perhitungan cooling load pada ruang kelas
DESCRIPTION
Perhitungan Cooling Load Pada Ruang Kelas berdasarkan standar Menteri Lingkungan HidupTRANSCRIPT
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 1/22
PerhitunganCooling Load Pa
Ruang Kelas
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 2/22
Geometri Bangunan
Tinggi Bangunan (t) = 3,78 m = 4,!8! "t
Pan#ang $inding %e&elah 'tara (Pu) = 7,44 m = 8,"t
Pan#ang $inding %e&elah %elatan (P%) = 4,! m *,8
m = 7,44 m = 8,8 "t Pan#ang $inding %e&elah Timur (P T) = +,78 m ,*
= 7 m = 7+,34+ "t
Pan#ang $inding %e&elah Barat (P&) = 7 m = 7+,34+
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 3/22
Proerti dan Komonen Ruang Temeratur -indoor = 3*,8 oC = .,4 o/
outdoor = 34 oC = .3,* o/
Lamu Phili TL 3! 0att = ! &uah
Kursi = 4. &uah
1e#a = 4. &uah
Kaasitas Ruangan = 4. orang LC$ = &uah
La2ar LC$ = &uah
Paan Tulis = * &uah
Lantai Kerami dengan Te&al = * m
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 4/22
Luas 5endela Timur (6 5T)
A JT = Luas total seluruh jendela pada dinding t
= A JT1 + A JT2 + A JT3
= (,38 m ,87 m) 8 &uah9 (,38 m ,43+ m&uah9 (,!8 m ,8*+ m) &uah9
= *,!448 m* ,!!* m* ,38! m*
= 4,!.* m*
A JT = 50,503 ft2
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 5/22
Luas 5endela Barat (6 5B)
A JB = Luas total seluruh jendela pada dinding B
= A JB1 + A JB2
= (,.+ m ,37 m) &uah9 (,8 m ,*+ m&uah9
= ,7*+ m* !,!4* m*
= 7,3!3+ m*
A JB = 5!3""ft2
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 6/22
Luas Pintu (6P)
A# = Luas total seluruh pintu
= (,7 m *,4 m) * &uah9
= *,8+! m*
A# = 30,"1 ft2
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 7/22
Luas Ruangan
Luas $uangan = L % + L%%
= 7 m 4,! m9 +,78 m *,84 m9= 3*,* m* !,4+ m*
= 48,!+ m*
Luas $uangan = 523,22 ft2
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 8/22
Luas $indingLuas &inding 'tara (A&')
6$B = 3,78 m 7,44 m = *8,*4* m* = 302,0 ft2
Luas &inding *elatan (A&T)
6$%: = 3,78 m 7,44 m = *8,*4* m* = 302,0 ft2
Luas &inding *elatan (A&*)
= Luas *eluruh &inding *elatan (A&*) Luas Jendela
(A J*)
6$% = 3*,77 "t* ; *+, "t*
A&* = 2,5-. ft2
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 9/22
A&T: = 3,78 m 7 m = *!,4! m* = 2/",/05 ft2
Luas &inding Timur (A&T)
= Luas *eluruh &inding Timur (A&T+) Luas Jendela
(A JT)
6$T = *84,8+ "t* ; +,+3 "t*
A&T = 23",302 ft2
$inding Timur
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 10/22
Luas $inding Barat (6$B)
A&*: = 3,78 m 7 m = *!,4! m* = 2/",/05 ft2
Luas &inding Barat (A&B)
= Luas *eluruh &inding Barat (A&B) Luas Jendela (A JB) Barat (A#)
6$B = *84,8+ "t* ;7+<3474"t* ; 3,74 "t*
A&B = 1/,1.. ft2
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 11/22
Be&an Panas 6ta
CLTDCORR = {CLTDroof + (78 – t r ) + (t o – 85)}
CLT$CORR = *3 (78 ; .,4) (.3,* ; 8+)9 = 1/,1. o
ATot= 523,22 ft2
' = 0,0/2 Btuhr!ft2!oQatap = U . ATot . CLTDCORR
= ,8* Btuhr<"t*<o/ . +*3,*7* "t* < 8,! o/
Qatap = 779,215 t!"# $ 228,%88 &att
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 12/22
Be&an Panas $inding Perhitungan Resistansi Termal
$Total = Lapisan 'dara #ermuaan &inding Luar + #
(*emen dan #asir) + Batu Bata + #lester (*emen danLapisan 'dara #ermuaan &inding &alam
R Total = ,44 ,!! ,3 ,!! ,*9 m*< oK<0at
$Total = 0,"2.
' = 1$Total
= *,347 0attm*<oK
= 0,"23 Btuhr! ft2!o
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 13/22
'ntu dinding 2ang tertutui oleh &a2angan
Q = U . A . 'T (1!.)
dimana - ΔT = temeratur outdoor ; temeratur ind
sehingga nilai ΔT = 34 oC >.3,* o/? ; 3*,8 oC >.,4 o/?
= ,* oC >*,! o/?
'ntu dinding 2ang terena sinar matahari seara lang
Q = U . A . CLTDCORR
CLTDCORR = {CLTDa + (78 – t R ) + (t o – 85)} (1!
dimana - CLTDwall = Cooling Load Temerature $i@erene
dinding (Ta&le 7 Chater *+ 6%AR6 /undamental Aand&.77)
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 14/22
$inding Bagian %elatan (QD*), dinding
tertutui &a2angan
6$% = *77,+.! "t*
ΔT = *,! o/
$% = U . ADT . ΔT $% = ,4*3 Btuhr<"t*<o/ < *77,+.! "t* <
*,! o/
Q&* = 25%,% t!"# $ 7,%- &att
$inding Bagian Timur (QD*),
sinar matahari
6$T = *34,3* "t*
CLT$CORR = 7 (78 ; .,4)
CLT$CORR = *,! o/
$T = U . ADT . CLTDCORR
$T = ,4*3 Btuhr<"t*<o/ . *3
Q$T = 21,-77 t!"# $ 2,
$inding Bagian Barat(QD), dinding tertutui
&a2angan
6$B = !4,3*8 "t*
ΔT = *,! o/
$B = U . ADB . ΔT
$B = ,4*3 Btuhr<"t*<o/ . !4,3*8 "t* *,! o/
Q$B = 15-,15% t!"# $ %.9-7&att
&inding To
Q/0/0 = QD* + QD
Qdinding = *+3,!34 Btuh *
15!15" Btuh
Qdinding = .1,/." t!"# $
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 15/22
#erhitungan Be4an #anas &ari Jendela 5endela ter&uat dari &ahan single #at glass dengan te&al mm = ,+m< 'ntu erhitungan ada aa digunaan
se&agai &eriut - tida ada "ator eneduhan ada aa
aa terleta di se&elah timur, utara dan selatan
Perhitungan Resistansi Termal
$Total = $aa
R Total = ,7. m*< oK<0att
' = 1$Total = ,+! 0attm*< oK = 3,105 Btuhr! ft2!o
$am%ar & Penam'ang (endela Ru
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 16/22
Perhitungan Be%an Panas Dari (endela 5endela Bagian %elatan (Q *), tertutui
&a2angan
6 5% = *+, "t*
ΔT = *,! o/
5% = U . A (T . ΔT
5% = ,+! Btuhr<"t*<o/ < *+, "t* <
*,! o/
Q (* = %-,%7 t!"# $ 8,9-% &att
5endela Bagian Timur (Q T ),
sinar matahari
AJT = 50,503 ft2
CLTDCORR = {9 + (78 – 91,04) +
CLTDCORR = 4,16 oF
QJT = U . A JT . CLTDCORR
QJT = 0,56 Bt!"#$ft2$oF . 50,503
Q JT = 117,652 Btu/h ≈ 34,484 W
5endela Bagian Barat (Q ), tertutui &a2angan
6 5B = 7+<3474"t*
Q = U . A . 'T
5B = ,+! Btuhr<"t*<o/ . 7+<3474 "t* < +,! o/
Q (B = 91,1 t!"# $ 2.95 &att
5adi total erhitungan &e&an ana
Q jendela= QJS + QJT + QJB
= 30,374 Bt!" + 117,652 Bt!" + 9
Q %&'&* = 329,166 Btu/h ≈ 96.4204 Watt
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 17/22
#erhitungan Be4an #anas &ari $adiasi 6atahari6ele7ati 8aa
Q = A . *C . *34 .CL
o*# *-*t-o' B*.-*' T-/# (QST ),t-* ** f*to# &'&"
AJT = 50,503 ft2
F = 31 Bt!"$ft2
CLF = 0,89
C = 0,76
QST = A JT . SC . SHGF . CLF
= 50,503 ft2 $ 31 Bt!"$ft2 $ 0,89 $0,76
QST = 1058,967 Btu/ h ≈ 310,383 Watt
o*# *-*t-o' B*.-*' B*#*t (QSB ), &'&"
AJB = 75$347ft2
F = 180 Bt!"$ft2
CLF = 0,83
C = 0,76
QSB = A JB . SC . SHGF . CLF
= 75$347 ft2 $ 180 Bt!"$ft2 $ 0,8QSB = 8555,199 Btu/ h ≈ 2507,22W
Qrad = Q ST + Q SB
= 1058,967 Bt!" + 8555,199 Bt!"
Qrad = 9614,166 Btu/h ≈ 2817,59Watt
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 18/22
#erhitungan Be4an #anas &ari *istem #enerangan(Lighting)
Q = W. BF. CLF.
Q = 3! 0att < ,*+ < . 1, lam'uQ = 72- &att $ 25,99 t!"#
B/ = ,*+ (untu lamu TL, sum&er -6%AR6 /undamental Aand&oo)CL/ = (arena temeratur ruang set
oint2 mengalami enaian ada malamhari, sum&er - 6%AR6 /undamentalAand&oo)
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 19/22
#erhitungan Be4an #anas &ari ator8e4eradaan 6anusia &alam $uangan
Penam&ahan Panas %ensi&leQ* = 6- . *34 . CL
Q* = 4. < *3 Btuhr < ,8
Q* = 9-1 t!"# $
22,%%&att
Penam&ahan Panas LatenQL = 6- . L34
QL = 4. < . Btuh
QL = 9%1- t!"# $
2728.9&att
Q ,people = QS + Q L
= 9016 Bt!" +
Q !"#!$" = 18326 Btu/h ≈
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 20/22
#erhitungan Be4an #anas &ari ator #eralatan$uangan
&'*/*"*' *'* &'-&
Q S = S%& . CLF Q* = * 0att < ,8
Q* = 1- &att $ 55,889 t!"#
Penam&ahan Panas Laten
Q L = 0,32 ' Q ( )10*QL = ,3* < . Btuh
QL = -,8 t!"# $ 17,82- &at
Q"u!-"t = Q S + Q L
= 545,889 Bt!" + 17,820 Bt!"
Q"u!-"t = 563,709 Btu/h ≈ 165,223 Watt
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 21/22
#erhitungan Be4an #anas Ai4at &ari %n9
&'*/*"*' *'* &'-&
Q* = 1,1 . f . 'T
Q* = , < **7,43! "m < *,! o/Q* = 291,%88 t!"#r $ 85,79% &att
ΔT = 34 oC 93,2 oF – 32,8 oC 91,04 oF = &'*/*"*' *'* L*t&'
QL = -,7 : f : '&
QL = ,7 < **7,43! "m < ,+ o/
QL = 29,-% t!"# $ 125,917 &att Q$tra = Q S + Q L
= 2916,388 Bt!" + 42
Qn!lt"a# = 3345,991 Btu/h ≈ 980
4,122760
3,0238,245487 3
=×
= ft
cfm
7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas
http://slidepdf.com/reader/full/perhitungan-cooling-load-pada-ruang-kelas 22/22
Total Be4an #endinginan (:ooling Load) $uang 8elas :12"
Do<Komonen
Be&an Panas
sensi&le
Btuh >0att?
latent
Btuh >0att?
< 6ta77.,*+>**8,388? ;
*< $inding!7,8!4>8,8*+?
;
3< 5endela3*.,!!>.!,4*4?
;
4< Radiasi 1atahari.!4,!!>*87,+.?
;
+<Penerangan(Lamu)
*4+!,4.. >7*? ;
!<Ke&eradaan1anusia
.! >*!4<+!? .3 >*7*!<!7?
7< Peralatan +4+,88. >!? !,8>7,8*?
8< EnFltrasi*.!,388>8+4,7.3?
4*.,!3>*+,.7?
Total CoolingLoad
*33+8<7..>!84!<4!?
.8>*87<47?
J*- Tot* &*' &'-'.-'*' (oo-'.
***" ;
Q = Q + QL
= 23358.$99 + 9800 Btu/h
= 33158,$99 Btu
= 3,68 %&"#e '&(e"
= 3.$3 ')
*&te
1 HP = 9000 BT<
1 = 735$5 >*tt = 0$986 (o#&
a-a julah % " %andln nt an de"lu-an un
endnnan "uan adalah %/ denan daa 3,68 %'/').
da#a"an an ada u-u"an 7, , 1, 1 7, dan 2 '). a-a
da#an 2 ') #eana- 2 nt.