Question

2. An aluminum cylinder has a radius (ra) of 3 cm and al n cylinder has a radius (11) of 3 cm and a length (L) of 10 cm. The cylinder is covered with an additional insulation laver which has the critical insulation thickness The insulation surface is exposed to ambient air with the convection heat trans coefficient of h = 5 W/m2.°C. The surface temperature (T1) of aluminum cylinder is 100 The ambient air temperature (T..) is 25 °C. T1=100°C Given) K1 = 204 W/m°C. K2 = 0.16 W/ m°C. Aluminum (K1) Assumption) 1-D, steady state heat transfer in radial direction h = 5 W/m2K To = 25 °C Insulation (K2) Find) a) Find the radius of insulation layer, 12 (10 pts). b) Sketch a total thermal resistance circuit from T1 to To including the convection effect. Label all resistances in terms of relevant variables (10 pts). c) Find the heat loss rate of the cylinder, q (10 pts). d) Find the insulation surface temperature, T2 (10 pts).

Answer 1

Given that, Tag = 2508 unu Alumin K22 Insulation naclius of aluminium, mi H T, = 100°C 9,- 3cm 09, = 0.03 m length of the cylinder, I = 10 cm L=00lm Convection heat transfer coefficient es ,h=5w/m²c. T Assumption : -D, steady, state heat transfer en radial direction, egy ki = 204 W/mic K2 = 0.16 w/mic a) rachius of insulation, r, e Pole m n = 0 1 0 32 . r = 32 mm

b) Thermal resistance circuit :- å uzun to In 1966) Latinh) 2 ITL K m Rt Where where , ennula) - = Conduction resistance ln (relai) 21 KL - 21092hL = Convection resistance a → R = total resistance.

rate and are equal. here , conduction heat transfer Convection heat transfer rate ů = -K2TTL (T3-T) 0 In( ) Q = haz (T2-Too) ☺ Inca2. > - K2272 ( Tz-T,) = hA2 (T2-To) en de and KZTY (T, -T2) - 12%) (T2-To) en ( 72 ) - 5 (0.032) (T2-25) ✔ 0.16 (100 -T2) en (2603) - => 0.16 ( 100-T2) = 0.0103 (12-25) [ Ir 16 - 0:16T2 = 0.0103T2 - 002575 => 0.1703T2 = 16:2575 16. 2575 - T2 = 701703 | T2 = 950 463 °c

in the insulation surface temperature is T2 = 95.463 °c Then å = naz (T2-Too) 25(21772 L) (T2-Too) = 5(2*3.14 X 0.032 X 0.1) (95.463 – 25) a = 7.08 W]