Question

(10 Points) To find the heat conduction coefficient of Iron: an Iron rod with diameter of 4 cm and length of 50 cm used for the following tests.

1: The rod heated to 650 ^oC (Uniform Temperature) at first and immersed in water bath at 50 ^oC and heat transfer coefficient of h=16.3 kW/m^2oC. The surface temperature of the rod measured to be 250 ^oC after 3.45 seconds.

2: Then we removed the rod and immersed into a new water bath at 10 ^oC. The surface temperature of the rod found to be 40 ^oC after 8.095 seconds. (no heat lost between Test one and two). (Use α=23.1×10^-6m²/s, ρ=7870 kgm^-3).

1. Determine the heat conduction coefficient of the Iron.
2. Determine the center temperature of the rod in first and second experiments.
3. Determine the maximum energy in the process

Answer 1

paget - Ang Guiven data, diameter of = 4cm length of = 90cm * bath at the rod heated to at fixt = 680°c * immersed in Later = 50c * heat transfer coefficient of(b) =1613 kalmace * the surface temperature of the rod measured 2500 after 3.45 seconds. * then we removed the rod and immesed in to a to be new water bath at loc. * the surface temperature of the rod found to be 40°c. after 8.095 seconds. * X = 23.1x106mals * P = 7870 kgms. a) To determine the heat Conduction coefficient of the iron from thermophysical property of metal from data table, K= Heat conduchre coefficient = 80.26 K * At Cp=447 J/kgik Scanned with CamScanner

b) To detemine the center temperature of the rod the rod Center temperature of first experiment, pages in the first and second experirress, the *the flere, Ti=65000 To=50°c, h=1613 Kuforstoe To 250c It=3:45 sec. Bi = hro K = 16:3X103 x2x102 80.2 326 80.2 Bi =4.064 ㅗ Bi 0-246 Now, - T=xt 802 23.1x10 6 X 3.45 (2x10²)² = 0.000079695 0.0004 = 0.199 a No.2 from heisler chart, To-To Ti-TD -0.5 CSScanned with CamScanner

where, To page-3 Centre temperature. To-50 =0 650-50 = 0.7 To-50 600 To-50 =0.7x600 To-50 = 420 To = 420750 To = 470c * the center temperatur of the rod in the second kseperiment) - 250°c ts=40°c Here, Ti = TSI yer ） Too-loc F Cage t=8.095 sec r=at 23.1x10678.095 ro (2x162) .: P=0.5 from heisler chart, To-To = 0.3 CScanned with CamScannTi - Too

page-4 To slo = 0.3 &50-10 Torlo = 0.3 240 Toto = 0.3 X240 To 50:37 240 to To=82°c To determine the in maximum the Energo Proceus! Marrimum Q = Prap (Tao-ti) = 7870 X447hly C4x10^)(0.5) + ( 50-650) =-33155332.1404 . Qormasi =-33155.332kw CScanned with CamScanner