Outline Unsteady Heat Transfer - California State University, Northridge
Unsteady Conduction February 28 and March 7, 2007 Outline Unsteady Heat Transfer Review material on fins Lumped parameter model Larry Caretto Mechanical Engineering 375 – Basis for and derivation of model – Solving lumped-parameter problems Unsteady solutions using charts Heat Transfer – Differential equation as basis for charts – Problem solving with charts – Semi-infinite solutions – Product solutions February 28 and March 7, 2007 2 Fin review Lumped Parameter Model Adds surface to enhance heat transfer Analysis for single fin linked to analysis of surface with multiple fins Equations for simple fins and charts for fin efficiency and effectiveness Rectangular fin equations: m (hp/kAc)1/2 T T cosh m( Lc x) Tb T cosh mLc Q& fin Tb T kAc hp tanh mLc Simplified model of unsteady heat transfer for a particular problem – Solid object, with constant k and a uniform initial temperature, Ti – Placed in fluid environment with constant temperature, T , at zero time (t 0) – Convection to the solid with constant heat transfer coefficient, h – Under certain conditions the temperature in the solid is assumed uniform 3 Parallel Resistance Uniform Temperature Basis Look at solid object initially at an initial temperature, Ti, placed into a medium at another temperature T with heat transfer coefficient h Figure 3-6) in Convective resistance is 1/hA, Çengel, Heat and Mass conductive resistance is L/kA Transfer 1 hA ME 375 Heat Transfer L kA 4 Figure 3-6) in Çengel, Heat and Mass Transfer 1 hA T T T T Q& 1 1 1 2 L 1 hA kA L kA L T1 T2 hL kA 1 T 1 T1 k hA T1 – T2 will be small compared to T 1 – T1 when hL/k is small 5 6 1
Unsteady Conduction February 28 and March 7, 2007 Application to Unsteady Case Lumped Parameter Model Basic model says that convection energy into solid hA(T – T) goes to increase uniform solid temperature, T, giving energy change ρcpVdT/dt dT dT hA (T T ) ρc pV hA(T T ) dt dt ρc p V Unsteady case: temperatures change Special case: convection resistance is much larger than conduction in solid Result: temperature differences in the solid are almost negligible Idealization: Assume that solid is at uniform temperature, T Model: ρcpVdT/dt hA(T – T) Define characteristic length, Lc V/A, and inverse time constant, b hA/ρcpV hA/ρcpLc – V is volume 7 Lumped Parameter Solution Lumped Parameter Analogy Have first-order differential equation with initial condition that T Ti at t 0 Combine solution with b definition dT hA (T T ) b(T T ) T Ti at t 0 dt ρc p V Solution is known to be exponential (T T ) (Ti T )e bt or 8 T (Ti T )e bt T You can show that solution satisfies differential equation and initial condition (T T ) (Ti T )e bt (Ti T )e hAt ρc pV 1/hA is convection resistance and ρcpV is capacity of solid to absorb heat hAt t t 1 ρc pV ρc pV Rthermal Cthermal hA Result same as RC electrical circuit 9 10 When can we use this? When can we use this? II Saw that solid temperature becomes closer to uniform as hL/k (known as Biot number, Bi) becomes smaller Criterion for application of lumped parameter solution is Bi hLc/k 0.1 Problem: what is Lc for a cylinder with a diameter of 0.1 m and a length of 0.5 m? So Lc 0.02273 m; can we use lumped parameter model if h 80 W/m2·K and k 240 W/m2·K? Criterion for application of lumped parameter solution is Bi hLc/k 0.1 π 2 D L V 1 1 4 0.02273 m Lc 2 4 A 2 π D 2 πDL 2 4 L D 0.5 m 0.1 m 4 11 ME 375 Heat Transfer 80 W (0.02273 m) hLc m 2 K 0.0076 0.1 Bi 240 W k m K So lumped parameter model is valid 12 2
Unsteady Conduction February 28 and March 7, 2007 Application of the Model Application of the Model II If the cylinder analyzed previously has cp 900 J/kg·K, ρ 2700 kg/m3, an initial temperature of 350OC and an environmental temperature of 30oC, how long will it take to reach 50oC? Given: Lc 0.02273 m, h 80 W/m2·K, cp 900 J/kg·K, ρ 2700 kg/m3, Ti 350OC, T 30oC, T 50oC Find: time, t, to reach T 50oC Given: h 80 W/m2·K, cp 900 J/kg·K, ρ 2700 kg/m3, Ti 350OC, T 30oC, T 50oC; Find: t (T T ) (Ti T )e bt t 1 ln T T b Ti T 80 W 1 J 2 0.001449 hA h K W s m b s ρc p V ρc p Lc 2700 kg 900 J (0.02273 m ) m3 kg K 13 14 Application of the Model III Review Conduction Equation 50 C 30 C 1 T T s 1914 s t ln ln b Ti T 0.001449 350o C 30o C T T T T k e&gen k k t x x y y z z T T T T ρc p k e&gen k k t x x y y z z o o Lumped parameter approach easy to use and does not require application of more complex calculations Applicable to general geometry Requires Bi hLc/k 0.1 Next consider unsteady problems with spatial variation ρc p For constant k, bring k 0 for one dimensional 0 for no heat generation outside the derivative heat transfer 2 2 ρc p T T k 2 t x T k T t ρc p x 2 15 16 1D, constant k, e& gen 0 What is 1D? Define the thermal diffusivity, α k/ρcp E L M Θ L T L Θ M E T 3 Dimensions: 2 Typical units are m2/s or ft2/s T k 2T t ρc p x 2 T 2T α 2 t x Solution requires initial (t 0, all x) and boundary conditions Figure 4-11(a) in Çengel, Heat and Mass Transfer 17 ME 375 Heat Transfer In theory, one dimensional heat transfer implies that the slab is infinite (or insulated) in the y and z directions Practically this means that the y and z dimensions are so large that end effects are not important 18 3
Unsteady Conduction February 28 and March 7, 2007 Specific Problem Specific Problem II Problem: at t 0, a large slab initially at Ti is placed in a medium at temperature T with a heat transfer coefficient, h Coordinates: Choose x 0 as center of slab (which runs from –L to L) for this symmetric problem Figure 4-11(a) in Çengel, Heat and Mass Transfer Initial condition: at t 0, T Ti for all x Boundary conditions: At x 0, T/ x 0 for symmetry. At x L an energy balance gives –k T/ x h(T – T ) Dimensionless form: shows important combi20 nations of variables Figure 4-11(a) in Çengel, Heat and Mass Transfer 19 Specific Problem Conditions Dimensionless Equation Form Differential equation with boundary and initial condition for T(x,t) T 2T T α 2 T ( x,0) Ti 0 t x x 0 x T k h[T (L, t ) T ] x x L Define dimensionless Θ T T Ti T temperature ratio Get dimensionless differential equation and initial/boundary conditions for Θ(ξ,τ) Θ 2Θ τ ξ 2 Define dimensionless distance, ξ x/L and dimensionless time, τ αt/L2, called the Fourier number Θ(ξ,0) 1 Θ 0 ξ ξ 0 Θ hL Θ(1, τ ) Bi Θ(1, τ) ξ ξ 1 k 21 Dimensionless Result T T Ti T αt τ 2 L x ξ L Θ Θ 2Θ τ ξ 2 Θ(ξ,0) 1 Equation Solution Θ 0 ξ ξ 0 Θ hL Θ(1, τ ) Bi Θ(1, τ) ξ ξ 1 k We started with the following variables to solve for T: Ti, T , x, L, t, α, h, k We now see that T (Ti – T )Θ T , where Θ depends on ξ, τ, and Bi Solution is infinite series with an infinite set of dimensionless parameters λn that depends on Bi 2 4 sin λ n Θ e λ n τ cos λ n ξ 2λ n sin 2λ n n 1 The values of λn are the roots of the equation λn/Bi cot λn – Next chart shows first seven λn values for Bi 1 23 ME 375 Heat Transfer 22 24 4
Unsteady Conduction February 28 and March 7, 2007 Finding λ Dimensionless Temperature Difference in a Slab with hL/k 0.1 25 1 λ cot λ Bi Bi 1 20 15 Functions 10 5 Line Cotangent Intersections 0 -5 0.9 tau 0.01 0.8 tau 0.1 tau 0.2 T T 0.6 Ti T 0.5 tau 0.3 -20 λ4 9.52933 0.1 8 10 12 14 16 18 20 T T Ti T1 0 0 λ6 15.7713 λ7 λ tau 4 tau 5 λ5 12.6453 6 tau 2 tau 3 λ3 6.43730 4 tau 1 tau 1.5 -15 2 tau 0.6 0.3 0.2 0 tau 0.5 tau 0.8 λ2 3.42561 -25 tau 0.4 0.4 λ1 0.86033 -10 tau 0.05 0.7 0.1 0.2 0.9 1 3 Dimensionless Temperature .00 tau 2 0 0.3 02 tau tau 1.5 0. 5 00 0. 01 0. tau 1 tau 0.75 0.6 u ta 0.7 u ta ta u ta u tau 0.2 tau 0.5 Dimensionless Temperature 0.8 Figure 7. Convection Boundary Condition, hL/k 10 0.8 0.4 0.7 tau 0.1 0.8 0.5 0.6 0.9 tau 0.05 tau 0.1 0.5 26 tau 0.02 tau 0.2 0.4 Dimensionless distance x/L T T Ti T1 Figure 6. Convection Boundary Condition, hL/k 1 0.9 0.3 25 18.9023 0. 05 0.7 0.6 tau 0.35 0.5 tau 0.5 0.4 0.3 tau 0.75 0.2 0.2 tau 1 tau 3 0.1 0.1 tau 4 tau 1.5 0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 x/L x/L 27 28 Slab Center-line (x 0) Temperature Chart Center-line Temperature, T0 Figure 4-15(a) in Çengel, Heat and Mass Transfer Rewrite general result to introduce An Θ T T Ti T 4 sin λ 2λ n sin n2λ n e λ τ cos λ nξ Ane λ τ cos λ nξ 2 n n 1 2 n n 1 Result for center-line (x 0) Θ0 2 2 T0 T An e λ n τ cos λ n 0 An e λ n τ Ti T n 1 n 1 Heisler charts show Θ0 as a function of τ αt/L2 with 1/Bi k/Lh as a parameter – An and λn depend on Bi 29 ME 375 Heat Transfer 30 5
Unsteady Conduction February 28 and March 7, 2007 Problem Solution Find time required to cool the centerline temperature of 0.3 m thick plate to 50oC if k 50 W/m·K, α 15x10-6 m2/s and initial temperature is 400oC. The heat transfer coefficient is 80 W/m2·K and the environmental temperature 20oC. Given: T0 50oC, Ti 400oC, T 20oC, L 0.3/2 0.15 m, k 50 W/m·K, α 15x10-6 m2/s, h 80 W/m2·K Find: t Chart relates (T0 – T )/(Ti – T ), k/hL, and αt/L2. From given data we can find T0 T 50o C 20 oC 0.0789 Ti T 400o C 20oC 50 W k 1 4.167 m K hL 80 W 0.15 m m K Find τ αt/L2 11.9 (see next chart) t 31 τL2 11.9(0.15 m )2 1.79 x10 4 s 4.96 h α 15 x10 6 m 2 32 s Θ T T Θ 0 T0 T Chart II Can find T at any x/L from this chart once T at x 0 is found from previous chart See basis for this chart on the next page Input Θ0 0.079 Input k/hL 4.167 Figure 4-15(a) in Çengel, Heat and Mass Transfer Find τ 11.9 Temperature Approximations Θ T T A e λ1τ cos λ1ξ A2e λ 2τ cos λ 2ξ L 1 2 2 Θ 0 T0 T A e λ1τ A e λ 2τ L 2 2 2 For “large” τ ( 0.2) series in numerator and denominator converge in one term T T A e λ1τ cos λ1ξ A2e λ 2τ cos λ 2 ξ L Θ cos λ1ξ 1 2 2 Θ 0 T0 T A e λ1τ A e λ 2τ L 2 2 1 2 35 ME 375 Heat Transfer 34 Temperature Approximations II For τ 0.2 Previous equation for Θ/Θ0 1 Figure 4-15(b) in Çengel, Heat and Mass Transfer Θ T T cos λ1ξ Θ 0 T0 T Depends only on ξ x/L and λ1 which depends on Bi hL/k Must first determine Θ0 as in previous example to get Θ What is Θ at x L in that example? 36 6
Unsteady Conduction February 28 and March 7, 2007 Θ T T Θ 0 T0 T Answer Alternative Approximation Example had T0 50oC, T 20oC, and k/hL 4.167 This chart gives (T – T )/(T0 – T ) 0.91 for x/L 1 So T (0.91)· (50oC – 20oC) 20oC 47.3oC Return to series for “large” τ ( 0.2) that converges in one term Figure 4-15(b) in Çengel, Heat and Mass Transfer Θ 2 2 T T An e λ n τ cos λ n ξ A1e λ1τ cos λ1ξ Ti T n 1 Can use this approximation to find Θ and can also solve for τ and ξ when τ 0.2 τ 1 λ21 T T T T 1 1 1 ξ cos 1 ln λ21τ λ1 T T Ti T A1 cosλ1ξ i A1e 37 38 Table Extract Approximate Equations Use Table 4-2 in text to find λ1 and A1 for given Bi Find T at any ξ x/L and τ αt/L2 from Repeat example with T0 50oC, Ti 400oC, L 0.3/2 0.15 m, T 20oC, k 50 W/m·K, α 15x10-6 m2/s, h 80 W/m2·K to find t for T0 50oC and T at surface for this time Recall 1/Bi 4.167 so Bi 0.24 Interpolate in table to find λ1 0.4684 and A1 1.0367 for Bi 0.24 Θ 2 T T A1e λ1τ cos λ1ξ Ti T – Last two slides have interpolation details 2 T (Ti T ) A1e λ1τ cos λ1ξ T First find time for T0 50oC 39 40 Approximate Equations II Data: T0 50oC, Ti 400oC, L 0.15 m, T 20oC, k 50 W/m·K, α 15x10-6 m2/s, h 80 W/m2·K, λ1 0.4476 and A1 1.0334 1 T T 1 1 τ 2 ln λ1 Ti T A1 cosλ1ξ 0.46842 50oC 20oC 1 11.74 ln 400oC 20oC (1.0367) cos0 t ( ) τL2 11.74 0.15 m 2 1.76 x10 4 s 4.83 h α 15 x10 6 m 2 41 s ME 375 Heat Transfer Approximate Equations III Find surface temperature for τ 11.74 2 T T (Ti T )Θ T (Ti T ) A1e λ1τ cos λ1ξ 20o C 2 400o C 20o C 1.0334e 0.4684 (11.74 ) cos[(0.4684 )(1)] 46.8o C ( ){ } Results similar to charts – 4.83 h vs. 4.96 h to reach T0 50oC – Surface temperature 46.8oC vs. 47.3oC For full solution T0 50oC in 4.83 h and surface temperature 46.7oC at that time 42 7
Unsteady Conduction February 28 and March 7, 2007 1D Cylinder, constant k, e& gen 0 Cylinder and Sphere T t 1 T 1 T kr k r r r r 2 φ φ T k e&gen z z ρc p Figure 4-11 in Çengel, Heat and Mass Transfer Same problem has similar chart solutions 43 Figure 2-3 from Çengel, Heat and Mass Transfer ρc p T k T r t r r r T α T r t r r 44 r Cylinder Center (r 0) Temperature Chart Same Problem for Cylinder Figure 4-16(a) in Çengel, Heat and Mass Transfer Constant initial temperature, Ti, and convection at outer radius r0 T α T T r T (r ,0) Ti 0 t r r r r r 0 T k h[T (r0 , t ) T ] r r r0 Define dimensionless distance, ξ r/r0 and dimensionless time, τ αt/r02, 45 1D Sphere, constant k, e& gen 0 T t 1 2 T kr r r 2 r T 1 k sin θ 2 θ r sin θ θ T 1 e&gen k 2 2 φ φ r sin θ ρc p Figure 2-3 from Çengel, Heat and Mass Transfer ρc p T k 2 T 2 r t r r r ME 375 Heat Transfer T α 2 T r t r 2 r 47 r 46 Same Problem for Sphere Constant initial temperature, Ti, and convection at outer radius r0 T α 2 T T r T (r ,0) Ti 0 r r 0 t r 2 r r T k h[T (r0 , t ) T ] r r r0 Define dimensionless distance, ξ r/r0 and dimensionless time, τ αt/r02, 48 8
Unsteady Conduction February 28 and March 7, 2007 Sphere Center (r 0) Temperature Chart Figure 4-17(a) in Çengel, Heat and Mass Transfer 49 More Approximate Solutions (T – T )/(T0 – T ) Charts Figures 4-16(b) and 4-17(b) in Çengel, Heat and Mass Transfer 50 More Approximate Solutions II Cylinder and sphere also have approximate solutions for τ 0.2 – Values of A1 and λ1 still depend on Bi and are different for each geometry 2 Cylinder Θ T T A1e λ1τ J 0 λ1 r Ti T Sphere Θ r0 2 r r T T A1e λ1τ 0 sin λ1 λ1r r0 Ti T 51 More Approximate Solutions III Table 4-2 in Çengel, Heat and Mass Transfer Semi-Infinite Solids The function J0(x) is the zero order Bessel function Plane that extends to infinity in all directions Practical applications: large area for short times – Tables in text or get value from Excel function besselj(x,0) or Matlab function besselj(0,x) Excel function requires analysis tool pack addin to be installed Here Bi hr0/k for cylinder and sphere is different from Bi hV/(kA) for lumped parameter solution 53 ME 375 Heat Transfer 52 Figure 4-24 in Çengel, Heat and Mass Transfer – Example: earth surface locally54 9
Unsteady Conduction February 28 and March 7, 2007 Two Semi-infinite Solid Results Initial temp Ti, surface temperature set to Ts at t 0 T Ts x erf Ti Ts 4αt hx h 2αt k k 2 Defined integrals erf(x) is called error function and erfc(x) 1 – erf(x) is the complementary error function – Excel/ Matlab functions erf(x) and erfc(x) Initial temp Ti, convection starts at t 0 to T with heat transfer coefficient, h T Ti x erf e T Ti 4αt Error Functions Excel requires analysis took pack add-in See plots on next chart x h αt erfc k 4α t x erf ( x) 2 2 e t π0 erfc ( x) 1 erf ( x) 2 2 e t π x 55 56 Semi-Infinite Medium Results Error Function and Complement 1 0.9 x 0.8 erf ( x ) Functions 0.7 2 2 e t π0 0.6 0.5 erfc( x) 1 erf ( x) 0.4 0.3 2 2 e t π x 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 1.2 x 1.4 1.6 1.8 2 57 Figure 4-29 in Çengel, Heat and Mass Transfer 58 Problem Solution How deep should water pipes be buried in a soil that is initially at 20oC to avoid freezing if the surface is at –15oC for 60 days? (Assume soil α 1.4x10-7 m2/s) Given: Soil with α 1.4x10-7 m2/s and Ti 20oC must not freeze for t 60 days if Ts -15oC. Find: depth required Assumption: Required depth will set T 0oC at t 60 days Use equation for semi-infinite medium with fixed surface temperature T Ts 0o C 15o C x o 0.42857 erf Ti Ts 20 C 15o C 4α t 59 ME 375 Heat Transfer ( ( ) ) From tables or erfinv function of Matlab find erf-1(.42857) 0.40019 x 4αt x 0.40019 αt 0.40019 1.4 x10 7 m 2 (60 d ) 86400 s s d x 0.682 m 60 10
Unsteady Conduction February 28 and March 7, 2007 Multidimensional Solutions Multidimensional Example Solution for finite cylinder is product of solution for infinite cylinder and infinite slab Can get multidimensional solutions as product of one dimensional solutions – All one-dimensional solutions have initial temperature, Ti, with convection coefficient, h, and environmental temperature, T , starting at t 0 – General rule: ΘtwoD ΘoneΘtwo where Θone and Θtwo are solutions from charts for plane, cylinder or sphere Figure 4-35 in Çengel, Heat and Mass Transfer 61 62 Multidimensional Example II T (r , x, t ) T Ti T x a/2 Multidimensional Problem finite cylinder T (r , t ) T Ti T infinite cylinder x -a/2 Figure 4-35 in Çengel, Heat and Mass Transfer Slab solutions have thickness 2L so we use L a/2 in this case T ( x, t ) T Ti T infinite A cylinder is 0.3 m high, with a radius of 0.1 m, k 50 W/m·K, α 15x10-6 m2/s, and initial temperature of 400oC. The heat transfer coefficient is 80 W/m2·K and the environmental temperature is 20oC. Find temperature in the centre of the cylinder and at its corner after one hour. slab 63 64 Multidimensional Solution Multidimensional Solution II We can apply the following equation: T (r , x, t ) T Ti T finite cylinder T (r , t ) T T ( x, t ) T Ti T infinite Ti T infinite cylinder slab First get the temperatures at the center T (0,0, t ) T Ti T finite cylinder T (0, t ) T T (0, t ) T infinite T T i Ti T infinite cylinder slab We find infinite quantities from charts 65 ME 375 Heat Transfer We have separate Biot and Fourier numbers for the two infinite geometries 15 x10 6 m 2 50 W (3600 s ) k 1 αt s m K 6.25 τ 2 5.40 W 80 hr0 0.1 m r0 (0.1 m )2 m K 50 W 15 x10 6 m 2 (3600 s ) k t 1 α s 4.167 τ 2 2.40 m K hL 80 W 0.15 m L (0.15 m )2 m K 66 11
Unsteady Conduction February 28 and March 7, 2007 τ 5.4 τ 2.4 Θ0 Slab chart uses τ 2.4 and k/hL 4.167 to find Θ0 0.62 Enter cylinder chart with values of τ 5.4 and k/hr0 6.25 to find Θ0 0.22 0.22 67 68 Figure 4-16(a) in Çengel, Heat and Mass Transfer Figure 4-15(a) in Çengel, Heat and Mass Transfer Multidimensional Solution III Multidimensional Solution IV Use Θ0 values just found T (0,0, t ) T Ti T finite cylinder T (0, t ) T T (0, t ) T Ti T infinite Ti T infinite cylinder T (0,0,3600 s ) T Ti T slab Edge temperature (x L 0.15 m and r r0 0.1 m) found as follows T (L, r0 , t ) T T (r , t ) T T (L, t ) T 0 Ti T finite Ti T cylinder finite (0.22 )(0.62 ) 0.1364 cylinder Ti T cylinder ) 20o C 400o C 20o C (0.1364) 72o C k/hL 4.167 Use auxiliary charts for this ratio Slab For k/hL 4.167 and x/L 1, Θ/Θ0 0.875 Previously found that Θ0 0.62 So slab component of product solution is (0.62)(0.875) Figure 4-15(b) in Çengel, 71 0.543 Heat and Mass Transfer ME 375 Heat Transfer slab Each term on right is product of center solution times ratio of point to center 69 Θ T T Θ 0 T0 T infinite – Must find this for each one dimensional solution separately T0 T (0,0,3600 s ) T (Ti T )(0.1364) ( infinite 70 Θ T T Θ 0 T0 T k/hr0 6.25 Cylinder For k/hr0 6.25 and r/r0 1, Θ/Θ0 0.93 Previously found that Θ0 0.22 So cylinder component of product solution is (0.22)(0.93) Figure 4-15(b) in Çengel, 0.205 72 Heat and Mass Transfer 12
Unsteady Conduction February 28 and March 7, 2007 Multidimensional Solution V T (L, r0 , t ) T Ti T Repeat Using One-term Values T (r , t ) T T (L, t ) T 0 infinite T T i Ti T infinite finite cylinder cylinder T (L, r0 , t ) T Ti T finite slab finite T (r , t ) T T (L, t ) T 0 infinite T T i Ti T infinite cylinder cylinder slab 2 T (L, t ) T x A1e λ1τ cos λ1 L Ti T infinite (0.205)(0.543) 0.111 cylinder Tedge T ( L, r0 ,3600 s ) T (Ti T )(0.111) ( T (L, r0 , t ) T Ti T slab 2 T (r , t ) T r A1e λ1τ J 0 λ1 Ti T infinite r0 ) 20o C 400o C 20o C (0.111) 62o C cylinder 73 74 Repeat Using One-term Values II Repeat Using One-term Values III For infinite slab, τ 2.40 and Bi 0.24; interpolation in Table 4-2 for Bi 0.24 gives λ1 0.4684 and A1 1.0367 For infinite cylinder, τ 5.40 and Bi 0.16; interpolation in Table 4-2 for Bi 0.24 gives λ1 0.5469 and A1 1.0388 2 T (r , t ) T r A1e λ1τ J 0 λ1 Ti T infinite r0 2 T (L, t ) T x A1e λ1τ cos λ1 L Ti T infinite cylinder slab 1.0367e (0.4684) 2 ( 2.40) 0.15 m 0.546 cos 0.4684 0.15 m 1.0388e (0.5469) 2 (5.40) 0.1 m 0.191 J 0 0.5469 0.1 m 75 76 Repeat Using One-term Values IV Now have individual terms in product solution T (L, r0 , t ) T Ti T finite T (r , t ) T T (L, t ) T 0 infinite T T i Ti T infinite cylinder cylinder slab (0.546 )(0.191) 0.1045 ) 20o C 400o C 20o C (0.1045) 60o C y y1 y2 y1 (x x1 ) y1 x x1 ( y2 y1 ) x2 x1 x2 x1 λ1 λ1,1 77 ME 375 Heat Transfer Given table with two data pairs (x1,y1) and (x2,y2) we want to find y for some value of x between x1 and x2 Here known x is Bi and we are trying to find y λ1 so interpolation formula is Tedge T ( L, r0 ,3600 s ) T (Ti T )(0.1045) ( Interpolation λ1, 2 λ1,1 Bi2 Bi1 (Bi Bi1 ) 78 13
Unsteady Conduction February 28 and March 7, 2007 Interpolation II Find λ1 for Bi 0.24 – Bi 0.24 is between Bi1 0.2 (λ1,1 .4328) and Bi2 0.3 (λ1,2 .5218) λ1 λ1,1 .4328 .4684 ME 375 Heat Transfer λ1, 2 λ1,1 Bi2 Bi1 (Bi Bi1 ) .5218 .4328 (.24 .2) .3 .2 79 14
ME 375 Heat Transfer 4 19 Specific Problem Problem: at t 0, a large slab initially at T i is placed in a medium at temperature T with a heat transfer coefficient, h Coordinates: Choose x 0 as center of slab (which runs from -L to L) for this Figure 4-11(a) in symmetric problem Çengel, Heat and Mass Transfer 20 Specific Problem II
Basic Heat and Mass Transfer complements Heat Transfer,whichispublished concurrently. Basic Heat and Mass Transfer was developed by omitting some of the more advanced heat transfer material fromHeat Transfer and adding a chapter on mass transfer. As a result, Basic Heat and Mass Transfer contains the following chapters and appendixes: 1.
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surface to transfer heat under the temperature difference conditions for which it was designed. Fouling of heat transfer surfaces is one of the most important problems in heat transfer equipment. Fouling is an extremely complex phenomenon. Fundamentally, fouling may be characterized as a combined, unsteady state, momentum, mass and heat transfer
Feature Nodes for the Heat Transfer in Solids and Fluids Interface . . . 331 The Heat Transfer in Porous Media Interface 332 Feature Nodes for the Heat Transfer in Porous Media Interface . . . . 334 The Heat Transfer in Building Materials Interface 338 Settings for the Heat Transfer in Building Materials Interface . . . . . 338
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 8 Rectangular Steady Conduction Figure 2-63 from Çengel, Heat and Mass Transfer Figure 3-2 from Çengel, Heat and Mass Transfer The heat .
Holman / Heat Transfer, 10th Edition Heat transfer is thermal energy transfer that is induced by a temperature difference (or gradient) Modes of heat transfer Conduction heat transfer: Occurs when a temperature gradient exists through a solid or a stationary fluid (liquid or gas). Convection heat transfer: Occurs within a moving fluid, or
Young I. Cho Principles of Heat Transfer Kreith 7th Solutions Manual rar Torrent Principles.of.Heat.Transfer.Kreith.7th.Sol utions.Manual.rar (Size: 10.34 MB) (Files: 1). Fundamentals of Heat and Mass Transfer 7th Edition - Incropera pdf. Principles of Heat Transfer_7th Ed._Frank Kreith, Raj M. Manglik Principles of Heat Transfer. 7th Edition
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