Question

A thin flat plate of length L, thickness t, and width W> L is thermally joined to two large heat sinks that are maintained at a temperature T.. The bottom of the plate is well insulated, while the net heat flux to the top surface of the plate is known to have a uniform value of Heat sink T. Heat sink T. a) Derive the differential equation that determines the steady-state temperature distribution T(x) in the plate. b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks.

Answer 1

Answer:

Given that,

Length of the thin flat plate L

Thickness t

- Lets take a small contowl volume at a distancese of de length. NO anetda -> Assumptions ; 1 steady state 2. Pacoperties core not rooming with temperature 3. Negligible contact resistance ' wow applying consenation of enorgy in a conterol volume de tod=ause the .. heat transfer q=q" xda is que xwxdx' 2x +9" wd x = het dax C- kA d e J de atq wdx = x + d (KA 1 Joze = qo'w ta (KA d ) = 0 k d e (t. w o T L J + %" w=0 Asea, A = txw) 쐞

t. + 9 dx = 0 d2T + 90" lo dy2 By intergrating above equation, kt By again integ sreding the above equation, we get, T=-9" x ² + axtla .. Boundary condition © at, x=0, Tato Boundary condition ® at, x=L, T2To >> Foom boundary condition o, . : Ca = To =) From borindary condition , . To = -2 0 2 2 +G.L+TO To + mva cih -56 =0 TECIL = 0 2 Kt 2kt 902 no L 2 kt (1 = 0 =) =- akt - Temperature distibution, 1. TERJE Age Eartat To 2kt

+ 90'L To 2K+ + 2kt x y T(x) = got (22- LR + To 2K+ ". Temperature distrübution, TO) = -90 L - (x² Last To ? - Heat transfer at x 20 960) E-K (wit) =-kw-t. (-20) (x - ²) x = 0 90"...(0-4). =) Heat transfer at 2 = q(2) = -K (wt). It real :-KW.E G (2-) = +90 WL - 2 Heat loss from the plate fs, q = 2011-20) - Zo" wm -190"we Heat loss.- 90' WL