Cooling Load Calculation Procedure Engineering Essay

chill system warhead Calculation result Engineering EssayThe total amount of warming energy that essential be removed from a dodging by a change system mechanism in a unit clipping, enough to the roam at which fire up is generated by people, machinery, and processes, plus the authorize flow of mania into the system not associated with the cool machinery.1 The sensible and latent fondness convert in the midst of the musculus quadriceps femoris port and the surroundings heap be classified as follows1. place shake up bring in qe, in Btu/h (W), represents the rate at which cacoethes enters a well-read topographic pointfrom an foreign source or is released to the quadriceps from an internal source during a given time interval.2. Space cool down thin, often simply called the modify charge Qrc, Btu /h (W), is the rate at whichheat must be removed from a conditi adeptd space so as to maintain a constant temperature and refreshing relative humidity. The sen sible chill pack is embody to the sum of the convective heat conveying from the surfaces of the building envelope, furnishings, occupants, thingumajigs, and equipment.4. Space heat extraction rate Qex, Btu /h (W), is the rate at which heat is actually removed fromthe conditioned space by the air system. The sensible heat extraction rate is equal to the sensible alter gist only when the space air temperature stay constant.5. Coil consign Qc, Btu /h (W), is the rate of heat transfer at the coil. The cooling system coil commit Qcc,Btu/h (W), is the rate at which heat is removed by the chilled water flowing by means of and through the coil or isabsorbed by the refrigerant inside the coil.Cooling excite unremarkably can be classified into two categoriesexternal and internal.External Cooling Loads1These laden argon formed be set about of heat wins in the conditioned spacefrom external sources through the building envelope or building rag and the expositition ramparts.Sou rces of external stacks include the following cooling excites1. Heat pucker entering from the exterior surrounds and roofs2. Solar heat evolve transmitted through the fenestrations3. Conductive heat strive coming through the fenestrations4. Heat gain entering from the partition walls and interior doors5. Infiltration of outdoorsy(prenominal)(prenominal) air into the conditioned spaceInternal Cooling Loads1These ladles are formed by the release of sensible and latent heat from the heat sources inside the conditioned space. These sources contribute internal cooling turn ons1. mass2. Electric lights3. Equipment and appliancesFor 1-1 1 take in the section of referencesCHAPTER 2 change LOAD numeration PROCEDURE2The estimation of cooling load for a space involves calculate a surface by surface conductive, convective, and radiative heat equilibrize for severally populate surface and a convective heat balance for the room air. base on the analogous underlying principl es, the following systems have been developed for conniving the cooling load.Cooling Load by ecstasy Function method (TFM). fare Equivalent Temperature Difference (TETD) method.Cooling Load Temperature Difference (CLTD) method.Transfer Function Method (TFM)1The transfer function method or burden actor method is a simplification of the laborious heat balance method. The wide application of the TFM is due to the user-friendliness of the inputs and outputs of the TFM software and the saving of computing time. In the transfer function method, interior surface temperatures and the space cooling load were first metrical by the exact heat balance method for many representative constructions. The transfer function coefficients (weighting factors) were then calculated which qualify the heat gains to cooling loads. Sometimes, transfer function coefficients were too developed through test and experiments.Calculation ProcedureThe slowness of space cooling load using the transfer funct ion method consists of twosteps. First, heat gains or heat loss from exterior walls, roofs, and stages is calculated using reply factors or conduction transfer function coefficients and the solar and internal heat gains are calculated directly for the inscriptiond hour. Second, room transfer function coefficients or room weighting factors are used to convert the heat gains to cooling loads, or the heat losses to heating loads. As described in Sec. 6.2, the sensible infiltration heat gain is the instantaneous sensible cooling load. All latent heat gains are instantaneous latent cooling loads.The TFM is limited because the cooling loads thus calculated depend on the value of transferfunction coefficients as well as the characteristics of the space and how they are varied from those used to generate the transfer function coefficients. In addition, TFM assumed that the total cooling load can be calculated by simply adding the individual components-the superposition principle. Howeve r, this assumption can cause round errors. complete Equivalent Temperature Difference (TETD) Method1In the total identical temperature difference (TETD)/time-averaging (TA) method, heat gains of a number of representative exterior wall and roof assemblies are calculated. The internal heat gains and conductive heat gain are calculated in the same manner as in the TFM.The radiant fraction of each of the sensible heat gains is then allocated to a period including the current and successive hours, a total of 1 to 3 h for light construction and 6 to 8 h for big(p) construction. The TETD/ TA method is also a member of the TFM family and is developed generally for manual calculation. TETD/TA is simpler in the conversion of heat gains to cooling loads. However, the time-averaging calculation procedure is subjective-it is more an art than a rigorous scientific method. withal the TETD/TA method inherits the limitations that a TFM possesses if the TFM is used to calculate the TETD.Cooling Load Temperature Difference (CLTD) Method 2CLTD is the method we used to calculate the cooling load of the project we were assigned. The CLTD method floors for the thermic response in the heat transfer through the wall or roof, as well as the response due to radiation of part of the energy from the interior surface of the wall to objects and surfaces within the space. The CLTD method makes use of (a) the temperature difference in the case of walls and the roofs and (b) the cooling load factors (CLF) in the case of solar heat gain through windows and internal heat sources, i.e ,Q = U x A x CLTDCWhereQ is the net room conduction heat gain through roof, wall or grouch (W)A is the raise of the roof , wall or water ice (m2)U is the overall heat transfer co efficient (kW/m2.K)CLTDC is the cooling load temperature difference (oC)For 1 2 see the section of referencesCHAPTER 3CALCULATING COOLING LOADS USING CLTDOutdoor be after Conditions2ASHRAE dining table A 2, F1980, provides th e outdoor pattern conditions for various locations in many countries including India, Malaysia and Singapore. The summer design mainstay lists hourly temperature which is exceeded by 1%, 2.5 % and 5% of all the hours in the year.selection of Indoor Conditions2In private homes, the indoor conditions may be chosen by the owner. But in public premisis, various codes and regulations and ordinances dictate the limits of the special(prenominal) indoor design conditions. For some critical occupancy, such as, hospitals, nursing homes, human bodyurer rooms, clean rooms, etc. limited indoor design conditions allow unremarkably be established by the regulating authorities or the owners. transmission system gains2Heat transfer through the different components of the building envelop occurs in general the process of conduction and convection and is generally referred to as transmission load. Transmission heat transfer is given by the following equationQ = = (U) (A) (TD)Where,Q is the he at transfer rate (W)Rt is the sum of the individual caloric opponents (m2.oC/W)A is the surface sports stadium perpendicular to heat flux (m2)TD is the design temperature difference amidst inside and outdoorsU = 1/Rt is the overall heat transfer co efficient (W/m2.oC)With, Rt = R1 + R2 + + Rm for resistance in series.The values of thermal resistances are provided for specific weightiness for typical building materials usually designated by U. For materials that vary in thickness according to the application, specific conductivity k is listed in footing of unit thickness. The relation between the two isR =Wherek is the coefficient of thermal conductivity (W/m.K)L is the length of the conduction path (m).CLTD/CLF calculation2To account for the temperature and the solar variations, the concept of cooling load temperature difference (CLTD) is introduced. The CLTD is a steady state representation of the complex heat transfer involving actual temperature difference between indoors and outdoors, mass and solar radiation by the building materials, and the time of mean solar daytime.The following relation makes corrections in the CLTDs for walls and roofs for deviations in design and solar conditions are as followsCLTDc = (CLTD + LM)k + (25.5 Tr) + (To 29.4) fWhereCLTDc is the corrected value of CLTD.LM is the glossiness adjustment for light coloured roof.Tr is the design room temeperatureTo is average outdoor temperature, computed as the design temperature less half the daily range.f is noggin fan factorSolar heat gain2When solar rays impinge on a glass surface, some of the radiation is reflected back outside forrader penetrating the glass. Of that radiation which is not reflected, some is transmitted through the glass and some is absorbed by the glass. The remain radiation is refracted slightly and goes on to heat the contents of the room. If there is external shading, such as with blinds or drapes or shades, a portion of t radiation entering the room is confined to the subject immediately adjacent to the window and has a modest effect on the conditioning of the room.All of these effects are accounted for to some degree by the following relation for calculation of cooling loads due to solar radiationQSHG = A(SC)(SHGF)(CLF)WhereQSHG is the solar radiation cooling load (W)A is the open glass subject (m2)SC is the shading co efficient for various types of glasses and shadingsSHGF is the maximum solar heat gain factor for specific orientation of surface, latitude and month (W/m2)CLF is the cooling load factor, dimensionless.Internal Loads2Lighting is often is the major space cooling load component. The rate of heat gain at any instant, however, is not the same as the heat equivalent of power supplied instantaneously to these lights. Only part of the energy from lights is transferred to the room air by convection, and thus becomes the cooling load. The remaining portion is the radiant heat that affects the conditioned space only by and by having been absorbed by walls, floor, furniture, upholstery, etc. and released after a time lag. The cooling load imposed by these sources is given byQ appliances= P(CLF)WhereQ appliances is the cooling load due to equipment of appliances (W)P is the input direct power rating of the appliance or equipment (W)CLF is the cooling load factor (dimensionless) depending on operating hours, room construction, and air circulation. line2The people who occupy the building give off thermal energy continuously, the rate of which depend on the level and type of military action in which they are engaged. For the sensible portion of the heat released, a cooling load factor similar the one applied to lights and appliances has been developed to account for the lag in time between occupancy and the observed cooling load. The sensible cooling load due to people is therefore,QS = (N)(GS)(CLFS)WhereQS is the sensible cooling load due to occupants (W)N is the number of occupantsGS is the sens ible heat gain depending on use and time for entry (W)CLFS is the cooling load factor (dimensionless) for people.The latent heat gain from occupants is found byQl = (N)(Gl)WhereQl is the latent heat gainN is the number of occupantsGl is the latent heat gains from occupants depending on activity and time from entryVentilation/Infiltration2Heat gain from ventilation and infiltration needs to be considered in the cooling load calculations.General Design Guidelines2The general procedure required to calculate the space cooling load is as followsBuilding configuration an characteristics Determine the building location, orientation and external shading, building materials, external surface colour and shape. These details are usually obtained from building plans and specifications.Outdoor design conditions Obtain the outdoor weather selective information for the building location and select the outdoor design conditions.Indoor design conditions Specify temperature, humidity, air velocity, etc.Operating schedules obtain a schedule of lighting, occupancy, internal equipment, appliances and processes generating heat load.Date and time Select the time of the day and month to estimate the cooling load. Several different times of the day and several different months need to be analyzed to determine the superlative degree load time. The particular day and month are often set by peak solar conditions.For 2 see the section of referencesCHAPTER 4 info FOR THE MAIN MOSQUEMosque 1st Floor elaborate 4.1 First floorFigure 4.2 Window type 1 (WT1)Figure 4.3 Window type 2 (WT2)Figure 4.4 entrances 1 (D1)4.2 Mosque Ground FloorFigure 4.5 Ground floorFigure 4.6 Window shell 3 (WT3)Figure 4.7 Window Type 4 (WT4)Figure 4.8 Door 2 (D2)General Information analog 32, Longitude 72 7 important Mosque, College of Electrical and Mechanical Engineering, Rawalpindi, Pakistan circumvents33cm brick, 1.5 cm cementum on both sidesRoof15 cm concretetwistCream color paint on both sidesGlass0.5cm d our shadedLighting 176 tubes each 18W, 8hrs per dayLighting 24 tubes each 40W, 8hrs per dayOccupancyccc people reasonably activeCeiling fan18 ceiling fans each 75 W, 8hrs per dayBracket fan9 bracket fans, each 40W, operating 8hrs per dayVentilation7.5 x 300 = 2200 liters/secNote hang up ceiling was broken therefore we considered it as the refrigerating space.U jimmy CalculationRoofComponentsL/km/(W/m.K)RW/m2.KReference put off remand titleOutside air0.044 skirt A6 rise up conductance (W/m2.oC) and resistance (m2.oC/W) for airConcrete 10cm0.15/0.510.2943Inner air0.160Table A6Surface conductance (W/m2.oC) and resistance (m2.oC/W) for air natural0.498U = 1/R = 1/0.498 = 2.01WallComponentL/km/(W/m.K)RW/m2.KReference TableTable titleOuter air0.044Table A6Surface conductance (W/m2.oC) and resistance (m2.oC/W) for airBrick0.33/0.321.0315Table A7Typical thermal properties of prevalent building and insulating materialsCement20.015/0.720.0417Table A7Typical thermal properties of common b uilding and insulating materialsInside air film0.120Table A6Surface conductance (W/m2.oC) and resistance (m2.oC/W) for airTotal1.237U = 1/R = 1/1.237 = 0.808For 3 see referencesGlassComponentL/km/(W/m.K)RW/m2.KReference TableTable titleGlass material0.005/0.050.1Table A7Typical thermal properties of common building and insulating materialsOuter air0.044Table A6Surface conductance (W/m2.oC) and resistance (m2.oC/W) for airInner air0.12Table A6Surface conductance (W/m2.oC) and resistance (m2.oC/W) for airTotal0.264U =1/R = 1/0.264 = 3.79Description of appliances breaker pointsGround floor (Qty)First Floor (Qty)Ceiling fans126Tube lights76 (small)4 (large)Wall fan9Area CalculationsEastern Wall AreasWall champaign108.11514 m2Door area26.3 m2Windows area12.76 m2 true(a) wall area69.055 m2Actual glass area30.4 m2Note The area for aluminum in the windows is not accounted for.4.6.2 Western Wall AreasWall area88.4816 m2Windows area6.583 m2Actual wall area81.899 m2Actual glass6.003 m2Note Th e area for aluminum in the windows is not accounted for.4.6.3 Union Wall AreasWall area52.45 m2Windows area12.61 m2Actual wall area39.84 m2Actual glass area11.62 m2Note The area for aluminum in the windows is not accounted for.4.6.4 siemensern Wall AreasWall area52.45 m2Windows area6.58 m2Door area1.86 m2Actual wall area44.01 m2Actual glass area7.58 m2Note The area for aluminum in the windows is not accounted for.4.6.5 Roof AreasTotal roof aream24.7 CLTD Correction CalculationTo = 47 16/2 = 39To is the average outside temperature on design day equal to our design temperature minus half of daily temperature range.ExposureCLTDLM4K525.5TrTo29.4F6CLTDc labor union70.50.525.525.53929.4113.35East1500.525.525.53929.4117.1South11-2.20.525.525.53929.4114West1100.525.525.53929.4115.1Roof361.10.525.525.53929.4127.05For 4, 5 6 see references.CHAPTER 5COOLING LOAD CALCULATIONS FOR THE MAIN MOSQUEBuildingMain mosqueLocationEME College, RWPMonthJune Day 22 Time 0200hrsPsychrometric analysisIte mTdbTwbRHEnthalpy (h)Sp. Humidity (w)Outside474170%183.870.0528Inside25.517.8250%57.350.012Difference21.523.1820%126.520.041Daily range = 16Transmission LoadItemDescriptionArea(m2)U factorCLTDc (oC)Qtransmission(W)Walls northeastern39.840.80813.35429.75South44.010.80814497.8East69.0050.80817.1953.4West81.8990.80815.1999.2Roof194.152.0127.0510556.03GlassNorth11.623.7920.1885.2South7.583.7920.1577.44East30.43.7920.12315.84West6.0083.7920.1457.3Total transmission cooling load (W)17671.89Solar RadiationDescriptionArea(m2)SCSHGFCLFQSHGNorth0South7.580.941890.36484.8East0West6.0030.946950.552228.6Total solar radiation gain cooling (W)2713.4Internal loadsItemInput (W)CLFQapplianceLights15280.08122.2Appliances171011710Total internal gain cooling (W)1832.24OccupancyNumberSHG/LHGCLFQlQsSensible300750.4911025Latent30055116500Total Occupancy gains, Qoccupancy (W)27525Ventilation/Infiltrationm3/sCFMT/wQlQsSensible2.21.23T=21.558.18Latent2.23010w=.0417180Total Infiltration / Ventilation load (W)7 238.2 terrific total cooling loadsQlQsTotal latent load (W)23680Total sensible load (W)33300.61Total load (kW)56.98Tons of refrigeration16.3CHAPTER 6RESULTS AND RECOMMENDATIONSResultsFollowing the CLTD method we calculated cooling load to be 16.3 Tons. In which main contribution was from people present in the mosque (almost half the contribution) and heat conduction through walls and windows glass. The contribution from each mode is shown in fig 6.1.Fig 6.1 Contribution from each mode of heat transfer in cooling loadRecommendationsUse 6 ACs each of 2.5 Ton and one of 1.5 ton, we will need all the ACs switched ON during Jumma prayers only. On normal long time we will switch ON 3 or 4 ACs depending on the number of occupants. As the number of occupants decreases the required cooling load also decreases. For different values of occupants required cooling load has been calculated and shown in the fig 6.2.Fig 6.2 Relation between number of occupants and cooling loadNormally 30 people ar e present in the mosque at prayer times so we need only 9.2 tons of refrigeration. We will switch ON three ACs of 2.5 tons and one of 1.5 tonsWe can minimize the Cooling load byIncreasing the glass thicknessBy using opaque sheets on the outer side of the windows and doorsBy using reflecting and insulating material on the roof, reflective material will thin solar radiation and insulating material will minimize conduction emergence trees on southern side