Refrigeration Cycles and Devices MCQ Quiz - Objective Question with Answer for Refrigeration Cycles and Devices - Download Free PDF

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

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.

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:

\( W = \frac{1.667}{2} = 0.8335~\text{kW} \)

\( Q_H = 1.667 + 0.8335 = 2.5~\text{kW} \)

Concept:  

p - h and T-s diagram for ideal vapour compression refrigeration cycle are shown below,

The decrease in the evaporator pressure will not affect condensing temperature because the condensing temperature is a function of condensing pressure not of evaporator pressure. So, there is no change in the temperature of condenser. 

From the p-h diagram:

Let pE1 and p­E2 be the initial evaporator pressure and decreased evaporator pressure respectively.

At pE1,

W (Compressor work) = h2 – h­1

RE (Refrigeration Effect) = h1­ ­­– h4

At pE2,

W’ (Compressor work) = h2’ – h1

RE’ (Refrigeration Effect) = h1’ – h4’

From the p-h diagram,

W’ > W, Compressor work is increased

RE’ < RE, Refrigeration effect is decreased

There is an increase in compressor work and a decrease in the refrigeration effect.

The coefficient of performance of a vapor compression refrigeration system decreases with the decrease in evaporator temperature at a fixed condenser temperature.

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 (Q_L) 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.

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.

Concept:

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

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

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:

η_Carnot = 50 % = 0.5, T_H = 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,

\((COP)_{carnot}=\frac{T_L}{T_H-T_L}=\frac{1}{\frac{T_H}{T_L}-1}=\frac{1}{2-1}=1\)

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

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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.

Concept:

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

Calculation:

Given:

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

Now

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

Concept:

COP (Coefficient of Performance) of ideal refrigerator = T₁ / (T₂ – T₁) = Q₁ / W_R

\(COP = \frac{{{Q_1}}}{W} = \frac{{{T_1}}}{{{T_2} - {T_1}}}\)

T₂ = Temperature of the hot reservoir

T₁ = Temperature of the cold reservoir

Q₁ = Heat rejected from the cold reservoir

W_R = Work required to pump the heat out

Calculation:

Given:

T₁ = -23 + 273 = 250 K

T₂   = 27+ 273 = 300 K

Q₁ = 0.5 kJ/s

\(COP = \frac{{{Q_1}}}{W} = \frac{{{T_1}}}{{{T_2} - {T_1}}} \Rightarrow \frac{{0.5}}{W} = \frac{{250}}{{300 - 250}} \Rightarrow \frac{{0.5}}{W} = 5\)

W_in = 0.1 kJ/s

Concept:  

W (Compressor work) = h2 – h­1

RE (Refrigeration Effect) = h1­ ­­– h4

The implementation of subcooling increases the COP of the vapour compression refrigeration system.

Process 3-3’ is the subcooling process at constant pressure.

\(COP = \frac{{refrigeration\;effect}}{{work\;input}} = \frac{{{h_1} - {h_4}}}{{{h_2} - {h_1}}} = \frac{{{h_1} - {h_{3'}}}}{{{h_2} - {h_1}}}\)

h₃ – h₃' = C_p (T₃ – T₃') ⇒ h₃' = h₃ – C_p (T₃ – T_3’)

Calculation:

Given:

h₁ = 220 kJ/kg, h₂ = 257 kJ/kg, h₃ = 87 kJ/kg

h₃' = h₃ – C_p (T₃ – T₃') = 87 – (5 × 5) = 62 kJ/kg.

\(COP = \frac{{220 - 62}}{{258 - 220}} = 4.15\)

Explanation:

Vapour Compression Refrigeration System:

The four processes in simple VCRS include:

1. Isentropic compression (1-2) in Compressor.
2. Constant Pressure heat removal (2-3) in Condensor.
3. Isenthalpic expansion (3-4) in a Throttling device.
4. Constant pressure heat removal (4-1) in Evaporator.

From the Above T-s diagram, we can conclude that the lowest temperature during the cycle in a vapour compression system occurs after expansion and during the evaporation process(4-1) in the evaporator.

In the Question the exact location of the lowest temperature has not been asked here they have asked after which process the lowest temperature will be obtained. The lowest temperature surely is obtained in the evaporator which comes after the expansion process hence the expansion is the correct answer. 

Explanation:

Capillary tube:

• A capillary tube is a long, narrow tube of constant diameter.
• The word “capillary” is a misnomer since surface tension is not important in the refrigeration application of capillary tubes.
• Typical tube diameters of refrigerant capillary tubes range from 0.5 mm to 3 mm and the length ranges from 1.0 m to 6 m.
• The main objective of the capillary tube is to reduce the pressure.
• The pressure reduction in a capillary tube occurs due to the following two factors:
  1. The refrigerant has to overcome the frictional resistance offered by tube walls. This leads to some pressure drop, and
  2. The liquid refrigerant flashes (evaporates) into a mixture of liquid and vapor as its pressure reduces. The density of vapor is less than that of the liquid. Hence, the average density of refrigerant decreases as it flows in the tube. The mass flow rate and tube diameter (hence area) being constant, the velocity of refrigerant increases since mass flow rate = ρVA. The increase in velocity or acceleration of the refrigerant also requires pressure drop. 

Explanation:

Dry air rated temperature (DART):

The concept of Dry Air Rated Temperature is used to compare different aircraft refrigeration cycles. Dry Air Rated Temperature is defined as the temperature of the air at the exit of the cooling turbine in the absence of moisture condensation. For condensation not to occur during expansion in the turbine, the dew point temperature and hence moisture content of the air should be very low, the air should be very dry. The aircraft refrigeration systems are rated based on the mass flow rate of air at the design DART. The cooling capacity is then given by:

Q = mc_p(T_i - T_DART)

where, m = Mass flow rate, T_DART =  Dry Air Rated Temperature, T_i = Cabin temperature.

Important Points

• DART increases monotonically with the Mach number for all the systems except the reduced ambient system.
• The simple system is adequate at low Mach numbers.
• At high Mach numbers, either a bootstrap system or a regenerative system should be used.
• A reduced ambient temperature system is best suited for very high Mach number, supersonic aircrafts.