More Related Content More from A.M. ATIQULLAH (20) 67243 cooling and heating & calculation2. G.O: 9. Understand the Calculation of Different Building Load.
wewfbœ fe‡bi †jvW K¨vjKy‡jkb m¤ú‡K©
aviYv
S.O: 9.1. Calculate the Cooling Load of a Class Room.
S.O: 9.2 Calculate the Cooling Load of a Library.
S.O: 9.3 Calculate the Cooling Load of Two Storage Building.
S.O: 9.4 Calculate the Cooling Load of Cold Storage.
Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE, 01711-056403
4. D`vniY t K¬v‡ki mvBR 6.5m6m3.8m ZvcgvÎvi cv_©K¨ 10℃, 20 Rb QvÎ I 1 Rb wkÿK,40w Gi 4 wU evwZ,`wÿY
†`qv‡ji ˆ`N©¨ 6.5m2 ‡Lvjv eviv›`v Ges 2m1.2m gv‡ci 2wU Kv‡Vi `iRv| c~e© †`qv‡j 1.67m1.2m 2wU Rvbvjv,hvi
Dc‡i 1/4Ask KvP| cÖvß Z_¨ Ges wb‡¤œ msM„nxZ DcvËmg~‡ni wfwˇZ K¬vkiægwUi Kywjs†jvW wbb©qKi|
mgvavbtDËi†`qv‡ji†ÿÎdjA=(6.53.8) = 24.7m2, cwðg†`qvjA = (63.8) = 22.8m2 , `wÿY†`qv‡ji, A
=(6.53.8)-4.8 = 19.9m2 , c~e©†`qv‡ji,A=(63.8)-3.006 = 19.8m2 , `wÿY w`‡KiKv‡Vi`iRvi
†ÿÎdj,A=(21.2)2 = 4.8m2, c~e©w`‡Ki Rvbvjvi†ÿÎdjA=(1.671.2)2 = 4.008m2, c~e©w`‡Ki
Kv‡PiRvbvjvi†ÿÎdjA=(4.0081/4) = 1.002m2 , c~e©w`‡Ki Kv‡ViRvbvjvi†ÿÎdjA=(4.008-1.002)
= 3.006m2
KvVv‡gvRwbZZvct1)Qs = A.U.TD
1) `wÿY†`qv‡ji‡jvWQ1 =19.91.98(32℃ − 𝟐𝟐℃) = 𝟑𝟗𝟒. 𝟎𝟐 Watt
2) cwðg†`qvjQ2 = (22.8 1.98 10) = 451.44 watt.
Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE, 01711-056403
U-factor w/m2 ‡iwW‡qkb
d¨v±র
DS
Factor
kixi n‡Z
wbM©ZZvc
‡fbwU‡jkb
‡iURbcÖwZ
‡fbwU‡jkb
d¨v±i
Qv` B‡Ui
‡`qvj
Kv‡Vi
`iRv
Kv‡Pi
Rvbvjv
QvÎ wkÿK
3.00 1.98 2.38 6.4 737 w/m2 0.43 117 j/sec 205 j/sec 3.5 lit/sec 60 j/lit
5. 3)c~e© †`qv‡ji †jvW, Q3 =(19.8 1.98 18) = 705.67 watt.
4)DËi †`qv‡ji †jvW, Q4 =(24.7 1.98 10) = 489.06 watt.
5) Qv` KZ©„K†jvW, Q5 =(6.5 6) 3 32 = 3744 watt.
6)`wÿ‡Y Kv‡Vi `iRvi ‡jvW, Q6 = (4.8 2.38) 10 = 114.24 watt.
7)cy‡e© Kv‡Vi Rvbvjvi ‡jvW, Q7 =(3.006 2.38) 10 = 71.54 watt.
8)cy‡e© Kv‡PiRvbvjvi ‡jvW, Q8 = (1.002 6.4) 10 = 64.128 watt.
KvVv‡gvRwbZZvc,QT = (Q1+ Q2 +Q3 +Q4 +Q5 +Q6 +Q7 +Q8) = 6034.098w
(L) Kv‡Pi wewKiYRwbZ Zvc= †gvU Kv‡Pi ‡ÿÎdj ‡iwW‡qkb d¨v±i DS d¨v±i
(M) wccj†jvW = gvby‡li msL¨v kixi n‡ZwbM©Z Zvc
Q = (20 117) + (1 205) =2545 Watt.
(N) †fw›U‡jkb †jvW =gvby‡li msL¨v †fw›U‡jkb †iU †fw›U‡jkb d¨v±i
Q = (20 1) 3.5 60= 4410 Watt.
Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE, 01711-056403
6. (O) ˆe`y¨wZK evwZRwbZ †jvW =evwZi msL¨v IqvU
e¨eü N›Uv
𝟐𝟒 N›Uv
Q = (4 40) =160 Watt.
myZivs †gvUKywjs †jvW= (QS + QG + QP +QV + QL ) = 13466.64Wat = 13.46 Kw.
wbivcËvRwbZ Kzwjs †jvW (Safety Factor) =(13.46 10%) = 1.35 Kw.
∴ me©‡gvU Kzwjs †jvW = (13.46 + 1.35) Kw. = 14.81 Kw.
= 4.14 TR. [∵ 1 TR = 3.57 Kw ]
A_©vr H †kÖYxK‡ÿ 4.14 U‡biGqvi KwÛkbvijvMv‡Zn‡e|
Page-03
Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE, 01711-056403
9. D`vniY n¨vÛ †vU ‡_‡K GKwU jvB‡eªwi K‡ÿi gvc 10.97m 6.7m 3.04m | Dˇi 3
wU, `wÿ‡b 4 wU Ges c~‡e© 2 wU K‡i 1.21 1.82m2 Gi Kv‡Pi Rvbvjv Av‡Q|
KÿwU‡Z GKwU 2.13 1.5m2 Gi Kv‡Vi `iRv Av‡Q| G Qvov KÿwU‡Z 12 wU
wUDejvBU, 8wU wmwjs d¨vb Ges 1wU K‡i †iwWI Ges †Uwjwfkb Av‡Q| GKmv‡_ 20
Rb wkÿv_x© covïbv Ki‡Z cv‡i | evB‡ii ZvcgvÎv 44℃ (WªvB evj¦ ZvcgvÎv) Ges wfZ‡ii
cÖvß ZvcgvÎv 18℃ ivL‡Zn‡e| wb‡¤œv³ Z_¨n‡Z Kzwjs †jvWwbY©q Ki|
Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE, 01711-056403
U-Factor W/m2
kixi n‡Z
wbM©Z Zvc wUDe jvBU wmwjs d¨vb
‡iwWI +
‡UwjwfkbQv` BU KvV KvP
3 1.98 2.38 6.4 117 J/Sec 60 Watt 65 Watt 1200Watt
10. mgvavbt
KvVv‡gviweeiY t
Dˇii Rvbvjvi †ÿÎdj = (1.21 1.82) 3 = 6.60m2
`wÿ‡Yi Rvbvjvi †ÿÎdj =(1.21 1.82) 4 = 8.80m2
c~‡e©i Rvbvjvi †ÿÎdj = (1.21 1.82) 2 = 4.40m2
∴ †gvURvbvjvi †ÿÎdj = (6.60 + 8.80 + 4.40) = 19.8m2
∴ †gvU`iRvi †ÿÎdj = (2.13 1.52) = 3.24m2
Dˇii ‡`qv‡ji †ÿÎdj =(10.97 3.04) – 6.60 = 26.75m2
`wÿ‡Yi ‡`qv‡ji †ÿÎdj = (10.97 3.04) – 8.80 = 24.55m2
c~‡e©i ‡`qv‡ji †ÿÎdj =(6.7 3.04) – 4.40 = 18.38m2
cwð‡gi ‡`qv‡ji †ÿÎdj = (6.7 3.04) = 20.37m2
∴ †gvU‡`qv‡ji †ÿÎdj = (26.75 + 24.55 + 18.38 + 20.37) = 90.05m2
∴ †gvUQv‡`i †ÿÎdj = (10.97 6.7) = 73.50m2
ZvcgvÎvicv_©K¨, = (44℃ – 18℃ ) = 26℃
Page-01
Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE, 01711-056403
11. Avgiv Rvwb,
(K) KvVv‡gvRwbZ Zvc, Qs = A.U.TD …….(1)
‡`qvj KZ©„K †jvW, Q1 = 90.05 1.98 (𝟒𝟒℃ − 𝟏𝟖℃) = 𝟒. 𝟔𝟑 Kw
Rvbvjv KZ©„K †jvW, Q2 = 19.8 6.4 26℃ = 𝟑. 𝟐𝟗 Kw
`iRvKZ©„K †jvW, Q3 =3.24 2.38 26℃ = 𝟎. 𝟐 Kw
Qv` KZ©„K †jvW, Q4 =73.50 3.00 26℃ = 𝟓. 𝟕𝟑 Kw
∴ ‡gvU KvVv‡gvRwbZ ‡jvW, Qs = (4.63 + 3.29 + 0.2 + 5.73) = 13.85 Kw
(L) gvbyl KZ©„K †jvW, Qp =(20 117)= 𝟐. 𝟑𝟒 Kw
(M) 12 টিটিউবলাইট কর্তৃ ক ললাড, QL =(12 60) = 𝟎. 𝟕𝟐 Kw
(N)8টি সিসলিংফ্যানকর্তৃ ক ললাড, QF =(8 65) = 𝟎. 𝟓𝟐 Kw
(O)‡iwWI Ges wUwf কর্তৃ ক ললাড, QR =1200W = 𝟏. 𝟐 Kw
∴ ‡gvU Kzwjs ‡jvW, QT = (13.85 + 2.34 + 0.72 + 0.52 + 1.2) = 18.63 Kw
Page-02
Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE, 01711-056403
12. 10% ‡mdwU d¨v±i = 18.63 10% = 1.863 Kw
myZivs me©‡gvU Kzwjs †jvW =(18.63 + 1.863)
= 20.493 Kw
=
𝟐𝟎.𝟒𝟗𝟑
𝟑.𝟓𝟕
[∵ 1TR = 3.57 Kw ]
=5.7 TR. [ TR = TON OF REFRIGERATION]
A_©vrH jvB‡eªwiK‡ÿ 5.7 U‡bi Gqvi KwÛkbvijvMv‡Zn‡e|
Page-03
Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE, 01711-056403
13. িমিযাবসলিঃ একটি দুইর্ল সবসিষ্ট লটাররজ ভবরনর অভযন্তরীণ মাপ
10m 8m 4m । H ভবরনর সনরের র্লা সিমাগার এবিং উপররর র্লা অসফ্ি কক্ষ সিরিরব বযবিার করা িয়।
সিমাগারর 1.5kg ওজরনর 2000 টি সেিংসি মারের পযারকট -20℃ র্াপমাত্রায় জমাট রাখরর্ িয়। লিই করক্ষ 10
জন শ্রসমক ও 10টি 45 ওয়ারটর বাসর্ জ্বাসলরয় দৈসনক 8 ঘন্টা কাজ করর। অপরসৈরক অসফ্ি করক্ষ 5 জন ললাক
4 টি 60 Watt এর বাসর্ ও 1 টি 150 Watt এর TV িবৃৈা োসলরয় কাজ করর। বাইররর র্াপমাত্রা 38℃
লেরক 25℃ সনরর্ িরল H দুই র্ল ভবরনর কুসলিং ললাড কর্ িরব? লেখারন অসফ্ি করক্ষ পূবৃ-পসিম সমরল
(1.67m 1.2m) মারপর দুটি কারের জানালা আরে এবিং উত্তররর লৈয়ারল সিমাগার ও অসফ্ি কক্ষ সমরল
(2m 1.2m ) মারপর দুটি কারের ৈরজা আরে।
প্ররয়াজনীয় র্েযাসৈিঃ
Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE, 01711-056403
কারের
লরসডরয়িন
ফ্যাক্টর
িরীর িরর্
সনগৃর্ র্াপ
সেিংসির
জমাঙ্কইট কাঠ কাচ
3.00 1.98 6.4 2.38 737 w/m2 117 j/s -2.25℃
14. িমাধানিঃ
০১। সিমাগাররর লক্ষরত্রিঃ
ৈরজার লক্ষত্রফ্ল= (2 1.2) = 2.4m2
লমাট লৈয়ারলর লক্ষত্রফ্ল = (10 4) 2 + (8 4) 2 – 2.4 = 141.6m2
∴লৈয়াল িরর্ সনগৃর্র্াপ, Q1 = 141.6 1.98 {38-(-20)}
= 16261.34 Watt.
কাে লেরক সনগৃর্র্াপ, =
S.O:9.3 Calculate the Cooling Load of Two Storage Building.
GKwU AvevwmK fe‡bi Kzwjs †jvW wba©viY
18. mgvßPresented By : A.M.Atiqullah INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE, 01711-056403
20. 9g- Aa¨v‡qi cÖkœvejx t
1. GKwU †kÖYx K‡ÿi Kzwjs †jvW wbY©q KiY cÖbvjx †jL?
2. GKwU jvB‡eªwii Kzwjs †jvW wbY©qKiY cÖbvjx avivevwnK fv‡e†j_?
3. GKwU AvevwkKfe‡bi Kzwjs †jvW wba©viY cÖbvjx †jL?
4. GKwU wngvMv‡ii Kzwjs †jvW wba©viY c×wZwjL ?
Presented By : A.M.Atiqullah INSTRUCTOR(Tech) RAC DHAKA POLYTECHNIC INSTITUTE,TEJGAON I/A Dhaka-1208
PresentedBy :A.M.AtiqullahINSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE,M # 01711-056403
22. S.O: 9.1 Survey a commercial building to be air conditioned.
S.O: 9.2 Calculate the following factor: Transmission load,
People load, solar heat load, Ventilation load, Infiltration load,
miscellaneous load, Equipment load, Equipment capacity.
Presented By : A.M.Atiqullah INSTRUCTOR(Tech)RACDHAKA POLYTECHNIC INSTITUTE, 01711-056403