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সপ্তম-অধ্যায়
হিটিং ল াড সম্পর্কে ধ্ারণা
আবু লমািাম্মদ আহিকু যা
ইন্সট্রাক্টর (লেক্) আর এহস
ঢাকা পহ র্েকহিক ইন্সটটউে,লিজগাাঁ ও,ঢাকা।
িািং 23-08-2020
কাহরগহরহিক্ষাঅহধ্দপ্তর্ররিত্ত্বাবধ্ার্িপহরচাহ ি
লেসবুক াইভ ক্লার্ি স্বাগি
OnlineTVET.com
Facebook.com/onlinetvet
আবু লমািাম্মদ আহিকু যা
ইন্সট্রাক্টর (লেক্) আর এহস
ঢাকা পহ র্েকহিক ইন্সটটউে,
লিজগাাঁ ও,ঢাকা -১২০৮
atiqullahrac@gmail.com
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
wnwUs †jvW কযাjKz‡jkb m¤ú‡K© aviYv
সপ্তম- অধ্যায়
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
“ক্লািলির্ে আমরাযা হিখর্বা”
১।kxZKvjxb kxZvZc wbqš¿Y বযবস্থার wnU jmmg~n
২।DB›Uvi Gqvi KwÛkwbs হসর্ের্মর wnU jm cÖwZ‡iv‡ai †KŠkj
৩।kxZKvjxb kxZvZc wbqš¿Y বযবস্থার wnU j‡mi েযাক্টর mg~n
৪। একট K¬vm iæ‡gi wnwUs †jvW কযা কুর্ িিপ্রণা ী
৫।
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
“আজর্কর পাঠ”
7.1ListtheSources of Heat Lossin
Winter Air Conditioning
7.2DescribetheTechnique of
Reducing Heat Loss in Winter
AirConditioning
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
7.3ExplaintheFactorsInvolved
WiththeHeatLossinWinter
Air-Conditioning
7.4 Calculate Heating Load of a
ClassRoom
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
“আজর্করপাঠ”
৭.১ kxZKvjxb kxZvZc wbqš¿Y বযবস্থার wnU jmmg~n
৭.২DB›Uvi Gqvi KwÛkwbs হসর্ের্মর wnU jm cÖwZ‡iv‡ai †KŠkj
৭.৩ kxZKvjxb kxZvZc wbqš¿Y বযবস্থারহিে র্সরেযাক্টরসমূি
৭.৪একট K¬vm iæ‡gi wnwUs †jvW
কযা কুর্ িি
7.1List theSources ofHeatLossinWinter AirConditioning
kxZKvjxb kxZvZc wbqš¿Y বযবস্থার wnU jmmg~n
kxZKvjxb kxZvZc wbqš¿Y বযবস্থার wnU jmmg~n wewfbœ
fv‡e n‡q _v‡K , †hgb-
১। †fw›U‡jk‡bi (VENTILATION) মাধ্যর্ম
২। `iRv-Rvbvjv Nb Nb †Lvjvi Kvi‡Y|
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
৩। Nb Nb †jvKRb iæ‡gi মর্ধ্য Avmv-hvIqvi Kvi‡Y
৪। iæ‡gi wj‡K‡Ri Kvi‡Y Ges
৫। wbqwš¿Z iæg BÝy‡jkb (INSULATION)
mwVKfv‡e KvR bv Ki‡j, A_©vr Ach©vß BÝy‡jkb ইিযাহদ
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
kxZKvjxb kxZvZc wbqš¿Y বযবস্থারহিে সসমূর্িরস্থািসমূি
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
kxZKvjxb kxZvZc wbqš¿Y বযবস্থারহিে সসমূর্িরস্থাি ওপহরমািসমূি
ললার বা লমর্ে কিতে ক
হসহ িংবা ছাদ কিতে ক
ওয়া বা লদয়া কিতে ক
উইর্ডা বা জািা া কিতে ক
এয়ার হ র্কজ কিতে ক
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
7.2Describe theTechnique of Reducing Heat Loss inWinter Air Conditioning
DB›Uvi Gqvi KwÛkwbs হসর্ের্মর wnU jm cÖwZ‡iv‡ai †KŠkj
DB›Uvi Gqvi KwÛkwbs wbqš¿Y বযবস্থায় wb‡¤œv³fv‡e wnU jm †iva Kiv hvq
১। `iRv-Rvbvjv (Door/Window) `„pfv‡e AvUwK‡q
২। Nb Nb gvbyl (People) cÖ‡ek bv K‡i
৩। Kÿ/ iæ‡gi BÝy‡jkb (Insulation) mwVK gvÎvq ¯’vcb কর্র
৪। iæ‡gi wj‡KR (Leakage) mg~n cÖwZ‡iva K‡i
৫। wnUvi/wi-wnUvi (Heater/Re-Heater) K‡qj wVKg‡Zv KvR bv Ki‡j
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
৬। wdëvi(Filter) eøK/eÜ n‡q †M‡j|
৭।েযার্ির w¯úW (Fan Speed) K‡g †M‡j|
৮।থার্মোেযাে (Thermostat) mwVKfv‡e KvR bv Ki‡j
৯। mwVKfv‡e ডার্ক্টর (Duct) মর্ধ্য Mig
evZvm cÖevwnZ bv n‡j,A_vr ডার্ক্টরDci
BÝy‡jkb Kiv bv n‡j ইিযাহদ।
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
DB›Uvi Gqvi KwÛkwbs হসর্ের্মর wnU jm cÖwZ‡iv‡ai †KŠkj
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
10 Ways to Prevent Heat Loss at Home
1. Insulation- ইন্সুর্ িি
2. Chimneys - হচমিী
3. Windows- জািা া
4. Doors- দরজা
5.TimedExtractorFan–
োইম্ড এক্সজেরেযাি
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
6.BleedYourRadiators–
লবল্ডলরহডর্য়ের
7. A Radiator Shelf – লরহডর্য়েরিাক
8. Disused Vents–হডসউস্ডলভি
9. Gaps Between Floor boards – গযাপ্স ললার
লবাড।
10. Curtains –কার্েে ি ইিযাহদ।
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
1. Insulation ইন্সুর্ িি
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
2. Chimneys- হচমিী
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
3. Windows জািা া
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
4. Doors - দরজা
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
5. Timed Extractor Fan- োইম এক্সোক্টর েযাি
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
6. Bleed Your Radiators
7. A Radiator Shelf- লরহডর্য়ের লসল্ফ বািাক
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
8. Disused Vents- অবযবহৃিলভি সমূিবন্ধ করণ
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
9. Gaps Between Floorboards
10. Curtains - কার্েে ন্স
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
7.3ExplaintheFactorsInvolvedWiththe Heat Lossin Winter Air-Conditioning
kxZKvjxb kxZvZc wbqš¿Y বযবস্থারহিে র্সরেযাক্টর সমূি
Air Conditioners are Complex
Machines that can Lose Their
Efficiency if Some Factors are not
Kept in Mind. These Factors are:
(i) Poor Airflow/ `~e©j evZvm cÖevn
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
(ii) Lack of Maintenance/ iÿYv-‡eÿ‡Y NvU©wZ
(iii) Clogged or Dirty Ducts/ বদ্ধ ev Acwi¯‹vi
ডাক্ট
(iv) Faulty Thermostat/ সমসযাযুক্তথার্মোেযাে
(v) Dirty Coils/ Acwi¯‹vi Kzwjs K‡qj
If your AC is not Performing as it usedto bethenit is
best measure to Call an HVAC Service Provider as
they will assess the Problem and will be able to
1| BÝy‡jkb t iæ‡gi BÝy‡jkb wVKg‡Zv
KvR bv Ki‡j wnU wj‡KR n‡q evB‡i †ei
n‡q Avm‡e| BÝy‡jkb mg~n mwVKfv‡e
†`Iqv‡ji m‡½ AvUKv‡bvi বযবস্থা Ki‡Z n‡e
cÖ‡qvR‡b weUzwgbmn Zv‡ii Rvwj বযবিারকরর্িির্ব
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
2| wi-wnUvi/wcÖ-wnUvi/wnUvi t wbqwš¿Z K‡ÿi মর্ধ্য
Mig evZvm cÖ‡e‡ki c~‡e© wnUvi n‡q evZvm Mig n‡q
cÖ‡ek K‡i| wnUvi hw` bó nq ev wVK g‡Zv KvR bv
K‡i Zvn‡j Zv‡ci AcPq NU‡e| myZivs iæ‡gi
ZvcgvÎv বতহদ্ধ Ki‡Z n‡j অবিযই wi-wnUvi বা
wcÖ-wnUvi/wnUvi(Heater) fv‡jv n‡Z n‡e|
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
3| †eøvqvi/েযাি t †eøvqvi ev েযার্ির(Fan)
w¯úW ev MwZ K‡g †M‡j wbqwš¿Z K‡ÿ
Mig evZv‡mi cwigvb K‡g hv‡e|
mwVK
cwigvb evZvm †c‡Z n‡j অবিযই
েযার্ির MwZi cÖwZ ক্ষয ivL‡Z n‡e|
4| cvwbi cv¤ú/হের্মর ZvcgvÎv K‡g †M‡j t
mieivnK„Z হের্মর ZvcgvÎvi cÖwZ ক্ষয
ivL‡Z n‡e, †h mKj Kvi‡Y cvwbi ev®ú বা
হের্মর ZvcgvÎv K‡g hvq Zvi mwVK
বযবস্থা cÖnY Ki‡Z n‡e|
5| wnwUs থার্মোেযাে wVK g‡Zv KvR bv Ki‡j t
wnwUs থার্মোেযােট mwVKfv‡e KvR
bv Ki‡j Zv †givgZmn cÖ‡qvRbxq
বযবস্থা cÖnY Ki‡Z n‡e|
7.4 Calculate Heating Load of A Class Room
একট K¬vm iæ‡gi wnwUs †jvW
কযা কুর্ িিkxZKv‡j evZv‡mi Av`ª©Zvi cwigvb LyeB Kg _v‡K,
hv gvbeRxe‡b A‡bK A¯^w¯ÍKi, Avi hw` evwn‡ii
ZvcgvÎvi K_v ej‡Z nq Zv Avgiv mevB Rvwb
ZvcgvÎv GZUvB Kg _v‡K, hv evB‡i †ei nIqv Am¤¢e n‡q c‡i
kxZcÖavb †`k¸‡jvi w`‡K ZvKv‡j Avgiv
Av‡iv eyS‡Z cvwi †h, ZvcgvÎv িূণয
wWwMÖi wb‡¤œ Ae¯’vb K‡i hv Ae¯’vb
cwieZ©b K‡i cvwb ei‡d cwiYZ nq|
GB Ae¯’vq N‡ii wfZ‡i `iRv- Rvbvjv eÜ
K‡i ZvcgvÎvi Lye †ewk cwieZ©b Kiv m¤¢e nq bv
cwieZ©b Ki‡Z n‡j wnUvi w`‡q ZvcgvÎv
Ges cvkvcvwk wnDwgwWwU (Humidity) বা
Av`ª©Zvও cwieZ©b Ki‡Z n‡e।
GKwU K¬vkiæ‡gi এহস/এয়ার কহডিঃ হডজাইি †`Lv‡bv n‡jv
‡hgb- 50 Rb QvÎ-QvÎxi aviY ÿgZv m¤úbœ GKwU
K¬vkiæg-এ Gwm Ki‡Z n‡e, hv wb‡¤œi িথযDcvË
অিুসার্র
†`qv n‡jv
1| evB‡ii Ae¯’v t 10 wWMÖx ‡mjwmqvm, WªvB evj¦ ZvcgvÎv (DBT)
Ges I‡qU evj¦ ZvcgvÎv (WBT ) = 8 wWMÖx ‡mjwmqvm
2| cÖ‡qvRbxq wfZ‡ii Ae¯’v t 20 wWMÖx ‡mjwmqvm, WªvBevj¦ ZvcgvÎv (DBT)
3| evZvm mvK©y‡jk‡bi cwigvb : 0.3 Nb.wgUvi/wg./QvÎ
Ges K‡q‡ji evBcvm েযাক্টর (BPF) = 0.32
D³ িথয DcvË Abymv‡i wbY©q Ki Ki‡Z n‡e t
1) K‡q‡ji wnwUs ÿgZv (wK‡jv-IqvU)
2) K‡q‡ji mvi‡d‡mi ZvcgvÎv wW.‡m.
3) wnDwgwWdvqv‡ii ÿgZv (CapcityofHumidifire)
mgvavb t †`qv Av‡Q, WªvB evj¦ ZvcgvÎv (DBT), td1 = 10℃
wfZ‡ii Ae¯’vi WªvB evj¦ ZvcgvÎv (DBT), td2 = 20℃
aviY ÿgZv (Seating capacity) = 50 student
I‡qU evj¦ ZvcgvÎv (WBT) tw1 = 8℃ ,
wi‡jwUf wnDwgwWwU(RH%), 𝝋 𝟐 = 60%
K‡q‡ji evBcvm েযাক্টর (BPF) = 0.32
GLv‡b, v1 = 0.3  50 m2/min = 15 m3/min
∴td3 = 27.5℃, Vs1 = 0.81 m3/kg of dry air
myZivs mieivnK…Z evZv‡mi fi, ma =
v1
Vs1
=
𝟏𝟓
𝟎.𝟖𝟏
= 18.82 kg/min
∴ mvB‡µv‡gwUªK PvU© n‡Z Gb_vjwc, h1= 24.8 kj/kg da.
Ges h2 = 42.6 kj/kg da
Avgiv Rvwb,
1| K‡q‡ji wnwUs ÿgZv ={ma.(h2-h1)}
=18.52(42.6–24.8) [gvb ewm‡q]
= 329.66 Kj/min, =
329.66
𝟔𝟎
Kj/Sec
= 5.5 Kw (Ans.)
2| evBcvm েযাক্টর (BPF) =
𝐭𝐝 𝟒
−𝐭𝐝𝟑
𝐭𝐝 𝟒
−𝐭𝐝𝟐
Or 0.32 =
𝐭𝐝 𝟒
−𝟐𝟕.𝟓
𝐭𝐝 𝟒
−𝟏𝟎
, [ gvb ewm‡q ]
Or, (0.32 𝐭𝐝 𝟒 - 3.2) = (𝐭𝐝 𝟒 - 27.5)
 𝐭𝐝 𝟒 =
𝟐𝟒.𝟑
𝟎.𝟔𝟖
= 35.7℃ (Ans.)
GLb, mvB‡µv‡gwUª©K PvU© n‡Z, †¯úwmwdK wnDwgwWwU,
w1 = 0.0058 kg/kg dry air. Ges
w2 = 0.0088 kg/kg dry air.
আবার আমরাজাহি,
3| wnDwgwWdvqv‡ii ÿgZv = {ma.(w2-w1)}
= {18.52(0.0088 – 0.0058)} [ gvb ewm‡q ]
= 0.055 kg/min = 0.55  60 kg/hr.
= 3.3 kg/hr. ( Ans.)
সুিরািংwnDwgwWdvqv‡ii ÿgZv = 3.3 kg/hr.
Heat Load or Heat Gain (হিেলগইি)
A Buildingor Room GainsHeat from ManySources
Inside Occupants, Computers, Copiers, Machinery,
andLightingallProduce Heat.WarmAirfromOutside
Enters Through Open Doors and Windows, or as
‘Leakage’ Though the Structure. However the Biggest
Source ofHeatisSolarRadiation fromtheSun, Beating
downon the Roof andWalls,andPouringThrough
the Windows, Heating Internal Surfaces.
The Sum of All These Heat Sources is
Know as the Heat Gain (or Heat Load) of
the Building, and is Expressed Either
in BTU (British Thermal Units) or Kw (Kilowatts)
For an Air Conditioner to Cool a Room
or Building its Output Must be Greater
Than the Heat Gain.
It is Important before Purchasing
an Air Conditioner that a Heat Load
Calculation isPerformed to Ensure
it is big Enough for the Intended Application
@
HeatLoadCalculations/ হিটিংল াডকযা কুর্ িিপ্রণা ীিঃ
Thereare Several Different Methodsof Calculating
the Heat Loadfor a Given Area: Quick Calculation for
Offices. For offices with Average Insulation and
L;ighting, 2/3 Occupants and 3/4 Personal Computers
and a Photocopier, the Following Calculations
will Suffice:
Heat Generated by Lighting By Calculatingthe
Heat Gain from Each Individual Item and
Adding them Together, an Accurate Heat
Load Figure can be Determined.
েু র্ের (ft) লক্ষর্ে,
Heat Load (BTU) = Length (ft.) x Width
(ft.) x Height (ft.) x 4
হমোর্রর (m)লক্ষর্ে,
Heat Load (BTU) = Length (m) x Width
(m) x Height (m) x 141
For Every Additional Occupant add 500 BTU
IfThere areanyAdditionalSignificantSources ofHeat,
or Instance Floor to Ceiling South Facing
Windows,or Equipmentthat ProducesLots of Heat,
the Above Method Will Under Estimate the Heat Load
In Which Case the Following Method Should be
Used Instead. A More Accurate Heat Load
Calculation for any Type of Room or Building
The Heat Gain of a Room or Building Depends on:
The Size of the Area Being Cooled
The Size and Position of Windows,
and Whether They Have Shading
The Number of Occupants
HeatGeneratedbyEquipmentandMachinery
Step One(1) ধ্াপবাপযোয়-১
Calculate the Area in Square Feet(ft2) of the
Space to be Cooled, and Multiply by 31.25
Area BTU = Length (ft.) x Width (ft.) x 31.25
Step Two(2) ধ্াপ বাপযোয়-২
Calculate the Heat Gain Through the
Windows. If the Windows Don’t have
Shading Multiply the Result by 1.4
North Window (BTU) উত্তর জািা ারজিয-১৬৪
=Area of North Facing Windows(m.sq)x164
South Window (BTU) দহক্ষণ জািা ারজিয-৮৬৮
=AreaofSouth Facing Windows(m.sq)x868
If No Shading, North Window BTU
= North Window BTU x 1.4 লকার্িালিডিাথাকর্ -১.৪
If No Shading, South Window BTU
=South Window BTU x 1.4
Add the Results Together.
Total Window BTU =
(North Window + South Window)
Step Three(3) ধ্াপবাপযোয়-৩(হপপ ল াড)
Calculate the Heat Generated by Occupants, Allow
600 BTU Per Person.
Occupant BTU = Number of People x 600
Step Four(4) ধ্াপবা পযোয়-৪ (ইকুইপর্মন্ট ল াড)
Calculate the Heat Generated by Each Item of
Machinery - Copiers, Computers, Ovens etc.
Find the Power in Watts for Each Item, Add Them
Together and Multiply by 3.4
Equipment BTU = Total Equipment Watts x 3.4
Step Five(5) ধ্াপ বা পযোয়-৫
Calculate the Heat Generated by Lighting.
Find the Total Wattage for All Lighting and
Multiply by 4.25
Lighting BTU = Total Lighting Watts x 4.25
Step Six(6) ধ্াপ বা পযোয়-৬
Addthe AboveTogether to Findthe Total Heat Load
Total Heat Load BTU = (Area BTU + Total Window
BTU + Occupant BTU + Equipment BTU + Lighting
BTU)
Step Seven(7) ধ্াপবাপযোয়-৭
Divide the Heat Load by the Cooling
Capacity of the Air Conditioning Unit
in BTU, to Determine how Many Air
Conditioners are Needed.
Number of a/c Units Required = Total
Heat Load BTU / Cooling Capacity BTU
evwoi KvR (Home Work)
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5| AsK t evB‡i 10℃ 𝐃𝐁𝐓 I 8℃ WBT
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ÿgZvm¤úbœ K¬vkiæ‡gi 25℃ DBT I 50 % RH
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0.3m3/min/Student Ges K‡q‡ji BPF n‡jv 0.32
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
এইক্লাসট পূিরায় লদখর্ি হভহজেকর্রা
onlinetvet.com
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
ধ্িযবাদ
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
সপ্তম- অধ্যায়
হবেয়িঃকুহ িংঅযাডহিটিংল াড
কযা কুর্ িি
CoolingAndHeatingLoadCalculation
হডর্লামা-ইি-ইহিহিয়াহরিংলরহজজার্রিি
অযাডএয়ারকহডিহিিংলেকর্িা হজ
৪থে-পবে
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka
PresentedBy:A.M.ATIQULLAH,INSTRUCTOR(Tech)RACDHAKAPOLYTECHNICINSTITUTE,TEJGAON,Dhaka

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Chapter 7(Understand the concept of heating load calculation)