Shell Balances Spherical Geometry ChE B S.S. Heat conduction with source term: Try spherical geometry using a shell balance: Input Output Source Qr r Qr rδr Sr 4 π r Δ r r Remember, we can substitute Fourier's Laws anytime to get: dt Qr k A dr A 4πr dt dt k r k r S ( r Δ r) dr dr dt dt r rδr r r dr dr Sr lim r Δ r Δr k 4 π r 4 π rδr r 4 π r or d T S r r r dr r k Integrating once we get dt S 3 r r r C dr 3k, rearranging we get: dt rsr C or dr 3k r We can use the obvious boundary condition that nothing real goes to : as r dt unless C dr dt rs r dr 3k Alternative: We can also say that the center o a sphere is a point o symmetry, so T ; we get the same result. r r -
ChE B Integrating again rs T( r) r C 6k I the surace temperature is T then RS ( ) r T T R C 6k RS r C T and rearranging 6k RS r r Tr () T 6k R The lux leaving the sphere Q r dt ka is dr R S 4 R k r π 3k 3 4π R Sr or simply the total heat generated by the sphere! 3 which makes a lot o sense. -
ChE B Conduction with Convection Example Heat Conduction in a Cooling Fin: How to simpliy a multi-dimensional problem with judicious approximations. T True Situation ) T is a unction o x,y,z ) For B «L,W T T(z) only Model ) A small amount o heat is lost rom edges 3) Heat transer coeicient is a unction o temperature, position ) o heat is lost rom edges LW» (BW 4 BL) 3) Heat lux rom in is given by ewton's Law o cooling q h(t-t ) h constant TT(z) only Starting rom these assumptions, we can model the in as a -D problem: A balance across an area o a segment Δz o the in gives Input by conduction (Output by conduction Output by convection) Q Q h WΔz T T ( ) z z z zδz - 3
ChE B dt dt Inserting Fourier's Law: Q z k Az k BW. Inserting this into the previous equation: dt dt k BW z k BW zδz h BWΔz ( T T ) dividing by WΔ z KB and taking limit as Δ z dt dt z zδz h ( T T ) Δz kb dt h ( T T ) kb and the boundary conditions: z T Tw wall temperature dt z L no heat lows out o in end Introduce dimensionless variables: ( ) T T t T T w ; z η z Lη ; L Tw T dt dt hl Make dimensionless to get: kb ) important groups o parameters ) bounds on variables 3) Remember this Writing the equation in terms o these variables dt h ( T T ) kb ( w ) d( Lη ) d T T t h ( T w T ) t kb ChE B dt Lh t t dη kb The solutions o the above equation are t C cosh η C sinh η Remember this ow, we need to evaluate C and C using the boundary conditions: - 4
dt t ( η ) η dη dt Csinh η Ccosh η dη evaluated at η C sinh C cosh sinh C C Ctanh cosh t ( η ) C [ cosh η tanh sinh η ] η t η at ( ) ( ) cosh( ) tanh( ) sinh( ) t C C ( η ) cosh ( η) tanh ( ) sinh ( η) t which can be rearranged to give (recall that cosh, sinh ) ChE B t ( η ) ( η ) cosh cosh The in eectiveness is deined by actual heat lost γ heat lost i in at Tw γ wl wl ( ( ) ) ht z T dy ( ) ht T dy w ( ) t z dη w integrals cancel dη cosh η sinh ηtanh [ sinh η tanh cosh η] tanh / So what makes or an eective in? - 5
ChE B Look at Lh kb / To maximize γ : γ tanh tanh take tanh tanh so tanh tanh or tanh tanh The only choices are, or solutions. is most eicient in So, straight ins o constant cross section on a straight wall aren t so good. We could see this more easily by looking at tanh as, Both numerator and denominator go to zero, to see i there is a limit. Using L'hopitals rule d lim ( tanh ) tanh tanh lim d d lim ( ) d and we see that the eiciency is at and cannot be higher. means η. This means many small ins are better than one or two big ins! Under the constraint o constant in weight, what is the optimal length / thickness ratios o a rectangular in? mk ( Tw T ) Q tanh rom analysis o the eiciency o in ρl Mass o in is: m ρ BLW hl And we can write as or k B / m h 3/ B Wρ k So that we can eliminate L as a unction o B - 6
and Q becomes Q C /3 tanh Q 4/3 /3 tanh /3 cosh From trial and error.49 or opt / ChE B L h.49 B opt kb Fins: Dierent geometries There are a multitude o shapes or ins. Your text has the eectiveness actors, η, or a variety o conigurations. For inned suraces, the total rate o heat loss is Q h T A η A A original area Δ A in area η eectiveness rom tables A A A base Water and air are separated by a steel sheet (radiator?) o.5 in thickness. We need to increase heat transer by adding straight triangular ins o.95 inch base thickness (t), in long (L) along the sheet with.5 in center to center spacing. The sheet is very side. air H O h water 45 Btu/hr t F h air Btu/hr t F.5 in.5 in What side do we put the ins on to maximize heat transer? How much do we increase Q? To determine which side to put the ins on we examine a plain sheet H O air and realize that we have 3 resistances in Sevier - h 45 s h e e t h HO sheet by convection sheet by conduction sheet air by convection - 7
ΔT tconduction Q R conduction R K R convection h (below) like electrical resistors t k h h sheet R sheet air water i sheet t.5in in 6 Btu/hr t F Btu/hr t F 45 Btu/hr t F hr t F (.4.5.) Btu R.5 ChE B ow we can examine where the major resistance is: R R sheet HO Rair.5.4. R.5 R.5 R.5.96.4 So almost all resistance none in sheet little in H O is on air side In cases like this, we want to see where the ins will do the most good. Since the air provides the most resistance, it makes sense to try to decrease this by increasing the area by adding ins to this side. Fins added to the water side will not do too much! To show this, look at square oot o sheet. At ½ in spacing, this means 4 ins. A t, but what is A? A.5 in x t is the base Area.5 in t A t 4 ins 9 t in/t in A (4 ins) ( t width) ( x / t) 4 t. top & bottom Length - 8
ChE B eglect cosine o angle since ins are long and narrow. Calculate η or ins rom value o - tables or η t Lc L.5 inches.85 eet tlc 4 A m (.5)(.5).56 in.78 x t Fins are steel with k 6 Btu/hr t F Air side h Water side h 45 Tables in handbooks / 3/ k.44 Lc.55 kam From homework problem., the solution η we get η air.89 η water.39 ow to compare the heat lows o in baseline A ΔT R Q.9 AΔT h air hwater 45 # no ins # with ins, we have T w T T Q ha( T Tw) no ins side ( ) η ( ) Q h A T T h A T T w w A A A b We don t know T w in this case so: q q ( w) ( w )( η ) ha T T h T T A A T w Solving or T w gives hta ht Aη A ha h Aη A ( ) ( ) And Q is Q ha T T ( ) w - 9
ChE B ( T T ) A Q A h A Aη h Enhancement due to change in area or decrease in R or Q to get to A ΔT A h A Aη h Air side with ins Water side Q Q 7.4 ΔT A Δ.96 A T Relative change Q Q ( 7.4.9) A Δ T Q Q (.96.9) A Δ T Q / Q 3.87 Q / Q.3 air Water So putting ins on the side o the greatest resistance is best. On the least resistance side, it doesn t help at all. We want A h A γ A hnon-in To get near linear improvement 5 45 Check! 45 ote that Q / Q is not A A A ~5, but less because o γ and resistance on air side. -