Question

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

What they said up there above me is true :)

Answer 2

(1) 5.74

(2) 5.09

(3) 3.05×10⁻⁵ kg/s

(4) 0.00573 kW

Step-by-step explanation:

The parameters given are;

Working temperature,
`T_C` = -15°C = 258.15 K

Temperature of the cooling water,
`T_H` = 30°C = 303.15 K

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

`\gamma_(Max) = (T_C)/(T_H - T_C) = (258.15)/(303.15 - 258.15) = 5.74`

(2) For ammonia refrigerant, we have;

`h_2 = h_g = 1466.3 \ kJ/kg`

`h_3 = h_f = 322.42 \ kJ/kg`

`h_4 = h_3 = h_f = 322.42 \ kJ/kg`

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

`h_1 = h_(f1) + x_1 * h_(gf)`

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

`\gamma = (h_1 - h_4)/(h_2 - h_1)`

`\gamma = (1278.5 - 322.42)/(1466.3 - 1278.5) = 5.09`

(3) The circulation rate is given by the mass flow rate,
`\dot m` as follows

`\dot m = (Refrigeration \ capacity)/(Refrigeration \ effect \ per \ unit \ mass)`

The refrigeration capacity = 105 kJ/h

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

Therefore;

`\dot m = (105)/(956.08) = 0.1098 \ kg/h`

`\dot m` = 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×
`\dot m`

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