During experiments on laser etching of microchips for IBM, a nuclear engineer places a discshaped sample 5 mm in diameter in a v
acuum chamber with surface temperature of 300 K. A 500 mW laser is then used to irradiate the sample uniformly. The sample surface can be considered ‘gray’ with an emissivity of 0.9 and can be considered ‘thin’ so that the disk temperature is uniform. In addition, a non-participating gas at a temperature of 27⁰C is used to cool the sample with a convection coefficient of 50 W/(m2K). Find the steady-state temperature of the disk
1 answer:
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Answer:
A) n = 0.6143, c ≈ 640m/min
B) n = 0.6143 , c = 637.53m/min
Explanation:
using the given data
A) A plot of flank wear as a function of time and also A plot for tool when
Flank wear is 0.75 and cutting edge speed is 100m/min, Time of cutting edge is said to be 20.4 min also for cutting edge speed of 155m/min , time for cutting edge is 10 min
is attached below
calculate for the constant N from the second plot
note : the slope will be negative because cutting speed decreases as time of cutting increase
V1 = 100m/min , V2 = 155m/min, T1 = 20.4 min, T2 = 10 min
= - N =
therefore - N =
= - 0.6143
THEREFORE ( N ) = 0.6143
Determine for the constant C from the second plot as well
note : C is the intercept on the cutting speed axis in 1 min tool life
connecting the two points with a line and extend it to touch the cutting speed axis and measure the value at that point
hence C ≈ 640m/min
B) Calculate the values of N and C in the Taylor equation solving simultaneous equations
using the above cutting speed and time of cutting values we can find the constant N via Taylor tool life equation
Taylor tool life equation = vT = C ------------- equation 1
cutting speed = v = 100m/min and 155m/min
tool life = T = 20.4 min and 10 min
also constant n and c are obtained from the previous plot
back to taylor tool life equation = 100 * 20.4 = C
therefore C = (100)(20.4)^n ---------------- equation 2
also using the second values of v and T
taylor tool life equation = 155 * 10 = C
therefore C = ( 155 )(10)^n ----------------- equation 3
Equate equation 2 and equation 3 and solve simultaneously
(100)(20.4)^n = (155)(10)^n
To find N
take natural log of both sides of the equation
= In ((100)(20.4)^n) = In((155)(10)^n)
= In (100) + nIn(20.4) = In(155) + nIn(10)^n
= n(3.0155) - n (2.3026) = 5.043 - 4.605
= 0.7129 n = 0.438
therefore n = 0.6143
To find C
substitute 0.6143 for n in equation 2
C = (100)(20.4) ^ 0.6143
C = 637.53 m/min
Attached are the two plots for solution A
