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Page 1: Bahan Ajar Forced Convection

Internal Forced Convection

Heat Transfer15/11/2013

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Mean Velocity and Mean Temperature

οΏ½Μ‡οΏ½=πœŒπ‘‰π‘š 𝐴𝑐

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Laminar and Turbulent Flow In Tubes

𝑅𝑒=πœŒπ‘‰π‘šπ·h

πœ‡ =π‘‰π‘šπ·h

𝑣

Re < 2300 laminar flow2300 < Re < 10000 transitional flowRe > 10000 turbulent flow

Turbulency

πΆπ‘–π‘Ÿπ‘π‘’π‘™π‘Žπ‘Ÿβ†’π·h=𝐷

π‘π‘œπ‘›βˆ’πΆπ‘–π‘Ÿπ‘π‘’π‘™π‘Žπ‘Ÿβ†’π·h=4 𝐴𝑐

𝑝

Hydraulic Diameter

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Entrance Length

Hydrodynamic / thermalEntrance region

Fully Developed Region

Lh,laminar β‰ˆ 0.05 ReD

Lt,laminar β‰ˆ 0.05 RePrDLh,turbulent β‰ˆ Lt,turbulent β‰ˆ 10D

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Thermal AnalysisTi Te

Ts

𝑄=�̇�𝐢𝑝 (π‘‡π‘’βˆ’π‘‡π‘– )𝑄=h 𝐴𝑠 βˆ†π‘‡ π‘™π‘š

βˆ† 𝑇 π‘™π‘š=(π‘‡π‘ βˆ’π‘‡π‘’ )βˆ’(π‘‡π‘ βˆ’π‘‡π‘–)

ln (π‘‡π‘ βˆ’π‘‡π‘’π‘‡π‘ βˆ’π‘‡π‘– )𝑇𝑒=π‘‡π‘ βˆ’(π‘‡π‘ βˆ’π‘‡π‘–)𝑒

(βˆ’ h 𝐴 𝑠

�̇�𝐢𝑝)

- Constant Surface Temperature- Constant Surface Heat Flux

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Laminar Flow

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Constant Surface Heat Flux 𝑁𝑒=hπ·π‘˜ =4.36

Constant Surface Temperature 𝑁𝑒=hπ·π‘˜ =3.66

Fully Developed

Entry Region

𝑁𝑒=3.66+0.065(𝐷𝐿 )π‘…π‘’π‘ƒπ‘Ÿ

1+0.04 [(𝐷𝐿 )π‘…π‘’π‘ƒπ‘Ÿ ]2 /3

Laminar Flow

*non-circularSee Table 8-1Heat Transfer a Practical Approach, Cengel Y.

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Laminar FlowPressure Drop

βˆ†π‘ƒ= 𝑓 𝐿𝐷

πœŒπ‘‰π‘š2

2

Friction Factor

𝑓 =64𝑅𝑒

Pump Power

π‘Šπ‘=Λ™π‘šβˆ† π‘ƒπœŒ

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Turbulent Flow

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Turbulent Flow

𝑁𝑒=0.023𝑅𝑒0.8π‘ƒπ‘Ÿ 1/3

𝑁𝑒=0.023𝑅𝑒0.8π‘ƒπ‘Ÿπ‘›

Colburn equation

Dittus-Boelter equation

n = 0.4 for heatingn = 0.3 for cooling

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Turbulent FlowPressure Drop

βˆ†π‘ƒ= 𝑓 𝐿𝐷

πœŒπ‘‰π‘š2

2

Friction Factor

1𝑓 =βˆ’2.0 log ΒΏΒΏ

Relative roughness

πœ€π·β†’π‘€π‘œπ‘œπ‘‘π‘¦ π‘‘π‘–π‘Žπ‘”π‘Ÿπ‘Žπ‘š

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Latihan

Minyak dengan temperatur 20oC mengalir pada sebuah pipa dengan diameter 30-cm dan kecepatan rata-rata 2 m/s. Sebagian pipa dengan panjang 200 m mengalir melalui danau beku dengan temperatur 0oC. Pengukuran menunjukan temperatur permukaan pipa berada pada nilai konstan 0oC. Mengabaikan nilai hambatan termal dari pipa, tentukan

a) Temperatur keluar minyak saat mengalir meninggalkan danau

b) Laju perpindahan kalor dari minyak ke luarc) Daya pompa untuk mengatasi pressure drop

Properti minyakρ = 888 kg/m3

k = 0.145 W/moCv = 901 x 10-6 m2/sCp = 1880 J/kg.oCPr = 10400

200 m

20oC2 m/s

0oCTe

D= 0.3 m

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Latihan

0.2 m

0.2 m

8 m

Udara1 atm80oC

Te

Ts = 60oC

Udara panas dengan temperatur 80oC memasuki duct dengan panjang 8-m dengan potongan penampang 0.2 x 0.2 m dengan laju 0.15 m3/s. Duct berada pada kondisi isothermal dengan permukaan temperatur 60oC. Tentukan temperatur keluar dari udara dan laju kehilangan kalor dari duct ke lingkungan

Properti udaraρ = 0.9994 kg/m3

k = 0.02953 W/moCv = 2.097 x 10-5 m2/sCp = 1008 J/kg.oCPr = 0.7154