Please solve part two
1 answer:
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Answer:
Following are the program to the given question:
#include <stdio.h>//header file
double square_root(double N, double initialGuess)//defining a method square_root that takes two variable in parameters
{
double NG, LG = initialGuess,diff;//defining double variable
while(1)//use loop to calculate square root value
{
NG = 0.5 * (LG + N / LG);//using given formula
diff = NG - LG;//calculating difference
if(diff < 0)//use if to check difference is less than 0
diff = -diff;//decreaing difference
if(diff < 0.005)//use if that check difference is less than 0.005
break;//using break keyword
else//defining else block
{
LG = NG;//holding value
}
}
return NG;//return value
}
int main()//defining main method
{
double ans, n,initialguess = 1.0;//defining double variable
n = 4;//use n to hold value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
n = 120.5;//use n to hold value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
n = 36.01;//use n to hold value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
n = 0.25;//use n to hold value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
printf("\nEnter a number: ");//print message
scanf("%lf", &n);//input value
ans = square_root(n, initialguess);//calculating the square root value and print its value
printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number
}
Output:
Please find the attachment file.
Explanation:
- In this code, a method "square_root" is declared that takes two variable "N, initialGuess" in its parameters, inside the method a three double variable is declared.
- It uses the given formula and uses the diff variable to hold its value and uses two if to check its value is less than 0 and 0.005 and return its calculated value.
- In the main method, three double variables are declared that use the "n" to hold value and "ans" to call the method that holds its value and print its value.
Answer:
T = 167 ° C
Explanation:
To solve the question we have the following known variables
Type of surface = plane wall ,
Thermal conductivity k = 25.0 W/m·K,
Thickness L = 0.1 m,
Heat generation rate q' = 0.300 MW/m³,
Heat transfer coefficient hc = 400 W/m² ·K,
Ambient temperature T∞ = 32.0 °C
We are to determine the maximum temperature in the wall
Assumptions for the calculation are as follows
- Negligible heat loss through the insulation
- Steady state system
- One dimensional conduction across the wall
Therefore by the one dimensional conduction equation we have
During steady state
= 0 which gives
From which we have
Considering the boundary condition at x =0 where there is no heat loss
= 0 also at the other end of the plane wall we have
hc (T - T∞) at point x = L
Integrating the equation we have
from which C₁ is evaluated from the first boundary condition thus
0 = from which C₁ = 0
From the second integration we have
From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows
→ C₂ =
T(x) = and T(x) = T∞ +
∴ Tmax → when x = 0 = T∞ +
Substituting the values we get
T = 167 ° C