active sidewalk weather management

7/30/2019 Active Sidewalk Weather Management

1/18

Active Sidewalk

Weather ManagementHeat Transfer Final Project

Joseph Cooper, Tyler Ludwig, Stephen Wess


2/18

Abstract

We are looking to analyze a system to whichwill prevent snow accumulation on a given

surface. In our system the surface of analysis

consists of a concrete sidewalk withinsulating base structure (CEMATRIXinsulating cement, gravel, and soil). Two

main options to consider are heating thesystem using buried electrical wires or we

may utilize a working fluid that flowsthrough internal pipes. Due to the high cost

of electricity, we elected to analyze a systemconsisting of hydronic fluid and buried

pipes. The goal of our analysis is to find theamount of heat flux needed to operate in

Rochester winter conditions, as well asidentifying the fluid temperature, depth, and

spacing of the pipes.

Problem Statement

During winter, snow accumulation bringsalong hassles as well as danger. Snow and

ice can cause people to slip and hurtthemselves on a number of different levels.

Removal of snow and the melting of icebecomes time consuming, labor intensive,

and also costly due to the need to own ashovel or snow blower, and maintain a stock

of salt to keep the walkway of ice and snowbuildup. The downfall of salt is that it

creates a mess, is harsh on the environment,and becomes ineffective at temperatures

below - 9C.

Our proposed solution to the problem athand is a heated sidewalk comprised of

working fluid flowing through pipes that lay

within the concrete. The working fluid is tobe heated, and pumped through the pipeswith the goal of maintaining a sidewalk

surface temperature above 0C, and runningat steady state with a given snowfall rate.

We will not consider melting a certain depthof accumulated snow as our analysis will

focus on an active system.

System Model

Below is a figure depicting the cross-section

of the system followed by a table of thermalconductivity values used in our analysis.

Assumptions:

Steady State.

Negligible contact resistances in

substrate Constant properties

No heat loss through sides of

sidewalk section (symmetry) Assume known temperature at frost

line (0C). Heat loss through surface includes

convection, radiation, and heat flux

to melt the snow at given rate Non-uniform surface temperature

Working fluid temperature drop isnegligible

Pipe wall temperature = working

fluid temperature

Material ThermalConductivity

(W/mK)

Cement 0.72

Cematrix 0.10

Gravel 2.40

Soil 0.52


3/18

Theory

Before the heat required from our system to

melt snow can be determined, the conditionsfor which the system is designed to

successfully handle must be set forth.

Ideally, the system will be able to effectivelymeet the demands of a severe winter storm

in Rochester. Conditions were chosen basedon a heavy snowfall during the coldest

period of the year. Meteorological datashows that Rochester experiences its coldest

temperatures during January, with anaverage low temperature of -7.5C. The

average wind speed in January is 5.2m/s.The National Weather Service defines a

snowfall rate of 1.3cm/hr as a heavysnowfall rate that garners a winter storm

warning. Our analysis will take place underthese conditions.

The heat flux required by the system

consists of the flux necessary to preventsnowfall accumulation and the flux lost to

surroundings through conduction,convection, and radiation.

The total heat flux is represented by thefollowing expression:

" " " " " "Tot s m h e bq q q q q q= + + + +

The first term, "s

q ,is the heat flux required

raise the temperature of the falling snowfrom -7.5C to 0C, plus the heat flux

required to raise the temperature of themelted snow from 0C to 0.56C (a

generally accepted as the film temperature atwhich the liquid must be before evaporation

occurs).

2 2, ," [ ( ) ( )]

s H O P ice s a P H O f aq S C t t C t t = +

Where tsis the melting temperature of snow,ta is the ambient air temperature, and tf =

0.56C.

S is the snowfall rate in equivalent depth ofwater.

2( )H O

Rate

snow

S s

=

An expression used for determining the

density of snow, given the ambienttemperature is great than -13C is:

Where Ta is the ambient air temperature, andUis the wind velocity. The density of snow

at our defined condition: 114.02 kg/m3. Thisresults in a snowfall rate in equivalent depth

of water (S) = 4.118e-07 m/s. Assuming

constant specific heats (,P ice

C = 2100 J/kgK,

2,P H OC = 4217 J/kgK) and constant density

of water (1000 kg/m3), "s

q =7.46 W/m2.

"m

q is the heat flux required to melt the

snow.

2

"m H O sq Sh=

Where hs is the heat of fusion of snow

(334 kJ/kg). For the conditions specified,"m

q =137.53 W/m2.

"hq is the heat flux lost to surroundings

through convection and radiation. Becausewe are not assuming the surface to be

isothermal due to the spacing of the tubing,we must use a finite difference method to

determine this heat loss. However, if weassume for now that the surface is in fact

isothermal, we can arrive a baseline valuefor this heat flux. This value represents the

heat flux lost if the tubing below the surfaceof the concrete had no spacing (impractical

due to installation cost!) thus causing theentire surface to be at 0.56C. The

convection coefficient calculated in order toevaluate this flux will be used as the

1.15 1.7500{1 0.951exp[ 1.4(278.15 ) 0.008 ]}snow a

T U

=


4/18

convection coefficient in the finite

difference model. The expression for "hq is:

4 4" ( ) ( )h c f a s f aq h t t T T = +

The convection coefficient hc must bedetermined by applying an appropriatecorrelation function. In order to select the

correct correlation function, it is necessaryto first determine the Reynolds number.

L

air

ULRe =

Where Uis the wind speed (5.2m/s),L is the

width of the sidewalk (1.5m), andair

is the

air kinematic viscosity (13.17e-6 m2/s). Thisgives a Reynolds number of 5.92e5(turbulent) with the transition from laminar

to turbulent flow occurring near 1.27macross the sidewalk surface.

Because the airflow over the sidewalk is not

entirely laminar or turbulent, the correlationfunction used will be for a mixed condition.

In our finite difference analysis, we expectto see a varying temperature across the

surface of the sidewalk. However, thecorrelation function we will use to determine

hcis valid for an isothermal plate. In reality,the temperature differences over the surface

will affect the heat flux, but we will assumethat the temperature difference is not

significant enough to invalidate thecorrelation function. The correlation

function used is:

4/5 1/3(0.037 871)L L

Nu Re Pr=

Pris the Prandtl number for air (0.715). Theaverage Nusselt number = 594.35.

air L

c

k Nuh

L=

The conductivity of the surrounding air =0.02385 W/mK. This results in a convection

coefficient of 9.45 W/m2K. This is the value

that will be used in the finite difference

model. The baseline flux due to convectionis simply hc(0.56 - -7.5) = 76.17 W/m

2.

The contribution of radiation to "hq

is

4 4( )s f aT T , where the emissivity of

concrete (s ) = 0.88. The temperature of the

surrounding (Ta) can be assumed equal tothe ambient air temperature while

precipitation is occurring.The baseline heatflux due to radiation = 31.56 W/m

2. With

the radiation and convection losses

combined, "hq = 107.73. We expect the

results from the finite difference model to

exceed this value due to points on thesurface of the sidewalk at temperatures

greater than 0.56C.

The fourth heat flux term contributing to the

total heat flux is "e

q , the heat flux necessary

to evaporate the melted snow. The followingexpression can be used to calculate this flux.

" ( )e air m f a wq h W W h=

In this expression, hw

is the heat of

vaporization for water (2502 kJ/kg). Themass transfer coefficient, hm for the water

evaporating into the air.

2/3

,

cm

air p air

hPrh

Sc c

=

Sc is the Schmidt number, a dimensionlessparameter that depends on the kinematic

viscosity of the air and the mass diffusion

coefficient of water vapor into air.

airSc

D

=

The mass diffusion coefficient can be

plotted as a function of temperature. For airat a temperature of -7.5C, D 2.0e-5 m

2/s.

Therefore, Sc = 0.6583.


5/18

The density of air at -7.5C = 1.3042 kg/m3,

the specific heat of the air = 1006.4 J/kgK.

Recalling that the Prandtl number = 0.715and hc = 9.45 W/m

2K, we calculate hm to be

0.0076 m/s.

Wf and Wa represent the humidity ratios at

the surface of the evaporating water and thesurrounding air, respectively.

0.622V

V

PW

P P

=

WherePis the atmospheric pressure and PV

is the saturation pressure of air a giventemperature. Wais to be evaluated at the dew

point temperature of the ambient air. For thecalculation, the average dew point in

Rochester during January was used (-7.3C).Wf is evaluated at the liquid film

temperature, 0.56C.

PVat -7.3C = 381.3 N/m2, andPVat 0.56C

= 639.9 N/m2. This gives Wf = 0.003953 &

Wa = 0.002350.

Collecting these terms and solving for "e

q

gives a heat flux required to evaporate the

melted snow equal to 47.9 W/m2.

The final term adding to the total heat flux is

the back loss, "bq . This is the loss through

the concrete down to the ground below.

There is no standard method for calculatingthis value. Generally, this heat flux is

between 5% and 20% for a well insulatedsystem. This will be verified with the finite

difference model.

Summing all of the heat flux terms(excluding "bq ), we arrive at a total heat

flux of 300.6 W/m2. Once again, this is a

baseline, preliminary heat flux calculation.We expect the actual heat flux found using

the finite difference model to be greater thanthis value.

Finite Difference Derivation & Modeling

To analyze the temperature distribution in

the system, a finite difference model wascreated. The analysis of the system was

done at a steady state condition as ourinterest in the characteristic behavior of the

system occurs at steady state.

To perform this analysis, a small section ofthe overall system was selected taking

advantage of symmetry to simplify andexpedite the solution algorithm.

The simplified section

bisected one pipe in thevertical plane and

extended to the midpointbetween two pipes, thus

symmetry can be used onboth the right and left

edges of the grid. Thesurveyed section was

divided into a gridconsisting of equally

spaced x and y indices.Due to the scale of our

system, a grid size of 5

units/cm was used. Itmust be noted that thenumber of grid points in

the x direction wasdependent on the spacing

between the pipes, and was accounted forwhen writing the finite difference algorithm.

An energy balance was performed to deriveequations representing the steady state

temperature at each node. Specialconsiderations were employed whenencountering symmetry and material

junctions where a more complex equationwas required. The following displays the

progression and derivation of each of thefoundational equations used in our analysis.

Y

X


6/18

x =y

+k2 (Tm,n1) (3k1 + k2 )Tm,n = 0

k1(T

m,n+1+T

m1,n+T

m+1,n)

Foundational Equations

(1)

(2)

(3)

(4)

Surface Nodes

(5)

(6)

To facilitate the linear solver used in thisanalysis, the radiation term was reduced to a

linear approximation with Tm,napproximated as the film temperature of

water before it evaporates, 0.56C.

(7)

(8)

(9)

Interior Nodes

(10)

(11)

* For symmetry, nodes at left or rightboundary account for symmetry by

multiplying the temperature of the nodeopposite the symmetry line by a factor of 2.

This approach applies to all symmetrynodes, including the top surface.

Material Junction Nodes

(12)

Where k1, k2 are the conduction coefficientsfor the two materials at the junction.

Pipe Boundary

For the pipe boundary, the actual known

pipe temperatures (the nodes represented ata constant temperature, Tpipe) were

subtracted from the solution matrix and werere-infused in their original location after the

surrounding node temperatures werederived. This allowed the final contour plot

to represent the known pipe temperature aswell as the resulting surrounding

temperatures.

To infuse the presence of the pipe into the

finite difference matrix, the numerical pipetemperature was added to the right hand side

of the finite difference equation for the nodeequations that bordered the pipe, thus

allowing for a boundary condition thatrepresented the actual pipe temperature. This

entire process was automated using analgorithm based upon the pipe geometry and

location.

The Matlab Solution

The implementation of the finite differencemethod in Matlab provided an efficient

solution and sufficient parameter variationcapabilities. The solution script utilized

Matlabs matrix expertise to intelligentlysort through each nodal location and apply

the correct nodal formula. This wasaccomplished by writing nodal equations

based upon the matrix indices and a positionalgorithm. The subtraction of the pipe

temperatures mentioned above were thenemployed and the resulting matrix was

simplified with a solver function. Due to the

q = qsnow + h(T Tm,n )+ (T4

T4

m,n )

q = qsnow + h(T Tm,n )+ hrad(T Tm,n )

Ein = Eout

q(i)(m,n) + q(x y) = 0i=14

q(node)(m,n) = k()

Tnode Tm,n

q = 0Tm,n+1

+Tm,n1

+Tm+1,n

+Tm1,n

4Tm,n

= 0

Tm,n1 +Tm+1,n +Tm1,n 3Tm,n +

x

kq = 0

q = qsnow +U(T Tm,n )

q = qsnow + h(T Tm,n )

+(T

2T

4

m,n )(T Tm,n )(T Tm,n )


7/18

small grid size, the variable count of oursystem reached upwards of 12,000, yet

Matlab was able to provide a solution thataveraged a 5-minute run time.

It must be mentioned that during thisanalysis, an initial assumption was made

concerning the frost line temperature. Wechose to assume that the temperature at the

frost line was known, thus completing ourboundary conditions. After successful

solutions were performed, we investigatedthe veracity of this assumption by varying

the analysis depth and comparing theresulting change in temperature at the

baseline. After several trials, we determinedthat any extension past the 0.8 m frost line

yielded approximately a 0.01% change intemperature at the baseline. Thus, we

validated our assumption and focused ourefforts on other areas of our analysis.

After a final solution was reached, creating a

contour plot of the system temperatureprofile was quite simple. The pipe

temperatures that were removed earlier werere-inserted to the temperature matrix and the

simplified system was propagated to

construct the entire system.

Results

We ran several configurations of our system,varying depth, pipe, separation and fluid

temperature. Our primary goal was to

achieve a minimum temperature of 0.56Cat the surface, which occurs at the surface

midpoint between two pipes. To beginnarrowing our result, the depth was

decidedly fixed to 6 cm, while the depth andfluid temperature were varied. As expected,

the farther the pipe separation, the more heatflux was lost to the surroundings. This is

caused by an increased temperature gradientacross the surface as the minimum surface

temperature is maintained. The results ofseveral trials can be seen in Figure 3 in the

Figures & Plots section of this report.

While many configurations exist for thissystem, our final solution was taken at a

depth of 5cm, pipe separation of 12 cm, and

fluid temperature of 28.5C. The resulting

heat flux lost through the surface and ground

were 315.85 W/m2

K and 16.77 W/m2

Krespectively. Studies have shown thatsimilar systems expect approximately

5-20 % loss through the ground; our well-insulated system loses 5.04 % and

corroborates this benchmark.

Taking into account the earlier derived valueof 192.89 W/m

2K of heat flux required to

melt the snow, the total heat flux lost viaconvection and radiation in our system

amounts to 122.96 W/m2K. This correlates

to our earlier baseline calculation of 107.7

W/m2K, and is reasonably higher given the

existing temperature gradient along the

surface. The graphical results of our finalsolution are plotted below in figure 1.

With our analysis complete, we can estimate

the heat loss in any given length ofsidewalk. Our total heat flux lost, through

the surface and the ground is 332.62W/m

2K.

Just for good measure, let us consider asidewalk of length 9 meters and 1.5 meters

wide. Our heat lost would be:

Watts

Qlost =1.5L *332.62

Qlost =1.5(9)*332.62 = 4490.37


8/18

Figure 1Contour of 1m Sidewalk Section

Figures & Plots


9/18

Figure2

AveragedIsothermsof1mSidewalkSection

Figure 3Pipe Spacing vs Q

C


10/18

Discussion of Results, Possible

Improvements

Improvements of our analysis would involvesolving for the heat, energy, and time

required to get the system up and running atsteady state from a starting point of ground

temperature. Another large case would be tostart from ground temperature with an

accumulation already standing on thesidewalk surface, which may increase start

up to steady state time and energyexponentially.

Conclusion

In conclusion, our resulting values from the

analysis discussed, compare favorably torealistic values estimated by printed articles

and informational sources. A system such asthe one analyzed is a real world possibility

and is already implemented for variouscircumstances requiring heating. For

example, the heating of floors in some up-scale homes are done using a heated

working fluid flow through piping imbedded

into the floor.

Citations

Cematrix.com. Edited by Cematrix

Corporation. Accessed 10/27/11.

http://www.cematrix.com/docs/technical/physical_properties2009mar04.pdf

Diffusion Coefficient for Air-Water Vapor

Mixtures Web. 02 Nov. 2011.

Galloway, Kevin; Landolt, Scott;

Rasmussen, Roy. Using liquid-equivalent snow gauge

measurements to determine snowdepth - Preliminary Results Nov.

2006

Incropera, Frank P., David P. Dewitt,Theodore L. Bergman, and Adrienne

D. Lavine. Fundamentals of Heatand Mass Transfer. 6th ed. Hoboken,

NJ: John Wiley & Sons, 2007.

Ramsey, James W. Development of SnowMelting Load Design Algorithms and

Data for Locations Around theWorld Vol. 1,ASHRAE1999

"Water Vapor and Vapor Pressure." Web.

07 Nov. 2011..


11/18

A1:FiniteDifferenceScript% Active Sidewalk Heating Finite Difference

clcclear all

close all

ticDepth_of_Field=.8; % mradius=5; % gridsrad_vec=[2 3 4 5 5 5];emissivity=.88;steff_boltz=5.67e-8;Pipe_Depth=5; % cmPipe_Separation=12; % cmDelta_x=.002; % m

Ysize=Depth_of_Field*500;Xsize=Pipe_Separation/2*5+1;

Tpipe=28.5; % CTpipe2=-Tpipe; % CTsurfC=-7.5; % CTsurfK=TsurfC+273.15; % KFrostLine=0; % CHsurf=9.451; % W/m^2KRsurf=3.9153; % W/m^2KUsurf=Hsurf+Rsurf; % W/m^2KQsnow=192.8894; % W/m^2Kkc=.72;kcm=.1;kgr=2.4;

ks=.52;Dcematrix=15; % cmDgravel=Dcematrix+5; % cmDsoil=Dgravel+30; % cm

% Allocate MemoryFDM=zeros(Xsize*Ysize,Xsize*Ysize);Q=zeros(Xsize*Ysize,1);

% Write the Eqautions to a Matrixfor j=1:Xsize

for i=1:Ysizepos=(i+Ysize*(j-1));

% Top Left Nodeif i==1 && j==1

FDM(pos,pos)=-(3+(Delta_x*Usurf)/kc);FDM(pos,pos+Ysize)=2;FDM(pos,pos+1)=1;Q(pos)=(Qsnow -Usurf*TsurfC)*(Delta_x/kc);


12/18

% Top Right Nodeelseif i==1 && j==Xsize

FDM(pos,pos)=-(3+(Delta_x*Usurf)/kc);FDM(pos,pos-Ysize)=2;FDM(pos,pos+1)=1;Q(pos)=(Qsnow -Usurf*TsurfC)*(Delta_x/kc);

% Top Middle Nodeselseif i==1

FDM(pos,pos)=-(3+(Delta_x*Usurf)/kc);FDM(pos,pos+Ysize)=1;FDM(pos,pos-Ysize)=1;FDM(pos,pos+1)=1;Q(pos)=(Qsnow -Usurf*TsurfC)*(Delta_x/kc);

% Top Left Node of Material Junctionelseif (i==5*Dcematrix | i==5*Dgravel | i==5*Dsoil) & j==1

if rem(5*i,Dcematrix)==0k1=kc; k2=kcm;

elseif rem(5*i,Dgravel)==0

k1=kcm; k2=kgr;elseif rem(5*i,Dsoil)==0k1=kgr; k2=ks;

end

FDM(pos,pos)=-(3*k1+k2);FDM(pos,pos+Ysize)=2*k1;FDM(pos,pos+1)=k2;FDM(pos,pos-1)=k1;

% Top Right Node of Material Junctionelseif (i==5*Dcematrix | i==5*Dgravel | i==5*Dsoil) & j==Xsize


elseif rem(5*i,Dgravel)==0k1=kcm; k2=kgr;

elseif rem(5*i,Dsoil)==0k1=kgr; k2=ks;

end

FDM(pos,pos)=-(3*k1+k2);FDM(pos,pos-Ysize)=2*k1;FDM(pos,pos+1)=k2;FDM(pos,pos-1)=k1;

% Top Middle Nodes of Material Junctionelseif (i==5*Dcematrix | i==5*Dgravel | i==5*Dsoil)


elseif rem(5*i,Dgravel)==0k1=kcm; k2=kgr;

elseif rem(5*i,Dsoil)==0


13/18

k1=kgr; k2=ks;end

FDM(pos,pos)=-(3*k1+k2);FDM(pos,pos+Ysize)=k1;FDM(pos,pos-Ysize)=k1;FDM(pos,pos+1)=k2;FDM(pos,pos-1)=k1;

% Bottom Left Nodeelseif i==Ysize && j==1

FDM(pos,pos)=-4;FDM(pos,pos+Ysize)=2;FDM(pos,pos-1)=1;Q(pos)=-FrostLine;

% Bottom Right Nodeelseif i==Ysize && j==Xsize

FDM(pos,pos)=-4;FDM(pos,pos-Ysize)=2;

FDM(pos,pos-1)=1;Q(pos)=-FrostLine;

% Bottom Middle Nodeselseif i==Ysize

FDM(pos,pos)=-4;FDM(pos,pos-1)=1;FDM(pos,pos-Ysize)=1;FDM(pos,pos+Ysize)=1;Q(pos)=-FrostLine;

% Left Side Nodeselseif j==1

FDM(pos,pos)=-4;FDM(pos,pos+Ysize)=2;FDM(pos,pos-1)=1;FDM(pos,pos+1)=1;

% Right Side Nodeselseif j==Xsize

FDM(pos,pos)=-4;FDM(pos,pos-Ysize)=2;FDM(pos,pos-1)=1;FDM(pos,pos+1)=1;

% Interior Nodeselse

FDM(pos,pos)=-4;FDM(pos,pos+1)=1;FDM(pos,pos-1)=1;FDM(pos,pos+Ysize)=1;FDM(pos,pos-Ysize)=1;

end

rad_vec=[2,3,4,5,5,5];


14/18

DfromTop=5*Pipe_Depth+1;LfromEdge=(Pipe_Separation/2)*5+1-radius;

% Insert Pipe Temperaturesif i==DfromTop-2 && j==(LfromEdge-1)

Q(pos:(pos+2*rad_vec(1)))=-Tpipe;

elseif i==(DfromTop-3) && j==(LfromEdge)Q(pos)=-2*Tpipe;Q(pos+2*rad_vec(2))=-2*Tpipe;

elseif i==(DfromTop-4) && j==(LfromEdge+1)Q(pos)=-2*Tpipe;Q(pos+2*rad_vec(3))=-2*Tpipe;

elseif i==(DfromTop-5) && j==(LfromEdge+2)Q(pos)=-2*Tpipe;Q(pos+2*rad_vec(4))=-2*Tpipe;

elseif i==(DfromTop-6) && j==(LfromEdge+3)Q(pos)=-Tpipe;Q(pos+2*rad_vec(5)+2)=-Tpipe;



end

endend

% Subtract Known Temperaturesfor j=Xsize:-1:1

for i=Ysize:-1:1pos=(i+Ysize*(j-1));

if i==(DfromTop-rad_vec(6)) && j==XsizeFDM(:,pos:(pos+2*rad_vec(6)))=[];FDM(pos:(pos+2*rad_vec(6)),:)=[];Q((pos):(pos+2*rad_vec(6)))=[];Qpos(6)=pos-1;

elseif i==(DfromTop-rad_vec(5)) && j==(Xsize-1)FDM(((pos):(pos+2*rad_vec(5))),:)=[];FDM(:,((pos):(pos+2*rad_vec(5))))=[];Q((pos):(pos+2*rad_vec(5)))=[];Qpos(5)=pos-1;

elseif i==(DfromTop-rad_vec(4)) && j==(Xsize-2)FDM(((pos):(pos+2*rad_vec(4))),:)=[];FDM(:,((pos):(pos+2*rad_vec(4))))=[];


15/18

Q((pos):(pos+2*rad_vec(4)))=[];Qpos(4)=pos-1;



elseif i==(DfromTop-rad_vec(1)) && j==(Xsize-5)FDM(((pos):(pos+2*rad_vec(1))),:)=[];FDM(:,(pos):(pos+2*rad_vec(1)))=[];Q((pos):(pos+2*rad_vec(1)))=[];Qpos(1)=pos-1;

endend

end

%SolveTmat=FDM\Q;T_pipe_ref=zeros(size(Tmat,1),1);

% Format the ResultsT_rem=54;for i=length(rad_vec):-1:1

Tadd=rad_vec(i)*2+1;

Adjust=T_rem-Tadd;T_rem=T_rem-Tadd;

Tmat1=[(Tmat((1:(Qpos(i)-Adjust)),1))'];Tmat2=[(Tpipe.*ones(1,2*rad_vec(i)+1))];Tmat3=[(Tmat((Qpos(i)-Adjust+1):length(Tmat)))'];Tmat=[Tmat1 Tmat2 Tmat3];Tmat=Tmat';

T_pipe_ref1=[(T_pipe_ref((1:(Qpos(i)-Adjust)),1))'];T_pipe_ref2=[(Tpipe2.*ones(1,2*rad_vec(i)+1))];T_pipe_ref3=[(T_pipe_ref((Qpos(i)-Adjust+1):length(T_pipe_ref)))'];T_pipe_ref=[T_pipe_ref1 T_pipe_ref2 T_pipe_ref3];T_pipe_ref=T_pipe_ref';

end

T_Plot1=[];T_Plot2=[];for i=1:XsizeT_Plot1(:,i)=Tmat(1+(i-1)*Ysize:i*Ysize);T_Plot2(:,i)=T_pipe_ref(1+(i-1)*Ysize:i*Ysize);


16/18

end

T_Plot=T_Plot1;x=[T_Plot flipdim(T_Plot,2)];x=[x x x x x x x x];

figure(1)contourf(flipdim(x,1))colorbarset(gcf,'position',[13 611 1084 1295])figure(2)imagesc(x)colorbarset(gcf,'position',[13 611 1084 1295])toc

% Heat Flux Through Top & Bottom SurfacesQ_Top=[];Q_Bot=[];for j=1:Xsize

Q_Top=[Q_Top (Usurf*(x(1,j)-TsurfC)+Qsnow)];Q_Bot=[Q_Bot ((T_Plot(70,j)-T_Plot(400,j))/(.002*5/kc+.002*25/kcm+.002*150/kgr+.002*150/ks))];endQ_Top=mean(Q_Top)Q_Bot=mean(Q_Bot)Q_Tot=Q_Top+Q_Bot;

load gongsound(y,Fs)


17/18

A2:Misc.Calculations% Heat Transfer Project scriptclcclear

% finding the density of snowTa = 273.15 - 7.5; % KelvinU10 = 5.2; % m/srhoSnow = 500*(1 - 0.951*exp(-1.4*(278.15 - Ta)^(-1.15) -0.008*U10^1.7)); % kg/m^3

% Finding equivalent rate of liquid precipitation.Snow_Rate = 0.013; % m/hrrhoH2O = 1000; % kg/m^3s = Snow_Rate/(3600*(1000/rhoSnow)); %m/hr

% Calculating

CpIce = 2100; % [J/kg*K]CpH2O = 4217; % [J/kg*K]tf = 273.15 + .56; % liquid film temperature [K] for melted snowts = 273.15; % melting temperature [K]

qs = rhoH2O*s*(CpIce*(ts - Ta) + CpH2O*(tf - ts)); % [W/m^2]

hf = 334000; % heat of fusion for water [J/kg]

qm = rhoH2O*s*hf; % [W/m^2]

% calculating Convection coefficient

% calculate Reynolds number based on worst case (wind across short edgeofsidewalk.)

kin_Visc = 13.1666e-06; % kinematic vis. @ walkway film temp(269.4K)[m^2/s]

w = 1.5; % sidewalk width [m]ReL = U10*w/kin_Visc; % Reynolds NumberPr = 0.715; % Prandtl NumberNuL = (0.037*ReL^(4/5) - 871)*Pr^(1/3); % Nusselt Numberkair = 0.023852; % thermal conductivity of air [W/m*K]hc = kair*NuL/w; % convection coefficient

CpAir = 1006.4; % [J/kgK]rhoAir = 1.3042; % kg/m^3

D = 2.0e-5; % mass diffusion coefficient of H2O into airSc = kin_Visc/D; % Schmidt Number

hm = (Pr/Sc)^(2/3)*hc/(rhoAir*CpAir); % mass transfer coefficient [m/s]

Wf = 0.00398; % Humidity ratio, air at 0.56 deg C assumed tf ofliquidWa = 0.00205; % Humidity ratio, ambient airhfg = 2501525; % Heat of vaporization at 0 deg C for H2O


18/18

qe = rhoAir*hm*(Wf - Wa)*hfg

e = 0.88; % emissivitySB = 5.67e-8; % Stefan-Boltzman Constant

% Radiation coefficient (estimation)hr = e*SB*(273.71^2 + Ta^2)*(273.71 + Ta);

% Baseline heat lost through convection and coefficientqh = hc*(0.56 - -7.5) + SB*e*((273.15+0.56)^4 - Ta^4)

% Baseline heat fluxqtot = qs + qm + qe + qh% Baseline heat lossQ = qtot*1.5*9

active sidewalk weather management

Documents