Question

A long rectangular copper bar of a cross section of 2cm by 3cm and a long cylindrical copper rod of 2.2 cm in diameter are initially at a uniform temperature of 120 °C. They are then exposed to a cooling air stream at 22 °C with a heat transfer coefficient of 180 W/m-K. Taking the properties of the copper to be k = 400 W/m-K and p-8930 kg/m', and C -0.385 kJ/kg-K, determine: a How long would it take for the rectangular copper bar and cylindrical copper rod to reach an average temperature of 30 °C.? b. Which one reaches the average temperature of 30 °C faster? Why? Explain your reasons.

Answer 1

Griven data Rectangulars bar = 2 cm x 3 cm : Copper rod, dia , d = 2-2cm . at initial temperature, To = 12000 Air temperature, To = 22oC. h = 180 Wlonak ka 400 W/mk i tolog 828930 kg/m3 C=0.385 ks Auge. Required final temperature, I = 30°C. We have to check whether it is possible to solve using Lumped slm analysis Lumped method is valid only if Bist no, Bi So. 1

We know that Bi = hle & Volume he a characteristic length it Surface area . for long bas, heal 2lb + bh+th) we can neglect axb) teams for long has because it is wery small compared to other areas. ad - ebbe lbh bas 2 (bhath) 2K (bad) 1b = 2x3 = 0.6cm 2 (6+1) 2 (2+3) - hic 180 N 0:6 X10 * Bi = 0.0027 = Lumped system is - valid Por long Cylindric Rod, (10 - 30-55 m = 0.0055 M Tohle C Bi <0.1 = . 180 x 0.55810 a 0.002475 400 Lumpled system analysis is valid

PCLc According to Lumped system IT-Too +/+ To-To for rectangular bar, ** Sele - 8930 x 0.385210 x 0.00 h 400 - 51.57 seconds -461.57 . e T- To To-To & 30-22 120-22 t = 129.2 seconds for Cylindrical Rod, p*= fche : 8930x 0.38*103 x0.0055 400 - 47.27 seconds = 47.20 -967.27 e 4 T- To / → 30-22 30-22 120-22 t = 118.44 seconds. (b) from above we can conclude that cylindrical rod reaches faster to 30°C (t = 118.445) than rectangulas bas (t = 129.25).