Question

Question You are studying heat transfer through a spherical shell container with a thermal conductivity k. The inner and outer radii are identified as a and b, respectively. The inside surface of the shell is exposed to a constant heat flux in the outward direction. The outside surface temperature of the container is measured at Note that only the variables values provided in the problem statement are known. Assume steady one-dimensional radial heat transfer a. Give the mathematical formulation of the problem. b. Obtain an expression of the temperature as a function of the radial position - Tir) in variable form. 6. Obtain an expression for the rate of heat transfer at both red and reb. Discuss your answers. Question 2 [Answers: a) 0.0429 m, 25.7°C; b) 21.2°C; c)39,600 W/m; d) h=113 W/m."C), 379 W/m..*)] At steady state, the temperature distribution in a large wall of thickness 0.15 m with a rate of heat generation (W/m) is given by: T(x) = -396r+34x + 25 where x[=] mand 7 [=] "C The wall exchange heat by convection (neglect radiation) on each side with an environment at 7. -10°C. The thermal conductivity of the wall is SO W/m-KL a. Find the location (x) and the value of the maximum temperature in the wall. b. Calculate (L) and sketch the temperature profile in the wall. c. Calculate the rate of heat generation in the wall é. d. Calculate the convective heat transfer coefficient h, and hz. 25C 015

Answer 1

Rate of heat input to shult atru nr² arly. Rate of heat Input from shell reor uw²a orto Rate of heatginuation : 418? Ar 4, (%). Hat G. Rate of heat accumulation : 0 477? Quw-4709 xly post yar?a Y Yo Corso (729x) av? Vula). q = -K (T) IT/ dr. = -K dT/ der' =) Patting an egn. y [ v>«(r) .) **, (X) = 0 Boundary cond" alr=a dlar o al. =b T = To. finite) 3K r2 dT =- Bu vrc dr 3K & const Atr=adt/ dr=o. So we have c = a 34v New dT = -4W a 34v -0. dr 3K 3K r²9 T= 4 + 2 + a 4v { - 1} + c 36 2 34 lr.

To+tu b²+ au 11. си 6K T- T -- W (+*+)-0() ) finally From ①. dT - "virtas tva der 3K 3K 7² Now qata KAI ara 4 (vnđ) - 194 ) q=0. a at b = -K A(AT) at q = (40) | Mobile 4 = < (476) obes)..

0.2 T = _396 7²344+25 New Masimp. dT(W) > 396(2)x +34 dT(x)=0 at Marn. 7-34 396(2): -> = 0.0429 m. 6). the T(L) T(L) = -396 L² +34 L-125. T( 9-15).-34 (6-IS) + 3 ( -1 ) + 25 T10.15)= 2 1. 19°c. () For Rate of heat Geneation whave a General formula from plane. wall T= -4v x² + x + C ₂. Ju 2K Comparing this with T = -396x² & 34 x+25.. 4 = 396 = 4 = 3968 21 50 W/m3 v = 39600W/3 d). dT - - 7928434. da TO) = 25°c T(0.15) = 21-19°c.

Now at Surface мом (т-то) * ( 4 ), to = 34. Is-Too = 25-10=15°c. (h) (15) = 50(34) k hi113 W/moc. Now at a=offm. ha (Tso-To) = x ( 2 ) End en(21.19-10) jk(84.8) k=50 W/m (d/dx) Ising hell. h (11.19) = k184.0) hy = 378.909 w/ ms