1 steel balls that are 18 mm in diameter are annealed by...

Name: 1 steel balls that are 18 mm in diameter are annealed by heating to 1200 k and then slowly cooled to 400 k in an air environment of tinfty 350 k and ho 20 wm2 cdotk the properties of steel 51847
Uploaded: 2023-07-01T06:14:36-08:00
Duration: 4 min 43 s
Channel: Adi S
Description: 1 steel balls that are 18 mm in diameter are annealed by heating to 1200 k and then slowly cooled to 400 k in an air environment of tinfty 350 k and ho 20 wm2 cdotk the properties of steel 51847

1. Steel balls that are 18 mm in diameter are annealed by heating to 1200 K and then slowly cooled to 400 K in an air environment of $T_{\infty} = 350$ K and $h_o = 20$ W/m$^2 \cdot$K. The properties of steel are $k = 60$ W/m$\cdot$K, $\rho = 7800$ kg/m$^3$ and $c = 600$ J/kg$\cdot$K. i. Estimate the time required for the cooling process. ii. Determine the instantaneous rate of heat transfer at the end of the cooling process (when $T = 400$ K). iii. Determine the total amount of heat transferred for a box of 100 steel balls.

Transcript

00:01 So let us start with the concept which we are going to use here for this question.

00:06 So during the annealing process, the time required for the cooling is given by t is equals to pvc times of rho divided by the has times of the ln of into the bracket ti minus of t infinity by t minus of t infinity.

00:30 So according to the question, we are given the information that is the steel bar at 12 mm in diameter, that is t is equals to 12 mm, heating annealed by heating the 1100s of kelvin.

00:49 So basically this will be the initial temperature that is 1100s of kelvin and final temperature t would be equals to 360 kelvin, this is given and t infinite temperature we have been given 325 of kelvin, while the value of h we have been given 20 watt per meter square kelvin.

01:19 So while assuming the properties of steel, we are given some values that is k is equals to 40 watt per meter, the value of density rho is equals to 7800 kg per meter cube, this is also given and c is equals to 600 joules per kg kelvin, that is specific heat capacity.

01:43 So we need to find the time required for the cooling.

01:45 So for that first of all, we need to find out for sphere lc would be equals to anode battery.

01:57 So from here, the biot number which is denoted by bi would be equals to la times of h by k.

02:07 So this should be equals to h times of anode by thrice of k by substituting the value of lc here.

02:14 So that will turns out equals to this is lc, 20 into anode is 0 .002 divided by 3.

02:28 So it will be 40 here, it will be 40...

Question

Please give Ace some feedback