\amarginfignocaption{b}{
\begin{center}
\begin{tabular}{@{}c@{}}
\lowres{\epsfxsize=53mm\epsfbox{../../images/BWBldgNYS.jpg.eps}}%
{\epsfxsize=53mm\epsfbox{../../images/BWBldgNY.jpg.eps}} \\
\end{tabular}\label{Claire2}
\end{center}
%\caption[a]{ }
}
% \section{What determines your heating bill?}
A perfectly sealed and insulated building would
hold heat for ever and thus would need no heating.
The two dominant reasons why buildings lose heat are:\index{building!heat consumption}
\ben
\item
{\bf Conduction} --
heat flowing directly through walls, windows and doors;
\item
{\bf Ventilation} --
hot air trickling out through cracks, gaps, or deliberate ventilation
ducts.
\een
In the standard model for heat loss, both these heat flows
are proportional to the temperature difference between the
air inside and outside.
For a typical British house, conduction is the bigger
of the two losses, as we'll see.
\subsection{Conduction loss}
%\margintab{
The rate of conduction of heat through a wall, ceiling, floor, or window
is the product of three things: the area of
the wall, a measure of conductivity of the wall
known in the trade as the ``U-value'' or \ind{thermal transmittance},
and the temperature difference --
\[
\mbox{power loss} =
\mbox{area}
\times
U
\times
\mbox{temperature difference} .
\]
The \ind{U-value} is usually measured in
\Wmm/K\@. (One \ind{kelvin}
(1\,K) is the same as one degree \ind{Celsius} (1\degreeC).)
Bigger U-values mean bigger losses of power.
The thicker a wall is, the smaller its U-value.
%Single-glazed window
%5 \Wmm/K, while an unfilled cavity wall would be about
%1.4 \Wmm/K\@.
Double-glazing is about as good as a solid brick wall.
%% my new front door is going to be 1.8, and its glass will be 1.6.
(See \tabref{tab.Uv}.)
The U-values of objects that are ``in series,''
such as a wall and its inner lining, can be combined
in the same way that electrical
conductances combine:
\[
u_{\rm series\ combination} = 1\left/ \left( \frac{1}{u_1} + \frac{1}{u_2} \right)
\right. .
\]
There's a worked example using this rule
on page \pageref{pUseries}.
{%%%%%%%%%%%%%%%%%%%%%% troublesome
% put me BEFORE troublesome heating. diagram page % see AFTER
\renewcommand{\floatpagefraction}{0.8}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Ventilation loss}
To work out the heat required to warm up incoming cold air,
we need the heat capacity of air:
% The heat capacity of air per unit volume is
% 1\,kJ/kg/\ndegreeC $\times$ 1.2\,kg/m$^3 = 1.2\,\kJ/\m^3/\ndegreeC$.
$1.2\,\kJ/\m^3/$K\@.
In the building trade, it's conventional to
describe the power-losses caused by \ind{ventilation} of a space
as the product of
the number of changes $N$ of the air per hour,\index{air changes}
the volume $V$ of the space in cubic metres,
\margintab{
\begin{center}
% http://www.diydata.com/planning/ch_design/sizing.php
\begin{tabular}{lc} \toprule
kitchen & 2\\
bathroom & 2\\
lounge & 1\\
bedroom & 0.5 \\
\bottomrule
\end{tabular}
\end{center}
\caption[a]{Air changes\index{air changes}\index{ventilation}
per hour: typical values of $N$
for draught-proofed rooms.\index{draught proofing}
The worst draughty rooms might have $N=3$ air changes per hour.
% One cubic metre
The recommended minimum
rate of air exchange is between 0.5 and 1.0 air changes per hour,
providing adequate fresh air for human health, for safe combustion of fuels and to prevent
damage to the building fabric from excess moisture in the air (EST 2003).
}
}%
the heat capacity $C$, and the temperature difference $\Delta T$
\begin{table}\figuremargin{
\begin{tabular}{lrrr} \toprule
& \multicolumn{3}{c}{U-values (\Wmm/K)} \\ \cmidrule{2-4}
& old & modern & best \\
& \makebox[0in][r]{buildings} & standards & methods \\
\midrule
Walls & & 0.45--0.6 & 0.12 \\
% Single-glazed window & 5\,\Wmm/K\\
% Unfilled cavity wall & 1.4\,\Wmm/K \\
\ \ solid masonry wall & 2.4 \\
% http://www.diydata.com/information/u_values/u_values.php
\ \ outer wall: 9\,inch solid brick & 2.2\\%.
\ \ 11\,in brick-block cavity wall, unfilled & 1.0\\%.
\ \ 11\,in brick-block cavity wall, insulated & 0.6\\%.
%%
\midrule
Floors & & 0.45 & 0.14 \\
\ \ suspended timber floor& 0.7\\%.
\ \ solid concrete floor & 0.8\\%.
\midrule
Roofs & & 0.25 & 0.12 \\
\ \ flat roof with 25\,mm insulation& 0.9\\%.
\ \ pitched roof with 100mm insulation & 0.3\\%.
\midrule
Windows & & & 1.5 \\
\ \ single-glazed & 5.0\\%.
\ \ double-glazed & 2.9\\%.
\ \ double-glazed, 20\,mm gap & 1.7\\%.
\ \ triple-glazed & 0.7--0.9\\%. sources: http://www.building.co.uk/story.asp?sectioncode=482&storycode=3110977&c=3
%% + Sustainable Solar Housing By Robert Hastings, Maria Wall, International Energy Agency Solar Heating
%% my new front door is going to be 1.8, and its glass will be 1.6.
\bottomrule
\end{tabular}}{
\caption[a]{U-values of walls, floors, roofs, and windows.}\label{tab.Uv}
%% Useful U-values:
}
\end{table}
%}
\begin{figure}
\figuremargin{%
\mbox{\epsfbox{metapost/heating.101}}
}{
\caption[a]{U-values required by British
and Swedish building regulations.\index{building!regulations}
}
\label{BuildingRegFig}
}
\end{figure}
%% /home/mackay/images/071201
between the inside and outside of the building.
% in appropriate units, which turns out to be $1/3$:
\beqan
\begin{tabular}{c} \mbox{power}\\
{(watts)} \\
\end{tabular}
&=& C \frac{N}{1\, \h} V (\m^3) \Delta T (\degreesK)\\
&=& (1.2\,\kJ/\m^3/\degreeK) \frac{N}{3600\, \s} V (\m^3)
\Delta T (\degreesK)
\\
&=&
%\ = \
\frac{1}{3} N V \Delta T .
\eeqan
\subsection{Energy loss and temperature demand (degree-days)}
% Again, the total energy lost is proportional to a
% property of the building times the
% temperature demand.
Since energy is power $\times$ time, you
can write the energy lost by {\em{conduction}\/} through an area
in a short duration as
\[
\mbox{energy loss} =
\leakcol{\mbox{area}
\times
U}
\times
(\Delta T \times \mbox{duration}) ,
\]
\index{convective heat loss rate}and the energy lost by {\em{ventilation}\/} as
\[
\leakcol{\frac{1}{3} N V} \times
(\Delta T\times \mbox{duration}).
\]
Both these energy losses have the form
\[
\leakcol{\mbox{Something}} \times (\Delta T\times \mbox{duration}),
\]
where the ``\leakcol{Something}'' is measured in watts per \ndegreeC\@.
As day turns to night, and seasons pass, the
temperature difference $\Delta T$ changes; we can think of a long period
as being chopped into lots of small durations, during each of which
the temperature difference is roughly constant.
From duration to duration, the temperature difference changes,
but the Somethings don't change.
When predicting a space's
total energy loss due to conduction and ventilation over a long period
we thus need to multiply two things:
\begin{enumerate}
\item
the sum of all the \leakcol{Somethings} (adding
\leakcol{$\mbox{area} \times U$}
for all walls, roofs, floors, doors, and windows,
and \leakcol{$\frac{1}{3} N V$} for the volume);
and
\item
the sum of all the
$\mbox{Temperature difference} \times \mbox{duration}$
factors (for all the durations).
\end{enumerate}
%% I got 15.5 1869.41 (explanation -- I use day averages, not half-hour)
%% made by daily.pl and gnudd
\begin{figure}[htbp]
\figuremargin{\small
\begin{center}
\begin{tabular}[b]{l}
\begin{tabular}[b]{ccc}
\raisebox{2cm}{(a)}&
\raisebox{2.653cm}{\makebox[0in][l]{\sf temperature (\ndegreesC)}}%
\mono%
{\epsfxsize=4in\epsfbox{../data/cambridge/mono/Cam2006Temp2020.eps}}%
{\epsfxsize=4in\epsfbox{../data/cambridge/Cam2006Temp2020.eps}}
\\
\raisebox{2cm}{(b)}&
\raisebox{2.653cm}{\makebox[0in][l]{\sf temperature (\ndegreesC)}}%
\mono%
{\epsfxsize=4in\epsfbox{../data/cambridge/mono/Cam2006Temp3017.eps}}%
{\epsfxsize=4in\epsfbox{../data/cambridge/Cam2006Temp3017.eps}}
\\
% \raisebox{2cm}{(c)}&\mbox{\epsfxsize=4in\epsfbox{../data/cambridge/Cam2006Temp3015.eps}} \\
\end{tabular}
\end{tabular}
\end{center}
}{
\caption[a]{The temperature demand
in Cambridge, 2006, visualized
as an area on a graph of daily average temperatures.
(a) Thermostat set to 20\degreesC, including
cooling in summer;
(b) winter thermostat set to 17\degreesC\@.
% ; (c) Winter thermostat set to 15\degreesC\@.
} \label{degreeday}
}
\end{figure}%
The first factor is a property of the building measured in
watts per \degreeC\@. I'll call this the {\sl\ind{leakiness}\/}
of the building.
%\[
% \mbox{Leakiness} = \sum
%\]
%
(This leakiness is sometimes called the building's
{\dem\ind{heat-loss coefficient}}.)
The second factor is a property of the weather; it's
often expressed as a number of ``\ind{degree-day}s,''
since temperature difference is measured in degrees, and
days are a convenient unit for thinking about durations.
For example, if your house interior is at 18\degreesC,
and the outside temperature is 8\degreesC\ for a week, then
we say that that week contributed
$10 \times 7= 70$ degree-days to the $(\Delta T
% \mbox{Temperature difference}
\times \mbox{duration})$ sum.
I'll call the sum of all the $(\Delta T
\times \mbox{duration})$ factors the {\dem{\ind{temperature demand}}\/} of a period.
\amarginfig{t}{\small
\begin{center}
\begin{tabular}[b]{l}
\raisebox{6.51cm}{\makebox[0in][l]{\sf\small \begin{tabular}{l}{temperature demand}\\{(degree-days per year)}\\ \end{tabular}}}%
\mono%
{\epsfxsize=50mm\epsfbox{../data/cambridge/mono/Cam2006dd.eps}}%
{\epsfxsize=50mm\epsfbox{../data/cambridge/Cam2006dd.eps}}
\\
\end{tabular}
\end{center}
\caption[a]{Temperature demand in Cambridge, in degree-days per year,
% for the year 2006 in Cambridge, in degree days,
as a function of thermostat setting (\degreesC).
Reducing the winter thermostat from 20\degreesC\ to
17\degreesC\ reduces the temperature demand of heating
by 30\%, from 3188 to 2265 degree-days.
Raising the summer thermostat from 20\degreesC\ to
23\degreesC\ reduces the temperature demand of cooling
by 82\%, from 91 to 16 degree-days.
}
\label{degreedayc}
}
\[
\mbox{energy lost } = \mbox{leakiness} \times \mbox{temperature demand}.
\]
We can reduce our energy loss by reducing
the leakiness of the building, or
by reducing
our temperature demand, or both.
The next two sections look more closely
at these two factors, using a house in Cambridge as a case-study.
There is a third factor we must also discuss.
The lost energy is replenished by the building's heating system,
and by other sources of energy such as the occupants,
their gadgets, their cookers, and the sun.
Focussing on the heating system,\index{heating!efficiency}
the energy {\em{delivered}\/} by the heating is not the same as the
energy {\em{consumed}\/} by the heating. They are related by
the {\dem\ind{coefficient of performance}\/} of the heating system.
\[
\mbox{energy consumed} =
\mbox{energy delivered} / \mbox{\ind{coefficient of performance}} .
\]
% For some simple heaters, such as electric bar fires,
% the coefficient of performance is 1.
For a \ind{condensing boiler} burning natural gas, for example,
the coefficient of performance
is 90\%, because 10\% of the energy is lost up the chimney.
To summarise, we can reduce the
energy consumption of a building in three ways:
\begin{enumerate}
\item
by reducing temperature demand;
\item
by reducing leakiness; or
\item
by increasing the coefficient of performance.
\end{enumerate}
%\[
%\mbox{energy lost } = \mbox{leakiness} \times \mbox{temperature demand}.
%\]
We now quantify the potential of these options.
% potential of each of these modifications.
(A fourth option -- increasing the building's
incidental heat gains, especially from the sun -- may also
be useful, but I won't address it here.)
\subsection{Temperature demand}
We can visualize the temperature demand nicely on a
graph of external temperature versus time (\figref{degreeday}).
For a building held at a temperature of 20\degreesC,
the total temperature
demand is the {\em{area}\/} between the
horizontal line at 20\degreesC\ and the external temperature.
In \figref{degreeday}a, we see that, for one year
in Cambridge, holding the
temperature at 20\degreesC\ year-round had
a temperature demand of
3188 degree-days of heating and
91 degree-days of cooling.
These pictures allow us easily to assess the effect of
turning down the thermostat and living without air-conditioning.
Turning the winter thermostat down to 17\degreesC,
the temperature demand for heating drops from
3188 degree-days to
2265 degree-days (\figref{degreeday}b), which corresponds
to a 30\% reduction in heating demand.
Turning the thermostat down to 15\degreesC\ reduces the temperature
demand from 3188 to 1748 degree days,
% (\figref{degreeday}c),
a 45\% reduction.
These calculations give us a ballpark indication of the
benefit of turning down thermostats, but will
give an exact prediction only if we take into account two details:
first, buildings naturally absorb energy from the sun, boosting the
inside above the outside temperature, even
\amarginfig{b}{
\begin{center}
\begin{tabular}{l}
\mono%
{\epsfxsize=50mm\epsfbox{../data/cambridge/mono/Cam2006ddd.eps}}%
{\epsfxsize=50mm\epsfbox{../data/cambridge/Cam2006ddd.eps}} \\
\end{tabular}
\end{center}
%}{
\caption[a]{The temperature demand
in Cambridge, 2006, replotted in units of degree-days per day,
also known as degrees.
In these units, the temperature demand is just the average
of the temperature difference between inside and outside.
}
\label{degreedayd}
}%
%% gnudd
without any heating; and second, the occupants and their gadget companions
emit heat, so further cutting down the artificial heating requirements.
% The Carbon Trust suggest correcting
% The base temperature used to calculate temperature demand
% degree days
% in the UK is 15.5\degreesC\@.
%% http://www.carbontrust.co.uk/resource/degree_days/what_are.htm
%% double check their figures for 2006:
%% SouthEast pr 335+334+321+ 191+96+39+7+27+22+82+207+268 = 1929
%% EastAnglia pr 350+325+343+222+109+ 54+ 11+ 31+ 19+ 75+217+290 =2046
%% gnudd
The temperature demand of a location, as conventionally expressed
in \ind{degree-day}s, is a bit of an unwieldy thing. I find it hard to
remember numbers like ``3500\,degree-days.'' And \ind{academics}
may find
the degree-day a distressing unit, since they already have another
meaning for degree days (one involving dressing up in \ind{gown}s
and \ind{mortar board}s).
We can make this quantity more meaningful and perhaps easier
to work with by dividing it by 365, the number of days in the year,
obtaining the temperature demand in ``degree-days per day,''
or, if you prefer, in plain ``degrees.'' \Figref{degreedayd}
shows this replotted temperature demand.
Expressed this way, the temperature demand is simply the
{\em{average}\/} temperature difference between inside and outside.
The highlighted temperature demands are:
8.7\degreesC, for a thermostat setting of 20\degreesC;
6.2\degreesC, for a setting of 17\degreesC;
and 4.8\degreesC, for a setting of 15\degreesC\@.
\subsection{Leakiness -- example: my house}
\amarginfig{c}{
%\begin{figure}
%\figuremargin{
\begin{center}
\begin{tabular}{@{}c@{}}
\lowres{\mbox{\epsfxsize=53mm\epsfbox{../../images/HouseSnowS.eps}}}%
{\mbox{\epsfxsize=53mm\epsfbox{../../images/HouseSnow.jpg.eps}}} \\
\end{tabular}
\end{center}
%}{
\caption[a]{My house.}\label{fig.MyHo}
%% , yesterday.}
}%
% case study
My house is a three-bedroom semi-detached house built
about 1940 (\figref{fig.MyHo}). By 2006, its kitchen had been slightly extended,
and most of the windows were double-glazed. The front
door and back door were both still single-glazed.
My estimate of the leakiness in 2006 is built up as shown in
\tabref{tab.LeakiBreak}.
\begin{table}
\figuremargin{
\begin{tabular}{lrrr} \toprule
{\sc{Conductive leakiness}}
& area & U-value & leakiness \\
& (m$^2$) & (\Wmm$\!$/\ndegreeC) & (W$\!$/\ndegreeC) \\
\midrule
\multicolumn{4}{l}{Horizontal surfaces}\\%Floor and ceilings} \\
% \midrule
\ \ \ Pitched roof & 48 & {\modcol{0.6}} & 28.8\\
\ \ \ Flat roof & 1.6 & 3 & 4.8\\
\ \ \ Floor & 50 & 0.8 & 40 \\
\midrule
\multicolumn{4}{l}{Vertical surfaces}\\% Walls \\
%\midrule
\ \ \ Extension walls & 24.1 &0.6 & 14.5\\
\ \ \ Main walls & 50 &{\modcol{1}}&50 \\
\ \ \ Thin wall (5\inch) & 2 &3 & 6 \\
\ \ \ Single-glazed doors and windows & 7.35 &{\modcol{5}} & 36.7\\
\ \ \ Double-glazed windows & 17.8 &2.9& 51.6\\
\midrule
\multicolumn{3}{l}{Total conductive leakiness} & 232.4\\
\bottomrule
\end{tabular}
\begin{tabular}{lccr}\toprule
{\sc{Ventilation leakiness}}
%\\
% Space
& volume & $N$ & leakiness \\
& (m$^3$) & (air-changes per hour) & (W$\!$/\ndegreeC) \\ \midrule
\ Bedrooms & 80 & 0.5 & 13.3\\
\ Kitchen & 36 & 2 & 24 \\
\ Hall & 27 & {\modcol{3}} & 27 \\
\ Other rooms& 77& 1 & 25.7 \\
\midrule
\multicolumn{3}{l}{Total ventilation leakiness} & 90 \\
\bottomrule
\end{tabular}\smallskip
% total volume of house : 80+36+27+77 = 220
}{
\caption[a]{
Breakdown of my house's conductive leakiness, and its
ventilation leakiness, pre-2006.
I've treated the central wall of the semi-detached house as
a perfect insulating wall, but this may be wrong if the
gap between the adjacent houses is
actually well-ventilated.\smallskip
I've highlighted the parameters that I altered
after 2006, in modifications to be described shortly.
% argon is used because it has conductivity 0.67 x that of air (especially
% because of its low heat capacity)
\index{leakiness}\index{wall}\index{door}\index{window}\index{double glazing}\index{glazing}\index{U-value}
}\label{tab.LeakiBreak}
}
\end{table}
The total leakiness of the \ind{house} was 322\,\WC\ (or 7.7\,kWh/d/\ndegreeC),
with conductive leakiness accounting for 72\%
and ventilation leakiness for 28\% of the total.
The conductive \ind{leakiness} is roughly equally divided into three parts:
windows; walls; and floor and ceiling.
% pr 322 * 24 / 1000.0
% We can also express the leakiness as .
To compare the leakinesses of two buildings that have different
floor areas,\index{building!leakiness}\index{building!heat-loss parameter} we can
% One way of presenting the leakiness that allows one building to
% be compared with another is to
divide the leakiness
by the floor area; this gives the
{\dem\ind{heat-loss parameter}\/}
of the building, which is measured in W/\ndegreeC/m$^2$.
The \ind{heat-loss parameter} of this house (total floor area 88\,m$^2$)
is
\[
% 322 / 88
3.7\,\mbox{W/\ndegreeC/m$^2$}.
\]
Let's use these figures
% from the previous section
to estimate the house's daily energy consumption on a cold winter's day,
and year-round.
On a cold day, assuming an external temperature of $-1$\degreeC\ and
%Standard external temperature: $-1$\degree C\@.
%http://www.diydata.com/planning/ch_design/sizing.php
%Standard internal temperature: 16--21\degree C\@.
an internal temperature of 19\degreeC,
the temperature difference is $\Delta T = 20 \degreeC$.
If this difference is maintained for 6 hours per day
then the energy lost per day is
\[
322\,\W/\ndegreeC \times 120\,\mbox{degree-hours}
\simeq 39\,\kWh .
\]
If the temperature is maintained at 19\degreesC\ for 24 hours per day,
the energy lost per day is
\[
155 \,\kWh/\d.
\]
To get a year-round heat-loss figure, we can take the temperature demand
of Cambridge from \figref{degreedayc}.
With the thermostat at 19\degreesC, the temperature demand
in 2006 was 2866 degree-days.
The average rate of heat loss, if the house is always held
at 19\degreesC, is therefore:
\[
\mbox{7.7\,kWh/d/\ndegreeC} \times 2866\,\mbox{degree-days/y}
/ (\mbox{365\,days/y}) = 61\,\kWh/\d.
\]
Turning the thermostat down to 17\degreesC,
the average rate of heat loss drops to 48\,kWh/d.
Turning it up to a tropical 21\degreesC, the average rate of heat loss
is 75\,kWh/d.
%Background
%
%The model says losses on a cold winter day are about 37 kWh/d by
%conduction and about 13 kWh/d by ventilation/draughts.
\subsubsection{Effects of extra insulation}
During 2007, I made the following modifications to the house:
\ben
\item
Added \ind{cavity-wall insulation} (which was missing in the main walls of the house) -- \figref{fig.Cavit}.
\item
Increased the insulation in the roof.
\item
Added a new front door outside the old -- \figref{frontD}.
\item
Replaced the back door with a double-glazed one.
\item
Double-glazed the one window that was still single-glazed.
\een
%% TODO *** change order to match table (merge 4/5)
What's the predicted change in heat loss?
The total leakiness before the changes was 322\,\WC\@.
Adding cavity-wall insulation (new U-value 0.6) to the main walls reduces
the house's leakiness by 20\,\WC\@.
% their leakiness from 50 to 30\,\WC\@.
The improved loft insulation (new U-value 0.3) should reduce
the leakiness by 14\,\WC\@.
The glazing modifications (new U-value 1.6--1.8) should reduce the conductive
leakiness by 23\,\WC, and the ventilation
leakiness by something like 24\,\WC\@.
%% 51-27 -- kitchen and hall
%% 24+23+14+20
That's a total reduction in leakiness of 25\%,
% 81\,W/\ndegreeC,
from roughly 320 to 240\,W/\ndegreeC\ (7.7 to
6\,kWh/d/\ndegreeC).
\Tabref{tabPREDhr} shows the predicted savings from each of
the modifications.
The \ind{heat-loss parameter} of this house (total floor area 88\,m$^2$)
is thus hopefully reduced by about 25\%, from 3.7 to
$2.7\,\mbox{\WC/m$^2$}$. (This is a long way from the 1.1\,\WC/m$^2$
required of a ``sustainable'' house in the new building codes.)
% put me AFTER troublesome heating. diagram page % see BEFORE
% END \renewcommand{\floatpagefraction}{0.8}
}
%%%
\begin{table}[!b]
\figuremargin{
\begin{center}
\begin{tabular}{lp{3in}r} \toprule
--& Cavity-wall insulation (applicable to two-thirds of the wall area) & 4.8\,kWh/d\\
-- & Improved roof insulation & 3.5\,kWh/d \\
--& Reduction in conduction from double-glazing two
doors and one window & 1.9\,kWh/d \\
% 1.2 front, 0.7 kitchen
--& Ventilation reductions in hall and kitchen from improvements
to doors and windows & 2.9\,kWh/d \\
% 2.1 hall, 0.8 kitchen
\bottomrule
\end{tabular}
\end{center}
% see /home/mackay/sustainable/girton/myhouse/Uvalues.gnumeric
% and _heating.tex
%%% SEE insulation.tex for further details added Mon 2/5/11
}{
\caption[a]{
Break-down of the predicted reductions in heat loss from my house, on a cold winter day.
}\label{tabPREDhr}
}
\end{table}
It's frustratingly hard to make a really big dent in the
leakiness of an already-built house!
As we saw a moment ago,
a much easier way of achieving a big dent in heat loss
is to turn the thermostat down. Turning down from 20 to 17\degreesC\
gave a reduction in heat loss of 30\%.
%% I find that 16 is too low when wearing shorts and sitting still
%% for a long time. Though it is fine for getting up,
%% having breakfast, and leaving.
%%
Combining these two actions -- the physical modifications
and the turn\-ing-down of the thermostat -- this model predicts that
heat loss should be reduced by nearly 50\%.
% 47.5
Since some heat is generated in a house by
sunshine, gadgets, and humans,
the reduction in gas consumption
should be more than 50\%.
% Put Eden and Bending material here.
I made all these changes to my house
and monitored my meters every week.
I can confirm that my heating
bill indeed went down by more than 50\%.
% In fact, owing to my slow writing pace, I can tell you here and now:
As \figref{fig.gas0} showed, my gas consumption
has gone down from 40\,kWh/d to 13\,kWh/d -- a reduction of 67\%.
%% RHFriend heat pump numbers
%% Source LG website. S18AW.
%%efficiency 380% or 360%
%% ground-source heat pump
% 1m below ground, temp is almost constant, about 7-13C\@.
%% source: www.iceenergy.co.uk/
%% use heat pump from 7-13 to 50C, ok for underfloor heating (but not conventional
%% radiators)
%% efficiency would be 700%.
%% 80C radiators would be less efficient.
%% ``I am saving 75%''
%Estimated loss, when temperature difference is 20\degreesC:
%{\em what's missing here?}
\subsubsection{Leakiness reduction by internal wall-coverings}
Can you reduce your walls' leakiness by covering the {\em{inside}\/} of the
wall with \ind{insulation}?
The answer is yes, but there may be two complications.
First, the thickness of internal covering is bigger than you might expect.
To transform an existing nine-inch solid brick wall (U-value 2.2\,\Wmm/K)
into a decent 0.30\,\Wmm/K wall, roughly 6\,cm of insulated lining board
is required
\tinyurl{65h3cb}{http://www.dorset-technical-committee.org.uk/reports/U-values-of-elements-Sept-2006.pdf}.
Second, \ind{condensation} may form on the hidden surface of
such internal \ind{insulation} layers, leading to \ind{damp} problems.
If you're not looking for such a big reduction in wall leakiness,
you can get by with a thinner internal covering.
For example, you can buy 1.8-cm-thick
% thermaline REVEAL phenolic foam and wallboard
insulated wallboards with a U-value of 1.7\,\Wmm/K\@.
With these over the existing wall, the
U-value would be reduced from 2.2\,\Wmm/K
to:\label{pUseries}
\[
1 \left/ \left( \frac{1}{2.2} + \frac{1}{1.7} \right) \right. \: \simeq \: 1\,\Wmm/K.
% 0.95
\]
Definitely a worthwhile reduction.
% see also Celotex tuff-R GA3000 and Celotex T-Break TB3000
\section{Air-exchange}
Once a building is really well insulated, the principal loss of heat will be
through \ind{ventilation} (\ind{air changes}) rather than through conduction.
The heat loss through ventilation can be reduced by transferring the heat
from the outgoing air to the incoming air.
Remarkably, a great deal of this heat can indeed be transferred without
any additional energy being required.
The trick is to use a \ind{nose},\index{nostril} as discovered by natural selection.
A nose warms incoming air by cooling down outgoing air.
There's a temperature gradient along the nose;
the walls of a nose are coldest near the nostrils.
% losest to the external temperature
\index{counter-current heat exchange}The longer your nose, the better it works as
a counter-current heat exchanger.
% Noses also use the same principle to reduce water-loss.
In nature's noses, the direction of the air-flow usually alternates.
Another way to organize a nose is to have two air-passages, one
for in-flow and one for out-flow, separate from the point of view of air,
but tightly coupled with each other so that heat can
easily flow between the two passages. This is how the noses work in buildings.
It's conventional to call these noses heat-exchangers.\index{heat exchanger}
{%%%%%%%%%%%%%%%%%%%%%% troublesome
% put me BEFORE troublesome heating. diagram page % see AFTER
\renewcommand{\floatpagefraction}{0.8}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%*** could move the house description into the margin; need to get a photo;
%% could put the heat exchang photo here.
\section{An energy-efficient house}
In 1984,
\marginfig{
\begin{center}
\begin{tabular}{@{}c@{}}
{\mbox{\epsfxsize=53mm\epsfbox{../../images/Serrekunda.eps}}}
\\
\end{tabular}
\end{center}
\caption[a]{The \ind{Heatkeeper} Serrekunda.
}\label{Heatkeeper}
}%
an energy consultant, Alan Foster, built
an energy-efficient house near Cambridge; he kindly gave me
his thorough measurements.
% Serrekunda Heatkeeper
The house is a timber-framed\index{building!Heatkeeper}
bungalow based on a Scandinavian ``\ind{Heatkeeper} Serrekunda''
design (\figref{Heatkeeper}),
with a floor area of 140\,m$^2$,
composed of three bedrooms, a study, two bathrooms,
a living room, a kitchen, and a lobby.
% the garage not included.
The wooden outside walls were
supplied in kit form by a Scottish company, and the main parts
of the house took only a few days to build.
% James Walker Ltd, cost was 25K pounds
% Timber
% Wall area 90.51, window area 31.69
% roof area 137.
% Venitlated volume 300 m**3, 40% loss in heat exchanger 0.38 air changes
% per hour. Specific heat 0.33. (gives 296 watts at 20C temp diff)
The walls are 30\,cm thick and have a U-value of 0.28\,\Wmm/\ndegreeC.
From the inside out, they consist of 13\,mm of plasterboard, 27\,mm
airspace, a vapour barrier, 8\,mm of plywood, 90\,mm of rockwool, 12\,mm
of bitumen-impregnated fibreboard, 50\,mm cavity, and 103\,mm of brick.
% The external bricks are not load-bearing.
% These walls
The ceiling construction is similar with 100--200\,mm of rockwool insulation.
The ceiling has a U-value of 0.27\,\Wmm/\ndegreeC, and the floor,
0.22\,\Wmm/\ndegreeC\@.
% calculated U values 0.168 round edges and 0.285 in centre.
% Floor = 50mm styrofoam with 40mm bearers supporting 19mm chipboard flooring.
% above a standard concrete floor.
The windows are double-glazed (U-value 2\,\Wmm/\ndegreeC),
with the inner panes' outer surfaces
specially coated to reduce radiation.
The windows are arranged to give substantial solar gain,
contributing about 30\% of the house's space-heating.
% In summer the orientation
% of the windows can be reversed to reduce solar gain.
% hermetically
The house is well sealed, every door and window lined with
neoprene gaskets.
The house is heated by warm air pumped through floor grilles;
in winter, pumps remove used air from several rooms, exhausting
it to the outside, and they take in air from the loft space.
The incoming air and outgoing air pass through a heat exchanger
(\figref{HeatExchanger}),
\amarginfig{t}{
\begin{center}
\begin{tabular}{@{}c@{}}
{\mbox{\epsfxsize=40mm\epsfbox{../../images/HeatEx.eps}}}
\\
\end{tabular}
\end{center}
\caption[a]{The \ind{Heatkeeper}'s heat-exchanger.
}\label{HeatExchanger}
}%
which saves 60\% of the heat in the extracted air.
The heat exchanger is a passive device, using no
energy: it's like a big metal nose, warming the incoming
air with the outgoing air.
On a cold winter's day, the outside air temperature was $-8$\degreesC,
the temperature in the loft's air intake was 0\degreesC, and the air
coming out of the heat exchanger was at $+8$\degreesC\@.
For the first decade,
the heat was supplied entirely by electric heaters, heating
a 150-gallon heat store during the overnight economy period.
% using four 3kW immersion heaters, to 92C
More recently a gas supply was brought to the house, and the space
heating is now obtained from a condensing boiler.
% Domestic hot water: big 56 gallon cylinder, electric heated to
% 60C overnight.
% for Dom HW, 13297 kWh in 8.333 y which is 1595 per y
The heat loss through conduction and ventilation is
% 3.51 kW * 24 / 20 h/degree-day
% 4.2\,kWh/degree-day.
4.2\,kWh/d/\ndegreeC\@.
The {\dem\ind{heat loss parameter}\/}
(the leakiness per square metre
of floor area)
is 1.25\,\Wmm/\ndegreeC\
(\cf\ my house's
$2.7\,\mbox{W/\ndegreeC/m$^2$}$).
\begin{figure}
\figuremargin{\small%
\begin{tabular}{@{}cccc@{}}
\multicolumn{4}{c}{%
\mbox{\epsfbox{metapost/heating.123}}
}\\
\raisebox{36mm}{\epsfig{width=36mm,angle=270,file=../../images/OldSchool.eps}}
&
\includegraphics[height=36mm]{../../images/Rutherford.eps}
&
\includegraphics[height=36mm]{../../images/LawFaculty.eps}
&
\includegraphics[height=36mm]{../../images/GatesCL2.eps}
\\
Old Schools
&
Rutherford building
&
Law faculty
&
Gates building\\
\end{tabular}
}{
\caption[a]{Building benchmarks.\index{Heatkeeper}\index{building!Heatkeeper}
Power used per unit area in various homes and offices.
}
\label{FigHeating123}
}
\end{figure}
With the house occupied by two people, the
average space-heating consumption, with the thermostat set at
19 or 20\degreesC\ during the day, was 8100\,kWh per year,
or 22\,kWh/d;
% actual consumption analysed to 3.62 kWh per degree day
% using I think 15.5 as the definition of a zero-degree-day
the total energy consumption for all purposes was
% 1584 kWh for hot water ,
% 14731 kWh for everything
% 16000 in the first year
about 15\,000\,kWh per year, or 40\,kWh/d.
% 8100 kWh/year / 140 m**2 in W/m**2
% 15000 kWh/year / 140 m**2 in W/m**2 ----> 12.2 Wmm
Expressed as an average power per unit area, that's
\Red{6.6\,\Wmm} for heating and
% . TYPO
\Red{12.2\,\Wmm} in total.
% 1.5m fluorescent tubes use 70W.
% Duncan Foster was T Blair's roommate
% His company worked on sizewll A and Wylfa
% In 2002 added an electric storage heater to enhance living room
% which was otherwise 2C lower than rest of house.
% At night (10.30pm to 7pm) thermostat is at 10C\@.
% thermostate settings now are 21 until 9am, 20 until 5pm, 21 until 10.30.
% At an average winter temp difference of 12C, the loss rate is 104.5 kWh/d
% and gains are 65.7 (from solar, cooking, lights, motors),
% so net losses required for heating are 38.78
% breakdown of heat losses (%): roof 44, floor 19, walls 27.5, windows 5.5,
% ventilation 4.
% In condensing gas boiler
\Figref{FigHeating123} compares the power consumption per unit area of
this Heatkeeper house with my house (before and after my efficiency push)
and with the European average. My house's post-efficiency-push consumption
is close to that of the Heatkeeper, thanks to the adoption of lower thermostat
settings.
\section{Benchmarks for houses and offices}
The German \ind{Passivhaus} standard aims for power consumption
for heating and cooling\index{building!Passivhaus}
of 15\,kWh/m$^2\!$/y, which is \Red{1.7\,\Wmm};
and total power consumption
of 120\,kWh/m$^2\!$/y, which is \Red{13.7\,\Wmm}.
% http://www.zerocarbonhouse.org.uk/passivhaus/
% http://www.passiv.de/English/PassiveH.HTM
% European average is 285 kWh / m2 / y which is 32.51Wmm
% My house before: 44 kWh per day per whole house 40+4
% which is 21\Wmm
% My house, now: 15 kWh per day per whole house 13+2
% which is 7.1\Wmm
% 13 kWh per day / 88 m**2 in W/m**2 -> 7.1 total (and 6.2 for heating)
%% SERVICE SECTOR ALL ENERGY:
% floor area of service sector buildings: 854 km**2
% 2005, total energy consumption: 19.320 M toe / y
% energy per person: 10.3 kWh/d.
% area per person: 14 sq m
% energy per unit area : 0.724 kWh/d/sq m = 30 \Wmm
The average energy consumption
of the UK service sector,
per unit floor area, is 30\,\Wmm.
% source \cite{ECUK}
% *** find out the area of their PV panels
\subsection{An energy-efficient office}
The National Energy Foundation built themselves
a low-cost low-energy building.
It has solar panels for hot water, solar photovoltaic (PV)
panels
generating up to 6.5\,kW of electricity,
and is heated by a 14-kW ground-source heat pump
and occasionally by a wood stove.
The floor area is 400\,m$^2$ and the number of
occupants is about 30.
It is a single-storey building.\index{building!energy-efficiency office}
The walls contain 300\,mm of rockwool insulation.
The heat pump's coefficient of performance in winter was 2.5.\label{pPoorCOP}
The energy used is 65\,kWh per year
per square metre of floor area
(\Red{7.4\,\Wmm}). The PV system
delivers almost 20\% of this energy.
% 1050 kWh/y per employee
% put me AFTER troublesome heating. diagram page % see BEFORE
% END \renewcommand{\floatpagefraction}{0.8}
}
\subsection{Contemporary offices}
New office buildings are often hyped up as being amazingly environment-friendly.
Let's look at some numbers.
% about the same time.
The
%% http://www.cabe.org.uk/default.aspx?contentitemid=1192&aspectid=10
William Gates building at Cambridge University\index{building!Cambridge University}
holds computer science researchers, administrators, and a small caf\'e.
% Roughly 274 people work there.
Its area is
11\,110\,m$^2$, and its energy consumption is\index{Cambridge University}
% 1982\,MWh/y.
2392\,MWh/y.
That's a power per unit area of
215\,kWh/m$^2$/y,
% *** check these numbers
or \Red{25\,\Wmm}.
% And 24\,kWh/d per person!
This building won a RIBA award in 2001
for its predicted energy consumption.
``The architects have incorporated many environmentally friendly features into the building.''
\tinyurl{5dhups}{http://www.arct.cam.ac.uk/UCPB/Place.aspx?rid=943658&p=6&ix=8&pid=1&prcid=27&ppid=201}
%%
%% 15 million
But are these buildings impressive?
Next door, the
Rutherford building, built in the 1970s without
any fancy eco-claims -- indeed without even double glazing --
%% Physics Rutherford Building 1595688.00352312 319.265306827356 4998.00000000
%pt9 := p0 yscaled 319.265306827356 xscaled 4.998000000 shifted point 1 of p8 ;
%p9 := p0 yscaled 219.204314866871 xscaled 4.998000000 shifted point 1 of p8 ;
%pg9 := p0 yscaled 92.3427068245512 xscaled 4.998000000 shifted point 3 of p9 ;
% 219.2+92.3 311.5 kWh/m**2/y
has a floor area of 4998 m$^2$
and consumes 1557\,MWh per year;
that's
0.85\,kWh/d/m$^2$, or
36\,\Wmm.
% Roughly 200? people work there.
% So that's 15\,kWh/d per person.
So the award-winning building is just 30\% better, in terms of power per unit area,
than its simple 1970s cousin.
% american eco house company
% http://www.enertia.com/
% ``Just below the surface, within reach of the average basement, is an infinite reservoir of heat that never drops below 50 degrees F.''
% Further reading: \cite{SolarHo}.
\Figref{FigHeating123}
compares these buildings
% Gates building, the Rutherford building,
and another new building, the Law Faculty,
with the Old Schools, which are ancient offices built pre-1890. For all the fanfare,
the difference between the new and the old is really quite disappointing!
% *** get photo of rutherford building or maybe cut reference to it.
Notice that the building power consumptions, per unit floor area,
are in just the same units (\Wmm) as the
renewable powers per unit area that we discussed
on pages \pageref{fig.plants155}, \pageref{plants99}, and \pageref{figW2}.
Comparing these consumption and production numbers helps us
realize how difficult it is to power modern buildings
entirely from \ind{on-site renewables}. The power per unit area of
biofuels (\figref{fig.plants155}, \pref{fig.plants155})
is 0.5\,\Wmm; of {\windfarm}s, 2\,\Wmm;
of solar photovoltaics, 20\,\Wmm\ (\figref{plants99},
\pref{plants99});
only solar hot-water panels come in at the right
sort of power per unit area, 53\,\Wmm\ (\figref{viridian}, \pref{viridian}).
%\marginfig{
%\caption[a]{Ancient and modern buildings}
%}
% So I think it makes sense for planning regulations to insist that
% solar water-heating should be incorporated into new buildings. But
% insisting that other renewable power should be generated on site
% seems a bit odd.
%% put all foster stuff above here
\begin{figure}
\figuredangle{\small
\begin{tabular}[b]{lr}
\multicolumn{1}{c}{\sf Cooling}&
\multicolumn{1}{c}{\sf Heating}\\
\mbox{\epsfbox{metapost/heatpump.22}}&
\mbox{\epsfbox{metapost/heatpump.21}}\\
\mbox{\epsfbox{metapost/heatpump.122}}&
\mbox{\epsfbox{metapost/heatpump.121}}\\
\end{tabular}
}{
\caption[a]{Ideal \ind{heat pump} efficiencies.\index{efficiency!heat pump}
Top left:
ideal electrical energy required, according to the
limits of thermodynamics, to pump heat {\em{out}\/} of a
place at temperature $T_{\rm{in}}$ when the heat is being pumped
to a place at temperature $T_{\rm{out}} = 35\degreesC$.
Right:
ideal electrical energy required to pump heat {\em{into}\/} a
place at temperature $T_{\rm{in}}$ when the heat is being pumped
from a place at temperature $T_{\rm{out}} = 0\degreesC$.
Bottom row: the efficiency is conventionally expressed
as a ``\ind{coefficient of performance},'' which is
the heat pumped per unit electrical energy.
In practice, I understand that well-installed
ground-source heat pumps and the best
air-source heat pumps usually have a
coefficient of performance of 3 or 4; however, government
regulations in Japan have driven the coefficient of performance
as high as 6.6.
}
\label{fig.gshptheory}
}\end{figure}
\section{Improving the coefficient of performance}
% While we are on the topic of heating, we should discuss
% the ultimate energy cost of two alternative styles
% of heating: electricity, or direct combustion.
% Also other options including heat pumps and
% combined heat and power.
You might think that the coefficient of performance
of a condensing boiler, 90\%, sounds pretty hard to beat.
But it can be significantly improved upon, by heat pumps.
Whereas the condensing boiler takes chemical energy and turns
90\% of it into useful heat, the
heat pump takes some electrical energy and uses it to {\em{move}\/}
heat from one place to another (for example, from outside
a building to inside). Usually the amount of useful
heat delivered is much bigger than the amount of electricity
used. A coefficient of performance of 3 or 4 is normal.
\subsection{Theory of heat pumps}
% A full explanation would require a discussion of entropy.
% {\em more here}
% For the time being, I'll just give
Here are the formulae
for the ideal efficiency of a \ind{heat pump}, that is, the electrical energy required
per unit of heat pumped. If we are pumping heat from an outside
place at temperature
$T_1$ into a place at higher temperature $T_2$, both temperatures being expressed
relative to absolute zero (that is, $T_2$, in kelvin,
is given in terms of the Celsius temperature $T_{\rm{in}}$,
by $273.15 + T_{\rm{in}}$), the ideal efficiency is:
\[
\mbox{efficiency} = \frac{T_2}{T_2 - T_1} .
% 1 - \frac{T_1}{T_2} .
\]
If we are pumping heat out from a place at temperature $T_2$ to
a warmer exterior at temperature $T_1$, the ideal efficiency is:
\[
\mbox{efficiency} = \frac{T_2}{T_1 - T_2} .
% \frac{T_1}{T_2} - 1 .
\]
These theoretical limits could only be achieved by systems that pump
heat infinitely slowly.
Notice that the ideal efficiency is bigger, the closer the inside temperature $T_2$
is to the outside temperature $T_1$.
% *** MOVE this sentence
% Ground-source heat pumps are discussed more in
% \chref{ch.gshp}.
While in theory
ground-source heat pumps might have better
performance than air-source, because the ground temperature is usually
closer than the air temperature to the indoor temperature,
in practice an air-source heat pump might be the best
and simplest choice.
In cities, there may be uncertainty about the future
effectiveness of ground-source heat pumps,
because the more people use them in winter, the colder the
ground gets; this \ind{thermal fly-tipping} problem may also
show up in the summer in cities where too many buildings
use ground-source (or should I say ``ground-sink''?)
heat pumps for air-conditioning.
\section{Heating and the ground}
Here's%
\margintab{\small
\begin{tabular}{@{}ll@{}}\toprule
Heat capacity: &$C= 820$\,J/kg/K \\
Conductivity: & $\kappa = 2.1$\,W$\!$/m/K\\
Density: & $\rho = 2750$\,kg/m$^3$ \\
\multicolumn{2}{@{}l}{Heat capacity
per unit
volume:}\\
&
$C_{\rm V} =
% \rho C =
2.3$\,MJ/m$^3$/K\\ \bottomrule
\end{tabular}
\caption[a]{Vital statistics for granite.\index{granite}
(I use \ind{granite} as an example of a typical rock.)
}
}
an interesting calculation to do.
Imagine having \ind{solar heating panels} on your roof,
and, whenever the water in the panels gets above
50\degrees\C, pumping the water through a large rock under
your house.
When a dreary grey cold month comes along, you could
then use the heat in the rock to warm your house.
% Normal ground-source heat pumps are different from this scheme in two ways:
% they don't usually bother with the solar heating panels
% and they use the cunning back-to-front refrigerator trick
% to boost
% the heat flow from the rock into the building.
% But before we discuss them, let's do this simple
% calculation for the solar scheme: r
Roughly
how big a 50\degree\C\ rock would you need to hold
enough energy to heat a house for a whole month?
Let's assume we're after 24\,kWh per day for 30 days
and that the house is at 16\degrees\C\@.
% the rock is perfectly insulated
% from the surrounding ground
The heat capacity of granite is
%% 0.195 from page 281 of Marks' Handbook.
%% Mechanical Engineers' Handbook
%% edited by Lionel S Marks. McGraw-Hill London 1951
%% and density is 2.5 (2.4-2.7) steel is 7.8. p 523
$0.195 \times
4200\,\J/\kg/\K = 820$\,J/\kg/K\@.
The mass of granite required is:
% pr 24*30*3.6e6 / 820 / 34
% 92969.8708751793
\begin{eqnarray*}
\mbox{mass} &=& \displaystyle\frac{ \mbox{energy} }{ \mbox{heat capacity} \times
\mbox{temperature difference} }
\\ &=& \displaystyle
\frac{ 24\times 30\times 3.6\,\MJ }{ ( 820\,\J/\kg/\ndegreeC ) (50{\degreesC}-16{\degreesC}) }
\\ &=&
100\,000\,\kg,
%% volume 37,000 litres, 37\,m^3 or 3.33**3 or 1*6*6
\end{eqnarray*}
100 tonnes, which corresponds to a cuboid of rock of size
$6\,\m \times 6\,\m \times 1\,\m$.
% cut material on area to _heating.tex
\subsection{Ground storage without walls}
\label{chSH}
OK, we've established the size of a useful ground store. But is it
difficult to keep the heat in? Would you need to surround your rock cuboid
with lots of insulation? It turns out that the ground itself is a pretty
good insulator.
\margintab{
\begin{center}
\begin{tabular}{lr}
\toprule
% \multicolumn{2}{r}{watts per metre-kelvin (\WmK)}
\multicolumn{2}{r}{(\WmK)}
\\ \midrule
water& 0.6 \\
% $ watt / (metre^2 \times kelvin metre^{-1}) $
%that is
quartz & 8\index{quartz} \\
%% http://dictionary.laborlawtalk.com/Thermal_conductivity
% Conductivity of
\ind{granite} &
2.1 \\ % \,{\WmK} \\
% 2.1\,Wm$^{-1}$K$^{-1}$.
earth's crust & 1.7 \\
dry soil & 0.14 \\
\bottomrule
\end{tabular}
\end{center}
\caption[a]{Thermal conductivities.\index{thermal conductivity}
For more data see \tabref{tab.condS}, \pref{tab.condS}.}
\label{ta.minicond}
}%
A spike of heat put down a hole in the ground will spread as
\[
\frac{1}{ \sqrt{4 \pi \kappa t} }\exp\left(
- \frac{x^2}{4 (\kappa/(C\rho)) t}
\right)
\]
where $\kappa$ is the conductivity of the ground,
$C$ is its \ind{heat capacity},
and $\rho$ is its density.
This describes a bell-shaped curve with
width
\[
\MidnightBlue{ \sqrt{ 2 \frac{\kappa}{C \rho} t } };
\]
for example, after six months ($t = 1.6 \times 10^7\,\s$),
using the figures for granite ($C = 0.82$\,kJ/\kg/K, $\rho = 2500$\,kg/m$^3$,
$\kappa = 2.1$\,{\WmK}), the width is
% sqrt( 2 2.1\,J/s/m/K * 1.6 \times 10^7\,\s
% / 820\,J/kg/C 2500$\,kg/m$^3$
% sqrt( 2 2.1 * 1.6 \times 10^7
% / 820 2500$/m$^2$
% = 6m
\MidnightBlue{6\,m}.
Using the figures for water ($C = 4.2$\,kJ/kg/K, $\rho = 1000$\,kg/m$^3$,
$\kappa = 0.6$\,{\WmK}), the width is
% sqrt( 2 0.6\,J/s/m/K * 1.6 \times 10^7\,\s
% / 4200\,J/kg/C 1000$\,kg/m$^3$
% = 6m
2\,m.
So if the storage region is bigger than $20\,\m\times20\,\m\times20\,\m$
then most of the heat stored will still be there in six months
time (because 20\,m is significantly bigger than 6\,m and 2\,m).
% The same formula also helps us decide how close together the
% pipes have to be in the storage region.
% When we get some hot water, we want to
% Two other constraints force the dimensions of the storage region:
% it must be big enough to hold all the energy that we want to
% leave there, and it must have enough surface area exposed to our pipes for
% us to be able to get the heat in.
\subsection{Limits of ground-source heat pumps}
The low thermal conductivity of the ground is a double-edged sword.
Thanks to low conductivity,
the ground holds heat well for a long time. But on the other hand,
low conductivity means that it's not easy to shove heat in and out of the ground
rapidly.
We now explore how the conductivity of the ground limits the use of
ground-source heat pumps.
\begin{figure}[!b]
\figuremargin{\small
\begin{center}
\begin{tabular}{l}
\raisebox{3cm}{\makebox[0in][l]{\sf temperature (\ndegreesC)}}%
\mbox{\epsfxsize=4in%
\mono%
{\epsfbox{../data/cambridge/mono/TempFit.eps}}%
{\epsfbox{../data/cambridge/TempFit.eps}}%
}
\end{tabular}
\end{center}
}{
\caption[a]{The temperature
in Cambridge, 2006, and a cartoon, which
says the temperature is the sum of an annual
sinusoidal variation
% of amplitude 8.33\degreesC
between $3$\degreesC\
and
$20$\degreesC,
and a daily sinusoidal variation
with range up to 10.3\degreesC\@.
% amplitude 5.15
The average temperature is 11.5\degreesC\@.
}
\label{TempFit}
}
\end{figure}
Consider a neighbourhood with quite
a high population density. Can {\em{everyone}\/} use ground-source heat pumps,
without using active summer replenishment (as discussed on \pref{pSummerReplen})?
The concern is that if we all sucked heat from the ground at the same
time, we might freeze the ground solid.
I'm going to address this question by two calculations.
First, I'll work out the natural flux of energy in and out of the ground in summer
and winter.
If the flux we want to suck out of the ground in winter is much bigger than
these natural fluxes then we know that our sucking is going to significantly
alter ground temperatures, and may thus not be feasible.
For this\label{pHP1} calculation, I'll assume the ground just below the surface
is held, by the combined influence of sun, air, cloud, and
night sky, at a temperature that varies slowly up and down
during the year (\figref{TempFit}).
{%%%%%%%%%%%%%%%%%%%%%% troublesome
% put me BEFORE troublesome heating. diagram page % see AFTER
\renewcommand{\floatpagefraction}{0.8}
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\subsection{Response to external temperature variations}
Working out how the temperature inside the ground responds,
and what the flux in or out is,
requires some advanced mathematics, which I've cordoned off
in box \ref{box.fluxtheory} (\pref{box.fluxtheory}).
The payoff from this calculation is a rather beautiful
diagram (\figref{TempDep}) that shows how the temperature varies
in time at each depth. This diagram shows the answer for
any material in terms of the {\dem{characteristic length-scale}\/}
$z_0$ (\eqref{eq.charleng}), which depends on the
conductivity $\kappa$ and heat capacity $C_{\rm V}$ of the material,
and on the frequency $\omega$ of the external temperature variations.
(We can choose to look at either daily and yearly variations using the same
theory.)
At a depth of 2$z_0$, the variations in temperature
are one seventh of those at the surface, and
lag them by about one third of a cycle (\figref{TempDep}).
At a depth of 3$z_0$, the variations in temperature
are one twentieth of those at the surface, and
lag them by half a cycle.
% At depth $z_0$, the temperature oscillations are reduced to
% 37\%, and at depth $2z_0$, they are reduced to just 14\%
% of the external amplitude.
For the case of daily variations and solid granite,
the characteristic length-scale is
$z_0 = 0.16\,\m$. (So 32\,cm of rock is the thickness you need to ride out
external daily temperature oscillations.)
For yearly variations and solid granite, the characteristic length-scale is
$z_0 = 3\,\m$.
% sustainable/data/cambridge
% gnuplot < gnuDep ; gv TempDepth.eps
\marginfig{\small
\begin{center}
\begin{tabular}{@{}l@{}}
\mbox{\epsfxsize=53mm%
\mono%
{\epsfbox{../data/cambridge/mono/TempDepthS.eps}}%
{\epsfbox{../data/cambridge/TempDepthS.eps}}%
}\\
\end{tabular}
\end{center}
% used to be - rather beautiful
%{\epsfbox{../data/cambridge/mono/TempDepth.eps}}%
%{\epsfbox{../data/cambridge/TempDepth.eps}}%
\caption[a]{Temperature (in \ndegreesC)
versus depth and time.
% in \ind{granite}.
The depths are given in units of the
characteristic depth $z_0$, which
for granite and annual variations is 3\,m.\smallskip
% \begin{oldcenter}
% \begin{tabular}{ll} \toprule
% depth 0 & surface \\
% depth 1 & 3\,m \\
% depth 2 & 6\,m \\
% depth 3 & 9\,m \\ \bottomrule
% \end{tabular}
% \end{oldcenter}
At ``depth 2''
% of $2z_0$
(6\,m),
the temperature is always about 11 or 12\degreeC\@.
At ``depth 1'' (3\,m), it wobbles
between 8 and 15\degreeC\@.
}
\label{TempDep}
%% gnuDep
}
\begin{table}[!b]
\figuremargin{\small
\begin{tabular}{lllll} \toprule
& thermal & heat & length-scale & flux \\
& conductivity & capacity & & \\
& $\kappa$ & $C_{\rm V}$ & $z_0$ & $A \sqrt{ { C_{\rm V} \kappa \omega } }$
% z_0 = \sqrt{ \frac{2 \kappa }{ C_{\rm V} \omega } }.
\\
& (\WmK) & (MJ/m$^3$/K) & (m) & (\Wmm) \\
\midrule
Air & 0.02 & 0.0012 \\
% see data/soil.m
Water & 0.57 & 4.18 & 1.2 & 5.7 \\
Solid granite & 2.1 & 2.3 & 3.0& 8.1 \\
Concrete & 1.28 & 1.94 & 2.6& 5.8\\[0.1in]
{\em Sandy soil} \\
dry & 0.30 & 1.28 & 1.5 & 2.3\\
50\% saturated & 1.80 & 2.12 & 2.9 & 7.2\\
100\% saturated & 2.20 & 2.96 & 2.7 & 9.5\\[0.1in]
{\em Clay soil} \\
dry & 0.25 & 1.42 & 1.3 & 2.2\\
50\% saturated & 1.18 & 2.25 & 2.3 & 6.0\\
100\% saturated & 1.58 & 3.10 & 2.3 & 8.2\\[0.1in]
{\em Peat soil} \\
dry & 0.06 & 0.58 & 1.0 & 0.7\\% 69\\
50\% saturated & 0.29 & 2.31 & 1.1 & 3.0\\
100\% saturated & 0.50 & 4.02 & 1.1 & 5.3\\
\bottomrule
\end{tabular}
}{\caption[a]{Thermal conductivity and heat capacity
of various materials and soil types, and the
deduced length-scale
$z_0 = \sqrt{ \frac{2 \kappa }{ C_{\rm V} \omega } }$
and
peak flux
$A \sqrt{ { C_{\rm V} \kappa \omega } }$
associated with annual temperature variations
with amplitude $A=8.3\degreesC$.
The sandy and clay soils have porosity 0.4;
the peat soil has porosity 0.8.
}\label{tab.condS}}
\end{table}
Let's focus on annual variations and discuss a few other materials.
Characteristic length-scales for various materials are in the third
column of \tabref{tab.condS}.
For damp sandy soils or concrete, the characteristic length-scale
$z_0$ is similar to that of granite -- about 2.6\,m.
In dry or peaty soils, the length-scale
$z_0$ is shorter -- about 1.3\,m.
That's perhaps good news because it means you don't have to dig so deep to find
ground with a stable temperature. But it's also coupled with some bad news:
the natural fluxes are smaller in dry soils.
The natural flux varies during the year
% , and lags the external temperature
% by 8 weeks. The flux into the ground is biggest in
and has a peak value (\eqref{eqAmpFlux})
that is smaller, the smaller the \ind{conductivity}.
For the case of solid granite, the peak flux is 8\,\Wmm.
For dry soils, the peak flux ranges from 0.7\,\Wmm\ to 2.3\,\Wmm.
For damp soils, the peak flux ranges from 3\,\Wmm\ to 8\,\Wmm.
What does this mean?
I suggest we take a flux in the middle of these numbers, 5\,\Wmm,
as a useful benchmark, giving guidance
about what sort of power we could expect to extract, per unit area, with a
ground-source heat pump.
If we suck a flux significantly smaller than 5\,\Wmm, the perturbation
we introduce to the natural flows will be small.
If on the other hand
we try to suck a flux bigger than 5\,\Wmm, we should expect that
we'll be shifting the temperature of the ground
significantly away from its natural value, and such fluxes
may be impossible to demand.
The population density of a typical English suburb corresponds
to 160\,m$^2$ per person (rows of semi-detached houses with about 400\,m$^2$ per
house, including pavements and streets).\label{pPickDen}
% 6200/km$^2$
% that's
At this density of residential area, we can deduce that
a ballpark limit for heat pump power delivery is
\[
5\,\Wmm \times 160\,\m^2 = 800\,\W = 19\,\kWh/d\mbox{ per person}.
\]
This is uncomfortably close to the sort of power we would like to
deliver in winter-time: it's plausible that our peak winter-time
demand for hot air and hot water, in an old house like mine, might be
40\,\kWh/d per person.
This calculation suggests that in a typical suburban area, {\em not
everyone can use ground-source heat pumps}, unless they are careful to
actively dump heat back into the ground during the summer.
Let's do a second calculation, working out how much power we could
steadily suck from a ground loop at a depth of $h=2$\,m.
Let's assume that we'll allow ourselves to suck the temperature at the ground
loop down to
$\Delta T = 5\degreesC$ below the average ground temperature at the surface,
and let's assume that the surface temperature is constant.
We can then deduce the heat flux
from the surface. Assuming a conductivity of 1.2\,\WmK\ (typical of damp clay soil),
\[
\mbox{Flux} =
%\mbox{Conductivity}
\kappa \times \frac{\Delta T}{h} = 3\,\Wmm.
% 1.2 * 5 / 2
\]
If, as above, we
% on \pref{pPickDen}, we
assume a population density corresponding to
160\,m$^2$ per person,
then the maximum power per person deliverable by ground-source
heat pumps, if everyone
in a neighbourhood has them,
is 480\,W, which is 12\,kWh/d per person.
% Add in the electrical contribution, assuming a COP of 3: 17\,kWh/d per person.
So again we come to the conclusion that in a typical suburban area
composed of poorly insulated houses like mine, {\em not
everyone can use ground-source heat pumps}, unless they are careful to
actively dump heat back into the ground during the summer.
And in cities with higher population density, ground-source heat pumps
are unlikely to be viable.
I therefore suggest air-source heat pumps are the best heating choice for
most people.
\section{Thermal mass}
\nocite{SolarHo}Does increasing the \ind{thermal mass} of a building\index{building!thermal mass}
% *** undefined?
help reduce its heating and cooling bills?
It depends. The outdoor temperature can vary during the
day by about 10\degreesC\@.
A building with large thermal mass -- thick stone walls, for example --
will naturally ride out those variations in temperature, and, without
heating or cooling, will have a temperature close to the average outdoor
temperature. Such buildings, in the UK, need neither heating nor cooling
for many months of the year.
In contrast, a poorly-insulated
building with low thermal mass might be judged too hot
during the day and too cool at night, leading to greater expenditure
on cooling and heating.
However, large thermal mass is not always a boon. If a room is
occupied in winter for just a couple of hours a day (think of a
lecture room for example),
the energy cost of warming the room up to a comfortable temperature will
be greater, the greater the room's thermal mass.
This extra invested heat will linger for longer in a thermally massive
room, but if nobody is there to enjoy it, it's wasted heat.
%% occupying the room
So in the case of in\-fre\-quent\-ly-used rooms it makes
sense to aim for a structure with low thermal mass,
and to warm that small mass rapidly when required.
% put me AFTER troublesome heating. diagram page % see BEFORE
% END \renewcommand{\floatpagefraction}{0.8}
}
\begin{boxfloat}
\figuremargin{
\begin{framedalgorithm}\small
If we assume the ground is made of solid homogeneous
material with conductivity $\kappa$ and heat capacity $C_{\rm V}$,
then the temperature at depth $z$ below the ground
and time $t$
responds to the imposed temperature at the surface
in accordance with the diffusion equation
\beq
\frac{ \partial T(z,t)}{\partial t} = \frac{ \kappa }{ C_{\rm V} }
\frac{ \partial^2 T(z,t) }{\partial z^2 } .
\eeq
For a sinusoidal imposed temperature with frequency $\omega$
and amplitude $A$ at depth $z=0$,
\beq
T(0,t) = T_{\rm surface}(t) = T_{\rm average} + A \cos( \omega t ) ,
\eeq
the resulting temperature
at depth $z$ and time $t$ is a decaying and oscillating function
\beq
T(z,t) = T_{\rm average} + A \,e^{-z/z_0} \cos( \omega t - {z}/{z_0} ),
\eeq
where $z_0$ is the characteristic length-scale of both the decay and
the oscillation,
\beq
z_0 = \sqrt{ \frac{2 \kappa }{ C_{\rm V} \omega } }.
\label{eq.charleng}
\eeq
The flux of heat (the power per unit area) at depth $z$ is
\beq
\kappa
\frac{ \partial T }{\partial z }
= \kappa \frac{A }{ z_0 } \sqrt{2} e^{-z/z_0}
\sin ( \omega t - z/z_0 - \pi/4 ) .
\eeq
For example, at the surface, the peak flux is
\beq
\kappa \frac{A }{ z_0 } \sqrt{2} =
A \sqrt{ { C_{\rm V} \kappa \omega } }.
\label{eqAmpFlux}
\eeq
% which
%% omega = 2*pi / (365.25*24*3600.0)
%% A=8.336 ; CV = 2255.0e3 ; kappa = 2.1 ; pr A*sqrt( CV*kappa* omega)
%% 8.09
% for the yearly driving signal of amplitude $A=8.3$\degreesC\
% corresponds to 8\,\Wmm.
\end{framedalgorithm}
}{
\caption[a]{Working out the natural flux caused by sinusoidal temperature variations.}
\label{box.fluxtheory}
}
\end{boxfloat}
\small
\section{Notes and further reading}
\begin{notelist}
\item[page no.]
% \item[]
% [K-value is thermal conductivity, usually measured in W$\!$/m/K, for example
% 0.023\,W$\!$/m/K\@. Sometimes also called $\lambda$-value.
% R-value is thermal resistance (the inverse of U-value), usually
% measured in m$^2$K/W.]
%\item[air-changes]
\item[\npageref{tab.condS}]
{\nqs{\Tabref{tab.condS}.}}
Sources:
\cite{SoilCond},
% http://books.google.co.uk/books?id=3lV1PFb-vz0C&pg=PA188&lpg=PA188&dq=conductivity+and+heat+capacity+of+clay&source=web&ots=UJN8kBqWfX&sig=dC_axYg-892RhY4HHFnZwl95FA4&hl=en&sa=X&oi=book_result&resnum=3&ct=result
\par
\myurlb{www.hukseflux.com/thermalScience/thermalConductivity.html}{http://www.hukseflux.com/thermalScience/thermalConductivity.html}
\end{notelist}
\normalsize
% http://www.ukace.org/pubs/reportfo/BuildIgn.pdf