Cooling and heating load calculation

A.M.ATIQULLAH(Avey†gvnv¤§`
AvwZKzj¨v)
INSTRUCTOR(TECH)
RAC(BÝUªv±i/AviGwm)
DHAKA POLYTECHNIC INSTITUTE,TEJGAON Dhaka.
XvKv cwj‡UK‡UKwbK BÝwUwUDU ‡ZRMuvI, XvKv-
67243 - Cooling and Heating Load Calculation
welq†KvW t 67243 (4_© ce©)
ÒAa¨vq-8gÓÒAóg-Aa¨vqÓÒAa¨vq-
8gÓ

S.O: 8.1 List the Components of Total Cooling Load for
Air Conditioning.
S.O: 8.2 State the Necessity of Safety Factor to
Calculate the Plant Capacity.
S.O: 8.3 State the Meaning of Plant Capacity of
Refrigeration And Air Conditioning Field.
S.O: 8.4 Describe the Procedure of Calculation of
Refrigeration And Air Conditioning Plant Capacity.
S.O: 8.5 Solve Problems Related to Total Cooling Load
of Air Conditioning Plant.
PresentedBy :A.M.Atiqullah,INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,M: 01711-056403

wngvqb I kxZvZc
wbqš¿Y cøv‡›Ui
K¨vcvwmwU m¤ú‡K©
aviYv
Aa¨vq -08

উপর োক্ত উৎরে তোরপ েমষ্টি ষ্টির্ণরে প এ উপ 5% হরত 10% সেফটি ফযোক্ট স োগ
ক রত হে। েকল তথ্য েঠিকভোরে ষ্টির্ণে ক ো িো সগরল সেফটি ফযোক্ট েোধো র্ত 10% ধ রত
হরে।
ষ্টহমোেি ন্ত্র োরত ষ্টপকরলোড আওেোর এেং পূর্ণ শীতলীক র্ ক্ষমতোে চলরত পোর সেজিয সেফটি ফযোক্ট স োগ
কর স পষ্ট মোি তোপ পোওেো োে, তোরক 24 ঘন্টো দ্বো ো গুর্(×) কর ষ্টহমোেি রন্ত্র দৈষ্টিক চোলি েমে দ্বো ো
ভোগ(÷) ক রল ষ্টহমোেি রন্ত্র ক্ষমতো(Capacity of refrigeration plant) পোওেো োে।
myZivs wngvqb h‡š¿i ÿgZv =
‡gvUKzwjs†jvW (Kw) 24 N›Uv
CwáZ wngvq‡bi mgq (N›Uv)
A_ev,UbAe†iwd«Rv‡ikb(TR)wngvqbh‡š¿iÿgZv=
‡gvU Kzwjs†jvW (wK‡jvIqvU)
3.516wK‡jvIqvU/Ub
Gwm cøv‡›Ui†gvUKzwjs†jv‡WiK‡¤úv‡b›Ub
mg~n

wewfbœ ai‡bi Drm †_‡K AwR©Z Zv‡ci mgwóB †gvU Kzwjs
†jvW (Total Cooling Load) | mvaviYZ †gvU Kzwjs †jv‡Wi
mv‡_ Avc`Kvjxb mg‡qi Rb¨ 5% ‡_‡K 10% ‡hvM Kiv nq|
‡iwd«Rv‡ikb I Gqvi KwÛkwbs cøv›U‡K mwVK I myôfv‡e
cwiPvjbviRb¨†mdwUd¨v±iAek¨Bai‡Zn‡e|
 wcK†jvWAvIqv‡icøv›UcwiPvjbvKi‡Z†mdwUd¨v±iwe‡klf‚wgKvcvjbK‡i|
 †mdwU d¨v±i bv ai‡j wcK AvIqv‡i cøv›U Pj‡ebv| mwVK gvÎvq
Kvh©Kwi†c‡Z n‡j†gvU†jv‡WiDci 10%†mdwUd¨v±i ai‡Z n‡e|
cøv›UK¨vcvwmwU‡Z†mdwU d¨v±‡ii
cÖ‡qvRbxqZv

Avi Gwm cøv‡›Uicøv›U K¨vcvwmwU K¨vjKz‡jkb wb‡¤œv³fv‡e Kiv nqt-
01| cÖ_g cøv‡›Ui†gvU Kzwjs †jvW wbY©q Ki‡Zn‡e|
02| †gvU Kywjs †jv‡Wi 10% †mdwU d¨v±i wn‡m‡e MY¨Ki‡Z n‡e|
03| Ae‡k‡l wngvqb h‡š¿iK¨vcvwmwU wbY©q Ki‡Z n‡e|
Avi Gwm cøv‡›Uicøv›U K¨vcvwmwU K¨vjKz‡jkb cÖ_‡g cÖ‡qvRb Heat
Transmission c×wZmg~n †ei Kiv|
1. Pvi †`qv‡ji gva¨‡g †jvW|
2. Qv‡`i gva¨‡g †jvW |
3. ‡g‡Si gva¨‡g †jvW|
4. `iRvI Rvbvjvi gva¨‡g †jvW|
5. cÖWv±†jvW (Avjy,gvQ,gvsm,kvKmewR BZ¨vw`)|
6. k¦vmcÖk¦vmRwbZ †jvW|
S.O: 8.3 State the Meaning of Plant Capacity of Refrigeration And Air Conditioning Field
AviGwmcøv‡›Uicøv›U K¨vcvwmwUK¨vjKz‡jkb

wewKiYRwbZ ZvcmÂvj‡b Zv‡cicwigvb ,
QR = Kv‡Pi†ÿÎdj  wewKiYRwbZ Zv‡ci¸Yv¼ D.S d¨v±i|
evqy Pjb ev AbycÖ‡ek †jvW (Ventilation or Infiltration Load),
Q = wbqwš¿Z iæ‡giRb¨cÖ‡qvRbxq gy³evZv‡mi cwigvb(L/S)  cÖwZwjUvievZv‡miRb¨ ewa©ZZv‡ci cwigvb(Kj/L)
A_ev,
Q =
K‡ÿAe¯’vbiZ RbmsL¨v  0.00354m3/sec/Rb
gy³ evZv‡miAv‡cwÿKAvqZb (m3/kg)
 ∆𝐡 [  ∆𝐡 = (T2-T1) cv_©K¨|]
wccj†jvW/Rbmgv‡ek†jvW t
Q = K‡ÿKg©iZRbmsL¨v IqvUGK‡KcÖZ¨‡Kikixi n‡ZwbM©ZZv‡cicwigvb|
cÖWv±†jvW,Q =
m.c.TD
24  3600
wK‡jvIqvU(Kw) , wPwjs †iUd¨v±i=
𝟏
𝟏.𝟓
= 𝟎. 𝟔𝟕
wPwjs †jvW,QC =
m.c.TD
wPwjs mgq(‡m‡KÛ) wPwjs†iU d¨v±i
,
m¤ú~iK†jvW(Supplementtary Load) 1.jvBU.2.†gvUi3.Ab¨vb¨cøvM†jvWmg~n|
Ae‡k‡l,CwáZev Kvw•LZkxZjwngvqbh‡š¿iÿgZvwbiƒcb
myZivswngvqbh‡š¿iÿgZv(Kw)=
‡gvU Kzwjs†jvW (Kw) 24
CwáZkxZjmgq (N›Uv)
Avi Gwm cøv‡›Ui cøv›U K¨vcvwmwU K¨vjKz‡jkb

WªvBevj¦ ZvcgvÎvi cwieZ©b(DBT) bvK‡iRjxqev®úhy³(Add)Kivi
c×wZ‡K Av`ª©KiYev wnDwgwWwd‡Kkb(Humidification) e‡j|
wnDwgwWwd‡Kkb`yBfv‡e Kiv nq, †hgbt-
1| cÖZ¨ÿc×wZ(Direct Method)Ges
2| c‡ivÿ ev AcÖZ¨ÿc×wZ(Indirect Method)|
1|cÖZ¨ÿ c×wZ(Direct Method) tG ai‡Yi c×wZ‡Z Gqvi KwÛkwbs ¯’v‡b
m¤úyb©iæ‡c myÿèvwZm~ÿè KYviAvKv‡i(Atomised State) cvwb †¯úª Kiv nq |
GUvLye†ewk Kvh©Ki bq|
2|c‡ivÿev AcÖZ¨ÿ c×wZ (Indirect Method) tG c×wZ‡Z Gqvi Iqvkv‡ii gva¨‡g Gqvi
KwÛkwbs cøv‡›UevZv‡m cvwb cÖ‡ek Kiv‡bvi ev mieiv‡ni e¨e¯’v _v‡K|

wnDwgwWwd‡Kkb cøv‡›Ui wnU †jvW
AR©b c×wZ
2
1
W2
W1
H2
H1
(H2 - H1) = L
(W1 – W2 )
L = td1 WªvB
evj¦ ZvcgvÎvq
ev®úxfe‡bi myß
Zvc
𝛟 = আরপষ্টক্ষক
আর্দ্ণ তো
W = স্পেসিসিক
সিউসিসিটি
H = এিথ্োলষ্টপ। 𝛟 𝟏
𝛟 𝟐

AsKt- 6m5m4m mvB‡RiGwm iæ‡gi Rb¨KZwK.IqvU†gwkbcÖ‡qvRb|hw`BDwbUˆ`wbK16 N›UvPvjy
ivL‡Z nq|Z_¨vw`t1|†Ìqv‡ji RbÜ= 0.346 W/m2 ° 𝐊,TD = 15 ° 𝐊,2|Qv‡ìRbÜ = 0.383 W/m2 ° 𝐊,
TD = 18° 𝐊,3|‡g‡SiRbÜ = 0.26 W/m2 ° 𝐊,TD = 12° 𝐊,4|Ae¯’vbiZ †jv‡KimsL¨v4 Rb|kixi†_‡K
wbM©Z Zvc=0.211KW, Bbwdj‡Uªkb†iU=8 wjUvi/‡m‡KÛ,evZvm cwieZ©bd¨v±i = 0.0186 kj/L, 40 W
Gi 2wU evwZ I 40w Gi 1wU†gvUi eëüZ nq|GKwU Kv‡ViìRv 2m  1.2m, U = 0.587 W/m2° k
mgvavbt- ìRvi †ÿÎdj =(2m  1.2m) = 2.4m2 ,
Pvi†`qv‡ji†ÿÎdj=2{(64) + (54)} – 2.4m2 = 85.6m2
‡g‡Si‡ÿÎdj = (65) = 30m2 ,
ছোরৈ ‡ÿÎdj= (65) = 30m2
1।কোঠোরমো মোধযরমপষ্ট েোষ্টহত তোপ, Q =A .U. TD
সৈেোল দ্বো ো = (85.6  0.346  15°) = 𝟒𝟒𝟒. 𝟐𝟔 𝐤𝐰
ìRv Øviv = (2.4  0.58  15 ) = 21.13 𝐤𝐰
Qv`Øviv= (30  0.383  18 ) = 206.82 𝐤𝐰
‡gvU=768.33 𝐤𝐰 Next Page …(02)

02| Bbwdj‡Uªkb †jvW =(8  0.186) = 1.488 kw
০৩।ষ্টপপল সলোড =(4  0.211)= 0.844 kw
০৪।েোষ্টিরমন্টোষ্ট সলোড =(40  2 + 40  1) = 120 watt.
= 0.12 k.watt
সমোট ষ্টিগণত তোপ, QT = (768.33 + 1.488 + 0.844 + 0.12) = 770.78 kw
10% ‡mdwU d¨v±i a‡i †gvU wbM©Z Zvc(770.78 + 77.07) = 847.85kw
∴ wngvqb h‡š¿i ÿgZv= (847.85 
𝟐𝟒
𝟏𝟔
)
=1271.77 wK‡jvIqvU (kw) Answer
evwK Ask

mgvß

QUESTION ?

8g- Aa¨v‡qi cÖkœvejx t
1.Gwmcøv‡›Ui †gvU Kzwjs†jv‡WiK‡¤úv‡b›Ub mg~n My‡jv †jL?
2. cøv›U K¨vcvwmwU‡Z †mdwU d¨v±‡ii cÖ‡qvRbxqZv †jL?
3. Avi Gwm cøv‡›Ui cøv›UK¨vcvwmwU K¨vjKz‡jkb c×wZ eY©bv
Ki?
4.wnDwgwWwd‡Kkbcøv‡›Ui wnU†jvW AR©bc×wZ
‡jL|
Presented By : A.M.Atiqullah INSTRUCTOR(Tech) RAC DHAKA POLYTECHNIC INSTITUTE,TEJGAON I/A Dhaka-1208PresentedBy :A.M.AtiqullahINSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,M # 01711-056403

01 FEBRUARY'2020Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RACD.P.I
25

1.1 Find out the value of K of common brick, wood, cellular
glass, corkboard, glass, expanded polyurethane, mineral wool
etc from the chart.
1.2 Find out the C value of sand aggregate, cinder aggregate,
tiles, plywood, and glass of different thickness.
1.3 Solve problems relating to conductance.

Cooling and heating load calculation

Recommended

Recommended

More Related Content

Similar to Cooling and heating load calculation

Similar to Cooling and heating load calculation (20)

More from A.M. ATIQULLAH

More from A.M. ATIQULLAH (20)

Recently uploaded

Recently uploaded (20)

Cooling and heating load calculation