8. S.O: 5.4 Calculate the Heat Gain Due
to Infiltration and Ventilation Load
of Door and Window.
S.O: 5.5 Solve the Problems
Relating to the External HeatLoad
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
12. Primary Heat Transport Modesare:
1.Conduction(পরিবহণ): Heatflow
onaMolecularScale.Mediumat
RestorMoving.
2.Convection(cwiPjb): 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
18. 3. Radiant Heat Exchange Radiation Heat Transfer is Very
Important in Building Application in the Following a Reason:
(a) Short Wave Length Radiation:
Solar Heat Absorption on Opaque Exterior
Surfaces,
Solar Heat Transmission Through
Transparent Surfaces,
Solar Heat Absorption and Reflection by
Interior Buildings Unlaces,
Absorptionand Reflection ofSolar Heat
by Window Glass.
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
19. (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
23. 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
24. 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
25. 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
32. ফসািাি এয়াি তাপ াত্রা (Solar Air Temp):
In Summer the Solar Radiation Affects
the Outside Surface of Wall and Roof.
The Absorbed Radiation Increases the
Temperature of the Outside Surface to a Value
that is Greater than Outside Air Temperature.
This Outside Surface Temperature is
Called Solar Air Temperature.
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
43. 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 theTransmissivity of the Glass,
“A” is its Absorptivity, and “N” is the
Fraction of Absorbed Energy 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
44. Solar Heat Gain Co-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
47. Infiltration is Described as Outside Air
that Leaks into a Building Structure.
These Leaks Could be Through the Building
Construction or Through Entry Doors.
Infiltration Heat Gains are Found by the
Following Equations.
These Equations are Discussed More in
the Psychrometrics Section
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
48. The First Equation is the Total Heat
Gains Using Enthalpy.
In this Equation, the Volumetric flow
Rate of the Infiltration or Ventilation
Air Must be Known.
This Value is Converted and
Multiplied by the Difference in
Enthalpy Between the Outdoor Air
Conditions and the Indoor Air-
Conditions.
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
49. The Following Two Equation Split
the Total Heat Gain into the
Sensible and Latent Heat Loads.
Sensible Heat Gains are Calculated
by Multiplying the CFM of the
Infiltrated Air by the Difference in
the Temperatures of the Indoor and
Outdoor Air
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
51. How to Calculate Heat Load :
Calculating aHeat Load is Necessary
Before theInstallation of aRadiant
HeatingSystemcanBegin, Since
Different Typesof Radiant Heating
Systems have Different BTU
Output Values.
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
52. A Typical Heat Load Calculation
Consists of Surface HeatLoss
CalculationandHeat Lossdueto
Air Infiltration.BothShouldbedone
Separatelyfor Every Room in the
House, So Having a FloorPlan with
Dimensions of allWalls, Floors,
Ceiling, as Well as Doors and
Windows is a Good Placeto Start.
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208
53. Belowis 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
54. 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
55. 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
56. 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
57. 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
58. 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
59. 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
60. 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
61. Latent Heat Gains areCalculatedby
Multiplying the CFM of Infiltrated
Airby the Differencein 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
62. 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
63. 4m উচ্চ একতিা একটি অরেস রবরডিং 30°N িযাটিটিউড বা
উত্তি অক্ষািংরিঅবরস্থত।
১।কক্ষঅেুর য়তাপফিাড(RoomSensibleHeatnLoad)RSHL
২।কক্ষসুপ্ততাপফিাড(RoomLatentHeat)RLH
৩।সবথর ােতাপফিাড(GrandTotalHeatLoad)
প্ররয়াজেীয়তর্যারদঃ
রভতরিিফদওয়ারিিউপিলাস্ট্াি = 1.25ফস.র .
বাইরিিফদওয়ারিিকন্সস্ট্রাকিে =20ফস.র .কিংরিেব্লক
=10ফস.র .ফেসরিক
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,TEJGAONI/ADhaka-1208