A 50 ton ammonia vapor compression refrigeration system operates at an evaporating temperature of -10°C and a condensing temperature of 30°C. There is no sub-cooling of liquid refr... A 50 ton ammonia vapor compression refrigeration system operates at an evaporating temperature of -10°C and a condensing temperature of 30°C. There is no sub-cooling of liquid refrigerant and superheating of suction vapor. Determine (i) the theoretical C.O.P. of the plant, and (ii) operating cost if the cost of electricity is Rs. 6.00 per kWh.
Understand the Problem
The question is asking to determine certain performance metrics of an ammonia vapor compression refrigeration system, including the theoretical coefficient of performance (COP) and the costs associated with electricity use, based on given temperatures for evaporation and condensation, alongside the lack of sub-cooling and the application of superheating.
Answer
The theoretical C.O.P. is approximately $6.58$ and the operating cost is Rs. $160.2$ per hour.
Answer for screen readers
(i) The theoretical C.O.P. of the plant is approximately $6.58$.
(ii) The operating cost is approximately Rs. $160.2$ per hour.
Steps to Solve
- Calculate the Theoretical Coefficient of Performance (COP)
The theoretical COP for a refrigeration cycle can be calculated using the formula:
$$ \text{COP} = \frac{T_{L}}{T_{H} - T_{L}} $$
where:
- ( T_{L} ) is the absolute temperature of the evaporator (in Kelvin),
- ( T_{H} ) is the absolute temperature of the condenser (in Kelvin).
Convert the temperatures from Celsius to Kelvin:
- ( T_{L} = -10°C + 273.15 = 263.15 , K )
- ( T_{H} = 30°C + 273.15 = 303.15 , K )
Now substitute ( T_{L} ) and ( T_{H} ) into the COP formula:
$$ \text{COP} = \frac{263.15}{303.15 - 263.15} = \frac{263.15}{40} $$
- Calculate the COP Value
Carry out the division:
$$ \text{COP} = \frac{263.15}{40} \approx 6.5788 $$
- Convert the System Load from Tons to kW
Convert the refrigeration capacity from tons to kilowatts. The relationship is: $$ 1 , \text{ton} = 3.517 , \text{kW} $$
So, for a 50-ton system:
$$ \text{Power} = 50 \times 3.517 = 175.85 , \text{kW} $$
- Calculate the Power Input to the System
Using the COP to calculate the power input ( W_{in} ):
$$ W_{in} = \frac{\text{Refrigeration Capacity}}{\text{COP}} $$
Substituting the values:
$$ W_{in} = \frac{175.85}{6.5788} \approx 26.7 , \text{kW} $$
- Calculate the Operating Cost
To find the operating cost per hour based on electricity cost:
Operating cost per hour = Power input (in kW) × Cost of electricity (in Rs. per kWh)
$$ \text{Operating Cost} = 26.7 \times 6 = 160.2 , \text{Rs. per hour} $$
(i) The theoretical C.O.P. of the plant is approximately $6.58$.
(ii) The operating cost is approximately Rs. $160.2$ per hour.
More Information
The coefficient of performance (COP) reflects the efficiency of the refrigeration cycle—the higher the COP, the more efficient the system is at transferring heat. The operating cost is a crucial factor for evaluating the long-term feasibility of the system.
Tips
- Not converting temperatures from Celsius to Kelvin correctly.
- Forgetting to convert the refrigeration tonnage into kilowatts.
- Using an incorrect formula or miscalculating the COP.
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