4th semester diploma in engineering refrigeration and air conditioning technology(RAT) Chapter-05(understand the external heat load for cooling load calculation)
8. 5.Understand the External HeatLoad for Cooling Load
Calculation
কুল িংল োডক্যো কুল শলেঅলিলরক্তলিটল োডসম্পলক্ে ধোরণো
“আজলক্র পোঠ”
S.O:5.1Describe the Heat Gain by
ConductionThroughBuildingStructure
5.2ExplaintheSolarHeatGainThroughout
Side Wall and Roof.
5.3 DescribeSolarHeatGainThrough
GlassArea
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
13. Primary Heat Transport Modesare:
1.Conduction(পলরবিণ): Heatflow
onaMolecularScale.Mediumat
RestorMoving.
2.Convection(পলরচ ে): Heat Conveyed
as Internal Thermal Energy of
Mass that is Displaced by Mean
or Turbulent Motion
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
19. 3. Radiant Heat Exchange Radiation Heat Transfer is Very
Important in Building Application in the Following a Reason:
(a) Short WaveLength Radiation:
Solar Heat Absorption on Opaque Exterior
Surfaces,
Solar Heat Transmission Through
Transparent Surfaces,
Solar HeatAbsorption and Reflection by
Interior Buildings Unlaces,
Absorption and Reflection ofSolar Heat
by Window Glass.
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
20. (b) Long-WaveLength Radiation:
Heat Emission by theExterior
Surfaces to the Sky,
HeatExchangeAmongInteriorSurfaces
Heat Exchange Between Interior
Surfaces and Occupants,
Heat Exchange Between the
LightingFixtureandInteriorSurfaces
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
24. Heat TransferThroughConduction:
Heat Transfer Through aMaterialTakes
Place byConduction fromWarmtoColdSide
TheSameProcessTakesPlaceinaBuilding
Generallythe Thermal Conductivityof the
Building Materials will be Much Lower.
In Solid Bodies IncludingBuilding
Components,Thermal Conduction Takes
PlacewhenOne Part of the Component is
Subjected to HigherTemperature and the
OtherPart toaLower TemperatureCondition.
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
25. Most Cases of Thermal Conduction are
UsuallyAnalysedand Treatedin Their
Simplifiedform as One DimensionalHeat
Flow Cases, i.e. Heat Flowin Directions
Other Thanthe Main Direction isNeglected.
Similarly, if the Changes in Atmospheric
Conditions(Inside orOutside)are Assumed
to be Very Slow,NeglectingThese
Changes, the Process of
Heat Transfer Can be Assumedto be
"SteadyStateHeat Transfer" in its
Simplified Form.
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
26. Conduction Heat Flow Rate Through a Wall of given Area can be
Calculatedusing
Qc = A×U× ∆𝐭 𝐄
Qc= Conduction Heat FlowRatein W
A = SurfaceArea in m2
U = TransmittanceValue in W/m2℃
∆𝐭 𝐄= TemperatureDifference
(Effective Temperature)
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
33. লসো োর এয়োর িোপ োত্রো (Solar Air Temp):
InSummertheSolarRadiationAffectstheOutsideSurface
ofWallandRoof.
TheAbsorbedRadiationIncreasestheTemperatureoftheOutside
SurfacetoaValuethatisGreaterthanOutsideAirTemperature.
This Outside Surface Temperature is Called Solar Air
Temperature.
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
44. Shading Coefficient(SC)
∴ SC = F(𝝀, 𝜽) = T(𝝀, 𝜽) + N × A (𝝀, 𝜽)
Here, 𝝀(Lamda) is the Wave Lengh of
Radiation and 𝛉 is the Angle of Incidence.
“T” is the Transmissivity of the Glass,
“A” is its Absorptivity,and “N” is the
Fraction of AbsorbedEnergy that is
Re-Emitted into the Space.
The Overall Shading Coefficient is thus
given by the Ratio:
S.C = F(𝛌, 𝛉)1 /F(𝛌, 𝛉)0
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
45. SolarHeatGainCo-efficient(SHGC)
T= 𝟑𝟓𝟎𝐧𝐦
𝟑𝟓𝟎𝟎𝐧𝐦
𝐓(𝛌)𝐄 𝛌 𝐝 𝛌
Here ’’T’’ (𝝀) is the Spectral
Transmittance at a given Wave Length
In Nanometers and E(𝝀) is the Incident
Solar Spectral Irrandiance
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
50. TheFollowing Two Equation Split
the Total Heat Gain into the
Sensible andLatent Heat Loads.
Sensible HeatGains areCalculated
by Multiplying the CFMofthe
Infiltrated Air by theDifference in
the Temperaturesof the Indoor and
Outdoor Air
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
52. How to Calculate Heat Load :
Calculatinga Heat Load is Necessary
Before theInstallation of aRadiant
Heating System can Begin, Since
Different Typesof Radiant Heating
Systems have Different BTU
Output Values.
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
53. A Typical Heat Load Calculation
Consists ofSurface Heat Loss
Calculation and Heat Loss due to
Air Infiltration.BothShouldbedone
Separately for Every Room inthe
House,So Having aFloor Plan with
Dimensions of all Walls, Floors,
Ceiling, as Well as Doors and
Windows is a Good Place to Start.
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
54. Below is a Sample5-step Manual to
Surface Heat Loss Calculation:
Step1–CalculateDeltaT(DesignTemperature)
DeltaTisaDifferenceBetweenIndoor
DesignTemperature(T1)andOutdoor
Design Temperature (T2),Where Indoor
Design Temperature is Typically 68-72℉
Depending onYour Preference,and
Outdoordesigntemperatureisatypical
Low Duringthe Heating Season.
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
55. The Former can be Obtained by
Calling yourLocal Utility Company
Assuming that T1is72℉and T2 is
–5℉, Delta T = 72℉- (-5℉)
=(72℉+5℉) = 77℉
Step 2 – Calculate Surface Area:
If the Calculation is done for an
OutsideWall,with Windows and Doors,
theCalculationsfortheWindowand
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
56. DoorHeatLoss Should be doneSeparately
Wall Area = Height x Width –Door
Surface-Window Surface
If, Wall Area = 8 ftx 22 ft – 24 sqft –
14sqft=176sqft–38sqft=138sqft
Step 3 – Calculate U-value:
Use "Typical R-Values and U-Values"
Guide to Obtain the Wall R-Value.
U-Value = 1 / R-Value
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
57. U-Value = 1 / 14.3 = 0.07
Step4–CalculateWallSurfaceHeatLoss:
SurfaceHeat Loss can be Calculated
Using the Formulabelow:
Surface Heat Loss = U-Value× Wall
Area× Delta T
If, SurfaceHeat Loss = 0.07× 138 sq
ft × 77 °F = 744 BTUH is
(U-valueisbasedonassuminga2×4woodframewallwith3.5"fiber
glass insulation)
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
58. Step5–CalculateTotal Wall Heat Loss:
Follow the Steps 1 through 4 to
Calculate Heat LossSeparately
for Windows, Doors, and Ceiling.
If, Door Heat Loss= 0.49× 24sq
ft× 77F = 906 BTUH
(U-value is based on assuminga solid wood door)
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
59. If, WindowHeat Loss = 0.65 × 14sqft
× 77F = 701 BTUH
(U-valueis based on assuminga double-panelwindow)
If, Ceiling Heat Loss = 0.05 × 352sq
ft × 77F = 1355 BTUH
(U-valueis based on assuminga 6"fiber glass insulation.Ceiling
surface is 22ft ×16ft)
Now, Add All the Number Together:
Total Wall Heat Loss = (WallLoss +
WindowLoss+DoorLoss+CeilingLoss)
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
60. Total Wall Heat Loss = (744 BTUH +
906 BTUH + 701 BTUH + 1352 BTUH)
= 3,703 BTUH
Air Infiltration Rates Should
Always be taken into Consideration
The Following Formula can beUsed
toCalculateHeatLossforaRoom
due to Air Infiltration:
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
61. Air Infiltration Heat Loss= Room
Volume × Delta T × Air Changes per
Hour × 0.018
Where Room Volume = Length ×
Width × Height
Air Changes per Hour Accounts forAir
Leakage into the Room.
ForExample: Air Infiltration Heat Loss =
(22ft × 16ft × 8ft) × 77F × 1.2 × 0.018 =
4,683 BTUH
Note:“ForActualCalculations, Contact YourContractor orSystemDesigner”.
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
62. Latent Heat Gains are Calculatedby
Multiplyingthe CFM of Infiltrated
Air by the Difference in the Humidity
RatiooftheIndoorAirandtheOutdoor Air
Qlatent = 4,840 × 𝐂𝐅𝐌 × (𝐖𝐨𝐮𝐭𝐝𝐨𝐨𝐫-Windoor)
W=Humidityratio[lbmwet/lbmdry]
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
63. It is Important to notethat these
Loads are not Seen Directly by
theCoolingCoil.TheseareIndirect
Loads that Occurin Each Air
ConditionedSpace. Ventilation
Air is Seen Directlyat the Coil
and Thus this Air Must be Cooled
DowntotheSupplyAirDistribution
Temperature Which is Much
LowerthantheRoomConditionAir
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208