All questions of Refrigeration & Air conditioning for Mechanical Engineering Exam

The correct answer is option 'B': ha whv. Let's understand why this is the right answer.

Enthalpy of Dry Air (ha):
Enthalpy is the total energy content of a substance and is given by the sum of its internal energy and the product of its pressure and volume. In the case of dry air, the enthalpy is solely dependent on its temperature. Therefore, the enthalpy of dry air can be represented as ha.

Enthalpy of Water Vapor (hv):
Similar to dry air, the enthalpy of water vapor is also dependent on its temperature. However, water vapor has an additional enthalpy due to its phase change. This enthalpy is known as the latent heat of vaporization. Therefore, the enthalpy of water vapor can be represented as hv.

Specific Humidity (w):
Specific humidity is the mass of water vapor present in a unit mass of moist air. It is a measure of the moisture content in the air. Specific humidity is given by the ratio of the mass of water vapor to the sum of the mass of dry air and the mass of water vapor. It can be represented as w.

Enthalpy of Moist Air:
When dry air and water vapor are combined, they form moist air. The enthalpy of moist air is the sum of the enthalpy of dry air and the enthalpy of water vapor. Since the enthalpy of dry air is ha and the enthalpy of water vapor is hv, the enthalpy of moist air can be represented as ha + hv.

However, we need to consider the specific humidity (w) as well. The specific humidity represents the ratio of the mass of water vapor to the sum of the mass of dry air and the mass of water vapor. Therefore, we need to multiply the enthalpy of water vapor (hv) by the specific humidity (w) to account for the mass of water vapor present in the moist air. Hence, the final expression for the enthalpy of moist air is ha + whv.

Therefore, option 'B' (ha whv) is the correct answer.

Introduction:
The Carnot cycle is a theoretical thermodynamic cycle that consists of four processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. It is an idealized cycle that represents the maximum possible efficiency for a heat engine or the maximum possible coefficient of performance for a refrigerator.

Given:
- Temperature of the low-temperature reservoir (T1) = 250 K
- Temperature of the high-temperature reservoir (T2) = 300 K

Coefficient of Performance:
The coefficient of performance (COP) is a measure of the efficiency of a refrigerator or a heat pump. For a refrigerator, the COP is defined as the ratio of the desired cooling effect (Q2) to the work input (W). Mathematically, it can be expressed as:

COP = Q2 / W

Calculating the COP:
In a Carnot cycle refrigerator, the work input (W) is given by:

W = Q2 - Q1

where Q1 is the heat rejected to the low-temperature reservoir.

The Carnot efficiency (η) is given by:

η = 1 - (T1 / T2)

For a Carnot cycle refrigerator, the COP can be expressed as:

COP = Q2 / (Q2 - Q1)

Since Q1 is equal to T1 multiplied by the heat absorbed from the low-temperature reservoir, and Q2 is equal to T2 multiplied by the heat rejected to the high-temperature reservoir, we can rewrite the COP equation as:

COP = T2 / (T2 - T1)

Substituting the given values, we have:

COP = 300 K / (300 K - 250 K)
= 300 K / 50 K
= 6

Conclusion:
Therefore, the coefficient of performance (COP) of the Carnot cycle refrigerator operating between 250 K and 300 K is 6, which corresponds to option B.

Is always greater than (COP)vapour compression refrigerator.

b) (COP)air is always equal to (COP)vapour compression refrigerator.

c) (COP)air is always less than (COP)vapour compression refrigerator.

d) (COP)air can be greater or less than (COP)vapour compression refrigerator depending on the operating conditions.

1 ton of refrigeration refers to the amount of heat transfer required to freeze one ton (2,000 pounds) of water into ice at 32°F in 24 hours. It is a unit used to measure the cooling capacity of refrigeration systems.

The correct answer is option 'A' - 210 kJ/min.

Let's break down the explanation:

1. Understanding the Concept:
Refrigeration is the process of removing heat from a space or substance to lower its temperature. The cooling capacity of a refrigeration system is measured in tons of refrigeration. One ton of refrigeration is equal to the amount of heat transfer required to freeze one ton of water in 24 hours.

2. Conversion Factors:
To determine the heat transfer rate in different units, we need to use appropriate conversion factors. Here are the conversion factors involved in this question:

- 1 ton of refrigeration = 12,000 British Thermal Units per hour (BTU/hr)
- 1 BTU/hr = 0.29307107 watts (W)
- 1 watt (W) = 1 joule per second (J/s)
- 1 joule (J) = 0.001 kilojoules (kJ)

3. Calculating the Heat Transfer Rate:
To convert the cooling capacity from tons of refrigeration to kilojoules per minute, we can use the following steps:

- Convert 1 ton of refrigeration to BTU/hr:
1 ton of refrigeration = 12,000 BTU/hr

- Convert BTU/hr to watts:
12,000 BTU/hr * 0.29307107 W/BTU = 3,517.65 W

- Convert watts to kilojoules per second (kJ/s):
3,517.65 W * 1 J/s / 1 W * 0.001 kJ/J = 3.51765 kJ/s

- Convert kilojoules per second to kilojoules per minute:
3.51765 kJ/s * 60 s/min = 210.96 kJ/min

Rounding off the answer to the nearest whole number, we get 210 kJ/min, which is the correct answer (option 'A').

To summarize, 1 ton of refrigeration implies a heat transfer rate of 210 kJ/min. This means that a refrigeration system with a cooling capacity of 1 ton can remove 210 kilojoules of heat from a space or substance every minute.

Throttling operation in a refrigeration cycle is carried out in the Expansion Valve.

Explanation:
The expansion valve is a key component in a refrigeration cycle that controls the flow of refrigerant from the high-pressure side to the low-pressure side of the system. It is responsible for creating a pressure drop in the refrigerant, which allows it to expand and cool down.

The throttling operation in the expansion valve can be explained in the following steps:

1. Pressure reduction: The high-pressure refrigerant from the condenser enters the expansion valve. The valve restricts the flow of refrigerant, causing a significant pressure drop. This pressure reduction is necessary for the refrigerant to change from a high-pressure liquid to a low-pressure mixture of liquid and vapor.

2. Temperature drop: As the refrigerant passes through the expansion valve, it undergoes a rapid expansion due to the pressure drop. This expansion results in a decrease in the refrigerant's temperature. The refrigerant absorbs heat from its surroundings, which leads to a cooling effect.

3. Phase change: The refrigerant undergoes a phase change from a saturated liquid to a saturated vapor during the throttling process. This phase change occurs as a result of the reduced pressure and temperature drop in the expansion valve.

4. Quality control: The expansion valve also helps to regulate the quality of the refrigerant leaving the valve. It ensures that the refrigerant leaving the expansion valve is in the desired state, with the correct ratio of liquid to vapor. This is important for the proper functioning of the evaporator and the overall refrigeration cycle.

The expansion valve plays a crucial role in the refrigeration cycle by controlling the flow and pressure of the refrigerant. It allows for the efficient transfer of heat from the conditioned space to the evaporator, where the refrigerant absorbs heat and cools down the space. Without the expansion valve, the refrigeration cycle would not be able to operate effectively, and the desired cooling effect would not be achieved.