Steady Heat Transfer With Conduction and Convection

Steady Heat Transfer Outline Steady Heat Transfer With Conduction and Convection • Review last lecture • Equivalent circuit analyses – Review basic concept – Application to series circuits with conduction and convection – Application to composite materials – Application to other geometries • Two-dimensional shape factors 2 Review Steady, 1-D, e& gen = 0 • Rectangular k (TL − T0 ) Q& q& = = − A L Review Heat Generation • Cylindrical shell Q& 2πk (T2 − T1 ) =− L ln (r2 r1 ) Figure 2-21 from Çengel, • Various Heat and Mass Transfer phenomena in solids can generate heat • Define e& gen as the heat I 2 ρL generated per I 2R I 2ρ = A = 2 unit volume e&gen = V LA A per unit time • Spherical shell k is an average thermal 4πk (T2 − T1 ) conductivity (or a constant Q& = value) if k is constant 1 / r1 − 1 / r2 T0, TL = temperatures at x = 0,L; T1, T2 = temperatures at inner (r1) and outer(r2) radii 3 4 Plot of (T - T0)/(TL - T0) for Heat Generation in a Slab Review Heat Generation II 2 1.8 • Temperature and heat flux equations 2k + e&genxL 2k TemperatureDifference Ratio T = T0 − e&genx2 1.6 (T − T )x − 0 L L e&gen(2x − L) k (T0 − TL ) q& = + 2 L 1.4 H=0 1.2 H = .01 H = .1 H=1 1 H=2 H=5 0.8 H = 10 0.6 2 H= 0.4 0.2 Q& x =0 + E& gen = Q& x = L L e&gen k (TL − T0 ) 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x/L 5 6 1 Steady Heat Transfer Steady Heat Transfer Definition Rectangular Steady Conduction Figure 3-2 from Çengel, Heat and Mass Transfer • In steady heat transfer the temperature and heat flux at any coordinate point do not change with time • Both temperature and heat transfer can change with spatial locations, but not with time • Steady energy balance (first law of thermodynamics) means that heat in plus heat generated equals heat out The heat transfer is constant in this 1D rectangle for both constant & variable k dT Q& = q& = − k A dx 7 Thermal Resistance Figure 2-63 from Çengel, Heat and Mass Transfer 8 Thermal Resistance II • Heat flow analogous to current • Temperature difference analogous to potential difference • Both follow Ohm’s law with appropriate resistance term • Conduction k A(T1 − T2 ) T −T ⇒ Q& = 1 2 Q& = L Rcond • Convection ( Q& = hA Ts − T f ) ⇒ Q& = Ts − T f Rconv ⇒ Rcond = L kA ⇒ Rconv = 1 hA • Radiation Rrad = ( 1 1 A1F12σ T13 + T23 + T22T1 + T12T2 )= Ah 1 rad 9 Where Does the Heat Go? 10 Where Does the Heat Go? II Energy conservation requires that conduction heat through wall equals the heat leaving the wall by convection and radiation Figure 1-18 from Çengel, Heat and Mass Transfer Q&1 = Q& 2 + Q& 3 Figure 1-18 from Çengel, Heat and Mass Transfer 11 Figure 3-5 from Çengel, Heat and Mass Transfer 12 2 Steady Heat Transfer Parallel Resistances (T∞ = Tsurr) 1 1 1 = + Rtotal Rconv Rrad T∞ = • 1 Tsurr Rtotal = As hconv + As hrad Define total heat transfer coefficient, htotal Figure 3-5 from Çengel, Heat and Mass Transfer htotal = 1 = hconv + hrad As Rtotal Combined Modes q& = h(T∞1 − T1 ) Convection or convection plus radiation L Q& = T∞1 − T∞ 2 1 L 1 + + Ah1 kA Ah2 ⇒ q& = q& = h(T2 − T∞ 2 ) 14 Combined Modes III Figure 3-6 from Çengel, Heat and Mass Transfer T −T T∞1 − T∞ 2 Q& = ∞1 ∞ 2 = Rtotal Rconv,1 + Rwall + Rconv,1 k (T1 − T2 ) L All q& values are the same Combined Modes II L q& = Figure 3-6 from Çengel, Heat and Mass Transfer 13 A is area normal to heat flow Convection or convection plus radiation Conduction Series Resistance Formula Q& T −T = ∞1 ∞ 2 A 1 +L+ 1 h1 k h2 A is area normal to heat flow Q& T −T = ∞1 ∞ 2 1 L 1 A + + h1 k h2 If you know h1, h2, L, k, T∞1, and T∞2, but you do not know T1 and T2, can you find the heat flux? Once you found the heat flux from the information give, can you find T1 and T2? q& = T1 = T∞1 − 15 Problem A house has a 4 in thick brick wall with k = 0.6 Btu/hr·ft·oF. The interior temperature is 70oF and the exterior temperature is 0oF. The inside and outside convection plus radiation coefficients are 3 Btu/hr·ft2·oF and 4 Btu/hr·ft2·oF, respectively. Find the heat flux through the wall. Given: Wall with L = 4 in = 4/12 ft and k =0.6 Btu/hr·ft·oF has convection on two sides. T∞1 = 70oF, T∞2 = 0oF, h1 = 3 Btu/hr·ft2·oF and h2 = 4 Btu/hr·ft2·oF. & Q Find: q& = 17 A Figure 3-6 from Çengel, Heat and Mass Transfer L q& h1 T2 = T∞ 2 + q& h2 16 Solution A is area normal to heat flow q& = L Figure 3-6 from Çengel, Heat and Mass Transfer Q& T −T 70o F −0o F = ∞1 ∞ 2 = 2 o A 1 +L+ 1 hr ⋅ ft ⋅ F 4 hr ⋅ ft ⋅o F hr ⋅ ft 2 ⋅o F + + ft h1 k h2 3 Btu 12 0.6 Btu 4 Btu q& = Q& 61.5 Btu = A hr ⋅ ft 2 Find values of T1 and T2. Can you check these values? 18 3 Steady Heat Transfer Solution II A is area normal to heat flow T −T q& = ∞1 1 1 h1 T −T q& = 2 ∞ 2 1 h2 Solution III Figure 3-6 from Çengel, Heat and Mass Transfer L q& ⇒ T1 = T∞1 − = 70o F − h1 q& ⇒ T2 = T∞ 2 + =0o F − h1 61.5 Btu hr ⋅ ft 2 = 49.5o F 3 Btu hr ⋅ ft 2 ⋅o F 61.5 Btu hr ⋅ ft = 15.4o F 4 Btu 19 hr ⋅ ft 2 ⋅o F A is area normal to heat flow Figure 3-6 from Çengel, Heat and Mass Transfer L How can we check results below found from analysis of overall problem and convection processes? q& = 61.5 Btu hr ⋅ ft 2 T1 = 49.5o F T2 = 15.4o F Analyze conduction step for consistency. 2 q& = 61.5 Btu hr ⋅ ft 2 Composite Materials ( 0.6 Btu 49.5o F − 15.4o F Q& k (T1 − T2 ) hr ⋅ ft ⋅o F = = = 4 A L ft 20 12 ) Composite Materials II How would you analyze this problem? Figure 3-9 from Çengel, Heat and Mass Transfer 21 Review Cylindrical Shell For constant k Q& r 2πk (T1 − T2 ) = L ⎛r ⎞ ln⎜⎜ 2 ⎟⎟ ⎝ r1 ⎠ Figure 250 from Çengel, Heat and Mass Transfer R= Q& r = 22 Cylindrical Shell with Convection Q& = A1 = 2πr1L Rconv,1 = ⎛r ⎞ 1 ln⎜ 2 ⎟ 2πkL ⎜⎝ r1 ⎟⎠ T1 − T2 T −T = 1 2 R ⎛r ⎞ 1 ln⎜ 2 ⎟ 23 2πkL ⎜⎝ r1 ⎟⎠ T∞1 − T∞ 2 Rconv,1 + Rcond + Rconv, 2 Rconv,2 = A2 = 2πr2 L Q& = 1 1 = h1 A1 h1 2πr1L 1 1 = h2 A2 h2 2πr2 L T∞1 − T∞2 ⎛r ⎞ 1 1 1 ln⎜ 2 ⎟ + + h1 2πr1L 2πkL ⎜⎝ r1 ⎟⎠ h2 2πr2 L Figure 3-25 from Çengel, Heat and Mass Transfer 24 4 Steady Heat Transfer Cylinder plus Convection Result Q& = Problem T∞1 − T∞2 ⎛r ⎞ 1 1 1 ln⎜ 2 ⎟ + + h1 2πr1L 2πkL ⎜⎝ r1 ⎟⎠ h2 2πr2 L • A hot-water pipe (k = 35 Btu/hr·ft·oF) in a house, made of ¾ inch schedule 40 pipe (OD = 1.050 in; ID = 0.824 in) is 40 ft long and contains water at 120oF. The air around the pipe is at 60oF. The heat transfer coefficients inside and outside the pipe are, respectively, 200 and 3 Btu/hr·ft2·oF. Determine the heat loss Q& 2π(T∞1 − T∞2 ) from the pipe. = We can rearrange this equation as shown below Q& 2π(T∞1 − T∞2 ) = L 1 1 ⎛ r2 ⎞ 1 + ln⎜ ⎟ + h1r1 k ⎜⎝ r1 ⎟⎠ h2r2 Figure 3-25 from Çengel, Heat and Mass Transfer L 25 Solution 26 Solution II 60oF, Given: T∞2 = T∞1 = 120oF, r1 = ID/2 = 0.412 in, r2 = OD/2 = 0.525 in, k =35 Btu/hr·ft·oF, L = 40 ft, h1= 200 Btu/hr·ft2·oF, h2= 3 Btu/hr·ft2·oF Figure 3-25 from Çengel, Heat and Mass Transfer 1 1 ⎛ r2 ⎞ 1 + ln⎜ ⎟ + h1r1 k ⎜⎝ r1 ⎟⎠ h2r2 & Find: Q Q& 2π(T∞1 − T∞2 ) = L 1 1 ⎛ r2 ⎞ 1 + ln⎜ ⎟ + h1r1 k ⎜⎝ r1 ⎟⎠ h227 r2 Given: T∞2 = 60oF, T∞1 = hr ⋅ ft 2 ⋅o F 1 1 12 in = 120oF, r1 = ID/2 = 0.412 in, h1r1 200 Btu 0.412 in ft r2 = OD/2 = 0.525 in, k =35 Btu/hr·ft·oF, L = 40 ft, 1 ⎛ r2 ⎞ hr ⋅ ft⋅o F ⎛ 0.525 in ⎞ ⎟ ln⎜ ⎟ = ln⎜ h1= 200 Btu/hr·ft2·oF, k ⎜⎝ r1 ⎟⎠ 35 Btu ⎜⎝ 0.412 in ⎟⎠ 2 o h2= 3 Btu/hr·ft · F hr ⋅ ft 2 ⋅o F 1 1 12 in 1,940 Btu = Find: Q& = hr h2r2 3 Btu 0.525 in ft Q& = ( Composite Cylindrical Shell II Composite Cylindrical Shell ) 2π(T∞1 − T∞2 )L 2π 120oF − 60oF (40 ft ) = o 1 1 ⎛ r2 ⎞ 1 + ln⎜⎜ ⎟⎟ + (0.146 + 0.007 + 7.619) hr ⋅ ft⋅ F h1r1 k ⎝ r1 ⎠ h2r2 Btu 28 Figure 3-26 from Çengel, Heat and Mass Transfer 1 1 = h2 A4 h2 2πr4 L Figure 3-26 from Çengel, Heat and Mass Transfer 29 1 ⎛ r2 ⎞ 1 ⎛ r3 ⎞ ln⎜ ⎟ ln⎜ ⎟ 1 1 = k1L ⎜⎝ r1 ⎟⎠ k2 L ⎜⎝ r2 ⎟⎠ h1 A1 h1 2πr1L 1 ⎛ r4 ⎞ ln⎜ ⎟ k3 L ⎜⎝ r3 ⎟⎠ 30 5 Steady Heat Transfer Composite Cylindrical Shell III Figure 3-26 from Çengel, Heat and Mass Transfer Another Problem • Insulation with k = 0.2 Btu/hr·ft·oF is to be added to the pipe in the previous example problem. Determine the heat transfer if the insulation is one inch thick. Q& = Q& = T∞1 − T∞2 ⎛r ⎞ ⎛ r2 ⎞ ⎛r ⎞ 1 1 1 1 1 + ln⎜ ⎟ + ln⎜ 3 ⎟ + ln⎜ 4 ⎟ + h1 2πr1L 2πk1L ⎜⎝ r1 ⎟⎠ 2πk2 L ⎜⎝ r2 ⎟⎠ 2πk3 L ⎜⎝ r3 ⎟⎠ h2 2πr4 L Q& = T∞1 − T∞ 2 ⎛ r2 ⎞ ⎛r ⎞ 1 1 1 1 + ln⎜ 3 ⎟ + ln⎜ ⎟ + h1 2πr1L 2πk1L ⎜⎝ r1 ⎟⎠ 2πk2 L ⎜⎝ r2 ⎟⎠ h2 2πr3 L 2πL(T∞1 − T∞ 2 ) 1 1 ⎛ r2 ⎞ 1 ⎛ r3 ⎞ 1 + ln⎜ ⎟ + ln⎜ ⎟ + h1r1 k1 ⎜⎝ r1 ⎟⎠ k2 ⎜⎝ r2 ⎟⎠ h2r3 31 Another Problem II Another Problem III Q& = Unchanged resistances from previous example 1 1 ⎛ r2 ⎞ 0.153 hr ⋅ ft⋅o F h1r1 + ln⎜ ⎟ = k1 ⎜⎝ r1 ⎟⎠ Know all terms from previous example except 32 these two Btu = New and modified resistances 1 ⎛ r3 ⎞ hr ⋅ ft⋅o F ⎛ 1.525 in ⎞ 5.332 hr ⋅ ft⋅o F ⎟= ln⎜ ⎟ = ln⎜ k2 ⎜⎝ r2 ⎟⎠ 0.2 Btu ⎜⎝ 0.525 in ⎟⎠ Btu 2πL(T∞1 − T∞2 ) 1 1 ⎛ r2 ⎞ 1 ⎛ r3 ⎞ 1 + ln⎜ ⎟ + ln⎜ ⎟ + h1r1 k1 ⎜⎝ r1 ⎟⎠ k2 ⎜⎝ r2 ⎟⎠ h2r3 ( ) 2π 120oF − 60oF (40 ft ) (0.146 + 0.007 + 5.332 + 2.623) hr ⋅ ft⋅ o F = 1860 Btu hr Btu • Insulation and outer convection resistances are largest 1 1 12 in 2.623 hr ⋅ ft⋅o F hr ⋅ ft 2 ⋅o F = = 3 Btu 1.525 in ft h2r3 Btu – Inner convection and pipe conduction negligible – Outer convection resistance less with insulation 33 34 Effect of Insulation Thickness Insulation Increases Q& ? 2500 • Why does initial amount of insulation increase heat transfer? 2000 1500 Heat loss (Btu/hr) 1000 500 0 0 1 2 3 4 5 6 7 8 9 10 – Tradeoff of two resistances – Added insulation adds conduction resistance – Added insulation also increases outer radius which decreases the outer convection resistance 1/(houterAouter) = 1/(houter2πrouterL) Thickness (in) 35 36 6 Steady Heat Transfer Resistances for Pipe Insulation Radius for Maximum Q& 12 10 Q& = 8 Resistance 6 (hr·ft·F/Btu) 2 0 0 0.5 1 1.5 2 ro = outer radius ko = thermal conductivity of outer layer ⎛1 1 1 ⎞⎟ 2πL(T∞1 − T∞2 )⎜ − ⎜ k r h r2 ⎟ dQ& 2 o ⎠ ⎝ o o =0 =− &: For maximum Q 2 dro ⎛ ⎛ ro ⎞ 1 ⎞ 1 ⎟ ⎜ Rother + ln⎜ ⎟ + ⎜ ko ⎜⎝ ri ⎟⎠ h2ro ⎟⎠ ⎝ Insultation Resistance Convection Resistance Total Resistance 4 2πL(T∞1 − T∞ 2 ) 1 ⎛r ⎞ 1 Rother + ln⎜⎜ o ⎟⎟ + ko ⎝ ri ⎠ h2 ro 2.5 3 & ro = ko/h2 for maximum Q Insulation Thickness (in) 37 Radius for Maximum Q& Spherical Shell with Convection & • ro = ko/h2 for maximum Q • In the example problem h2 = 3 Btu/hr·ft2·oF, and ko = 0.2 Btu/hr·ft·oF so & ro = 0.0667 ft = 0.8 in for maximum Q • Pipe radius was 0.525 in; ro = 0.8 in gives an insulation thickness of 0.275 in • Note that ro = ko/h2 does not depend on ri and is usually larger than ri & • There is no radius for minimum Q 39 Spherical Shell Result Rsph Q& = A2 = 4πr22 L 4π(T∞1 − T∞ 2 ) 1 r −r 1 + 2 1+ 2 kr1r2 h2 r22 h1r1 Figure 3-25 from Çengel, Heat and Mass Transfer Q& = A1 = 4πr12 L Rsph T∞1 − T∞ 2 Rconv,1 + Rsph + Rconv, 2 Rconv,1 = 1 1 = h1 A1 h1 4πr12 Rconv,2 = 1 1 = h2 A2 h2 4πr22 A2 = 4πr22 L Figure 3-25 from Çengel, Heat and Mass Transfer Rsph 1 1 − r1 r2 = 4πk 40 Conduction Shape Factors T∞1 − T∞ 2 Q& = 1 1 − 1 1 r1 r2 + + 2 4 π k h1 4πr1 h2 4πr22 A1 = 4πr12 L 38 41 • Simplified analysis – for multidimensional geometries with each surface at a uniform temperature – Use shape factor, S, whose equation is found from tables like Çengel Table 3-7 & = kS(T – T ) – Basic equation: Q 1 2 – S must have dimensions of length • Equations for S depend on parameters in the different geometries 42 7 Steady Heat Transfer Buried Pipe Shape Factor Example Shape Factor 3.5 3.0 S/L 2.5 2.0 1.5 1.0 0.5 0.0 0 From Table 7-1 in Çengel, Heat and Mass Transfer 10 20 30 40 50 60 70 80 90 100 z/D 43 44 8