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.​

Respuesta :

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.

What they said up there above me is true :)