A refrigerated space is maintained at -15℃， and cooling water is available at 30℃， the refrigerant is ammonia. The refrigeration capacity is 105 kJ/h. If the compressor is operated reversibly:

(1) What is the value of ε for Carnot refrigerator?

(2) Calculate the ε for the vapor-compression cycle;

(3) Calculate the circulation rate for the refrigerant;

(4) Calculate the rating power of the compressor.​

Answer:

(1) 5.74

(2) 5.09

(3) 3.05×10⁻⁵ kg/s

(4) 0.00573 kW

Explanation:

The parameters given are;

Working temperature, [tex]T_C[/tex]  = -15°C = 258.15 K

Temperature of the cooling water, [tex]T_H[/tex] = 30°C = 303.15 K

(1) The Carnot coefficient of performance is given as follows;

[tex]\gamma_{Max} = \dfrac{T_C}{T_H - T_C} = \dfrac{258.15}{303.15 - 258.15} = 5.74[/tex]

(2) For ammonia refrigerant, we have;

[tex]h_2 = h_g = 1466.3 \ kJ/kg[/tex]

[tex]h_3 = h_f = 322.42 \ kJ/kg[/tex]

[tex]h_4 = h_3 = h_f = 322.42 \ kJ/kg[/tex]

s₂ = s₁ = 4.9738 kJ/(kg·K)

0.4538 + x₁ × (5.5397 - 0.4538) = 4.9738

∴ x₁ = (4.9738 - 0.4538)/(5.5397 - 0.4538) = 0.89

[tex]h_1 = h_{f1} + x_1 \times h_{gf}[/tex]

h₁ = 111.66 + 0.89 × (1424.6 - 111.66) = 1278.5 kJ/kg

[tex]\gamma = \dfrac{h_1 - h_4}{h_2 - h_1}[/tex]

[tex]\gamma = \dfrac{1278.5 - 322.42}{1466.3 - 1278.5} = 5.09[/tex]

(3) The circulation rate is given by the mass flow rate, [tex]\dot m[/tex] as follows

[tex]\dot m = \dfrac{Refrigeration \ capacity}{Refrigeration \ effect \ per \ unit \ mass}[/tex]

The refrigeration capacity = 105 kJ/h

The refrigeration effect, Q = (h₁ - h₄) = (1278.5 - 322.42) = 956.08 kJ/kg

Therefore;

[tex]\dot m = \dfrac{105}{956.08} = 0.1098 \ kg/h[/tex]

[tex]\dot m[/tex] = 0.1098 kg/h = 0.1098/(60*60) = 3.05×10⁻⁵ kg/s

(4) The work done, W = (h₂ - h₁) = (1466.3 - 1278.5) = 187.8 kJ/kg

The rating power = Work done per second = W×[tex]\dot m[/tex]

∴ The rating power = 187.8 × 3.05×10⁻⁵ = 0.00573 kW.

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The CountLetters application is limited to counting letters in a single word. Modify the CountLetters application to count the letters in an entire phrase, which contains spaces. Care must be taken to ignore the spaces and any other non alphabetic character found in the phrase. Be sure to change comments and variable names appropriately so that the reader of the application code understands that the letters in a phrase are counted.


import java.util.Scanner;

public class CountLetters {

public static void main(String[] args) {
final int LOW = 'A'; //smallest possible value
final int HIGH = 'Z'; //highest possible value
int[] letterCounts = new int[HIGH - LOW + 1];
Scanner input = new Scanner(System.in);
String phrase;
char[] wordLetters;
int offset; //array index

/* prompt user for a phrase */
System.out.print("Enter a phrase: ");
phrase = input.nextLine();
input.close();

/* convert word to char array and count letter occurrences */
phrase = phrase.toUpperCase();
wordLetters = phrase.toCharArray();
for (int letter = 0; letter < wordLetters.length; letter++) {
offset = wordLetters[letter] - LOW;
letterCounts[offset] += 1;
}

/* show letter occurrences */
for (int i = LOW; i <= HIGH; i++) {
System.out.println((char)i + ": " + letterCounts[i - LOW]);
}
}

}

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