(SEM VI) THEORY EXAMINATION 2024-25 REFRIGERATION AND AIR CONDITIONING

BME601 – REFRIGERATION AND AIR CONDITIONING

Section-Wise Solved Answers (2024–25)

SECTION A

Attempt all questions in brief (2 × 7 = 14 marks)

(a) Define Refrigeration Effect and Ton of Refrigeration (TR).

Refrigeration effect is the amount of heat removed from a refrigerated space or substance to lower its temperature below that of the surroundings. It is usually expressed in kJ/kg of refrigerant.
Ton of Refrigeration (TR) is a unit of refrigeration capacity and is defined as the rate of heat removal required to freeze one ton of water at 0°C into ice at 0°C in 24 hours.

1 TR=3.517 kW=211 kJ/min1 \, TR = 3.517 \, kW = 211 \, kJ/min1TR=3.517kW=211kJ/min 

(b) Differentiate between Refrigerator and Heat Pump.

A refrigerator is designed to remove heat from a low-temperature region and reject it to the surroundings to maintain cooling. The main objective is cooling.
A heat pump also removes heat from a low-temperature source, but its primary purpose is to deliver heat to a high-temperature region for heating applications. Both operate on similar cycles, but their objectives differ.

(c) Why is sub-cooling done in a vapour compression cycle?

Sub-cooling is done to reduce the enthalpy of liquid refrigerant leaving the condenser. This increases the refrigeration effect, reduces the formation of vapour bubbles before expansion, and improves the overall COP of the vapour compression refrigeration system.

(d) What is the significance of DART (Dry Air Rated Temperature) in aircraft refrigeration?

DART represents the effective temperature of dry air entering the aircraft refrigeration system. It accounts for pressure and temperature changes due to high-altitude flight and helps in accurate estimation of cooling requirements for cabin conditioning.

(e) Define refrigerant and absorbent in vapour absorption refrigeration system.

A refrigerant is the working fluid that absorbs heat during evaporation and releases heat during condensation.
An absorbent is a substance that absorbs the refrigerant vapour, forming a strong solution. In the ammonia-water system, ammonia is the refrigerant and water is the absorbent.

(f) What are Ozone Depletion Potential (ODP) and Global Warming Potential (GWP)?

ODP indicates the ability of a refrigerant to deplete the ozone layer relative to CFC-11.
GWP measures the potential of a refrigerant to contribute to global warming compared to CO₂ over a specific time period. Lower ODP and GWP refrigerants are environmentally preferred.

(g) Define Sensible Heat Factor (SHF) and Apparatus Dew Point (ADP).

Sensible Heat Factor is the ratio of sensible heat to total heat load in an air-conditioning process.
Apparatus Dew Point is the effective coil surface temperature at which air leaving the cooling coil would be saturated if cooling were ideal.

SECTION B

Attempt any three (7 × 3 = 21 marks)

(a) Reversed Carnot Refrigeration Cycle and its limitations.

The reversed Carnot cycle consists of two isothermal and two isentropic processes operating between two temperature limits. It is the most efficient refrigeration cycle theoretically. However, it is not practical due to difficulty in maintaining isothermal heat transfer, requirement of wet compression, and complex control of processes in real systems.

(b) Comparison of Boot-Strap and Reduced Ambient aircraft refrigeration systems.

The boot-strap system uses an additional heat exchanger and compressor to improve cooling capacity, making it suitable for high-speed aircraft.
The reduced ambient system uses ram air to reduce air temperature before expansion. Boot-strap systems provide higher COP and better control, while reduced ambient systems are simpler.

(c) Effect of superheating and sub-cooling on VCRS performance.

Superheating increases the refrigerating effect but also increases compressor work.
Sub-cooling increases refrigeration effect without increasing compressor work, thereby improving COP. Both processes enhance system performance when properly controlled.

(d) Working of vapour absorption refrigeration system (Ammonia-Water).

In this system, ammonia evaporates in the evaporator and absorbs heat. The vapour is absorbed by water in the absorber, forming a strong solution. The solution is pumped to the generator, where ammonia is separated using heat input. The ammonia vapour is condensed, expanded, and reused. Mechanical work is minimal, and thermal energy drives the cycle.

(e) Secondary refrigerants and comparison with primary refrigerants.

Secondary refrigerants are substances like brine or glycol used to transfer cooling from the primary refrigeration system to the load.
Primary refrigerants directly undergo phase change. Secondary refrigerants are safer and reduce refrigerant charge but add pumping losses.

SECTION C

Attempt any one (7 marks)

(a) Numerical on simple VCRS

Given:
Refrigeration capacity = 5 kW
Enthalpy at evaporator inlet = 75 kJ/kg
Enthalpy at evaporator outlet = 183 kJ/kg
Enthalpy at compressor discharge = 210 kJ/kg

Refrigeration effect:

qL=183−75=108 kJ/kgq_L = 183 - 75 = 108 \, kJ/kgqL​=183−75=108kJ/kg

Mass flow rate:

m˙=5108=0.0463 kg/s\dot{m} = \frac{5}{108} = 0.0463 \, kg/sm˙=1085​=0.0463kg/s

Compressor work:

W=210−183=27 kJ/kgW = 210 - 183 = 27 \, kJ/kgW=210−183=27kJ/kg

Power consumption:

P=0.0463×27=1.25 kWP = 0.0463 \times 27 = 1.25 \, kWP=0.0463×27=1.25kW

COP:

COP=51.25=4COP = \frac{5}{1.25} = 4COP=1.255​=4

Heat rejected:

QH=QL+W=5+1.25=6.25 kWQ_H = Q_L + W = 5 + 1.25 = 6.25 \, kWQH​=QL​+W=5+1.25=6.25kW 

(b) Ice plant numerical (outline explanation)

Total heat removal includes cooling water from 35°C to 0°C, freezing at 0°C, cooling ice to −8°C, and 10% heat leakage. Using given data, refrigeration load is calculated first, then compressor power using R-12 properties.