Answer:
a) q = 7671 W
T0 = 47.6°C
b) ΔP = 202.3 N/m²
P = 58.2 W
c) hDarray = 2 times hD of an isolated element.
Explanation:
see the image for the solution.
(a) For BCC iron, calculate the diameter of the minimum space available in an octahedral site at the center of the (010) plane,
1 answer:
Response:
a) The diameter available is 0.0384 nm
b)This space is less than the size of a carbon atom, which has a radius of 0.077 nm, indicating that the carbon atom won't occupy these sites.
Clarification:
For BCC iron
According to the information in Appendix B, the lattice parameter (a) is determined to be 0.2866 nm
BCC iron encompasses 4 atomic radii, thus the body diagonal length =
which represents the atomic radius of BCC iron
4r =
Substituting the value of (a) from Appendix B, set as 0.2866 nm
4r =
leading to r = 0.4964 nm / 4 = 0.1241 nm
Refer back to Appendix C, where the atomic radius of BCC iron is stated as 0.1241 nm, assuming the atomic sizes for iron remain consistent.
Thus, the radius ratio = 0.62
According to Figure 3.2, the space necessary for an interstitial at the BCC site exists between atoms located at the FCC site, containing two atoms, each equal to a radius of 0.2482 nm
The diameter of the minimum space available
= 0.2482 nm
With a = 0.2666 nm
therefore
d = 0.2866 nm - 0.2482 nm = 0.0384 nm
When comparing with the diameter of a carbon atom
This space is smaller than that of a carbon atom which has a radius of 0.077 nm, confirming that the carbon atom will not be able to occupy these positions.
You might be interested in
Response:
Refer to the explanation
Clarification:
Code:
import java.util.Scanner;
public class NumberPattern {
public static int x, count;
public static void displayNumPattern(int num1, int num2) {
if (num1 > 0 && x == 0) {
System.out.print(num1 + " ");
count++;
displayNumPattern(num1 - num2, num2);
} else {
x = 1;
if (count >= 0) {
System.out.print(num1 + " ");
count--;
if (count < 0) {
System.exit(0);
}
displayNumPattern(num1 + num2, num2);
}
}
}
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int num1;
int num2;
num1 = scnr.nextInt();
num2 = scnr.nextInt();
displayNumPattern(num1, num2);
}
}
See attached example output
Given Information:
Frequency = f = 60 Hz
Complex rated power = G = 100 MVA
Inertia constant = H = 8 MJ/MVA
Mechanical power = Pmech = 80 MW
Electrical power = Pelec = 50 MW
Number of poles = P = 4
No. of cycles = 10
Required Information:
(a) stored energy =?
(b) rotor acceleration =?
(c) change in torque angle =?
(c) rotor speed =?
Answer:
(a) stored energy = 800 Mj
(b) rotor acceleration = 337.46 elec deg/s²
(c) change in torque angle (in elec deg) = 6.75 elec deg
(c) change in torque angle (in rmp/s) = 28.12 rpm/s
(c) rotor speed = 1505.62 rpm
Explanation:
(a) Calculate the rotor's stored energy at synchronous speed.
The stored energy is represented as
Where G stands for complex rated power and H signifies the inertia constant of the turbo-generator.
(b) If we suddenly increase the mechanical input to 80 MW against an electrical load of 50 MW, we shall find the rotor's acceleration while ignoring mechanical and electrical losses.
The formula for rotor acceleration is given by
Where M is defined as
(c) If the acceleration derived in part (b) persists over 10 cycles, we will calculate both the change in torque angle and the rotor speed in revolutions per minute at the end of this duration.
The change in torque angle is expressed as
Where t is determined from
Consequently,
The change in torque in rpm/s is provided by
The rotor speed in rpm at the culmination of this 10-cycle period is calculated as
Where P indicates the number of poles on the turbo-generator.
Answer:
Refer to the attached document
1512 ft
Explanation:
Because the acceleration is constant or zero, the acceleration-time graph consists of horizontal segments. The values for t2 and a4 are derived as follows:
Acceleration - Time
0 < t < 6: Velocity change = area beneath the a–t graph
V_6 - 0 = (6 s)(4 ft/s²) = 24 ft/s
6 < t < t2: The velocity rises from 24 to 48 ft/s,
Velocity change = area beneath the a–t graph
48 - 24 = (t2 - 6) * 6
t2 = 10 s
t2 < t < 34: The velocity remains constant, meaning acceleration is zero.
34 < t < 40: Velocity change = area beneath the a–t graph
0 - 42 = 6*a4
a4 = - 8 ft / s²
A negative acceleration shows the area lies below the t-axis, indicating a decrease in speed.
Velocity - Time
Since acceleration remains constant or zero, the v−t graph is made up of linear segments connecting the calculated points.
Position change = area beneath the v−t graph
0 < t < 6: x6 - 0 = 0.5*6*24 = 72 ft
6 < t < 10: x10 - x6 = 0.5*4*(24 + 48) = 144 ft
10 < t < 34: x34 - x10 = 48*24 = 1152 ft
34 < t < 40: x40 - x34 = 0.5*6*48 = 144 ft
Summing these position changes yields the distance from A to B:
d = x40 - 0 = 1512 ft
