The capacity of a refrigerator is 200 TR when working between -6°C and 25°C. Determine the mass of ice produced per day from water at 25°C. Also find the power required to drive th... The capacity of a refrigerator is 200 TR when working between -6°C and 25°C. Determine the mass of ice produced per day from water at 25°C. Also find the power required to drive the unit. Assume that the cycle operates on Reverse Carnot Cycle and latent heat of ice is 335 kJ/kg. Read more

Steps to Solve

1. Calculate the Coefficient of Performance (COP) The COP of a refrigeration system is given by the formula: $$ \text{COP} = \frac{Q_c}{W} $$ where ( Q_c ) is the refrigeration effect and ( W ) is the work input. Here, ( Q_c = 40 , \text{kg} ) of ice produced per hour, which can be converted to energy using the latent heat of fusion for ice (335 kJ/kg): $$ Q_c = 40 , \text{kg} \times 335 , \frac{\text{kJ}}{\text{kg}} = 13400 , \text{kJ/hour} = \frac{13400}{3600} , \text{kJ/s} \approx 3.72 , \text{kW} $$ Since it consumes 1 kW of energy, the COP becomes: $$ \text{COP} = \frac{3.72 , \text{kW}}{1 , \text{kW}} = 3.72 $$

2. Convert TR to kJ/hr 1 TR (ton of refrigeration) is equivalent to 3.517 kW, hence: $$ \text{TR} = 200 , \text{TR} \times 3.517 , \text{kW/TR} \approx 703.4 , \text{kW} $$

3. Calculate the mass of ice produced per day To find the mass of ice produced per day, use the hourly production: $$ \text{Mass of ice/day} = 40 , \text{kg/hr} \times 24 , \text{hr/day} = 960 , \text{kg/day} $$

4. Calculate the power required to drive the unit Using the COP derived earlier, we can find the power required using: $$ W = \frac{Q_c}{\text{COP}} $$ Substituting in the values gives: $$ W = \frac{3.72 , \text{kW}}{3.72} = 1 , \text{kW} $$

5. Determine the latent heat and specific heat used For each kg of ice produced, we know the latent heat is 335 kJ/kg, which has been accounted in the previous calculations.