Question

3. A spherical reactor made of glass (Pyrex) has inside and outside diameters of 0.15m and 0.30m, respectively. The inner surface is maintained at TS,1=150°C and the outer surface is maintained at Ts.2=30°C. (a) Determine the temperature at the midpoint of the glass wall. (b) For the same surface temperatures and reactor dimensions, how does the midpoint temperature differ if the reactor were made of 316 stainless steel?

Answer 1

To= outer sphere temperature Tma Temperature at midpoint T = inner surface. temperature. en a sphere as shown in the Consider figure a sphere as sh. Given inside diameter= 0.15 m. Inner nadius ro - 0.15 = Outen diameter do= 0.3 m Duter radius= do ro= 0.3=0. At the midpoint the radius is. m- rotri = 0.15 +0.075 = 0.1125m He at conduction equation fora sphene between nadii r and raisi Q = (OTA TT k Crir₂ - where o is heat transfer rate and kis thermal conductivity, AT = T₂-T, Teacher's Signature

Q = 40 k (rorm) CT-To) At steady state this is egne to the heat transfer between midpoint and inner surface VO Q =90k Crorm) (Tom-T6) = 471 rin) 55:- In more Tom-To) = rm CT: -Tm) - Given (rm-r;) Ti = 150°C. To= 30°c 0-15x 0.125 XCTM-30) 50.0 75x0.1125 (o. 15 0.1125 x 615o-In). Co.1125 -0.075) 0.45(Tm-30) = 0.2250150-Tm). 2(Tm- 30 -150-Im 2Tm- 60 =150-Tm Teacher's Signature

3 Im= 150+60=210 Tmo 70°C From equation 2 wesee that This not dependent on thermal conductivity k. Therefore inthis case also Tm- 70°C