Question

Consider the base plate of a 800-W household iron hav- ing a thickness of L = 0.6 cm, base area of A = 160 cm², and thermal conductivity of k = 20 W/m - °C. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached the outer surface temperature of the plate is mea- sured to be 85°C. Disregarding any heat loss through the upper part of the iron and taking the nodal spacing to be 0.2 cm, (a) obtain the finite difference formulation for the nodes and (b) determine the inner surface temperature of the plate by solving those equations. Answer: (b) 100°C Insulation Resistance heater, 800 W 85 °C Base plate -000000000 Ax = 0.2 cm 0 2 31 200000000003

Answer 1

Resistance heates Base plate k 850 8oow 2 31 Assumptions: 1 ) Heat Transfer through the base plate is steady 2) Heat Transfer is one-dimensional since the plate is 19890. zelative to thickness 3) There is no heat genccation in the plate 4) Radiation heat transferis negleded. pooperties Theomal conductivity Ka 20 W/m°c Analysis: Ases 6.20m: 0.002m The no of nodes m 0.6+134 bet m +1: 0.2 The right suffale surfale Temp 73=85c is given.

in this problem, 3 unknown nodal Temp, and 3 equations, Hodes 1 & 2 are interior pe nodes. finte dirference relation can be expressed as Tm-1-2Tm & Tmti em + Ax? Tm-i-2 Tmf Tm+1=0 cė0), fer molt2 The finite diffesence equations for nodeo on the left surface subjeded to a uniform heat flure & is obtained by an energy balance on the halfvolume element about nodeo and taking the direction of all heat toansfeos to be towaods the node Node o (left sysfall - heat flux): 90 tk Ti-To=0 / AL Model Cintesico) To - 2Ti + Toro (2) Node 2 cinteriors Ti-2T2 t To ad (3)

Ose: 0.002 m, K: 20 wlmc, T3:85 85c, 90: So 800(W) 50,000 W/m² (0.0160m m², b) The nodal Temp put all these values in nodal equations (1), (2), (3) 50,000 t 20x Ti-To (1) 0.002 TO - 2T1 + T2 (2) T1-2 T2 Ý 85 (3) from eay ( ) - 50000 TI- TO 20 0.002 Tla To a -5 Ta - 5+ To (4) put TI in eam (2) 4 (3)

To -21T12 + T2 a TO-2 ( -5 + To ) + T2 = 0 To +10-2To + T2 -Tot Tz +10 (5) € TI-252 +85 -5+To-272 +85 To-2 T2 t go (6) adding eay (5)¢ (6) - TO + T2+ 100 To 2724 800 + - T2 + 90-0 T2= qoc 7 froom ea 4 (3) Tl-2x T2 +850 T 1-2x90 + 85 co Tl - 180 +8520 T1: 95°c

foom eq 4 14) TI? -5 t To 952 - 5+ To To = 95+5 To: 100 To 100 1000, TIE 950 T22 90°c ANS. -