A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at 2308C by rejectin
g its waste heat to cooling water that enters the condenser at 188C at a rate of 0.25 kg/s and leaves at 268C. The refrigerant enters the condenser at 1.2 MPa and 658C and leaves at 428C. The inlet state of the compressor is 60 kPa and 2348C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator, and (d) the theoretical maximum refrigeration load for the same power input to the compressor.
1 answer:
Answer:
hello your question is incomplete attached below is the missing part and also attached is the solution
answer: a) 0.4801
b) 5.398 kw
c) 2.14
d) 12.72
Explanation:
The quality of the refrigerant at the evaporator inlet
h4 = hf4 + x4(hfx4)
Refrigeration load
Ql = m(h1-h4)
COP of the refrigerator
Ql / m(h2-h1) - Qm
Theoretical maximum refrigeration load
( Ql )max = COPr.rev * [m(h2-h1) - Qin]
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Explanation:
find attached the figure
Answer:
See explaination
Explanation:
Code;
import java.util.Scanner;
public class NumberPattern {
public static int x, count;
public static void printNumPattern(int num1, int num2) {
if (num1 > 0 && x == 0) {
System.out.print(num1 + " ");
count++;
printNumPattern(num1 - num2, num2);
} else {
x = 1;
if (count >= 0) {
System.out.print(num1 + " ");
count--;
if (count < 0) {
System.exit(0);
}
printNumPattern(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();
printNumPattern(num1, num2);
}
}
See attachment for sample output
