HEAT TRANSFER MECHANISMHEAT TRANSFER MECHANISM

1.1. CONDUCTIONCONDUCTION

Q = k.A.x

T

(W)

Rcond = Ak

x

.

(oC/W)

HEAT TRANSFER MECHANISMHEAT TRANSFER MECHANISM

2.2. CONVECTIONCONVECTION

Q = h.A(Ts - T) (W)

Rconv = Ah.

1 (oC/W)

HEAT TRANSFER MECHANISMHEAT TRANSFER MECHANISM

3.3. RADIATIONRADIATION

Q = As 4sT (W)

CONDUCTIONCONDUCTION

CONDUCTIONCONDUCTION

CONDUCTION (SOLUTION)CONDUCTION (SOLUTION)The roof of an electrically The roof of an electrically heated home is 6 m long, heated home is 6 m long, 8 m wide, and 0.25 m 8 m wide, and 0.25 m thick, and is made of a flat thick, and is made of a flat layer of concrete whose layer of concrete whose thermal conductivity is k thermal conductivity is k == 0.8 W/m.°C. The 0.8 W/m.°C. The temperatures of the inner temperatures of the inner and the outer surfaces of and the outer surfaces of the roof one night are the roof one night are measured to be 15°C and measured to be 15°C and 4°C, respectively. 4°C, respectively. Determine the rate of Determine the rate of heat loss through the roof heat loss through the roof that night.that night.

CONVECTION, Q = h A (TCONVECTION, Q = h A (Tss – – TT))

CONVECTION (EXAMPLE)CONVECTION (EXAMPLE)A 2-m-long, 0.3-cm-diameter electrical wire extends A 2-m-long, 0.3-cm-diameter electrical wire extends across a room at 15°C, as shown in Fig. 1–33. Heat across a room at 15°C, as shown in Fig. 1–33. Heat is generated in the wire as a result of resistance is generated in the wire as a result of resistance heating, and the surface temperature of the wire is heating, and the surface temperature of the wire is measured to be 152°C in steady operation. Also, the measured to be 152°C in steady operation. Also, the voltage drop and electric current through the wire voltage drop and electric current through the wire are measured to be 60 V and 1.5 A, respectively. are measured to be 60 V and 1.5 A, respectively. Disregarding any heat transfer by radiation, Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient determine the convection heat transfer coefficient for heat transfer between the outer surface of the for heat transfer between the outer surface of the wire and the air in the room.wire and the air in the room.

CONVECTION (SOLUTION)CONVECTION (SOLUTION)

• P = VI = (60 V)(1.5 A) =P = VI = (60 V)(1.5 A) = 90 W 90 W

• AAss = = DL = DL = (0.003 m)(2 m) =(0.003 m)(2 m) = 0.01885 0.01885

• Q = h A (TQ = h A (Tss – T – T))

RADIATION RADIATION

)( 44surrs TTQ

4... ss TAQ

RADIATION (EXAMPLE)RADIATION (EXAMPLE)Consider a person standing in Consider a person standing in a room maintained at 22°C at a room maintained at 22°C at all times. The inner surfaces all times. The inner surfaces of the walls, floors, and the of the walls, floors, and the ceiling of the house are at an ceiling of the house are at an average temperature of 10°C average temperature of 10°C in winter and 25°C in in winter and 25°C in summer. Determine the rate summer. Determine the rate of radiation heat transfer of radiation heat transfer between this person and the between this person and the surrounding surfaces if the surrounding surfaces if the exposed surface area and the exposed surface area and the average outer surface average outer surface temperature of the person temperature of the person are 1.4 mare 1.4 m22 and 30°C. and 30°C.

RADIATION (SOLUTION)RADIATION (SOLUTION)

SIMULTANS MODESIMULTANS MODE

Konveksi = Q1, Q5, Q8Konveksi = Q1, Q5, Q8

Konduksi = Q2, Q6Konduksi = Q2, Q6

Radiasi = Q3, Q4, Q7Radiasi = Q3, Q4, Q7

COMPOSITE WALLCOMPOSITE WALL• Q = Q = TTtottot/R/Rth,totth,tot

• RRth,totth,tot = R = Rth,Ath,A + R + Rth,Bth,B + R + Rth,Cth,C

COMPOSITE WALLCOMPOSITE WALL

Q = Q = TTtottot/R/Rth,totth,tot

RRth,totth,tot = Jabarkan (tugas) = Jabarkan (tugas)

COMPOSITE WALL + COMPOSITE WALL + CONVEKSICONVEKSI

COMPOSITE WALL + COMPOSITE WALL + CONVEKSICONVEKSI

SILINDER dan BOLASILINDER dan BOLA

SILINDER dan BOLASILINDER dan BOLA

Bola :Bola :

Silinder Silinder ::

SILINDER dan BOLA SILINDER dan BOLA KompositKomposit

SILINDER KompositSILINDER Komposit

= (T= (T11 – T – T22)/R)/Rtotaltotal

SOAL LATIHANSOAL LATIHAN

• Sebuah lapisan serat kaca tebalnya Sebuah lapisan serat kaca tebalnya 13 cm. Beda suhu antara kedua 13 cm. Beda suhu antara kedua permukaan adalah 85 permukaan adalah 85 ooC. C. Konduktivitas termal serat kaca Konduktivitas termal serat kaca adalah 0.035 W/m.adalah 0.035 W/m.ooC. Hitunglah C. Hitunglah kalor yang dipindahkan melalui kalor yang dipindahkan melalui bahan itu per jam per satuan luas.bahan itu per jam per satuan luas.

JAWABJAWAB

• Diketahui serat kacaDiketahui serat kaca

• X = 13 cm; delta T 85 X = 13 cm; delta T 85 ooC; C; Konduktivitas termal 0.035 W/m.Konduktivitas termal 0.035 W/m.ooC. C.

• Q/A = 22.8 W/mQ/A = 22.8 W/m22 = 22.8 W.h/m = 22.8 W.h/m22

•1.4 W/m.oC.

SOAL LATIHANSOAL LATIHAN• Sebuah dinding dilapisi isolasi Sebuah dinding dilapisi isolasi

yang memiliki konduktivitas yang memiliki konduktivitas termal 1.4 W/m.termal 1.4 W/m.ooC setebal 2,5 C setebal 2,5 cm. Suhu bagian dalam isolasi cm. Suhu bagian dalam isolasi adalah 315 adalah 315 ooC. Suhu udara luar C. Suhu udara luar 38 38 ooC. Dinding melepaskan kalor C. Dinding melepaskan kalor ke lingkungan secara konveksi. ke lingkungan secara konveksi. Hitunglah nilai koefisien Hitunglah nilai koefisien perpindahan panas konveksi (perpindahan panas konveksi (hh) ) agar suhu permukaan luar agar suhu permukaan luar lapisan isolasi menjadi 41 lapisan isolasi menjadi 41 ooC.C.

JAWABJAWAB

• Q = Q konduksi = Q konveksiQ = Q konduksi = Q konveksi

• Q konduksi = k.delta T/ x = 15344 Q konduksi = k.delta T/ x = 15344 W/mW/m22

• h = (Q/A)/T = 5115 W/mh = (Q/A)/T = 5115 W/m2 O2 OC.C.

SOAL LATIHANSOAL LATIHAN• Salah satu sisi dinding mempunyai Salah satu sisi dinding mempunyai

suhu 100 suhu 100 ooC, sedang sisi lainnya C, sedang sisi lainnya berada dalam lingkungan konveksi berada dalam lingkungan konveksi dengan T = 10 dengan T = 10 ooC dan h = 10 C dan h = 10 W/mW/m22..ooC. Dinding memiliki nilai k = C. Dinding memiliki nilai k = 1.6 W/m.1.6 W/m.ooC dan tebalnya 40 cm. C dan tebalnya 40 cm. Hitunglah laju perpindahan kalor Hitunglah laju perpindahan kalor melalui dinding.melalui dinding.

JAWABJAWAB

• R1 (konduksi) = 0.4/1.6 = 0.25R1 (konduksi) = 0.4/1.6 = 0.25

• R2 (koveksi) = 1/10 = 0.1R2 (koveksi) = 1/10 = 0.1

• R tot = R1 + R2 = 0.35R tot = R1 + R2 = 0.35

• Q = delta T/ R tot = 90/0.35 = 257 Q = delta T/ R tot = 90/0.35 = 257 WW

KONDUKSI DG SUMBER KONDUKSI DG SUMBER KALORKALOR• BIDANG DATARBIDANG DATAR

• SILINDERSILINDER

• BOLABOLA

KONDUKSI DG SUMBER KONDUKSI DG SUMBER KALORKALOR

KONDUKSI DG SUMBER KONDUKSI DG SUMBER KALORKALOR

APLIKASIAPLIKASI• RADIUS KRITIS ISOLASIRADIUS KRITIS ISOLASI

SIRIP (NEXT)SIRIP (NEXT)

SIRIP PADA RADIATORSIRIP PADA RADIATOR

CONTOH SIRIPCONTOH SIRIP

• LIHAT TABEL 3-4LIHAT TABEL 3-4

SIRIP SANGAT PANJANGSIRIP SANGAT PANJANG

ADIABATIC FIN TIPADIABATIC FIN TIP

PANJANG SIRIP PANJANG SIRIP TERKOREKSITERKOREKSI

EFISIENSI SIRIPEFISIENSI SIRIP

EFISIENSI SIRIPEFISIENSI SIRIP

EFISIENSI SIRIPEFISIENSI SIRIP

EFISIENSI SIRIPEFISIENSI SIRIP

EFECTIVITAS SIRIPEFECTIVITAS SIRIP

EFISIENSI VS EFEKTIVITASEFISIENSI VS EFEKTIVITAS

KONDUKSI DUA DIMENSIKONDUKSI DUA DIMENSI

• FAKTOR BENTUK KONDUKSI (S) : FAKTOR BENTUK KONDUKSI (S) : LIHAT TABEL 3-5LIHAT TABEL 3-5

• Q = k.S.Q = k.S.TT

KONDUKSI TRANSIENKONDUKSI TRANSIEN

• Steady (Tunak): variabel tidak Steady (Tunak): variabel tidak berubah sebagai fungsi waktu.berubah sebagai fungsi waktu.

• Transient (Fana): variabel berubah Transient (Fana): variabel berubah sebagai fungsi waktu.sebagai fungsi waktu.

• Konduksi transien: konduksi yang Konduksi transien: konduksi yang berubah menurut waktu.berubah menurut waktu.

PENDEKATAN SOLUSIPENDEKATAN SOLUSI

• Menggunakan metode sistem Menggunakan metode sistem tergabung (lumped system), dimana tergabung (lumped system), dimana “temperature of such bodies can be “temperature of such bodies can be taken to be a function of time only, taken to be a function of time only, T(t)”T(t)”

• Parameter: Angka Biot (Bi)Parameter: Angka Biot (Bi)

• Kriteria Bi Kriteria Bi 0,1 0,1

PRINSIPPRINSIP

• TT sebagai sebagai fungsi fungsi tt

Prosedur Prosedur

• Hitung panjang karakteristik, LcHitung panjang karakteristik, Lc

• Hitung Angka Biot, BiHitung Angka Biot, Bi

• dandan

• CONTOH: CONTOH: Bola Bola TembagaTembaga

A 7.5 cm diameter orange is subjected to A 7.5 cm diameter orange is subjected to a cold environment. Assuming that the a cold environment. Assuming that the orange has properties similar to those of orange has properties similar to those of water at 20 °C and that h = 11 W/mwater at 20 °C and that h = 11 W/m2 o2 oC, C, determine the suitability of a lumped determine the suitability of a lumped analysis for predicting the temperature analysis for predicting the temperature of the orange during cooling. of the orange during cooling.

• Solution: From Table B-3 (SI), the Solution: From Table B-3 (SI), the thermal conductivity of water at 20 thermal conductivity of water at 20 ooC is C is 0.597 W/m-K. Also, for a sphere 0.597 W/m-K. Also, for a sphere