11_Heat_Transfer tmb rev

11_Heat_Transfer tmb rev - Heat Transfer ENES 101 / 101H /...

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Unformatted text preview: Heat Transfer ENES 101 / 101H / 101Y Introductory Engineering Science Modes of Heat Transfer There are three modes to transfer energy: CONDUCTION CONVECTION RADIATION All heat transfer processes involve one or more of these three modes Conduction Heat transfer results from a temperature gradient within a homogeneous substance (solid, liquid or gas) If the temperature profile is linear within the medium then T T2 - T1 = x x 2 - x1 Temperature T1 T2 x1 x2 Distance Such linearity always exists in a homogeneous medium of fixed k during steady state heat transfer k experimental measurement of thermal conductivity Conductive heat transfer rate The rate of heat transfer via conduction is given by the Fourier Law qx heat transfer rate in the xdirection (W or Btu/hr) o K thermal conductivity (W/m*K or Btu/ft* F) A area normal to the direction of the heat flow (m2 or ft2) T temperature gradient in the xdirection (K/m or x o F/ft) T q x = -kA x The minus sign is required from the second law of thermodynamics which states that heat transfer must take place from warmer to colder. Fourier's Law T - kA(T2 - T1 ) - kA(T2 - T1 ) q x = -kA = = x x 2 - x1 x T1 - T2 qx = thermal potential difference x thermal resistance kA Resistance to heat flow Directly proportional to material thickness Inversely proportional to thermal conductivity Inversely proportional to area perpendicular to heat transfer Example #1a Consider onedimensional, steady state heat transfer across a material whose thermal conductivity is 0.058 Btu/(hr ft F) a) What material is this? Btu 1 W 1hr 1ft 1. 8 o F 0.058 o hr ft F 9.486 x10 - 4 Btu 3600s 0.3048m 1K s = 0.1003 W/m*K material is wood, pine perpendicular to grain (p.56) o Example #1b a) Suppose A = 5ft , T =150 F, T =30 F. If x = 0.1 ft, how much heat is being transferred per - (T2 - T1 ) - (30 - 150 )o F Q= = hour? x 0.1ft kA Q = 348 T2 2 1 o 2 o Btu 2 0.058 5ft o hr ft F ( ) T1 Btu hr Q A to Q x Example #1c a) Suppose A = 5ft , T =150 F and x = 0.1 ft. What value of T corresponds to a - (T2 - T1 ) 2 x value of Q = 400 Btu/hr? kA Qx T2 = T1 - kA Btu 400 ( 0.1ft ) hr T2 = 150 o F - Btu 2 0.058 5ft o hr ft F T2 = 12.1o F Q= 2 1 o ( ) Example #1d a) Suppose A = 5ft , T =150 F, T =30 F. What thickness of material will result in a heat transfer rate of 1000 BTU/hr? - (T2 - T1 ) 2 1 o 2 o x kA kA(T1 - T2 ) x = Q Btu 2 o 0.058 5ft (150 - 30 ) F hr ft o F x = Btu 1000 hr x = 0.0348ft Q= ( ) Example #2 Consider a onedimensional steady state heat transfer across two materials in series, as shown below: T = 18 C Q = 3.0 Btu/hr A B T T =6C A 2" slab of wood grain perp. to heat flow B 6" brick, k = 2 x kwood Example #2a a) 1 o If T = 18 C and T = 6 C what is T ? 2 o R 1 i R Electrical circuit: T1 Q= T T T = = R R1 + R2 x1 + x2 k A Ax k B Ax x R1 = k A Ax Ti x R2 = k B Ax 2 T2 1ft 2in x1 hr o F 12in R1 = = = 2.87 Btu k A Ax 0.058 Btu 1ft 2 hr ft o F 1ft 6in x 2 hr o F 12in R2 = = = 4.31 Btu 2 k B Ax Btu 2 0.058 1ft hr ft o F ( ) ( ) Example #2a o 1 .8 o F (18 - 6) C o C T = 3.0 Btu Q= = hr R (2.87 + 4.31) hr o F Btu Btu 1ft - 3. 0 ( 2in ) ( - Q12 )( x A ) = hr 12in = -8.64 o F T1 = Btu 2 ( k A Ax ) 0.058 1ft o hr ft F Ti = T1 + T1 = 18 o C ( 9 / 5 ) + 32 - 8.64 = 55.8o F ( ) [ Check by calculating T 2 Example #2b a) If T = 18 C and T = 10 C what is T ? 1 o i o o F 1 .8 (18 - 10) C o o 2 C T T = 5.01 Btu Q12 = = = hr o F hr R x A 2.8736 k A Ax Btu Btu ( 6in ) 1ft - 5.011 ( - Q12 )( x2 ) = hr 12in = -21.6o F T2 = Btu 2 ( k 2 Ax ) 2 0.058 1ft o hr ft F T2 = Ti + T2 = 10 o C ( 9 / 5 ) + 32 - 21.6o F = 28.4o F ( ) [ Example #2c a) What thickness of brick will be required (assuming wood remains at 2")? T1 = 18oC, Ti = 10oC, T2 = 0oC from part (b) Q12 = 5.011 Btu/hr o o 1 .8 o F = -18 o F T2 = -10 C = -10 C 1 oC Btu 2 - 18 o F 2 x 0.058 1ft o T2 k 2 Ax hr ft F x 2 = = Btu - Q12 - 5.011 hr 12in x 2 = 0.4167ft = 5inches 1ft ( ) ( ) ...
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