SATHEE: Shortcut Methods

Numerical for Heat Engines:

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 $$T_c = 127 °C + 273 = 400 K,$$ $$T_h = 27 °C + 273 = 300 K$$
Calculate the efficiency $$e_{max} = 1- \frac{T_c}{T_h} = 1 - \frac{400}{300} \approx 0.3627= 36.27%$$

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 $$T_h = 750 K − 273 = 477 °$$

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 $$V_1 = 250 cm^3 = 0.25 L,$$
The compression ratio is given by $$\frac{V_1}{V_2} = 10,$$ Solving for $$V_2, V_2 = \frac{V_1}{10} = 0.025 L$$
The adiabatic index for air is about (1.4)$
Now we can use the formula $$P_2 = P_1\left(\frac{V_1}{V_2}\right)^\gamma = (101.3 \text{ kPa})(10)^{1.4} \approx 1.987 \text{ MPa}$$

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 $$COP = \frac{Q_c}{W_{net}} = \frac{400 \text{ J}}{800 \text{ J}} = 2$$

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 $$COP_{carnot} = \frac{T_c}{T_h-T_c},$$ 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 $$T_c = 8°C + 273 = 281 K,$$ $$T_h = 32 °C + 273 = 305 K$$
Substitute the values and calculate $$COP_{carnot} = \frac{T_c}{T_h-T_c} = \frac{281 K}{305 K -281 K} = \frac{281 K}{24K} \approx 5$$

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 $$COP = \frac{Q_c}{W_{in} } = \frac{Q_c}{\left(\frac{Q_h-Q_c}{ COP_{th}}\right)}$$ 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 $$T_c = - 8 °C+ 273 = 265 K,$$ $$T_h = 117°C + 273 = 390 K$$
Calculate the theoretical COP $$COP_{th} = \frac{T_c}{T_h - T_c } = \frac{265 K}{390 K - 265 K } \approx 1.8$$
Assuming the ratio (Q_h /Q_c = 2), we can finally calculate the COP as $$COP = \frac{Q_c}{\left(\frac{Q_h-Q_c}{ COP_{th}}\right)} = \frac{Q_c}{\left(\frac{2Q_c-Q_c}{ 1.8}\right)} $$ Solving for $$Q_c$$ and then we can find $$COP = \frac{Q_c}{\frac{Q_c}{1.8}} = 1.8,$$ $$COP \approx 3.93$$

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.