Heat Loss & Insulation

Leaving Cert Higher Level Construction Studies revision notes with diagrams.

10 min readHigher LevelBy Studytok
Practise this topic — free →

Heat Transfer Fundamentals, Fabric Loss, and Thermal Comfort

Heat always flows from a warmer place to a colder place. In winter, that means from inside the house to outside. Heat moves through the building envelope by three mechanisms:

  • Conduction: Heat transfer through solid materials by particle vibration, such as heat passing through concrete blocks, timber framing, or metal lintels.
  • Convection: Heat transfer through the circulation of liquids or gases, such as warm air rising in an open wall cavity or air currents tumbling down a cold windowpane.
  • Radiation: Heat transfer across open space by electromagnetic waves without heating the air in between, such as heat radiating from a stove or sunlight streaming through south-facing glass.

In an older, uninsulated house, heat escapes in these approximate typical proportions:

  • External walls: ~35%
  • Roof and attic: ~25%
  • Ground floor: ~15%
  • Draughts and ventilation: ~15%
  • Windows and external doors: ~10%

Thermal Comfort and Radiant Temperature

Comfort indoors involves more than air temperature alone. Mean radiant temperature is the average temperature of the surfaces surrounding a person, including walls, windows, floor, and ceiling. Because the human body radiates heat directly to colder surroundings, a room with warm air can still feel chilly if the walls or glazing are cold. Well-insulated walls raise interior surface temperatures, giving much better comfort at the same air temperature.

Thermal comfort depends on five key factors:

  1. Air (dry bulb) temperature (typically 18 °C to 21 °C in living areas)
  2. Mean radiant temperature of surrounding surfaces
  3. Relative humidity (ideally 40% to 60%)
  4. Air movement (draughts cause chills, but fresh air prevents stuffiness)
  5. The occupant's activity level and clothing

Thermal Properties, U-Values, and Temperature Gradients

Calculating how a structural element resists heat flow relies on four core thermal measurements:

  1. Thermal conductivity (kk-value): The rate of heat flow through a 1 m thickness of a material per degree temperature difference across its faces, measured in W/m ∘C\text{W/m }^\circ\text{C} or W/m K\text{W/m K}. Dense materials have high values (concrete block is about 1.330 W/m ∘C1.330\text{ W/m }^\circ\text{C}), while good insulators have very low values (PIR board is about 0.022 W/m ∘C0.022\text{ W/m }^\circ\text{C}).
  2. Thermal resistivity (rr-value): The resistance to heat flow of one metre thickness of a material, per metre squared (r=1/kr = 1/k, measured in m ∘C/W\text{m }^\circ\text{C/W} or m K/W\text{m K/W}). Some exam questions supply rr instead of kk for plasters and screeds.
  3. Thermal resistance (RR-value): The actual resistance provided by a specific material layer of thickness TT in metres:
R=TkorR=T×rR = \frac{T}{k} \quad \text{or} \quad R = T \times r

For example, 100 mm of mineral wool with k=0.040 W/m ∘Ck = 0.040\text{ W/m }^\circ\text{C} gives:

R=0.1000.040=2.500 m2 ∘C/WR = \frac{0.100}{0.040} = 2.500\text{ m}^2\,^\circ\text{C/W}

Boundary layers of still air clinging to building surfaces also resist heat flow. These are the internal surface resistance (RsiR_{si}) and external surface resistance (RseR_{se}). Any unventilated air cavity provides a cavity resistance (RaR_a). Note that a ground-floor slab cast on subsoil has an internal surface resistance (RsiR_{si}) but no external surface resistance (RseR_{se}).

  1. Thermal transmittance (UU-value): The rate of heat loss through one square metre of a complete structural element per degree temperature difference between inside and outside:
Rt=Rse+R1+R2+Ra+⋯+RsiR_t = R_{se} + R_1 + R_2 + R_a + \dots + R_{si}U=1RtU = \frac{1}{R_t}

Measured in W/m2 ∘C\text{W/m}^2\,^\circ\text{C}, a lower U-value means better thermal performance.

Temperature Gradients Through a Wall

The temperature drop across any individual layer is directly proportional to its share of the total thermal resistance:

Temperature drop across layer=(RlayerRt)×Total ΔT\text{Temperature drop across layer} = \left(\frac{R_{\text{layer}}}{R_t}\right) \times \text{Total } \Delta T

To plot a gradient:

  1. Calculate the total resistance (RtR_t) and overall temperature difference (ΔT\Delta T).
  2. Start at the indoor air temperature.
  3. Subtract the temperature drop across RsiR_{si}, then subtract the drop across each layer in turn, moving outwards to the cold face.

Because high-performance insulation contributes the vast majority of the total resistance, almost the entire temperature drop occurs inside that insulation layer. The cold outer face of the insulation is where temperatures plummet, showing why indoor water vapour must be stopped before it can reach this cold zone.

Wall section aligned with a schematic temperature profile showing the largest temperature drop across the insulation.
Wall section aligned with a schematic temperature profile showing the largest temperature drop across the insulation.

Steady-State Heat Loss, Ventilation Loss, and Annual Fuel Costs

Heat loss through the building fabric is calculated under steady-state conditions using:

Q=U×A×ΔTQ = U \times A \times \Delta T

where QQ is the heat loss rate in Watts (Joules per second), AA is the exposed surface area in m2\text{m}^2, and ΔT\Delta T is the temperature difference across the element. For walls and roofs, ΔT=Tinside−Toutside\Delta T = T_{\text{inside}} - T_{\text{outside}}. For ground floors, ΔT=Tinside−Tsubsoil\Delta T = T_{\text{inside}} - T_{\text{subsoil}}.

Ventilation Heat Loss

Heat is also carried away when warm indoor air escapes and is replaced by cold outdoor air:

Qv=0.33×N×V×ΔTQ_v = 0.33 \times N \times V \times \Delta T

where NN is the ventilation rate in air changes per hour (ACH), VV is the room volume in m3\text{m}^3, and 0.33 Wh/m3 ∘C0.33\text{ Wh/m}^3\,^\circ\text{C} is the volumetric heat capacity of air. For a living room of 48 m348\text{ m}^3 with N=1.5 ACHN = 1.5\text{ ACH} and ΔT=15 ∘C\Delta T = 15\,^\circ\text{C}, ventilation heat loss is:

Qv=0.33×1.5×48×15=356.4 WQ_v = 0.33 \times 1.5 \times 48 \times 15 = 356.4\text{ W}

Total building heat loss equals total fabric loss plus total ventilation loss.

Five-Step Space-Heating Cost Method

To calculate the annual running cost of replacing fabric heat loss:

  1. Calculate heat loss rate (QQ): Find QQ in Watts (Q=U×A×ΔTQ = U \times A \times \Delta T).
  2. Calculate total heating time (tt) in seconds:
t=hours/day×7 days/week×weeks/year×3600 s/hrt = \text{hours/day} \times 7\text{ days/week} \times \text{weeks/year} \times 3600\text{ s/hr}
  1. Calculate total energy lost in kilojoules (kJ):
Energy (kJ)=Q×t1000\text{Energy (kJ)} = \frac{Q \times t}{1000}
  1. Calculate fuel quantity consumed: Divide total energy lost by fuel calorific value (oil is typically 37,350 kJ/litre37,350\text{ kJ/litre}; wood pellets are typically 17,350 kJ/kg17,350\text{ kJ/kg}):
Fuel Quantity=Energy (kJ)Calorific Value\text{Fuel Quantity} = \frac{\text{Energy (kJ)}}{\text{Calorific Value}}
  1. Calculate annual cost: Multiply fuel quantity by the unit price (€).

Costs-in-Use, Payback, Statutory Standards, and Heat Gains

Costs-in-use means evaluating a building element over its entire working life: initial purchase and installation costs plus annual space-heating running costs.

Payback Period

The payback period is the time in years required for heating cost savings to repay the capital cost of an insulation upgrade:

Annual Saving (€)=Annual Cost Before Upgrade−Annual Cost After Upgrade\text{Annual Saving (€)} = \text{Annual Cost Before Upgrade} - \text{Annual Cost After Upgrade}Payback Period (years)=Capital Cost of Upgrade (€)Annual Saving (€/year)\text{Payback Period (years)} = \frac{\text{Capital Cost of Upgrade (€)}}{\text{Annual Saving (€/year)}}

If upgrading an uninsulated external wall costs €4,200 and reduces the annual oil bill from €352 to €96, the annual saving is €256. The payback period is 4200/256=16.4 years4200 / 256 = 16.4\text{ years}. If fuel prices rise over time, the annual financial saving increases, which shortens the payback period.

Statutory Thermal Transmittance Values

Technical Guidance Document L (TGD Part L: Conservation of Fuel and Energy – Dwellings) sets mandatory maximum U-values for new dwellings:

  • External walls: 0.18 W/m2 ∘C0.18\text{ W/m}^2\,^\circ\text{C}
  • Ground floors: 0.18 W/m2 ∘C0.18\text{ W/m}^2\,^\circ\text{C}
  • Pitched roof (insulated at ceiling or rafters): 0.16 W/m2 ∘C0.16\text{ W/m}^2\,^\circ\text{C}
  • Flat roof: 0.20 W/m2 ∘C0.20\text{ W/m}^2\,^\circ\text{C}
  • External windows, doors, and rooflights: 1.40 W/m2 ∘C1.40\text{ W/m}^2\,^\circ\text{C}

By comparison, the voluntary Passive House standard demands a much stricter wall U-value of 0.15 W/m2 ∘C0.15\text{ W/m}^2\,^\circ\text{C} or lower. To check compliance in an exam question, compare your calculated U-value against the statutory maximum: the element complies only if its U-value is equal to or lower than the regulation figure.

Internal and Solar Heat Gains

A house gains heat from internal and external sources, reducing the net heat the boiler or heat pump must supply:

  • Solar heat gain: Short-wave solar radiation passes through glazing and is absorbed by high thermal mass elements like concrete floors and plastered masonry, which slowly re-radiate warmth into the room. Gains are highest through large south-facing windows.
  • Internal gains: Heat released inside the dwelling by occupants (roughly 80 W to 100 W per person), electric lighting, cooking, and household appliances.
Total Heat Gain (W)=Solar Gain+Occupant Gain+Appliance Gain+Lighting Gain\text{Total Heat Gain (W)} = \text{Solar Gain} + \text{Occupant Gain} + \text{Appliance Gain} + \text{Lighting Gain}Net Space Heating Demand (W)=Total Heat Loss−Total Heat Gain\text{Net Space Heating Demand (W)} = \text{Total Heat Loss} - \text{Total Heat Gain}

Insulation Materials, High-Void Structures, and Retrofit Techniques

High-void insulating materials consist of small, motionless pockets of air or blowing gas enclosed within lightweight cellular or fibrous structures. Because still air is an exceptionally poor conductor of heat and tiny pockets prevent air from circulating by convection, heat transfer is severely suppressed. Common high-void materials include mineral wool, expanded polystyrene, and cellulose. If these materials become wet on site, their performance collapses because liquid water conducts heat roughly 25 times faster than still air.

Reflective insulating materials (such as aluminium foil facings and multi-foil membranes) work differently. Their shiny, low-emissivity surface reflects radiant heat back towards the warm side and gives off very little radiation itself. They only work when they face a still air gap, because touching a solid surface lets heat conduct straight through.

MaterialTypical kk-value (W/m ∘C\text{W/m }^\circ\text{C})Typical Application
Mineral Wool (Rock/Glass)0.035 – 0.040Loft quilts, timber-frame studs, acoustic partitions
Expanded Polystyrene (EPS)0.031 – 0.037External wall insulation (EWI), floor slab insulation
Extruded Polystyrene (XPS)0.028 – 0.034Heavy-load ground floors, foundation perimeter edges
Polyisocyanurate (PIR)0.022 – 0.026Partial-fill cavity walls, cut-rafter warm roofs
Cellulose Fibre0.038 – 0.040Blown attic insulation, timber-frame cassettes
Reflective Foil BarriersVariableCavity membranes (requires adjacent still air gap)

Retrofitting Existing External Walls

  1. Cavity Bead Injection: Pumping EPS beads bonded with an adhesive emulsion through holes drilled in outer mortar joints. Ideal for uninsulated cavity walls without disturbing interior finishes.
  2. External Wall Insulation (EWI): Fixing rigid insulation (EPS or mineral wool) to the outside face of masonry in five ordered steps: (a) mechanically fix an aluminium base starter track at DPC level; (b) bond and mechanically anchor insulation boards tightly to the wall; (c) apply a cementitious base coat render; (d) embed an alkali-resistant glass-fibre mesh into the wet base coat; (e) apply an external primer followed by a weatherproof, breathable polymer finish.
  3. Internal Dry-Lining: Fixing composite insulated plasterboard to the inside wall face with adhesive dabs or timber battens. It keeps the external facade unaltered, but slightly reduces internal room area and requires careful sealing around sockets and joists.

Thermal Bridging, Detailing, and Thermal Looping

A thermal bridge (or cold bridge) happens where a material with higher thermal conductivity penetrates the insulation, or where the insulation layer is broken. Typical locations include concrete window lintels, window sills and jambs, junctions between ground-floor slabs and external walls, and eaves junctions.

Cavity-wall and ground-floor section showing AAC blocks, overlapping insulation and a continuous DPM-to-DPC connection.
Cavity-wall and ground-floor section showing AAC blocks, overlapping insulation and a continuous DPM-to-DPC connection.

Thermal bridges can account for a significant share of fabric heat loss in an otherwise well-insulated house. They also create cold interior surface spots, causing surface condensation and mould growth, which damages finishes and can aggravate asthma and other respiratory problems.

Faulty cavity insulation permits air to rise behind the boards and return through the outer cavity; tightly fitted boards remove this bypass.
Faulty cavity insulation permits air to rise behind the boards and return through the outer cavity; tightly fitted boards remove this bypass.

Detailing the Wall-to-Floor Junction

To eliminate thermal bridging where a ground floor meets an external cavity wall, assemble the junction in these ordered steps:

  1. Lay autoclaved aerated concrete (AAC) thermal-break blocks in the inner leaf at and below finished floor level to slow downward conduction.
  2. Continue the cavity wall insulation down below finished floor level, so that it overlaps the perimeter edge insulation and the floor insulation and leaves no uninsulated gap at the junction.
  3. Install a continuous layer of rigid sub-floor insulation horizontally beneath the concrete floor slab or screed.
  4. Fit vertical perimeter edge insulation (minimum 25 mm thick) between the edge of the floor slab and the inner block leaf, ensuring it laps tightly against the sub-floor insulation.
  5. Dress the damp proof membrane (DPM) under the slab up and over the perimeter insulation, linking seamlessly with the wall damp proof course (DPC).

Attic Insulation Detailing

When insulating an attic floor with mineral wool, roll the first layer between the ceiling joists to the joist depth. Then lay the second layer perpendicular (at right angles) across the joists. This cross-laying covers the timber joists, eliminating thermal bridging through the wood and sealing any gaps between the first-layer batts.

Thermal Looping

Thermal looping occurs when an unsealed air gap is left between the inner masonry leaf and cavity insulation boards. Cold air enters at the base of the cavity, absorbs heat escaping from the inner leaf, warms up, rises by convection, and spills over the top of the insulation into the cold outer air space. This continuous convective loop pulls heat straight out of the inner masonry leaf, which can seriously reduce the insulation's actual performance. It is prevented by bedding boards tightly against smooth blockwork and taping or tongue-and-grooving all board edges.

You've read the theory
Now turn it into exam marks.

Practise heat loss & insulation as questions and flashcards in Studytok, with explanations when you get stuck.

Continue for free →
  1. ✓
    Read the notes
    6 sections
  2. 2
    Test yourself
    Questions marked instantly
  3. 3
    Keep revising
    Flashcards and exam-style practice