67243 -cooling and heating load calculation

A.M.ATIQULLAH(আবু ম োহোম্মদ আতিকুল্যো)
INSTRUCTOR(TECH) RAC(ইন্সট্রোক্টর/আর এতি)
DHAKA POLYTECHNIC INSTITUTE,TEJGAON I/A Dhaka
ঢোকো পতল্টেকতিকইন্সটিটিউে,মিজগোাঁ ও তি/এ ঢোকো-
১২০৮
atiqullahrac@gmail.com
তবষয় মকোডঃ৬৭২৪৩,৪র্থ পবথ©
67243-CoolingandHeatingLoadCalculation
৬ষ্ঠ-অধ্যোয় 6th

1.BasicRefrigerationandAirconditioning-P.N.Ananthanarayanan
2.PrincipalofRefrigeration-RoyJ.Dossat
3.AtextbookofRefrigerationandAir-conditioning-R.SKhurmi&JKGupta
4.AcourseinRefrigerationandairconditioning-S.C.Arora&SDomkundwar
5.Modernrefrigerationandairconditioningforengineers-Prof.PSDesai
6.Refrigerationandairconditioning-CPArora
7.Refrigerationandairconditioning-AhmadulAmeen
PresentedBy :A.M.AtiqullahINSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,M #

িীিোিপ তিয়ন্ত্রণ কটের
তিিটরর িোপ িম্পটকথ ধ্োরণো
G.O:6.UnderstandtheInternalHeatLoadforAir
Conditioning.
অধ্যোয়-০৬
PresentedBy :A.M.AtiqullahINSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,M # 017-056403

Specific Objectives (S.O)
S.O: 6.1. Describe the Heat Gain from The Products.
(e¯‘ ev†cÖvWv±KZ…©K AwR©Z Zvcmg~n)
S.O: 6.2 Describe the Heat Gain from the Occupants/ Peoples.
(kxZvZc wbqwš¿Z ¯’v‡b Ae¯’vbKvix Ges Kg©iZ gvbyl †_‡K ZvcAR©b)
S.O: 6.3 Describe the Heat Gain from the Appliance.
(e¨eüZ Dcvsk ev hš¿vsk †_‡K AwR©Z Zvc)
S.O: 6.4 State the Importance of Chilling Rate Factor.
wPwjs †iU d¨v±i -Gi ¸iæZ¡
S.O: 6.5 Solve the Problems Related to Product,
People and Miscellaneous Load.
(e¯y‘, gvbylGes Avbylw½K †jvWm¤úwK©Z AsKevmgm¨vewjI mgvavb)

S.O:6.1. Describe the heat gain from the
products.
B›Uvibvj A_© n‡jv wfZ‡i, myZivs Gqvi KwÛkwbs B›Uvibvj wnU †jvW ej‡Z g~jZt
AeKvVv‡gvi wfZ‡ii †jvW mg~n‡K eySv‡bv n‡q‡Q| wbqwš¿Z K‡ÿi g‡a¨ †hgb-
wewfbœ ai‡bi Lv`¨mvgMÖx KZ©©„K , RbMb ev †jvKmsL¨vi †jvW , wewfbœ ai‡bi
Dcvsk, kw³ Drm/ hš¿vsk mg~n BZ¨vw` KZ©„K †jvWmgyn B›Uvibvj wnU †jv‡Wi
AšÍfz©³|

Gwm/kxZvZc wbqš¿Y K‡ÿ e¯‘ ev †cÖvWv±
KZ…„©K †h Zvc AwR©Z nq g~jZ Zv‡KB cÖWv±
†jvW e‡j| hLb †Kvb cÖWv± wngvwqZ ¯’v‡b
cÖ‡ek K‡i Ges hvi ZvcgvÎv wngvwqZ ¯’v‡bi
ZvcgvÎv A‡cÿv †ewk, ZLb D³ cÖWv± wngvwqZ
¯’v‡b Zvc cwiZ¨vM Ki‡Z _vK‡e, hZÿY bv ch©šÍ
cÖWv± VvÛv n‡q wngvwqZ ¯’v‡bi ZvcgvÎvi
mgvbnq|PresentedBy :A.M.AtiqullahINSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,M # 01711-056403

Gwm/kxZvZc wbqš¿Y K‡ÿ e¯‘ ev †cÖvWv± KZ„©K
†h Zvc AwR©Z nq g~jZ Zv‡KB cÖWv± ‡jvW
(Products Load) e‡j| cÖWv± ‡_‡K AwR©Z
Zv‡ci cwigvbwb‡¤œv³ mgxKiY Øviv wbY©qKiv nq
Q = m.c.TD
Q = Zv‡ci cwigvb(wK.IqvU)
m = e¯‘i fi (†KwR)
c = Avt Zvc (wK‡jvRyj/‡KwR)
TD = ZvcgvÎvi cv_©K¨ (Temperature Different)

Gwm iæ‡g Ae¯’vbKvix Ges Kg©iZ gvby‡li kixi
†_‡K wbM©Z Aby‡gq Zvc(‡mwÝej wnU) I myß
Zv‡ci(‡j‡U›U wnU) mgš^‡q Kzwjs †jvW MwVZ
nq|
GKwU Gwm/kxZvZc wbqš¿xZ ¯’v‡b KZRb †jvK
Dcw¯’Z _v‡K Zvi Dci wfwË K‡i Zv‡`i †_‡K
AwR©Z Zv‡ci cwigvb wbY©q Kiv nq| cÖwZwU
†jvK KZ…©K Drcvw`Z wnU †jv‡Wi cwigvb wbf©i
K‡iZv‡`i Kg©Kv‡ÛiDci|PresentedBy :A.M.AtiqullahINSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,M # 01711-056403

Office

kxZvZc wbqš¿xZ ¯’v‡b mPivPi †QvU Lv‡Uv
M„n¯’vwj hš¿cvwZe¨eüZn‡q _v‡K|‡hgb-
B‡jKwUªK AvqiY(Bw¯¿), †KUjx, wnUvi,
†nqvi Wªvqvi, Kwd †gKvi, †eøÛvi †gwkb,
gvB‡µvI‡qf I‡fb, wUwf, wd«R mn wewfbœ
cÖKvi B‡jKwUªK I B‡jK‡UªvwbK&ª
mvgMÖx|PresentedBy :A.M.AtiqullahINSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,M # 01711-056403

S.O: 6.3 Describe the Heat Gain from the Appliance
COMPUTER
DRYERS
WASHING
MACHING
TVs

wPwjs †iU d¨v±it †Kv‡bv e¯Íi kxZjvq‡bi mgq cÖ_g K‡qK N›Uv †cÖvWv±
†jv‡Wi nvi A‡cÿvK…Z †ewk _v‡K| KviY mvgMÖxi Avw` ZvcgvÎv Ges
wPwjsiæ‡giZvcgvÎvig‡aëëavbLye†ewk n‡q_v‡K| myZivswPwjs -Gi
cÖv_wgKch©v‡q†cÖvWv± †jvWcÖvq1.5¸Yev †ò¸Y†ewk_v‡K|
wPwjs †iU †jvW, Qc =
𝐦.𝐜.𝐂𝐋𝐓𝐃
wPwjsGi mgq × wPwjs †iU d¨v±i
wPwjs †iU d¨v±i=
𝟏
𝟏.𝟓
= 0.67
myZivs mvgMÖxi Zvc myÿfv‡e wbY©q Ki‡Z n‡j wPwjs †iU d¨v±i we‡ePbv Kiv DwPZ|
wPwjs†iU d¨v±i we‡ePbv bv Ki‡j†Kv‡bv e¯Í kxZjKivic~‡e©cÖPzimg‡q cÖ‡qvRb nq|
†mRb¨ e¯Í `ªæZ kxZj Ki‡Z mvgMÖxi Zv‡ci mv‡_ wPwjs †iU d¨v±i we‡ePbv Ki‡Z nq|
GUvig~jD‡Ïk¨n‡jvm‡e©v”PKzwjs†jvW, hv‡ZkxZjKi‡YinviwVKiv‡L|

chilling rate factor msµvšÍ mgm¨vejx t
300 †KwR gvQ GKwU wPjv‡i 35 wW.‡m. ZvcgvÎvq cÖ‡ek Kwi‡q -10 wW.‡m. ZvcgvÎvq 10
N›Uv kxZj Kiv n‡j gvQ †_‡K AcmvwiZ Zv‡ci cwigvb wbY©q Ki| (kxZjKi‡Yi mgq 15 N›Uv,
gv‡Qi Av‡cwÿKZvc315 wK‡jv Ryj/ ‡KwR wW. †Kjwfb, wPwjs †iU d¨v±i =0.68) .
‡`qv Av‡Q,
fi (m) = 300 kg, gv‡Qi Av‡cwÿK Zvc(c) = 315 kj/kg°k, mgq T = 15 N›Uv,
wPwjs †iU d¨v±i C.R,F = 0.68, ZvcgvÎvqcv_©K¨ TD = (T1 - T2 ) = 45℃
mgvavb t
Avgiv Rvwb,AcmvwiZZv‡ci cwigvb, Qc =
𝐦.𝐜.𝐓𝐃
wPwjs Gi mgq × wPwjs †iU d¨v±i
∴ Qc =
𝟑𝟎𝟎 ×𝟑𝟏𝟓 × {𝟑𝟓−(−𝟏𝟎)}
15 ×𝟔𝟎×𝟔𝟎 ×𝟎.𝟔𝟖
[ gvb ewm‡q]
myZivs AcmvwiZZv‡ci cwigvb = 115.81 kw (Ans)
S.O: 6.4 State the importance of chilling rate factor
wPwjs †iU d¨v±i - Gi ¸iæZ¡

1| cÖWv± ‡jvW/mvgMÖxi Zvc t cÖWv± †jvW ev mvgMÖxi Zvc ej‡Z
Avgiv mvaviYZ eywS †h, Ò‡Kv‡bv e¯‘y ev Lv`¨mvgMÖx wbw`©ó
ZvcgvÎvq VvÛv K‡i msiÿY Ki‡Z †h cwigvb Zvc Acmvi‡Yi cÖ‡qvRb nq,
Zv‡K cÖWv± ev mvgMÖxi Zvc e‡jÓ| wb‡¤œv³ m~‡Îi gva¨‡g cÖWv±
†jvWwbY©qKivhvqt-
01) cÖWv± †jvW,Q=
wbqwš¿ZK‡ÿcÖwZw`b cÖ‡ekK„ZmvgMÖxi cwigvb(‡KwR)× mvgMÖxi ZvcgvÎv
Kgvi cwigvb ( n«v‡mi)(wW.‡m.)×mvgMÖxi Av‡cwÿK Zvc(Ryj/‡KwR°‡Kjwfb) †Uwej Gi mvnv‡
86400 (𝟔𝟎 𝟔𝟎 𝟐𝟒)
02) e¯Zz/Product Load: Q = m.c.∆𝐓, A_ev Q =
m.c.∆𝐓
CwáZ kxZjxKiY mgq(‡m.)
‡bvUt hw` mvgMÖx‡K/cÖWv±‡K GK w`‡bi Kg mg‡q RgvU
Kivi K_v D‡jøL _v‡K Z‡e†jvW †m Abymv‡i evov‡Z n‡e|
mvgMÖxc¨v‡KUKivi c`v‡_©i GesmvgMÖx¯’vbvšÍ‡iih‡š¿i (†hgb-Uªwj)†jvW(Zvc)AšÍfz©³Ki‡Z n‡e|
PresentedBy : A.M.AtiqullahINSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,M # 01711-056403
S.O: 6.5 Solve the problems related to product, people and miscellaneous load
e¯y‘, gvbyl Ges Avbylw½K†jvWm¤úwK©Zmgm¨vewj

2| gvby‡li †jvW/ Poeople Load: wbqwš¿Z K‡ÿ hviv KvR K‡i Zv‡`i kixi †_‡K
wbm„Z Zvc G †jv‡WiAšÍf©y³| cÖwZwUgvby‡li Zvc Acmvi‡Yi ÿgZvmgvb bq|
Kv‡Riai‡YiDci Zvc Acmvi‡Yicwigvbwbf©iK‡i, hv†Uwe‡j†`Lv‡bvn‡q‡Q-
∴ Ae¯’vbKvixgvby‡li†jvW(IqvU)t
=
Ae¯’vbKvixi msL¨v × Ae¯’v‡bi mgq (N›Uv) × cÖwZ Ae¯’vbKvixi Zvc wbM©g‡bi cwigvb (IqvU)
24 N›Uv
A_ev,wccj †jvW= Ae¯’vbKviximsL¨v × cÖ‡Z¨‡Ki kixin‡Z wbM©ZZv‡ci cwigvb(IqvU)
GBm~‡Îigva¨‡gAwZ mn‡RB wbY©qKiv hvq|
∴ Q = Kg©iZ gvby‡limsL¨v× wnUBKzBf¨v‡j›U(wK.IqvU)×
Ae¯’vbKvj(N›Uv)
24 N›Uv

3| wewea/Avbylw½K †jvW/Miscellaneous Load: hLb BKzBc‡g›U
mg~n Aweivg bv P‡j ZLb wb‡Pi m~Î eënvi K‡i Avbylw½K †jvW wbb©q Kiv
nq-
Avbylw½K †jvW mgq =
BKyBc‡g›U AvDUcyU(IqvU) × BKzBc‡g›UPvjy _vKvi mgq (N›Uv)
24 N›Uv
‡bvU t hw`BKzBc‡g›U †jvW Ab¨vb¨†jv‡Wi Zzjbvq †ewknq, Z‡e-
Avbylw½K †jvW = BKzBc‡g›U AvDUcyU (IqvU)
1. ê`yÿwZK evwZKZ©„K AwR©ZZvc, Q =
WATT × cÖwZN›Uv
24 N›Uv
2.†gvUiKZ…©KAwR©ZZvc,Q = ‡gvU‡ii AvDUcyU(wK.IqvU) × d¨v±i ×
eë.mgq(N›Uv)
24 N›Uv
3. Kg©iZ gvby‡li kixi n‡Z wbM©Z Zvc wbb©‡qi m~ÎQ = RbmsL¨v Zv‡ci Zzj¨v¯‹(Kw) 
𝐡𝐨𝐮𝐫𝐬 𝐨𝐜𝐜𝐮𝐩𝐢𝐞𝐝
𝟐𝟒 𝐡𝐫𝐬
4. wngvqb h‡š¿i ÿgZv(Kw) = =
‡gvU Kywjs †jvW× 24 N›Uv
CwáZ evKvsw•LZ kxZj mgq(N›Uv)
(wK‡jv IqvU)

mgm¨vewjt 12m × 9m × 4m mvB‡Ri GKwU wngvMvi/‡Kvì †÷v‡iR 2.25kg IR‡bi
2,000 wU c¨v‡KU AvswkK wngvwqZ gyiwMi †MvkZ (3℃) n‡Z ( -20℃) ZvcgvÎvq Rwg‡q
msiÿ‡Yi Rb¨ 3 (wZb) Rb kÖwgK 200 watt Gi 2wU evwZ R¡vwj‡q ˆ`wbK 6 N›Uv KvR K‡i|
ˆ`wbK8 (AvU) N›UvqCwáZ/Kvsw•LZwngvq‡biRb¨kxZjh‡š¿i ÿgZvKZ n‡e wbY©qKi|
cÖ‡qvRbxq Z_¨vw` t
01.wngvMv‡ii evB‡iiZvcgvÎv = 33 wW.‡m.
02.gyiwMi Av‡cwÿK Zvc wngv‡¼i Dc‡i = 3.18wK‡jv Ryj/‡KwR-wW.‡Kjwfb|
03.gyiwMi Av‡cwÿK Zvc wngv‡¼i bx‡P = 1.55wK‡jv Ryj/‡KwR-wW.‡Kjwfb|
04.myß Zvc = 246 wK‡jv Ryj/‡KwR-wW.‡Kjwfb|
05 †`Iqvj,Qv` I †g‡SiBD d¨v±i = 0.383IqvU/ eM© wgUvi-wW.‡Kjwfb|
06.gy³ evZv‡mi nvi = 14 wjUvi/‡m‡KÛ|
07.evZv‡miewa©Z Zv‡ci cwigvb = 0.12 wK.Ryj/‡KwR|
08.gvby‡li kixi n‡Z wbM©Z Zvc = 407 IqvU/Rb|
09.gyiMxi RgvUv¼ ( - 2.25wW.‡m.)
S.O: 6.5 Solve the problems related to product, people and miscellaneous load.
e¯y‘, gvbyl GesAvbylw½K†jvW m¤úwK©Z

mgvavbt wngvMvi mvBRt (12m × 9m× 4m)
‡gvU cwigvb = 2.25 kg IR‡bi 2000 wU c¨v‡KU
1.†`Iqv‡ji†ÿÎdj={2(124)+2(94)}=(96+72)=168eM©wgUvi|
2. Qv` I †g‡Si †ÿÎdj = 2(129) = 216 eM© wgUvi|
3. †g‡S, Qv` IPvi ‡`qv‡ji ‡gvU †ÿÎdj = (168 + 216) = 384 eM©wgUvi|
4. †gvU †Mvk‡Zicwigvb = (2.25 × 2000) = 4,500 †KwR|
Avgiv Rvwb,wngvqbh‡š¿iÿgZv (Capacity of TR) =
‡gvU Kywjs†jvW× 24 N›Uv
CwáZ evKvsw•LZ kxZj mgq(N›Uv)
(wK‡jvIqvU)
e¯y‘, gvbyl Ges Avbylw½K †jvW m¤úwK©Z mgm¨vejx t

e¯y‘, gvbyl GesAvbylw½K †jvW m¤úwK©Z
(1)Zvccwien‡bi gva¨g (2)cwievwnZ Zvc (3)Zv‡cicwigvb
1|‡`Iqvj,Qv`,‡g‡S Q = A.U. CLTD
= 384 ×0.383 ×[33°-(-20°)]
= 7794.816 Watt =
𝟕𝟕𝟗𝟒.𝟖𝟏𝟔
𝟏𝟎𝟎𝟎
= 7.795 kw
7.795 Kwatt
2| gyiwMi cÖWv± †jvW
QT = Q1+Q2+Q3
1.Q = 4500×3.18×[3°-(2.25°)] = 10732.5 KJ {m~Ît Q=m.s.td}
2.Q = 4500×246 = 1107000.00 KJ {m~Ît Q= m.L}
3.Q = 4500×1.55[-2.25°-(-20°)] =123806.25 KJ{m~ÎtQ=
m.s.td}
Total QT = 1241538.75, QT =
𝐦.𝐜.𝑻𝑫
𝟐𝟒×𝟑𝟔𝟎𝟎
=
𝟏𝟐𝟒𝟏𝟓𝟑𝟖.𝟕𝟓
𝟐𝟒×𝟑𝟔𝟎𝟎
= 14.36 KW
14.36 KW
3|gvby‡li †`n n‡Z wbM©ZZvc Q =
𝟑×𝟒𝟎𝟕×𝟔
𝟐𝟒
= 305 Watt {m~Ît Q =
𝐏𝐞𝐨𝐩𝐥𝐞  𝐟𝐚𝐜𝐭𝐨𝐫 𝐓𝐢𝐦𝐞
𝟐𝟒 𝐡𝐫𝐬
} 0.305 KW
4| evwZ n‡Z wbM©ZZvc Q =
200×2×6
24
= 100 Watt {m~Ît Q =
Total Watt  Use of Time
24 hrs
} 0.1 KW
5| evZvm cwieZ©b RwbZZvc Q = 14× 0.12 = 1.68 KW {m~Ît Q = Liter/Sec 
factor}
1.68 KW
6|†mdwU d¨v±i ‡gvU †jv‡Wi Dci10% nv‡i = 2.42 KW ‡gvU=( 24.24+2.42)
me©‡gvUZvc= 26.66 KW
হিgvqb h‡š¿i ÿgZv =
ম োট কুহ িং ম োড  ২৪ ঘন্টো
কোিংহিত শীত স য় (ঘন্টো)
=
𝟐𝟔.𝟔𝟔 𝟐𝟒
𝟖
=
𝟔𝟑𝟗.𝟖𝟒
𝟖
= 79.98 Kwatt. Ans.

QUESTION ?

6ô- Aa¨v‡qi cÖkœvejx t
1. e¯‘ ev†cÖvWv± KZ…©K AwR©ZZvcmg~n eY©bvKi|
2. kxZvZc wbqwš¿Z¯’v‡b Ae¯’vbKvix GesKg©iZ gvbyl †_‡K ZvcAR©b †jL?
3. e¨eüZDcvsk ev hš¿vsk†_‡K AwR©ZZvcmg~n †jL?
4. wPwjs †iU d¨v±i -Gi ¸iæZ¡ †jL?
5. 5m × 5m× 3m mvB‡Ri GKwU wngvMvi/‡Kvì †÷v‡iR 2.25kg IR‡bi 3000wU
c¨v‡KUAvswkK wngvwqZ gyiwMi †MvkZ(3℃)n‡Z ( -20℃)ZvcgvÎvqRwg‡q msiÿ‡Yi
Rb¨ 3 (wZb) Rb kÖwgK 400 watt Gi 2wU evwZ R¡vwj‡q ˆ`wbK 7 N›UvKvR K‡i| ˆ`wbK
8 (AvU) N›UvqCwáZ/Kvsw•LZ wngvq‡bi Rb¨ kxZj h‡š¿i ÿgZvKZ n‡ewbY©q Ki|
Pjgvb... ..... .. ... ... .. .. .. .. ..
| 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

cÖ‡qvRbxq Z_¨vw` t
1. wngvMv‡ii evB‡ii ZvcgvÎv= 33 wW.‡m.
2. gyiwMi Av‡cwÿK Zvcwngv‡¼i Dc‡i = 3.18 wK‡jv Ryj/‡KwR-wW.‡Kjwfb|
3. gyiwMi Av‡cwÿK Zvcwngv‡¼i bx‡P = 1.55 wK‡jv Ryj/‡KwR-wW.‡Kjwfb|
4. myß Zvc= 246 wK‡jv Ryj/‡KwR-wW.‡Kjwfb|
5 †`Ihvj,Qv` I †g‡Si BDd¨v±i = 0.383 IqvU/eM© wgUvi-wW.‡Kjwfb|
6. gy³ evZv‡mi nvi = 14 wjUvi/‡m‡KÛ|
7.evZv‡mi ewa©Z Zv‡ci cwigvb = 0.12 wK.Ryj/‡KwR|
8. gvby‡li kixi n‡Z wbM©Z Zvc = 407 IqvU/Rb|
9. gyiMxi RgvUv¼ (- 2.25 wW.‡m.)
5. 5m × 5m× 3m mvB‡RiGKwU wngvMvi/‡Kvì †÷v‡iR 2.25kg IR‡bi 3000wU
c¨v‡KUAvswkK wngvwqZ gyiwMi †MvkZ(3℃)n‡Z ( -20℃)ZvcgvÎvqRwg‡q msiÿ‡Yi
Rb¨ 3 (wZb) Rb kÖwgK 400 watt Gi 2wU evwZ R¡vwj‡q ˆ`wbK 7 N›UvKvR K‡i| ˆ`wbK
8 (AvU)N›UvqCwáZ/Kvsw•LZ wngvq‡bi Rb¨ kxZj h‡š¿i ÿgZvKZ n‡ewbY©q Ki|

01 FEBRUARY'2020Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RACD.P.I
30

S.O: 6.1 Calculate the people’s load for a general office.
S.O: 6.2 Calculate the people’s load for a
Gymnasium/ Auditorium/ Restaurant.

67243 -cooling and heating load calculation

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