Refrigeration and Air Conditioning MCQ Quiz - Objective Question with Answer for Refrigeration and Air Conditioning - Download Free PDF
Last updated on Jun 27, 2025
Latest Refrigeration and Air Conditioning MCQ Objective Questions
Refrigeration and Air Conditioning Question 1:
The reversed Carnot cycle is used as a basis for which of the following?
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 1 Detailed Solution
Explanation:
The Reversed Carnot Cycle:
- The reversed Carnot cycle is a theoretical thermodynamic cycle that forms the basis for understanding the operation of refrigerators and heat pumps. This cycle is essentially the Carnot cycle operating in reverse. While the Carnot cycle describes the most efficient process for converting heat into work (as used in heat engines), the reversed Carnot cycle describes the most efficient process for transferring heat from a colder area to a hotter area using work input, which is the principle behind refrigeration and heat pumping systems.
In a reversed Carnot cycle, the main goal is to transfer heat from a low-temperature reservoir to a high-temperature reservoir. The cycle consists of four thermodynamic processes:
- Isentropic Compression: In this process, the working fluid (often a refrigerant) is compressed isentropically, meaning there is no change in entropy. This compression increases the temperature and pressure of the fluid.
- Isothermal Heat Rejection: The working fluid, now at a higher temperature, releases heat to the high-temperature reservoir (e.g., the surrounding environment). This process occurs at a constant temperature.
- Isentropic Expansion: The fluid undergoes isentropic expansion, reducing its pressure and temperature. This process prepares the fluid for absorbing heat in the next step.
- Isothermal Heat Absorption: Finally, the working fluid absorbs heat from the low-temperature reservoir (e.g., the space to be cooled). This process also occurs at a constant temperature.
The reversed Carnot cycle is idealized and assumes no irreversibilities, which means it represents the maximum possible efficiency for any refrigeration or heat pump system. Real-world systems, such as vapor-compression refrigeration cycles, are modeled on the reversed Carnot cycle but include inefficiencies like friction, pressure drops, and non-ideal gas behavior.
Applications:
The reversed Carnot cycle is the theoretical foundation for:
- Refrigerators: Devices that transfer heat from the interior of a refrigerator (low temperature) to the surrounding environment (high temperature) to maintain a cool interior.
- Heat Pumps: Systems that transfer heat from a cold source (e.g., the ground or air in winter) to a warm space (e.g., the interior of a building) for heating purposes.
Refrigeration and Air Conditioning Question 2:
The Bell-Coleman cycle is also known as:
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 2 Detailed Solution
Concept:
- Bell Coleman cycle is also known as Reversed Brayton cycle or Reversed Joule cycle
- The working fluid of the Bell Coleman refrigeration cycle is Air.
- This system of refrigeration is used for Air Craft refrigeration and it has light weight.
where
- Process 1 2: isentropic compression
- Process 2 3: constant pressure heat rejection
- Process 3 4: isentropic expansion
- Process 4 1: constant pressure heat absorption
Air Refrigeration System and BellColeman Cycle or Reversed Brayton Cycle:
- In the air refrigeration system the air is taken into the compressor from the atmosphere and compressed.
- The hot compressed air is cooled in a heat exchanger up to the atmospheric temperature (in ideal conditions).
- The cooled air is then expanded in an expander. The temperature of the air coming out from the expander is below the atmospheric temperature due to isentropic expansion.
- The lowtemperature air coming out from the expander enters into the evaporator and absorbs the heat. The cycle is repeated.
Refrigeration and Air Conditioning Question 3:
Which of the following does a cascade refrigeration system use?
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 3 Detailed Solution
Explanation:
Cascade Refrigeration System
Definition: A cascade refrigeration system is a specialized type of refrigeration system that uses two or more refrigeration cycles with different refrigerants to achieve very low temperatures. Each cycle operates at different temperature ranges and is connected in series, where the evaporator of one cycle serves as the condenser for the next cycle.
Working Principle: In a cascade refrigeration system, multiple refrigeration circuits are used in stages, each with its own refrigerant that is best suited for the specific temperature range it operates within. The first stage (high-temperature stage) uses a refrigerant with a relatively higher boiling point, which condenses at a temperature that serves as the evaporator temperature for the second stage (low-temperature stage). The second stage uses a refrigerant with a much lower boiling point, capable of achieving extremely low temperatures.
For example, in a two-stage cascade system, the high-temperature stage might use a refrigerant such as R-134a, while the low-temperature stage might use a refrigerant like R-23. The high-temperature stage absorbs heat from the environment and transfers it to the low-temperature stage, where the second refrigerant absorbs the heat and evaporates at a much lower temperature, providing the desired refrigeration effect.
Advantages:
- Ability to achieve extremely low temperatures that are not possible with a single refrigeration cycle.
- Enhanced efficiency and performance by using refrigerants that are optimized for specific temperature ranges.
- Flexibility in selecting refrigerants that are environmentally friendly and suitable for the desired temperature range.
Disadvantages:
- Increased complexity in design and operation due to multiple refrigeration circuits and components.
- Higher initial cost and maintenance requirements compared to single-stage systems.
- Requires precise control and coordination between the different stages to ensure optimal performance.
Applications: Cascade refrigeration systems are commonly used in applications requiring very low temperatures, such as in cryogenics, pharmaceutical industries, and scientific research laboratories.
Analysis of Other Options:
Option 2: A single refrigerant in both cycles
This option is incorrect because a cascade refrigeration system specifically utilizes multiple refrigerants with different boiling points to achieve the desired temperature ranges. Using a single refrigerant in both cycles would not allow the system to reach extremely low temperatures effectively.
Option 3: Only ammonia as a refrigerant
This option is incorrect because while ammonia is a common refrigerant used in various refrigeration systems, a cascade refrigeration system requires multiple refrigerants with different boiling points to operate efficiently. Relying solely on ammonia would not provide the necessary temperature differential between the stages.
Option 4: Only air as a working fluid
This option is incorrect because air is not used as a refrigerant in cascade refrigeration systems. Refrigeration systems typically use refrigerants that undergo phase changes (liquid to vapor and vice versa) to absorb and reject heat. Air does not have the necessary properties to function effectively as a refrigerant in this context.
Refrigeration and Air Conditioning Question 4:
The humidity ratio of atmospheric air at 28°C [] dry bulb temperature and 760 mm Hg pressure is 0.018 kg/kg of dry air. What will be the relative humidity of air?
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 4 Detailed Solution
Concept:
Relative humidity is given by:
\( \phi = \frac{P_v}{P_{sat}} \times 100 \)
Where partial vapor pressure is calculated using:
\( P_v = \frac{\omega \times P}{0.622~ + ~\omega} \)
Given:
- \( \omega = 0.018 \)
- \( P = 760 \, \text{mm Hg} \)
- \( P_{sat} = 28.34 \, \text{mm Hg} \)
Calculation:
\( P_v = \frac{0.018 \times 760}{0.622 + 0.018} \approx 21.375 \, \text{mm Hg} \)
\( \phi = \frac{21.375}{28.34} \times 100 \approx {75.42\%} \)
Refrigeration and Air Conditioning Question 5:
A refrigerator with a COP of 2 removes heat from the refrigerated space at the rate of 100 kJ/min. The amount of heat rejected to the surroundings will be:
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 5 Detailed Solution
Concept:
The COP of a refrigerator is defined as:
\( COP = \frac{Q_L}{W} \Rightarrow W = \frac{Q_L}{COP} \)
Total heat rejected: \( Q_H = Q_L + W \)
Given:
- \( COP = 2 \)
- \( Q_L = 100~\text{kJ/min} = 1.667~\text{kW} \)
Calculation:
\( W = \frac{1.667}{2} = 0.8335~\text{kW} \)
\( Q_H = 1.667 + 0.8335 = 2.5~\text{kW} \)
Top Refrigeration and Air Conditioning MCQ Objective Questions
A Carnot heat pump works between 27° C and 327°C. What will be its COP?
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 6 Detailed Solution
Download Solution PDFConcept:
\({\rm{COP\;of\;Carnot\;Heat\;pump}} = \frac{{{T_1}}}{{{T_1} - {T_2}}}\)
Calculation:
Given:
T1 = 327° C = 600 K, T2 = 27° C = 300 K
\(\therefore COP = \frac{{{T_1}}}{{{T_1} - {T_2}}}\)
\(\therefore COP = \frac{{600}}{{600 - 300}} = \frac{{600}}{{300}} = 2\)
∴ COP of Carnot heat pump = 2
A Carnot engine receiving heat at 400 K has an efficiency of 50 %. What is the COP of a Carnot refrigerator working between the same temperature limits?
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 7 Detailed Solution
Download Solution PDFConcept:
\(η_{carnot}=\frac{T_H-T_L}{T_H}=1-\frac{T_L}{T_H}\)
\((COP)_{carnot}=\frac{T_L}{T_H-T_L}\)
Calculation:
Given:
ηCarnot = 50 % = 0.5, TH = 400 K
\(η_{carnot}=\frac{T_H-T_L}{T_H}=1-\frac{T_L}{T_H}\)
\(\Rightarrow\frac{T_L}{T_H}=0.5=\frac{1}{2}\)
Now,
\((COP)_{carnot}=\frac{T_L}{T_H-T_L}=\frac{1}{\frac{T_H}{T_L}-1}=\frac{1}{2-1}=1\)
If the volume of moist air with 50% RH is isothermally reduced to half its original volume then relative humidity of moist air becomes
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 8 Detailed Solution
Download Solution PDFConcept:
Relative humidity is given by
\(ϕ = \frac{{{P_v}}}{{{P_{vs}}}}\)
where P v is pressure of water vapour and Pvs is pressure of water vapour at saturation point.
For isothermal process:
PV = constant
Calculation:
Given:
ϕ1 = 50%, Pv1
Since volume is reduced to half so the pressure has become twice.
Pv2 = 2Pv1
\(\frac{{{ϕ _2}}}{{{ϕ _1}}} = \frac{{{P_{{v_2}}}}}{{{P_{v1\;}}}}\)
\(\frac{{{ϕ _2}}}{{{ϕ _1}}} = 2\)
ϕ2 = 2ϕ1
ϕ2 = 2 × 50 ⇒ 100%.
Additional InformationHere, the initial condition is given as Moist air, but we cannot apply ideal gas to it. Technically the wording is incorrect but here we have to assume it as ideal and solve accordingly.
The commonly used refrigerant in domestic refrigerators is:
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 9 Detailed Solution
Download Solution PDFExplanation:
- Domestic refrigerators generally run on the vapor compression cycle. In this cycle, a circulating refrigerant such as R134a enters a compressor as low-pressure vapour at or slightly below the temperature of the refrigerator space.
- Ammonia is generally used in the vapour absorption cycle.
- Nitrogen and CO2 are not used as refrigerants.
The R 134a is an eco-friendly refrigerant.
In a refrigeration system, why are expansion devices located closer to the evaporator?
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 10 Detailed Solution
Download Solution PDFExplanation:
Refrigeration system:
A refrigeration system contains a minimum of four key components i.e., compressor, condenser, expansion valve, and evaporator.
Expansion device:
- The purpose of the expansion device is to rapidly reduce the pressure of the refrigerant in the refrigeration cycle.
- This allows the refrigerant to rapidly cool before entering the evaporator.
- Expansion device located closer to the evaporator in order to minimize the heat gain.
- The most common devices are capillary tube, thermal expansion valve, electronic expansion valve
In a psychometric chart, what does a vertical downward line represent
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 11 Detailed Solution
Download Solution PDFExplanation:
Basic Processes in Conditioning of Air:
Sensible heating:
The moisture content of air remains constant so specific humidity is constant, temperature increases as it flows over a heating coil.
Sensible cooling:
The moisture content of air remains constant so specific humidity is constant, but its temperature decreases as it flows over a cooling coil.
Dehumidification:
- When the temperature remains constant but specific humidity decreases.
- It is represented by a vertical line.
Humidification:
- When in a process, the temperature remains constant but specific humidity increases.
- It is represented by a vertical line.
Cooling and dehumidification:
- This process involves lowering both the air temperature and specific humidity.
- This process is commonly used in summer air conditioning in which air passes over a cooling coil.
- When moist air is cooled below its dew point, the vapor is condensed from the air resulting in simultaneous cooling and dehumidification.
Heating and Humidification:
- During winter it is essential to heat and humidify the room air for comfort.
- It is done by first sensibly heating the air and then adding water vapor to the air stream through steam nozzles, as a result, both the temperature and humidity ratio of air increases.
Cooling & humidification:
Air temperature drops and its humidity increases.
Heating and de-humidification:
This process can be achieved by using a hygroscopic material, which absorbs the water vapor from the moisture. If this process is thermally isolated, then the enthalpy of air remains constant, as a result, the temperature of air increases.
Additional Information
Dry Bulb Temperature: Actual temperature of gas or mixture of gases.
Wet Bulb temperature: Temperature obtained by an accurate thermometer having a wick moistened with distilled water.
Dew point temperature: Temperature at which the liquid droplets just appear when the moist air is cooled continuously.
Relative humidity along the saturation line is 100%.
Domestic refrigerator working on vapour compression cycle uses the following type of expansion device
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 12 Detailed Solution
Download Solution PDFConcept:
- The capillary tube is one of the most commonly used throttling devices in the domestic refrigerators, deep freezers, water coolers, and air conditioners.
- Capillary tubes have very small internal diameters and very long length and they are coiled to several turns so that it would occupy less space (compact).
- They are easy to manufacture, cheap and compact.
Working:
- When the refrigerant leaves the condenser and enters the capillary tube, its high pressure drops down suddenly due to the very small diameter of the capillary tube and the long length gives more friction head and drops pressure further.
- The decrease in pressure leads to cooling of refrigerant and the low-temperature refrigerant can take the heat from the room.
Moist air at 35°C and 100% relative humidity is entering a psychometric device and leaving at 25°C and 100% relative humidity. The name of the device is
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 13 Detailed Solution
Download Solution PDFConcept:
A psychrometric chart is represented as shown in the figure.
Dry Bulb Temperature: Actual temperature of gas or mixture of gases
Wet bulb temperature: Temperature obtained by an accurate thermometer having a wick moistened with distilled water
Dew point temperature: Temperature at which the liquid droplets just appear when the moist air is cooled continuously.
Relative humidity along saturation line is 100%.
Basic Processes in Conditioning of Air:
Sensible heating: Moisture content of air remains constant so specific humidity is constant, temperature increases as it flows over a heating coil
Sensible cooling: Moisture content of air remains constant so specific humidity is constant, but its temperature decreases as it flows over a cooling coil
Cooling and dehumidification: When moist air is cooled below its dew-point by bringing it in contact with a cold surface, some of the water vapour in the air condenses and leaves the air stream as a liquid, as a result, both the temperature and humidity ratio of air decreases.
Heating and Humidification: During winter it is essential to heat and humidify the room air for comfort. It is done by first sensibly heating the air and then adding water vapour to the air stream through steam nozzles, as a result, both the temperature and humidity ratio of air increases.
Cooling & humidification: Air temperature drops and its humidity increases.
Heating and de-humidification: This process can be achieved by using a hygroscopic material, which absorbs the water vapour from the moisture. If this process is thermally isolated, then the enthalpy of air remains constant, as a result, the temperature of air increases.
Calculation:
Inlet (State 1): 35°C and 100% RH
Outlet (State 2): 25°C and 100% RH
From the above figure, relative humidity is same but as the temperature is decreasing and specific humidity is also decreasing. So the process 1-2 is cooling and dehumidification process.
The name of the device is dehumidifier.
Which of the following is a secondary refrigerant, when used above 0°C?
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 14 Detailed Solution
Download Solution PDFExplanation:
There are two types of refrigerants
- Primary Refrigerant: Primary refrigerants are substances that undergo a cyclic process and produce lower temperatures. There is a latent heat transformation for the refrigerants. For e.g. R-11, R-12, R-22, R-134a, R-1150 etc.
- Secondary Refrigerants: There are working substances that are first cooled by primary refrigerants and then used for cooling at desired places. e.g.H2O, Brine.
Additional Information
- R-11 – Large central air conditioning plant
- R-12- Domestic refrigerator, water cooler etc.
- R-22- Window AC
- NH3- Cold storage or Icing plants
- CO2- Transportation of dry ice
- Air- Air-craft refrigeration system
- Brine- Milk-chilling plants
Mistake Points
Both water and Brine are secondary refrigerants, but here he asks above the 0°C, so water will be the correct answer.
A heat pump works on a reversed Carnot cycle. The temperature in the condenser coils is 27° C and that in the evaporator coils is - 23° C. For a work input of 1 kW, how much is the heat pumped?
Answer (Detailed Solution Below)
Refrigeration and Air Conditioning Question 15 Detailed Solution
Download Solution PDFConcept:
\(CO{P_{HP}} = \frac{{Desired\;Output}}{{Required\;Input}} = \frac{{{Q_1}}}{W} = \frac{{{Q_1}}}{{{Q_1} - {Q_2}}} = \frac{{{T_1}}}{{{T_1} - {T_2}}} = \frac{{{Q_H}}}{{{Q_H} - {Q_L}}}\)
Calculation:
Given:
T1 = 27°C = 300 K, T2 = -23°C = 250 K, W = 1 kW
Now
\(\frac{{{Q_1}}}{1} = \frac{{{300}}}{{{300} - {250}}} \Rightarrow Q = 6 \ kW\)