Heat Engine, Heat Pump and Refrigerator MCQ Quiz - Objective Question with Answer for Heat Engine, Heat Pump and Refrigerator - Download Free PDF

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:

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

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

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.

Concept:

The coefficient of performance (COP) of a refrigerator is defined as the ratio of the heat removed from the cold medium (Qc) to the work input (W) to the refrigerator.

∴ COP = Qc / W

Where, Qc = Heat removed from the cold medium, W = Work input or power consumed by the refrigerator.

Calculation:

Given:

Qc = 7200 kJ/hr,

COP = 2

First, we need to convert Qc into kW:

∴ Qc = 7200 kJ/hr = 7200 / 3600 kJ/s = 2 kW

Now, using the formula for COP:

COP = Qc / W

∴ 2 = 2 kW / W

∴ W = 2 kW / 2 = 1 kW

The power consumed by the refrigerator is 1 kW.

Explanation:

Heat Pump:

• In a basic heat pump cycle, there are several key components that work together to transfer heat from one location to another.
• A heat pump is an essential part of HVAC (heating, ventilation, and air conditioning) systems, used to either heat or cool a space by moving thermal energy.
• The primary components typically found in a basic heat pump cycle include the evaporator, compressor, condenser, and expansion valve.

Evaporator:

• The evaporator is a critical component in the heat pump cycle. It is responsible for absorbing heat from the surrounding environment. In the heating mode, the evaporator absorbs heat from the outside air, ground, or water source. In cooling mode, it absorbs heat from the indoor air. The refrigerant inside the evaporator evaporates (changes from liquid to gas) as it absorbs heat, thereby extracting thermal energy from the environment.

Compressor:

• The compressor is the heart of the heat pump system. It compresses the refrigerant, increasing its pressure and temperature. The high-pressure, high-temperature gas then moves to the condenser. The compressor's role is crucial because it enables the transfer of heat by moving the refrigerant through the system. Without the compressor, the cycle would not be able to function effectively.

Condenser:

• The condenser is where the refrigerant releases the absorbed heat. In the heating mode, the condenser releases heat into the indoor space, warming it up. In the cooling mode, it releases heat to the outside environment. The refrigerant condenses (changes from gas to liquid) as it releases heat, which is then circulated back to the evaporator.

Expansion Valve:

• The expansion valve regulates the flow of refrigerant into the evaporator. It reduces the pressure of the refrigerant, causing it to cool down before entering the evaporator. This pressure reduction allows the refrigerant to absorb heat more effectively in the evaporator. The expansion valve is essential for maintaining the proper balance of refrigerant flow and ensuring the efficiency of the heat pump cycle.

Concept:

According to Carnot cycle the efficiency of heat engine can be defined as the ratio of net work done per cycle and the total amount of heat absorbed per cycle by the working substance from the source

i.e. \(\eta =~\frac{Net~work~done~per~cycle~}{Total~amount~of~heat~absorbed~per~cycle}=\frac{W}{{{Q}_{1}}}\)

Here W= Q1 Q2 

\(\therefore \eta =\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}\) or in terms of % \(\eta =\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}\times 100\)

Where

T1 = temperature of heat source

T2 = temperature of the sink

Q1 = Heat supplied by the source

Q2 = Heat supplied to the sink

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,

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

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:

The coefficient of performance is the ratio of heat extracted in the refrigerator to the work done on the refrigerant.

COP (Coefficient of Performance) of the ideal refrigerator

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

where T2 = Temperature of the hot reservoir, T1 = Temperature of the cold reservoir, Q1 = Heat rejected from the cold reservoir, WR = Work required to pump the heat out.

Calculation:

Given:

T1 = 5°C = 278 K, T2 = 100°C = 373 K

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

COP = 278/95 = 2.93

Mistake Points

Here ambient temperature and heat rejection temperature both provided. So while calculating COP use T_H = Temperature at which heat is rejected.

Concept:

Where, T₁ = Low-temperature reservoir, T₂ = High-temperature reservoir, Q₁ = Heat supplied to refrigerator, Q₂ = Heat rejected by refrigerator, W_R = Work input given to refrigerator

Heat rejected to the higher temperature in Carnot refrigeration = Refrigeration effect + Work input

Q₂ = Q₁ + W_R

Calculation:

Given:

T_L= - 30°C ,  W_input = 1.28 kW , Q_S = 1 TR = 3.5167 kW

Q_R = Q_S + W_input  = 3.51 + 1.28 = 4.79 kW

Concept:

As there is no mention that it works on the Carnot cycle, therefore the COP of Heat Pump is given by,

\(COP = \frac{{Heat\;rejection}}{{Heat\;rejection - Heat\;addition}}\)

Calculation:

Given:

Q_add = 750 kW, Q_rej = 1000 kW, T_H = 30°C, T_L = 15°C 

∴ \(COP~=~\frac{1000}{1000-750}\)

COP = 4

Concept:

A heat pump is a device used to warm places by transferring thermal energy from a cooler space to a warmer space using the refrigeration cycle.

The coefficient of performance or COP of a heat pump is a ratio of useful heating provided to work (energy) required.

\({COP_{HP}={Heat\: Rejected \over Work\: Provided}} =\frac{Q_1}{W}\)

Calculation:

Given:

Rate of heat rejected, Q₁ = 360 kJ/min = \( {360 \over 60}\) kJ/sec = 6 kW, power supplied, W = 2 kW

coefficient of performance of a heat pump

COP = \({Q_1 \over W}\) 

= \( {6\over 2}\)

= 3

Additional Information

The coefficient of performance or COP of a refrigerator is a ratio of heat absorbed to work (energy) required. 

\(\mathbf{COP_{RE}={Heat\: Absorbed \over Work\: Provided}}\)\(\mathbf{= {Q_2 \over W}}\)

∴ COP_HP = COP_RE + 1

Concept:

The COP of the ideal Vapour Compression Refrigeration System (VCRS):

It is the ratio of refrigerating effect (R.E) to the work done by the compressor (Wc).

\(COP=\frac{Refrigerating\;effect}{Work\;done\;by\;compressor}\Rightarrow \frac{R.E}{W_c}\)

Process 1-2: Isentropic work done by Compressor

Process 2-3: Heat removal at constant pressure by Condensor

Process 3-4: Isenthalpic expansion by a Throttling valve

Process 4-1: Heat addition at constant pressure by Evaporator.

Calculation:

Given:

h3 = h4 = 90 kJ/kg, h1 = 180 kJ/kg, h2 = 200 kJ/kg

\(COP = \frac{{Refrigeration\ effect}}{{Work\ done}} = \frac{{{h_1}\;- \;{h_4}}}{{{h_2}\;-\;{h_1}}}\)

\(\therefore\frac{{180\;- \;90}}{{200\;-\;180}} \Rightarrow 4.5\)

Concept:

The coefficient of performance is the ratio of heat extracted in the refrigerator to the work done on the refrigerant.

COP (Coefficient of Performance) of the ideal refrigerator

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

where T2 = Temperature of the hot reservoir, T1 = Temperature of the cold reservoir, Q1 = Heat rejected from the cold reservoir, WR = Work required to pump the heat out.

Calculation:

Given:

A 10-tonne refrigeration system

1 tonne = 3.5 kW hence, 10 tonne = 35 kW

Q1 = 35 kW

Refrigeration system consumes 10 kW of electrical energy.

W_R = 10 kW

\(COP =\frac{Q_1}{W_R} = \frac{{35}}{{10}} = 3.5\)