Heat Transfer Exam Review? Heat Transfer Mechanisms Conduction Convection Radiation

Heat Conduction 1 Transfer of molecular motion No net mass transfer Iron bar – Hot end, Cold end Cold gas pump handle Hot aluminum foil Other examples

Heat Conduction 2 Relates Heat Flow (Q/t) to ΔT across material Similar problems: Stress/strain - relates strain (ΔL/Lo) to stress (F/A) and elastic modulus Thermal expansion - relates expansion relate ΔL/Lo to temperature and coefficient of thermal expansion Electrical conduction – relates current (I) to voltage (V) and conductivity/resistivity Equation involves usual suspects: Temperature difference Length Cross-sectional area Material property Power = Q/t = (kA/l)ΔT

Heat conduction equation 1 Heat Flow across material by conduction: 𝑃𝑜𝑤𝑒𝑟= 𝑄 𝑡 =𝑘 𝐴 𝑙 ∆𝑇 Power in Watts or J/s Area A in m2 Length l in m ΔT in C° or K Thermal conductivity in 𝐽 𝑠−𝑚−𝐶° 𝑜𝑟 𝑤𝑎𝑡𝑡𝑠 𝑚−𝐶° Always from hot to cold

Heat conduction equation 2 Heat Flow across material by conduction: 𝑃𝑜𝑤𝑒𝑟= 𝑄 𝑡 =𝑘 𝐴 𝑙 ∆𝑇 Power in Watts or J/s Area A in m2 Length l in m ΔT in C° or K C° or K Thermal conductivity in 𝐽 𝑠−𝑚−𝐶° 𝑜𝑟 𝑤𝑎𝑡𝑡𝑠 𝑚−𝐶°

Example 14-10 Heat conduction through glass 𝑄 𝑡 =𝑘 𝐴 𝑙 ∆𝑇 =(0.84 𝐽 𝑠 𝑚 𝐶°) 3 𝑚 2 0.0032 𝑚 (15°𝐶−14°𝐶) =788 𝐽 𝑠=788 𝑤𝑎𝑡𝑡𝑠 Heat conduction through “dead air layer” =(0.023 𝐽 𝑠 𝑚 𝐶°) 3 𝑚 2 0.0032 𝑚 (1 𝐶°) =2.2 𝐽 𝑠=2.2 𝑤𝑎𝑡𝑡𝑠 “Thermopane” – dry air sandwiched between double glass plate “R values” at hardware store

Problem 37 What must be heat flow across wall to maintain 20 C temperature difference 𝑄 𝑡 =𝑘 𝐴 𝑙 ∆𝑇 =(0.84 𝐽 𝑠 𝑚 𝐶°) 16 𝑚 2 0.12 𝑚 (30° 𝐶 −20° 𝐶) =2240 𝐽 𝑠=2240 𝑤𝑎𝑡𝑡𝑠 About 22-23 bulbs

Problem 39 – (1) Heat flow must be same, temperature differences must add. 𝑄 𝑡 𝐶𝑢 = 𝑄 𝑡 𝐴𝑙 𝑘 𝐴 𝑙 ∆𝑇 𝐶𝑢 = 𝑘 𝐴 𝑙 ∆𝑇 𝐴𝑙

Problem 39 – (2) Since lengths and cross-sectional areas same. (380 𝑤𝑎𝑡𝑡𝑠 𝑚 𝐶°) 250 °𝐶−𝑇 = (200 𝑤𝑎𝑡𝑡𝑠 𝑚 𝐶°) 𝑇 95,000−380 𝑇 𝐶°=200 𝑇 𝐶° 580 𝑇 𝐶°=95,000 𝑇=163.8° 𝐶

Convection Convection mechanisms Wind blowing in window Warm air rising in room Convection oven Land/sea breeze http://en.wikipedia.org/wiki/Sea_breeze

Electromagnetic Radiation Transfer by electromagnetic radiation Sun on a warm day Hand near a radiator Campfire Light bulb Electromagnetic Radiation Narrow broadcast radiation (microwave, radio, TV) Broad thermal radiation (blackbody radiation law)

Thermal Radiation – frequency vs temperature Variation with wavelength/frequency and temperature

Thermal Radiation animation

Thermal radiation imaging Boston Marathon Bomber - April 2013

Blackbody Radiation Law Object radiates across spectrum (animation) Total power radiated 𝑃𝑜𝑤𝑒𝑟= 𝑄 𝑡 =𝑒𝜎𝐴 𝑇 4 T (K) A (m2) σ 5.67 x 10-8 W/m2T4 e is emissivity ( shiny = 0, black =1) Good emitter = good absorber

Example 14-11 2-way radiated heat given off and absorbed 𝑃𝑜𝑤𝑒𝑟=𝑟𝑎𝑑𝑖𝑎𝑡𝑒𝑑 ℎ𝑒𝑎𝑡 𝑔𝑖𝑣𝑒𝑛 𝑜𝑓𝑓 −𝑟𝑎𝑑𝑖𝑎𝑡𝑒𝑑 ℎ𝑒𝑎𝑡 𝑎𝑏𝑜𝑠𝑟𝑏𝑒𝑑 =𝑒𝜎𝐴 𝑇 𝑎𝑡ℎ𝑙𝑒𝑡𝑒 4 −𝑒𝜎𝐴 𝑇 𝑤𝑎𝑙𝑙𝑠 4 = 0.7 5.67∙ 10 −8 𝑊 𝑚 2 𝑇 4 1.5 𝑚 2 307 𝐾 4 − 288 𝐾 4 =120 𝑊𝑎𝑡𝑡𝑠

Example 14-12 (radiation part) Estimate 0.75 L pot as cube 0.1 m on side Five 0.1 x 0.1 m sides exposed = area 0.05 m2 𝑃𝑜𝑤𝑒𝑟=𝑟𝑎𝑑𝑖𝑎𝑡𝑒𝑑 ℎ𝑒𝑎𝑡 𝑔𝑖𝑣𝑒𝑛 𝑜𝑓𝑓 −𝑟𝑎𝑑𝑖𝑎𝑡𝑒𝑑 ℎ𝑒𝑎𝑡 𝑎𝑏𝑜𝑠𝑟𝑏𝑒𝑑 =𝑒𝜎𝐴 𝑇 𝑝𝑜𝑡 4 −𝑒𝜎𝐴 𝑇 𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 4 =𝑒 5.67∙ 10 −8 𝑊 𝑚 2 𝑇 4 0.05 𝑚 2 368 𝐾 4 − 293 𝐾 4 =𝑒∙30 𝑊𝑎𝑡𝑡𝑠 For ceramic pot (e = 0.7) 𝑒∙30 𝑊𝑎𝑡𝑡𝑠=21 watts For shiny pot (e = 0.1) 𝑒∙30 𝑊𝑎𝑡𝑡𝑠=3 watts

Example 14-12 (calorimetry part) A pot losing a heat Q drops a temperature ΔT: 𝑄=𝑚𝑐∆𝑇 A pot losing a heat Q per second drops a temperature ΔT per second: 𝑄 ∆𝑡 =𝑚𝑐 ∆𝑇 ∆𝑡 ∆𝑇 ∆𝑡 = 𝑄 ∆𝑡 𝑚𝑐 = 𝑒 30 𝐽 𝑠 0.75 𝑘𝑔 4186 𝐽 𝑘𝑔 𝐶° =𝑒 ∙ 0.01 𝐶° 𝑠 The ceramic pot loses 0.007 C°/s and cools about 12° C in a half hour. (1800 s) The shiny pot loses 0.001 C°/s and cools about 2° C in a half hour. (1800 s)

Celsius vs. Kelvin Degree sizes same! Kelvin = Celsius with zero shifted to absolute zero. 𝑇 𝐾 =𝑇 𝐶 +273.15 May use Celsius when only relative changes important. Thermal Expansion Specific and Latent Heat problems Thermal Conduction Must use Kelvin when absolute temperature important. Ideal Gas Law Kinetic Theory Thermal Radiation First and Second Law Thermodynamics Similar to Gauge vs. Absolute Pressure