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

IntroductionHeat Transfer:

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.

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

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

0,06 % C0,5 % C0,9 % C0,09 % C

Calor calculation

Q = mcT

Q : calor m : massc : specific heatT : temperature differenceCalculation

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

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

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

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

Result

QAISI 1006AISI 1050AISI 1090Q1211,375207,5202,5Q2108,186116,8122,202Q311,111,2511,59Qtotal330,661335,55336,292

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