Heat & U-Value Calculations

Leaving Cert Higher Level Construction Studies revision notes with diagrams, key terms and self-check questions.

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Heat loss and U-value calculations appear regularly on the Higher Level Construction Studies paper, usually as Question 5. This topic covers the thermal properties of building materials, calculating the thermal transmittance (U-value) of composite external elements, sizing insulation retrofits, and determining annual heating costs. It also covers steady-state heat loss across multiple building elements, costs-in-use and payback periods, internal and solar heat gains, and statutory standards under Building Regulations Part L.

Thermal Properties: Conductivity, Resistivity, and Resistance

Heat flows naturally from warmer spaces indoors to colder conditions outside through the building envelope. This occurs by conduction through solid materials, convection across air cavities, and radiation across surfaces and spaces. Three physical properties determine how materials handle this heat flow:

  • Thermal conductivity (kk): The rate of heat flow through 1 m21\text{ m}^2 of a material 1 m1\text{ m} thick when the temperature difference across its faces is 1C1^\circ\text{C} (or 1 K1\text{ K}). It is measured in W/mC\text{W/m}^\circ\text{C}. Lower values mean better insulation. For example, mineral wool has a low conductivity (k0.040 W/mCk \approx 0.040\text{ W/m}^\circ\text{C}), whereas dense concrete conducts heat rapidly (k1.440 W/mCk \approx 1.440\text{ W/m}^\circ\text{C}).
  • Thermal resistivity (rr): The reciprocal of thermal conductivity (r=1/kr = 1/k). Measured in mC/W\text{m}^\circ\text{C/W}, it tells you how strongly one metre of the material resists heat flow.
  • Thermal resistance (RR): The actual resistance to heat flow provided by a specific layer of thickness TT (always measured in metres). Depending on whether the exam provides kk or rr, calculate RR using:
R=TkorR=T×rR = \frac{T}{k} \quad \text{or} \quad R = T \times r

Resistance is measured in m2C/W\text{m}^2\,^\circ\text{C/W}. Note that C^\circ\text{C} and K\text{K} are interchangeable when dealing with temperature differences, so W/m2K=W/m2C\text{W/m}^2\text{K} = \text{W/m}^2\,^\circ\text{C}.

Given Resistances

Sometimes the paper gives a layer's resistance (RR, in m2C/W\text{m}^2\,^\circ\text{C/W}) directly instead of providing kk or rr. Typical examples are hollow concrete blocks (e.g. R=0.210 m2C/WR = 0.210\text{ m}^2\,^\circ\text{C/W}), unventilated cavities (e.g. R=0.170 m2C/WR = 0.170\text{ m}^2\,^\circ\text{C/W}), and surface resistances (RsoR_{so} and RsiR_{si}).

Check the unit on the exam paper to decide how to treat each value:

  • m2C/W\text{m}^2\,^\circ\text{C/W} \rightarrow thermal resistance (RR): use as given; do not multiply or divide by thickness.
  • W/mC\text{W/m}^\circ\text{C} \rightarrow thermal conductivity (kk): use R=T/kR = T / k.
  • mC/W\text{m}^\circ\text{C/W} \rightarrow thermal resistivity (rr): use R=T×rR = T \times r.

How Insulating Materials Work

High-void insulation (such as mineral wool, expanded polystyrene, and PIR board) works by trapping still air or inert gas inside tiny closed cells or fibres. Because still air is an exceptionally poor conductor, and the small pockets prevent air movement by convection, these materials achieve very low kk-values (0.022 to 0.040 W/mC0.022\text{ to } 0.040\text{ W/m}^\circ\text{C}).

Reflective insulation (such as aluminium foil facings on quilt or boards) targets radiant heat. The low-emissivity shiny surface reflects radiant heat back into the building and radiates very little heat across the adjacent void. For its reflective benefit, a foil face must face an air space. Pressed tightly against another solid material, it loses almost all of that radiant-barrier benefit.

Closed cells trap still gas; a separate foil face reflects radiant heat across an adjacent air space.
Closed cells trap still gas; a separate foil face reflects radiant heat across an adjacent air space.

Calculating the U-Value of Composite Elements

The U-value (thermal transmittance) measures the overall rate of heat transfer through one square metre of a complete structure when the temperature difference between internal and external air is 1C1^\circ\text{C}. Its unit is W/m2C\text{W/m}^2\,^\circ\text{C}:

U=1RTU = \frac{1}{R_T}

RTR_T is the total thermal resistance found by adding the resistances of every layer, surface air film, and cavity:

RT=Rso+R1+R2++Rn+RsiR_T = R_{so} + R_1 + R_2 + \dots + R_n + R_{si}
A cavity wall section shows heat flowing from warm indoor air through plaster, blockwork, insulation, cavity, outer blockwork and render to colder outdoor air.
A cavity wall section shows heat flowing from warm indoor air through plaster, blockwork, insulation, cavity, outer blockwork and render to colder outdoor air.
Heat passes downward from indoor air through screed, insulation and a concrete slab into the ground; the internal surface resistance is marked above the floor.
Heat passes downward from indoor air through screed, insulation and a concrete slab into the ground; the internal surface resistance is marked above the floor.

Ground Floor Boundary Conditions

When calculating the U-value of a ground-supported (solid) concrete floor, heat passes down into the earth rather than into open outdoor air:

  • There is no external surface resistance (RsoR_{so}) because the slab sits directly against sand blinding, hardcore, and subsoil.
  • The temperature difference (ΔT\Delta T) uses the ground or subsoil temperature (typically given as 5C5^\circ\text{C} or 6C6^\circ\text{C}), not the outside air temperature.
  • Suspended timber or beam-and-block floors over ventilated crawl spaces do have an exposed underside and require an underfloor surface resistance.

Tabulated Layout and Rounding

Marking schemes reward clear tabulation and method. A table with the columns Layer / Thickness (TT in m) / Property (kk or rr) / Resistance (RR in m2C/W\text{m}^2\,^\circ\text{C/W}) is the safest format. Carry at least three decimal places through intermediate calculations, and round only at the final U-value.

Thermal Bridging

A thermal bridge occurs wherever insulation is interrupted or bridged by a material with higher thermal conductivity, such as concrete lintels, window sills, or wall ties. This creates an easy path for heat loss, cooling the internal wall surface and leading to condensation and mould. In exam answers, pair named thermal bridges with concrete construction solutions:

  • Window and door jambs: insert insulated cavity closers.
  • Openings over doors/windows: use thermally broken or insulated lintels.
  • Timber frame walls: fix continuous rigid insulation over the studs to eliminate framing bridging.

Annual Heat Loss and Heating Costs

The cost part of Question 5 calculates the annual expense of heat escaping through an element. Work through these five steps in order:

  1. Rate of heat loss (QQ in Watts):
Q=U×A×ΔTQ = U \times A \times \Delta T

where UU is in W/m2C\text{W/m}^2\,^\circ\text{C}, AA is the element area in m2\text{m}^2, and ΔT=TinternalTexternal\Delta T = T_{\text{internal}} - T_{\text{external}} in C^\circ\text{C}. Because 1 W=1 J/s1\text{ W} = 1\text{ J/s}, QQ tells you how many Joules escape every second. The paper often notes that 1000 Watts=1 kJ per second1000\text{ Watts} = 1\text{ kJ per second}.

  1. Total annual heating duration (tt in seconds):
t=weeks×7 days×hours per day×3600 seconds/hourt = \text{weeks} \times 7\text{ days} \times \text{hours per day} \times 3600\text{ seconds/hour}
  1. Total annual energy lost in kilojoules (EE in kJ):
E=Q×t1000E = \frac{Q \times t}{1000}
  1. Quantity of fuel consumed:
Fuel required=ECalorific Value of Fuel\text{Fuel required} = \frac{E}{\text{Calorific Value of Fuel}}

Check the fuel specified:

  • Heating oil: calorific value 37,350 kJ/litre\approx 37,350\text{ kJ/litre} \rightarrow volume in litres.
  • Wood pellets: calorific value 17,350 kJ/kg\approx 17,350\text{ kJ/kg} \rightarrow mass in kilograms.
  • If electricity is priced per kWh\text{kWh}, use E (in kWh)=Q (W)×heating hours/1000E\text{ (in kWh)} = Q\text{ (W)} \times \text{heating hours} / 1000.
  1. Annual running cost (€):
Annual Cost=Fuel quantity×Unit price in euro\text{Annual Cost} = \text{Fuel quantity} \times \text{Unit price in euro}

Divide prices stated in cent by 100 (e.g. 55 cent=0.5555\text{ cent} = €0.55).

Exam Checklist for Costing

  • Check ΔT\Delta T calculation (TintTextT_{\text{int}} - T_{\text{ext}} or TsubsoilT_{\text{subsoil}}).
  • Calculate time in full seconds.
  • Divide Joules by 1000 to get kJ.
  • Divide by calorific value to obtain litres or kg.
  • Convert cent to euro before the final multiplication.
  • State answer to two decimal places with the € symbol.

Insulation Retrofits and Costs-in-Use

Sizing Retrofit Insulation Thickness

A follow-on part often asks for the thickness of additional insulation required to achieve an upgraded U-value (such as a Passive House standard of 0.15 W/m2C0.15\text{ W/m}^2\,^\circ\text{C}). Use this three-step method:

  1. Target total resistance (RtargetR_{\text{target}}):
Rtarget=1UtargetR_{\text{target}} = \frac{1}{U_{\text{target}}}
  1. Additional resistance needed (ΔR\Delta R):
ΔR=RtargetRexisting\Delta R = R_{\text{target}} - R_{\text{existing}}
  1. Additional thickness (TT in metres):
T=ΔR×kT = \Delta R \times k

Multiply by 1000 to state the final thickness in millimetres (mm).

Costs-in-Use and Payback Periods

Costs-in-use evaluates building choices based on long-term operating costs rather than just initial capital expense:

  1. Calculate annual heating cost before the upgrade.
  2. Calculate annual heating cost after the upgrade using the new U-value (cost decreases in direct proportion to U-value).
  3. Determine annual financial saving:
Annual Saving=Cost beforeCost after\text{Annual Saving} = \text{Cost before} - \text{Cost after}
  1. Calculate the payback period in years:
Payback Period=Capital Cost of UpgradeAnnual Saving\text{Payback Period} = \frac{\text{Capital Cost of Upgrade}}{\text{Annual Saving}}

Upgrading poorly insulated elements yields short payback times, whereas adding more insulation to an already well-insulated wall shows diminishing returns and a longer payback period.

A schematic U-value curve falls steeply at first and then flattens as added insulation thickness increases.
A schematic U-value curve falls steeply at first and then flattens as added insulation thickness increases.

Whole-Building Heat Loss and Heat Gains

Steady-State Heat Loss for a Whole Structure

When evaluating an entire room or building, calculate heat loss for each external element separately and sum them:

Total Fabric Heat Loss=(U×A×ΔT)\text{Total Fabric Heat Loss} = \sum (U \times A \times \Delta T)

Always use the net wall area: calculate the gross external wall area and subtract window and door openings. Openings are calculated separately with their own higher U-values.

A wall elevation distinguishes opaque wall, window and door areas within one gross rectangular outline.
A wall elevation distinguishes opaque wall, window and door areas within one gross rectangular outline.

Calculating Heat Gain

A building gains heat from three internal and environmental sources, reducing the thermal load on the heating system:

  1. Occupants: Body heat released (use the figure provided on the paper, typically 80100 W80\text{--}100\text{ W} per person).
  2. Lighting and appliances: Electrical equipment indoors converts electrical energy into heat.
  3. Solar gain: Solar radiation through windows and glazed doors.
Total Heat Gain=Occupant Gain+Appliance Gain+Solar Gain (all in Watts)\text{Total Heat Gain} = \text{Occupant Gain} + \text{Appliance Gain} + \text{Solar Gain (all in Watts)}

To calculate solar gain:

Solar Gain (W)=Glazed Area (m2)×Solar Radiation Intensity (W/m2)×Solar Gain Factor\text{Solar Gain (W)} = \text{Glazed Area } (\text{m}^2) \times \text{Solar Radiation Intensity } (\text{W/m}^2) \times \text{Solar Gain Factor}

Statutory Standards and Passive Design

Table 1 of the 2022 Technical Guidance Document L (Conservation of Fuel and Energy – Dwellings) gives these maximum average elemental U-values for new dwellings:

Building ElementTGD Part L Maximum Elemental U-Value (W/m2C\text{W/m}^2\,^\circ\text{C})
Pitched roof (ceiling level)0.160.16
Pitched roof (on slope / rafter level)0.160.16
Flat roof0.200.20
External walls0.180.18
Ground floors0.180.18
External windows, doors, and rooflights1.401.40

These are the general Table 1 values; a floor with underfloor heating has a stricter limit of 0.15 W/m²K. Check the edition and assumptions specified in your question.

Passive solar design uses building layout to maximise free solar energy. Main living areas and primary glazing should face south (ideally within 3030^\circ of due south). Appendix C of the 2022 guidance uses windows, doors and rooflights equal to 25%25\% of floor area in its notional dwelling for DEAP calculations. This is a reference assumption, not a universal cap on glazing area.

Key terms

Thermal Conductivity (k)
The rate of heat flow through 1 m² of a material 1 m thick when the temperature difference across its faces is 1 °C, measured in W/m°C.
Thermal Resistivity (r)
The reciprocal of thermal conductivity (1/k), expressing a material's intrinsic opposition to heat flow per unit thickness, measured in m°C/W.
Thermal Resistance (R)
The opposition to heat flow provided by a specific layer thickness, calculated as R = T/k or R = T × r, measured in m²°C/W.
Thermal Transmittance (U-value)
The overall rate of heat loss through 1 m² of a building element when the temperature difference between inside and outside is 1 °C, calculated as U = 1/Rt in W/m²°C.
Calorific Value
The amount of heat energy released when one unit of fuel is completely burned, measured in kJ/litre for oil or kJ/kg for wood pellets.
Thermal Bridge
A localized area of the building envelope where insulation is broken or bridged by higher-conductivity materials, creating a direct path for heat loss and condensation.
Surface Resistance
The thermal resistance offered by the thin film of stationary air clinging to internal (Rsi) and external (Rso) building surfaces.
Heat Gain
Heat added to an interior space from internal sources (occupants and appliances) and solar radiation through glazing, measured in Watts.

Check yourself

  1. A 100 mm thick insulation board has a thermal resistance of 2.50 m²°C/W. Calculate its thermal conductivity (k).

    Rearrange R = T / k to give k = T / R. Convert 100 mm to 0.10 m: k = 0.10 / 2.50 = 0.040 W/m°C.

  2. A building element loses 1,000,000 kJ of heat annually. The heating system uses wood pellets with a calorific value of 17,350 kJ/kg priced at 52 cent per kg. Calculate the annual heating cost.

    Mass of pellets = 1,000,000 / 17,350 = 57.64 kg. Cost = 57.64 kg × €0.52 = €29.97.

  3. A wall measures 8 m by 3 m and contains a window of 3 m². The wall has a U-value of 0.20 W/m²°C and the window has a U-value of 1.40 W/m²°C. If ΔT is 15 °C, calculate the total steady-state heat loss.

    Net wall area = (8 × 3) - 3 = 21 m². Wall loss = 0.20 × 21 × 15 = 63 W. Window loss = 1.40 × 3 × 15 = 63 W. Total heat loss = 63 + 63 = 126 W.

  4. A living room contains 3 occupants (heat output 100 W each) and lighting and equipment totalling 250 W. South-facing glazing of 4 m² receives 300 W/m² of solar radiation with a solar gain factor of 0.60. Calculate the total heat gain.

    Occupant gain = 3 × 100 = 300 W. Equipment gain = 250 W. Solar gain = 4 × 300 × 0.60 = 720 W. Total heat gain = 300 + 250 + 720 = 1,270 W (1.27 kW).

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