perhitungan cooling load pada ruang kelas

7/21/2019 Perhitungan Cooling Load Pada Ruang Kelas

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PerhitunganCooling Load Pa

Ruang Kelas



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+



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



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



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



Luas Pintu (6P)

A# = Luas total seluruh pintu

= (,7 m *,4 m) * &uah9

= *,8+! m*

A# = 30,"1 ft2



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



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



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



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



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



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



'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)



$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!"# $



#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



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



#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



#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)



#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 ≈



#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



#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



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.

perhitungan cooling load pada ruang kelas

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