This question is incomplete, the complete question is;
For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.
Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.
Answer:
the required time to achieve the same concentration at a 4.9 is 83.733 hrs
Explanation:
Given the data in the question;
treatment time t₁ = 11.3 hours
Carbon concentration = 0.444 wt%
thickness at surface x₁ = 1.8 mm = 0.0018 m
thickness at identical steel x₂ = 4.9 mm = 0.0049 m
Now, Using Fick's second law inform of diffusion
/ Dt = constant
where D is constant
then
/ t = constant
/ t₁ =
/ t₂
t₂ = t₁
t₂ = t₁
/ 
t₂ = (
/
)t₁
t₂ =
/
× t₁
so we substitute
t₂ =
0.0049 / 0.0018
× 11.3 hrs
t₂ = 7.41 × 11.3 hrs
t₂ = 83.733 hrs
Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs
Answer:
public static int average(int j, int k) {
return (int)(( (long)(i) + (long)(j) ) /2 );
}
Explanation:
The above code returns the average of two integer variables
Line 1 of the code declares a method along with 2 variables
Method declared: average of integer data type
Variables: j and k of type integer, respectively
Line 2 calculates the average of the two variables and returns the value of the average.
The first of two integers to average is j
The second of two integers to average is k
The last parameter ensures average using (j+k)/2
Answer:
Noise or sound is the energy produced by sounding object which can be felt by our hearing organs .
Explanation:
Noise is produced by vibration in a body.
Answer: l = 2142.8575 ft
v = 193.99 ft/min.
Explanation:
Given data:
Thickness of the slab = 3in
Length of the slab = 15ft
Width of the slab = 10in
Speed of the slab = 40ft/min
Solution:
a. After three phase
three phase = (0.2)(0.2)(0.2)(3.0)
= 0.024in.
wf = (1.03)(1.03)(1.03)(10.0)
= 10.927 in.
Using constant volume formula
= (3.0)(10.0)(15 x 15) = (0.024)(10.927)Lf
Lf = (3.0)(10.0)(15 x 15)/(0.024)(10.927)
= 6750 /0.2625
= 25714.28in = 2142.8575 ft
b.
vf = (0.2 x 0.2 x 3.0)(1.03 x 1.03 x 10.0)(40)/(0.024)(10.927)
= (0.12)(424.36)/0.2625
= 50.9232/0.2625
= 193.99 ft/min.