Perpindahan Panas Pada Dinding Lapis Bola - Operasi Perpindahan Kalor
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Presentasi mengenai materi perpindahan panas pada dinding lapis berbentuk bola. Pada presentasi ini memuat contoh-contoh kasus dari textbook dan sumber dari internet.
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Perpindahan Panas Pada Dinding Lapis Bola - Operasi Perpindahan Kalor
- 1. SLIDESMANIA.COM PERPINDAHAN PANAS KONDUKSI PADA DINDING LAPIS RANGKAP BERBENTUK BOLA (HEAT CONDUCTION THROUGH SPHERICAL WALLS) Kelompok VII Faiprianda Assyari Rahmatullah 1807111319 Niken Triani Klaway Reza 1807111284 Novia Yolanda 1807111316
- 2. SLIDESMANIA.COM CONTOH SOAL I Determine the heat transfer rate through a spherical copper shell of thermal conductivity k = 386 W/moC, inner radius ri = 2 cm, and outer radius ro = 6 cm if the inner surface is kept at Ti = 200oC and the outer surface at To = 100oC JAWAB π = βπ βπ π‘β = 4π(ππ β ππ) 1 ππ β 1 ππ π π = 386 π πππΆ ππ = 2 ππ = 0.02 m ππ = 6 ππ = 0.06 π ππ = 200β ππ = 100β π = 4(3.14)(200 β 100) 1 0.02 β 1 0.06 386 π = 14544.48 π Sumber: Necati Ozisik, 1985, Heat Transfer: A Basic Approach ri ro Ti To π = 14.54448 ππ
- 3. SLIDESMANIA.COM CONTOH SOAL III Consider a spherical container of inner radius ri = 8 cm, outer radius ro = 10 cm, and thermal conductivity k = 45 W/moC. The inner and outer surfaces of the container are maintained at constant temperatures of Ti = 200oC and To = 80oC, respectively, as a result of some chemical reactions occurring inside. Determine the heat transfer ! JAWAB π = βπ βπ π‘β = 4π(ππ βππ) 1 ππ β 1 ππ π π = 45 π πππΆ ππ = 8 ππ = 0.08m ππ = 10 ππ = 0.1 π ππ = 200β ππ = 80β π = 4(3.14)(200 β 80) 1 0.08 β 1 0.1 45 π = 27129.6 π π = 27.1296 ππ Sumber: Yunus A. Cengel, 2015, Heat and Mass Transfer
- 4. SLIDESMANIA.COM CONTOH SOAL IV A spherical container with thermal conductivity (k = 12.5 Btu/hr.ft.oF) with 10 in inside diameter (di) and 12 outside diameter (do) is covered by 3 in layer of asbestos insulation have thermal conductivity (k = 0.14 Btu/hr.ft.oF). If the inside wall temperature is 550oF and outside temperature is 100oF, calculate the heat transfer! JAWAB π1 = 12.5 π΅π‘π’ βπ. ππ‘. ππΉ π2 = 0.14 π΅π‘π’ βπ. ππ‘. ππΉ π1 = 5 ππ = 0.42 ππ‘ π2 = 6 ππ = 0.5 ππ‘ π3 = 6 ππ + 3 ππ = 9 ππ = 0.75 ππ = 550ππΉ ππ = 100ππΉ π = βπ βπ π‘β = 4π(ππ β ππ) 1 π1 β 1 π2 π1 + 1 π2 β 1 π3 π2 π = 4(3.14)(550 β 100) 1 0.42 β 1 0.5 12.5 + 1 0.5 β 1 0.75 0.14 π = 1179.372 π΅π‘π’/βπ Sumber: Frank Keith, 2011, Principles of Heat Transfer