Heating and cooling loads

Every heating and cooling system starts with a number: the load. Before an engineer can choose a heat pump, size a duct or decide how big the electrical service must be, someone has to estimate how much heat the building loses on the coldest day and how much it gains on the hottest. In the second week, Carvalho brings Priyanka Sethi, the energy modeler who worked with Ingrid on the community hub last winter, to show the students how a load calculation is built. 'An energy model tells you how much energy the building uses in a year,' Sethi says. 'A load calculation tells you how big the equipment must be for the worst hour. Both matter, and they are different questions.' The students spend the week calculating the hub's design heating load and the cooling load of its hall on a crowded summer evening.

Loads, energy and design conditions

A load is a rate: the heat that must be added to or removed from a space, per unit of time, to hold it at the chosen indoor conditions, measured in British thermal units per hour (Btu/h), or tons of cooling, where one ton is 12,000 Btu/h. Energy is a quantity: the total heat or electricity used over a period, measured in kilowatt-hours or thousands of Btu. Equipment is sized to the peak load; energy bills depend on how loads vary over the year and how efficiently the equipment meets them.

Peak loads are calculated for design conditions, which are not the most extreme weather ever recorded but conditions that are exceeded only rarely. Codes and design guides publish them for each city. For Minneapolis, design guides give a winter design temperature in the range of about −12 to −16 °F, depending on the data set and the probability chosen, and a summer design dry-bulb temperature of about 89 °F. Indoor design conditions are typically 70 °F in winter and 75 °F with moderate humidity in summer.

Where heat goes in winter

In winter, a heated building loses heat in three ways: by conduction through its walls, roof, floor and windows; by air leakage, as cold outside air enters through cracks and warm air leaves; and by ventilation, as fresh air is brought in deliberately and stale air exhausted. Some heat is regained from people, lights, equipment and sunshine, but the design heating load usually ignores these gains, because the coldest hour may come on a dark, empty night.

Here U is the thermal transmittance of each element, A its area and ΔT the difference between indoor and outdoor design temperatures. The constants in the air equation come from the density and specific heat of air. With a heat recovery ventilator, only the fraction of ventilation heat that is not recovered counts as a load; with a recovery effectiveness of 0.85, only fifteen hundredths of it remains.

The hub's design heating load

Sethi asks the students to calculate the hub's design heating load from the envelope Ingrid designed: walls of R-31 effective, a roof of R-51, triple glazing, a slab edge insulated to cut the thermal bridge, an airtight envelope and a heat recovery ventilator.

# Design heating load of the community hub (studio values)
t_in, t_out = 70, -14
dT = t_in - t_out
elements = {                    # name: (area ft2, U-value Btu/h ft2 F)
    "walls":   (13450, 0.0323),
    "roof":    (16150, 0.0196),
    "windows": (4520, 0.16),
}
loads = {k: a * u * dT for k, (a, u) in elements.items()}
loads["slab edge"] = 720 * 0.289 * dT          # perimeter (ft) x linear loss per unit length
volume = 388000                                     # heated volume (ft3)
loads["air leakage"] = 0.018 * 0.1 * volume * dT   # 0.1 air changes per hour
vent = 5085                                      # ventilation rate (cfm)
loads["ventilation"] = 1.08 * vent * dT * (1 - 0.85)  # heat recovery effectiveness 0.85
total = sum(loads.values())
for k, v in loads.items():
    print(f"{k:12s} {v / 1000:7.1f} kBtu/h")
print(f"total        {total / 1000:7.1f} kBtu/h  ({total / 3412:.0f} kW)")
print(f"load per unit floor area: {total / 27232:.1f} Btu/h ft2")

walls           36.5 kBtu/h
roof            26.6 kBtu/h
windows         60.7 kBtu/h
slab edge       17.5 kBtu/h
air leakage     58.7 kBtu/h
ventilation     69.2 kBtu/h
total          269.2 kBtu/h  (79 kW)
load per unit floor area: 9.9 Btu/h ft2

The total is about 270,000 Btu/h (about 80 kW), roughly 10 Btu/h per square foot of floor, which is low for a public building in this climate. Three items are each about a quarter of the total: ventilation, even after heat recovery, the windows and air leakage. The windows lose more heat than all the walls even though they are only a quarter of the wall area, because even triple glazing has a U-value about five times that of the insulated wall. Air leakage is nearly as large, which is why the airtightness target Ingrid set last winter matters so much. Ventilation would be by far the largest item without heat recovery.

Sethi runs the same calculation for a version of the hub with double glazing, an envelope three times leakier and a less effective heat recovery ventilator. The load roughly doubles. 'That is the difference between one heat pump and three,' she says, 'and between an electrical service the utility can give you easily and one it has to build a new transformer for.' The envelope decisions the students made in earlier courses are, in the end, mechanical decisions.

Where heat comes from in summer

Cooling loads are more complicated than heating loads, because heat arrives from more sources and at different times. Sunshine through windows is often the largest envelope gain; it peaks on west windows in the late afternoon and on east windows in the morning. Heat conducted through walls and roofs arrives late, because the mass of the construction delays it. People give off heat: about 250 Btu/h of sensible heat and 200 Btu/h of latent heat each when seated. Lights and equipment add heat equal to the power they use. And outdoor air brought in for ventilation must be cooled and dried.

Engineers distinguish sensible heat, which raises the air temperature, from latent heat, which is moisture that must be condensed out of the air. Both must be removed by the cooling equipment. Latent loads come from people, cooking, showers and humid outdoor air; in the Prairies and the upper Midwest, summer air can be humid enough that dehumidification matters as much as cooling.

The hall on a summer evening

The hub's hall is used for community dinners, dances and meetings of up to three hundred people. Joon-ho calculates its cooling load at the design condition on a July evening with the room full, the lights on and the late sun on the west windows.

# Cooling load of the hub's hall, full, on a July evening (studio values)
people = 300
area = 6460                         # floor area (ft2)
sens, lat = {}, {}
sens["people"] = people * 250
lat["people"] = people * 200
sens["lights"] = area * 1.9
sens["equipment"] = 6800
sens["sun on west windows"] = 646 * 0.30 * 143 * 0.5     # area x SHGC x sun x external shading
sens["walls and roof"] = 14530 * 0.021 * 22          # equivalent temperature difference
oa = people * 7.5 + area * 0.06                          # outdoor air (cfm)
sens["outdoor air"] = 1.08 * oa * 14 * (1 - 0.70)        # energy recovery, sensible 0.70
lat["outdoor air"] = 0.68 * oa * 21 * (1 - 0.50)         # latent recovery 0.50
S, Lt = sum(sens.values()), sum(lat.values())
for k, v in sens.items():
    print(f"sensible  {k:20s} {v / 1000:6.1f} kBtu/h")
for k, v in lat.items():
    print(f"latent    {k:20s} {v / 1000:6.1f} kBtu/h")
print(f"outdoor air {oa:,.0f} cfm")
print(f"total {(S + Lt) / 1000:.1f} kBtu/h, of which people {(sens['people'] + lat['people']) / (S + Lt):.2f}")
print(f"sensible heat ratio {S / (S + Lt):.2f}; in tons of cooling, {(S + Lt) / 12000:.1f}")

sensible  people                 75.0 kBtu/h
sensible  lights                 12.3 kBtu/h
sensible  equipment               6.8 kBtu/h
sensible  sun on west windows    13.9 kBtu/h
sensible  walls and roof          6.7 kBtu/h
sensible  outdoor air            12.0 kBtu/h
latent    people                 60.0 kBtu/h
latent    outdoor air            18.8 kBtu/h
outdoor air 2,638 cfm
total 205.4 kBtu/h, of which people 0.66
sensible heat ratio 0.62; in tons of cooling, 17.1

The result surprises the students. The people in the room are more than half the cooling load; the sun and the envelope are small by comparison, because Ingrid's design shaded the west windows and insulated the walls well. The outdoor air needed for three hundred people is large too, and without energy recovery it would add much more. About a third of the load is latent, moisture from breath and from humid outdoor air, so the system must dehumidify as well as cool. The hall's peak cooling load is about 60 kW, about seventeen tons, for a few evenings a year; most of the time the room is empty or lightly used.

That pattern, a large peak that occurs rarely, is typical of assembly rooms, and it shapes the system design. A system sized for the peak runs most of the year at a small fraction of its capacity, so it must be able to turn down efficiently. Ventilation that varies with the number of people, using carbon dioxide sensors, lesson 5, saves a great deal of energy in such rooms.

Diversity and the building peak

The building's peak load is less than the sum of the peaks of its rooms, because they do not all peak at once. The hall peaks on summer evenings; the library reading room on hot weekday afternoons; the daycare in the late morning. West rooms peak in the afternoon and east rooms in the morning. Engineers call this diversity. They size each room's terminal unit for that room's peak, but the central plant for the building's simultaneous peak, which may be a quarter or a third smaller.

Heating behaves differently: on the coldest night, every room loses heat at once, so there is little diversity in the design heating load. In a climate as cold as Minneapolis's, the heating load usually decides the size of a heat pump plant, while the cooling load decides the size of the ducts in crowded rooms.

Exercise 2: loads for the Riverlot building

This exercise takes one week.

• Calculate the design heating load of one typical residential floor of the Riverlot building, with the envelope values from your urban studio design.
• Calculate the cooling load of the grocery on a July afternoon, including people, lights, refrigerated cases (which remove heat from the room), equipment, sun and outdoor air.
• Present each load as a bar chart by component and identify the largest items.
• Propose two design changes that reduce each load, and recalculate.

Self-critique checklist

• Have I used design temperatures rather than average temperatures?
• Have I separated sensible and latent loads?
• Have I counted ventilation air after heat recovery, not before?
• Is my building peak smaller than the sum of the room peaks, and do I know why?