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Heat Transfer with Phase TransformationBy:Kikin Hermanto
2711100014Frizka Vietanti 2712100022Rahmandhika Firdauzha Hary
Hernandha 2712100040Panji Akbar Prasetya 2712100114Soni Aji Pradana
2712100126Muhyidin Abdulkhodir Malikurrahman 2712100146Case:
Influence of Carbon Contents and Phase Transformation to Total
Calor
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IntroductionHeat Transfer:
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Phase Transformation:Perubahan wujud zat adalah perubahan
termodinamika dari satu fase benda ke keadaan wujud zat yang
lain.Perubahan wujud zat ini bisa terjadi karena peristiwa
pelepasan dan penyerapan kalor. Perubahan wujud zat terjadi ketika
titik tertentu tercapai oleh atom/senyawa zat tersebut yang
biasanya dikuantitaskan dalam angka temperatur.
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Specific Heat CapacityThe amount of heat energy required to
raise the temperature of the body per unit of mass.In SI unit,
specific heat capacity(c) is the amount of heat in joules required
to raise 1 gram of a substance 1 K
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CaseM kg of mild steel AISI 1006, AISI 1050 and AISI 1090 are
heated until it reach 1250C, then it cooled until 100 C. Calculate
the total calor (Q) for each transformation based on cooling
curve
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0,06 % C0,5 % C0,9 % C0,09 % C
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Calor calculation
Q = mcT
Q : calor m : massc : specific heatT : temperature
differenceCalculation
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Specific Heat of Steel
Carbon SteelTemperature (C)Specific Heat (W/mK)
AISI 100612500,16917300,14821000,111
AISI 105012500,1667300,16001000,1125
AISI 109012500,16207300,16741000,1159
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AISI 1006 calor calculationQ1= mcT = M. 0,1691.1250 = 211,375 M
JQ2= mcT = M. 0,1482.730 = 108,186 M JQ3 = mcT = M.0,111.100 = 11,1
M J Qtotal = Q1 + Q2 + Q3 = 211,375M J + 108,186M J + 11,1M J =
330,661M Joule
Ferrit + PearlitT oF
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AISI 1050 calor calculationQ1= mcT = M. 0,166.1250 = 207,5 M
JQ2= mcT = M. 0,1600.730 = 116,8 M JQ3 = mcT = M.0,1125.100 = 11,25
M J Qtotal = Q1 + Q2 + Q3 = 207,5M J + 116,8M J + 11,25M J =
335,55M Joule
T oF
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AISI 1090 calor calculationQ1= mcT = M. 0,1620.1250 = 202,5 M
JQ2= mcT = M. 0,1674.730 = 122,202M JQ3 = mcT = M.0,1159.100 =
11,59M J Qtotal = Q1 + Q2 + Q3 = 202,5M J + 122,202M J + 11,59M J =
336,292M Joule
2T oFT oF
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Result
QAISI 1006AISI 1050AISI
1090Q1211,375207,5202,5Q2108,186116,8122,202Q311,111,2511,59Qtotal330,661335,55336,292
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ConclusionSo after we compare the result from AISI 1006, AISI
1050 and AISI 1090 total calor, we can conclude that Specific heat
capacity and carbon composition affected the value of the total
calor which released by steel. AISI 1090 have highest Calor to
transform from austenit to ferit+pearlite / pearlite+sementit