Cooling load analysis by CLTD and RLF Sher ali nawaz 1 Muzammil shahbaz 2 Waqas Javid, M.Umer Sohail Department of mechanical engineering Wah engineering college University of wah sheralinawaz15@gmail.com, muzammilkhokha881@gmail.com ABSTRACT Control of indoor environment is very important for human comfort. Air conditioning is very necessary for controlling inside room temperature, humidity and for ventilation. Cooling loads are affected by the outdoor weather conditions. Different techniques were used for finding the cooling load. The objective of this research is to calculate the cooling load calculation of WEC CAD lab by using Cooling load temperature difference and residential load factor method. Mathematical modelling is carried out of CAD lab for internal and external loads by CLTD and RLF, which is the real-time application of HVAC system. The result shows 6.98 ton of refrigeration is required for CAD lab at comfort zone. Keywords: Cooling load temperature difference method (CLTD), Heating ventilating air conditioning (HVAC).
1 INTRODUCTION: Buildings are built to provide a safe and comfortable internal environment despite variations in the external conditions. Human body is an amazingly adaptable organism. It is very difficult to achieve the desire cooling effect by HVAC system without cooling load calculations. As human comfort is major priority now a day so HVAC systems get a lot of importance. Estimated cooling load is required for installing proper sized HVAC system. [1] Long time ago cooling load estimation has started in India but now a day it is necessary to improve the methods for estimation of cooling load. So, in this paper it is shown that how to calculate the cooling load by CLTD and RLF method. The methods used here CLTD and RLF to compare results and to show which is the appropriate method. 2 LITERATURE REVIEW: A lot of researches on cooling load estimation is done by people till now. In 1758, Cambridge University professor John Hadley discover and Franklin discover evaporation of volatile liquids and alcohol, which can be evaporate rapidly than water that can enough to cool and freeze water. Michael Faraday compresses liquified ammonia in 1820. Freon-12 is used as a refrigerant in 1970.First attempt of CLTD method was done by Elite Software in 1979.RLF and Heat balance method is introduced by ASHRAE in 2005. [2] In 2016, a research paper publish in IRJET on a topic of comparison of cooling load estimation by CLTD method and computer software (Elite CHVAC) written by Ujwal Kumar Sen, Anil punia and Rajesh rana at RPS college of engineering and technology. [3] Now, we work on a topic of comparison of results of RLF and CLTD method for cooling load estimation. 3 COOLING LOAD CALCULATION METHODS: There are many methods for calculating cooling load but most accurate methods are given below. 3.1 Total Equivalent Temperature Difference: This method is used for cooling load calculation before CLTD and RLF methods. This method is introduced by ASHRAE(American Society Of Heating Refrigerating and Air-conditioning Engineers). 3.2 Transfer function method This method includes advanced spreadsheet so that this method is very tough. 3.3 Radiant Time Series: This method is developed by ASHRAE and follow the given steps. 1: We get Radiant and convective parts by the splition of heat gains. 2: We have to calculate twenty-four hours different readings of heat gains in various components (conduction time series are consider for conduction time delay).
3: Apply appropriate RTS to different (radiant) parts of heat gain components to consider for time delay that converts to cooling load. 4: Sum all heat gain components and this heat gain part is used to find out cooling load for each hour of cooling load component. [4] 3.4 Cooling Load Temperature Difference/Solar Cooling Load / Cooling Load Factor(CLTD/SCL/CLF): Basically, this method is originated from TFM and it uses the tabulated data in calculation that makes the calculation easy but this method needs a lot of time and of course there will be some error during calculation. 3.5 Residential Load Factor(RLF): It is an old and short form of method that gives an approximate valve for cooling load estimation but it is not accurate than CLTD method. 3.6 Heat Balance: ASHRAE developed this method that are in program language that gives more accuracy. [5] 4 FACTORS EFFECTING THE THERMAL COMFORT. Operative temperature: 20 0 C to 26 0 C Humidity: A dew point temperature of 2 to 17 0 C Average air velocity: 0.25m/s 4.1 Building loads are divided into the following four categories: 4.1.1 Transmission: Across a building elements heat loss or heat gain due to the temperature difference. 4.1.2 Solar: Through transparent building components or absorption by the opaque components heat gain due to the transmission of solar energy. 4.1.3 Infiltration: In the air-conditioned room heat loss or heat gain due to the infiltration of outside air. 4.1.4 Internal: Within the space (light, people, equipment s) heat gain due to the release of energy.
5 PROCEDURE FOR ESTIMATING COOLING LOADS: For outdoor summer we to design values wet bulb temperature, dry bulb temperature and average temperature. The term which is appropriate for the activities to be carried out in the space is indoor design temperature. We have to compute the heat transfer coefficient for the building components and calculating the U values here on the basis of building plans and specifications. We have to estimate the rate of infiltration and ventilation of outside air from the building plans and specifications, system operating schedule, and designed values of wind velocity and temperature difference. Latent load also included for cooling load. Orientation of building and Determination of the location is required. Determine the cooling load temperature difference, solar heat gain factors and cooling load factors on the basis of building components and design conditions. On the basis of heat-transfer coefficients, areas and temperature differences determined the rate of heat gain to space. The amount of heat gain from internal sources (light, people and equipment s) applying the cooling load factor when appropriate. For finding the maximum capacity required for cooling we have to sum all the load components. We have to adopt the above procedure for calculating the cooling load of lab. Objective of this research is to obtain desired cooling in CAD LAB.Various methods are available to do this but we are looking for the precise and accurate method to obtain desired cooling in university CAD LAB. For this, we focused on RLF and CLTD methods to obtain desired accuracy. [6] 6 DRAWING OF CAD LAB: Height of wall= 13ft Roomlength=37ft Room width = 32 ft Window = 6ft 6ft=36 ft 2 Door = 7ft 5ft=36 ft 2
7 DESIGN CONDITIONS: Cad lab is located in WAH ENGINEERING COLLEGE TAXILA having 40 0 north. Time: 3pm May 2017 Month: Internal temperature of Lab = t r = 25 0 C (77 0 F) External temperature of Lab=t e = 43 0 C (109.4 0 F) 7.1 Wall: 4in. face brick+ 2in. insulation +4in. common brick. 7.2 Door: 11/2 in. thick, mineral fibre core. 7.3 Window: Aluminium frame with single glass. 7.4 Roof: 6 in. Light weight concrete. 8 COOLING LOAD TEMPERATURE DIFFERENCE METHOD (CLTD): 8.1 Important formulas: 8.1.1 For exterior surfaces: Exterior surfaces are roof, wall and glass. Average room temperature can be obtained from the formula, ta= to- (DR/2) DR= Daily temperature range. to= outside dry bulb temperature. 8.1.2 For interior surfaces: Interior surfaces are partition, floor and ceiling. Q=U A TD TD= Temperature difference between the conditioned and unconditioned space, 0 F 8.1.3 Heat gain from lights: Q=3.4 W BF CLF W= Light capacity, watts BF= ballast factor CLF= cooling load factor for light 3.4 factor converts watts to BTU/hr Value of BF for fluorescent light=1.2 8.1.4 Heat gain through glass: (solar) Q=U A CLTD C Q= Cooling load for roof, wall and glass, BTU/hr U= Overall heat transfer coefficient, BTU/hr-ft 2-0 F A= Area, ft 2 CLTD C= Corrected cooling load temperature difference, 0 F CLTD C= CLTD + LM+ (78- t r) + (t a-85) CLTDc= Corrected value of CLTD, 0 F LM= Correction of latitude and month. t a= Average room temperature at design day, 0 F Q= SHGF A SC CLF SHGF= Maximum solar heat gain factor for glass, BTU/hr-ft 2 SC= Shading coefficient CLF= cooling load factor for glass 8.1.5 Heat gain from people: Here we considered 50 people sitting in CAD lab. Q S= q s n CLF Q l= q l n
Q S, Q l = Sensible and latent heat gain q s, q l= Sensible and latent heat gain per person n= Number of people Value of CLF=1 8.2 Calculations: 8.2.1 Heat gain from lights: No of lights= 48 Power of light= 20 watts Total power= 48 20= 960 watts Value of CLF for latent load= 1.0 Light load= Q light=3.4 W BF CLF =3.4 960 1.2 1.0 =3916.8 BTU/hr. 8.2.2 Heat gain from equipment s: Fan: No of fans= 6 Power of fan= 80 watts Total power= 80 6=480 watts Fan load= 3.4 W CLF =3.4 480 1 = 1632 BTU/hr. 8.2.3 Heat gain from computer Desktop: No of computer= 50 Power of computer= 155 watts Total power= 155 50=7750 watts Computer load=7750 CLF 3.4=7750 1 3.4 = 26350 BTU/hr. 8.2.4 Heat gain from people: No of people= 50 q s=250 q l=200 Value of CLF= 1.0 Q people= (q s+q l) n CLF = (250+200) 50 1.0 = 22500 BTU/hr. 8.2.5 Conduction through roof: Area of roof= 37 32= 1184 ft 2 6 in. Light weight concrete. U= 0.158 BTU/hr-ft 2-0 F LM= 1 CLTD=51 tr= 77 0 F ta= to- (DR/2) =109.4-(71.6/2) = 73.6 0 F CLTD C= CLTD + LM+ (78- t r) + (t a-85) =51+ 1+ (78-77) + (73.6-85) =41.6 0 F Conduction through roof: Q roof=u A CLTD c =0.158 1184 41.6 =7782.19BTU/hr. 8.2.6 Conduction through west wall: Area of wall= 37 13= 481 ft 2 4in. face brick+ 2in. insulation +4in. common brick. LM= 0 CLTD=14 tr= 77 0 F ta= to- (DR/2) =109.4-(71.6/2) = 73.6 0 F CLTD C= CLTD + LM+ (78- t r) + (t a-85)
=14+ 0+ (78-77)+ (73.6-85) =3.6 0 F Conduction through west wall: Q W,Wall=U A CLTD c =0.111 481 3.6 =192.2BTU/hr. 8.2.7 Conduction through south wall: Area of wall= 32 13-5 7= 381 ft 2 4in. face brick+ 2in. insulation +4in. common brick. LM= 1 CLTD=14 tr= 77 0 F ta= to- (DR/2) =109.4-(71.6/2) = 73.6 0 F CLTD C= CLTD + LM+ (78- t r) + (t a-85) =14+ 1+ (78-77)+ (73.6-85) =4.6 0 F Conduction through south wall: For door: Q South, Wall=U A CLTD c =0.111 381 4.6 =194.5 BTU/hr. Below calculation based on two doors in CAD lab. TD= 109.4-77= 32.4 0 F A= 7ft 5ft= 35 ft 2 U= 0.58 BTU/hr-ft 2-0 F Q door=u A TD =0.58 35 32.4 =657.72 BTU/hr. Q S,WALL= Q wall + Q door = 194.5+ 657.72 =852.22 BTU/hr. 8.2.8 Conduction through east wall: Area of wall= 37 13-36= 445 ft 2 Aluminium frame with single glass. LM= 0 CLTD=24 tr= 77 0 F ta= to- (DR/2) =109.4-(71.6/2) = 73.6 0 F CLTD C= CLTD + LM+ (78- t r) + (t a-85) =24+ 0+ (78-77)+ (73.6-85) =13.6 0 F Conduction through east wall: For window: Q wall=u A CLTD c =0.111 445 13.6 =671.772 BTU/hr. CAD lab is ventilated by four windows. A= 6ft 6ft= 36ft 2 SHGF= 220 BTU/hr-ft 2 SC= 0.94 (without shading) CLF= 0.26 (without interior shading) Q window= SHGF A SC CLF =220 36 0.94 0.26 =1935.64 BTU/hr. Q E,WALL= Q wall + Q window = 671.772+ 1935.6 =2607.372 BTU/hr.
8.2.9 Conduction through north wall: There is a big room but it is divided into two. Two rooms are separated by glass. One for lab and other is empty in the north side. So there is no heat through the north wall. 8.3 Total cooling load of Lab: Q= Q Light+ Q Equ.+Q Computer+Q People +Q Roof +Q W,Wall +Q S,Wall +Q E,Wall +Q N,North Q=3916.8+1632+26350+22500+7782.19+19 2.2+852.22+2607.372+0 = 65832.782BTU/hr. Adding 15% in terms of FOS, Infiltration, Duct heat gain factor. Q total= Q 1.15= 65832.782 1.15= 75707.6993 1 ton of refrigeration= 12000 BTU/hr. Q total=6.30 ton of refrigeration 9 RESIDENTIAL LOAD FACTORS (RLF): It is simplest procedure for calculating the cooling load. 9.1 Important formulas: Heat gain through walls, floors, ceiling, and doors is caused by 1) Air temperature difference across the surfaces 2 2) Solar gain incident on the surfaces. CF opq= U (OF t TD+ OF b+of r DR) q opq = opaque surface cooling load, BTU/hr. U= Overall heat transfer coefficient, BTU/hr-ft 2-0 F A= Area, ft 2 TD= Cooling design temperature difference, 0 F OFt, OFb, OFr = Opaque suface cooling factors. DR= Cooling daily range, 0 F 9.1.1 Internal gain: The contributions of occupants, lighting, and appliance gains to peak sensible and latent loads can be estimated as q ig,s = 464+ 0.7A cf +75N oc q ig,l = 68+ 0.07A cf +41N oc Noc= number of occupants Acf= conditioned floor area of building, ft 2 qig,l, qig,s= latent and sensible cooling load from internal gains, BTU/hr. 9.1.2 Transparent fenestration surfaces: Cooling load for windows. q fen=a CF fen CF fen= U (TD-0.46DR)+PXI SHGC IAC FF S q fen = fenestration cooling load, BTU/hr. CF fen =surface cooling factor, BTU/hr-ft 2 PXI = peak exterior irradiance, including shading modifications, BTU/hr-ft 2 PXI= T xe t (unshaded fenestration) T x= transmission of exterior attachment (insect screen, shade screen) E t= peak total irradiance, BTU/hr-ft 2 SHGC= solar heat gain coefficient IAC= interior shading attenuation coefficient FF S= fenestration solar load factor 9.2 Calculations: 9.2.1 Heat gain from lights: No of lights= 48 Power of light= 20 watts q ig,l = 68+ 0.07A cf +41N oc =68+0.07 1184+41 48
Q light =q ig,l = 2118.8 BTU/hr. 9.2.2 Heat gain from equipment s: Fan: No of fans= 6 Power of fan= 80 watts Total power= 80 6=480 watts q ig,l = 68+ 0.07A cf +41N oc =68+0.07 1184+41 6 Q fan =q ig,l = 396.88 BTU/hr. Heat gain from computer Desktop: No of computer= 50 Power of computer= 155 watts Total power= 155 50=7750 watts Computer load= q ig,l =68+ 0.07A cf +41N oc =68+0.07 1184+41 50 Q computer =q ig,l = 2200.8 BTU/hr. 9.2.3 Heat gain from people: No of people= 50 q ig,s = 464+ 0.7A cf +75N oc =464+0.7 1184+75 50= 5042.8BTU/hr. q ig,l = 68+ 0.07A cf +41N oc =68+0.07 1184+41 50 = 2200.88BTU/hr. Q people =q ig,s +q ig,l= 5042.8+2200.88 =7243.68 BTU/hr. 9.2.4 Conduction through roof: Area of roof= 37 32= 1184 ft 2 U= 0.158 BTU/hr-ft 2-0 F 6 in. Light weight concrete. CF opq= U (OF t TD+ OF b+of r DR) =0.158(0.62 32.4 + 25.7 0.8-8.1-0.19 71.6)=2.993152 Q roof=q opq=1184 2.99=3540.16 BTU/hr. 9.2.5 West wall: Area of wall= 37 13= 481 ft 2 4in. face brick+ 2in. insulation +4in. common brick. CF opq= U (OF t TD+ OF b+of r DR) =0.111(1 32.4 + 14.8-0.36 71.6) =2.378 Q W,Wall= q opq=481 2.378=1143.818 BTU/hr. 9.2.6 North wall: There is a big room but it is divided into two. Two rooms are separated by glass. One for lab and other is empty in the north side. So there is no heat through the north wall. 9.2.7 South wall: Area of wall= 32 13-5 7= 381 ft 2 4in. face brick+ 2in. insulation +4in. common brick. CF opq= U (OF t TD+ OF b+of r DR) =0.111(1 32.4 + 14.8-0.36 71.6) =2.378 Q Wall= q opq=381 2.378=906.018 BTU/hr. For door: A= 7ft 5ft= 35 ft 2 U= 0.58 BTU/hr-ft 2-0 F CF opq= U (OF t TD+ OF b+of r DR) =0.58 (1 32.4 + 14.8-0.36 71.6) =12.4
Q door= q opq=35 12.4= 434 BTU/hr. Q S,WALL= Q wall + Q door = 906.018 + 434 =1340.018 BTU/hr. 9.2.8 East wall: Area of wall= 37 13-36= 445 ft 2 Aluminium frame with single glass. CF opq= U (OF t TD+ OF b+of r DR=0.111(1 32.4 + 14.8-0.36 71.6)=2.378 Q Wall= q opq=445 2.378= 1058.21 BTU/hr. For window: A= 36ft 2 U= 1.27 BTU/hr-ft 2-0 F Q total= Q 1.15= 20287.246 1.15= 23330.33BTU/hr. 1 ton of refrigeration= 12000 BTU/hr. Q total= 2 ton of refrigeration 10 RESULTS AND COMPARISON: 10.1 Comparison between CLTD and RLF method: Loads CLTD (BTU RLF (BTU/hr.) Light 3916.8 2118.8 People 22500 7243.68 Computer 26350 2200.8 Equipment 1632 396.88 Roof 7782.19 3540.16 West wall 192.2 1143.818 North wall 0 0 East wall 2607.372 2303.09 South wall 852.22 1340.018 Total load Adding 15% in terms of FOS, Infiltration, Duct heat gain factor. q fen=a CF fen CF fen= U (TD- 0.46DR) + PXI SHGC IAC FF S PXI= T x E t (unshaded fenestration) =1 237=237 Total load (BTU/hr.) Total load (tons) 75707.6993 23330.33 6.30 TR 2 TR CF fen=1.27(32.4-0.46 71.6) +237 0.75 0.64 0.31=34.58 q fen=a CF fen=36 34.58=1244.88 BTU/hr. Q E,WALL= Q wall + Q Window = 1058.21 + 1244.88 =2303.09 BTU/hr. 9.3 Total cooling load of Lab: Q= Q Light+ Q Equ. +Q Computer +Q People +Q Roof +Q W,Wall +Q E,Wall +Q S,Wall +Q N,North Q=2118.8+396.88+2200.8+7243.68+3540.1 6+1143.818+1340.018+2303.09+0 = 20287.246BTU/hr. Adding 15% in terms of FOS, Infiltration, Duct heat gain factor. 11 CONCLUSION: ASHARE standards provides 1.5 TR for 1000 feet square of floor area under normal conditions. After consideration of external and internal loads in above calculations, CLTD provides 6.30TR for 1184 ft 2 of floor area. In CLTD method we consider the watts/ power in formulas but RLF method doesn t use watts in formulas. In CLTD method 24-hour procedure is required to accurately determine the cooling load profile of the residence but RLF doesn t. We consider the various loads in CLTD and RLF methods. Above obtained results are showing the accuracy level of the CLTD method because 6.30TR is acceptable in our researched area for human comfort.
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