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Prof.Dr.Essam E.Khalil
Fellow ASHRAE ,Fellow ASME and Fellow AIAA
Faculty of Engineering, Cairo University Cairo- Egypt
Outdoor Cooling for Thermal Comfort around Spectators
In Air Conditioned Sport Facilities
Sunday, November 20, 2016 1 47. Međunarodni kongres i izložba o KGH, Beograd, 30.11–2.12.2016
47th International HVAC&R Congress and Exhibition, Belgrade, Nov.30–Dec.2 2016
Introduction:
In the Middle East, the hot outdoor climate induces a need for cooling
to reach the human comfortable. Especially, the solar radiations are
significant throughout the year. Cooling system of outdoor spaces is
rather complicated and its complexity is increasing with the
advancement of technology.
Nowadays, FIFA World Cup bid is the successful proposal from the
government of Qatar to host 2022 FIFA World Cup and since this
moment Qatar proceeded to the Air Conditioning in the stadiums for
both the players and spectators and dependence on solar energy to
save electric energy that used in cooling system.
The major objective of this study is made on simplified simulation model for analyzing the results of
Qatar stadium by CFD Software and distribute the air inlets to achieve the best distribution of cold
air and specify influence of the sun radiation and wind in the distribution of cold air and study some
parameters such as: The effects of ambient temperature, ambient relative humidity, forced air
velocity and solar radiation intensity.
Sunday, November 20, 2016 2
Case Study: Lekhwiya Sports Stadium is based in the
city of Doha and exposed to sun rays. The
official capacity is 12,000 spectators, and
the stadium is located within the complex of
the Internal Security Forces in the Duhail
district of the capital Doha.
The simulation results were used to validate the simplified simulation model and find solutions
and proposals to reach the human comfortable by using several kinds of technologies such as:
convection cooling and radiative cooling to remove the load capacity and save energy. Wind
speed and direction can influence the amount of natural ventilation and infiltration to a large
extent, but numerical studies of natural ventilation and infiltration in which the wind conditions are
varied.
Sunday, November 20, 2016 3
Estimation Problems That Facing Outdoor Air Conditioning:
•The outdoor spaces are always
exposed to the hot sun rays which
makes it more complicated because the
sun rays raise the temperature which
increases the cooling load that affects
the energy consumed in air conditioning.
•The outdoor spaces are exposed to
external factors and the most important
of these factor is wind speed and
direction that cannot be neglected as it
that affects the cooling load
calculations.
•There are many sources of energy
such as heat from lighting, machines
and people and is difficult to specify
people number and their activities in
outdoor spaces and the amount of
energy generated from machines.
Sunday, November 20, 2016 4
•The outdoor spaces are exposed to humidity
causing two problems, first is increasing the
temperature by latent heat
•The cooling load in the outdoor spaces is too
large if compared to same area for indoor
spaces.
•The temperature is very high in outdoor
spaces because of many factors such as the
sun's heat and exposed to external hot air and
therefore deal with open cycle where cold air
can't make recirculation.
and second is air
condensation leads to
there is quantity of
water droplets, which
must be disposed
them.
Sunday, November 20, 2016 5
•There is difficult to clean and sterilize air in
the outdoor spaces, especially if there is
dust in outdoor spaces.
•The energy consumed in outdoor air
conditioning is very large because the cooling
load in outdoor spaces is large.
•The outdoor spaces are exposed to
external factors such as noise and vibrations
which are difficult to control.
•When outdoor spaces become wide spaces
that are exposed to the sunlight. it can be
difficult to specify the locations of air exits
distribution to cover the zone that need to be
cooled as well as it can be difficult to specify
the locations of air intakes distribution to work
recirculation cycle.
Sunday, November 20, 2016 6
•Cooling Load Estimation in Outdoor Spaces
• Calculation of Solar Radiation:
Solar radiation on an exposed surface involves the determination of the beam and diffuse radiation.
Where, IDN: The total incident solar radiation
A1: The extraterrestrial solar intensity
B: The atmospheric extinction coefficient
PL/P0: The pressure ratio at location concerned
relative to the standard atmospheric pressure
β: The sun altitude angle
Halt: the altitude in meters above sea level.
The values A1, B and sky-diffused factor C1, were estimated by Joudi (1988) :
A1= 1158 *(1+ 0.066*cos ((360*n)/370))
B= 0.175* (1- 0.2*cos (0.93*n))-(0.0045*(1-cos (1.86*n))
C1=0.0965*(1- 0.42*cos ((360/370)*n)-(0.0075*(1-cos (1.95*n)) Sunday, November 20, 2016 7
The beam irradiance (IB) can be calculated as:
IB = IDN*cos (θ)
The diffuse irradiance (Id) is given by (Lunde,1980):
Id = IDN *[(C1*(1+cos (Σ))/2)+(s*(C1+sin(β))*(1-cos (Σ))/2)]
s: The reflectivity of ordinary ground taken as 0.2
Finally, the total incident solar radiation can be calculated by the sum of the beam and diffuse intensities. It = IB + Id
Sunday, November 20, 2016 8
To estimate the heat transfer rate due to solar radiation. it is assumed that the values of
the average solar radiation intensity for the spectator area It1 = 800 W/m2 and the
maximum or the peak solar radiation intensity for the playground area It2= 1000 W/m2.
QI = A1*It1 + A2*It2
where,
QI: The heat transfer due to solar radiation.
A1: The spectator area.
It1 : Total radiation incident on sitting area.
A2: Playground area of the surface exposed to
radiation.
It2 : Total radiation incident on playground area.
QI = 8940*800 + 10965*1000 = 18117000 watt =
18117 KW
Sunday, November 20, 2016 9
Internal Load Calculation
Space heat gain is the rate at which heat enters a space, or heat generated within a
space by people, lighting, and equipment. Lighting and most equipment loads are
sensible heat, while the heat generated by people bodies are a combination of
sensible and latent loads. ASHRAE Fundamentals Handbook - gives representative
the thermal load of people depends on the number of people and their activity level.
Heat Gain from Occupants at Various Activities Sunday, November 20, 2016 10
Number of people = 12000 Person and there are 100 Person will have their athletics activity and
others will have their light bench work activity
SHG1 = 210 W & LHG1 = 315 W, for athletics activity
SHG2 = 80 W & LHG2 = 140 W, for light bench work activity
Adjusted values for total heat shown in Table (1) is for normal percentage of men, women and
children of which heat released from adult female is 85% of adult male, and that from child is 75%
QPS = 100*210 + 12000*80 = 981000 W = 981 KW
QPL = 100*315 + 12000*140 = 1711500 W = 1711.5 KW
QPT = QPS + QPL = 981 + 1711.5 = 2692.5 KW
Sunday, November 20, 2016 11
There are the heat gain from electric lights, equipment and appliances must be taken into account
because they may contribute to increase cooling load.
Autodesk Wikihelp can be extremely useful for approximately calculating from the following table
Example Internal Loads for Different Space Types
Sunday, November 20, 2016 12
Assumptions:
During the midday:
•The spectator area A1= 8940m2
•Lighting Power Density of spectator area I1= 8 W/m2
•Playground area A2= 10965m2
•Lighting Power Density of Playground area I2= 0 because of
depending on the sun light at midday.
QLight1 = 8940*8 = 71520 Watt = 71.52 KW
During the night:
•The spectator area A1= 8940m2
•Lighting Power Density of spectator area
I1= 13 W/m2
•Playground area A2= 10965m2
•Lighting Power Density of Playground area
I2= 8 W/m2
QLight2 = 8940*13 + 10965*8
QLight2 = 203940 Watt = 203.94 KW
Sunday, November 20, 2016 13
The loads due to ventilation and infiltration The infiltration and ventilation is the uncontrolled inward flow of outdoor air and have a significant
impact in cooling load due to the pressure difference between the air inside the stadium and the hot
air outside the stadium.
Ventilation is one of the major contributors to cooling load and it is important select the ventilation
requirements suitably. The amount of air required for ventilation purposes depends on the
application, so standards for minimum ventilation requirements guideline have been formulated
Form ASHRAE standard 62.1-2007.
Sunday, November 20, 2016 14
Estimate the amount of air required for ventilation (V̊o):
To calculate the amount of air required for ventilation in stadium that can be divided into
two regions, The first region is the sport arena or play area and its area is about 10965
m2 and the air rate of this region is 1.5 L/s.m2 from above table and the second region is
the spectator area that accommodates 12000 spectators and its area is about 8940 m2
and the air rate of this region 3.8 L/s for each person and 0.3 L/s.m2.
V̊o = 10965*1.5 + 8940 * 0.3 + 12000*3.8 = 64729.5 L/s = 65 m3/sec .
Sunday, November 20, 2016 15
Estimate the infiltration in Stadium:
The infiltration in stadium is different from the infiltration in building, because the Infiltration
in the building occurs from outdoor air through cracks and openings in the building
envelope due to the pressure difference so the infiltration is a very small and sometimes
can be neglected but infiltration in stadium is the major contributors to cooling load and
can't be neglected.
To find out the amount of air due to infiltration necessary to know the density of the air
inside and outside the stadium due to the temperature difference.
The pressure difference may be caused
by any of the following:
1) Wind pressure.
2) The buoyancy effects due to the
temperature difference between the air
inside the stadium and the hot air outside
the stadium.
3) Mechanical ventilation.
Sunday, November 20, 2016 16
• The pressure difference from wind
The magnitude of the external wind pressure depends on the wind speed, direction and the
shape of the obstacle. The formula for the wind pressure can be calculated with the following
equation:
ΔPw = Ch*Cp(θ)*ρ*V2/2
Ch = A02*(H/Href)
2a
Cp(θ) = 0.5*[((Cp(0º) + Cp(180º)) * cos2(θ))1/4 + ((Cp(0º) - Cp(180º)) * cos(θ))3/4
+ ((Cp(90º) + Cp(270º)) * sin2(θ))2 + ((Cp(90º) - Cp(270º)) * sin(θ))]
where, ΔPw: The pressure from wind [Pa]
Ch: The wind pressure coefficient considering sheltering [-]
Cp(θ): The wind pressure coefficient for the wind angle θ [-]
θ: The angle between the face direction and the wind direction [°]
ρ: The ambient air density [kg/m3]
V: The wind speed at the surface of the object [m/s]
A0: The wind shelter coefficient [–]
a: The wind shelter exponent [–]
H: The height of the stadium [m] Href: The height of the measurement equipment [m] Sunday, November 20, 2016 17
At Face (1) Ch = 1.069
CP (90) = -0.35
At Face (2) Ch = 1.069
CP (90) = -0.35
At Face (3) Ch = 1.069
CP (90) = 0.4
At Face (4) Ch = 1.069
CP (90) = -0.3
The magnitude of the external wind
pressure at face (3):
ΔPw = Ch*Cp(θ)*ρ*V2/2
ΔPw = (1.069*0.4*1.0698*(8)2)/2
ΔPw = 14.638 pa
•The direction of wind at face 3, because
the faces 3 and 4 are open to outside and
the effect of wind is big at these faces as it
is shown in
•The wind speed V = 8 m/s
•The height of the stadium H = 15 m
•The wind shelter coefficient A0 = 0.35
• The wind shelter exponent a = 0.4
Sunday, November 20, 2016 18
•The buoyancy effects:
Air contains a mixture of dry air and water vapor. The amount of water vapor is a function of
the relative humidity; it is also related to the dew point temperature of the air. The
information required for the air density calculation is pressure and temperature and from
the ideal gas law:
P*V = m*R*T
The density of air contains a mixture of dry air molecules and water vapor molecules
may be simply written as:
ρ = (Pd/(Rd*T)) + (Pv/(Rv*T))
The formula for determining the density of air is substituted and rearranged:
ρ = (P/(Rd*T))*(1-(0.378*PV/P))
The pressure decreases by about 1.2 kPa for every 100 meters. The following
equations describe the pressure and altitude.
Sunday, November 20, 2016 19
The empirical formula for estimating the saturation vapor pressure (PSV) is a polynomial
developed by Herman Wobus:
Outside stadium:
Temperature T = 49ºC
Relative humidity RH= 55%
Saturation pressure of water
vapor (PSV1):
Psv1 = 117.4034 mb= 11740.34 Pa
The actual vapor pressure (PV1):
Pv1 = RH% * PSV1 = 6457.18 Pa
Inside stadium:
Temperature T = 25ºC
Relative humidity RH= 50%
Saturation pressure of water
vapor (PSV2):
Psv2 = 31.6701 mb= 3167.01 Pa
The actual vapor pressure (PV2):
Pv2 = RH% * PSV2 = 1583.5 Pa Sunday, November 20, 2016 20
The density of air outside stadium:
ρ1 = (P/(Rd*T1))*(1-(0.378*PV1/P))
ρ1 = (101325/(287.05 *(49+273)))*(1-(0.378*6457.18/101325)) = 1.0698 Kg/m3
The pressure difference from buoyancy or stack effect (ΔPb):
ΔPb = H*(ρ2-ρ1)*g
ΔPb = H*(ρ2-ρ1)*g =15*(1.1775 -1.0698)*9.81 = 15.85 Pa
The density of air inside stadium:
ρ2 = (P/(Rd*T2))*(1-(0.378*PV2/P))
ρ2 = (101325/(287.05 *(25+273)))*(1-(0.378*1583.5/101325)) = 1.1775 Kg/m3
The total pressure difference ΔPTot:
ΔPTot = ΔPb + ΔPw = 15.85 + 14.638 = 30.488 Pa
Sunday, November 20, 2016 21
The amount of air required form
infiltration (V̊inf):
V1 = (2 * ΔPTot / ρ) ^ 0.5
V1 = (2*30.488/1.0698)^0.5 = 7.55 m/s
V̊inf = V1 *Area = 7.55*160*2 = 2416 m3/sec
Outside stadium:
Outside temp.To = 49 oC
Relative humidity RHo= 55 %
Dry Bulb Temp. (DBo) = 49 oC
Wet Bulb Temp. (WBo) = 39.233 oC
Humidity Ratio (Wo) = 0.04261
Specific Volume (vo) = 0.975 m3/kg
Enthalpy (ho) = 159.6 kj/kg
Dew Point Temp. (DPo) = 37.533 oC
Inside stadium:
Outside temp.Ti = 25 oC
Relative humidity RHi= 50 %
Dry Bulb Temp. (DBi) = 25 oC
Wet Bulb Temp. (WBi) = 17.879 oC
Humidity Ratio (Wi) = 0.00992
Specific Volume (vi) = 0.858 m3/kg
Enthalpy (ho) = 50.324 kj/kg
Dew Point Temp. (DPi) = 13.867 oC
Sunday, November 20, 2016 22
The sensible heat transfer rate due to ventilation and infiltration, Qs is given by:
Qs = mͦT*CP*(To-Ti) = VͦT*ρo*CP*(To-Ti)
Qs = 2481*(1/0.975)*1.0216*(49-25) = 62390 kW
The latent heat transfer rate due to ventilation and infiltration, Ql is given by:
QL = mͦT*hfg*(Wo-Wi) = VͦT*ρo*hfg*(Wo-Wi)
QL = 2481*(1/0.975)*2501*(0.04261-0.00992) = 208042 kW
Hence total heat transfer rate due to ventilation and infiltration, QT1 is:
QT1 = Qs + QL = 62390 + 208042 = 270432 KW
The total Cooling Load in stadium, QT:
1) The heat transfer due to solar radiation: QI = 18117 KW
2) The heat transfer due to people: QPT = 8830.5 KW
3) The heat transfer due to light: QLight = 71.52 KW
4) The heat transfer due to ventilation and infiltration: QT1 = 270432 KW
QT = 18117 + 8830.5 +71.52 + 270432 = 297451.02 KW = 84578.8 TR
Sunday, November 20, 2016 23
CFD Modeling and Proposed Solutions
The aim CFD Modeling is to explain the model and boundary conditions that effects in the
model. In addition, clarification the main problem and the available solutions by using
CFD. The study illustrates the influence of different parameters and it is used to determine
the diffuser inlet, velocity, temperature, the turbulence model and the importance of
radiation in air simulations. The values of these parameters which provide the best
solution to achieve the human comfortable.
•Covering the stadium by using sunshade:
The first step to achieve the air conditioning of the stadium depends on trying to cover the
largest part possible to reduce the impact of the sun's rays, and so it will be the proposed
idea using a large sunshade to block out as much of
the sun's rays and maintain the
stadium's temperature, this sunshade
will be of transparent material so as to
allow the passage of sunlight. The
following figure illustrates the proposed
sunshade in Lekhwiya Sports Stadium.
Sunday, November 20, 2016 24
The advantages using transparent glass or plastic that covers the stadium:
1) The transparent glass or plastic can allow sunlight to permeate and reach the playground
area to reduce industrial light.
2) The transparent glass or plastic can reduce the concentration of the sun's rays, because
the glass will reflect and absorbs part of radiation and allows the passage of the rest of
radiation.
3) The transparent glass or plastic can reduce the air leakage and infiltration rate due to
reduce the cooling load in stadium and saving energy consumption.
Sunday, November 20, 2016 25
•The strategy of Air conditioning in stadium:
To choose the locations of the air inlets and outlets due to achieve the human
comfortable, so the stadium will be divided into two zones one of them is special to the
spectators zone and the other is special to the play ground zone at stadium. Spectators
zone is worked air conditioning systems by four different ways and then it will be inserted
in the CFD program to determine the best ways to achieve the human comfortable.
Case (1) supply the cold air from behind the seats adjacent to the return air diffusers:
In this case will be distributed the locations of supply
air diffusers adjacent to the locations of return air
diffusers behind the spectators chairs as shown in
the following figure.
No. of supply Air Diffusers = 300 Diffusers
No. of Return Air Diffusers = 200 Diffusers
The size of each one = 300X300mm
So the flow rate of each one
V̊Diff = 87.5 / 300 = 0.29 m3/sec = 290 L/sec
m̊Diff = 105 / 300 = 0.35 Kg/sec
Sunday, November 20, 2016 26
Case (1) supply the cold air from behind the seats adjacent to the return air diffusers
Sunday, November 20, 2016 27
Case (2) supply the cold air from behind the seats and pulls the return air under the seats:
No. of supply Air Diffusers = 500 Diffusers
No. of Return Air Diffusers = 150 Diffusers
The size of each one = 300X300mm
So the flow rate of each one
V̊Diff = 87.5 / 500 = 0.175 m3/sec = 175 L/sec
m̊Diff = 105 / 500 = 0.21 Kg/sec
In this case will be distributed the locations of supply air diffusers behind the spectator
chairs and the locations of return air diffusers under the chairs as shown in the
following figure and this distribution is allowed to increase the numbers of supply air
diffusers behind the spectator chairs due to a larger area of distribution and this leads
to reduce the flow rate of supply air diffusers which will reflect on the human
comfortable by reducing noise.
Sunday, November 20, 2016 28
Case (2) supply the cold air from behind the seats and pulls the return air under the seats
Sunday, November 20, 2016 29
Case (3) supply the cold air from above and air return from behind the seats:
In this case will be distributed the locations of supply air by using jet nozzles that are
selected by throw and air flow rate and the locations of return air diffusers behind the
spectator chairs as shown in the following figure and this distribution is allowed to
improve the distribution of supply air, because there is no obstructions such as chairs
affect on the movements of air.
The flow rate of jet air nozzles in the
first three rows with throw (30m) = 400 L/sec
The flow rate of jet air nozzles in the
middle four rows with throw (20m) = 300 L/sec
The air flow rate of jet air nozzles in the
last rows with throw (10m) = 150 L/sec
No. of jet Air Nozzles for each row=30 nozzles
The total air flow = (90*400) + (120*300)
+ (90*175) = 87750 L/sec = 87.75 m3/sec
No. of Return Air Diffusers = 500 Diffusers
The size of each Jet Air Diffusers = 500mm
The size of each Return Air Diffusers
= 300X300mm Sunday, November 20, 2016 30
Case (3) supply the cold air from above and air return from behind the seats
Sunday, November 20, 2016 31
In this case is like the previous case (3), but will be added the chilled water pipes inside
the concrete structure to improve the air conditioning inside the spectators zone to take
advantage radiant cooling and remove solar loads from structural elements.
Case (4) supply the cold air from above and air return from behind the seats and
radiant cooling in the slab of stands
The flow rate of jet air nozzles in the
first three rows with throw (30m) = 400 L/sec
The flow rate of jet air nozzles in the
middle four rows with throw (20m) = 300 L/sec
The air flow rate of jet air nozzles in the
last rows with throw (10m) = 150 L/sec
No. of jet Air Nozzles for each row =
30 nozzles
The total air flow = (90*400) + (120*300)
+ (90*175) = 87750 L/sec = 87.75 m3/sec
No. of Return Air Diffusers = 500 Diffusers
The size of each Jet Air Diffusers = 500mm
The size of each Return Air Diffusers
= 300X300mm Sunday, November 20, 2016 32
Case (4) supply the cold air from above and air return from behind the seats and radiant
cooling in the slab of stands
Sunday, November 20, 2016 33
The air-conditioning of the playground area:
When conditioning-air in the stadium playground will be distributed jet air nozzles at the
lower part of the stadium at height of 1m from the ground, as well as the distribution of
jet air nozzles at the top of the stadium at height 8m from the ground as shown in figure
and the supply air temperature 12C with velocity 10m/s at lower part of the stadium and
12m/s at the top of playground.
Jet Air Nozzles at The Top
Jet Air Nozzles at The Bottom
Sunday, November 20, 2016 34
•Results and Discussion Case (1): Supply the cold air from behind the seats
The volume is meshed into
almost 81239 Nodes and
450021 Elements
Boundary conditions Values
Supply Cold Air From
Behind The Seats
Mass flow rate = 0.35 Kg/sec
Temperature = 20°C
Species O2 = 21%
Species N2 = 79%
Return Air Pressure outlet
Outside Air Velocity = 0.05 m/s
Temperature = 49°C
Species O2 = 21%
Species N2 = 73%
Species H2O = 6%
Human Body Temperature = 30°C
Human Faces Velocity = 0.08 m/s
Temperature = 30°C
Species O2 = 14%
Species N2 = 70%
Species H2O = 12%
Species CO2 = 4%
Sunday, November 20, 2016 35
Analyzing The Results of Case1
Temperature contours (°C) of case (1)
Sunday, November 20, 2016 36
Velocity contours (m/s) of supply air behind the
seats Sunday, November 20, 2016 37
Relative humidity contours of case (1)
Sunday, November 20, 2016 38
Radiation temperature (°C) contours
of case (1)
Sunday, November 20, 2016 39
Sunday, November 20, 2016 40
PMV and PPD Calculations:
The PMV and PPD are calculated from knowledge of the six basic variables :
(1) Activity is the effect on the metabolism and the metabolic rate of the spectators equal to 1.2
met
(2) Clothing and the total thermal resistance of clothing was taken equal to 0.5 clo
(3) Four environmental variables and can estimate these variables by ANSYS CFD results:
a- air temperature
b- air velocity
c- mean radiant temperature
d- air humidity
The equations are in accordance with
Fanger and ISO 7730
Sunday, November 20, 2016 41
Point Air
Temp.
(°C)
Mean Radiant
Temp.
(°C)
Velocity
(m/s)
Relative
Humidity
(%)
PMV PPD Sensation
1 22.54 37.29 11.89 8.01 -2.94 99% Cold
2 23.79 38. 52 3.90 26.6 -1.94 74% Cool
3 23.89 38.55 3.81 26.9 -1.88 71% Cool
4 23.92 38.57 2.54 27.86 -1.52 52% Cool
5 23.99 38.68 2.01 32.05 -1.25 38% Slightly Cool
6 24.52 37.45 2.21 32.55 -1.22 36% Slightly Cool
7 23.68 36.44 2.78 24.05 -1.76 65% Cool
8 23.65 33.95 1.85 23.33 -1.53 53% Cool
9 23.86 33.69 1.71 22.50 -1.42 47% Slightly Cool
10 24.02 34.72 0.92 23.08 -0.76 17% Slightly Cool
11 23.77 34.87 0.59 21.51 -0.47 10% Neutral
12 21.49 25.14 5.36 6.50 -3.43 100% Cold
13 21.26 24.88 5.45 6.482 -3.54 100% Cold
14 21.01 24.80 11.00 6.25 -4.03 100% Cold
15 20.95 24.82 11.55 6.95 -4.08 100% Cold
16 22.78 24.81 1.06 7.20 -1.99 76% Cool
17 22.77 24.40 1.012 7.20 -1.99 76% Cool
18 22.96 24.54 8.054 7.23 -3.05 99% Cold
PMV and PPD Calculations of Case(1):
1- The temperature is low at points (1,16,17,18) behind the back of spectators
and at points (12,13,14,15) at spectator’s feet. 2- Air velocity at points (1,18) behind the back of spectators is higher than
the spectator's head.
3- the chairs and spectators become the obstacle of the supply air movement.
Sunday, November 20, 2016 42
Case (2) supply the cold air from behind the seats and the return air
under the seats:
The volume is meshed into
almost 81807 Nodes and
451896 Elements
Boundary conditions Values
Supply Cold Air From Behind
The Seats
Mass flow rate = 0.21
Kg/sec
Temperature = 20°C
Species O2 = 21%
Species N2 = 79%
Return Air under Seats Pressure outlet
Outside Air Velocity = 0.05 m/s
Temperature = 49°C
Species O2 = 21%
Species N2 = 73%
Species H2O = 6%
Human Body Temperature = 30°C
Human Faces Velocity = 0.08 m/s
Temperature = 30°C
Species O2 = 14%
Species N2 = 70%
Species H2O = 12%
Species CO2 = 4%
Sunday, November 20, 2016 43
Temperature contours (°C) of case (2)
Analyzing The Results of Case2
Sunday, November 20, 2016 44
Velocity contours (m/s) of Case (2)
Sunday, November 20, 2016 45
Relative humidity contours of case (2)
Sunday, November 20, 2016 46
Sunday, November 20, 2016 47
PMV and PPD Calculations of Case(2):
Point Air
Temp.
(°C)
Mean Radiant
Temp.
(°C)
Velocity
(m/s)
Relative
Humidity
(%)
PMV PPD Sensation
1 22.46 24.81 7.31 26.01 -3.08 99% Cold
2 24.81 38.54 1.95 7.50 -1.13 32% Slightly Cool
3 24.92 38.15 1.98 26.9 -0.99 26% Slightly Cool
4 25.12 38.12 1.38 25.86 -0.68 15% Slightly Cool
5 25.23 38.39 1.62 32.05 -0.71 16% Slightly Cool
6 25.33 38.41 1.63 13.50 -0.76 17% Slightly Cool
7 24.78 34.99 1.14 24.05 -0.72 16% Slightly Cool
8 24.03 33.24 1.45 23.33 -1.26 38% Slightly Cool
9 23.73 35.46 0.977 22.50 -0.85 20% Slightly Cool
10 23.95 34.50 1.89 23.08 -1.44 48% Slightly Cool
11 23.71 33.76 1.67 21.51 -1.45 48% Slightly Cool
12 22.72 26.85 3.752 6.50 -2.72 97% Cold
13 22.19 25.65 3.012 5.482 -2.82 98% Cold
14 21.61 25.12 7.68 6.25 -3.58 100% Cold
15 21.84 25.03 8.05 11.95 -3.47 100% Cold
16 23.73 25.85 0.045 7.20 -0.26 6% Neutral
17 23.20 24.39 0.17 7.20 -0.69 15% Slightly Cool
18 23.86 24.97 7.053 23.24 -2.50 93% Cold
1- The temperature is low at points (1,16,17,18) behind the back of spectators and at points (12,13,14,15) at
spectator’s feet.
2- This case provides the required comfort because the air velocity behind the seats is lower than case (1).
3-The air velocity at the back of spectators is higher than the spectator's head, but the air velocity in case (2) is lower
than case (1) due to the flow rate in supply diffusers in case (1) is lower than case (2).
Sunday, November 20, 2016 48
Case (3) supply the cold air from above and air return from behind the seats:
Boundary conditions Values
Supply air diffusers above The
Seats
The first three rows:
Mass flow rate of each = 1.92 Kg/sec
Temperature = 20°C
The middle fourth rows:
Mass flow rate of each = 1.44 Kg/sec
Temperature = 20°C
The Latest three rows:
Mass flow rate of each = 0.84 Kg/sec
Temperature = 20°C
Where each row consists of four jet air nozzles
Return air Pressure Outlet
Human Body Temperature = 27°C
Human Faces Velocity = 0.08 m/s
Temperature = 30°C
Species O2 = 14%
Species N2 = 70%
Species H2O = 12%
Species CO2 = 4%
Outside Air Velocity = 0.05 m/s
Temperature = 49°C
Species O2 = 21%
Species N2 = 73%
Species H2O = 6%
The volume is meshed into almost 114929 Nodes and 621556
Elements Sunday, November 20, 2016 49
Analyzing the Results of Case3 :
Temperature contours (°C) of case (3)
Sunday, November 20, 2016 50
Cross section of jet air nozzles temperature
Sunday, November 20, 2016 51
Sunday, November 20, 2016 52
Cross section of jet air nozzles velocity
Sunday, November 20, 2016 53
Relative Humidity contours of case (3)
Sunday, November 20, 2016 54
Radiation temperature (°C) contours of case (3)
Sunday, November 20, 2016 55
PMV and PPD Calculations of Case(3): Point Air
Temp.
(°C)
Mean Radiant Temp.
(°C)
Velocity
(m/s)
Relative
Humidity
(%)
PMV PPD Sensation
1 24.12 36.55 0.04 13.90 1.38 45% Slightly Warm
2 22.44 32.89 2.08 10.42 -2.13 82% Cold
3 21.37 32.06 2.28 8.63 -2.61 95% Cold
4 21.96 32.06 3.23 9.69 -2.70 97% Cold
5 21.46 32.91 2.62 8.55 -2.67 96% Cold
6 21.92 33.52 2.73 10.18 -2.52 94% Cold
7 22.90 33.42 1.75 8.43 -1.83 69% Cool
8 22.92 35.27 1.80 8.60 -1.76 65% Cool
9 22.89 35.05 1.96 8.75 -1.85 70% Cool
10 22.88 35.74 1.05 8.61 -1.23 37% Slightly Cool
11 22.92 36.25 1.10 8.82 -1.23 37% Slightly Cool
12 23.87 36.52 1.12 9.75 -0.96 24% Slightly Cool
13 24.11 36.68 1.01 9.37 -0.77 18% Slightly Cool
14 25.86 36.53 1.02 9.80 -0.45 9% Neutral
15 25.01 36.76 0.05 9.19 1.53 53% Warm
16 25.06 36.34 0.04 9.15 1.48 50% Slightly Warm
17 25.60 36.97 0.04 9.10 1.57 55% Warm
18 24.97 36.82 0.04 9.88 1.54 53% Warm
1- The distribution of temperature in the stands is more suitable than case (1) and case (2), because the air is
directed through the jet air nozzles above the spectators and put the return air diffusers behind the back of
spectators, which gives a better distribution of air.
2-The minimum value of PMV is -2.70 at point 4 and this value is considered cold sensation and the maximum value
of PMV is 1.54 at point 18 and this value is considered warm sensation. the thermal comfort value in this case is
better than case (1) and (2).
Sunday, November 20, 2016 56
Case (4) supply the cold air from above and air return from behind the seats and radiant
cooling in the slab of stands:
The volume is meshed into almost
2462041 Nodes and 2272414 Elements
Boundary conditions Values
Supply air diffusers above The Seats The first three rows:
Mass flow rate of each = 1.92 Kg/sec
Temperature = 20°C
The middle fourth rows:
Mass flow rate of each = 1.44 Kg/sec
Temperature = 20°C
The Latest three rows:
Mass flow rate of each = 0.84 Kg/sec
Temperature = 20°C
Where each row consists of four jet air nozzles
The Radiant Cooling Pipes Temperature = 6°C
Return air Pressure Outlet
Human Body Temperature = 30°C
Human Faces Velocity = 0.08 m/s
Temperature = 30°C
Species O2 = 14%
Species N2 = 70%
Species H2O = 12%
Species CO2 = 4%
Outside Air Velocity = 0.05 m/s
Temperature = 49°C
Species O2 = 21%
Species N2 = 73%
Species H2O = 6%
Sunday, November 20, 2016 57
Analyzing the Results of Case (4):
Temperature contours (°C) in case (4)
Sunday, November 20, 2016 58
Velocity contours (m/s) of jet air nozzles
Sunday, November 20, 2016 59
Relative humidity contours in case (4)
Sunday, November 20, 2016 60
Radiation temperature (°C) contours in
case (4)
Sunday, November 20, 2016 61
PMV and PPD Calculations of Case(4): Point Air
Temp.
(°C)
Mean
Radiant
Temp.
(°C)
Velocity
(m/s)
Relative
Humidity
(%)
PMV PPD Sensation
1 22.21 26.13 0.80 13.28 -1.81 67% Cool
2 27.33 37.56 1.30 23.31 -0.13 5% Neutral
3 22.48 38.74 1.93 13.11 -1.90 72% Cool
4 22.25 37.66 2.76 13.19 -2.25 87% Cold
5 22.81 38.46 1.58 14.77 -1.48 50% Slightly Cool
6 22.92 36.47 1.49 35.27 -1.37 44% Slightly Cool
7 24.54 36.03 1.52 36.06 -0.91 23% Slightly Cool
8 23.42 34.05 1.51 24.42 -1.42 46% Slightly Cool
9 22.47 35.26 0.90 14.93 -1.17 34% Slightly Cool
10 22.64 34.46 1.05 10.83 -1.36 43% Slightly Cool
11 22.50 34.50 1.05 15.06 -1.37 44% Slightly Cool
12 25.17 28.02 0.52 15.45 -0.61 13% Slightly Cool
13 23.77 27.07 0.97 16.58 -1.44 48% Slightly Cool
14 24.65 27.76 0.79 15.50 -1.04 28% Slightly Cool
15 26.07 28.35 0.08 15.55 0.52 11% Slightly Warm
16 25.15 27.08 0.25 16.59 -0.21 6% Neutral
17 26.85 26.59 0.07 15.40 0.39 8% Neutral
18 24.13 26.07 0.92 17.93 -1.37 44% Slightly Cool
1-The temperature at points 1, 14, 15, 16, 17 and 18 in this case is lower than the temperature at the same
points in case (3) due to the cold water pipe embedded in the slab of stands.
2-It is observed that the minimum value of PMV is -2.25 at point 1 and this value is considered cold sensation
and the maximum value of PMV is 0.39 at point 17 and this value is considered neutral sensation. From these
results in the previous figure, we note that the thermal comfort value in this case is the best case.
Sunday, November 20, 2016 62
Supply the cold air in the play ground area
The volume is meshed into almost 1605524 Nodes and
8725914 Elements
Boundary conditions Values
The above supply air diffusers to Playground Velocity = 15 m/s
Temperature = 12°C
Species O2 = 21%
Species N2 = 79%
The below supply air diffusers to Playground Velocity = 12 m/s
Temperature = 12°C
Species O2 = 21%
Species N2 = 79%
Outside Air Velocity = 8m/s
Temperature = 49°C
Species O2 = 21%
Species N2 = 73%
Species H2O = 6%
Sunday, November 20, 2016 63
Analyzing The Results of the play ground area
Temperature contours (°C) at plane (Y-X) Temperature contours (°C) at plane (Y-Z)
Sunday, November 20, 2016 64
Velocity contours (m/s) at plane (Y-X) Velocity contours (m/s) at plane (Y-Z) Sunday, November 20, 2016 65
PMV and PPD Calculations of play ground area:
Point Air
Temp.
(°C)
Mean
Radiant
Temp.
(°C)
Velocity
(m/s)
Relative
Humidity
(%)
PMV PPD Sensation
1 29.30 39. 38 0.85 60.51 2.41 91% Warm
2 28.38 36.47 3.09 68.44 0.99 26% Slightly Warm
3 28.29 38.89 3.18 68.54 1.01 27% Slightly Warm
4 28.21 38.91 2.14 68.34 1.06 29% Slightly Warm
5 28.21 39.01 2.05 68.33 1.09 30% Slightly Warm
6 29.44 39.64 0.83 60.23 2.49 93% Warm
7 29.43 39.19 0.82 63.83 2.58 95% Warm
8 28.69 37.11 3.29 68.51 1.05 28% Slightly Warm
9 28.02 38.98 3.15 68.45 0.94 24% Slightly Warm
10 28.59 38.09 2.49 68.67 0.96 24% Slightly Warm
11 28.07 39.16 2.18 68.31 1.01 27% Slightly Warm
12 29.59 39.21 0.84 63.78 2.59 95% Warm
1- The temperature is slightly warm because there are two reasons, the first reason is increasing
the activity of the body or metabolic rate of players where the metabolic rate of the players equal to
2 met and the other reason is the area of playground is very large
2-The air velocity is low at the players and the air velocity is high at jet air nozzles, so the maximum
velocity at the player's head at point 8 in previous figure and its value 3.29 m/s and the minimum
velocity at the player's legs at point 7 in previous figure and its value 0.84 m/s.
Sunday, November 20, 2016 66
Sunday, November 20, 2016 67
CONCLUSIONS AND RECOMMENDATIONS
1) Using the large sunshade to block out as much of the sun's rays and maintain the
stadium's temperature is very important, this sunshade will be of transparent material
so as to allow the passage of sunlight.
2) it was found that installing supply air from above the seats with radiant cooling could
improve the HVAC performance in the Stadium. It also can be seen from the CFD
simulation results and installing positions of the air supply can affect the human
comfortable.
3) If we used the first Case (1), we shall be used the chairs or furniture made from
materials that do not keep the temperature such as: wood or plastic because the
materials that keep the temperature such as: metals may be harmful the spectator's
back.
4) The importance of CFD in HVAC design because it allows engineers to visualize flow
velocity, density, thermal impact and chemical concentrations for any region where the
flow occurs. This in-turn helps engineers analyze the problem areas and suggest the
best solutions. CFD is widely used across the construction industry for analysis and
design optimization of an HVAC system.
Sunday, November 20, 2016 68