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1
Environmental Control
Systems I- Temperature and
Humidity, Arch 353
Lecture # 8
Dr. Hussain Alzoubi
2
Total energy in spaces:
1- Solar radiation from windows
2-Heat gain by conduction from walls and windows
3-Appliances and devices
4-Human activities.
-In winter: solar radiation contributes positively
to heating loads in spaces.
-In summer, solar radiation contributes negatively
to cooling loads.
-Heat transfer through walls increases heating loads
in winter, and cooling loads in summer.
3
Example of software for energy calculations and
Thermal comfort:
Ecotect (http://usa.autodesk.com/ecotect-analysis/)
Enegyplus(http://apps1.eere.energy.gov/buildings/energyplus/)
DEROB-LTH(http://apps1.eere.energy.gov/buildings/tools_directory/software.cfm/ID=211/pagename=alpha_list
Energy10(http://apps1.eere.energy.gov/buildings/tools_directory/software.cfm/ID=36/pagename=alpha_list
Trnsys (http://apps1.eere.energy.gov/buildings/tools_directory/software.cfm/ID=58/pagename=alpha_list
Ecotect Tutorialecotect tutorial
4
How to convert solar radiation into heat:
1- From tables, find the total incident energy
on the targeted window.
2-Specify the transmittance factor of the window glass.
3-The net solar heat raises the temperature of the air mass as follows
Solar Heat= m c T or:
Solar Heat= m c (T2-T1)
Where :
m : mass of the heat storage (lb or kg)
c : specific heat
T1 : temperature before mass heated by solar heat (c or F
T2 : temperature after mass heated by solar heat
6
H= Solar heat,
Wh, or
Calorie, or
Joule,
Btu,…etc
Thermal
storage
Mass, mSpecific heat, c
The temperature of the thermal storage will be
Raised by T;
T = H/mc
For example :
T = Joule / Kg . Joule/Kg. C
T = Joule / Kg . Joule/Kg. C
7
Example :
A 20 ft x 20 ft x 10 ft room has a south window of 20 ft2
with a transmittance factor of 90%,
if the total solar radiation on south
façades in one day is 200 Btu/ft2. How much air temperature
change will be attributed to this solar energy.(assume :
1-No heat loss through walls
2- No infiltration
3-The room is empty and the solar energy will be embodied in
the air mass. (assume air density is 1 kg per 1m3).
Solution:
Solar heat = 20 x 0.90 x 200= 3600 Btu
Air mass = [20 x 20 x10/35.28]*1 kg =113.337kg= 249 lb.
Specific heat from table 3-1 is 0.24Solar Heat= m c T
3600 = 249 x 0.24 x T
T = 3600/ (249 x 0.24) = 60F
Air mass ? Depends on:
-Humidity
- Air temp.
- Altitude
(above sea level)
8
• Temperature Degree Celsius,
Centigrade, [ºC]
Kelvin [ºK]
Fahrenheit(English units)
• Energy = Joule, [J]
= Calorie [cal]
1 [cal] = 4.2 [J]
KWh,
Btu (English units)
• Power = Rate of Energy flow = Watts,
[w], [J/s]
Examples of thermal units and energy
19
Example : in the following example, you are
given the energy required to heat the cylinder
and you need to find the required rate of energy
Per unit time (power), if you want to heat the
cylinder in 1 hour.
20
Temp
( °C )
Density
pure
water
( g/cm3 )
Density
pure water
( kg/m3 )
Density
pure
water
lb/cu.ft
0 (solid) 0.9150 915.0 -
0 (liquid) 0.9999 999.9 62.42
4 1.0000 1000 62.42
20 0.9982 998.2 62.28
40
0.9922
992.2 61.92
60 0.9832 983.2 61.39
80 0.9718 971.860.65
For heat storage, you need some material densities:
highest
27
•In Series
When materials are placed in series, their thermal resistances are added so
that the same area will conduct less energy for a given temperature
difference. An example of this is a cavity-brick wall, with two layers
of brick, an air gap and 12mm of plasterboard - all in series.
Heat transfer by conduction
•In Parallel
When materials are placed in parallel, their thermal conductance are added
and the total energy flow is increased for a given temperature difference.
An example of this would be a cavity-brick wall with a window inserted
within it.
Less conductivity
with multi-layers
28
The rate at which heat flows through a homogenous
material under steady-state conditions is given by:
Q =A . T/R OR Q =A C T (C = 1/ R )where:
Q = the resultant heat flow (Watts or Btu/unit time)
A = the surface area through which the heat flows (m² or ft²)
T = the temperature difference between the warm and cold sides
of the material (K , C, or F), and
R = the thermal resistance per unit area of the piece of
material (m²K/W).
C: conductance of material ((W /m²K).
For a composite building element made up of a number of layers
of different materials, its total resistance is given as:
Rt = Rso + SRn + Rsi
where the resistance of the nth layer is:
Rn = tn/kn
where:
Rt = the total overall resistance of the element (m²K/W),
Rn = the resistance of the nth
material within a composite element (m²K/W),
tn = the thickness of the nth
material in a composite element (m), and
kn = is the conductivity of the nth
material in a composite element.
Rso : the resistance of the outside air film
Rsi : the resistance of the inside air film
29
Example1: a wall is composed of two common brick layers of Thickness =
15 cm separated by an air gap of 5cm (air at rest). Find the U-value of the
wall (neglect the air films)
Solution :
From table 3-4,
-The K-value of brick is 0.72 W/(m2 ) (C/m)
-The K-value air is 0.024 W/(m2 ) (C/m)
Accordingly,
-The R-value for each brick layer is Rn = tn/kn
= 0.15 / 0.72
= 0.208 m2 C /W
The R-value for the air gap is Rn = tn/kn
= 0.05 / 0.024
= 2.08 m2 C /W
Rt = Rbrick 1 + Rbrick2 + Rair
Rt = 0.208 + 0.208 + 2.08 = 2.5 m2 C /W
U-value = 1/Rt = 1/(2.5) = 0.4 W/ m2 C
Example2: if the wall in example 1 has an area of 10 m2 and separates two
rooms with a temperature difference between them, T=5 C , what is the
rate of heat transfer through this wall in Watt.
Q = AU T = 10 m2 X 0.4 (W/ m2 C) X 5 C = 20 Watt.
b) What is the heat transfer in 10 hours
H = Q (Watt) x T(hour) = 20 X 10 = 200 Watt-hour
30
-Concept of conductivity(K), conductance(C), and U-value1- All of them are values for material conductivity
2-Conductivity is the ability of a material to conduct heat from side to
side per unit thickness (inch or meter).
3-Conductance is the ability of a material to conduct heat from side to
side for a specific thickness of a specific material.
4- U-value is a measure of a building elements ability to conduct heat;
the higher the U Value, the greater the conduction of heat ( it measures
the rate of heat transfer through multi-layer walls.)
Accordingly: the heat transfer through walls
could be expressed as follows:
Q = A* T * K /t ( for a wall made of one homogenous material)
Q = A* C* T (for a wall made of one homogenous material with a specific thickness)
Q = A* U* T (for a composite wall made of different materials)
Q = A T/R
Where :
Q: rate of heat transfer through a wall
A : area of wall
K: conductivity
T: Temperature difference between the two sides of the wall
t: thickness of wall
C: conductance
U : U-value of a multi-layer wall.
Notice that :
U-value = 1/(R1+R2+……..Rn)
but NOT EQUAL TO (1/R1) + (1/R2) +…. (1/Rn)
If you have a multi layer wall with specific conductance for each layer,
C1, C2, …Cn, then the U-value = 1/ [(1/C1) + (1/C2) + … (1/Cn)].
but NOT EQUAL TO C1+ C2 + …Cn
32
Remember:
Air and water are bad in conduction and good
in convection. Still air can be used as a good insulator.
There are three ways of heat transfer:
1-Radiation (example : solar radiation)
2-Convection (heating by water and air: both of them
are carried from heat source to other spaces.
3-Conduction ( example: heat transfer through walls)
EXAMPLE
The brick conductivity, K, is 0.72 W/(m2)(C/m) and the air conductivity,
K, is 0.024 W/(m2)(C/m). You are required to design a wall with two layers of
brick and air cavity. If each layer of brick has a thickness of 10 cm,
find the minimum width of the air cavity so that you keep
a U-value of 0.5 W/(m2)(C). (Neglect the air films on both sides).
Solution:
Assume air gap is (Ag) m
Rtotal = R(air) + R brick1+R brick2
Rtotal = (Ag/0.024) +(0.1/0.72) +(0.1/0.72)
As Rtotal = 1/U-value, Rtotal = 1/0.5 = 2
2 = Ag/.024 + 0.2/.72
Ag = 0.024(2-1/3.6)
Ag= .0413 meter, or 4.13 cm.
33
HEAT,Q = UA(Ti-To)
For varying indoor Temperature:
dQ= U(Ldy)(Ti-To)
Ti is a function of y
In this example:
Ti= 70+(5/h)y
Then dQ = UL(70+(5/h)y-To) dy
HEAT (Q)=UL(70+(5/h)y-To) dy
Assumptions:
1-(To) is constant with
height.
2-Inside temperature
has linear relation
with height y.
L
dy
h
70
75
Outside
wall
section in a wall
T=
T=
Btu or joule
0
h
34
L=6m
H=3m
20c
21.5c
T Outside
=10 c
wall
section in a wall
Ti=
Ti=
Example 1:The internal temperature at a certain wall varies from 20 C at
the skirt of the wall to 21.5 C at the highest point. if the width of the wall is 6 m and the height
is 3 m, find the heat loss through this wall, (outside temperature is 10 C
and U-value of this wall is 1.2 W/ m2 C )
Solution:
Ti= 20+0.5y, y measured in meter.
dQ = UL(20+0.5y-To) dy
HEAT Loss (Q)=UL(20+0.5y-To) dy
HEAT Loss (Q)=1.2*6 (20+0.5y-10) dy
=7.2 (0.5y+10) dy
=(3.6y +72) dy
0
3
0
30
3
= [1.8y2 +72y]
= [1.8(3)2 +72(3)]- [1.8(0)2 +72(0)]
=232.2 Watt OR Joule/second
3
0
35
L=6m
H=3m
20c
21.5c
T Outside
=10 c
wall
section in a wall
Ti=
Ti=
Example 2: (using average temperature instead of integration)
The internal temperature at a certain wall varies from 20 F at
the skirt of the wall to 21.5 F at the highest point. if the width of the wall is 6 m and the height
is 3 m, find the heat loss through this wall. (U-value of this wall is 1.2 W/ m2 C
Alternative Solution:
Average Ti= (20+21.5)/2 = 20.75 C
Q = UA(Ti-To)
HEAT Loss (Q)= 1.2 * (6X3)(20.75-10)
= 1.2 * 18*10.75
= 232.2 Watt OR Joule/second
(the same answer)
Average = 20.75
Area= 18 m2
43
Time lag(1)The time delay due to the thermal mass is known as a time lag.The thicker and more
resistive the material, the longer it will take for heat waves to
pass through. The reduction in cyclical temperature on the inside surface compared
to the outside surface is knows and the decrement. Thus, a material with a decrement
value of 0.5 which experiences a 20 degree diurnal (daily) variation in external
surface temperature would experience only a 10 degree variation in internal
surface temperature.
Decrement factor = Ti (max) / To (max)
1- http://new-learn.info/learn/packages/clear/thermal/buildings/building_fabric/properties/time_lag.html
52
A 4m (W) x 5m (L) x 3m (H) room has an average U-value as follows:
U-value for walls = 0.8 W/(m2)(C)
U-value for roof = 0.5 W/(m2)(C)
Heat loss rate for floor = 6.3 W/(m2) (or 2 Btu/hour/ft2, see lecture 8)
U-value for windows = 1.2 W/(m2)(C)
U-value for door = 0.45 W/(m2)(C)
The area of the door is 2m2 and the area of the window is 1.5 m2.
The indoor temperature is 20 C for all walls and the outdoor temperature is –5 C.
1-Find the rate of heat loss through all surfaces of the room in Watt.
(Neglect heat loss from edges and from opening cracks).
2-Find the total heat loss from 8 am to 5 pm during a workday.
Solution:
-Find area of surfaces: as follows
Area of walls = 2x(4x3)+2x(5x3) – 2 (door) – 1.5(window) = 50.5m2
Area of roof = 4x5 = 20 m2
Area of floor = 4x5 = 20 m2
Area of door = 2 m2
Area of window = 1.5 m2
Heat loss from walls = AUT = AU(Ti – To) = 50.5 x 0.8x(20-(-5)) = 1010 Watt
Heat loss from roof = AUT = AU(Ti – To) = 20 x 0.5x(20-(-5)) = 250 Watt
Heat loss from floor = A X F = 20 x 6.3 = 126 Watt
Heat loss from door = AUT = AU(Ti – To) = 2 x 0.45x(20-(-5)) = 22.5 Watt
Heat loss from window = AUT = AU(Ti – To) = 1.5 x 1.2x(20-(-5)) = 45 Watt
1) The rate of heat loss from all surfaces is = 1010 +250+126+22.5+45 = 1453.5 Watt
OR(Joule/second)
= 1453.5 X 3.412 Btu/hour
2) The heat loss from 8 am to 5 pm = 1453.5 Watt x 9 hours = 13081.5 Watt-hour
= 13.081 Kilo Watt-hour
=13.0815x3412 Btu
5m
ft
4m
Plan
Example:
54
To calculate the heat gain:
1- Calculate all heat gain from windows by radiation
2-Calculate all heat gain from appliances and devices.
3-Calculate all heat from space occupants(human
being)
4-Subtract or add heat gain or loss by conduction
through Windows, walls, ceiling, and floor)
To minimize the heat loss through walls
1-Increase the R-value, decrease the U-value by
material choice or playing with material thickness.
2- For windows, you can increase the R-value by
using multi glass layer ( double pane windows or
triple pane windows, use weather stripped frames)
3-Use special gases between the glass layers.
(typically argon, krypton, or xenon)
55
Scaled R-value
Ti
To
OutsideInside
T1
T2
T3
T4
-Draw temperature profile to know the condensation
point
57
T4
T3
T2
T1
Inside Outside
To
Ti
Scaled R-value
To
Outside
Thickness
T1
Inside
Ti
T3
T4
T2
Rt
Ti-To
(Ti-To)/Rt = (Ti-Tx)/Rx
(Ti-To)(Rx/Rt)= Ti-Tx
Tx= Ti- (Ti-To)(Rx/Rt)
Rx
Tx
59
Example:
Draw the temperature profile for the following
wall
C1:
C1=2Btu/(hr)(ft2)(F)
C2=3Btu/(hr)(ft2)(F)
C3=0.3Btu/(hr)(ft2)(F)
C4=3Btu/(hr)(ft2)(F)
C5=2Btu/(hr)(ft2)(F)
Ti=72F
To= 32 F
c1 c2 c3 c4 c5
insulation
Air film
Air film
Solution :
Finding R-values for the wall layers as follows:
R1 = 1/C1 = 1/2 = 0.5
R2 = 1/C2 = 1/3 = 0.34
R3 = 1/C3 = 1/0.3 = 3.34
R4 = 1/C4 = 1/3 = 0.34
R5 = 1/C5 = 1/2 = 0.5
R1 R2 R3 R4 R5
0.50.5 3.34
0.34
0.34
Rt = .5 + .34+3.34+.34+.5 = 5 ft2F/(Btu/hr)
60
2- draw R-values
R1 R2 R3 R4 R5
R-value plot
32F
72F
32F
42F
52F
62F
72F T1T2
T3
T4
3-Find the internal temperature by calculations:
Using this equation from the previous slide
Ti at any layer = Tx= Ti- (Ti-To)(Rx/Rt
T1 = 72 – (40)(0.5/5) = 68 F
T2 = 72 – (40)((0.5+0.34)/5) = 65 F
T3 = 72 – (40)(0.5+0.34 +3.34/5) = 39 F
T4 = 72 – (40)(0.5+0.34 +3.34+0.34/5) = 36 F
Example 2: if you have a room with
61
Major greenhouse gases[5]
-Greenhouse gases are those gaseous constituents of the atmosphere,
both natural and anthropogenic(caused by human activity), that absorb and
emit radiation at specific wavelengths within the spectrum of infrared radiation
emitted by the Earth's surface
1-Water vapor (H2O)
2-carbon dioxide (CO2),
3-nitrous oxide (N2O),
4-methane (CH4) and
5-ozone (O3)
(Pollution increase it)
62
Wavelength and absorbance
-Some reflected waves cannot go through the gaseous layer
-It depends on wavelength of the reflected waves
The German physicist Wilhelm Wiens proposed the following equation:
63
Example
-as the temperature of the same waves become low
on the surface of the earth, the wavelength of the
reflected heat will be longer and will not go through
the layer of the greenhouse gases.
64
Glass pane acts as the greenhouse gases.
-a pane of glass is wavelength selective in that it
Transmits a great deal of the short-wavelength solar
radiation that strikes its surface , but absorbs most of the
long-wavelength radiations that emanate from low-temperature
surfaces inside buildings. So the glass is ideal material for trapping
solar energy.
Glass
Short waves
long waves
66
Effect of heat on building materials
1- Materials expand if heated which causes
structural problems.
2- Expansion depends on
a) Coefficient of linear expansion of material
b) length of building element ( beam, wall, column..etc
c) temperature degrees
68
-Effect of solar heat on structure1-temperature causes expansion and contraction
2- if the materials of the structure are different, the expansion causes cracks
column with walls
beams with slab
foundation with upper walls.
3-if the structure is long, the expansion causes cracks
4-if the wall has two layers with different temperature or different materials,cracks
happen if the wall is not reinforced.
plan
elevation
70
Linear expansion and temperature in buildings
-depends on temperature
-structure length
-material type, expansion factor
L = L0 T
Where:
= the coefficient of linear expansion
L =the change in length caused by a temperature change T
L0= the length at the initial temperature
At T1
At T2L
71
Material per F per C
Aluminum 13X10-6 24X10-6
Glass 4 to 5 X10-6 7 to 9.5 X10-6
Concrete 6.0X10-6 11X10-6
Steel 6.7X10-6 12 X10-6
Table 1: Coefficient of linear expansion for some
selected Materials () [6]
Example: a 100 m concrete beam is built in
an arid area and has temperature variation between
summer and winter of 40C . Find the expansion of
this beam.
73
T L
Material per C
Aluminum 24x10-6
Brass 18x10-6
Concrete 11x10-6
Copper 17.1 x10-6
Glass 7 to 9.5 x10-6
Steel 12x10-6
Table 3.2:Coefficient of Linear Expansion for Selected Building Materials
(Reference: Harris, Solar energy system design, page 64)
L L0
Cold weather
Hot weather
expansion
L = L0 T
Example: a 100 ft concrete wall is exposed to sunlight from 7 am to 4 pm
If the wall temperature was raised from 35F to 135F , how much it has
expanded in length (meter)?
Solution:
From table 3.2 below, of concrete is 11x10-6
T = (135 – 35 )x5/9 = 55.6 C
L0 = 100/3.28 = 30.5 meter
L = L0 T = 11x10-6 / C x 30.5mx 55.6 C = 0.01865m = 1.87 cm
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