A 50 ton ammonia vapor compression refrigeration system operates at an evaporating temperature of -10°C and condensing temperature of 30°C. There is no sub-cooling of liquid refrig... A 50 ton ammonia vapor compression refrigeration system operates at an evaporating temperature of -10°C and condensing temperature of 30°C. There is no sub-cooling of liquid refrigerant and superheating of suction vapor. Determine (i) theoretical C.O.P. of the plant and (ii) operating cost of the plant, if the cost of electricity is Rs. 6.00 per kWh. Write a short note on (any two): (a) Refrigerated transport system, (b) Vapour absorption refrigeration system, (c) Ice manufacturing.
Understand the Problem
The question is asking to calculate the theoretical coefficient of performance (C.O.P.) of an ammonia vapor compression refrigeration system given its operating conditions, as well as to determine the operating cost based on the electricity rate. Additionally, it requires short notes on two specific topics related to refrigeration systems.
Answer
(i) $6.58$; (ii) $\text{Rs. } 160.20 \text{ per hour}$.
Answer for screen readers
(i) The theoretical C.O.P. of the plant is approximately $6.58$.
(ii) The operating cost of the plant is about $\text{Rs. } 160.20 \text{ per hour}$.
Steps to Solve
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Convert Refrigeration Capacity To convert the refrigeration capacity from tons to kilowatts (kW): $$ 1 \text{ ton} = 3.517 \text{ kW} $$ So, $$ 50 \text{ tons} = 50 \times 3.517 = 175.85 \text{ kW} $$
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Determine the C.O.P. (Coefficient of Performance) The theoretical C.O.P. of a refrigeration system can be calculated using: $$ \text{C.O.P.} = \frac{T_L}{T_H - T_L} $$ where ( T_L ) is the absolute temperature of the evaporator (in Kelvin) and ( T_H ) is the absolute temperature of the condenser (in Kelvin).
Convert the temperatures from Celsius to Kelvin: $$ T_L = -10 + 273.15 = 263.15 \text{ K} $$ $$ T_H = 30 + 273.15 = 303.15 \text{ K} $$
Now substituting these values into the C.O.P. equation: $$ \text{C.O.P.} = \frac{263.15}{303.15 - 263.15} = \frac{263.15}{40} = 6.57875 $$
- Calculate Operating Cost To determine the operating cost of the plant, first find the power input (in kW) using the theoretical C.O.P.: $$ \text{Power input} = \frac{\text{Cooling Capacity}}{\text{C.O.P.}} $$ So, $$ \text{Power input} = \frac{175.85}{6.57875} \approx 26.70 \text{ kW} $$
Now, calculate the operating cost per hour: $$ \text{Operating Cost} = \text{Power input} \times \text{Cost of Electricity} $$ $$ \text{Operating Cost} = 26.70 \text{ kW} \times 6 \text{ Rs/kWh} \approx 160.20 \text{ Rs/hour} $$
- Short Notes on Refrigeration Topics (a) Refrigerated Transport System: This system is designed to preserve perishables during transportation. Refrigerated trucks, containers, and ships maintain a controlled temperature environment. They often use vapor compression or absorption refrigeration systems.
(b) Vapour Absorption Refrigeration System: This system utilizes a refrigerant and absorbent mixture to provide cooling. It is typically powered by gas or waste heat instead of electricity. It is quieter and more environmentally friendly, making it suitable for applications where noise is a concern.
(i) The theoretical C.O.P. of the plant is approximately $6.58$.
(ii) The operating cost of the plant is about $\text{Rs. } 160.20 \text{ per hour}$.
More Information
The theoretical Coefficient of Performance (C.O.P.) is a critical measure of the efficiency of refrigeration systems. A higher C.O.P. indicates better performance. The operating cost gives insight into the economic viability of running such a system.
Tips
- Ignoring unit conversions: Always ensure temperatures are expressed in absolute units (Kelvin) when using thermodynamic equations.
- Incorrect C.O.P. calculation: Make sure to double-check the formula used for C.O.P. and the input values.
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