Can any body helpme to get thermal conductivity of material ASME SA556 / ASTM A556... thank you for your help?
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I don't know what you plan on using this information for but I'm sure it it very similar to other steels. For many calculations small errors in the number have little or no practical effect on the answer.\n\nMy heat transfer book gives the following values:\n\nMild steel 36 Btu/(hr)(sqft)(degF/ft) @ 32 degs F\nMild steel 33 Btu/(hr)(sqft)(degF/ft) @ 212 degs F\n\n1% Carbon steel 26.2 Btu/(hr)(sqft)(degF/ft) @ 32 degs F\n1% Carbon steel 25.9 Btu/(hr)(sqft)(degF/ft) @ 212 degs F
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Just use the formula\n\nE = m*Cp*(Delta T)\n\nE=energy\nm=mass\nCp= Specific Heat\nDelta T = change in temperature\n\nCp= 0.51998 J/(C*g)\n\nGood Luck with the test!
545704
Well, get out your heat transfer text book and start calculating. You should have all the info you should need.
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i think wikipedia is the best for the purpose\n\nhttp://en.wikipedia.org/wiki/Thermal_energy
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http://lakeaccess.org/russ/conductivity.htm
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(m)\n\ngoto the following website for your question with neat diagrams\n\nhttp://hyperphysics.phy-astr.gsu.edu/HBASE/electric/gaulaw.html
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for an ohmic material:\n\nP=VI\nV=IR\nso\nP=I^2 * R\n\ncheers!
737811
You use the formal \nH= mcdeltat\n\nmanipulate it to be...\n\nc (specific heat capactiy) = H /(mdelta t)\n\nChange temperature to Kelvin,....4degrees = 277Kelvin\n\nC=221/(55X277)\n\n\nC= 221/15235\n\nC=0.0145 (in JOULES)
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go check http://www.asci.org/
44841
A good place to start will be to look at the following websites \n\nThermal Energy and States of Matter\nhttp://www.pschweigerphysics.com/thermenergy.html\n\nThermal Energy\nhttp://sol.sci.uop.edu/~jfalward/physics17/chapter6/chapter6.html\n\nThermodynamics and Thermal Physics guides\nhttp://physics.about.com/cs/thermodynamics/\n\nIf your doing a project you might look into "Ocean Thermal energy conservation" as a topic.\n\nOcean Thermal energy conservation is used to generate electricity using the temperature difference of seawater at different depths
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Try Thermo.com/neslab contact them with any questions you have.
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Q = m.s.ΔT\nHeat loss by the copper = heat gained by the metal.\nCopper (heat loss) = 200.0g x 0.38J/g.ºC x (195.5ºC-228.4ºC) = -2500.4J\ndivide the heat by the mass and ΔT of the metal to get the specific heat of the metal:\n2500.4J/(65.0g x (195.5ºC-18.3ºC)) = 0.2171 = 0.21 J/g.ºC (correct SF)
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Heat capacity = q/(delta)T\n\nC= 135J/(54°C-25°C)\nC= 135J/25°C\nC= 5.4 J/°C
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Read in Detail at this site\nhttp://books.google.com/books?vid=ISBN0768917425&id=Ca58uytAJoAC&printsec=toc&dq=mechanical+engineering%3F+CIVIL+%26+ELECTRICAL%3F
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Here is a link to a "Polymer Physical Property Comparison Chart."\nhttp://www.lauren.com/Technical_Info/physical_properties.asp
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Maybe this will help (Lexan LS2)\n\nCoefficient of linear thermal expansion (parallel) \n0.7 E-4/°C
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This has been here for a while without answer.....\nSoft PVC is a plastic, and therefore would not be a good conductor of heat.