Indoor Environment and Thermal Comfort

Environmental Engineering
1 Indoor Environment and Thermal
Comfort
Vladimír Zmrhal (room no. 814)
Master degree course
1 st semester
Dpt
Dpt. of
Environmental
Engineering
1
Environmental Engineering
CONTENT
Lecture no. Topic Lecturer
1 Indoor Environment and Thermal Comfort Dr. Zmrhal
2 – 4 Heat Transfer and Fluid Mechanics Dr. Barták
5 – 7 Heating Dr. Hojer
8 – 10 Ventilation and Air-conditioning Dr. Zmrhal
11 Noise reduction Dr. Kučera
Study Information System - KOS !!!
Timetable
2
1

Environmental Engineering
EXAM
Exchange Students (Erasmus)
4 tests (minimum 50 points from 100)
Students of Master Study Programme Mechanical Engineering
(Field of study: Environmental Engineering)
4 test (minimum 50 points from 100)
oral examination
State Final Exam !!!
(January)
3
Environmental Engineering
TESTS
Week no. Topic
5 Heat Transfer and Fluid Mechanics
8 Heating
11 Ventilation and Air-conditioning
12 Noise and Vibration, Indoor Environment and Thermal Comfort
13 Resit
http://www.users.fs.cvut.cz/~zmrhavla/EE/EE.htm
4
2

Environmental Engineering
LITERATURE
ASHRAE Handbook
2008 HVAC Systems and Equipment
2009 Fundamentals
2010 Refrigeration
2011 HVAC Application
5
Indoor Environment
Factors of Indoor Environment
Thermal environment
Indoor contaminants (CO 2 , combustion products, VOC, tobacco
smoke, …)
Outdoor pollutants
Odours
Acoustic environment
Lightning
Ionizing radiation
Electromagnetic waves
…
IAQ – Indoor Air Quality
6
3

Physiology
Heat Production
Q = Q −W : A
m
Q Qm
W
= −
A A A
D D D
m
D
[W]
[W/m 2 ]
q = q − w
[W/m 2 ]
q m
w
…total metabolic rate
…external mechanical work
A D …body surface area [m 2 ]
7
Physiology
DuBois surface area
A = 0.202m h
D
0.425 0.725
[m 2 ]
f
cl
A
A
= cl
D
… clothing area factor - correction factor for
clothed body
A cl … area of clothed body [m 2 ]
A D = 1.8 m 2
… for man 70 kg and 1.73 m
f cl = 1.1 ÷ 1.83
8
4

Physiology
Human Thermoregulation
metabolic activities of the body result almost completely in heat that
must be continuously dissipated and regulated to maintain normal
body temperature
internal temperature rise with the activity ∼ 37 °C
temperature regulatory centre in the brain (hypothalamus)
(36.8 °C at rest of comfort; 37.4 when walking)
skin temperature 33 to 34 °C
resting adults produces about 100 W of heat
9
Energy Balance
Figure on the board
q = qm − w = ± qc ± qr + qev ± qres ± q
{
con [W/m 2 ]
0
10
5

Energy Balance
Total metabolic rate
m
External mechanical work
w = η ⋅q m
[W/m 2 ]
η
…mechanical efficiency η = 0 …for most activities
η max = 0.2 …bicycle ergometer
11
Energy Balance
Metabolic heat generation
Activity
Sleeping
Reading, seated
Filing, seated
Walking 2 km/h
Dancing
q … 1 met 2 q m
[W/m 2 ] [met]
46 0.8
58
1
70 1.2
110 1.9
140 to 255 2.4 to 4.4
= 58.1 W/m
12
6

Energy Balance
Heat transfer by convection
( )
q = h f t − t [W/m 2 ]
c c cl cl a
Convective heat transfer coefficient h c [W/m 2 K]
h = 2.38( t − t ) 0.25
c cl a
… free convection
hc
= 12.1
w
… forced convection
13
Energy Balance
Heat exchange by radiation
( )
q = h f t − t [W/m 2 ]
r r cl cl r
Radiant heat transfer coefficient h r [W/m 2 K]
h
r
A T −T
= εσ
A t − t
4 4
r cl r
D cl r
≈ 4.7ε
[W/m 2 K]
ε …emissivity of the clothing, usually 0.95 [-]
σ …Stefan-Boltzmann constant 5.6710 -8 [W/m 2 K 4 ]
A r …effective radiation area of the body [m 2 ], A r /A D ≈ 0.7
14
7

Energy Balance
Conduction through clothing
1
qr + qc = ( tsk − tcl
)
[W/m 2 ]
R
cl
R
cl
⎛ s ⎞ ⎛ s ⎞
= ∑⎜ ⎟ + ∑⎜ ⎟
⎝ λ ⎠ ⎝ λ ⎠
cl
air
[W/m 2 K]
Rcl
= 0.155⋅ I
… 1 clo = 0.155 m 2 K/W
15
Energy Balance
Clothing insulation
Clothing ensembles
Trousers, long-sleeved shirt, longsleeved
sweater, T-shirt
Walking shorts, short-sleeved shirt
R [m 2 K/W]
0.155
0.055
I [clo]
1
0.36
…
…
…
16
8

Energy Balance
Respiration heat loss
q
res
( − h )
M hex
=
A
D
in
[W/m 2 ]
M
h ex
h in
…pulmonary ventilation rate [kg/s]
…enthalpy of exhaled air [J/kg]
…ethalpy of inspired (ambient) air [J/kg]
−
M = ⋅ 6 ⋅q ⋅ A
1.43 10
m D
[kg/s]
17
Energy Balance
t = 32.6 + 0.066⋅ t + 32⋅
x
ex in in
ϕ = 100
ex
[°C]
[%]
or
x = 0.0277 + 0.000065 ⋅ t + 0.2⋅ x
[kg w.v. /kg d.a. ]
t in
ex in in
… temperature of inspired (ambient) air [°C]
x in … humidity ratio [kg w.v. /kg d.a. ]
18
9

Energy Balance
Evaporative heat loss from the skin
q = q + q
ev ev , dif ev , rsw
[W/m 2 ]
natural diffusion of water through the skin
( )
q = ⋅ ⋅t − − p
−3
ev , dif
3.05 10 256
sk
3373
v
p v
t sk
… water vapor pressure in ambient air [Pa]
… temperature of skin [°C]
19
Energy Balance
heat loss by evaporation of sweat secretion
A
q = h p − p
"
( )
ev
ev , rsw ev v , sk v
AD
h ev … evaporative heat transfer coefficient [W/m 2 K]
A ev … effective evaporative area of the body [m 2 ]
p“ v,sk … saturated water vapor pressure at skin temperature [Pa]
h
ev
16.7hc
=
⎡ ⎛ 1 ⎞
1+ 2.22h ⎢R − ⎜1− ⎟ h + h
⎣ ⎝ fcl
⎠
( )
c cl c r
⎤
⎥
⎦
20
10

Parameters Influenced Thermal Comfort
Indoor Environment Parametres
air temperature t a [°C]
relative humidity (RH) ϕ [%]
mean radiant temperature (MRT) t r [°C]
air velocity w a [m/s]
turbulence intensity Tu [-]
Personal Parameters
metabolism q m and work w
thermal insulation of the clothing
and also age, sex (male/female), …
21
Thermal Comfort
condition of mind that expresses satisfaction with the thermal
environment
effect on health and performance
Prediction of Thermal Comfort
Rohles and Nevins (1971) indicate values that provides thermal
comfort (optimum)
required temperature of skin
t
,
= 35.7 − 0.0275( q − w )
sk req
required evaporative heat loss
m
( )
qev , rsw , req
= 0.42 qm
− w − 58.15
22
11

Thermal Comfort
Thermal comfort equation (TCE)
( ) ( )
q − w = h f t − t + h f t − t +
m c cl cl a r cl cl r
q c
( ⎡
( qm
w ) ⎤ pa
)
−3
+ 3.06⋅10 256 35.7 − 0.0275 − − − 3373 +
⎣
t sk,req
⎦
q ev,dif
−6
( q w ) q ( h h )
+ 0.42 − − 58.15 + 1.43⋅10
−
m m ex in
q res
23
q r
q ev,rsw,req
Thermal Comfort
where t cl calculates from the heat flow through clothing
1
qc + qr = ( tsk − tcl
)
Rcl
1
hcfcl ( tcl − ta ) + hrfcl ( tcl − tr ) = ⎡ 35.7 0.0275 ( qm w ) tcl
R
⎣ − − − ⎤⎦
cl
q c q r
t sk,req
tcl = 35.7 − 0.0275( qm − w ) − Rcl ⎡⎣
hcfcl ( tcl − ta ) + hrfcl ( tcl − tr
) ⎤⎦
iteration
24
12

Thermal Comfort
Example: Heat flow through clothing
???
25
Thermal Comfort
ASHRAE Thermal Sensation Scale
+3 hot
+2 warm
+1 slightly warm
0 neutral
-1 slightly cool
-2 cool
-3 cold
Thermal Sensation
=
Predicted Mean Vote
(PMV)
26
13

Thermal Comfort
Fanger‘s Model of Thermal Comfort (Standard Model)
thermal load on the body
L = actual heat flow from the body – heat loss to the
actual environment for a person hypothetically kept at
comfort values of t sk and q ev,rsw at the actual activity level
respectively
L = left side of theTCE – right side of the TCE
27
Thermal Comfort
Predicted Mean Vote – PMV index
PMV predicts the mean response of a large group of people
according to thermal sensation scale
−0.036q ( 0.303
m
0.028)
PMV = e + L
Predicted Percent of Dissatisfied – PPD index
PPD = − ⎡
⎣− PMV + PMV
4 2
100 95exp (0.03353 0.2179 )
⎤
⎦
28
14

Thermal Comfort
PMV = 0 → about 5 % of the people will be dissatisfied
29
Thermal Comfort
Category of global
thermal comfort
A
B
C
PPD [%]

Thermal Comfort
Operative temperature
( ) ( )
q + q = h + h f t − t
c r c r cl cl o
h = h + h
c
r
( − ) + ( − ) = ( + )( − )
h t t h t t h h t t
c cl a r cl r c r cl o
t
o
h t
=
h
c a
c
+ h t
+ h
r r
r
[°C]
31
Thermal Comfort
=
c
hc
+
r
A h h
= + ( 1−
)
t At A t
o a r
w [m/s]
A [-]

Thermal Comfort
Operative temperature necessary for comfort (PMV = 0) of person
in summer clothing at RH = 50 %
33
Thermal Comfort
Mean radiant Temperature
the uniform temperature of an imaginary enclosure in which radiant
heat transfer from the human equals the radiant heat transfer in the
actual nonuniform enclosure
t F T F T F T
4
4 4 4
r
=
p−1 1
+
p−2 2
+ ... +
p−n n
− 273.15
[°C]
F p-n
T n
…angle factor between a person and surface n
…surface temperature of the surface [K]
34
17

Thermal Comfort
35
Thermal Comfort
Simplification - elementary case (rectangle vs. point)
Eckert
F
1−2
1 1 c a + b + c
= − arctg
8 4π
ab
2 2 2
In the room:
∑F1 − 2
= 1
36
18

Thermal Comfort
Angle factor algebra
[°C]
F = F + F + F + F
B−A B−11 B−12 B−21 B−22
F = F − F
B−A B−11 B−12
( )
F = F − F + F + F
B−A B−11 B−12 B−21 B−22
37
Thermal Comfort
Example: Calculation of MRT and t o
Room: L = 6 m, W = 4 m, H = 2.7 m
t a = 28 °C
t w = t a
t ceiling = 18 °C
w = 0.15 m/s
Calculate MRT and t o in the middle of the room at hight h = 1.5 m.
MRT = ?
t o = ? 2 2 2
F
1−2
1 1 c a + b + c
= − arctg
8 4π
ab
38
19

Local Discomfort
Thermal non-uniform conditions
radiant temperature asymmetry
draft
vertical air temperature difference
warm or cold floors
Source: Thermal comfort. The booklet. INNOVA 2002.
39
Local Discomfort
Radiant temperature asymmetry
the difference between the
plane radiant temperature of
the opposite sides of a small
plane element
∆ t = t − t
pr pr 1 pr 2
40
20

Local Discomfort
Radiant temperature asymmetry ∆t pr [°C]
Category of thermal comfort
A,B
Warm ceiling

Local Discomfort
Draft Rate
DR = − t w − w ⋅ Tu +
0.62
(34
a
)(
a
0.05) (0.37
a
3.14)
Source: Thermal comfort. The booklet. INNOVA 2002.
43
Local Discomfort
Vertical air temperature difference
100
PD =
1 + exp 5.67 − 0.856 ∆ta
, v
( )
Source: Thermal comfort. The booklet. INNOVA 2002.
44
22

Local Discomfort
Warm or cold floors
Category
of thermal
comfort
A,B
C
Percentage
of
dissatisfied
PD [%]

Adaptive Model of Thermal Comfort
Adaptation – people can acclimatize themselves
changing posture and activity
clothing changing
leaving the space / move
opening a window, shading …
For buildings without mechanical cooling (air-conditioning) or with lowenergy
cooling systems (night ventilation, …)
47
Thank you for your
attention
48
24