FIGUE 3 STEADY HEAT CONDUCTION IN PLANE WALLS The energy balance for the wall can be expressed as ate of ate of heat trans fer heat trans fer into the wall out of the wall ate of change of the energy of the wall or Q & in Q & out de dt wall Heat flow through a wall is one dimensional when the temperature of the wall varies in one direction only x FIGUE 3 Separating the variables in the above equation and integrating from where T ( T, to x L, where T ( L T we get x L x. Q cond, wall dx T T T kadt Performing the integrations and rearranging gives T T L cond, wall ka (W Under steady conditions, the temperature distribution in a plane wall is a straight line.
x FIGUE 3 The Thermal esistance Concept Where cond, wall T T wall (W L wall / ka ( C W thermal resistance 3 FIGUE 3 3 Analogy between thermal and electrical resistance concepts. 4
FIGUE 3 4 Consider convection heat transfer from a solid surface of area As and temperature Ts to a fluid whose temperature sufficiently far from the surface is T, with a convection heat transfer coefficient h. Newton s law of cooling for convection heat transfer. rate Q ha ( T T can be rearranged as conv s s conv Where T s T conv Schematic for A s convection resistance at a surface. ha ( C W wall / s thermal resistance 5 FIGUE 3 5 rad εσa s Ts T 4 4 ( T T h A ( T T rad s sur surr rad rad rad s h A s s sur thermal resistance or, the radiation resistance ( T + T ( T T rad h rad εσ s surr s + As ( Ts Tsurr surr Schematic for convection and radiation resistance at a surface. 6 3
FIGUE 3 6 The thermal resistance network for heat transfer through a plane wall subjected to convection on both sides, and the electric analogy. 7 FIGUE 3 6 The esistance Network Under steady conditions we have ateof ateof heat convection heat conduction into the wall through the wall ateof heat convection from the wall or. Q conv T T h A T T L ( T ka h A( T 8 4
FIGUE 3 6 which can be rearranged as. T T Q / h A T T conv, T T T T L / ka / h A T T wall conv, Adding the numerators and denominators yields (Fig.3-7 T T. T T Q total ( W 9 FIGUE 3 7 Thermal esistance Network. A useful mathematical identity. 5
FIGUE 3 8 The temperature drop across a layer is proportional to its thermal resistance. L total conv + wall + conv + + C / W h A ka h A FIGUE 3 9 The thermal resistance network for heat transfer through a two-layer plane wall subjected to convection on both sides. T T T T / h A conv, 6
FIGUE 3 Once Q & is known, an unknown surface temperature T j at any surface or interface j can be determined from. T Q T + T T i j total, i j T T ( / h A + ( L / k A conv, wall, The evaluation of the surface and interface temperatures when T and T are given and Q & is calculated. 3 FIGUE 3 Schematic for Example 3-. 4 7
FIGUE 3 Assumptions Schematic for Example 3-..steady state heat transfer. (-D heat transfer 3. thermal conductivity is constant Properties. the thermal conductivity is given to be k.9 W/m*C Analysis A 3m x 5m 5 m 5 FIGUE 3 Example 3-. SOLUTION T T Q & ka L. wall Q ΔT where wall (.9W / m. C L wall / ka ( C W.3m (.9W / m. C( 5 m ( 6 ( 6 C 5m.3m. C Q 63 W. C / W 63W. C / W 6 8
FIGUE 3 Schematic for Example 3-. Assumptions.steady state heat transfer. (-D heat transfer 3. thermal conductivity is constant Properties. the thermal conductivity is given to be k.78 W/m*C 7 solution i glass o total conv, L ka conv.8333 h A (W / m C(.m.8m (.78W / m C(.m conv,.83 h A (4W / m C(.m + glass + conv.855 C / W C / W C / W.8333 +.855 +.83.7 C / W Then Steady rate of heat transfer through the window become T T total [ ( ].7 66 W 8 9
Knowing rate of heat transfer, the inner surface temperature of the winder can be determine from T T conv, T T Q & conv, (66(.8333. C Discussion Note that the inner surface temperature of the window glass will be even. C though the temperature of the air in the room is maintained at C Such T low T Q surface & conv, temperature (66(.8333. are Chighly undesirable since they cause the formation of fog or even frost on the inner surfaces of the glass when the humidity in the room is high. 9 FIGUE 3 3 Schematic for Example 3-3. Assumptions.steady state heat transfer. (-D heat transfer 3. thermal conductivity is constant Properties. the thermal conductivity of glass and air space is given to be k.78,.6 W/m*C
solution i o total conv, 3 air conv h A (W / m conv, glass glass, C(. m L.4m k A (.78W / m C(.m L.m k A (.6W / m C(.m h A (4W / m C(.m + + air + glass, conv.8333 +.47 +.35 +.47 +.83.433.8333 +.35 C / W.83 C / W.47 C / W C / W C /W Then Steady rate of heat transfer through the window become T T total [ ( ].433 69. W Knowing rate of heat transfer, the inner surface temperature of the winder can be determine from T T Q & conv, (69.(.8333 4. C
FIGUE 3 4 Temperature distribution and heat flow lines along two solid plates pressed against each other for the case of perfect and imperfect contact. 3 FIGUE 3 5 A typical experimental setup for the determination of thermal contact resistance. 4
FIGUE 3 6 Effect of metallic coatings on thermal contact conductance. 5 FIGUE 3 7 Schematic for Example 3-4. Assumptions.steady state heat transfer. (-D heat transfer 3. thermal conductivity is constant Properties. the thermal conductivity of aluminum at room temperature is k37 W/m*C 6 3
FIGUE 3 Example 3-4. SOLUTION The thermal contact resistance is 4 c.99x m. C / W h W / m. C L c For a unit surface area, the thermal resistance of a flat plate is deed as k c L k where L is the thickness of the plate and k is the thermal conductivity, Setting cr,the equivalent thickness is determined from the relation above to be 4 ( 37 W / m. C (.99 x m. C / W.5m. 5cm 7 FIGUE 3 8 Schematic for Example 3-5. Properties -The thermal conductivity of copper is given to be k386 W/m*C. -The contact conductance h c 4, W / m 8 4
FIGUE 3 8 Solution plate conv total L ka 4. C / W h A (5W / m C(.m h A (4W / m C(8x int erface 4 c c.m (386W / m C(.m int erface + plate + ambient.3 +.6 + 4. C / W.6.3 C / W m C / W 9 FIGUE 3 8 Solution The rate of heat transfer is determined to be ΔT Q total ( 7 C.4W 4.36 C / W The temperature jump at the interface is determined from ΔT int. (.4W (.3 C / W.37 C erface Q int erface 3 5
6 3 FIGUE 3 9 ( + + + T T T T T T Q Q Q & & & Thermal resistance network for two parallel layers.. T T Q total total + + 3 FIGUE 3 Thermal resistance network for combined series-parallel arrangement. conv conv total total T T Q + + + + + 3 3. and 3 3 3 3 3,, ha A k L A k L A k L conv
FIGUE 3 Schematic for Example 3-6. Assumptions.steady state heat transfer. (-D heat transfer 3. thermal conductivity is constant Properties. the thermal conductivity are given to be k.7 W/m*C for plaster layer, and k.6 W/m*C for the rigid foam 33 solution i 3 4 o conv, foam 6 5 brick conv, h A (W / m L.3m ka (.6W / m C(.5 xm plaster, side plaster, center L ka L ka L ka.6m (.7W / m C(. xm h A (5W / m.4 C / W C(.5 xm.m (.W / m C (.5xm.6m (.W / m C(.5 xm C(.5xm 4.6 C / W. C / W.6 C / W.36 C / W 48.48 C / W 34 7
solution The three resistance 3, 4 and 5 in the middle are parallel, and their equivalent resistance is determined from + + + +.3W / C mid 3 4 5 48.48. 48.48 Which give mid.97 C / W Now all the resistance are in series, and the total resistance mid i + + + mid + 6 +.4 + 4.6 +.36 +.97 +.36 +.6 6.85 C / W is 35 solution Then thesteady rate of Q Q tatal T T. total A 3mx5m 5m. [ ( ] C W ( 4.38 per.5 m surfacearea 6.85 C / W or 4.38/.5 7.5 W per m Then the rate of heat transfer throuugh theentire wallbecomes. ( 7.5W / m ( 5m 63W heat transfer through the wallbecomes area.the total area of the wallis 36 8
FIGUE 3 Alternative thermal resistance network for example 3-6 for the case of surfaces parallel to the primary direction of heat transfer being adiabatic. 37 FIGUE 3 3 Heat is lost from a hot water pipe to the air outside in the radial direction, and thus heat transfer from a long pipe is one-dimensional. 38 9
FIGUE 3 4 cond, cyl r r r r dt ka dr cond, cyl T T dr kdt A rl A π T T T T πlk ln( r / r cond, cyl A long cylindrical pipe (or spherical shell with specified inner and outer surface temperatures T and T. cyl ln(r /r /πlk & T T Qcond, cyl cyl 39 FIGUE 3 5 The thermal resistance network for a cylindrical (or spherical shell subjected to convection from both the inner and the outer sides. T T total 4
FIGUE 3 5 The rate of heat transfer under steady condition can be expressed as T T total where total conv, + cyl, + cyl, + cyl,3 + conv, ln( r / r ln( r3 / r ln( r4 / r3 + + + + h A πlk πlk πlk3 h A where A πr L and A4 πr4 L. T T T T Q ( r r conv, + ln cyl, / + h ( πr L πlk We could also calculate T from. T T T T Q ln( r r ( r r + 3 + conv, 3 / ln 4 / 3 + + πlk πlk h 3 o ( πr L 4 4 4 FIGUE 3 6 The thermal resistance network for heat transfer through a three-layered composite cylinder 4
FIGUE 3 7 The ratio ΔT/ across any layer is equal Q & to Which remains constant in one-dimensional steady conduction. 43 FIGUE 3 8 Schematic for Example 3-7. Assumptions.steady state heat transfer. (-D heat transfer and thermal symmetry at midpoint 3. thermal conductivity is constant Properties.the thermal conductivity of steel is given to be k5w/m*c.heat of fusion of water at atmospheric is h if 333.7 kj/kg 3.Emissivity outer surface of tank is ε 44
solution (a Inner and outer surface area of the tank are A A πd πd π (3 π (3.4 8.3m 9.m adiation heat transfer coefficient is given by h rad εσ ( T + T ( T + T [(95 + (78 ][ 95 + 78] 5.34 W / m C 8 h rad ( (5.67 * 45 Then the individual thermal resistance become i o rad total conv, sphere conv.44 h A (8W / m C(8.3m r r.5.5..47 4πkr r 4π (5W / m C(.5m(.5m h A (W / m.646 h A (5.34W / m C(9.m rad i + + + o rad C(9.m.345 C / W C / W.44 +.47 + C / W.345 + C / W.646.74 C / W 46 3
Then Steady rate of heat transfer through the window become [ ] T T 89 W or 8.7kJ / s.74 total We now determine the outer surface temperature from Which give equiv + o rad & T T Q equiv.345 +.646.5 T T Q & (89(.5 4 equiv C 47 Solution (b The total amount of heat transfer during a 4-h period is Q Q & Δt ( 8.9kJ / s(4x36s 673, 7 kj Note that it take 333.7kJ of energy to melt kg of ice at C, the amount of ice that will melt during a 4-h period is Q 673,7 kj m kg ice h 333.7 kj / kg 79 if.5 C / W combined Discussion h A ( 5.34W / m. C( 9m combined 48 4
FIGUE 3 9 Schematic for Example 3-8. 49 Assumptions. steady state heat transfer. (-D heat transfer and thermal symmetry at centerline and variation in the axial direction 3. thermal conductivity is constant 4. thermal contact resistance at interface is negligible Properties.the thermal conductivity of cast iron is given to be k 8W/m*C and k.5w/m*c for glass wool insulation 5 5
The area of the surface exposed to convection are determine to be A πr L π (.5m(m.57m A πr L π (.575m(m.36m Then the individual thermal resistance become i conv, pipe insulation h A (6W / m C(.57m.6 ln( r / r ln(.75/.5. πk L π (8W / m C(m ln( r / r ln(5.75 /.75.35 πk L π (.5W / m C(m C / W C / W C / W 5 o total conv h A + + i 3 (8W / m + o C(.36m.54 C / W.6 +. +.35 +.54.6 C / W Then Steady rate of heat loss from the steam become T T total [ 3 5].6 W Temperature drop across the pipe and the insulation are determine Δ T Δ T pipe insulation Q & Q & pipe ( W (. C / W. C insulation ( W (.35 C / W 84 C 5 6
FIGUE 3 3 An insulated cylindrical pipe exposed to convection from the outer surface and the thermal resistance network associated with it. T ins T + conv T T ln( r / r + πlk h(πr L 53 FIGUE 3 3 Critical radius of insulation for a cylindrical body to be k, ( m h r cr cylinder Critical radius of insulation for a spherical shell is k, ( m h r cr sphere where k is the thermal conductivity 54 7
FIGUE 3 3 Schematic for Example 3-9. 55 Assumptions. steady state heat transfer. (-D heat transfer and thermal symmetry at centerline and variation in the axial direction 3. thermal conductivity is constant 4. thermal contact resistance at interface is negligible 5. heat transfer coefficient incorporates the radiation effect Properties.the thermal conductivity of plastic is given to be k.5w/m*c 56 8
The rate of heat transfer become equal to the heat generate within the wire, which is determine to be Q & W & e VI ( 8V ( A 8W The values of these two resistances are determine to be A πr L π (.35m (5m.m therefore conv plastic total ha ln( r / r ln(3.5/.5.8 πkl π (.5W / m C(5m conv (W / m + plastic C(.m.76.76+.8.94 C / W C / W C / W 57 We now determine the interface temperature from T T Q & T T + Q & total total 3 + (8W (.94 C / W 5 C The critical radius of insulation of the plastic cover is determine from k.5w / m C r cr. 5m h W / m C 58 9
FIGUE 3 33 HEAT TANSFE FOM FINNED SUFACES The thin plate s of a car radiator greatly increase the rate of heat transfer to the air (photo by Yunus Cengel and James Kleiser. The rate of heat transfer from a surface at a temperature T s to the surrounding medium at T is given by Newton s law of cooling as conv ha ( T T S S 59 FIGUE 3 34 Some innovative designs. 6 3
FIGUE 3 35 Fin Equation The energy balance on this volume element can be expressed as ate of heat ate of heat ate of heat conduction in. to conduction from the + convection from the element at x element at x + Δx the element Q & & + cond, x Qcond, x+ Δx conv Volume element of a at location x having a length of Δx, cross-sectional area of A c, and perimeter of p. where conv ( pδx ( T h T 6 ΔxΔ x FIGUE 3 35 Fin Equation Substituting and dividing by Δx,we obtain cond, x+δx Δx Taking the limit as cond, x + hp( T T d cond + hp( T T dx From Fourier s law of heat conduction we have dt cond kac dx Gives the differential equation governing heat transfer in s, d dx ka c dt dx Δx gives hp( T T 6 3
FIGUE 3 36 Boundary conditions at the base θ ( T θ b T b Boundary conditions at the tip. Boundary conditions at the base and the tip. θ( L T( L T ; L 63 FIGUE 3 37 Very long : long T ( x T ax x hp / ka c T b T e e dt kac hpkac ( Tb T dx x Q & h[ T ( x T ] da hθ ( x da A A A long circular of uniform cross section and the variation of temperature along it. 64 3
FIGUE 3 38 Under steady conditions, heat transfer from the exposed surfaces of the is equal to heat conduction to the at the base. Boundary condition at tip: Adiabatic tip: T ( x T cosh a( L x T T cosh al b dθ dx xl Q & insulated tip dt kac dx x hpka ( T T tanh al c b 65 FIGUE 3 39 Corrected length: Lc L + A c p L c, rec tan gular L + L and L c, cylinder + D 4 t Corrected length L c is deed such that heat transfer from of length L c with insulated tip is equal to heat transfer from the actual of length L with convection at the tip 66 33
FIGUE 3 4 The heat transfer from the will be maximum in this case and can be expressed as, max ( T ha T b Fins enhance heat transfer from a surface by enhancing surface area. 67 FIGUE 3 4 η,max or Fin efficiency Actualheat transferrate from the Ideal heat transferrate from the ( T η η ha T, max b Temperature distribution in a. 68 34
FIGUE 3 4 Fin efficiency For the cases of constant cross section of very long s and s with insulated tips, the efficient can be expressed as η long and η insulated tip,max,max hpka ha c ( Tb T ( T T b hpka c ha kac L hp al ( Tb T tanh al tanh ( T T al b al 69 FIGUE 3 4 Efficiency of circular, rectangular, and triangular s on a plain surface of width w 7 35
FIGUE 3 43 Efficiency of circular of length L and constant thickness t. 7 FIGUE 3 44 ε Fin Effectiveness no ha b ( T T b Heat trans fer rate from the of base area Ab Heat trans fer rate from the surface area A b no b ( T T b b ( Tb T ( T T η ha A ε η ha ha A The effectiveness of such a long is determined to be hpkac ( Tb T kp ε long hab ( Tb T hac no b b 7 36
FIGUE 3 45 ε, overall total, total, no h ( Aun + η A ( Tb T ha ( T T no, b Various surface areas associated with a rectangular surface with three s. 73 FIGUE 3 46 Because of the gradual temperature drop along the, the region near the tip makes little or no contribution to heat transfer. 74 37
FIGUE 3 47 Schematic for Example 3-. T Δ Assumptions. Steady operating conditions 85 C exist. The transistor case is isothermal at 85 C Properties 85 C The case-to-ambient thermal resistance is given to be C / W case ambient Tc T case ambient ( 85 5 C 3W C/ W 75 FIGUE 3 48 Schematic for Example 3-. 76 38
Assumptions. steady operating conditions exist. The heat transfer coefficient is uniform over the entire surfaces 3. thermal conductivity is constant 4. heat transfer by radiation is negligible Properties.the thermal conductivity of the is given to be k8w/m*c 77 In the case of no, heat transfer from the tube per mater of its length is determined from Newton s law of cooling to be A no no πd L π (.3m(m.94m ha no ( T b T (6(.94( 5 537W The efficiency of the circular s attached to a circular tube is plotted in fig.3-43 Noting that L/(D -D /(.6-.3.5m in this case we have r + t (.3 + *..7 r.5 ( L + t h kt (.5 + *. * 6W / m C (8W / m C(.m (.7 ( 78 39
From (,( η. 95 A π ( r r + πr t [(.3m (.5m ] π + π (.3m(.m.46m Q & η Q & η ha ( T T, max b.95(6w / m 5.W C(.46m ( 5 C 79 Heat transfer from the unned portion of the tube is A un un πd S π (.3m(.3m.83m ha un ( T b T (6W / m C(.83m ( 5 C. 6W Noting that there are s and thus inter spacings per meter length of the tube, the total heat transfer from the ned tube becomes total, n( + un (5. +.6 53W 8 4
Therefore, the increase in heat transfer from the tube per meter of its length as a result of the addition of s is total, no, 53 537 4783W ( per m tube length increase Discussion The overall effectiveness of the ned tube is ε 53 W 537 W total,, overall total, no 9.9 8 FIGUE 3 49 Schematic for Example 3-3. The shape factor for this configuration S πl ln ( 4 z / D πx 3m S 6. 9m ln ( 4x.5/. 8 4
FIGUE 3 5 Schematic for Example 3-4. 83 FIGUE 3 5 Schematic for Example 3-5. 84 4
FIGUE 3 5 Ventilation paths for a naturally ventilated attic and the appropriate size of the flow area around the radiant barrier for proper air circulation. 85 FIGUE 3 53 Three possible locations for an attic radiant barrier. 86 43
FIGUE 3-54 Thermal resistance network for a pitched roof-attic-ceiling combination for the case of an unvented attic. 87 44