Numerical for Heat Engines:

1. Problem: A heat engine operates between a temperature of (127°C) and (27°C). Calculate the maximum efficiency of the engine? (JEE Main 2019)

Answer: 36.27% Shortcut:

• The maximum efficiency of a heat engine is given by $$e_{max}=1-\frac{T_c}{T_h},$$
  where $$T_c$$ and (T_h) are the temperatures of the cold and hot reservoirs, respectively.
• Convert the temperatures to Kelvin $$T_c=127°C+273=400K,$$ $$T_h=27°C+273=300K$$
• Calculate the efficiency $$e_{max}=1-\frac{T_c}{T_h}=1-\frac{400}{300}\approx 0.3627=36.27$$

2. Problem: A Carnot engine has an efficiency of 60%. If the low-temperature reservoir is at 27 °C, calculate the minimum temperature required for the engine to operate. Answer: − 194°C or 79 K

Shortcut:

• The efficiency of a Carnot engine is given by $$e_{carnot}=1-\frac{T_c}{T_h}$$
• To find the minimum temperature, rearrange the formula, $$T_h=\frac{T_c}{1-e_{carnot}}$$
• Convert the temperature to Kelvin $$T_c=27°C+273=300K$$
• Substitute the values and calculate $$T_h=\frac{300K}{1-0.6}=\frac{300K}{0.4}=750K$$
• Convert back to Celsius $$T_h=750K-273=477°$$

3. Problem: A 4-stroke engine has a displacement of (250 cc) and a compression ratio of (10:1). Calculate the pressure at the end of the compression stroke if atmospheric pressure is 101.3 kPa, assuming adiabatic compression. (CBSE 2018) Answer: 1.987 MPa

Shortcut:

• Since the process is adiabatic (no heat transfer), we can use $$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$$
• Convert the volume to liters $$V_1=250c{m}^3=0.25L,$$
• The compression ratio is given by $$\frac{V_1}{V_2}=10,$$ Solving for $$V_2,V_2=\frac{V_1}{10}=0.025L$$
• The adiabatic index for air is about (1.4)$
• Now we can use the formula $$P_2=P_1{\left(\frac{V_1}{V_2}\right)}^{\gamma }=(101.3\text{ kPa})(10{)}^{1.4}\approx 1.987\text{ MPa}$$

Numerical for Refrigerators:

1. Problem: Calculate the Coefficient of Performance (COP) of a refrigerator if it absorbs 400 J of heat energy from the cold chamber and releases 800 J of heat energy to the hot chamber. Answer: COP = 2 Shortcut:

• The Coefficient of Performance (COP) of a refrigerator is defined as $$COP=\frac{Q_c}{W_{net}},$$ where (Q_c) is the heat absorbed from the cold chamber, and (W_{net}) is the work done.
• Since the work done is equal to the heat released to the hot chamber we have $$W_{net}=Q_h=800\text{ J}$$
• Therefore $$COP=\frac{Q_c}{W_{net}}=\frac{400\text{ J}}{800\text{ J}}=2$$

2. Problem: A Carnot refrigerator maintains a temperature of (8 °C) inside the cold compartment while rejecting heat into a room at (32 °C). Calculate the ideal Coefficient of Performance (COP). Answer: COP = 5

Shortcut:

• The ideal COP for a Carnot refrigerator is given by $$CO{P}_{carnot}=\frac{T_c}{T_h-T_c},$$ where (T_c) is the temperature of the cold reservoir and (T_h) is the temperature of the hot reservoir.
• Convert the temperatures to Kelvin $$T_c=8°C+273=281K,$$ $$T_h=32°C+273=305K$$
• Substitute the values and calculate $$CO{P}_{carnot}=\frac{T_c}{T_h-T_c}=\frac{281K}{305K-281K}=\frac{281K}{24K}\approx 5$$

3. Problem: A refrigerator works on a vapour compression refrigeration cycle. If the temperature of the refrigerant inside the evaporator is – (8 °C) and the temperature of the refrigerant inside the condenser is (117 °C), what is the Coefficient of Performance? Answer: COP = 3.93 Shortcut:

• The COP for a vapour compression refrigerator is $$COP=\frac{Q_c}{W_{in}}=\frac{Q_c}{\left(\frac{Q_h-Q_c}{CO{P}_{th}}\right)}$$ where (Q_c) and (Q_h) are the heat absorbed by the cold chamber and released to the hot chamber, respectively, and (COP_{th}) is the theoretical COP (COP for a Carnot refrigerator operating between the same temperatures)
• Convert the temperatures to Kelvin $$T_c=-8°C+273=265K,$$ $$T_h=117°C+273=390K$$
• Calculate the theoretical COP $$CO{P}_{th}=\frac{T_c}{T_h-T_c}=\frac{265K}{390K-265K}\approx 1.8$$
• Assuming the ratio (Q_h /Q_c = 2), we can finally calculate the COP as $$COP=\frac{Q_c}{\left(\frac{Q_h-Q_c}{CO{P}_{th}}\right)}=\frac{Q_c}{\left(\frac{2Q_c-Q_c}{1.8}\right)}$$ Solving for $$Q_c$$ and then we can find $$COP=\frac{Q_c}{\frac{Q_c}{1.8}}=1.8,$$ $$COP\approx 3.93$$

Additional Tips:

• Understand the concepts and principles thoroughly before attempting numerical problems.
• Practice a variety of numerical problems for better proficiency in different types of calculations.
• Pay attention to units and conversions.
• Check your solutions and ensure that you are using the correct formulas and methods.