Question

4. A wall, 4 m wide in steady-state experiences uniform volumetric heat generation " = 300 , shown in Figure 1. The wall has constant thermal conductivity k=5 and is exposed to air T = 293 K on both sides. At x=L: T(L) = T2 = 300 K, and h2 = 120 At r= -L: T-L) = T. (a) Determine the temperature distribution T(x) in the wall. Partial Ans: C = -240 (b) What is the temperature (in K) at the center of the wall? (c) Determine the temperature T in K). Ans: T = 492 K. (d) Find the heat transfer coefficient hi in .). (e) Determine the value in K) and location in m) of the maximum temperature.

Answer 1

Answers:

a)

T(0) = -30.12 – 487 +516

b)

T1 = 0) = 516K

c)

T1 =T1= -1) = 492K

d)

h1 = 1.809W/mK

e)

-Ime = -0.8m

  

Tmar = 535.2K

Go through the pictures attached below to know the procedure

Tue, 2 L=am Given informations: Cet hom PSE qu" - 300 W/m? Қиял = 5 @[ms Too; ku * To - 293 k TCL) = 300 K = Tz t Eth h2= 120 W/m2K Watan General heat diffusim equation ; Hote... ) ( ) ( PL)+ 9"- sop Assumptions: 1. Steady stafe 4. uniform head generatim 2. Constant Kwall 3. I-D heat transfer (in x-dinection) The heat diffusion equation reduces to § T(X) = -9" x2 + 4x +62 TCL) = 300R 7 -9"12 + CL +62 = 300 ar a) - 300x2? -- 4x2 + Ca = 300 2x5 o 124762. =:4201 Now, total heat generated -q"XV = q (AX2L) Total heat lost = heat Loss from right side + heat loss from left side of the wall Total heat generated . Total heat lost q" (Axal) - Pright + Ploft ASSAT

Tatu, + x = tx ****9) - KO ) Sinui Rarely Rugt » - 4* (**) Now in fright sieße Qright = hq.A.( Ta - Tiw) AKM (A-G1). 42. $: (Ta - Tu) 95 * (2012 - 4) = 120 (30) – 293) = 120-41 = 120 (7) $ 41 =120-168 = Q = -48 Km 24 +22=420 * [C2 = 516 KI TXXX XXX98X_** Tix) = - "s? teix + ez 5 7.2 prill MT(X) = - 307 *x +(-48) X + 516 | 2X5

a> T(X) = -30x2 -48 x +516 4 Temperature distribution T = T (-1) Left = 360X A Watt w = -30(-22-48 (-2) +516 = 49.25 Temperature of the centre of the weell (x =0) T(x=0) = 516 K T Total heat generated = Q Lost + Qout meg left right » q" (A*2L) = a lost tha. A. (T2-T) is af (2026) - 120 (200-32)] sy boeren = 360 x 4 Watt Tontti shexal Ti-Towe) hu = 360/(492 -243) = 1-809 - Iw=1.809 Fore Location of maximum temperature dI -9"x+uco sa se Left Too tu co

- Uch | -48 XS xe 1 = -0.8m 300 Tmax = T ( Naimes) = -30 (-0.832 -48 (-0.8). +516 = 535.2K Tmax = 535.24 imat Ti = 4924 Tmax - 535.27 Tex:o) = 5164 T2 = 300k -2m M X=0 X=-0.8m Gent Vaniafim of temperaftenen on

Note: Lets verify the value of  $C_1$ in another way

T: =(+;8+ca

( after putting x= +L)

30+ 72 +(アター) = 1

( after putting x= -L)

50+ 70-17 ) = XL-

12-T1 = 2C1L

→C1=(73 - 71 21

300 – 492 C= 2x2