Refrigeration and Air Conditioning MCQ Quiz - Objective Question with Answer for Refrigeration and Air Conditioning - Download Free PDF

Last updated on May 30, 2025

Latest Refrigeration and Air Conditioning MCQ Objective Questions

Refrigeration and Air Conditioning Question 1:

The Bell-Coleman cycle is also known as:

  1. Reversed Brayton cycle
  2. Brayton cycle
  3. Rankine cycle
  4. Carnot cycle

Answer (Detailed Solution Below)

Option 1 : Reversed Brayton cycle

Refrigeration and Air Conditioning Question 1 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.

04.11.2017.020

where

  1. Process 1 2: isentropic compression
  2. Process 2 3: constant pressure heat rejection
  3. Process 3 4: isentropic expansion
  4. 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 2:

Which of the following does a cascade refrigeration system use?

  1. Two or more refrigerants with different boiling points
  2. A single refrigerant in both cycles
  3. Only ammonia as a refrigerant
  4. Only air as a working fluid

Answer (Detailed Solution Below)

Option 1 : Two or more refrigerants with different boiling points

Refrigeration and Air Conditioning Question 2 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 3:

COP of refrigerator based on reverse Carnot cycle decreases on

  1. increasing the higher temperature and keeping the lower temperature constant
  2. keeping the higher temperature constant and increasing the lower temperature
  3. increasing the higher temperature and decreasing the lower temperature
  4. decreasing the difference in operating temperatures

Answer (Detailed Solution Below)

Option 3 : increasing the higher temperature and decreasing the lower temperature

Refrigeration and Air Conditioning Question 3 Detailed Solution

Explanation:

COP of a Refrigerator:

  • The Coefficient of Performance (COP) of a refrigerator is a measure of its efficiency and is defined as the ratio of the heat absorbed from the refrigerated space (QL) to the work input (W) required to transfer that heat. Mathematically, for a reverse Carnot cycle:

The Coefficient of Performance (COP) of a Carnot refrigerator is given by:

\( \text{COP}_{\text{Carnot}} = \frac{T_L}{T_H - T_L} \)

Where, \( T_L \) is the lower temperature and \( T_H \) is the higher temperature, both in Kelvin.

Analysis:

If \( T_H \) is increased and \( T_L \) is decreased:

  • The denominator \( (T_H - T_L) \) increases
  • The numerator \( T_L \) decreases

As a result, the COP value decreases.

Refrigeration and Air Conditioning Question 4:

In a gas cycle refrigeration, an expander is used instead of a throttle valve for pressure drop of the refrigerant, because

  1. there is inadequate content of temperature
  2. there can be even heating of the gas if the temperature before throttling is not below the maximum inversion temperature
  3. there can be leakage of gas
  4. enough cooling or temperature drop is not obtained by throttling

Answer (Detailed Solution Below)

Option 4 : enough cooling or temperature drop is not obtained by throttling

Refrigeration and Air Conditioning Question 4 Detailed Solution

Explanation:

Expander in Gas Cycle Refrigeration

  • In gas cycle refrigeration systems, an expander is used instead of a throttle valve for the pressure drop of the refrigerant. The expander works by allowing the refrigerant to expand, reducing its pressure and temperature while recovering work from the expansion process. This contrasts with the use of a throttle valve, where the pressure drop occurs without work recovery, leading to inefficiencies.
  • In gas cycle refrigeration, the cooling effect is achieved by reducing the temperature of the refrigerant as it expands. When a throttle valve is used for this pressure drop, the process is essentially isenthalpic (constant enthalpy).
  • In an isenthalpic process, the temperature drop obtained is often inadequate for efficient cooling because the refrigerant does not perform work during the pressure drop. On the other hand, an expander allows the refrigerant to expand in a manner that performs work, reducing its temperature more significantly. This enhanced temperature drop makes the expander a more effective component for achieving the desired cooling effect in gas cycle refrigeration systems.
  • The expander also contributes to higher overall efficiency in the refrigeration system because the work recovered during expansion can be utilized elsewhere in the system, reducing the energy input required for operation. Thus, the inadequate cooling effect of throttling is the primary reason for using an expander instead of a throttle valve in gas cycle refrigeration.

Refrigeration and Air Conditioning Question 5:

Frosting on the evaporator coils

  1. causes increase in heat transfer rate
  2. is the sign of good cooling
  3. causes decrease in heat transfer rate
  4. is the sign of good compressor

Answer (Detailed Solution Below)

Option 3 : causes decrease in heat transfer rate

Refrigeration and Air Conditioning Question 5 Detailed Solution

Explanation:

Frosting on the Evaporator Coils

  • Frosting on evaporator coils refers to the accumulation of ice or frost on the surface of the evaporator coils in refrigeration or air conditioning systems. This phenomenon occurs when the temperature of the coils drops below the freezing point of water vapor in the surrounding air, causing condensation to freeze and form a layer of ice or frost.

Reasons Behind Frosting:

  • Low Temperature: The evaporator coils are designed to operate at low temperatures to facilitate heat absorption from the surrounding air. However, if the temperature drops excessively, water vapor in the air condenses and freezes on the coil surface.
  • High Humidity: In environments with high humidity, the amount of water vapor available for condensation increases, making frosting more likely to occur.
  • Improper Airflow: Restricted or inadequate airflow around the coils can lead to localized temperature drops, promoting frost accumulation.
  • Faulty Defrost Mechanism: Many systems are equipped with defrost cycles to periodically remove frost buildup. If the defrost system malfunctions, frosting can become excessive.

Impact on Performance:

  • Reduced Cooling Efficiency: The insulating properties of ice hinder the transfer of heat from the air to the refrigerant, leading to decreased cooling capacity.
  • Increased Energy Consumption: The system must work harder to achieve the desired cooling effect, resulting in higher energy usage and operational costs.
  • Potential System Damage: Prolonged frosting can lead to compressor strain, refrigerant flow issues, and other mechanical problems, potentially damaging the system.

Mitigation Strategies:

  • Regular Maintenance: Periodically inspect and clean the evaporator coils to prevent frost buildup.
  • Defrost Mechanism: Ensure the defrost system is functioning correctly to automatically remove frost at regular intervals.
  • Monitor Humidity Levels: Use dehumidifiers or other methods to control indoor humidity and reduce the likelihood of frosting.
  • Optimize Airflow: Maintain proper airflow around the coils by cleaning air filters and ensuring unobstructed vents.

Top Refrigeration and Air Conditioning MCQ Objective Questions

A Carnot heat pump works between 27° C and 327°C. What will be its COP?

  1. 0.09
  2. 1.00
  3. 1.09
  4. 2.0

Answer (Detailed Solution Below)

Option 4 : 2.0

Refrigeration and Air Conditioning Question 6 Detailed Solution

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Concept:

F2 Sumit Madhu 10.08.20 D 2

\({\rm{COP\;of\;Carnot\;Heat\;pump}} = \frac{{{T_1}}}{{{T_1} - {T_2}}}\)

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?

  1. 4
  2. 1
  3. 2
  4. 3

Answer (Detailed Solution Below)

Option 2 : 1

Refrigeration and Air Conditioning Question 7 Detailed Solution

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Concept:

\(η_{carnot}=\frac{T_H-T_L}{T_H}=1-\frac{T_L}{T_H}\)

\((COP)_{carnot}=\frac{T_L}{T_H-T_L}\)

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

  1. 25%
  2. 60%
  3. 70%
  4. 100%

Answer (Detailed Solution Below)

Option 4 : 100%

Refrigeration and Air Conditioning Question 8 Detailed Solution

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Concept:

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:

  1. Ammonia
  2. CO2
  3. Nitrogen
  4. R 134a

Answer (Detailed Solution Below)

Option 4 : R 134a

Refrigeration and Air Conditioning Question 9 Detailed Solution

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Explanation:

  • 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.

26 June 1

The R 134a is an eco-friendly refrigerant.

In a refrigeration system, why are expansion devices located closer to the evaporator?

  1. To avoid the flow of the refrigerant
  2. To minimise the heat gain
  3. To ease the flow of the refrigerant
  4. To maximise the heat gain

Answer (Detailed Solution Below)

Option 2 : To minimise the heat gain

Refrigeration and Air Conditioning Question 10 Detailed Solution

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Explanation:

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

F1 Sumit T.T.P Deepak 23.01.2020 D 1

Domestic refrigerator working on vapour compression cycle uses the following type of expansion device

  1. Electrically operated throttling valve
  2. Manually operated valve
  3. Thermostatic valve
  4. Capillary tube

Answer (Detailed Solution Below)

Option 4 : Capillary tube

Refrigeration and Air Conditioning Question 11 Detailed Solution

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Concept:

  • 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

  1. Humidifier
  2. Dehumidifier
  3. Sensible heater
  4. Sensible cooler

Answer (Detailed Solution Below)

Option 2 : Dehumidifier

Refrigeration and Air Conditioning Question 12 Detailed Solution

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Concept:

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%.

gateme 5jun 2018 3

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.

gateme 5jun 2018 2

Calculation:

Inlet (State 1): 35°C and 100% RH

Outlet (State 2): 25°C and 100% RH

GATE ME 2014 D Q17

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.

In a psychometric chart, what does a vertical downward line represent

  1. Sensible cooling process
  2. Adiabatic saturation process
  3. Humidification process
  4. Dehumidification process

Answer (Detailed Solution Below)

Option 4 : Dehumidification process

Refrigeration and Air Conditioning Question 13 Detailed Solution

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Explanation:

Basic Processes in Conditioning of Air:

gateme 5jun 2018 2

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%.

gateme 5jun 2018 3

Which of the following is a secondary refrigerant, when used above 0°C?

  1. Sodium chloride
  2. Glysols
  3. Brines 
  4. Water

Answer (Detailed Solution Below)

Option 4 : Water

Refrigeration and Air Conditioning Question 14 Detailed Solution

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Explanation:

There are two types of refrigerants

  1. 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.
  2. 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?

  1. 1 kW
  2. 5 kW
  3. 6 kW
  4. None of the above

Answer (Detailed Solution Below)

Option 3 : 6 kW

Refrigeration and Air Conditioning Question 15 Detailed Solution

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Concept:

RRB JE ME 24 17Q Mix RPSC Hindi - Final Satya ji Madhushri D9

\(CO{P_{HP}} = \frac{{Desired\;Output}}{{Required\;Input}} = \frac{{{Q_1}}}{W} = \frac{{{Q_1}}}{{{Q_1} - {Q_2}}} = \frac{{{T_1}}}{{{T_1} - {T_2}}} = \frac{{{Q_H}}}{{{Q_H} - {Q_L}}}\)

Calculation:

Given:

T1 = 27°C = 300 K, T2 = -23°C = 250 K, W = 1 kW

Now

\(\frac{{{Q_1}}}{1} = \frac{{{300}}}{{{300} - {250}}} \Rightarrow Q = 6 \ kW\)

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