Kelompok 5 Jawaban Soal Ujian Lama

2) G. Rosanta Margaretha Sitohang (05021181419003) 3) Amrina (05021381419071)

4) Ilham Hartono (05021281419015)

= 0,49

m2.℃

W

36m2

= 0,014℃

W

Total Pindah Panas Q” = ∆ T

Rtotal

=T∝−Ti R¿tal

=

(30−20)℃

0,49m2.℃

W

=20,4W

m2=20,4W _{ Per satuan Luas}

Q total = Q” × A = 20,4W

m2×36m 2

=734,4W

Q= T∝−Tdinding dlm RKonv. luar−Rdindingdlm

734,4W=30℃−Tdinding dlm

0,011℃

W

8,08℃=30℃−T_dindingdlm

T_dindingdlm=21,92℃  22℃

Dik : Ti = 300°C

KP = 20 W/m.°C

D1 = 5 cm  r1 = 0,025 m

D2 = 5,5 cm  r2 = 0,0275 m

D3 = 8,5 cm  r3 = 0,0425 m

KI = 0,04 W/m.°C

T∞ = 25°C

h2 = 15 W/m2.°C

h1 = 90 W/m2.°C

A1=2π r1L=2π(0,025m) (1m)=0,157m2

A3=2π r3L=2π(0,0425m)(1m)=0,267m 2

Tahanan Termal Total

Rtotal = Rkonv1+RP+RI+Rkonv2

Pindah Panas Total

Qtotal = Qkonduksi pipa = Qkonduksi insulasi = Qkonveksi luar & dalam

A = 0,525¿

2

=0,87m2 π D2

=π¿

Tahanan Termal Rtotal =

r2−r1

4πK r1r2+

1

hA

=

(0,525−0,5)m

4π

(

0,0017 W

m. K

)

(0,525m)(0,5m)

+ 1

(

20 W

m2. K

)

(0,87m 2

)

= (4,46+0,06) K W

= 4,52 K

W

Q = Qkonduksi insulasi = Qkonveksi luar =

T_∝−Ti R_total=

(300−77)K

4,52 K

W

=49,34W=0,04934kJ

s

= 0,225

m2.℃

W

32m2

= 7,03℃

W

Total Pindah Panas Q” = ∆ T

Rtotal

=T∝−Ti Rtotal

=

(30−20)℃

7,03m2.℃

W

=W

m2=1,42W _{ Per satuan Luas}

Q total = Q” × A = 1,42W

m2×32m 2

=45,44W

Q= T∝−Tdinding dlm RKonv. luar−Rdindingdl m

45,44W=30℃−Tdindingdlm

0,004℃

W

0,18℃=30℃−T_dindingdlm

T_dindingdlm=29,82℃