Chapter 4. Unsteady State Conduction

Transcription

1 Chapter 4 Unsteady State Cnductin Chapter 5 Steady State Cnductin Chee 318 1

2 4-1 Intrductin ransient Cnductin Many heat transfer prblems are time dependent Changes in perating cnditins in a system cause temperature variatin with time, as well as lcatin within a slid, until a new steady state (thermal equilibrium) is btained. In this chapter we will develp prcedures fr determining the time dependence f the temperature distributin Real prblems may include finite and semi-infinite slids, r cmplex gemetries, as well as tw and three dimensinal cnductin Slutin techniques invlve the lumped capacitance methd, exact and apprximate slutins, and finite difference methds. We will fcus n the Lumped Capacitance Methd, which can be used fr slids within which temperature gradients are negligible. Chapter 5 Steady State Cnductin Chee 318 2

3 4-2 Lumped Capacitance Methd Cnsider a ht metal that is initially at a unifrm temperature, i, and at t0 is quenched by immersin in a cl liquid, f lwer temperature he temperature f the slid will decrease fr time t>0, due t cnvectin heat transfer at the slid-liquid interface, until it reaches ( x,0) i x Chapter 5 Steady State Cnductin Chee 318 3

4 Lumped Capacitance Methd If the thermal cnductivity f the slid is very high, resistance t cnductin within the slid will be small cmpared t resistance t heat transfer between slid and surrundings. emperature gradients within the slid will be negligible, i.e.. the temperature f the slid is spatially unifrm at any instant. ( x,0) i x Chapter 5 Steady State Cnductin Chee 318 4

5 Lumped Capacitance Methd Starting frm an verall energy balance n the slid: ha ( s ) ρvc Chapter 5 Steady State Cnductin Chee d dt he time required fr the slid t reach a temperature is: t ρvc ha s θ θ i ln where θ θ he temperature f the slid at a specified time t is: θ θ i i has ρvc t exp (4.5) he ttal energy transfer, Q, ccurring up t sme time t is: Q t t q dt has 0 0 ( ρvc) θ [ 1 exp( t τ )] θ dt / i i i E & t & ut E st

6 ransient emperature Respnse Based n eq. (4.5), the temperature difference between slid and fluid decays expnentially. Let s define a thermal time cnstant 1 τ t ρvc ha ( ) s R t C R t is the resistance t cnvectin heat transfer, C t is the lumped thermal capacitance f the slid Increase in R t r C t causes slid t respnd mre slwly and mre time will be required t reach thermal equilibrium. t Chapter 5 Steady State Cnductin Chee 318 6

7 Validity f Lumped Capacitance Methd Need a suitable criterin t determine validity f methd. Must relate relative magnitudes f temperature drp in the slid t the temperature difference between surface and fluid. slid (due t cnductin) slid / liquid (due t cnvectin) ( L / ka) (1/ ha) R R cnd cnv hl k Bi? What shuld be the relative magnitude f slid versus slid/liquid fr the lumped capacitance methd t be valid? Chapter 5 Steady State Cnductin Chee 318 7

8 Example 4-1 A steel ball (c0.46 kj/kg. C, k 35 W/m.C) 5 cm in diameter and initially at a unifrm temp. f 450 C is suddenly placed in a cntrlled envirnment in which the temperature is maintained at 100 C. he cnvectin heat transfer cefficient is 10 W/m^2. C. Calculate the time required fr the ball t attain a temperature f 150 C. Chapter 5 Steady State Cnductin Chee 318 8

9 Example 4-1 We anticipate that the lumped-capacity methd will apply because f the lw value f h and high value f k. We can check by using Equatin (4-6): h( V / A) k ( 10 )[ ( 4 / 3) π ( 0.025) 2 4π ( 0.025) ( 35) 3 ] < 0.1 s we may use Equatin (4-5). We have C p 7800kg / m [486lb m / ft ] C h 10W / m. [1.76Btu / h. ft 2. F] C C c 460J / kg. C [ 0.11Btu / lb. F] m ha pcv ( ) 4π ( 0.025) 2 ( 7800)( 460)( 4π / 3)( 0.025) s 3 e [ ha / pcv ]τ τ e τ 5819 s 1. 62h Chapter 5 Steady State Cnductin Chee 318 9

10 Example 4-2 Semi-Infinite Slid with Sudden Change in Surface Cnditins A large blck f steel is initially at a unifrm temperature f 35C. he surface is expsed t a heat flux (a) by suddenly raising the surface temperature t 250C and (b) thrugh a cnstant surface heat flux f W/m2. Calculate the temperature at a depth f 2.5 cm after a time f 0.5 min fr bth these cases. Chapter 5 Steady State Cnductin Chee

11 Example 4-2 We can make use f the slutins fr the semi-infinite slid given as Equatins (4-8) and (4-13a). Fr case a / 2 2 ετ ( )( )( ) he errr functin id determined frm Appendix A as erf 2 ατ erf We have i 35 C and 250 C, s the temperature at x2.5cm is determined frm Equatin (4-8)as (, τ ) + ( i ) erf 2 ατ ( )( ) C Chapter 5 Steady State Cnductin Chee

12 Example 4-2 Fr the cnstant-heat-flux case b, we make use f Equatin (4-13a). Since q / A is given as 3.2 m W /, we can insert the numerical Values t give (, τ ) ( 2)( )[ ( )( 30) lπ ] 1/ (0.61) 35 + e 45 5 ( 0.025)( ) ( ) C 2.5cm, τ 30s Fr the cnstant-heat-flux case the surface temperature after 30 s wuld be evaluated with x0 in Equatin (4-13a). hus, ( 0) ( 2)( )[ ( )( 30) 45 / π ] 1/ C Chapter 5 Steady State Cnductin Chee

13 Bit and Furier Numbers he lumped capacitance methd is valid when hlc Bi < 0.1 where the characteristic length: k L c V/A s Vlume f slid/surface area We can als define a dimensinless time, the Furier number: F αt L 2 c Eq. (5.2) becmes: θ θ i i [ Bi F] exp (5.4) Chapter 5 Steady State Cnductin Chee

14 Example (Prblem 5.6 extbk) he heat transfer cefficient fr air flwing ver a sphere is t be determined by bserving the temperature-time histry f a sphere fabricated frm pure cpper. he sphere, which is 12.7 mm in diameter, is at 66 C befre it is inserted int an air stream having a temperature f 27 C. A thermcuple n the uter surface f the sphere indicates 55 C, 69 s after the sphere is inserted in the air stream. Calculate the heat transfer cefficient, assuming that the sphere behaves as a spacewise isthermal bject. Is yur assumptin reasnable? Chapter 5 Steady State Cnductin Chee

15 Other transient prblems When the lumped capacitance analysis is nt valid, we must slve the partial differential equatins analytically r numerically Exact and apprximate slutins may be used abulated values f cefficients used in the slutins f these equatins are available ransient temperature distributins fr cmmnly encuntered prblems invlving semi-infinite slids can be fund in the literature Chapter 5 Steady State Cnductin Chee