Sustainability Project Thermo Hector Ceja

7/27/2019 Sustainability Project Thermo Hector Ceja

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Markstein Sustainability ProjectThe Wolfpack

Team Members:

Hector G. CejaShouxun Wang

Phys 324

December 13, 2013

.

Abstract

The following project will be analyzing one of the campus buildings, Markstein Hall, and will con-struct a thermal model of its energy needs along with recommendations for minimizing its energyconsumption. This project will touch on different parts of thermodynamics that we can apply to helpengineering a sustainable building.

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Contents

Acknoledgements 3

Project Description and Goals 3

Markstein Hall Building at CSUSM 3

Thermal Radiation on Building 4

Sun intensity on Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Earths Albido . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6Weather Fluctuation in San Marcos. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Calculating Clear Sky Solar Radiation (ASHRAE, Inc. , 2009) . . . . . . . . . . . . . . . . . . . . 8

Thermal Model 15

Heat Sink (Central Plant-Cooling) 16what does it do? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Chiller System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Energy consumption 20Lighting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Desktop . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Projector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Laptops . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21HVAC System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21seasonal change in energy consumption . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

Thermal Comfort 25

HVAC system of building 25HVAC System Main Box (rooftop) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25Outside Air intake . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Psychrometric Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Mixed air from Return air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Heat Transfer to/from Radiators and air . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Supply Fan/Return fan. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28Air mixture with room . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

Heat Source and Heat transfers 29Thermal radiation Windows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30Thermal radiation Walls/roof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32Thermal radiation window rim. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34Projecture/ Light Fixture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Student Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35What heats up a room . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Hot air taken out of room. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Energy in vs out 36

the Computer Program- controls cooling/heating 36

improvements to the building 36building repairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37Motion Detectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37Computer program - Verdiem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38Solar Panels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38New lighting fixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

Bibliography 39

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Acknowledgements

Our special thanks are extended to the Steve Holbrook and Ed Johnson of the SustainabilityDepartment at CSUSM for their assistance in providing us with data for Markstein Hall. In addition,we would like to thank our professor Dr. Dominguez with providing us an opportunity to apply topicslearned in class on real life applications.

Project Description and Goals

Our main goal is for us to learn how to analyze any building and be able to change it and make itmore sustainable. After we started researching and looking at the building for this project we found outthat it can can be made very complex. Thus, our goal was slightly molded to intake as much informationregarding our building and thermodynamics and make it easier for our classmates to understandthe basic thermodynamics. After we were able to grasp the main thermodynamic applications to ourbuilding it would become easier for us to suggest improvements to make the building more sustainable.

Markstein Hall Building at CSUSM

Markstein Hall is CSUSMs Business building and was finished on January 2006. The building wasnamed after the single largest donor to this project Markstein Beverage Co. which is a local businessin San Marcos, CA (form CSUSM Website).

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Thermal Radiation on Building

Sun intensity on Earth

The first step is to calculating the solar constant on to the earth using our class notes.We will first start with the Stefan-Boltzman Law and assuming emmision is 1.0.Then

Jsun= bT4sun

whereJsun= Energy flux b = 5.67 10

8Wm2K4 Tsun= 6000K

So:

Jsun= (5.67 108Wm2K4)(6000K)4

Jsun= 73.5 106Wm2

Now the Total energy emitted by the suns photosphere can be calculated by

Wsun= Jsun (area of sun)= Jsun4r2sun

where

Wsun= (Total energy emitted form sun) Jsun73.5 106Wm2 rsun= 647 10

6m

Then

Wsun= Jsun 4r2sun

= 73.5 106Wm2 4 (647 106m)2

Wsun= 3.865 1026W

Now we will calculate the Earths Solar Constant which is the energy received by Earth. We knowthat the Suns energy radiates in all directions , then we can say that that energy is being spread out

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over the surface of sphere of every increasing volume and surface area. Then the surface area of thesphere of energy has a radius of the average Earth distance from the sun. We know that the distancefrom earth to the sun is rSE = 1.50 1011m. The surface of this sphere is now :

ASE = 4 rSE

= 4 (1.50 1011m)2

ASE = 2.83 1023m2

So, we are going to spread out our3.865

1026

Wover this area (ASHRAE, Inc. , 2009). To solve we willbe dividing the total energy by the area calculated to give us the average intensity flux of solar energy(solar constant) as it goes to earth (ASHRAE, Inc. , 2009).

ISE = W

ASE

=3.865 1026W

2.83 1023m2

ISE = 1367Wm2

Now it is obvious that the solar constant can not be the energy that falls on am2 of Earth. We canthink of the the solar constant as a measurement at a right angle to the suns ray. As seen on theimage most of the earth is set back at an angle giving us a lower intensity.

Figure 1: Earth picture webpage

As seen on the image able the total energy intercepted by Earth it must equal

WSE = ISE (Area of Earth Cross-section)

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http://www.uwmc.uwc.edu/geography/100/rad-temp.htmhttp://www.uwmc.uwc.edu/geography/100/rad-temp.htm


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Figure 2: Earth picture webpage

The cross-section is not a perfect circle but we will assume for out calculations. Therefore, the totalenergy that hits the Earths cross-section is:

WSE = ISE (r2E) whererE (the radius of earth)

Now the earth is spinning and we can estimate and average energy spread out across the surface ofearth (4r2E). Therefore, we can estimate intensity of energy hitting on an estimate of square meter by:

IE = WSE

(surface area of earth)

= WSE

4 r2E

= ISE

(r2

E)4r

2E

= 1

4ISE

IE = 341.75Wm2

Keep in mind that this is not exact but is an estimate around the world,IEaround the equator is higherthan in CSUSM.

Earths Albido

Another factor that we need to take into account is the planetary albedo (). Albedo is the ratio of theintensity of light reflected from an object, such as a planet, to that of the energy/light it receives from

the sun.

Figure 3: from NASA webpage

Albedo ratio will range from 0 (total absorber) to 1(total emitter). We know from several resource andHyperphysicsthat earths average planetary albedo is 0.367 and therefore it absorbs

absorbes = 1 = 1 0.367= 0.633

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http://www.uwmc.uwc.edu/geography/100/rad-temp.htmhttp://visibleearth.nasa.gov/view.php?id=60636http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/albedo.htmlhttp://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/albedo.htmlhttp://visibleearth.nasa.gov/view.php?id=60636http://www.uwmc.uwc.edu/geography/100/rad-temp.htm


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Therefore the incoming energy(IE)is

IE =IE0.633

IE = 216.33Wm2

Weather Fluctuation in San Marcos

Weather takes a big part in affecting our buiding. To analyze the change in our building when

weather changes it would take to much time. Some properties of weather that can make big changesto our building are:

Humidity Temperature wetness/dryness clouds (bloking sun)

Heres a link to Weather fluctuation in WeatherSpark.

On the next image we can compare the actual weather that was recorded to weather that can beseen for the next couple days.

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http://weatherspark.com/#!graphs;a=USA/CA/San_Marcoshttp://weatherspark.com/#!graphs;a=USA/CA/San_Marcos


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Now we recorded Humidity and Temperature of the roof of Marstein building and when plotted wegot

Calculating Clear Sky Solar Radiation (ASHRAE, Inc. , 2009)

Before we apply make the calculations of building we need to estimate what fraction of the Solarconstant actually hits our walls. Therefore we will now calculate the clear-sky Solar Radiation SanMarcos CA on 11/25/13. Keep in mind that we know the solar constantISE from above converted to[Btu h1 ft2]British Thermal Unit (BTU) is is the amount of energy needed to cool or heat one poundof water by one degree Fahrenheit.

ISE = 1367 Wm2 =433.3 Btu hr1 ft2

Now because the earths orbit is slightly elliptical,we can recalculate LE from before. Thus thesun radiant flux IE varies throughout the year, reaching a maximum of447.6 Btu hr

1 ft2 near the

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beginning of Januarry, when the earth is closest to the sun and a minimum of 419.1 Btu hr1 ft2

near the beginning of July, when the earth is farthest from the sun. Next we found an approximationequation for the IE perpendicular to the suns ray to be

IE = ISE

1+0.033 cos

360o

(n 3)

365

where n is the day of year (1 for january 1, and 32 for February 1, etc.) and the value inside the cos isin degrees.

For our day of the year we can calculate 11/25/13 where n = 330 and ISE = 433.3 Btu hr

1 ft

2.Thus makesIE to be

IE = ISE

1+0.033 cos

360o

(n 3)

365

= (433.3 Btu hr1 ft2)

1+0.033 cos

360o

(n 3)

365

= (433.3 Btu hr1 ft2)(1+0.033 0.7936)

= (433.3 Btu hr1 ft2) 1.026

IE = 444.65 Btu hr1 ft2

We can see that its bigger thanIE calculated before.

Next we can take into consideration declination. We can assume that the earths equatorial planeis tilted at an angle of 23.45 to the orbital plane. Then the solar declination (the angle between theearthsun line and the equatorial plane) varies throughout the year, as shown below.

This variation causes the changing seasons with their unequal periods of daylight and darkness.The declination can be calculated allow of way but one way we will calculate it is

= 23.45o sin

360o

n +284

365

where is in degrees and the value inside the sine is also in degrees. Thus for our day of the year

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(11/25/13,n= 330) we get

= 23.45o sin

360o

n +284

365

=23.45o sin

360o

330+284

365

=23.45o sin

360o

614

365

=23.45o sin (605.6)

=23.45o 0.9107

= 21.36o

Now we would think of the suns Position in a day and can be expressed in terms of the solaramplitude above the horizontal and the solar azimuth measured from the south as seen in figure above.The solar altitude angle is defined as the angle between the horizontal plane and a line coming fromthe sun. Its value ranges from 0o when the sun is on the horizon, to 90o if the sun is directly overhead.For our example we will calculate the sun radiation on our building at

solar azimuth directly south = 0

andsolar altitude = 35.82o

(calculated frompredication.com. ) which is at solar noon and the sun is due south.Next thing we will use is air mass from ASHREA Handbook 2009. The relative air maxmis the ratio

of the mass of the atmosphere in the actual earth/sun path to the mass that would exist if the sunwhere directly overhead. We can see the air mass is a function of solar altitude (in degrees) as seenbelow.

m= 1

sin +.050572(6.07995+ )1.6364

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http://pveducation.org/pvcdrom/properties-of-sunlight/sun-position-calculatorhttp://pveducation.org/pvcdrom/properties-of-sunlight/sun-position-calculator


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Then for our example when we calculated = 35.82o then

m= 1

sin +.050572(6.07995+ )1.6364

= 1

sin(35.82o) + .050572(6.07995+35.82o)1.6364

= 1

0.58524075402551+0.0001120262891

=

1

0.5853527803

m= 1.708371488

Now we are getting closer to finding the clear-sky solar radiation with a surface that is perpen-dicular/direct sun ray (IE) and a horizontal surface/diffused sun ray (IEHor) We can calculate bothwith

IE = IEexp[bmab] IEHor = IEexp[dm

ad]

Where

IE = beam normal irradiance (to rays of the sun)

IEHor = diffuse horizontal irradiance (on horizontal surfaces)

IE = Normal irradiance from equation before

m= air mass calculated aboveb andd = beam and diffused optical depths-a measure of how opaque a medium is to

radiation passing through

aband ad = beam and diffusive air mass exponents

The Air mass exponentsab and ad are correlated to b andd through the following relationships.

ab= 1.219 0.043b 0.151d 0.204bd

andad= 0.202 0.852b 0.007d 0.357bd

We we unable to find the exact values ofband dtherefore we will make an estimate from differentlocations in the state. Then we will letb 0.431and d 2.317.

Then calculatingab and adab= 1.219 0.043b 0.151d 0.204bd

= 1.219 0.043(0.431) 0.151(2.317) 0.204(0.431)(2.317)

ab= 0.6469

and

ad= 0.202 0.852b 0.007d 0.357bd

=0.202 0.852(0.431) 0.007(2.317) 0.357(0.431)(2.317)

ad= 0.5379

Now we can solve forIE andIEd. Then we can plug in and solve for

IE = IEexp[bm

ab

]= 444.65 Btu hr1 ft2 exp

(0.431)(1.708371488)0.6469

= 444.65 Btu hr1 ft2 0.5436525076

IE = 241.735 Btu hr1 ft2

And

IEd = IEexp[dmad]

=444.65 Btu hr1 ft2 exp(2.317)(1.708371488)0.5379

=444.65 Btu hr1 ft2 0.1760

IEd = 78.273 Btu hr1 ft2

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We have now calculated the solar radiance hitting our roof (white on figure )of a clear sky on 11/25/13at solar noon.

Next we will try to calculate solar radiance of our red and blue walls on figure above. Theorientation of receiving surface is best characterized by its tilt angle/slope (

P) between angle of surface

with horizontal plane. In our case we have our wallsP

= 90o. We also characterize by its wall surfaceazimuth () defined as the displacement from south of the projection, on the horizontal plane, of thenormal to the surface. Surfaces that face west have a positive surface azimuth and those face easthave a negative surface azimuth. We can estimate for our building model to have

blue = 45o and red = 45

o

The surface-solar azimuth angle is defined as the angular difference between the solar azimuthand the surface azimuth :

=

Values of greater than 90o or less than 90o indicate that the surface is in the shade. This is whywe do not calculate for the walls in the shade.For our example we have first for the blue wall and we know = 45o

blue=

=0 45o

blue= 45o

For our example we have now for the red wall and we know = 45o

red =

=0

(45o)

red = 45o

Now the angle between the line normal to the irradiated surface and the earth-sun line is calledthe angle of incident. This is important because the sun does not directly hit these surfaces thus thesun intensity is partially diffused. Its value is given by

cos = cos cos cosX

+sin cosX

Now that for our vertical surfaces (P

=90o) equation goes to

cos = cos cos

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For our examples we have first for the blue wall and we know = 35.82o andblue = 45o

cos blue = cos cos

=cos 35.82o cos 45o

= (0.8109)(0.7071)

cos blue = 0.57338739

blue = 55.01o

For the red wall and we know = 35.82o andred = 45o

cos red = cos cos

=cos 35.82o cos 45o

= (0.8109)(0.7071)

cos red = 0.57338739

red = 55.01o

We can see that both these wall will receive the same amount of sun radiation.Now when we are calculating the solar irradiance onto a wall it is a little more complicated than

the roof. Total clear-sky irradiance It reaching the receiving surface is the sum of three components:

It,b The beam component originating from the solar disc

It,d the diffusive component originating from sky dome

It,r the ground-reflected component originating from the ground in front of the receiving surface.

ThenIt = It,b+ It,d+ It,r

There are many different way to calculate this but we will follow one example given in ASHREA 2009.TheBeam Componentis obtained from a geometric relationship:

It,b = IEcos

where is the angle of incident.TheDiffusive Componentis more difficult but is given by

It,d = IEdY

withY= (0.55+.437cos +0.313cos2 )

Keep in mind that this is an approximation for clear sky not for cloudy days.TheGroundReflected Componentirradiance for surfaces of all orientations is given by

It,r = (IEsin + IEd)g1 cos

P

2

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where g is ground reflectance, which we can take to be 0.2 for a typical ground surface. Now we cancalculated for out building sides walls. For both our walls we can calculate total clear sky radiance.

Beam Component

It,b = IEcos

= 241.735 Btu hr1 ft2 cos (55.01o)

= 241.735 Btu hr1 ft2 (0.57343346)

It,b = 138.62 Btu hr1 ft2

Diffusive Component

Y= (0.55+.437cos +0.313cos2 )

= (0.55+.437 cos (55.01) +0.313 cos2 (55.01))

= (0.55+.437 0.57343346+0.313 (0.57343346)2)

= (0.55+0.2505904220+0.1029225170)

= (0.55+0.2505904220+0.1029225170)

Y= 0.9035

then

It,d = IEdY

= 78.273 Btu hr1 ft2 0.9035

It,d = 70.72 Btu hr1 ft2

GroundReflected Component

It,r = (IEsin + IEd)g1 cos

P

2

= ((241.735 Btu hr1 ft2) sin (35.82) + (78.273 Btu hr1 ft2))(0.2)1 cos 90o

2

= ((241.735 Btu hr1 ft2) (0.5852) + (78.273 Btu hr1 ft2))(0.2)1

2

=141.46 Btu hr1 ft2 +7.83 Btu hr1 ft2

It,r = 149.29 Btu hr1 ft2

Therefore It is for each the red and blue wall.

It = It,b+ It,d+ It,r

=138.62 Btu hr1 ft2 +70.72 Btu hr1 ft2 +149.29 Btu hr1 ft2

It = 358.63 Btu hr1 ft2

Now we can calculate the total sun radiation that hit the Markstein Hall building on 11/25/13 atnoon when the sun is south of the building. We only did an estimate of our building and our estimateswill be based on the rectangle box seen in out figure below and size estimated.

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The area of the Roof in estimates would be

Aroof = 270ft. 120ft.= 32,400 ft2

The area of the blue wall is estimated to be

Ablue= 120ft. 45ft.= 5, 400 ft2

The area of the red wall is estimated to be

Ared = 270ft. 45ft.= 12,150 ft2

Now total Sun radiation can be calculated by

U = It(Ared) + It(Ablue) + IEd(Aroof)

=358.63 Btu hr1ft2(12,150ft

2) +358.63 Btu hr1ft2(5,400ft

2) +78.273 Btu hr1ft2(32,400ft

2)

=4, 357, 354.5 Btu hr1 +1, 936,602 Btu hr1 +2, 536 Btu hr1

U = 6, 296,492 Btu hr1

Now this is what hits the building at this time but we will need to calculate how much goes inside

the building. Instead of calculating how many goes into the whole building we will do it for the roomthat we will be analyzing below on the HVAC System.

Thermal Model

A thermal model is helpful for a bigger building to be able to control temperatures for differentparts of the building. It would be inefficient if the you had to turn on the AC for a whole floor if onlyhalf of the floor needs to be cooled or heated up. Therefore our building has 8 HVAC (short for heating,ventilation, and air conditioning) systems that are in charge of regulating, cooling, and heating upeach of their thermal block. In the following picture we can see how the HVAC systems are located onthe roof and to what floor they correspond.

We also have below how the building is separated into three thermal block- the north side in greenthat HVAC-5 and south side HVAC-3 in blue. HVAC 8 is shut off and would supply the stairwell. Wecan also see the deferent temperature readings throughout the floor.

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Heat sink (Central Plant - Cooling)

Figure 4: CSUSM Central Plant (DMJIM Hariss, 2008)

Central Plant can be considered the heat sink and hot water supplier to our campus. Central Plantis our heat sink because most of the heat extracted from the outside air is being sent back to the chiller.

The chiller container seen above is a big container that holds 1.5 million gallons water that is chillednightly. During the day Central plant funnels cold/hot water up to campus through undergroundtunnels. Cold and Hot ,well insulated, pipes run under building and then up to the roof to be use forthe HVAC system or hot water.

what does it do?

As mentioned above Central Plant supplies all of campus with chilled water as seen below.

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Figure 5: discharge (DMJIM Hariss, 2008)

We can see that the chilled water is feed up to campus at 3500 gallons per minute and returns at3500 gallons per minute. This could not be possible by just gravity thus 4 pumps seen above help pushall the chilled water to campus. If we assume that during the day when we are using the AC systemson campus water will return warmer than the water supplied.

Figure 6: chiller (DMJIM Hariss, 2008)

Because it takes a a lot of energy to chill 1.5 million gallons during the day, Central Plant plantschills the water during off peak hours and saves money. The system can be seen above of the waterflow from the chilled reservoir to the chiller system above. It takes the warmer temperature water, chillsit and then returns it from the bottom of the reservoir.

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Chiller System

If you have driven on campus and you see white smoke from in back of C lot you are mistaken.Central Plant chills its reservoir during the nights then our next part is looking at a part of the chillingloop. A cooling tower is the big box with horizontal fan blades as seen on our main photo of CentralPlant above.

Figure 7: chiller (DMJIM Hariss, 2008)

This cooling tower is mainly used as a heat extractor to the water being passed through the chillingsystem seen above. the chiller loop can be seen below where cooing water return is feed to the coolingtowers and cooling water supply is brought down to the system.

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Figure 8: cooling tower (DMJIM Hariss, 2008)

Figure 9: cooling tower (DMJIM Hariss, 2008)

Hot water is supplied by the condenser then sent to the top of the cooling tower and trickled downwhile fans drive cold air form the surroundings and heat from the water is sent out through the top ofthe cooling tower as a cloud of condensed steam. The water that is at the bottom is then colder and isthen sent back to the condenser.

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Hot and Cold water Transport

If we had more time we could have done some estimates on heat lost or gained when hot and coldwater is getting transported to campus.

Energy consumption

According to reading from Facilities the total Energy consumption of Markstein Hall up to the

month of October is seen below.

For our analysis we where un able to fully estimate our energy consumed to match our readingsabove because their was to many locations we could not access and appliances in locked break rooms.Therefore we will focus on estimates from room 107 with data received from Nov. 25 2013.

LightingWe will estimate that their are 5light fixturesand in each one their is 3 Flouresent lights of 55 W

each. Therefore the amount of Watts per hour that the lights take in room 107 is

5 (fixtures) 3 (lights) 55 (Watts) =825Watts

which then825Wh 24= 19800 WH = 19.8kWH per day

Desktop

For our desktop computer found in the room we can average it would consume 250 Watts per hour.

Then in a 18 hours of usage day it would use250 W 18 hrs= 4500 WH = 4.5kWH per day

Projector

For a class projector we can estimate that it used a max of 322 W. In a day we can guess the useof a projector would be 7 hours. Then the amount of of usage in a day would be

322 W 7hr= 2254 WH = 2.3kWh per day

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Laptops

now for people that bring laptops we can estimate that each laptop uses 65 W. Then if we estimatetheir is 6 classed during that day, each class is 1 hour long,10 students are using the the outlets,and 3 students stay to study for 5 hours with computers plugged in we can calculate the amount ofconsumed by

(6 classes 1 hr 10 students 65 W laptops) + (3 students 5 hrs 65 W laptops) = 3900WH +975WH

= 4875WH

= 4.9KWHper day

HVAC System

We will now estimate the amount of energy that we can be used to keep our room at a good temperature.Facilities was able to help us out and estimated that in a week day the average total amount of energyused by the whole HVAC system is 5000 KWH. We can see below a graph of how many KWH whereused every 15 min in a week.

Next we were able to get data from the Steve Holbrook of the energy usage of the drive fan of allHVAC systems as seen below.

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We will add up the total of energy used for the supply fan and make all HVAC systems a fraction ofthe energy of the daily 5000 KWH.

104 KWH (AH.1) +80 KWH (AH.2) +92 KWH (AH.3)

+103 KWH (AH.4) +158 KWH (AH.5)

+37 KWH (AH.6) +16 KWH (AH.7) + 0 KWH (AH.8) = 590KWH

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Now we can estimate that our of the 5000 KWH per day of the whole HVAC system, AH.1 is

5000 KWH AH.1

Total = 5000 KWH

104 KWH

590 KWH =881 KWH per day

Now we can divide it into average hours of the day by

881 KWH

24 hr =37KWH

Now from the image below of the Thermal block of AH.1 that room 107 can be estimated to be 17

of

space if energy was evenly distributed.

Figure 10: First floor Markstein Building

Therefore an estimate energy used to keep this room fresh is

37 KWH

7= 5.3 KWH

Now from plot data of the RPMs of the fan usage we can estimate how many hours of the day HA.1 isin use.

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As seen above AH.1 is set to be on from 6:30 AM to 10:00 PM thus being in use for 15.5 hours.Therefore the estimated energy used to keep room at a good temperature is

5.3 KWH 15.5hrs= 79.5 KWHon a week day

Now we can take it a step further and add our estimates of energy usage by room 107 in a day.

lighting + Descktop + Projector + Laptops + AH.1=

19.8 KWH +4.5 KWH +2.3 KWH +4.9 KWH +79.5 KWH = 110.7 KWH per day

Now we can convert it to Btus Which we will use to compare energy used to keep building energycomfortable. We know

1kW= 3412.142 BTU/hr

Then110.7 KWH per day 3412.142 BTU/hr= 377,724 BTU/hr per day

seasonal change in energy consumption

As seen on the plot below we se some fluctuations during times of the year.

We can see that for this year shown in grayish blue their was a spike of usage that might have beencaused by the summer hot weather.

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Thermal Comfort

Thermal Comfort is a very interesting section that we did not have time to get to but we can estimatethat a person in thermal comfort at72Fo from a chapter to a text good we found online.

HVAC system of buildingWe have seen above in the thermal block model that our building is supplied by 8 HVA

HVAC System Main Box (rooftop)

introduce units of air volume. cubic ft per min

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example using Pshylometric fundament pg. 1.15 mixing of moist air. example 4 pg 1.17

Outside Air intake

The most important part in regulating a temperate in a room is to cycle fresh air into the building.

Psychrometric Charts

Now one thing that we need to learn more is the thermodynamics of air itself. Thus to do help us apsychometric chart graphically represents the thermodynamic properties of moist air. When ever we do

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calculation of changes to air from the exterior we need to take into account that air has moisture andI varies in different temperatures and pressures. To help us and Thermodynamics we found a usefultool to use called the Psychrometric charts. For our use in Markstein we will be using one that havethe following parameters and seen in Figure 1:

sea-level pressure Normal temperature between 32 to130F

Figure 11: Pscychrometric Chart for normal temperature and close to sea level

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We found interesting that there is many different Psychometric charts for different circum-stances. A psychometric chart graphically represents the thermodynamics properties of moist air.

Mixed air from Return air

We did not have time to calculate example for mixture properties which would have been really cool.I learned that we can use a Psychometric chart before to calculate different maid mixtures and thechange of its properties.

Heat Transfer to/from Radiators and air

We where unable to get information regarding the surface area of the radiator and how muchhot/chilled water is passed in the radiator and also un able to calculate the heat transfer from air tothe radiator. If we had more time for our project we could have done some close estimates.

Supply Fan/Return Fan

We have seen above when we estimated the amount of energy is use for room 107 to keep the roomaround a certain temperature how they vary in energy consumption and speed of the fans that varyaround the day.

Air mixture with room

Now to cool the room we have an air mixture of cold air being put into the room and mixed withthe warmed air from heat radiation in the room. We know that heated air rises to the top and cold airlowers to the bottom. As seen below we put the supplied air away form the window side to cool air thatis at lower temp than air next to the window.

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The air ducts are located next to the window because as sun radiation comes in and warm up theair, the air would rise and go to the return air duct to interact less with things inside the room. Fromthe Markstein Blue prints we are given an average of air supply/return of the air ducts in the room. The2 return air ducts take away up an average of 750 cubic feet per minute and 2 of the four air supply

supply 295 cubic ft per minute and other two close to the return air give 415 cubic feet per minute.Therefore a total of supply is

2 295+2 415= 590+830= 1420ft.3/min

and return2 750= 1500 ft.3/min.

If we had more time we would find a relation of temperature change as different speed of supply andreturn fans of the HVAC System.

Heat source and Heat transfers

There are 3 major ways their is heat being introduced into the room which are Cunduction, Convection,and Thermal Radiation.

Conduction can be said to be the transfer of thermal energy between regions of matter due to atemperature gradient. For example heat spontaneously flows from a region of higher temperature to aregion of lower temperature, temperature differences over time, approaching thermal equilibrium.

Convectioncan be said to be the movement of molecules within fluids (gases, liquids). For examplethe mixing in cold air into a room of hot air.

Thermal Radiation can be said to be the emission of electromagnetic waves from all matter thathas a temperature greater than absolute zero. For example, sunlight, incandescent lights, infared,computers and many more. Next we will be applying some to of the things in our room.

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Thermal radiation Windows

Windows in building Radiation transmission in and heat out.Given in the blue prints the windows have a U factor of0.290 U can be calculated by the following

relation of different materials next to each other

We can also interchange in units by

1h ft2 Fo/Btu= 0.176110 K m2/W, or 1 K m2/W= 5.678263 h ft2 Fo/Btu

can show a figure of temperature going from higher outside and getting lower as pass glass paneglass and then interior.

if no AC then the outside would heat the inside until equilibrium or switch and then give radiationout till equilibrium

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We can calculate the total amount of energy flow Q by

Q= Qthermo+ Qsolar

where

Qth = Steady-state heat transfer caused by in/out door temp. diff., [Btu/h]

Qsol = Steady-state heat transfer caused by solar radiation, [Btu/h]

And then

Q= UA(Tout

Tin) + (SHGC)AIt

where

Q= instantaneous energy flow [Btu/h]

U= overall coefficient of heat transfer (U-Factor),[Btu/h ft2 Fo]

A= Total projected area of fenestration , [ft2]

T =Indoor and outdoor temperature Fo

SHGC= solar heat gain coefficient, dimensionless

It = sun radiation calculated before [Btu/h ft2]

Now we need to figure out our SHGC and for out window it has a tint that helps it lower its heat radiatedin seen below

For our example we can solve for the amount of Q going in through window on 11/25/13 at noon

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when the sun is directly above on the south direction.

Q= UA(Tout Tin) + (SHGC)AIt

where

Q= instantaneous energy flow [Btu/h]

U = 0.290 Btu/h ft2 Fo

A= 2(windows) [2.5ft. 6ft] =30ft2

Tin = 74F

o

Tout = 73Fo

SHGC= 0.26 due to some tint

It = 358.63 Btu/hr1ft2

then

Q= (0.290 Btu/hft2

Fo)(30ft2)(73Fo 74Fo) + (0.26)(30ft

2)(358.63 Btu/hr1ft2)

= 8.700+2797.31

Q= 2, 789 Btuh1

Next We will estimate how much power goes into the room. We will make and estimate that the sungives this radiation from 7 in the morning till 3 in the afternoon ( 8 hr)when the sun does not radiate

through the winder directly.2,789 Btuh1 8hr= 22,312BT U

Thermal radiation Walls/roof

for the roof we are given from the blue prints of Markstein thatUroof = 0.053Btu/hft2 Fo and for ourwall U = 0.186Btu/h ft2 Fo

If the steady-state the fluxItand Iroofis in one direction (perpendicular to the building envelope)we can write the following equation for each material layer within the building envelope as

q= kmT

x

= CT = 1

R

T = UT

where

T = temperature difference between two interfaces of one material layer, [Fo]

x= layer thickness, [ft]

km = mean thermal conductivity of material layer with thickness x, [Btu/h ft2

Fo]

C= thermal conductance of layer with thickness x, [Btu/h ft2 Fo]

R= thermal resistance of layer with thicknessx [h ft2 Fo/Btu]

U= overall coefficient of heat transfer (U-Factor), [Btu/h ft2 Fo]

We can see by thermal photos below of what is the temperature outside the building.

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We can find the energy that can go through the wall and roof by

Q= UA(Tsolair Tinside)

But we need to calculate Tsolair which is how to take into account solar energy hitting the wall andoutside temperature. In this case we would need to find a temperature called Solar air temperature.We can findTsolair by

Tsolair = Tambient+sIsolar

ho

(T4ambient T4surr)

ho

We can estimate for our light colored walls

s

ho

light

= 0.45

17W/m2 Co =0.026m2 Co/W= 0.15h ft2 Fo/Btu

where s is the solar absorptivity and h0 is the combined convection and radiation heat transfer coeffi-cient

Now we can estimate that for out date our

Tambient

= 76Fo Tsurr

= 93.5Fo Isolarroof

= 78.272Btu/h ft2 Isolarwall

= 358.63Btu/h ft2

= 0.90estimate h o = 17W/m2

Co = 2.99Btu/ft2 Fo = 1.714 109Btu/h ft2

We can can estimate how much energy can go though ourroofby first calculating itsTsolairR

TsolairR = Tambient+sIsolarroof

ho

(T4ambient T4surr)

ho

=76Fo + (0.15h ft2 Fo/Btu)(78.272Btu/h ft2) 0.90 1.714 109Btu/h ft2 (764 93.54)

2.99Btu/ft2 Fo

=76Fo +11.74Fo 0.0222Fo

TsolairR = 87.7Fo

Next we can calculate the QR by

QR = UAroof(TsolairR Tinside)

= (0.053Btu/h ft2 Fo)(10ft 21ft)(87.7Fo 76Fo)

= (0.053Btu/h ft2 Fo)(210ft2)(11.7F0)

QR = 130.221Btu/h

Now just make a estimate of 8 hours of energy going into through the roof and we know we might beoff due to different intensities.

130.221BTU/hr 8hr= 1042 BTUs

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Now we can find it for awall and again first finding TsolairW

TsolairW =Tambient+sIsolarwall

ho

(T4ambient T4surr)

ho

=76Fo + (0.15h ft2 Fo/Btu)(358.63Btu/h ft2) 0.90 1.714 109Btu/h ft2 (764 93.54)

2.99Btu/ft2 Fo

=76F

o +53.79F

o

0.0222F

o

TsolairW =129.76 Fo

Next we can calculate the Qw by

QW=UAwall(TsolairW Tinside)

= (0.186Btu/h ft2 Fo)(15ft 21ft)(129.76Fo 76Fo)

= (0.186Btu/h ft2 Fo)(315ft2)(53.76Fo)

QW= 3150 Btu/h

Now just make a estimate of 8 hours of energy going into through the wall and we know we might beoff due to different intensities.

3150BTU/hr 8hr= 25,200 BTUs

We can see that less heat goes through the roof than in the wall. It makes sense because the roofnot only has a was but additional insulation than a 8 inch wall. In addition the wall has reflectionfactors the give it more energy as calculated above.

Thermal radiation window rim

Unfortunately we did not have enough time to compute data on how much energy can go in fromthe metal under the window.

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Video of heat through conduction.

Projecture/ Light Fixture

Student Radiation

We know that any person will have an average of100 Wand we can convert it into Btu/hr from

1 W= 3.412Btu/hr

So 100 W= 100 3.412Btu/hr= 342.2Btu/hr

Now we can show how many people can be in the room and for how many hours we can estimate. Wecan have

Qs = (6 classes (25 342.2BTU/hr) students 2 hrper class)

+ ((3 342.2 BTU/hr) students studying after hours 6 hours )

=102,660 BTUs +6, 160 BT U

Qs = 108,820 BT U

What heats up a room

Now we can summarize the things that introduce heat into a room. we have seen heat from:

Solar heat through walls, window, and walls

Electronics - computers, laptops, projector

Students

Lighting- fluorescent lights and spot lights

We can now estimate a total amount of heat going into the room adding most that are mentioned above.

Q= through window + Through roof + through wall + Students

= 22,312BT U + 1, 042BT U + 25,200 BT U + 108,820BT U

Qheat 157, 374BT U

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Hot air taken out of room

We can see in the link below how got air goes into the return air vent.Our video of heat going out the vent

We found it interesting how the heat still mixes with cold air and swirls into the return air duct.

Energy in vs out

Our last calculation we would like to do is to compare Energy used to keep the room in thermalcomfort and how much thermal energy can go into a room. Most of our calculations have been takenfrom data recording from 11/25/15 for Markstein Hall Building room 107. Energy used on a day wasestimated to be

Qelectricity 377, 724BT U = 110.7 KW

And Energy in heat that is put to the room is also calculated above as

Qheat 157, 374 BT U= 46 KW

We can see that it takes 2.41 times more Power to keep the power the room and and estimate tokeep the room cool. This Power might even be greater because it takes energy to bring chilled waterfrom Central Plant to the building.

Computer Programs that controls all cooling/heating in buildings

Facilities Takes care of all the controls by a computer program. We did not have enough timeto meet up with Steve Holbrook nor learn how the program works. One thing we understood was theeverything is controlled by perfected program that has it set to a high energy save. Therefore each roomon campus that has the thermostat box it does not work because the HVAC system is controlled bythe computer program. If their was a sudden need to cool a room Steve is able to access the computerprogram with his iPad and bypass the program to boost the AC in that room. We had privilege to seehim use this program when the CSU Chancellor gave a forum in the Arts Building and the class wasfull.

improvements to the building

Our building has already improved as seen in the from our energy consumption plot and we willtry to suggest improvements. We did not have enough time to make some calculation or cost forimprovements.

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Building repairs

Because during construction of Markstein Hall the workers where focused on finishing the buildingthe air ducts systems might not be completely sealed. If we can take some time to check the air ductsystem we can save energy by making the HVAC fans work better. Fans speeds would not have to beincreased due to the pressure lost in the air ducts.

Motion Detectors

Each of the classes are controlled with sensors then if their is no one in a room and the temp gainsa slight temp cool air is pushed in to the room. If we have less cold air being waisted then we can lowerthe energy consumption of the HVAC system. Then if someone walks into the room and temp is slightlyhigher than the HVAC system can push cold air in. Of course the program will not let the room go upto high temperatures.

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Computer program- Verdiem

The computer program Verdiem can be useful because it may help shut computers off at certainpoint of the night or when not in use. We had had the chance to walk into a computer room late atnight and the room has a high temperature.

Solar Panels

We thought for solar panels they might not be the best idea right now due to high price and takes along time to pay off. By the time we would pair it of there would be something better to have than theSolar panels used.

Using a website called solar estimate and we where able to find a cost of $ 56,000 dollars to installand pay it off in 30 years.

New Lighing fictures

We currently have our building on fluorescent lights that are better than regular lighting. The nextstep would be to go to LED lighting. We did not have enough time to to make some research but ourguess would be lower thermal energy with same lighting.

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Bibliography

American Society of Heating, Refrigerating and Air-Conditioning Engineers, Inc. . (2009). 2009ASHRAE HANDBOOK FUNDAMENTALS . Atlanta, GA. Online version available at: webpage.

AC Martin Partners (2006). 2006 CSUSM College of Business Blue Prints. ac martin partners, incPlanning Architecture Engineering.

DMJIM Hariss (2008). 2008 CSUSM Central Plant Upgrade. DMJIM Hariss, AECOM, Energy and PowerService Group.
http://app.knovel.com/hotlink/toc/id:kpASHRAE37/ashrae-handbook-fundamentalshttp://app.knovel.com/hotlink/toc/id:kpASHRAE37/ashrae-handbook-fundamentalshttp://app.knovel.com/hotlink/toc/id:kpASHRAE37/ashrae-handbook-fundamentals

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