Perhitungan HC

BAB IV

HASIL PENGAMATAN DAN PENGOLAHAN DATA

4.1. Hasil Pengamatan

4.1.1. Sistem Linier

Tabel 1. Hasil Pengamatan Sistem Linier

MaterialT

inQ T1 T2 T3 T4 T5 T6 T7 T8 T9 Diameter DX

Kuningan

Besar30 25 91,3 87,8 85,6 61 57,5 54,9 36 34,1 32 25 mm

10

mm

Kuningan

kecil30 25 91,3 87,8 85,6 78 76,5 74,9 37,1 35,2 33,2 13 mm

10

mm

Stainless

Steel30 25 91,3 87,8 85,6 58,7 57,8 56,9 34 32,5 31,1 25 mm

10 mm

4.1.2. Sistem Radial

Tabel 2. Hasil Pengamatan Sistem Radial

MaterialT

inQ T1 T2 T3

T

4

T

5

T

6

T7 T8 T9 DX

Kuningan 302

546,2 45,1 44,1 - - - 37,2 35,8 33,7

10

mm

Dimana :

Ro = 10 mm = 10-2 m

RL = 50 mm = 5. 10-2 m

L = 30 mm = 3. 10-2 m

4.2. Pengolahan Data

4.2.1. Sistem linear

a. Kuningan Besar

D = 25 mm = 25. 10-3 m

A = Luas (bernilai konstan)

A=14

π D2

A=14

π (25.10−3m)2

A=4 ,91. 10−4 m2

T4 = 61°C = 334 K

Ti1 = (T 3+T 4 )

2 =

85,6+612

= 73,3° = 346,3 K

Dengan rumus :

k=Q . ∆ xiA . ∆T

Maka harga K :

k 1= 25 W .10−2m4 , 91.10−4 m2 . (346,3−334 ) K

= 41,3955 W/m.K

k 2= 25 W .10−2 m4 , 91.10−4 m2 . (334−330,5 ) K

= 145,475 W/m.K

k 3= 25 W .10−2 m4 , 91. 10−4 m2 . (330,5−327,9 ) K

= 195,832 W/m.K

Ti2 = (T 6+T 7)

2 =

327,9+3092

= 318,45 K

k 4= 25 W .10−2m4 ,91.10−4 m2. (327,9−318,45 ) K

= 53,8798 W/m.K

Tabel 3. Hasil Perhitungan Menentukan Nilai k

No Q T1 (K) T2 (K)DT

(K)A (m2) DX (m) k (W/m.K)

1 25 334 346,3 12,3 4,9063. 10-4 10-2 41,3955

2 25 330,5 334 3,5 4,9063. 10-4 10-2 145,475

3 25 327,9 330,5 2,6 4,9063. 10-4 10-2 195,832

4 25 318,45 327,9 9,45 4,9063. 10-4 10-2 109,1455

k−

1 = k 1+k 2+k 3+k 4

4

k−

1 = 41,3955+145,475+195,832+109,1455

4

k−

1 = 109,1455 W/m.K

b. Kuningan Kecil

D = 13 mm = 13. 10-3 m


A=14

π D2

A=14

π (13.10−3m)2

A=1,327. 10−4 m2

T4 = 79°C = 351 K

Ti1 = (T 3+T 4 )

2 =

85,6+782

= 81,8°C = 354,8 K

Dengan rumus :

k=Q . ∆ xiA . ∆T

Maka harga K :

k 1= 25 W . 10−2 m1,327 .10−4 m2 . (354,8−351 ) K

= 495,7759W/m.K

k 2= 25 W . 10−2 m1,327 .10−4m2 . (351−349,5 ) K

= 1255,9658 W/m.K

k 3= 25W .10−2m1,327 .10−4 m2 . (349,5−347,9 ) K

= 1177,4679 W/m.K

Ti2 = (T 6+T 7)

2 =

347,9+310,12

= 329 K

k 4= 25 W . 10−2 m1,327. 10−4 m2 . (347,9−329 ) K

= 99,6798 W/m.K



(K)A (m2) DX (m) k (W/m.K)

1 25 351 354,8 3,8 1,327. 10-4 10-2 495,7759

2 25 349,5 351 1,5 1,327. 10-4 10-2 1255,9658

3 25 347,9 349,5 1,6 1,327. 10-4 10-2 1177,4679

4 25 329 347,9 18,9 1,327. 10-4 10-2 99,6798

k−

2 = k 1+k 2+k 3+k 4

4

k−

2 = 495,7759+1255,9658+1177,4679

4

k−

2 = 757,22 W/m.K

c. Stainless Steel

D = 25 mm = 25. 10-3 m


A=14

π D2

A=14

π (25.10−3m)2

A=4 ,91. 10−4 m2

T4 = 58,7°C = 331,7 K

Ti1 = (T 3+T 4 )

2 =

85,6+58,72

= 72,15°C = 345,15K

Dengan rumus :

k=Q . ∆ xiA . ∆T

Maka harga K :

k 1= 25 W .10−2 m4,91. 10−4 m2 . (345,15−331,7 ) K

= 37,856 W/m.K

k 2= 25W . 10−2m4,91. 10−4 m2 . (331,7−330,8 ) K

= 565,7388 W/m.K

k 3= 25W .10−2m4,91. 10−4 m2 . (330,8−329,9 ) K

= 565,7388 W/m.K

Ti2 = (T 6+T 7)

2 =

329,9+3072

= 318,45 K

k 4= 25W .10−2m4,91.10−4 m2 . (329,9−318,45 ) K

= 44,4685 W/m.K



(K)A (m2) DX (m) k (W/m.K)

1 25 331,7 345,15 13,45 4,91. 10-4 10-2 37,856

2 25 330,8 331,7 0,9 4,91. 10-4 10-2 565,7388

3 25 329,9 330,8 0,9 4,91. 10-4 10-2 565,7388

4 25 318,45 329,9 11,45 4,91. 10-4 10-2 44,4685

k−

3= k 1+k 2+k 3+k 4

4

k−

3 = 37,856+565,7388+565,7388+44,4685

4

k−

3 = 303,451 W/m.K


A = Luas (bernilai konstan) = (2)(3,14)(0,03m) = 0,1884 m

Q = bernilai konstan = 25 W

Ro = 10 mm = 10-2 = 0,01 m

RL = 50 mm = 5. 10-2 m = 0,05 m

L = 30 mm = 0,03 m

Dimana :

T4 = T 3−(T 3−T 7)

4= 44,1−

(44,1−37,2)4

= 42,375°C = 315,375 K

T5 = T 3−2 (T 3−T 7)

4= 44,1−

2(44,1−37,2)4

= 40,65°C = 313,65 K

T6 = T 3−3 (T 3−T 7)

4= 44,1−

3 (44,1−37,2)4

= 38,925°C = 311,925 K

Ti1 = (T 3+T 4 )

2 =

44,1+42,3752

= 43,2375°C = 316,2375 K

Dengan Rumus :

Q=2 πLk . ∆ T

ln [ RL

R0]

k=

Q ln [ RL

R0]

A . ∆ T

Maka harga K :

k 1=25W ln [ 0,05

0,01 ]0,1884 m. (316,2375−315,375 ) K

= 247,6076 W/m.K

k 2=25 W ln [ 0,05

0,01 ]0,1884 m. (315,375−313,65 ) K

= 123,8038 W/m.K

k 3=25 W ln [ 0,05

0,01 ]0,1884 m. (313,65−311,925) K

= 123,8038 W/m.K

Ti2 = (T 6+T 7)

2 =

38,925+37,22

= 38,0625°C = 311,0625 K

k 4=25W ln [ 0,05

0,01 ]0,1884 m. (311,925−311,0625) K

= 247,6076 W/m.K


No Q T1 (K) T2 (K) DT (K) A (2πL) DX (m) k (W/m.K)

1 25 315,375 316,2375 0,8625 0,1884 10-2 247,6076

2 25 313,65 315,375 1,725 0,1884 10-2 123,8038

3 25 311,925 313,65 1,725 0,1884 10-2 123,8038

4 25 311,062 311,925 0,8625 0,1884 10-2 247,6076

5

k−

4 = k 1+k 2+k 3+k 4

4

k−

4 = 247,6076+123,8038+123,8038+247,6076

4

k−

4 = 185,7057 W/m.K

Tabel 7. Hasil Perhitungan harga K setiap material dan sistem

Jenis Material Q (W) K (W/m.K)

1. Sistem Linier

a. Kuningan Besar

(A = 4,91 x 10-4 m2)

b. Kuningan Kecil

(A = 1,327 x 10-4 m2)

25

25

109,1455

757,22

c. Stainless Steel

(A = 4,91 x 10-4 m2)

25 303,451

2. Sistem Radial 25 185,7057

4.3. Mencari Harga Q secara Teori

4.3.1. Sistem Linier

Q= k . A . ∆ T∆ x

Dimana : ∆ T=T i 1−T i2

∆ x = 10 mm = 0,01 m

a. Kuningan Besar (A = 4,91 x 10-4 m2)

Untuk k = 244,6415 W/m.K dan ∆ T = (346,3-318,45)K = 27,85 K

Q=(244,6415

Wm . K )( 4,91. 10−4 m2 )(27,85 K )

0,01 m

= 334,5313 W

b. Kuningan Kecil (A = 1,327 x 10-4 m2)

Untuk k = 629,9865 W/m.K dan ∆ T=¿ (354,8-329)K = 25,8 K

Q=(629,9865

Wm . K )( 1,327.10−4 m2) (25,8 K)

0,01 m

= 215,6859 K

c. Stainless Steel (A = 4,91 x 10-4 m2)

Untuk k = 138,6605 W/m.K dan ∆ T = (345,15-318,45)K = 26,7 K

Q=(138,6605

Wm . K )( 4,91. 10−4 m2 )(26,7 K )

0,01 m

= 181,779 W


Q=2 πLk . ∆ T

ln [ RL

R0]

Dimana : L = 30 mm = 0,03 m

Untuk k = 254,7085 W/m.K dan T = 316,2375 – 311,0625 = 5,175 K

Q=2 (3,14 ) (0,03 m )(254,7085

Wm. K )(5,175 K )

ln [ 0,050,01 ]

= 154,3016 W

4.3.3. Tabel Hubungan k dan Q

Jenis Material T = Ti1 – Ti2 K (W/m.K) Q (W)

1. Sistem Linier

a. Kuningan Besar

A = 4,91 x 10-4 m2

27,85 109,1455 334,5313

b. Kuningan Kecil

A = 1,327 x 10-4 m2

25,8 757,22 215,6859

c. Stainless Steel

A = 4,91 x 10-4 m2

26,7 303,451 181,779

2. Sistem Radial 5,175 185,7057 154,3016

4.3.4. Persen Kesalahan Harga Q

% Kesalahan=|Q Teori−Q PraktekQTeori |x 100 %

a. Sistem Linier

Kuningan Besar

- Q praktek = 25 W

% Kesalahan=|334,5313 W −25 W334,5313W |x 100 % = 92,5268%

Kuningan Kecil

- Q praktek = 25 W

% Kesalahan=|215,6859 W −25 W215,6859W |x 100 % = 88,40907 %

Stainless Steel

- Q praktek = 25 W

% Kesalahan=|181,779 W −25 W181,779 W |x 100 % = 86,247 %

b. Sistem Radial

- Q praktek = 25 W

% Kesalahan=|154,3016 W −25 W154,3016W |x 100 % = 83,7979 %

4.4. Perhitungan Harga U untuk Tiap Logam

a. Sistem Linier

Kuningan Besar

A = konstan = π4

D2=π4

¿ = 4,91.10-4 m2

Tin = 30°C = 303 K

Tout = T 6+T 7¿ ¿2 =

54,9+362

= 45,45°C = 318,45 K

Xh = 3,5.10-2 m

Xs = Xc = 4,0.10-2 m

k h=Q x . X h

A ¿¿ =

25 W x3,5.10−2m(4,91. 10−4 m2 ) (364,3−303 ) K

= 29,0714 W/m.K

k s=Qx . X s

A ¿¿ =

25W x 4. 10−2 m(4,91. 10−4 m2 )(303−318,45)K

= - 131,8266 W/m.K

k h=Q x . X c

A ¿¿ =

25 W x 4.10−2m(4,91. 10−4 m2 ) (318,45−305 ) K

= 151,4245 W/m.K

1U

= Xhkh

+ Xsks

+ Xckc

=3,5. 10−2

29,0714− 4.10−2

131,8226+ 4. 10−2

151,4245

U = 490,13056 W/m.K

Kuningan Kecil

A = konstan = π4

D2=π4

¿ = 1,327.10-4 m2

Tin = 30°C = 303 K

Tout = T 6+T 7¿ ¿2 =

74,9+37,12

= 56°C = 329K

Xh = 3,5.10-2 m

Xs = Xc = 4,0.10-2 m

k h=Q x . X h

A ¿¿ =

25 W x3,5. 10−2 m(1,327.10−4 m2 ) (364,3−303 ) K

= 107,566 W/m.K

k s=Qx . X s

A ¿¿ =

25 W x 4. 10−2 m(1,327.10−4 m2 )(303−329) K

= - 289,838 W/m.K

k h=Q x . X c

A ¿¿ =

25W x 4. 10−2 m(1,327.10−4 m2 ) (329−306,2 ) K

= 330,5173 W/m.K

1U

= Xhkh

+ Xsks

+ Xckc

=3,5.10−2

107,566− 4.10−2

289,838+ 4. 10−2

330,5173

U = 1017,2898 W/m.K

Stainless Steel

A = konstan = π4

D2=π4

¿ = 4,91.10-4 m2

Tin = 30°C = 303 K

Tout = T 6+T 7¿ ¿2 =

56,9+342

= 45,45°C = 318,45 K

Xh = 3,5.10-2 m

Xs = Xc = 4,0.10-2 m

k h=Q x . X h

A ¿¿ =

25 W x3,5.10−2m(4,91. 10−4 m2 ) (364,3−303 ) K

= 29,0714 W/m.K

k s=Qx . X s

A ¿¿ =

25W x 4.10−2 m(4,91. 10−4 m2 )(303−318,45)K

= - 131,8266 W/m.K

k h=Q x . X c

A ¿¿ =

25 W x 4.10−2m(4,91. 10−4 m2 ) (318,45−304,1 ) K

= 141,9275 W/m.K

1U

= Xhkh

+ Xsks

+ Xckc

=3,5. 10−2

29,0714− 4.10−2

131,8226+ 4. 10−2

141,9275

U = 252,7056 W/m.K

b. Sistem Radial

k=

Q ln [ RL

R0]

A . ∆ T

L = 30 mm = 0,03 m

A = Luas = (2)(3,14)(0,03m) = 0,1884 m

Q = bernilai konstan = 25 W

Ro = 10 mm = 10-2 = 0,01 m

RL = 50 mm = 5. 10-2 m = 0,05 m

ln [ RL

R0] = ln [ 0,05

0,01 ] = 1,6094

Tin = 30°C = 303 K

Tout = T 6+T 7¿ ¿2 =

311,925+310,22

= 311,0625K

Xh = 3,5.10-2 m

Xs = Xc = 4,0.10-2 m

k h=

Q ln [ RL

R0]

A . (T 1−T ¿)=

25W .1,60940,1884 (319,2−303 ) K

= 13,1828 W/m.K

k s=

Q ln [ RL

R0]

A . ( T¿−T out )=

25 W .1,60940,1884 (303−311,0625 ) K

= -26,488 W/m.K

k c=

Q ln [ RL

R0]

A . (T out−T 9 )=

25 W . 1,60940,1884 (311,0625−306,7 ) K

= 48,9539 W/m.K

1U

= Xhkh

+ Xsks

+ Xckc

=3,5. 10−2

13,1828− 4.10−2

26,488+ 4. 10−2

48,9539

U = 561,685 W/m.K

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