Section 7 HEAT LOSS CALCULATIONS

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Section 7HEAT LOSS CALCULATIONS

• Dr. Congxiao Shang

Objectives - To estimate heat loss energy
consumption of a building over a period of time
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7.0 Review U - Values
Review in Section 6, we learned
In thermal transmittance, the U value is simply
defined as 1/R Unit WK-1m-2
(Remember R is unit area resistance, U is heat
transfer per unit area)
A higher R-Value means the materials are more
resistant to heat loss.
R U clear ?
A lower U-Value means the system will transmit
less heat.
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7.1 Heat Loss Rate
For heat loss in a building FIVE COMPONENT
PARTS- Losses through the walls Losses through
the windows Losses through the roof Losses
through the floor Losses by ventilation
The value obtained is the Heat Loss Rate, WK-1,
or the heat loss per unit temperature difference,
as seen from the definition of the U value,
WK-1m-2.
Ventilation is treated differently, as seen in
the next slide -
This is an intrinsic factor for a particular
building, it may be changed by varying the
insulation of one or more components.
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7.2 Ventilation
For ventilation, we talk in terms of air-change
rate (usually per hour) -
An air change rate of 1.0 means that the whole
volume of air within a building is replaced once
per hour by air from outside which then has to be
heated.
It is convenient to obtain a factor which is
dependant on temperature difference- in terms of
an equivalent U - value .
The specific heat of air is about 1300 J m-3 C-1
So what is the total energy required to heat
incoming air?
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7.2 Ventilation
Divided by 3600 to bring the value to Watts (J
S-1)
or,
building volume x ( air change rate x 0.361) x
temp diff.
the factor for ventilation heat loss
Equivalent to U - value
but the unit here is W m-3 C-1, i.e. its the
energy loss per unit volume per temperature
difference , whereas the U-value for the
surface cases is, W m-2 C-1 , per unit area
per temperature difference. So, what is the
total heat loss rate in a building ? - see 7.3
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7.3 Heat Loss Calculations
Hence, the total heat loss rate, H, (i.e. heat
loss per temperature difference) of a room (or
building) is-
H ? (Area x UValue) Volume x ach x
0.361
Unit W C-1
Sum of the heat loss from surfaces (walls,
roofs, floors, windows)
the heat loss by ventilation where ach is air
change rate (per hour)
In a steady state condition, heat loss must be
replaced by heat supplied by a heating system.
When you design the system, the size will depend
on
- the heat loss rate, H - the internal design
temperature - the external design temperature
(Note a design temperature is the one used as
a reference for designing purposes, e.g. of
heating devices)
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7.3 Heat Loss Calculations
Typically in the UK, the internal temperature is
taken as 20 or 21 C for residential properties,
while external temperatures may vary according to
external conditions - e.g. -1 C for the UK.
e.g. for a house with H 250 W C -1, and a
design temperature difference of 20 to -1 C
(which equals 21C) the heat loss from the house
will be , 21 x 250 W 5250 W.
The heating device must be capable of supplying
this amount of heat in order to maintain
comfortable living conditions in this house.
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7.4 Notes on sizing heating appliances
For design considerations, the following points
may be noted

1. the design cannot assume incidental gains for
   sizing purposes (we then take this into account
   in the overall consumption).
2. basic heat loss calculations assume a steady
   state, but the actual temperature may vary below
   the design temperature e.g. the designed heating
   system will not be able to cope with heat demand
   if the external temperature falls below -1C
   the internal temperature also drops.
3. most houses have a significant heat storage,
   which helps when the temperature initially falls
   to a low temperature.
4. the boiler must also be capable of supplying hot
   water. boiler sizes typically come in increments
   of 3 kW.
5. Hot water requirements vary with time of day, so
   the boiler must be able to cope with high demand
   (additional 3 kW).
6. At times of low hot water demand, this additional
   capacity can be used to boost the heating supply.

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7.5 Annual Energy Consumption for Heating
We need to evaluate the overall consumption to
judge the effectiveness of savings...
Energy used in heating varies according to

• external temperature
• incidental gains (e.g. heat from occupants,
  appliances, passive solar)
• dynamic considerations (e.g. timed heating,
  variable temperature, etc)

Detailed method- Assess incidental gains
temperature variations throughout the year and
then estimate the annual consumption.
Approximate Methods- There are various ways to
adequately judge savings, e.g. the Degree-Day
Method, as assumptions for each are consistent.
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7.5 Annual Energy Consumption for Heating
7.5.1 Degree - Day Method simple formula
As we know, heat requirement is proportional to
temperature difference.
Each day when there is 1 degree of temperature
difference based on the balance temperature, we
add 1 when there are 2 degrees of temperature
difference, we add 2 and so on. i.e. Each day
when there are n degrees of temperature
difference we add n
Degree-Days is the sum of these numbers - East
Anglia annual figure is 2430
Note the Degree-days assumes some incidental
gains . the unit of Degree-days is ºC-day
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7.5 Annual Energy Consumption for Heating
7.5.1 Degree - Day Method simple formula

• There is a "free" temperature rise from the
  incidental gains.
• The temperature at or above which no heating is
  required is called the neutral or balance
  temperature, and is normally taken as 15.5C.
  This means that the effective internal
  temperature is taken as 15.5 C.
• If the external temperature is 14.5 C we add 1
  and so on.
• However, if the temperature is greater than or
  equal to the neutral temperature, we add zero.

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7.5 Annual Energy Consumption for Heating
7.5.2 Degree - Day Method More Exact Formula
In a year, there are many days when the daily
temperature is partly above the neutral
temperature and partly below it. The consequence
is often that a small amount of heating may be
required, even though the mean daily external
temperature is above the neutral temperature.
Considering the above, the improved method of
Degree-Day
For each day,
DegreeDays Tneutral - 0.5 (Tmax Tmin), if
0.5 (TmaxTmin) lt Tneutral
0, if 0.5 (TmaxTmin)
gtTneutral
During the Heating Period Tmax -
maximum external temperature Tmin - minimum
external temperature Tneutral -
neutral or base temperature.
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7.5 Annual Energy Consumption for Heating
7.5.2 Degree - Day Method More accurate
Formula
For each day,
Tneutral
If Tneutral lt Tmin, Degree-Day 0
If Tneutral gt Tmax,
Degree-Day Tneutral - 0.5 (TmaxTmin)
If Tneutral lt Tmax, then
If Tneutral gt 0.5 (Tmax Tmin)
Degree-Day 0.5 (Tneutral-Tmin)- 0.25
(Tmax-Tneutral)
If Tneutral lt 0.5 (Tmax Tmin)
Degree-Day 0.25 (Tneutral-Tmin)
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7.5 Annual Energy Consumption for Heating
7.5.2 Degree - Day Method
Note
In both methods, Degree-Days are calculated over
a number of days in the period of consideration,
e.g. a week, month, quarter, year.
The same procedure may be used to determine the
amount of energy used for cooling. In this
case, there will be a cooling neutral
temperature, which will normally be different
from the heating one.
In the UK cooling neutral temperature 22C
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7.5 Annual Energy Consumption for Heating
7.5.2 Degree - Day Method for cooling
Tmax
Cooling Formula
Tneutral
Tmin
Approximately, for each day,
Degree-Day 0.5 (Tmax Tmin) - Tneutral, if 0.5
(Tmax Tmin) gt Tneutral
0, if 0.5 (Tmax Tmin)
lt Tneutral
This formula follows the same general format as
the simple heating one. There is a more accurate
cooling formula similar to the more accurate
heating one.
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7.5 Annual Energy Consumption for Heating
7.5.3 Degree - Day Method Degree Day Table
20-year average heating degree-days for the UK
1979 1998 (for reference only)
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
1 Thames Valley 337 303 256 190 111 47 19 22 51 130 234 302
2 South Eastern 356 323 280 217 136 66 32 38 75 155 255 321
3 Southern 342 310 277 221 138 68 37 42 77 150 244 309
4 South Western 289 268 250 198 124 59 24 26 51 116 199 255
5 Severn Valley 320 288 250 190 114 47 18 20 47 128 215 284
6 Midlands 373 333 290 230 152 76 39 43 83 176 269 342
7 West Pennines 360 317 286 219 139 73 35 40 78 165 259 330
8 North Western 370 323 303 239 162 90 47 53 98 183 272 346
9 Borders 363 319 306 258 199 112 58 60 101 182 267 333
10 North Eastern 380 328 298 234 163 83 41 46 87 178 274 346
11 East Pennines 371 327 287 225 152 77 39 41 80 170 267 340
12 East Anglia 374 336 291 228 145 72 36 36 69 157 266 341
13 West Scotland 376 327 311 239 164 92 54 62 111 200 285 358
14 East Scotland 386 332 314 252 189 103 57 63 110 199 289 362
15 NE Scotland 389 336 324 264 197 115 63 70 119 211 294 364
16 Wales 329 303 285 232 159 88 44 43 74 150 230 294
17 Northern Ireland 358 314 298 234 162 88 47 54 98 179 268 329
18 NW Scotland 323 291 313 256 208 129 84 77 118 206 254 328
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7.5 Annual Energy Consumption for Heating
7.5.4 Example using Degree-days
(see also previous exam questions)
Heat loss rate of a house 450 W C-1, and
incidental gains 2025 Watts. If there are
1100 degree-days in a 3 month period, and the
thermostat is set at 20 C, what is the balance
temperature and what is the energy consumption
over the period ? (Formula - see practical
hand-out)
Free temperature rise is 2025/450 4.5
C Thus actual balance temperature 20 -
4.5 15.5 C
So energy consumption over the period is (H, x
Degree-day) 450 x 1100 x 86400
42,768,000,000 W s 42.8 GJ
WC-1
C day
1 day 86400 seconds
Note Units
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Annual Energy Consumption for Heating
Energy used in heating varies according to

• external temperature
• incidental gains (e.g. heat from occupants,
  appliances, passive solar)
• dynamic considerations (e.g. timed heating,
  variable temperature, etc)

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7.6 Dynamic Heating
Dynamic heating refers to relatively detailed
heating variations with temperature variations
during a day (or a short period of time).
For example Variation in Boiler output with
external temperature for a house. The thermostat
temperature is 20C. The climatic data refers to
the period of 5th - 10th January 1985 in Norwich
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7.6 Dynamic Heating
Calculations are more complex in this case
Loss/gain of heat from/to a heat store during a
specific time period has to be accounted for.
Knowledge of the dynamic behaviour of heating is
necessary . The approach taken in the lectures
is a simple block model to illustrate the basic
principle. In the practical a much more refined
model is used which calculates changes each
minute and also divides the various components.
An initial assumption of the temperature profile
through the walls, etc. is required (see section
6.7, page 27). The profile shown in Fig.6.7 is
only true when steady state conditions exist,
which in the case of a house would represent
several days heating when the outside temperature
was constant.
In dynamic heating, the internal temperature
will fall exponentially once the boiler is turned
off. Alternatively if the internal temperature is
constant and the boiler is kept on, the heat
output from the boiler will vary if the external
temperature fluctuates.
Heat flows lag the change in temperature, by 6
9 hours in a heavy weight house ( in a brick
construction) by 1, in a lightweight insulated
structure of timber.
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7.6 Dynamic Heating
7.6.1 Response of typical house to external
temperature variations in summer
The internal temperature varies in a sinusoidal
shape with the external temperature, but with
lower amplitude and with the peaks shifted
(lagged..). The heavier the construction the
lower the amplitude of the internal temperature
and the greater the shift. In light weight
buildings - e.g. Sainsbury centre, the amplitude
of the internal temperature can be quite high,
though less than the external temperature.
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7.6 Dynamic Heating
7.6.2 Response of a typical house to external
temperature variations in winter
In winter, the temperature will always be below
the thermostat setting the boiler will cut in
to supply heat as required
The figure shows the situation for a non-time
switched case.
boiler output peaks when Tex is at its lowest
The output is at minimum when Tex is at a maximum

• The amplitude of variation in the boiler output
  is again much less than the temperature.
• Shaded area beneath the boiler curve represents
  the total energy consumed during the period.

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7.6 Dynamic Heating
7.6.3 Effect of time switching constant
external temperature
Boiler on periods 0700 - 1100 1700 - 2400
Troom
At 7am, the boiler cuts in then T rises to the
thermostat level, 20oC, shortly before it cuts
out in the late morning.
At 1700, a similar thing happens. But after
about 4 hours, the boiler throttles back as the
thermostat level is reached. Then, the boiler
output drops with time as the heat storage of the
house is replenished.
If the outside temperature had been constant at
0oC, the output of the boiler would have been
constant at 5.9 kW
However, at the end of the day, the boiler output
is actually above the steady state level due to
heat in the internal fabric still being
replenished.
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7.6 Dynamic Heating
7.6.3 Effect of time switching constant
external temperature
Troom
-A consequence of time switching is thus to
require a larger boiler in order to reach
thermostat level more quickly
- With a larger boiler, the saving would have
been less, because the mean T will be higher and
the shaded area (energy consumption) is even
larger.
Here, the output with time switching only leads
to a saving of just over 13
Key thing about saving with time-switching The
energy required is directly proportional to mean
internal temperature, If It is higher, with a
large boiler or better insulation, the saving
will be proportionally less .
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7.6 Dynamic Heating
7.6.3 Effect of time switching constant
external temperature
On the other hand, if a boiler is too small, the
temperature may never quite make thermostat level
and will continue to decline with time.
Clearly this is a non-viable option the
theoretical saving has no meaning here.
While the above graphs illustrate the situation
with "idealised" boilers, i.e. the boilers
throttle back, once the thermostat temperature is
reached. Real boilers have some specific
situations and ways to reduce the output to save
energy (See notes for examples) .
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7.6 Dynamic Heating
7.6.3 Effect of time switching constant
external temperature
Real boilers have some specific situations and
ways to reduce the output to save energy (See
notes for examples)
Domestic boilers are either on or off. What is
the situation with time switching ?
A further complication arise from the fact that
there is no single thermostat level. What will be
the actual temperature profile?

The actual energy supplied by the boiler will
reflect the external T some hours beforehand. How
do you deal with this?
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Annual Energy Consumption for Heating
based on

• steady state condition (designed internal and
  external temperature degree- day method based on
  a balance temperature)
• dynamic heating behaviour (e.g. timed heating,
  temperature varies with different time, etc)

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7.7 Dynamic Heat Loss - a worked example
To properly analyse the dynamic situation
required splitting the components of the
buildings into numerous sub-components, each with
its own thermal capacity.
E.g., a 100 mm thick brick could be divided into
layers 5mm thick, and each layer would be treated
as being of uniform temperature thermal
capacity.
Clearly, the thinner the layer the better, but
without the aid of finite element computer
modelling, it becomes an impossible manual task.

(The practical in week 7 uses a program written
by a former ENV student to analyse the situation
with such fine layers. The example here is just a
simple approach to illustrate the basic principle)
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7.7 Dynamic Heat Loss - a worked example

• Example - Waveney terrace
• a thermal capacity 4 GJ C-1
• a heat loss rate 45 kW C-1
• 700 students live there, with an average body
  heat output of 100 W,
• i.e. there is incidental gain of 700 x 100 70
  kW
• - Time span 2 hours

the heating goes off at midnight, the
thermostat temperature is 20C, the external
temperature is 0 C and the maximum heat supply
is 2000 kW.
The heat loss from the building at midnight
(20 - 0 ) x 45 900 kW. The net heat loss
900 - 70
830 kW. Over a period of 2 hours, the total
heat loss
830 2 3600 /1000000 5.98 GJ
Now, we can thus estimate the drop in the
temperature in the building 5.98/4 -1.49 C
the 4 is the thermal capacity of the building
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7.7 Dynamic Heat Loss - a worked example
We can thus work out the temperature at the end
of 2 hours to be 20 - 1.49 18.51C.
We can use a tabular form to calculate the net
heat gain or loss with different time
Time Inter. Temp (C) Exter. Temp (C) Heat loss (kW) Body Heat (kW) Heat from boiler (kW) Net Heat (kW) Heat gain/loss (GJ) Temp change (C)
2400 20.00 0.00 -900 70 0 -830 -5.98 -1.49
0200 18.51 0.00 -45x18.51 70 0 -
0400 17.13 0.00 70
0600 15.87 0.00 70 2000
0800 18.31 0.00 70 2000 1246 8.97 2.24
1000 20.55 0.00 70 2000 1145 0.32 0.08
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7.8 Energy issues with radiator on outside wall
See handouts only for those who are interested
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7.9 Method to determine air-change rate
The method is to measure the rise in temperature
above normal by providing supplementary heating
to the room (in question), then we can use basic
heat loss calculations to estimate the exchange
rate
we need to consider not only heat loss to the
outside, but also heat lost to adjoining rooms .
0ºC
0ºC
Roof
A room in a single storey building is 4m x 4m in
plan 2.4m high. It has one external wall with
a single glazed window 2.4 m2 in area. On a
day when the external temperature is 0C, the
rest of the building is heated uniformly to 20
C.
20ºC
20ºC
External wall
2.4
20ºC
Internal wall
Internal wall
Floor
Internal wall
4
0ºC
4
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7.9 Method to determine air-change rate
Assuming all the gaps at windows and doors were
sealed and supplementary heating of 1494 W was
supplied to the room. After two hours the room
temperature stabilised at 25C. The specific heat
of air is 1305 J m-3 C-1. How to Estimate the
intrinsic ventilation rate in air changes per
hour (ACH)?
0ºC
U-value data are given in Table 2.
0ºC
Roof
U-Values (W m-2 oC-1) U-Values (W m-2 oC-1) U-Values (W m-2 oC-1) U-Values (W m-2 oC-1)
External Walls 1.0 Floor 1.0
Internal Walls 2.5 Window 5.0
Roof 0.2
20ºC
20ºC
External wall
2.4
20ºC
Internal wall
25ºC
Internal wall
Floor
Internal wall
4
Assume radiator in the room is off.
0ºC
4
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7.2 Ventilation
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7.9 Method to determine air-exchange rate
Then by Continuity, supplementary heat (heat
losses to outside heat losses across internal
walls) ventilation losses
The solution is best in tabular form
Area (m2) U-value (W m-2 oC-1) Temp. difference heat lost (W)
internal heat transfers - 3 walls 4 x 2.4 m 28.8 2.5 5 360
external wall transfer 7.2 1.0 25 180
window transfer 2.4 5.0 25 300
roof transfer 16.0 0.2 25 80
floor transfer 16.0 1.0 25 400
Total conductive losses 1320 1320 1320 1320
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7.9 Method to determine air- change rate
So, the ventilation loss 1494-1320 174
W volume of room is 4 x 4 x 2.4 38.4 m3,
and temperature difference 25oC So,
ventilation loss 38.4 x 25 x 1305 / 3600 x ach
174
seconds in an hour So, number of
air changes per hour 0.500