outdoor cooling for thermal comfort around spectators...

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.


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.


•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.


•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.


•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


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


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


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


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


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.


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 .


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.


• 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


•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.


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


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


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


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.


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.


•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


Case (1) supply the cold air from behind the seats adjacent to the return air diffusers


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


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.


Case (2) supply the cold air from behind the seats and pulls the return air under the seats


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


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


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


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


first three rows with throw (30m) = 400 L/sec


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


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


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


•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%


Analyzing The Results of Case1

Temperature contours (°C) of case (1)


Velocity contours (m/s) of supply air behind the

seats Sunday, November 20, 2016 37

Relative humidity contours of case (1)


Radiation temperature (°C) contours

of case (1)


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


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.


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


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


Temperature = 49°C

Species O2 = 21%

Species N2 = 73%

Species H2O = 6%



Temperature = 30°C

Species O2 = 14%

Species N2 = 70%

Species H2O = 12%

Species CO2 = 4%



Analyzing The Results of Case2


Velocity contours (m/s) of Case (2)


Relative humidity contours of case (2)


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).


Case (3) supply the cold air from above and air return from behind the seats:


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:


Temperature = 20°C

The Latest three rows:


Temperature = 20°C

Where each row consists of four jet air nozzles

Return air Pressure Outlet



Temperature = 30°C

Species O2 = 14%

Species N2 = 70%

Species H2O = 12%

Species CO2 = 4%


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 :



Cross section of jet air nozzles temperature


Cross section of jet air nozzles velocity


Relative Humidity contours of case (3)


Radiation temperature (°C) contours of case (3)


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).


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


Supply air diffusers above The Seats The first three rows:


Temperature = 20°C

The middle fourth rows:


Temperature = 20°C

The Latest three rows:


Temperature = 20°C

Where each row consists of four jet air nozzles

The Radiant Cooling Pipes Temperature = 6°C

Return air Pressure Outlet



Temperature = 30°C

Species O2 = 14%

Species N2 = 70%

Species H2O = 12%

Species CO2 = 4%


Temperature = 49°C

Species O2 = 21%

Species N2 = 73%

Species H2O = 6%


Analyzing the Results of Case (4):

Temperature contours (°C) in case (4)


Velocity contours (m/s) of jet air nozzles


Relative humidity contours in case (4)


Radiation temperature (°C) contours in

case (4)


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.


Supply the cold air in the play ground area

The volume is meshed into almost 1605524 Nodes and

8725914 Elements


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%


Analyzing The Results of the play ground area

Temperature contours (°C) at plane (Y-X) Temperature contours (°C) at plane (Y-Z)


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.


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.



Questions

Prof.Dr.Essam E.Khalil [email protected]

outdoor cooling for thermal comfort around spectators...

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