A PROJECT REPORT ON

“DESIGN OF AR CONDITIONING SYSYTEM FOR CAD LAB”

Submitted ByRohit P. Jadhav

Raman R. VilhekarRavinayak D. ChatAmruta A. Gore

Guided By:Mr. M.A.Ali

Lecturer(Sel. Grade)

B.N.College Of Engineering, Pusad

CONTENTSSr. No. Topic Page No.

Abstract01 Introduction 01

02 Thermodynamic cycles 03

03 Main components of Air conditioning system

06

04 Human Comfort 08

05 Cooling load estimation 12

06 Duct design 19

07 Calculations 23

08 Components selection 49

09 Appendix 62

10 Conclusion 72

11 Reference 74

Introduction

• What is Air Conditioning?

• Need of Air Conditioning

• Applications

• Design importance

Sources of Heat generation

• External loads

• Internal loads

• Other loads

• Process loads

Cooling Load component

• Sensible heat load component

• Latent heat load component

1.Hour Angle at 12:00 pm=0º

2.Declination angle for 21 June. d=23.47 sin [360(284+N)]/365 =23.47 sin [360(284+172)]/365 d=23.46 ° (N=no. of days)

3.Latitude angle =19°N — from Net for pusad.

4.Altitude angle ,=[cosl.cosh.cosd+sinl.sind] = [ cos(19).cos(0).cos(23.46)+sin(19).sin(23.46)] = [(0.19)(1)(0.917)+(0.32)(0.99)] = [0.861+0.1248] = 80.33 -cross check it by page 1-24

5.zenith angle (ѱ)=(π/2) - (β) =9.676.solar azimuth angle(γ): = [cosl.sind- cosd.cosh.sinl]/cos β But l<d =0

7.surface azimuth angle:(ξ ) southo° west 90° North180° East-90° cross check it by ASHRAE handbook 09

8.wall solar azimuth angle(α ) = [π-(γ+ξ)].F F=-1 for noon and 1 for afternoon For south side; α =[π-(62.79+0)]×1 =117.21° For west side; α = [ 62.79+90)] =27.21° For North side; α =-62.79° For East side; α =[π-(62.79-90)] =207.21°

9.Direct Radiation from sun IDN =A exp ( -B/ sinβ)………..w/m2

A=apperant solar radiation =1080 for summer B=Atmospheric extinction coeff. =0.21 for summer IDN=1080 exp [(-0.21)/sinβ]

10.Diffuse Radition From sky. Id=C.IDN.FWS………………w/m2

C=const. for cloudness sky. C=0.135 for mid suffer. Fws=view factor or configuration factor Fws=(1+cos∊/2…………..(∊=tilt angle) [for horinzontal surface ∊ =o°,for vertical surface ∊ =90°] For horizontal surface; Fws=(1+coso)/2 =2/2 =1

For vertical surface ; Fws=(1+cos90)/2 =1/2 =0.5 Id =0.135.IDN . FWS

11.Reflected short wave(solar)radition(Ir) Ir =(IDN+ID)ρg.FWG

ρg=reflectivity of the ground FWG=View factor. FWG=(1-COS ∊ )/2 (∊=Tilt angle ) For horizontal ∊ =o° For vertical ∊ =90° For horizontal, FWG=(1-COS0)/2 =0 For vertical, FWG=(1-COS90)/2 =0.5

AMOUNT OF HEAT AVAILABLE (PRODUCE)FOR PUSAD: Latitude=19°NReflectivity=0.6Hour angle=0°Declination=+23.5° Cloudless condition:

Wall azimuth angle(ξ)= 0° south facing =180° North facing = -90°East facing North side wall: Altidude=80.33°………..(calculated) Solar azimuth angle(γ)=0 (as l<d) Now, Wall solar azimuth angle, α =180-(γ+ξ) =180-(0+180) α =0° Incidance angle ɵ ,= (cosβ.cosα) = (cos80.33.cos0) ɵ=80.33°

1.Direct Radiation = =IDN.cosɵ IDN=A.exp(-B/sinβ) =1080 exp(-0.21/sin80.33) IDN=872.78w/m2

Direct Radiation =IDN.cosɵ =872.78cos(80.33) Direct Radiation =146.6 w/m2

2.Diffuse Radiation(Id): View factor , FWS=(1+cos∊)/2 Vertical surface,∊=90° FWS=(1+0/2) =0.5 Id=C.IDN.FWS

=0.135×827.78×0.5 Id=55.87 w/m2

3.Reflected Radiation(Ir): ρ =0.6 =reflectivity View factor, FWG=(1-cos∊)/2 Vertical surface,∊=90°

FWG=(1-cos90)/2 =0.5 Ir=(IDN+ID)ρg.FWG

=(872.78+55.87)×0.6×0.5 Ir=278.595 w/m2

Total incident radiation(IT): =IDN.cosɵ+Id+Ir

=146.6+55.87+278.595 IT=481.065w/m2

Area of North side wall=520sq.ft =520/10.76sq.m IT=(481.065×520)/10.76

IT= 23.2 kw SIMILARLY, For EAST Side wall: Total incident radiation is

IT= 7.86 kw

ExternaL Load: Amount Of heat transmitted in Lab by following way, 1.Through wall . 2.Through Glass. 3.Through Roof. 4.Through Floor. 1. Through wall: CAD Lab wall =200 mm common brick,so take a wall of B type a) Heat transfer through North side wall: CLTDadj=CLTDtable+(Tav-29) (27 p.no,ch34,ASHRAE handbook) =8+(35-29) (ASHRAE hand book 1989,p no26.37) =14 A=36.91m2

Qnorth=Uwall.Awall.CLTDadj

=1.71×36.91×14 =883.625w b) Heat transfer through East side wall. A=22.4m2

CLTDadj=CLTDtable+(Tav-29) =15+(35-29) =21

Qeast=UwallAwallCLTDadj

=1.7×22.4×21 =804.384w c)Heat transfer through south side wall: A=52.65m2

Q=UA ∆T =1.71×52.65× (35-20) =1350.47w This is shaded wall hence CLTD is not used. d)Heat transfer through west side wall: A=21.96m2

Q=UAT =1.71×21.96× (35-20) =563.27w This is shaded wall hence CLTD is not used.

2. THROUGH WINDOW North Side: qconduction=UA(CLTD)

(U value taken from table5,p.no611,ASHRAE2001 fundamental) . (CLTD value taken from 1989ASHRAE handbook,Table29,p.no26.39,at 12 noon). SINGLE GLASS WITH INTERNAL SHADING 6mm. ASHARAE-2001 Fundamentals. =54.28w For one windows heat transfer through glass by conduction=54.28w For 4 window=454.28 =217.12w Heat transfer through fenestration Qsolar=A.SC.SHGF.CLF =2.36×0.5×186×.89 =195.34w (for SC refer table5,p.no611,ASHRAE2001 fundamental) (for SHGF refer chapter 26,table34) (for CLF refer chapter26,table39) There are 4 windows on North side Q=4 ×195.34 =781.36w

East Side: Heat transfer through glass due to conduction qconduction=U.A.CLTD =4.6×2.36×5 =54.28w (For U refer table5,p.no611,ASHRAE2001 fundamental) For CLTD refer chapter26 1989 ASHRAE handbook) There are 2 windows on east side qconduction=2× 54.28w =108.56w Heat transfer due to fenestration: Qsolar =A.SC.SHGF.CLF =2.36×0.5×663×0.27

(Depth of inset is zero hence total area of window is considered)(For SC refer table5,p.no611,ASHRAE2001 fundamental)(For SHGF Refer table34,chapter24,1989ASHRAE handbook)(For CLF refer chapter26,table39) =211.232w

There are 2 windows on east side. Qsolar=2×211.232 =422.464 w3. Through Roof: A=155 m2

Q=U.A.(CLTD) =0.761×155×(38-20) ( For U Refer Table29,p.no26.35,1989ASHRAE handbook,at solar time12pm) Q=1887.28 w4.Through Floor:

Q =U.A .∆T =4.101×155×(35-20) =9534.825 w (from above calculated) (Floor is the slab of room which is uncondition hence ∆T is considered.)

Internal Loads

• These include following factors-• People • Lighting• Appliances and equipment

People

Qsensible = N.(Sensible heat gain).CLF

N= 40 no. of people Sensible heat gain = 70 w (for SHG refer table 3, 1989 ASHRAE HANDBOOK , For people sited in office doing very

light work)

CLF=0.61 (for CLF refer table 40,1989 FUNDAMENTAL HANDBOOK) (Operation of CAD LAB from 10am to 6pm will be 8 hours and after a practical of 2

hour new bath students will enter in the LAB.)

Qsensible =40×70×0.61

= 1708 w

• Qlatent =N.(Latent heat gain).CLF

N= 40 Persons Latent heat gain=45 w (for LHG refer table 3,1989 ASHRAE HANDBOOK.)

CLF=1 (for CLF refer table 27,1989 ASHRAE HANDBOOK.)

• Qlatent =40×45×1

=1800 W

Light and Appliances

• LIGHT: Q= Input × CLF

= (40×26)×0.82(For CLF Refer table 48, 1989 ASHRAE HANDBOOK.)

= 852.8 W • APPLIANCES:• A)COMPUTERS: • Qsensible= Heat gain×CLF

Heat Gain=[ (CRT 15”×Watt)+(LCD 14”×Watt)+(LCD 19”×Watt) ] =[ (5×255)+(19×153)+(14×155)]

= 1275+2907+3825 =8007 W

• Qsensible= 8007 ×0.78 (for CLF refer table48,1989 FUNDAMENTAL ASHRAE HANDBOOK.)

=6245.46 W• B)FAN: Qsensible= Heat Gain ×CLF

=(12×55)×0.78(For CLF refer table 48,1989 FUNDAMENTAL ASHRAE HANDBOOK)

=514.8 W • C)PRINTER:Qsensible = Heat Gain× CLF

=(1 ×92)×0.78 (For CLF refer table48,1989 FUNDAMENTAL ASHRAE HANDBOOK)

=71.76 W

Ventilation Load :

For meeting place or office, ventilation required=3.5 l/sec/person.• Total outdoor air required : VO,V=3.5No.of person.

=3.5×40 =140 m3/s• Mass flow rate of ventilation air: MO,V=Volumetric flow rate /specific volume

For outside condition:DBT=44ºC,RH=20%VO =Specific volume

=0.915 m3/kgWO=Moisture content

=0.015 kg/kg dry air.Ho=specific enthalpy

=74 kJ/kg da Mo,v=0.14/0.915

=0.153 kg/s

• SENSIBLE HEAT TRANSFER DUE TO VENTILATION IS GIVEN BY: QS,V=MO,V.CPM(TO-TI)

=0.153×1.0216×(44-20)

QS,V=3.75 kw• Latent heat transfer due to ventilation is given by; QL,V=MO,V.hfg(wo-wi)

• For inside conditions:DBT=20ºC RH=50%WI=0.0075

WO=0.015

QL,V=0.153×2501×(0.015-0.0075)

QL,V=2.86 kw

Infiltration Load:

• Infiltration rate ,

MInf=Density of air (ACH×Volume of the room)/3600

=1.095(1×155×3)/3600MInf=O.14 kg/sec

• QS=MInfCpm(To-TI)

=0.14×1.0216×(44-20)=3.43 kw

• QL=MInf.hfg(WO-WI)

=0.14 =2.62 kw

• SENSIBLE HEAT TRANSFER DUE TO VENTILATION IS GIVEN BY: QS,V=MO,V.CPM(TO-TI)

=0.153×1.0216×(44-20)

QS,V=3.75 kw• Latent heat transfer due to ventilation is given by; QL,V=MO,V.hfg(wo-wi)

• For inside conditions: DBT=20ºC RH=50% WI=0.0075

WO=0.015

QL,V=0.153×2501×(0.015-0.0075)

QL,V=2.86 kw

SENSIBLE HEAT GAIN IN CAD LAB:

• Heat gain through North side wall = 883.625 w• Heat gain through East side wall = 804.384w• Heat gain through South side wall=1350.47w• Heat gain through West side wall=563.27 w• Heat gain through North side window=998.48w• Heat gain through East side window =531.02 w• Heat gain from roof =1887.28 w• Heat gain through floor =9534.825 w• Heat due to Occupants =1708 w• Heat due to Light =852.8 w• Heat due to Computers =6245.46 w• Heat due to Fan =514.8 w• Heat due to Printer =71.76 w• Heat due to ventilation =3750 w• Heat due to infiltration =3430 w • TOTAL SENSIBLE HEAT GAIN =33126 W

Latent Heat Gain :

• 1.Person =1800 watt

=1.8kw• 2.ventilation =2.86kw• 3.Infiltration =2.62kw• Total latent heat=7.28kw• Taking 5% fan power.• Taking 5% leakage factor.• Taking 3% duct gain.• Safety Heat Load =(33.126+7.28)×(13/100)

=5.25 kw

• Total heat load =33.126+7.28+5.25 = 45.658 kw• Total TR of system= 45.658/3.517

=12.98 =13

Duct calculation

• By velocity method:-• Cosider velocity of 8m/s

• Main duct area(position,K)=Quantity of air supplied/velocity=0.334/8=0.042m2

Diameter of the duct=0.231m

Comparison data

Rectangular Round Comparison

Installation time 70 hours 57 hours 81 %

Suspensions 120 pcs 60 pcs 50 %

Insulation 155 m2, 70 mm 136 m2, 50 mm 75 %

Flow leakage 540 l/s 60 l/s 11 %

Weight 1407 kg 973 kg 69 %

Frictional loss• PK1F=(0.022243.QAIR

1,852L)/D4.973

=(0.022243×0.334×2.26)1.852/(0.231.973)• PK1F=15.1 pa

• PH1F=(0.022243.QAIR1.852L)/D4.973

=(0.022243×0.3341.852×2.26)/(0.231)4.973

• ΔPH1F=9.64 pa

• Δ PU C=46.08 Pa

• Δ PUB=46.08 Pa

• Δ PUD=59.9 Pa

• Δ PUE=59.9 Pa

• Δ PUF=93.69 Pa

• Δ PUG=93.69 Pa

Dynamic Coefficient:• AS/AC=0.209/0.292 =0.715

• Ab1/Ac=0.042/o.292 =0.143

• Qb1/Qc=0.334/2.34 =0.14

• Cb=1.2 (ASHRAE HANDBOOK FUNDAMENTAL 2001,P.N.839)

• As/Ac=Aj/Ai=0.125/0.209=0.59

• Ab1/Ac =As/Ai=0.042/0.209=0.2

• Qb1/Qc=0.334/1.672=0.19

• Cb=1.56 (ASHRAE HANDBOOK FUNDAMENTAL 2001,P.N.839)

• As/Ac=Ak/Aj=0.042/0.125=0.336

• Ab1/Ac =As/Aj=0.042/0.125=0.336

• Qb1/Qc=0.334/1.004=0.332

• Cb=2.44 (ASHRAE HANDBOOK FUNDAMENTAL 2001,P.N.839)

• ROUND FITTING: • D=230 mm• r/D=1.5• Co=0.11 (ASHRAE HANDBOOK FUNDAMENTAL 2001,P.N.816)

• Calculation Of Pressure Drop:• Dynamic loss due to fire damper• Co=Dynamic loss coefficient=0.12• (ASHRAE HANDBOOK FUNDAMENTAL 2001,P.N.819)

• Dynamic loss (ΔPUFD) =Co .(ρv2/2)

=0.12(1.2×82/2) =4.608

• Dynamic loss due to Screen A0/A1=1 ( Same duct dia.)

Consider, n=0.9Where, n= free area ratio of screen A0= Cross sectional Area of duct

A1= Cross sectional Area of duct where screen is located.

(ASHRAE HANDBOOK FUNDAMENTAL 2001,P.N.819)

Dynamic loss coefficient Co=0.14

Dynamic loss = Co .(ρv2/2)

=0.14(1.22/2) = 5.376 Pa

• 7.Section A-I-J-K-H:-• Total pressure loss at A-I-J-K-H

=ΔPFD+ΔPSL+ΔPA1F+ΔP1F+ΔPJF+ΔPK1F+ Δ PH1F

+ΔPU C+ΔPUB+ΔPUD+ΔPUE+ΔPUF+ Δ PUG+ Δ PU H+ Δ Pelbow+ Δ Pdiffuser

• Δ PFD=4.608 Pa

• Δ PSL=5.37 Pa

• Δ PA1F=4.45 Pa

• Δ P1F=5.82 Pa

• Δ PJF=7.57 Pa

• PK1F=(0.022243.QAIR1,852L)/D4.973

=(0.022243×0.334×2.26)1.852/(0.231.973)• PK1F=15.1 pa

• PH1F=(0.022243.QAIR1.852L)/D4.973

=(0.022243×0.3341.852×2.26)/(0.231)4.973

• ΔPH1F=9.64 pa

• Δ PU C=46.08 Pa

• Δ PUB=46.08 Pa

• Δ PUD=59.9 Pa

• Δ PUE=59.9 Pa

• Δ PUF=93.69 Pa

• Δ PUG=93.69 Pa

• PU H=C (.ρv2/2)

=0.11[(1.2×82]/2 =4.22 pa

• ELEBOW loss, ΔPelbow =4.22 Pa

• DIFFUSER loss,ΔPdiffuser =15 Pa• Total pressure loss at A-I-J-K-H :

=4.608+5.37+4.45+5.82+7.57+15.1+9.64+46.08+59.9+59.9+93.69+93.69+4.22+4.22+15=475.33 Pa

• Thus run with maximum pressure drop in A-I-J-K-H is index run ,hence FTP required is;

• FTP=ΔPA-I-J-K-H

=475.33 pa =475.33 N/m2

• AMOUNT OF DAMPERING REQUIRED AT B & C. = 475.33-89.36 =385.97 Pa

• AMOUNT OF DAMPERING REQUIRED AT E & D. =475.33-201.168 =274.164 Pa

• AMOUNT OF DAMPERING REQUIRED AT G & F. =475.33-362.31 =113.02 Pa

• WFAN=FTP×aair/ηfan

=475.33×2.34/0.9 =1235.85 watt

• WFAN =1.235 KW

Components Selection

Conclusion

References• American Society of Heating Refrigerating Air Conditioning Engineers (ASHRAE)

HVAC Handbook, 2001 Fundamentals, Page No. 796, 816, 818, 819 chapter 26,table34.

• Handbook of Air Conditioning System Design by Carrier Air Conditioning

Compony, McGRAW-HILL Book Compony, Year of publication 1965, Page No.1.1-1.41 and 2.1 -2.65.

• American Society of Heating Refrigerating Air Conditioning Engineers (ASHRAE)

HVAC Handbook , Published on 1989, Page no. 22.5 – 22-13, 26.7- 26.47, 27.9- 27.29.

• Refigeration and air conditioning by C.P.Arora, TATA McGraw Hill Publications,

Second Edition 2000, Page no. 675- 690, 756-805.

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