321 / (101,325 − 1,321) = 0.00822; Δw = 0.03349 − 0.00822 = 0.02527 kg/kg |
| Infiltration | V̇ = 800.0 × 0.3 / 3600 = 0.0667 m³/s; sensible = 0.0667 × 18.0 × 1.23 = 1.476 kW; latent = 0.0667 × 0.02527 × 3010 = 5.070 kW |
| Total | Sensible 119.526 + latent 5.180 = 124.706 kW; largest contributor: solar |
| Capacity | Required = 124.706 × (1 + 20/100) = 149.648 kW; units = ⌈149.648 / 25⌉ = 6 active + 1 standby = 7 × 25 kW (installed 175.0 kW, duty 150.0 kW at 99.8 %) PASS |
| Result | 1/1 checks pass → PASS; required 149.65 kW → 7 units; governing: cooling capacity. Solar gain (82.0 kW) dominates the load; the simplified solar model is conservative — check the sunlit areas and solar factors, or shade the roof. |
| Reference | Hand calculation: equipment 500 × 2 × 0.02 + 75 × 4 × 0.03 = 29.000 kW; conduction ΔT 18 K × ΣUA 250.00 W/K = 4.500 kW (walls 240 m² × 0.5, roof 200 × 0.35, floor 200 × 0.3); solar (60 × 200 + 200 × 350)/1000 = 82.000 kW; lighting 200 m² × 12 W/m² = 2.400 kW; people 2 × (0.075 + 0.055) = 0.260 kW; infiltration V̇ = 800 × 0.3/3600 = 0.06667 m³/s, sensible 0.06667 × 18 × 1.23 = 1.476 kW, Tetens P_sat(40 °C) = 7395 Pa → w_out = 0.03349, P_sat(22 °C) = 2643 Pa → w_in = 0.00822, Δw = 0.02527 kg/kg, latent 0.06667 × 0.02527 × 3010 = 5.070 kW; sensible 119.526 + latent 5.180 = total 124.706 kW; required 124.706 × 1.2 = 149.648 kW → ⌈149.648/25⌉ = 6 active + 1 standby = 7 × 25 kW (175 kW installed, duty 150 kW at 99.8 %); largest contributor solar (validation case c0012-large-switchroom-mixed-equipment) |
Example 11 — MCC room with five equipment types, N+2 units
Five equipment lines with quantities summed item by item (including panels at 100 % heat), two standby units on 25 kW machines.
Given: Room 15 × 8 × 4 m; outside 42 °C / 40 % RH, inside 25 °C / 50 % RH. Equipment: VSD 160 kW 160 kW × 4 at 3 % heat; VSD 55 kW 55 kW × 8 at 3 % heat; Transformer 2,000 kW × 1 at 1 % heat; UPS 40 kW × 2 at 6 % heat; PLC panel 2 kW × 6 at 100 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 2 occupant(s), lighting 10 W/m², 0.5 air changes/h, safety factor 15 %, AC units of 25 kW with 2 standby.
| Geometry | Afloor = 15 × 8 = 120.0 m²; Awall = 2 × (15 + 8) × 4 = 184.0 m²; V = Afloor × H = 120.0 × 4 = 480.0 m³; ΔT = 42 − 25 = 17.0 K |
|---|---|
| Equipment | Qequip = 160 × 4 × 0.03 + 55 × 8 × 0.03 + 2,000 × 1 × 0.01 + 40 × 2 × 0.06 + 2 × 6 × 1 = 69.200 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 184.0 + 0.4 × 120.0 + 0.3 × 120.0) × 17.0 / 1000 = 3.305 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 120.0 m² × 10 W/m² / 1000 = 1.200 kW |
| People | Qpeople = 2 × (0.075 sensible + 0.055 latent) = 0.260 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(42 °C) = 8,224 Pa, Pv = 8,224 × 40/100 = 3,290 Pa, w = 0.622 × 3,290 / (101,325 − 3,290) = 0.02087; inside Psat(25 °C) = 3,167 Pa, Pv = 3,167 × 50/100 = 1,584 Pa, w = 0.622 × 1,584 / (101,325 − 1,584) = 0.00988; Δw = 0.02087 − 0.00988 = 0.01100 kg/kg |
| Infiltration | V̇ = 480.0 × 0.5 / 3600 = 0.0667 m³/s; sensible = 0.0667 × 17.0 × 1.23 = 1.394 kW; latent = 0.0667 × 0.01100 × 3010 = 2.207 kW |
| Total | Sensible 75.249 + latent 2.317 = 77.565 kW; largest contributor: equipment |
| Capacity | Required = 77.565 × (1 + 15/100) = 89.200 kW; units = ⌈89.200 / 25⌉ = 4 active + 2 standby = 6 × 25 kW (installed 150.0 kW, duty 100.0 kW at 89.2 %) PASS |
| Result | 1/1 checks pass → PASS; required 89.20 kW → 6 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 160 × 4 × 0.03 + 55 × 8 × 0.03 + 2000 × 1 × 0.01 + 40 × 2 × 0.06 + 2 × 6 × 1 = 69.200 kW; conduction ΔT 17 K × ΣUA 194.40 W/K = 3.305 kW (walls 184 m² × 0.6, roof 120 × 0.4, floor 120 × 0.3); solar (0 × 150 + 0 × 300)/1000 = 0.000 kW; lighting 120 m² × 10 W/m² = 1.200 kW; people 2 × (0.075 + 0.055) = 0.260 kW; infiltration V̇ = 480 × 0.5/3600 = 0.06667 m³/s, sensible 0.06667 × 17 × 1.23 = 1.394 kW, Tetens P_sat(42 °C) = 8224 Pa → w_out = 0.02087, P_sat(25 °C) = 3167 Pa → w_in = 0.00988, Δw = 0.01100 kg/kg, latent 0.06667 × 0.01100 × 3010 = 2.207 kW; sensible 75.249 + latent 2.317 = total 77.565 kW; required 77.565 × 1.15 = 89.200 kW → ⌈89.200/25⌉ = 4 active + 2 standby = 6 × 25 kW (150 kW installed, duty 100 kW at 89.2 %); largest contributor equipment (validation case c0012-mcc-room-many-items-n-plus-2) |
Example 12 — Net heat loss on a cold day: no cooling required
Outside −10 °C with only 0.5 kW of equipment: fabric and infiltration losses exceed the gains, so the total is negative, the required cooling capacity is zero, no AC units are counted (only the standby unit) and the tool warns that heating may be required.
Given: Room 8 × 5 × 3 m; outside -10 °C / 80 % RH, inside 20 °C / 40 % RH. Equipment: Control panel 10 kW × 1 at 5 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 0 occupant(s), lighting 10 W/m², 0.5 air changes/h, safety factor 15 %, AC units of 10 kW with 1 standby.
| Geometry | Afloor = 8 × 5 = 40.0 m²; Awall = 2 × (8 + 5) × 3 = 78.0 m²; V = Afloor × H = 40.0 × 3 = 120.0 m³; ΔT = -10 − 20 = -30.0 K |
|---|---|
| Equipment | Qequip = 10 × 1 × 0.05 = 0.500 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 78.0 + 0.4 × 40.0 + 0.3 × 40.0) × -30.0 / 1000 = -2.244 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 40.0 m² × 10 W/m² / 1000 = 0.400 kW |
| People | Qpeople = 0 × (0.075 sensible + 0.055 latent) = 0.000 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(-10 °C) = 287 Pa, Pv = 287 × 80/100 = 229 Pa, w = 0.622 × 229 / (101,325 − 229) = 0.00141; inside Psat(20 °C) = 2,337 Pa, Pv = 2,337 × 40/100 = 935 Pa, w = 0.622 × 935 / (101,325 − 935) = 0.00579; Δw = 0.00141 − 0.00579 = -0.00438 kg/kg |
| Infiltration | V̇ = 120.0 × 0.5 / 3600 = 0.0167 m³/s; sensible = 0.0167 × -30.0 × 1.23 = -0.615 kW; latent = 0.0167 × -0.00438 × 3010 = -0.220 kW |
| Total | Sensible -1.959 + latent -0.220 = -2.179 kW; largest contributor: equipment |
| Capacity | Net heat loss: required = max(0, -2.179 × (1 + 15/100)) = 0.000 kW — no cooling required, heating may be required; 0 active + 1 standby = 1 × 10 kW (installed 10.0 kW) PASS |
| Result | 1/1 checks pass → PASS; net heat loss 2.18 kW → no cooling required, heating may be required; governing: cooling capacity. Net heat loss of 2.18 kW: the room loses more heat than it gains, so no cooling is required at this design condition — heating may be required; check the summer design condition for the cooling load. |
| Reference | Hand calculation: floor 8 × 5 = 40 m², volume 40 × 3 = 120 m³; equipment = 0.500 kW; conduction ΔT -30 K × ΣUA 74.80 W/K = -2.244 kW (walls 78 m², roof 40, floor 40); solar = 0.000 kW (sunlit roof 0 m²); lighting 40 m² × 10 W/m² = 0.400 kW; people 0 × 0.130 = 0.000 kW; infiltration V̇ = 120 × 0.5/3600 = 0.01667 m³/s, sensible × -30 × 1.23 = -0.615 kW, Tetens w_out = 0.00141, w_in = 0.00579, latent × -0.00438 × 3010 = -0.220 kW; sensible -1.959 + latent -0.220 = total -2.179 kW; net heat loss → required max(0, -2.179 × 1.15) = 0 kW → 0 active units + 1 standby; duty capacity 0 so utilisation is not defined; largest contributor equipment (validation case c0012-net-heat-loss-cold-day) |
Example 13 — Default switchroom with a sunlit wall
The calculator's default room with 40 m² of sunlit wall: solar gain on the whole roof dominates the load.
Given: Room 11 × 7 × 3 m; outside 35 °C / 60 % RH, inside 22 °C / 50 % RH. Equipment: VSD 132 kW × 1 at 3 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 40 m² wall at 150 W/m², whole roof at 300 W/m². 1 occupant(s), lighting 10 W/m², 0.5 air changes/h, safety factor 15 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 11 × 7 = 77.0 m²; Awall = 2 × (11 + 7) × 3 = 108.0 m²; V = Afloor × H = 77.0 × 3 = 231.0 m³; ΔT = 35 − 22 = 13.0 K |
|---|---|
| Equipment | Qequip = 132 × 1 × 0.03 = 3.960 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 108.0 + 0.4 × 77.0 + 0.3 × 77.0) × 13.0 / 1000 = 1.543 kW |
| Solar | Qsolar = (40 × 150 + 77 × 300 + 0 × 250) / 1000 = 29.100 kW |
| Lighting | Qlight = 77.0 m² × 10 W/m² / 1000 = 0.770 kW |
| People | Qpeople = 1 × (0.075 sensible + 0.055 latent) = 0.130 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(35 °C) = 5,631 Pa, Pv = 5,631 × 60/100 = 3,379 Pa, w = 0.622 × 3,379 / (101,325 − 3,379) = 0.02146; inside Psat(22 °C) = 2,643 Pa, Pv = 2,643 × 50/100 = 1,321 Pa, w = 0.622 × 1,321 / (101,325 − 1,321) = 0.00822; Δw = 0.02146 − 0.00822 = 0.01324 kg/kg |
| Infiltration | V̇ = 231.0 × 0.5 / 3600 = 0.0321 m³/s; sensible = 0.0321 × 13.0 × 1.23 = 0.513 kW; latent = 0.0321 × 0.01324 × 3010 = 1.278 kW |
| Total | Sensible 35.961 + latent 1.333 = 37.294 kW; largest contributor: solar |
| Capacity | Required = 37.294 × (1 + 15/100) = 42.889 kW; units = ⌈42.889 / 10⌉ = 5 active + 0 standby = 5 × 10 kW (installed 50.0 kW, duty 50.0 kW at 85.8 %) PASS |
| Result | 1/1 checks pass → PASS; required 42.89 kW → 5 units; governing: cooling capacity. Solar gain (29.1 kW) dominates the load; the simplified solar model is conservative — check the sunlit areas and solar factors, or shade the roof. |
| Reference | Hand calculation: equipment 132 × 1 × 0.03 = 3.960 kW; conduction ΔT 13 K × ΣUA 118.70 W/K = 1.543 kW (walls 108 m² × 0.6, roof 77 × 0.4, floor 77 × 0.3); solar (40 × 150 + 77 × 300)/1000 = 29.100 kW; lighting 77 m² × 10 W/m² = 0.770 kW; people 1 × (0.075 + 0.055) = 0.130 kW; infiltration V̇ = 231 × 0.5/3600 = 0.03208 m³/s, sensible 0.03208 × 13 × 1.23 = 0.513 kW, Tetens P_sat(35 °C) = 5631 Pa → w_out = 0.02146, P_sat(22 °C) = 2643 Pa → w_in = 0.00822, Δw = 0.01324 kg/kg, latent 0.03208 × 0.01324 × 3010 = 1.278 kW; sensible 35.961 + latent 1.333 = total 37.294 kW; required 37.294 × 1.15 = 42.889 kW → ⌈42.889/10⌉ = 5 active + 0 standby = 5 × 10 kW (50 kW installed, duty 50 kW at 85.8 %); largest contributor solar (validation case c0012-outdoor-vsd-room) |
Example 14 — Partly shaded roof and sunlit wall
Only 40 m² of the 120 m² roof sees the sun (the rest is under a canopy): the sunlit roof area is given instead of defaulting to the whole roof.
Given: Room 12 × 10 × 4 m; outside 38 °C / 60 % RH, inside 24 °C / 50 % RH. Equipment: Switchboard 1,600 kW × 2 at 0.8 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 30 m² wall at 150 W/m², 40 m² roof at 300 W/m². 1 occupant(s), lighting 10 W/m², 0.5 air changes/h, safety factor 15 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 12 × 10 = 120.0 m²; Awall = 2 × (12 + 10) × 4 = 176.0 m²; V = Afloor × H = 120.0 × 4 = 480.0 m³; ΔT = 38 − 24 = 14.0 K |
|---|---|
| Equipment | Qequip = 1,600 × 2 × 0.008 = 25.600 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 176.0 + 0.4 × 120.0 + 0.3 × 120.0) × 14.0 / 1000 = 2.654 kW |
| Solar | Qsolar = (30 × 150 + 40 × 300 + 0 × 250) / 1000 = 16.500 kW |
| Lighting | Qlight = 120.0 m² × 10 W/m² / 1000 = 1.200 kW |
| People | Qpeople = 1 × (0.075 sensible + 0.055 latent) = 0.130 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(38 °C) = 6,639 Pa, Pv = 6,639 × 60/100 = 3,983 Pa, w = 0.622 × 3,983 / (101,325 − 3,983) = 0.02545; inside Psat(24 °C) = 2,983 Pa, Pv = 2,983 × 50/100 = 1,492 Pa, w = 0.622 × 1,492 / (101,325 − 1,492) = 0.00929; Δw = 0.02545 − 0.00929 = 0.01616 kg/kg |
| Infiltration | V̇ = 480.0 × 0.5 / 3600 = 0.0667 m³/s; sensible = 0.0667 × 14.0 × 1.23 = 1.148 kW; latent = 0.0667 × 0.01616 × 3010 = 3.243 kW |
| Total | Sensible 47.177 + latent 3.298 = 50.475 kW; largest contributor: equipment |
| Capacity | Required = 50.475 × (1 + 15/100) = 58.046 kW; units = ⌈58.046 / 10⌉ = 6 active + 0 standby = 6 × 10 kW (installed 60.0 kW, duty 60.0 kW at 96.7 %) PASS |
| Result | 1/1 checks pass → PASS; required 58.05 kW → 6 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 1600 × 2 × 0.008 = 25.600 kW; conduction ΔT 14 K × ΣUA 189.60 W/K = 2.654 kW (walls 176 m² × 0.6, roof 120 × 0.4, floor 120 × 0.3); solar (30 × 150 + 40 × 300)/1000 = 16.500 kW; lighting 120 m² × 10 W/m² = 1.200 kW; people 1 × (0.075 + 0.055) = 0.130 kW; infiltration V̇ = 480 × 0.5/3600 = 0.06667 m³/s, sensible 0.06667 × 14 × 1.23 = 1.148 kW, Tetens P_sat(38 °C) = 6639 Pa → w_out = 0.02545, P_sat(24 °C) = 2983 Pa → w_in = 0.00929, Δw = 0.01616 kg/kg, latent 0.06667 × 0.01616 × 3010 = 3.243 kW; sensible 47.177 + latent 3.298 = total 50.475 kW; required 50.475 × 1.15 = 58.046 kW → ⌈58.046/10⌉ = 6 active + 0 standby = 6 × 10 kW (60 kW installed, duty 60 kW at 96.7 %); largest contributor equipment (validation case c0012-partial-sunlit-roof) |
Example 15 — Site solar factors on a light-coloured roof, no margin
Measured solar factors (90 W/m² wall, 120 W/m² roof) replace the conservative defaults and the safety factor is set to 0 %, so required equals total.
Given: Room 10 × 8 × 3.5 m; outside 35 °C / 60 % RH, inside 24 °C / 50 % RH. Equipment: Transformer 800 kW × 2 at 1.2 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 25 m² wall at 90 W/m², whole roof at 120 W/m². 1 occupant(s), lighting 10 W/m², 0.5 air changes/h, safety factor 0 %, AC units of 15 kW with 0 standby.
| Geometry | Afloor = 10 × 8 = 80.0 m²; Awall = 2 × (10 + 8) × 3.5 = 126.0 m²; V = Afloor × H = 80.0 × 3.5 = 280.0 m³; ΔT = 35 − 24 = 11.0 K |
|---|---|
| Equipment | Qequip = 800 × 2 × 0.012 = 19.200 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 126.0 + 0.4 × 80.0 + 0.3 × 80.0) × 11.0 / 1000 = 1.448 kW |
| Solar | Qsolar = (25 × 90 + 80 × 120 + 0 × 250) / 1000 = 11.850 kW |
| Lighting | Qlight = 80.0 m² × 10 W/m² / 1000 = 0.800 kW |
| People | Qpeople = 1 × (0.075 sensible + 0.055 latent) = 0.130 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(35 °C) = 5,631 Pa, Pv = 5,631 × 60/100 = 3,379 Pa, w = 0.622 × 3,379 / (101,325 − 3,379) = 0.02146; inside Psat(24 °C) = 2,983 Pa, Pv = 2,983 × 50/100 = 1,492 Pa, w = 0.622 × 1,492 / (101,325 − 1,492) = 0.00929; Δw = 0.02146 − 0.00929 = 0.01216 kg/kg |
| Infiltration | V̇ = 280.0 × 0.5 / 3600 = 0.0389 m³/s; sensible = 0.0389 × 11.0 × 1.23 = 0.526 kW; latent = 0.0389 × 0.01216 × 3010 = 1.424 kW |
| Total | Sensible 33.899 + latent 1.479 = 35.377 kW; largest contributor: equipment |
| Capacity | Required = 35.377 × (1 + 0/100) = 35.377 kW; units = ⌈35.377 / 15⌉ = 3 active + 0 standby = 3 × 15 kW (installed 45.0 kW, duty 45.0 kW at 78.6 %) PASS |
| Result | 1/1 checks pass → PASS; required 35.38 kW → 3 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 800 × 2 × 0.012 = 19.200 kW; conduction ΔT 11 K × ΣUA 131.60 W/K = 1.448 kW (walls 126 m² × 0.6, roof 80 × 0.4, floor 80 × 0.3); solar (25 × 90 + 80 × 120)/1000 = 11.850 kW; lighting 80 m² × 10 W/m² = 0.800 kW; people 1 × (0.075 + 0.055) = 0.130 kW; infiltration V̇ = 280 × 0.5/3600 = 0.03889 m³/s, sensible 0.03889 × 11 × 1.23 = 0.526 kW, Tetens P_sat(35 °C) = 5631 Pa → w_out = 0.02146, P_sat(24 °C) = 2983 Pa → w_in = 0.00929, Δw = 0.01216 kg/kg, latent 0.03889 × 0.01216 × 3010 = 1.424 kW; sensible 33.899 + latent 1.479 = total 35.377 kW; required 35.377 × 1 = 35.377 kW → ⌈35.377/15⌉ = 3 active + 0 standby = 3 × 15 kW (45 kW installed, duty 45 kW at 78.6 %); largest contributor equipment (validation case c0012-solar-factors-override-no-margin) |
Example 16 — Training room: thirty occupants
No equipment and no solar: thirty people (0.075 kW sensible + 0.055 kW latent each) are the largest contributor and the latent share is significant.
Given: Room 10 × 8 × 3 m; outside 30 °C / 60 % RH, inside 24 °C / 50 % RH. Equipment: no equipment. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 30 occupant(s), lighting 8 W/m², 1 air changes/h, safety factor 15 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 10 × 8 = 80.0 m²; Awall = 2 × (10 + 8) × 3 = 108.0 m²; V = Afloor × H = 80.0 × 3 = 240.0 m³; ΔT = 30 − 24 = 6.0 K |
|---|---|
| Equipment | Qequip = 0 = 0.000 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 108.0 + 0.4 × 80.0 + 0.3 × 80.0) × 6.0 / 1000 = 0.725 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 80.0 m² × 8 W/m² / 1000 = 0.640 kW |
| People | Qpeople = 30 × (0.075 sensible + 0.055 latent) = 3.900 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(30 °C) = 4,246 Pa, Pv = 4,246 × 60/100 = 2,547 Pa, w = 0.622 × 2,547 / (101,325 − 2,547) = 0.01604; inside Psat(24 °C) = 2,983 Pa, Pv = 2,983 × 50/100 = 1,492 Pa, w = 0.622 × 1,492 / (101,325 − 1,492) = 0.00929; Δw = 0.01604 − 0.00929 = 0.00675 kg/kg |
| Infiltration | V̇ = 240.0 × 1 / 3600 = 0.0667 m³/s; sensible = 0.0667 × 6.0 × 1.23 = 0.492 kW; latent = 0.0667 × 0.00675 × 3010 = 1.354 kW |
| Total | Sensible 4.107 + latent 3.004 = 7.111 kW; largest contributor: people |
| Capacity | Required = 7.111 × (1 + 15/100) = 8.177 kW; units = ⌈8.177 / 10⌉ = 1 active + 0 standby = 1 × 10 kW (installed 10.0 kW, duty 10.0 kW at 81.8 %) PASS |
| Result | 1/1 checks pass → PASS; required 8.18 kW → 1 units; governing: cooling capacity |
| Reference | Hand calculation: equipment none = 0.000 kW; conduction ΔT 6 K × ΣUA 120.80 W/K = 0.725 kW (walls 108 m² × 0.6, roof 80 × 0.4, floor 80 × 0.3); solar (0 × 150 + 0 × 300)/1000 = 0.000 kW; lighting 80 m² × 8 W/m² = 0.640 kW; people 30 × (0.075 + 0.055) = 3.900 kW; infiltration V̇ = 240 × 1/3600 = 0.06667 m³/s, sensible 0.06667 × 6 × 1.23 = 0.492 kW, Tetens P_sat(30 °C) = 4246 Pa → w_out = 0.01604, P_sat(24 °C) = 2983 Pa → w_in = 0.00929, Δw = 0.00675 kg/kg, latent 0.06667 × 0.00675 × 3010 = 1.354 kW; sensible 4.107 + latent 3.004 = total 7.111 kW; required 7.111 × 1.15 = 8.177 kW → ⌈8.177/10⌉ = 1 active + 0 standby = 1 × 10 kW (10 kW installed, duty 10 kW at 81.8 %); largest contributor people (validation case c0012-training-room-occupants) |
Example 17 — Tropical site, 95 % RH outside, 2 ACH
Very humid outside air at a high air-change rate: the latent infiltration term from the humidity-ratio difference (Tetens) exceeds the sensible one.
Given: Room 10 × 6 × 3.5 m; outside 34 °C / 95 % RH, inside 24 °C / 50 % RH. Equipment: Switchboard 2,000 kW × 1 at 0.5 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 1 occupant(s), lighting 10 W/m², 2 air changes/h, safety factor 15 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 10 × 6 = 60.0 m²; Awall = 2 × (10 + 6) × 3.5 = 112.0 m²; V = Afloor × H = 60.0 × 3.5 = 210.0 m³; ΔT = 34 − 24 = 10.0 K |
|---|---|
| Equipment | Qequip = 2,000 × 1 × 0.005 = 10.000 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 112.0 + 0.4 × 60.0 + 0.3 × 60.0) × 10.0 / 1000 = 1.092 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 60.0 m² × 10 W/m² / 1000 = 0.600 kW |
| People | Qpeople = 1 × (0.075 sensible + 0.055 latent) = 0.130 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(34 °C) = 5,326 Pa, Pv = 5,326 × 95/100 = 5,060 Pa, w = 0.622 × 5,060 / (101,325 − 5,060) = 0.03269; inside Psat(24 °C) = 2,983 Pa, Pv = 2,983 × 50/100 = 1,492 Pa, w = 0.622 × 1,492 / (101,325 − 1,492) = 0.00929; Δw = 0.03269 − 0.00929 = 0.02340 kg/kg |
| Infiltration | V̇ = 210.0 × 2 / 3600 = 0.1167 m³/s; sensible = 0.1167 × 10.0 × 1.23 = 1.435 kW; latent = 0.1167 × 0.02340 × 3010 = 8.217 kW |
| Total | Sensible 13.202 + latent 8.272 = 21.474 kW; largest contributor: equipment |
| Capacity | Required = 21.474 × (1 + 15/100) = 24.696 kW; units = ⌈24.696 / 10⌉ = 3 active + 0 standby = 3 × 10 kW (installed 30.0 kW, duty 30.0 kW at 82.3 %) PASS |
| Result | 1/1 checks pass → PASS; required 24.70 kW → 3 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 2000 × 1 × 0.005 = 10.000 kW; conduction ΔT 10 K × ΣUA 109.20 W/K = 1.092 kW (walls 112 m² × 0.6, roof 60 × 0.4, floor 60 × 0.3); solar (0 × 150 + 0 × 300)/1000 = 0.000 kW; lighting 60 m² × 10 W/m² = 0.600 kW; people 1 × (0.075 + 0.055) = 0.130 kW; infiltration V̇ = 210 × 2/3600 = 0.11667 m³/s, sensible 0.11667 × 10 × 1.23 = 1.435 kW, Tetens P_sat(34 °C) = 5326 Pa → w_out = 0.03269, P_sat(24 °C) = 2983 Pa → w_in = 0.00929, Δw = 0.02340 kg/kg, latent 0.11667 × 0.02340 × 3010 = 8.217 kW; sensible 13.202 + latent 8.272 = total 21.474 kW; required 21.474 × 1.15 = 24.696 kW → ⌈24.696/10⌉ = 3 active + 0 standby = 3 × 10 kW (30 kW installed, duty 30 kW at 82.3 %); largest contributor equipment (validation case c0012-tropical-humid-latent) |
Example 18 — Uninsulated steel shed under a canopy
Single-skin steel walls (U 5.0) and roof (U 6.0) over a slab: fabric conduction at 17 K is the largest contributor even with no direct sun.
Given: Room 8 × 6 × 3 m; outside 42 °C / 30 % RH, inside 25 °C / 50 % RH. Equipment: VSD 75 kW × 2 at 3 % heat. U-values wall 5, roof 6, floor 0.5 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 1 occupant(s), lighting 10 W/m², 0.5 air changes/h, safety factor 15 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 8 × 6 = 48.0 m²; Awall = 2 × (8 + 6) × 3 = 84.0 m²; V = Afloor × H = 48.0 × 3 = 144.0 m³; ΔT = 42 − 25 = 17.0 K |
|---|---|
| Equipment | Qequip = 75 × 2 × 0.03 = 4.500 kW |
| Conduction | Qcond = Σ U·A·ΔT = (5 × 84.0 + 6 × 48.0 + 0.5 × 48.0) × 17.0 / 1000 = 12.444 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 48.0 m² × 10 W/m² / 1000 = 0.480 kW |
| People | Qpeople = 1 × (0.075 sensible + 0.055 latent) = 0.130 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(42 °C) = 8,224 Pa, Pv = 8,224 × 30/100 = 2,467 Pa, w = 0.622 × 2,467 / (101,325 − 2,467) = 0.01552; inside Psat(25 °C) = 3,167 Pa, Pv = 3,167 × 50/100 = 1,584 Pa, w = 0.622 × 1,584 / (101,325 − 1,584) = 0.00988; Δw = 0.01552 − 0.00988 = 0.00565 kg/kg |
| Infiltration | V̇ = 144.0 × 0.5 / 3600 = 0.0200 m³/s; sensible = 0.0200 × 17.0 × 1.23 = 0.418 kW; latent = 0.0200 × 0.00565 × 3010 = 0.340 kW |
| Total | Sensible 17.917 + latent 0.395 = 18.312 kW; largest contributor: conduction |
| Capacity | Required = 18.312 × (1 + 15/100) = 21.059 kW; units = ⌈21.059 / 10⌉ = 3 active + 0 standby = 3 × 10 kW (installed 30.0 kW, duty 30.0 kW at 70.2 %) PASS |
| Result | 1/1 checks pass → PASS; required 21.06 kW → 3 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 75 × 2 × 0.03 = 4.500 kW; conduction ΔT 17 K × ΣUA 732.00 W/K = 12.444 kW (walls 84 m² × 5, roof 48 × 6, floor 48 × 0.5); solar (0 × 150 + 0 × 300)/1000 = 0.000 kW; lighting 48 m² × 10 W/m² = 0.480 kW; people 1 × (0.075 + 0.055) = 0.130 kW; infiltration V̇ = 144 × 0.5/3600 = 0.02000 m³/s, sensible 0.02000 × 17 × 1.23 = 0.418 kW, Tetens P_sat(42 °C) = 8224 Pa → w_out = 0.01552, P_sat(25 °C) = 3167 Pa → w_in = 0.00988, Δw = 0.00565 kg/kg, latent 0.02000 × 0.00565 × 3010 = 0.340 kW; sensible 17.917 + latent 0.395 = total 18.312 kW; required 18.312 × 1.15 = 21.059 kW → ⌈21.059/10⌉ = 3 active + 0 standby = 3 × 10 kW (30 kW installed, duty 30 kW at 70.2 %); largest contributor conduction (validation case c0012-uninsulated-steel-shed) |
Example 19 — Required capacity just over two 10 kW units
The same room with 20 W more equipment heat: 20.01 kW tips the ceiling to a third unit at 66.7 % utilisation.
Given: Room 10 × 5 × 3 m; outside 35 °C / 60 % RH, inside 25 °C / 50 % RH. Equipment: Rectifier 931 kW × 1 at 2 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 0 occupant(s), lighting 10 W/m², 0 air changes/h, safety factor 0 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 10 × 5 = 50.0 m²; Awall = 2 × (10 + 5) × 3 = 90.0 m²; V = Afloor × H = 50.0 × 3 = 150.0 m³; ΔT = 35 − 25 = 10.0 K |
|---|---|
| Equipment | Qequip = 931 × 1 × 0.02 = 18.620 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 90.0 + 0.4 × 50.0 + 0.3 × 50.0) × 10.0 / 1000 = 0.890 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 50.0 m² × 10 W/m² / 1000 = 0.500 kW |
| People | Qpeople = 0 × (0.075 sensible + 0.055 latent) = 0.000 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(35 °C) = 5,631 Pa, Pv = 5,631 × 60/100 = 3,379 Pa, w = 0.622 × 3,379 / (101,325 − 3,379) = 0.02146; inside Psat(25 °C) = 3,167 Pa, Pv = 3,167 × 50/100 = 1,584 Pa, w = 0.622 × 1,584 / (101,325 − 1,584) = 0.00988; Δw = 0.02146 − 0.00988 = 0.01158 kg/kg |
| Infiltration | V̇ = 150.0 × 0 / 3600 = 0.0000 m³/s; sensible = 0.0000 × 10.0 × 1.23 = 0.000 kW; latent = 0.0000 × 0.01158 × 3010 = 0.000 kW |
| Total | Sensible 20.010 + latent 0.000 = 20.010 kW; largest contributor: equipment |
| Capacity | Required = 20.010 × (1 + 0/100) = 20.010 kW; units = ⌈20.010 / 10⌉ = 3 active + 0 standby = 3 × 10 kW (installed 30.0 kW, duty 30.0 kW at 66.7 %) PASS |
| Result | 1/1 checks pass → PASS; required 20.01 kW → 3 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 931 × 1 × 0.02 = 18.620 kW; conduction ΔT 10 K × ΣUA 89.00 W/K = 0.890 kW (walls 90 m² × 0.6, roof 50 × 0.4, floor 50 × 0.3); solar (0 × 150 + 0 × 300)/1000 = 0.000 kW; lighting 50 m² × 10 W/m² = 0.500 kW; people 0 × (0.075 + 0.055) = 0.000 kW; infiltration V̇ = 150 × 0/3600 = 0.00000 m³/s, sensible 0.00000 × 10 × 1.23 = 0.000 kW, Tetens P_sat(35 °C) = 5631 Pa → w_out = 0.02146, P_sat(25 °C) = 3167 Pa → w_in = 0.00988, Δw = 0.01158 kg/kg, latent 0.00000 × 0.01158 × 3010 = 0.000 kW; sensible 20.010 + latent 0.000 = total 20.010 kW; required 20.010 × 1 = 20.010 kW → ⌈20.010/10⌉ = 3 active + 0 standby = 3 × 10 kW (30 kW installed, duty 30 kW at 66.7 %); largest contributor equipment (validation case c0012-unit-boundary-just-above) |
Example 20 — Required capacity just under two 10 kW units
With no margin, no infiltration and no occupants the load is 19.99 kW: ⌈19.99/10⌉ = 2 units at 99.9 % utilisation.
Given: Room 10 × 5 × 3 m; outside 35 °C / 60 % RH, inside 25 °C / 50 % RH. Equipment: Rectifier 310 kW × 1 at 6 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 0 occupant(s), lighting 10 W/m², 0 air changes/h, safety factor 0 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 10 × 5 = 50.0 m²; Awall = 2 × (10 + 5) × 3 = 90.0 m²; V = Afloor × H = 50.0 × 3 = 150.0 m³; ΔT = 35 − 25 = 10.0 K |
|---|---|
| Equipment | Qequip = 310 × 1 × 0.06 = 18.600 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 90.0 + 0.4 × 50.0 + 0.3 × 50.0) × 10.0 / 1000 = 0.890 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 50.0 m² × 10 W/m² / 1000 = 0.500 kW |
| People | Qpeople = 0 × (0.075 sensible + 0.055 latent) = 0.000 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(35 °C) = 5,631 Pa, Pv = 5,631 × 60/100 = 3,379 Pa, w = 0.622 × 3,379 / (101,325 − 3,379) = 0.02146; inside Psat(25 °C) = 3,167 Pa, Pv = 3,167 × 50/100 = 1,584 Pa, w = 0.622 × 1,584 / (101,325 − 1,584) = 0.00988; Δw = 0.02146 − 0.00988 = 0.01158 kg/kg |
| Infiltration | V̇ = 150.0 × 0 / 3600 = 0.0000 m³/s; sensible = 0.0000 × 10.0 × 1.23 = 0.000 kW; latent = 0.0000 × 0.01158 × 3010 = 0.000 kW |
| Total | Sensible 19.990 + latent 0.000 = 19.990 kW; largest contributor: equipment |
| Capacity | Required = 19.990 × (1 + 0/100) = 19.990 kW; units = ⌈19.990 / 10⌉ = 2 active + 0 standby = 2 × 10 kW (installed 20.0 kW, duty 20.0 kW at 99.9 %) PASS |
| Result | 1/1 checks pass → PASS; required 19.99 kW → 2 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 310 × 1 × 0.06 = 18.600 kW; conduction ΔT 10 K × ΣUA 89.00 W/K = 0.890 kW (walls 90 m² × 0.6, roof 50 × 0.4, floor 50 × 0.3); solar (0 × 150 + 0 × 300)/1000 = 0.000 kW; lighting 50 m² × 10 W/m² = 0.500 kW; people 0 × (0.075 + 0.055) = 0.000 kW; infiltration V̇ = 150 × 0/3600 = 0.00000 m³/s, sensible 0.00000 × 10 × 1.23 = 0.000 kW, Tetens P_sat(35 °C) = 5631 Pa → w_out = 0.02146, P_sat(25 °C) = 3167 Pa → w_in = 0.00988, Δw = 0.01158 kg/kg, latent 0.00000 × 0.01158 × 3010 = 0.000 kW; sensible 19.990 + latent 0.000 = total 19.990 kW; required 19.990 × 1 = 19.990 kW → ⌈19.990/10⌉ = 2 active + 0 standby = 2 × 10 kW (20 kW installed, duty 20 kW at 99.9 %); largest contributor equipment (validation case c0012-unit-boundary-just-below) |
Example 21 — Leaky shed at 10 air changes per hour
A small, poorly sealed room with little equipment: infiltration (sensible plus latent) is the largest contributor.
Given: Room 6 × 4 × 3 m; outside 40 °C / 60 % RH, inside 25 °C / 50 % RH. Equipment: Battery charger 5 kW × 1 at 5 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 1 occupant(s), lighting 10 W/m², 10 air changes/h, safety factor 15 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 6 × 4 = 24.0 m²; Awall = 2 × (6 + 4) × 3 = 60.0 m²; V = Afloor × H = 24.0 × 3 = 72.0 m³; ΔT = 40 − 25 = 15.0 K |
|---|---|
| Equipment | Qequip = 5 × 1 × 0.05 = 0.250 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 60.0 + 0.4 × 24.0 + 0.3 × 24.0) × 15.0 / 1000 = 0.792 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 24.0 m² × 10 W/m² / 1000 = 0.240 kW |
| People | Qpeople = 1 × (0.075 sensible + 0.055 latent) = 0.130 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(40 °C) = 7,395 Pa, Pv = 7,395 × 60/100 = 4,437 Pa, w = 0.622 × 4,437 / (101,325 − 4,437) = 0.02848; inside Psat(25 °C) = 3,167 Pa, Pv = 3,167 × 50/100 = 1,584 Pa, w = 0.622 × 1,584 / (101,325 − 1,584) = 0.00988; Δw = 0.02848 − 0.00988 = 0.01861 kg/kg |
| Infiltration | V̇ = 72.0 × 10 / 3600 = 0.2000 m³/s; sensible = 0.2000 × 15.0 × 1.23 = 3.690 kW; latent = 0.2000 × 0.01861 × 3010 = 11.202 kW |
| Total | Sensible 5.047 + latent 11.257 = 16.304 kW; largest contributor: infiltration |
| Capacity | Required = 16.304 × (1 + 15/100) = 18.750 kW; units = ⌈18.750 / 10⌉ = 2 active + 0 standby = 2 × 10 kW (installed 20.0 kW, duty 20.0 kW at 93.8 %) PASS |
| Result | 1/1 checks pass → PASS; required 18.75 kW → 2 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 5 × 1 × 0.05 = 0.250 kW; conduction ΔT 15 K × ΣUA 52.80 W/K = 0.792 kW (walls 60 m² × 0.6, roof 24 × 0.4, floor 24 × 0.3); solar (0 × 150 + 0 × 300)/1000 = 0.000 kW; lighting 24 m² × 10 W/m² = 0.240 kW; people 1 × (0.075 + 0.055) = 0.130 kW; infiltration V̇ = 72 × 10/3600 = 0.20000 m³/s, sensible 0.20000 × 15 × 1.23 = 3.690 kW, Tetens P_sat(40 °C) = 7395 Pa → w_out = 0.02848, P_sat(25 °C) = 3167 Pa → w_in = 0.00988, Δw = 0.01861 kg/kg, latent 0.20000 × 0.01861 × 3010 = 11.202 kW; sensible 5.047 + latent 11.257 = total 16.304 kW; required 16.304 × 1.15 = 18.750 kW → ⌈18.750/10⌉ = 2 active + 0 standby = 2 × 10 kW (20 kW installed, duty 20 kW at 93.7 %); largest contributor infiltration (validation case c0012-ventilated-shed-high-ach) |
Example 22 — Winter design day: fabric losses offset transformer heat
Outside colder than the room makes the conduction and both infiltration terms negative; the 24 kW of transformer losses still leave a net cooling load.
Given: Room 10 × 6 × 4 m; outside 5 °C / 80 % RH, inside 22 °C / 40 % RH. Equipment: Transformer 1,000 kW × 2 at 1.2 % heat. U-values wall 0.6, roof 0.4, floor 0.3 W/m²·K. Sunlit: 0 m² wall at 150 W/m², 0 m² roof at 300 W/m². 0 occupant(s), lighting 10 W/m², 0.5 air changes/h, safety factor 15 %, AC units of 10 kW with 0 standby.
| Geometry | Afloor = 10 × 6 = 60.0 m²; Awall = 2 × (10 + 6) × 4 = 128.0 m²; V = Afloor × H = 60.0 × 4 = 240.0 m³; ΔT = 5 − 22 = -17.0 K |
|---|---|
| Equipment | Qequip = 1,000 × 2 × 0.012 = 24.000 kW |
| Conduction | Qcond = Σ U·A·ΔT = (0.6 × 128.0 + 0.4 × 60.0 + 0.3 × 60.0) × -17.0 / 1000 = -2.020 kW |
| Solar | Qsolar = (0 × 150 + 0 × 300 + 0 × 250) / 1000 = 0.000 kW |
| Lighting | Qlight = 60.0 m² × 10 W/m² / 1000 = 0.600 kW |
| People | Qpeople = 0 × (0.075 sensible + 0.055 latent) = 0.000 kW |
| Humidity | Tetens Psat = 611.2 × e17.67·T/(T+243.5): outside Psat(5 °C) = 872 Pa, Pv = 872 × 80/100 = 698 Pa, w = 0.622 × 698 / (101,325 − 698) = 0.00431; inside Psat(22 °C) = 2,643 Pa, Pv = 2,643 × 40/100 = 1,057 Pa, w = 0.622 × 1,057 / (101,325 − 1,057) = 0.00656; Δw = 0.00431 − 0.00656 = -0.00225 kg/kg |
| Infiltration | V̇ = 240.0 × 0.5 / 3600 = 0.0333 m³/s; sensible = 0.0333 × -17.0 × 1.23 = -0.697 kW; latent = 0.0333 × -0.00225 × 3010 = -0.225 kW |
| Total | Sensible 21.883 + latent -0.225 = 21.658 kW; largest contributor: equipment |
| Capacity | Required = 21.658 × (1 + 15/100) = 24.907 kW; units = ⌈24.907 / 10⌉ = 3 active + 0 standby = 3 × 10 kW (installed 30.0 kW, duty 30.0 kW at 83.0 %) PASS |
| Result | 1/1 checks pass → PASS; required 24.91 kW → 3 units; governing: cooling capacity |
| Reference | Hand calculation: equipment 1000 × 2 × 0.012 = 24.000 kW; conduction ΔT -17 K × ΣUA 118.80 W/K = -2.020 kW (walls 128 m² × 0.6, roof 60 × 0.4, floor 60 × 0.3); solar (0 × 150 + 0 × 300)/1000 = 0.000 kW; lighting 60 m² × 10 W/m² = 0.600 kW; people 0 × (0.075 + 0.055) = 0.000 kW; infiltration V̇ = 240 × 0.5/3600 = 0.03333 m³/s, sensible 0.03333 × -17 × 1.23 = -0.697 kW, Tetens P_sat(5 °C) = 872 Pa → w_out = 0.00431, P_sat(22 °C) = 2643 Pa → w_in = 0.00656, Δw = -0.00225 kg/kg, latent 0.03333 × -0.00225 × 3010 = -0.225 kW; sensible 21.883 + latent -0.225 = total 21.658 kW; required 21.658 × 1.15 = 24.907 kW → ⌈24.907/10⌉ = 3 active + 0 standby = 3 × 10 kW (30 kW installed, duty 30 kW at 83.0 %); largest contributor equipment (validation case c0012-winter-cold-climate) |
Standards & References
- ASHRAE Handbook — Fundamentals — Heat load calculation methodology, cooling load temperature differences, personnel heat gain values
- Tetens' Equation (1930) — Saturation vapour pressure approximation, valid −40°C to +60°C
- ASHRAE Standard 62.1 — Ventilation for Acceptable Indoor Air Quality (infiltration rates)
- ISO 6946 — Building components — Thermal resistance and thermal transmittance (U-values)
- IEC 62271-1 — High-voltage switchgear and controlgear — Common specifications (switchroom environmental requirements)
Tips & Limitations
Tips
- For VSDs and power electronics, the load factor is typically the inverse of the efficiency. A 97% efficient VSD has a load factor of 0.03 (3% losses dissipated as heat).
- Solar gains often dominate the heat load for exposed buildings. Consider roof insulation, light-coloured roofing, and shade structures to reduce cooling requirements.
- Set
wall_area,roof_area, andfloor_areato zero to auto-calculate from room dimensions. Override with actual values for non-rectangular layouts; a given floor area also sets the lighting load, the room volume (\(A_{floor} \times H\)) and the default roof area. - For switchrooms inside larger buildings (no external exposure), set solar areas to zero and use a smaller \(\Delta T\) based on the surrounding building temperature.
- The sensible heat ratio (SHR) helps in selecting the right type of cooling. SHR > 0.9 (typical for switchrooms) favours sensible cooling units; lower SHR may need dehumidification.
- Use N+1 redundancy (
redundant_units = 1) for critical installations to maintain cooling during maintenance or unit failure.
Limitations
- Steady-state only — transient loads (motor starting, fault events) and thermal mass effects are not modelled.
- Simplified solar model — uses constant factors (W/m²) rather than accounting for latitude, orientation, time of day, or season.
- Floor conduction — uses the full outside–inside \(\Delta T\), which is conservative. The ground temperature is typically between the two.
- Standard atmosphere — humidity calculations assume 101,325 Pa atmospheric pressure. For high-altitude installations, actual results will differ.
- No equipment diversity — all equipment is assumed to operate simultaneously at the specified load factor. No diversity reduction is applied.
- Mixing unit systems — all inputs must use the same unit system. The calculator will not detect mixed inputs.
Developer Guide
Architecture
File layout:
components/c0012_switchroom_heat_load/component.py— Component registration, parameter schema, calculator class, equation builder, summary cardscomponents/c0012_switchroom_heat_load/calculations.py— Pure thermodynamic functions (no Django dependencies)components/c0012_switchroom_heat_load/equations.py— LaTeX equation builder for report generationcomponents/c0012_switchroom_heat_load/excel_generator.py— Excel report generation with formatted tables
The calculator class is SwitchroomHeatLoadCalculator,
inheriting from BaseCalculator (not
ElectricalCalculator — this is an HVAC calculation).
All internal calculations use metric (SI) units with conversion
wrappers for imperial display.
Constants are defined at module level in
calculations.py:
ATMOSPHERIC_PRESSURE_PA= 101,325PERSONNEL_SENSIBLE_HEAT_KW= 0.075PERSONNEL_LATENT_HEAT_KW= 0.055AIR_SPECIFIC_HEAT_KJ_KG_K= 1.006AIR_DENSITY_KG_M3= 1.2LATENT_HEAT_VAPORIZATION_KJ_KG= 2501
Code Paths
Key functions in calculations.py:
| Function | Purpose |
|---|---|
calculate_switchroom_heat_load(data) |
Main entry point; orchestrates all sub-calculations and returns comprehensive results |
calculate_equipment_load(equipment_list) |
Sum equipment heat: rating × qty × load_factor |
calculate_conduction_load(surfaces, temp_diff) |
Envelope conduction: U × A × ΔT per surface |
calculate_solar_load(...) |
Solar gains through walls, roof, and windows |
calculate_lighting_load(floor_area, density) |
Lighting heat: area × density |
calculate_personnel_load(num_personnel) |
Sensible (75 W) + latent (55 W) per person |
calculate_infiltration_load(...) |
Sensible + latent infiltration heat |
calculate_saturation_pressure(temp_celsius) |
Tetens' equation for saturation vapour pressure |
calculate_humidity_ratio(temp, rh) |
Humidity ratio from temperature and relative humidity |
Unit conversion functions:
| Function | Purpose |
|---|---|
convert_temp_to_celsius(temp, unit_system) |
°F to °C conversion |
convert_length_to_meters(length, unit_system) |
Feet to meters conversion |
convert_area_to_m2(area, unit_system) |
ft² to m² conversion |
convert_u_value_to_metric(u_value, unit_system) |
BTU/hr·ft²·°F to W/m²·K |
convert_power_to_kw(power, unit_system) |
BTU/hr to kW conversion |
convert_power_to_display(power_kw, unit_system) |
kW to BTU/hr conversion for display |
Testing
Running tests:
pytest calcs/components/c0012_switchroom_heat_load/ -v
pytest calcs/test_c0012* -v
pytest -m validation mcp_public/tests/test_validation_corpus.py -k c0012 -q
pytest calcs/test_help_examples.py -k heat_load -qWorked examples: the examples above are generated from the validation corpus
(mcp_public/validation/cases/fe_calc_heat_load/) by
python manage.py generate_help_examples --tool fe_calc_heat_load, which writes
_switchroom_heat_load_examples.html. Add or change a case there, then regenerate; the
fragment is never edited by hand.
How to extend:
- Add new heat load sources by creating a new
calculate_*_load()function incalculations.pyand adding its result to the sensible or latent totals incalculate_switchroom_heat_load(). - For new equipment types with specific loss characteristics, extend
the
equipment_listschema or add a dedicated equipment library lookup. - To improve solar accuracy, replace the constant factor model with a location-aware calculation using latitude, orientation, and time of year.
- For multi-zone support, split the room into zones and call
calculate_switchroom_heat_load()for each zone, then aggregate. - Write tests covering edge cases: zero solar exposure, imperial units, high humidity, and zero infiltration.