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MSI Unit 3

MSI Unit 3

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  • 1. Thermal Property
  • 2. Thermal PropertyHeat capacityThermal Property – Response of material to application of heatManifestation – Rise in temperature and change in dimension.Temperature rise – Heat absorptionHeat capacity is the ability of a material to absorb heatHeat capacity, C, is defined as the amount of energy requiredto produce a unit temperature rise (J/mol-K or cal/mol-K), q is energy, T is temperature.Specific heat, c, is heat capacity per unit mass (J/kg-K orcal/g-K) Constant pressure or constant volume heat capacity, Cp andCv respectively.d Td qC 
  • 3. Thermal ConductivityIf there is a temperature gradient, heat will flow from higher tolower temperature region. This is Thermal conduction.The ability of a material to transfer the heat is the Thermalconductivity, k.d Xd Tkq q is steady state heat flux i.e.heat flow per unit area per unittime (W/m2)k is thermal conductivity of themedium (W/m-K)dT/dX is the thermal gradient inthe medium.
  • 4. Conduction MechanismAtoms vibrate about their equilibrium positions with highfrequency and low amplitudes. Amplitude increases with rise intemperature.The vibrations of adjacent atoms are coupled due to atomicbonding and this leads to generation of elastic waves whichmove through the lattice at the velocity of sound and thus carriesthe heat.Each quantum of the wave is known as phonon.
  • 5. Conduction MechanismFree electrons gain kinetic energy in the hotter region andmove towards the colder region thus transferring the heat.Therefore, thermal conductivity k = kl (lattice) + ke (electron)Since a large number free valence electrons are available inmetals, the electron mechanism is much more efficient. Thisimparts great thermal conductivity that metals are known for.Thermal and electrical conductivities in metals are related byWiedemann–Franz law: L= k/T, L is a constant,  is electricalconductivity.Ceramics do not have free electrons as all electrons are tightlybound in the atomic bonds and hence, are poor conductors ofheat.Polymers conduct heat by vibrational and rotational motion ofchain molecules and hence, are poor conductors of heat.
  • 6. Thermal ConductivityThermal conductivity of ceramics generally decreases withincreasing temperature due to phonon scattering. At very hightemperature it increases again due to change in heat transfermode from conduction to radiation.
  • 7. Determination of Thermal conductivityExperimental set upHeat generation/conduction unit with controlsThermocouples and Data loggerMaintain two ends of the sample at constant temperaturesInsert five to six thermocouples along the length at differentlocationsPlot temperature as a function of distance and find dT/dX fromthe slope.Find the thermal conductivity, k (q = - k dT/dX) [Experimentaltechniques to find q can be found athttp://www.physics.uoguelph.ca/~detong/phys3510_4500/Thermal%20expansion%20and%20conductivity.pdfhttp://www.the-three.net/documents/portfolio/thermal_conductivity_report.pdf
  • 8. Thermal ExpansionMost of the solids expand when heated. This is known asthermal expansion.It can be expressed asWhere, L is the change in length due to a temperature rise ofT. l is known as linear coefficient of thermal expansion (CTE).l (C-1) is a material property which depends on the type ofatomic bonding. The extent to which a material expands onheating will depend on its l .The atomic mechanism of thermal expansion can be viewedas an increase in the inter-atomic separation.Therefore, it will depend on the shape of the energy vs. inter-atomic distance curve.TLLol
  • 9. Thermal ExpansionThe energy and vibrational amplitude (width of the potentialenergy trough) increase with increasing temperature and sodoes the interatomic separation (indicated by the open circles).For a material with a broader potential curve, the increase inthe interatomic distance is more (Fig. a) and hence thermalexpansion is more.The increase in atomic distance and hence, the expansion ismuch lower for a deep and narrow potential trough (Ceramics).(a) (b)
  • 10. Low/Zero Thermal ExpansionCoefficient of thermal expansion (CTE) for metals is in therange of 5 x 10-6 – 25 x 10-6/C.For a typical ceramic like Al2O3 CTE = 7.6 x 10-6/CThere is a class of materials which have very low or near-zerothermal expansion.Invar (64Fe – 36Ni) has CTE of 1.6 x 10-6/C (Up to 230 C,the temperature can be increased by heat treatment).Super Invar (63Fe-32Ni-5Co) – 0.72 x 10-6/C.This is believed to be caused by magnetostriction aphenomena which lead to volume change on magnetization.
  • 11. Negative Thermal ExpansionSome materials contract on heating (negative CTE).Zirconium tungstate (ZrW2O8) for example contractscontinuously from 2 to 1050 K.A composite (mix) of positive and negative expansionmaterials may give rise to a zero expansion material.
  • 12. Thermal Properties of some materialsMaterial cp (J/kg-K) l (C-1 x 10-6) k (W/m-K)MetalsAlumnium 900 23.6 247Copper 386 17.0 398Silver 235 19.7 428Steel 502 16 15.9Super Invar 500 0.72 10CeramicsAlumina (Al2O3) 775 7.6 39Fused Silica (SiO2) 740 0.4 1.4Pyrex glass 850 3.3 1.4PolymersPolyethylene 1850 106 - 198 0.50Polystyrene 1170 90-150 0.13Teflon 1050 126-216 0.25
  • 13. Thermal StressThermal stresses arise due to –Constrained expansion or contraction e.g. heating orcooling a rod with fixed rigid ends.Uneven heating/coolingThermal expansion mismatch inside the solid.Thermal stress  due to temperature change from To to Tf = El(To – Tf) = ElTE is the elastic modulus.Upon heating, the stress is compressive and tensile whilecooling if the expansion/contraction is restrained.
  • 14. ExampleA steel rod is to be used with its ends held rigid. What is themaximum temperature the rod can be heated to without thecompressive stress in it exceeding 180 MPa. Elastic modulusof the rod E = 190 GPa.Solution:  -180 x 106 Pa (Compressive) = El(To – Tf)l for steel  14 x 10-6 C-1. To = room temperature = 25 CCT of9 21 01 41 01 9 01 01 8 02 5696 
  • 15. Thermal ShockThermal stresses might cause fracture in brittle materials likeceramics due to rapid heating or cooling if theexpansion/contraction is restrained. This is known as thermalshock.The ability of material to withstand such shocks is known asthermal shock resistance (TSR), f is the fracture stress.Thermal shock can be prevented by controlling the externalconditions like lowering heating and cooling rates andcontrolling the thermal/mechanical parameters such as CTEand fracture stress as per the equation above.lEkT S R f
  • 16. Referenceshttp://neon.mems.cmu.edu/rollett/27301/L8_therm_cond-Nov07.pdfhttp://www.engineersedge.com/properties_of_metals.htmhttp://www4.ncsu.edu/~pamaggar/403_Thermal.pdfwww.claisse.info/student/Powerpoints/1.3%20Thermal.ppthttp://www.cmse.ed.ac.uk/MSE3/Topics/ThermalProperties.pdfKey words: Thermal properties; Heat capacity; Thermalconductivity; Thermal expansion; Thermal shock
  • 17. Quiz1. What is heat capacity? What is specific heat?2. Briefly explain the mechanism of heat conduction in solids?3. What is phonon?4. Why do metals have good thermal conductivity?5. Why are ceramics poor conductors of heat?6. What is the origin of thermal expansion in solids?7. Why thermal expansion of ceramics is much lowercompared to metals?8. What kind of stresses will be developed if the ends of asolid are constrained while (i) heating (ii) while cooling?9. Is it possible to have zero or negative thermal expansion?10. What causes thermal shock?11. What is thermal shock resistance? How can it beimproved?
  • 18. Quiz12. A brass rod is to be used with its ends held rigid. What isthe maximum temperature the rod can be heated to fromroom temp without the compressive stress in it exceeding172 MPa. Elastic modulus of brass E = 100 GPa andl = 20 x 10-613. A 0.35 m long brass rod is heated from 15 to 85 C withits ends held rigid. Find out the magnitude and type of stressdeveloped if it was free of stress at 15 C. Elastic modulus ofbrass is 100 GPa and  of brass is 20 x 10-6/C