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 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

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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
𝑞 =
∆𝑇
∑𝑅𝑡ℎ
=
4𝜋(𝑇𝑖 − 𝑇𝑜)
1
𝑟𝑖
−
1
𝑟𝑜
𝑘
𝑘 = 386
𝑊
𝑚𝑜𝐶
𝑟𝑖 = 2 𝑐𝑚 = 0.02 m
𝑟𝑜 = 6 𝑐𝑚 = 0.06 𝑚
𝑇𝑖 = 200℃
𝑇𝑜 = 100℃
𝑞 =
4(3.14)(200 − 100)
1
0.02
−
1
0.06
386
𝑞 = 14544.48 𝑊
Sumber: Necati Ozisik, 1985, Heat
Transfer: A Basic Approach
ri
ro
Ti
To 𝑞 = 14.54448 𝑘𝑊

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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
𝑞 =
∆𝑇
∑𝑅𝑡ℎ
=
4𝜋(𝑇𝑖 −𝑇𝑜)
1
𝑟𝑖
−
1
𝑟𝑜
𝑘
𝑘 = 45
𝑊
𝑚𝑜𝐶
𝑟𝑖 = 8 𝑐𝑚 = 0.08m
𝑟𝑜 = 10 𝑐𝑚 = 0.1 𝑚
𝑇𝑖 = 200℃
𝑇𝑜 = 80℃ 𝑞 =
4(3.14)(200 − 80)
1
0.08
−
1
0.1
45
𝑞 = 27129.6 𝑊
𝑞 = 27.1296 𝑘𝑊
Sumber: Yunus A. Cengel, 2015,
Heat and Mass Transfer

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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
𝑘1 = 12.5
𝐵𝑡𝑢
ℎ𝑟. 𝑓𝑡. 𝑜𝐹
𝑘2 = 0.14
𝐵𝑡𝑢
ℎ𝑟. 𝑓𝑡. 𝑜𝐹
𝑟1 = 5 𝑖𝑛 = 0.42 𝑓𝑡
𝑟2 = 6 𝑖𝑛 = 0.5 𝑓𝑡
𝑟3 = 6 𝑖𝑛 + 3 𝑖𝑛 = 9 𝑖𝑛 = 0.75
𝑇𝑖 = 550𝑜𝐹
𝑇𝑜 = 100𝑜𝐹
𝑞 =
∆𝑇
∑𝑅𝑡ℎ
=
4𝜋(𝑇𝑖 − 𝑇𝑜)
1
𝑟1
−
1
𝑟2
𝑘1
+
1
𝑟2
−
1
𝑟3
𝑘2
𝑞 =
4(3.14)(550 − 100)
1
0.42
−
1
0.5
12.5
+
1
0.5
−
1
0.75
0.14
𝑞 = 1179.372 𝐵𝑡𝑢/ℎ𝑟
Sumber: Frank Keith, 2011, Principles of Heat Transfer

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Terima Kasih!