Answer:
a) 158.4 HP.
b) 1235.6 °F.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to set up an energy balance for the turbine's inlets and outlets:
[tex]m_{in}h_{in}=W_{out}+m_{out}h_{out}[/tex]
Whereas the mass flow is just the same, which means we have:
[tex]W_{out}=m_{out}(h_{out}-h_{in})[/tex]
And the enthalpy and entropy of the inlet stream is obtained from steam tables:
[tex]h_{in}=1860.7BTU/lbm\\\\s_{in}= 2.2096BTU/lbm-R[/tex]
Now, since we assume the 80% accounts for the isentropic efficiency for this adiabatic gas turbine, we assume the entropy is constant so that:
[tex]s_{out}= 2.2096BTU/lbm-R[/tex]
Which means we can find the temperature at which this entropy is exhibited at 15 psia, which gives values of temperature of 1200 °F (s=2.1986 BTU/lbm-K) and 1400 °F (s=2.2604 BTU/lbm-K), and thus, we interpolate for s=2.2096 to obtain a temperature of 1235.6 °F.
Moreover, the enthalpy at the turbine's outlet can be also interpolated by knowing that at 1200 °F h=1639.8 BTU/lbm and at 1400 °F h=174.5 BTU/lbm, to obtain:
[tex]h_{out}=1659.15BUT/lbm[/tex]
Then, the isentropic work (negative due to convention) is:
[tex]W_{out}=2500lbm/h(1659.15BUT/lbm-1860.7BUT/lbm)\\\\W_{out}=-503,875BTU[/tex]
And the real produced work is:
[tex]W_{real}=0.8*-503875BTU\\\\W_{real}=-403100BTU[/tex]
Finally, in horsepower:
[tex]W_{real}=-403100BTU/hr*\frac{1HP}{2544.4336BTU/hr} \\\\W_{real}=158.4HP[/tex]
Regards!
in c the square root of a number N can be approximated by repeated calculation using the formula NG = 0.5(LG + N/LG) where NG stands for next guess and LG stands for last guess. Write a function that calculates the square root of a number using this method. The initial guess will be the starting value of LG. The program will com- pute a value for NG using the formula given. The difference between NG and LG is checked to see whether these two guesses are almost identical. If they are, NG is accepted as the square root; otherwise, the next guess (NG) becomes the last guess (LG) and the process is repeated (another value is computed for NG, the difference is checked, and so on). The loop should be repeated until the difference is less than 0. 005. Use an initial guess of 1. 0. Write a driver function and test your square root function for the numbers 4, 120. 5, 88, 36.01, 10,000, and 0. 25
PLEASE İN C PROGRAMMİNG
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.