U-values, insulation and heat pumps
On a frosty January morning in Kilkenny, two houses on the same road have the same size and the same heating. One has a warm hall, the other has frost on the inside of the windows. Why does one house hold its heat so much better, and what does that cost each family over a winter?
What a U-value tells you
Heat always flows from the warm side of a wall to the cold side. How fast it flows depends on three things: the area of the wall, the temperature difference across it, and what the wall is made of. The U-value puts the last of these into one number.
Why does the formula look like this? Double the area and twice as much heat escapes. Double the temperature difference and the heat flows twice as fast. A low U-value means the element lets very little heat through, so a low U-value is what you want in a well-insulated house.
Adding up a whole house
Heat leaves through every element at the same time, so the losses add. For walls, roof, windows and floor with the same $\Delta T$:
$$P = (U_1A_1 + U_2A_2 + \dots)\,\Delta T$$The quantity $\sum UA$ is the heat loss coefficient in $\text{W K}^{-1}$. Air that leaks out through vents and gaps (ventilation) carries heat away too, and is given as another figure in $\text{W K}^{-1}$ that you simply add on.
The U-value of a wall is $0.35\ \text{W m}^{-2}\text{K}^{-1}$. a) State what a U-value measures. b) Explain whether a low or a high U-value is better for a…
The U-value of a wall is $0.35\ \text{W m}^{-2}\text{K}^{-1}$.
- a) State what a U-value measures.
- b) Explain whether a low or a high U-value is better for a well-insulated house.
- c) Calculate the rate of heat loss through $20\ \text{m}^2$ of this wall when the temperature difference between inside and outside is $15\ \text{K}$, using $P = UA\Delta T$.
A house in Waterford has the following elements, with U-values and areas: walls $90\ \text{m}^2$ at $0.30\ \text{W m}^{-2}\text{K}^{-1}$, roof $50\ \text{m}^2$ at…
A house in Waterford has the following elements, with U-values and areas: walls $90\ \text{m}^2$ at $0.30\ \text{W m}^{-2}\text{K}^{-1}$, roof $50\ \text{m}^2$ at $0.16\ \text{W m}^{-2}\text{K}^{-1}$, windows $18\ \text{m}^2$ at $1.4\ \text{W m}^{-2}\text{K}^{-1}$, and floor $50\ \text{m}^2$ at $0.25\ \text{W m}^{-2}\text{K}^{-1}$.
- a) Calculate the heat loss rate through each element for a temperature difference of $18\ \text{K}$.
- b) Calculate the total heat loss rate.
- c) State which element should be improved first to reduce heat loss, and justify your answer.
Insulation and how U-values are made
Heat leaves a roof by conduction (through the solid materials), convection (warm air rising and moving) and radiation (infrared given off by the warm surface). Insulation is designed to slow these down.
Materials such as fibreglass, mineral wool and foam are good insulators because they trap air in tiny pockets. Air is a very poor conductor, and in a small pocket it cannot circulate, so convection is stopped. The heat has to get through many thin layers of still air and fibre.
Good insulation also gives other benefits: a more comfortable and drier home (less condensation and mould), a better BER rating, and lower carbon dioxide emissions from burning fuel.
Layers in series
A wall is often made of several layers. Each layer of thickness $L$ (in m) and thermal conductivity $k$ (in $\text{W m}^{-1}\text{K}^{-1}$) resists the flow of heat by an amount $L/k$. The resistances add:
$$U = \frac{1}{\sum \dfrac{L}{k}}$$A thick layer or a layer with small $k$ adds a lot of resistance. This is why a thin layer of mineral wool ($k$ about $0.04$) helps far more than a thick layer of concrete ($k$ about $1$).
Seeing heat loss with a thermal camera
Every object gives off infrared radiation, and the hotter its surface, the more it emits. A thermal imaging camera detects this infrared and displays hotter surfaces as brighter colours. It needs no visible light, so it works in the dark. On a cold night, windows appear bright because their outer surface is warmer than the wall's. Glass has a high U-value, so it passes heat quickly and the temperature drop across it is small: its outer surface stays close to the inside temperature. A well-insulated wall resists the flow, so its outer surface is close to the outside temperature.
A homeowner in Kilkenny adds insulation to the attic of her house. a) Name two ways in which heat is lost through an uninsulated roof. b) Explain why trapped air in…
A homeowner in Kilkenny adds insulation to the attic of her house.
- a) Name two ways in which heat is lost through an uninsulated roof.
- b) Explain why trapped air in fibreglass insulation reduces heat loss.
- c) State one advantage of insulation, other than reducing a household's energy bills.
An attic conversion in Dundalk has a sloped roof of area $30\ \text{m}^2$. The roof consists of $12\ \text{mm}$ plasterboard ($k = 0.25\ \text{W m}^{-1}\text{K}^{-1}$),…
An attic conversion in Dundalk has a sloped roof of area $30\ \text{m}^2$. The roof consists of $12\ \text{mm}$ plasterboard ($k = 0.25\ \text{W m}^{-1}\text{K}^{-1}$), $150\ \text{mm}$ mineral wool ($k = 0.035\ \text{W m}^{-1}\text{K}^{-1}$) and $20\ \text{mm}$ of timber boarding ($k = 0.13\ \text{W m}^{-1}\text{K}^{-1}$). Ignore surface resistances. For a layered material, the U-value is $U=\dfrac{1}{\sum L/k}$.
- a) Calculate the U-value of the roof.
- b) Calculate the heat loss rate when the inside is at $20\,^\circ\text{C}$ and the outside at $-2\,^\circ\text{C}$.
- c) Building regulations require a U-value of at most $0.16\ \text{W m}^{-2}\text{K}^{-1}$. Calculate the minimum thickness of mineral wool needed to meet this, keeping the other layers unchanged.
- d) Explain why increasing the thickness gives diminishing returns in reducing heat loss.
A thermal imaging camera shows that the windows of a house in Letterkenny are much brighter (hotter) than the walls on a cold night. a) Explain how the camera forms an…
A thermal imaging camera shows that the windows of a house in Letterkenny are much brighter (hotter) than the walls on a cold night.
- a) Explain how the camera forms an image of the house, and why it can detect hot surfaces in the dark.
- b) A window has U-value $2.8\ \text{W m}^{-2}\text{K}^{-1}$ and area $2.0\ \text{m}^2$, while a wall section of the same area has U-value $0.25\ \text{W m}^{-2}\text{K}^{-1}$. For a temperature difference of $20\ \text{K}$, calculate the heat loss through each and the ratio of losses.
- c) Explain why the outer surface of the window is warmer than the outer surface of the wall.
- d) A heat pump is proposed to heat the house. Explain, with reference to the first and second laws of thermodynamics, how it can deliver more heat energy than the electrical energy it uses without violating conservation of energy.
Energy use and cost
Power is energy per second, so energy $= Pt$. In the SI system the energy is in joules, with $P$ in W and $t$ in s. Your electricity bill uses a bigger unit, the kilowatt-hour (kWh): the energy used by a $1\ \text{kW}$ appliance running for $1$ hour.
For kWh, use $P$ in kW and $t$ in hours. The conversion to joules is why a kWh is much bigger than the joule: $3{,}600{,}000\ \text{J}$.
Now you can see what insulation is worth. If a retrofit lowers the heat loss, the heating must supply less power, so the energy and the cost both fall in the same proportion.
A heating system uses $5.0\ \text{kW}$ of power for $4.0$ hours on a cold winter day in Sligo. a) Calculate the energy used in $\text{kWh}$. b) Calculate the energy used…
A heating system uses $5.0\ \text{kW}$ of power for $4.0$ hours on a cold winter day in Sligo.
- a) Calculate the energy used in $\text{kWh}$.
- b) Calculate the energy used in joules.
- c) At a cost of €0.35 per $\text{kWh}$, calculate the cost of running the system for this time.
A house in Athlone has external walls of total area $120\ \text{m}^2$ with U-value $1.20\ \text{W m}^{-2}\text{K}^{-1}$. After retrofitting with external insulation the…
A house in Athlone has external walls of total area $120\ \text{m}^2$ with U-value $1.20\ \text{W m}^{-2}\text{K}^{-1}$. After retrofitting with external insulation the U-value falls to $0.20\ \text{W m}^{-2}\text{K}^{-1}$. The average temperature difference between inside and outside over the heating season is $10\ \text{K}$.
- a) Calculate the rate of heat loss through the walls before and after retrofitting.
- b) The heating season lasts $200$ days with the heating on continuously. Calculate the energy saved in $\text{kWh}$.
- c) The heating system is oil-fired and delivers heat at a cost of €0.12 per $\text{kWh}$. Calculate the money saved over the season.
A school hall in Ennis has a floor area of $200\ \text{m}^2$. The building envelope (walls, roof, windows and floor) has a combined $UA$ value of $350\ \text{W K}^{-1}$.…
A school hall in Ennis has a floor area of $200\ \text{m}^2$. The building envelope (walls, roof, windows and floor) has a combined $UA$ value of $350\ \text{W K}^{-1}$. In addition, ventilation causes a heat loss of $1.2\times10^{2}\ \text{W K}^{-1}$. The hall is kept at $19\,^\circ\text{C}$ while the average outside temperature during a school term of $100$ days is $7\,^\circ\text{C}$. The hall is heated for $10$ hours on each of $100$ days.
- a) Calculate the total heat loss coefficient and the heating power required.
- b) Calculate the energy needed in $\text{kWh}$ over the term.
- c) The school considers installing a heat pump with an average COP of $3.2$ to replace a gas boiler of efficiency $90\%$. Gas costs €0.09 per $\text{kWh}$ and electricity €0.34 per $\text{kWh}$. Calculate the cost of each option over the term.
- c) Comment on the result and on one non-financial factor the school should consider.
Heat pumps
An electric heater turns every joule of electrical energy into one joule of heat. A heat pump does better. It uses electrical energy to move heat from a cold place (the outside air, the ground or water) to the inside of the house, like a fridge working in reverse. It delivers more heat than the electrical energy it uses, and the extra comes from the surroundings. This does not break the conservation of energy.
By conservation of energy, heat delivered $=$ electrical energy $+$ heat taken from outside. So the heat taken from outside $=$ heat delivered $-$ electrical input.
A heat pump takes in $1.0\ \text{kW}$ of electrical power and delivers $3.5\ \text{kW}$ of heating power to a house in Tralee. a) Calculate the coefficient of…
A heat pump takes in $1.0\ \text{kW}$ of electrical power and delivers $3.5\ \text{kW}$ of heating power to a house in Tralee.
- a) Calculate the coefficient of performance (COP) of the heat pump, defined as the heat delivered divided by the electrical energy input.
- b) Calculate the power the heat pump extracts from the outside air.
- c) Compare the running cost over $10$ hours with that of a resistive electric heater delivering $3.5\ \text{kW}$, if electricity costs €0.34 per $\text{kWh}$.
Back to the start: the house with the warm hall most likely has lower U-values (insulated walls, roof and double or triple glazing), so the same $\Delta T$ drives far less power through its walls. This means less energy, and a smaller bill, every winter.
Summary
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The U-value is the heat loss per $\text{m}^2$ per kelvin, in $\text{W m}^{-2}\text{K}^{-1}$. Low is better. The rate of heat loss is $P = UA\Delta T$.
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For a whole house, add $UA$ for each element, then add the ventilation loss: $P = (\sum UA + \text{ventilation})\,\Delta T$.
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Insulation traps still air, which cuts conduction and convection. For layers, $U = 1/\sum(L/k)$, with $L$ in metres.
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A thermal camera detects the infrared emitted by surfaces. Windows look brighter because their outer surface is warmer.
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$E = Pt$, with $1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$. Cost $=$ kWh $\times$ price per kWh.
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A heat pump has $\text{COP} = \dfrac{\text{heat delivered}}{\text{electrical input}}$. It takes the difference from the outside air or ground.