Answer:
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %
Explanation:
given data
pressure p1 = 1.4 MPa = 14 bar
temperature t1 = 32°C
exit pressure = 0.08 MPa = 0.8 bar
to find out
the quality of the refrigerant exiting the expansion valve
solution
we know here refrigerant undergoes at throtting process so
h1 = h2
so by table A 14 at p1 = 14 bar
t1 ≤ Tsat
so we use equation here that is
h1 = hf(t1) = 332.17 kJ/kg
this value we get from table A13
so as h1 = h2
h1 = h(f2) + x(2) * h(fg2)
so
exit quality = 
exit quality = 
so exit quality = 0.2337 = 23.37 %
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %
Answer:
import java.util.Scanner;
public class FindMatchValue {
public static void main (String [] args) {
Scanner scnr = new Scanner(System.in);
final int NUM_VALS = 4;
int[] userValues = new int[NUM_VALS];
int i;
int matchValue;
int numMatches = -99; // Assign numMatches with 0 before your for loop
matchValue = scnr.nextInt();
for (i = 0; i < userValues.length; ++i) {
userValues[i] = scnr.nextInt();
}
/* Your solution goes here */
numMatches = 0;
for (i = 0; i < userValues.length; ++i) {
if(userValues[i] == matchValue) {
numMatches++;
}
}
System.out.println("matchValue: " + matchValue + ", numMatches: " + numMatches);
}
}
Sorry bro people do this22.2 pls
Answer:
Option B
10,20,24,75,70,18,60,35
Explanation:
The first, second and third iteration of the loop will be as follows
insertion sort iteration 1: 20,24,10,75,70,18,60,35
insertion sort iteration 2:10,20,24,75,70,18,60,35
insertion sort iteration 3: 10,20,24,75,70,18,60,35