Constant specific heat at constant pressure c_p = .460 J/kg, degree C Thermal co
ID: 283051 • Letter: C
Question
Constant specific heat at constant pressure c_p = .460 J/kg, degree C Thermal conductivity k = 16.3 W/m degree C In the same mechatronic robotic manufacturing company in problem number one above, LARGE brass plates as it relates to its GEOMETRICAL LENGTH approximately 2L = 4 cm as it relates to its GEOMETRICAL WIDTH thick are utilized as press plates the forming of special robotic parts for the robotic systems. The large brass plates are initially at a uniform temperature of T_1 = 20 degree C and are heated by passing them through an oven that is maintained at T_infinity = 500 degree C. The plates remain in the oven for a period of time approximately t = ten (10) minutes for metallurgical reduction of the mechanical properties. You are to determine the following: a. The centerline temperature of the brass plate. b. The surface temperature at x = L which is the half length of the plate which is 2L wide Also recall from class lectures on the Heisler Charts and the Engineering Roadmap concepts I exposed the class to that by definition the following definitions apply; theta = T (x, t) - T, and theta_i = T_i - T_w and theta_0 = T_0 - T_infinityExplanation / Answer
half thickness of plate=0.02m
1/Bi=k/hL
k=110W/m degree celsius
h=120W/m degree celsius
110/120*0.02
=110/2.4=45.8
tau=alpha t/L^2
alpha=33.9*10^-6m2/s
L=0.02m
t=time=10min=10*60=600 sec
Now tau=33.9*10^-6 *600/(0.02)^2
=33.9*10^-6*600/4*10^-4
=20340*10^-6/4*10^-4
=5085*10^-2
=50.85
Now To-T infinite/T initial-T infinite=0.46
where To=centerline temperature=??
Tinfinite=500degree celsius
Also
1/Bi=K/hL=45.8
x/L=L/L=1
T-Tinfinte/To-Tinfinite=0.99
Therefore T-Tinfintie/Ti-Tinfinite=0.46*0.99
=0.455
T=Tinfinite+0.455(Ti-Tinfinite)=500+0.455(20-500)
=282degree celsius
Therefore surface temperature=282 degree celsius