Transcript Section 7 - University of East Anglia

Section 7
HEAT LOSS CALCULATIONS
Dr. Congxiao Shang
Objectives:
- To estimate heat loss & energy
consumption of a building over a period of time
7.0 Review: U - Values
Review – in Section 6, we learned
In thermal transmittance, the U value is simply defined as
1/R; Unit: W·K-1·m-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.
A lower U-Value means the system
will transmit less heat.
R
U
clear ?
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
Ventilation is treated differently,
as seen in the next slide -
Area  U  value
The value obtained is the Heat
Loss Rate, W·K-1, or the heat
loss per unit temperature
difference, as seen from the definition
of the U – value, W·K-1·m-2.
This is an intrinsic factor for a
particular building, it may be
changed by varying the insulation
of one or more components.
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?
7.2 Ventilation
Divided by 3600 to bring the
value to Watts (J S-1)
building volume x ( air change rate x
1300
) x temp diff.
3600
or,
building volume x ( air change rate x 0.361) x temp diff.
Equivalent to U - value
the factor for ventilation
heat loss
but the unit here is W m-3 °C-1, i.e. it’s 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
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 U–Value)
+ 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)
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 21°C) 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.
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 -1°C & 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.
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 DegreeDay Method, as assumptions for each are consistent.
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
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 15.5°C. 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.
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.
Tma
Considering the above,
the improved method of Degree-Day:
For each day,
DegreeDays
Tneutra
x
l
0.5 (Tmax+Tmin)
Tmin
= Tneutral - 0.5 (Tmax +Tmin), if 0.5 (Tmax+Tmin) < Tneutral
= 0,
if 0.5 (Tmax+Tmin) >=Tneutral
During the Heating Period:
Tmax - maximum external temperature; Tmin - minimum external temperature
Tneutral - neutral or base temperature.
7.5 Annual Energy Consumption for Heating
7.5.2 Degree - Day Method – More accurate Formula
For each day,
If Tneutral =< Tmin, Degree-Day = 0
Tneutra
l
If Tneutral > Tmax,
Degree-Day = Tneutral - 0.5 (Tmax+Tmin)
If Tneutral < Tmax, then
If Tneutral >= 0.5 (Tmax + Tmin)
Degree-Day = 0.5 (Tneutral-Tmin)- 0.25 (Tmax-Tneutral)
If Tneutral < 0.5 (Tmax + Tmin)
Degree-Day = 0.25 (Tneutral-Tmin)
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 = 22°C
7.5 Annual Energy Consumption for Heating
7.5.2 Degree - Day Method – for cooling
Tma
x
Cooling Formula:
Tneutra
l
Approximately, for each day,
Tmin
Degree-Day = 0.5 (Tmax + Tmin) - Tneutral, if 0.5 (Tmax + Tmin) > Tneutral
= 0,
if 0.5 (Tmax + Tmin) =< 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.
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
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
Thus actual balance temperature
= 4.5 °C
= 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
Note: Units
W°C-1
°C day
1 day = 86400 seconds
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…)
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 20°C. The climatic data
refers to the period of 5th 10th January 1985 in
Norwich
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.
7.6 Dynamic Heating
7.6.1 Response of typical house to external temperature variations in summer
V a ria tio n o f In te rn a l a n d Ex te rn a l T e m p e ra tu re s
25.00
T em p erat u r e
20.00
15.00
10.00
Ex ter nal Temper atur e
Inter nal Temper atur e
5.00
0.00
0
6
12
18
24
30
36
42
48
T im e
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.
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.
7.6 Dynamic Heating
7.6.3 Effect of time switching – constant external temperature
Boiler on periods:
07:00 - 11:00 & 17:00 - 24:00
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 17:00, a similar thing happens.
But after about 4 hours, the boiler throttles
back as the thermostat level is reached. Then,
If the outside temperature had been
the boiler output drops with time as the heat
constant at 0oC, the output of the boiler
storage of the house is replenished.
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.
7.6 Dynamic Heating
7.6.3 Effect of time switching – constant external temperature
-A consequence of time
switching is thus to require a
larger boiler in order to reach
thermostat level more quickly;
Troom
- 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 .
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) .
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?
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…)
14.5 kW
16 kW
Troom
16 kW
13 kW
Troom
7.7
Dynamic Heat Loss - a worked example
To properly analyse the dynamic situation required “splitting”
the components of the buildings into numerous subcomponents, 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)
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 20°C, 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]
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 = 18.51°C.
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)
-5.98
-1.49
24:00
20.00
0.00
-900
70
0
-830
02:00
18.51
0.00
-45x18.51
70
0
-
04:00
17.13
0.00
70
06:00
15.87
0.00
70
2000
08:00
18.31
0.00
70
2000
1246
8.97
2.24
10:00
20.55
0.00
70
2000
1145
0.32
0.08
7.8
Energy issues with radiator on outside wall
See handouts – only for those who are interested
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
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 0°C, the rest of the
building is heated uniformly to 20
°C.
Roof
Window
= 2.4 m2
20ºC
2.4
20ºC
External wall
20ºC
Internal wall
4
4
0ºC
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 25°C. 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
0ºC
U-value data are given in Table 2.
Roof
U-Values (W m-2 oC-1)
External
Walls
1.0
Internal
Walls
2.5
Roof
0.2
Floor
Window
= 2.4 m2
1.0
20ºC
Window
External wall
5.0
2.4
20ºC
20ºC
25ºC
Internal wall
4
Assume radiator in the room is off.
4
0ºC
7.2 Ventilation
Divided by 3600 to bring the
value to Watts (J S-1)
building volume x ( air change rate x
1300
) x temp diff.
3600
or,
building volume x ( air change rate x 0.361) x temp diff.
Equivalent to U - value
the factor for ventilation
heat loss
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
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