67243 cooling and heating & calculation

A.M.ATIQULLAH(Avey†gvnv¤§`
AvwZKzj¨v)
INSTRUCTOR(TECH.) RAC(BÝUªv±i/AviGwm)
DHAKA POLYTECHNIC INSTITUTE,TEJGAON, Dhaka.
XvKvcwj‡UKwbK BÝwUwUDU,‡ZRMuvI,XvKv-Presented By : A.M.ATIQULLAH, INSTRUCTOR(Tech) RAC DHAKA POLYTECHNIC INSTITUTE,TEJGAON I/A Dhaka-1208
welq†KvW t 67243 (4_© ce©)
ÒPZz_© -
Aa¨vqÓ
4th

G.O: 4. Understand the Aspect of “Building
Survey for Air Conditioning Heat Load”,
wewìsmv‡f©igva¨‡g kxZvZcwbqš¿Y e¨e¯’vi wnU†jvW wbY©q|
Presented By : A.M.Atiqullah INSTRUCTOR(Tech) RAC DHAKA POLYTECHNIC INSTITUTE,TEJGAON I/A Dhaka-1208Presented By : A.M.ATIQULLAH, INSTRUCTOR(Tech) RAC DHAKA POLYTECHNIC INSTITUTE,TEJGAON I/A Dhaka-1208

wewìs mv‡f©i
gva¨‡g kxZvZc wbqš¿Y
e¨e¯’vi wnU †jvW
wbY©q|
Aa¨vq -04
Presented By : A.M.Atiqullah INSTRUCTOR(Tech) RAC DHAKA POLYTECHNIC INSTITUTE,TEJGAON I/A Dhaka-1208

Specific Objectives (S.O-5) :-
S.O: 4.1. State the Meaning of Building Survey for
Cooling Load Calculation. Kywjs †jvW K¨vjKz‡jkb wewìs m‡f©i A_©
S.O: 4.2. Mention the Physical Factors Which Should
be Considered in Estimation of Cooling Load.
S.O: 4.3. Explain the Physical Factors Which Should be
Considered During Survey for Heat Load Estimation.
S.O: 4.4. Mention the Non-Residential Cooling Load.
S.O: 4.5. Mention the Residential Cooling Load.

wewìs m‡f© ev Rwi‡c Kywjs †jvW K¨vjKz‡jkb t
kxZvZc wbqš¿‡Yi(Air Conditioning) †ÿ‡Î Kzwjs
‡jvW wbY©‡q wewìs ev feb mg~n cÖv_wgK Ae¯’vq
Rwic ev mv‡f© Kivi cÖYvwj‡K g~jZt wewìs mv‡f©
Kzwjs †jvW K¨vjKz‡jkb e‡j| KviY †jv‡Wi wnmve-
wbKvk Ki‡Z n‡j fe‡Yi wewfbœ w`K mg~n we‡ePbvq
Avb‡Z nq|Presented By : A.M.Atiqullah INSTRUCTOR(Tech) RAC DHAKA POLYTECHNIC INSTITUTE,TEJGAON I/A Dhaka-1208

01|wewìsevfe‡bi †fŠ‡MvwjKAe¯’vbm¤ú‡K©|
02|wewìsevfebwU†Kvb gyLx|
03|Avkcv‡kiAe¯’vb|
04|wewìsevfe‡bi eënvi|
05|wewìsevfe‡bi cwigvc|
06|wewìsevfe‡bi evB‡iiweeiY|
07|wbg©vbmvgMÖx GesìRvI Rvbvjv|
08|wmuwoevGmKv‡jUi, Ges†jvKmsL¨v|
09|wfZ‡iiwÿZAvmevecÎ-hš¿cvwZiAe¯’vb|
10|jvBU, d¨vb I ‰e`ÿwZKmiÄvgI eënviKvj|
11| †fw›U‡jkb(Ventilation)ev Bbwdj‡Uªkb (Infiltration)|
12|mvwf©m(Service)my‡hvM-myweavBZ¨vw`m¤ú‡K©Rvb‡Zn‡e|

1
2
3
4
5
7
8
9
10
11
12
6

1
2
3
4
5
6
7
8
9
Qv`
jv&B
U
‡`Iqvj
‡g‡S
m~h©
¨
hš¿vsk

myc&Í
Zvc
Aby‡gq Zvc K‡qj¯’vb

wnU ev Zv‡ci bvbvwea Drm †_‡K AsK K‡l ÓwnU †jvWÓ wbY©q Kiv AZ¨šÍ RwUj
ev KwVb I mgq ¯^v‡cÿ e¨vcvi| Avi wewfbœ ms¯’v †hgb- A¨vk‡i(ASHARE)
GAviAvB(ARI) K¨vwiqvi(CARRIER) Gqvi KwÛkwbs (Gwm) †Kv¤úvbx
gyjëvb Z_¨ m¤^wjZ AmsL¨ PvU© I Mªvd ˆZix K‡i †i‡L‡Q, hv wWRvBb Kv‡R
AnineënviKi‡Z nq| GmeMÖvdevPv‡U©i g‡a¨ i‡q‡Q t
01| †fw›U‡jkbGqvi-PvU (Ventilation Air Chart).
02| evB‡ii evZv‡mi Zvc-PvU© (Out side Air Heat-Chart).
03| wfZ‡ii Zv‡ci Drm- jvBU(Light),d¨vb(Fan), ‡jvK(People),
cøvM(Pluge/Socket)†jvWBZ¨vw`|
04| wewfbœ cÖKvi I wewfbœ Ae¯’vq Møv‡mi wfZi w`‡q cÖevwnZ
†mŠiZvcPvU©|
05| †Ìqvj I Qv‡ì wfZi w`‡qwewfbœmg‡q cÖevwnZcwi‡ekMZ ZvcPvU©|Presented By : A.M.Atiqullah INSTRUCTOR(Tech) RAC DHAKA POLYTECHNIC INSTITUTE,TEJGAON I/A Dhaka-1208

µtbs Zvc cwien‡bi gva¨g ev Drmmg~n eM©dzU/cwi
gvb
% d¨v±i = wewUBD/
N›Uv
1|
2|
3|
4|
5|
6|
7|
8|
K. m~h©gyLxRvbvjv (wbU)
L. Rvbvjv-`iRv me©‡gvU
‡`Iqvj, m~h©gyLx-Qqvhy³
cvwU©kb
wmwjs
‡g‡S
iææ‡gi AvqZb NbdzU
‡jvK (People)
jvBU,d¨vb,hš¿cvwZ
9| mgwó(Subtotal ) †mbwmej wnU
=
10| j¨v‡U›U wnU I Wv±(Duct) jm × µwgK 9 mgwó
=
11| me©‡gvU (Total)
=
S.O: 4.3 Explain the Physical Factors Which Should be
Considered During Survey for Heat Load Estimation
wnU †jvW Gw÷‡gk‡b wewìs ev fe‡bi ˆZixK…Z bg~bv dg© †`Lv‡bv n‡jv- (weªwUk
c×wZ‡Z)

wnU †jvW Gw÷‡gk‡b wewìs ev fe‡bi ˆZixK…Z bg~bv dg© †`Lv‡bv n‡jv- (weªwUk
c×wZ‡Z)

µtbs Zvc cwien‡bi gva¨g ev Drmmg~n cwigvb × d¨v±i = wewUBD/N›Uv
1| K.m~h©gyLx RvbvjvwbU (`wÿ‡Y)
L.`iRv-Rvbvjv(me©‡gvU)
43.2
28.3
45
30
1,944
8,490
2| ‡`Iqvj,m~h©gyLx
- Qvqvhy³
825
304
5.7
3.2
4703
973
3| cvwU©kb 308 5.2 1,602
4| wmwjs 940 3.7 3,478
5| ‡g‡S - - -
6| jvBU - = 500IqvU
d¨vb - 65 ×3 =195IqvU
‡iwWI,wUwf,B‡jKwUªKAvqiYGK‡Î †gvU =1200IqvU
ÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑÑ
ÑÑÑ
‡gvU =1895IqvU
GK‡Î KvRKi‡e Ggb × 𝟕𝟓%
1421 3.4 4.831
7| mgwó †mbwmej wnU - 32,241
8| ‡j‡U›U wnU, Wv±jm BZ¨vw` - 33% × 32241 10,640
9| ‡jvK msL¨v 9 200 1,800
10| fe‡bi AvqZb 10,780 4,420
11| me©‡gvU - = 42,881

‡fw›U‡jkb =
32241
23
= 1402cfm
evZvmcwieZ©b =
10780
1402
= 7.7wgwb‡U 1 evi
Gqvi KwÛkbh‡š¿i K¨vcvwmwU/ÿgZv =
42881
12000
= 3.57Ub(TR )
myZivs wngvqb h‡š¿i ÿgZv n‡jv 3.5 Ub|

evBcvm d¨v±i (By Pass Factors-BPF): “K‡qj wbM©Z
evZv‡mi ZvcgvÎv I K‡q‡ji ZvcgvÎvq cÖweó evZv‡mi ZvcgvÎv I
K‡q‡ji ZvcgvÎvi we‡qvMd‡ji(-) AbycvZ‡KB evBcvm d¨v±i e‡j|
G‡KBPF Øviv cÖKvkKiv nq|
∴BPF =
K‡qj wbM©Z evZv‡mi ZvcgvÎv− K‡qj ZvcgvÎv
cÖweó evZv‡mi ZvcgvÎv− K‡qjZvcgvÎv
=
𝐓 𝟑
−𝐓𝟐
𝐓 𝟏
−𝐓𝟐
‡‡hgbt †`Iqv Av‡Q,
K‡q‡ji ZvcgvÎv T2 = 40℉ (wWMªx. dv.), cÖweó evZv‡mi ZvcgvÎv T1 = 85℉(wW.dv.)
wbM©Z evZv‡mi ZvcgvÎv, 𝐓 𝟑 = 58℉ (wW. dv.)myZivs evBcvm d¨v±i KZ n‡e?
Avgiv Rvwb, B.P.F =
𝐓 𝟑
−𝐓 𝟐
𝐓 𝟏
−𝐓 𝟐
= ( 𝟓𝟖 −𝟒𝟎
𝟖𝟓 −𝟒𝟎
) =
𝟏𝟖
𝟒𝟓
= 0.4 ANSWER.

(K.) AbvevwmK †ÿ‡Î Af¨šÍixYZvcAR©bcÖwµqv t
01. evq~ AbycÖ‡e‡ki gva¨‡g Gwm iy‡gi cÖweó evB‡ii evZv‡mi Av`©ªZvi Kvi‡Y Zvc
AR©b|
02. Gwm iæ‡gi gvby‡li kixi †_‡K wbM©Z Rjxq ev®ú Nbxfe‡bi Kvi‡b AwR©Z Zvc|
03. cvwU©k‡bi gva¨‡g wbqwš¿Z ¯’v‡b cÖZ¨ÿfv‡e cÖweóRjxq ev‡®úi Kvi‡Y AwR©ZZvc|
04.gvby‡lik¦vmcÖk¦v‡miKv‡Rwbqwš¿ZDcv`vb cÖ‡qvRbxq¯’v‡bevZvmmieiv‡ni Kvi‡Y AwR©ZZvc|
(L.) AbvevwmK †ÿ‡Î evZvm AbycÖ‡e‡ki Kvi‡Y ZvcAR©b cÖwµqv t
01. ¯Í‚cRwbZ cÖfve( STACK EFFECT)
02. evZv‡mi Pvc (WIND OR AIR PRESSURE )
03. evq~ cwieZ©b RwbZ(AIR- CHANGE )|
(M.) AbvevwmK †ÿ‡Î Av`ª©ZvRwbZ ZvcAR©b cÖwµqv t
(N.) ‡dÝ‡Uªkb(Fenestration) Zvc AR©b cÖwµqv t Rvbvjvi KvP †_‡K/Rvbvjvi
gva¨‡g†hZv‡ciD™¢eN‡U, ZvnvBFenestration Heat Gain wn‡m‡eaivnq|Presented By : A.M.Atiqullah INSTRUCTOR(Tech) RAC DHAKA POLYTECHNIC INSTITUTE,TEJGAON I/A Dhaka-1208

jv&BU †jvW
‡jvK
hš¿vsk
evZvm
AbycÖ‡ek
wewKiY
m~h©¨

1. Space Characteristics and Heat Load Sources.
2. Computer Application in Cooling Load Calculation:
3. MATERIALS AND APPLICABLE METHOD
4. Space Design Cooling Load Calculation by CLTD/SCL/CLF
5. External Cooling Loads: (i) Roofs, walls and glass: The basic cooling
load equation for exterior surfaces such as roofs, walls, and conduction
through glass is given as;
q = U × A × CLTD……….. (1)
(*) Fenestration: Space cooling load from fenestration include the heat
gain from conduction and solar radiation. The cooling load from
conduction and convection heat gain is calculated by;
q
cond = U × A × (CLTD)…… (2)
qrad = A × SC × SCL ............. (3)
6. Partitions, ceilings, and floors: The cooling load from
partitions, ceilings, and floors is calculated as follows;
q = U × A ta − trc ……….…..(4)

7. Internal Cooling Loads: (A). “People”: The total sensible heat
qs = N × SHGp CLFp ….. (5) And the latent cooling load due to people is:
ql = N × LHGp…………... (6)
(B). “Lights”: At any time, the space cooling load from lighting can be estimated as;
qel = W × Ful × Fsa × CLFel ….. (7)
(C). “Power”: The cooling load value of power-driven equipments can be calculated as:
qp = P × Ef × CLF ……….(8)
(D). “Appliances”: To calculate the cooling load capacity of appliances in a
particular circumstance either of the following equations can be applied;
q (sensible) = q input × FU × FR × CLF …..…. (9) Or,
q sensible = q input × FL × CLF…………... (10)
E. “Ventilation and Infiltration Air”: Space cooling load due to
ventilation and infiltration air is calculated as follows;
q sensible = 1.23 × Q to − ti………………... (11)
q latent = 3010 × Q W0 − Wi …………...… (12)
q = 1.20 × Q h0 − h …………………….(13)

AvevwkKKzwjs†jvWwb‡¤œv³fv‡eKivnqt-
(01.) AvevwkK †jvWd¨v±i (Residential Load Factor/RLF) wbiƒcY t
AvevwkK †ÿ‡Î gvby‡likix‡iiZvcgvÎv I kxZjZvi তিনটি তিষয়ের উপরতনর্ভ র কয়র-
01| ড্রাইিাল্ব িাপমাত্রা(D.B.T)
02| স্পেতিতিক তিউতমতিটি(SPECIFIC HUMIDITY/ SP.Volm)
03| এোর স্পর্য় াতিটি( AIR-VELOCITY)
(02.) AvevwkK wnUe¨v‡jwÝs (Residential Heat Balancing/RHB)cÖwµqv t
Wv‡±i AvK…wZ wbiæcY Ges Gwm e¨e¯’vq AšÍf©y³
cÖ‡Z¨KwU K‡ÿ mwVK cwigvb wbqwš¿Z evZvm mieivn
wbwðZ Kivi Rb¨ †eøvqvi , †W¤úvi Ges wbM©gb wMÖ‡ji
†h mgš^q mvab Kiv nq Zv‡K wnU wm‡÷g e¨v‡jwÝs ev
(RHB) e‡j|

S.O: 4.3 Residential Cooling Load (RCL)

4.3 Residential Cooling Load

mgvßPresented By : A.M.Atiqullah INSTRUCTOR(Tech) RAC DHAKA POLYTECHNIC INSTITUTE,TEJGAON I/A Dhaka-1208

QUESTION ?

4_©-Aa¨v‡qicÖkœvejx t
01| wewìs m‡f© ev Rwi‡c Kywjs †jvW K¨vjKz‡jkb ej‡Z Kx eySvq ?
02| Kzwjs †jvW wbY©‡q evwn¨Kd¨v±i mg~n ‡jL?
03| w¯’iI cwieZ©bkxj †jvW mg~n eY©bvKi ?
04| AbvevwmK Gi†ÿ‡Î Kzwjs †jvW wbiæcb ej‡Z Kx †evSvq?
05| AvevwkK Gi †ÿ‡Î Kzwjs †jvW wbiƒcY c×wZ eY©bv Ki?

01 FEBRUARY'2020Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RAC
37

4.1 Calculate the intensity of direct radiation on a surface on the
following criteria. a) The component of direct radiation normal
to a horizontal surface. b) The component of direct radiation
normal to a vertical plan.
4.2 Calculate the glass area in the sun.
4.3 Calculate steady state heat transfer through glass.

67243 cooling and heating & calculation

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