Heat engine (application of the second law of thermodynamics) – problems and solutions

30 Heat engine (application of the second law of thermodynamics) – problems and solutions

1. Heat input Q_H = 3000 Joule and heat output Q_L = 1000 Joule. What is the efficiency of the heat engine?

Known :

Heat input (Q_H) = 3000 Joule

Heat output (Q_L) = 1000 Joule

Work done by the engine (W) = 3000 – 1000 = 2000 Joule

Wanted: Efficiency of the heat engine (e)

Solution :

2. A heat engine produces 2000 Joule of mechanical work and the engine discharges heat to the environment at a rate of 500 Joule. What is the efficiency of the heat engine?

Known :

The work is done by the engine (W) = 2000 Joule

Heat output (Q_L) = 5000 Joule

Heat input (Q_H) = 2000 + 5000 = 7000 Joule

Wanted: Efficiency (e)

Solution :

3. A heat engine has an efficiency of 30%. If the engine produces 10,000 Joule of mechanical work, how much heat is discharged as waste heat from this engine?

Known :

Efficiency (e) = 30 % = 30/100 = 0.3

The work is done by engine (W) = 3,000 Joule

Wanted : Heat output (Q_L)

Solution :

Heat output (Q_L) = heat input (Q_H) – work is done by engine (W)

Q_L = 10,000 Joule – 3,000 Joule

Q_L = 7,000 Joule

4. Determine the efficiency of a Carnot engine operating between two reservoirs at temperatures of 500 K and 300 K.
Solution: Using the Carnot efficiency formula, \(\eta = 1 – \frac{T_{\text{cold}}}{T_{\text{hot}}}\), we get \(\eta = 1 – \frac{300}{500} = 0.4\) or \(40\%\).

5. A heat engine absorbs 2000 J of heat from a hot reservoir and expels 1200 J to a cold reservoir. Calculate the work done by the engine.
Solution: Using the first law of thermodynamics, \(Q_{\text{hot}} = W + Q_{\text{cold}}\), the work done is \(W = 2000 – 1200 = 800\ \text{J}\).

6. Calculate the efficiency of a heat engine that absorbs 1000 J from a hot reservoir and does 600 J of work.
Solution: \(\eta = \frac{W}{Q_{\text{hot}}} = \frac{600}{1000} = 0.6\) or \(60\%\).

7. A heat engine operates between 400 K and 200 K. Determine the maximum possible efficiency.
Solution: \(\eta_{\text{max}} = 1 – \frac{T_{\text{cold}}}{T_{\text{hot}}} = 1 – \frac{200}{400} = 0.5\) or \(50\%\).

8. A heat engine rejects 300 J to the cold reservoir while doing 200 J of work. Calculate the heat absorbed from the hot reservoir.
Solution: \(Q_{\text{hot}} = W + Q_{\text{cold}} = 200 + 300 = 500\ \text{J}\).

9. Determine the efficiency of an Otto cycle with a compression ratio of 8.
Solution: \(\eta = 1 – \frac{1}{r^{(\gamma – 1)}} = 1 – \frac{1}{8^{(1.4 – 1)}} \approx 0.564\) or \(56.4\%\), where \(\gamma\) is the heat capacity ratio.

10. A heat engine performs 150 J of work and rejects 100 J to the cold reservoir. Find the heat absorbed from the hot reservoir.
Solution: \(Q_{\text{hot}} = W + Q_{\text{cold}} = 150 + 100 = 250\ \text{J}\).

11. Determine the efficiency of a Diesel engine with a compression ratio of 18 and a cutoff ratio of 2.
Solution: \(\eta = 1 – \frac{1}{r^{(\gamma – 1)}} \left( \frac{r_c – 1}{r_c} \right) \approx 0.627\) or \(62.7\%\), where \(r\) is the compression ratio, \(r_c\) is the cutoff ratio, and \(\gamma\) is the heat capacity ratio.

12. Calculate the work done by a heat engine that absorbs 400 J and has an efficiency of 50%.
Solution: \(W = \eta \cdot Q_{\text{hot}} = 0.5 \cdot 400 = 200\ \text{J}\).

13. A heat engine operates between 600 K and 300 K. Find the Carnot efficiency.
Solution: \(\eta = 1 – \frac{T_{\text{cold}}}{T_{\text{hot}}} = 1 – \frac{300}{600} = 0.5\) or \(50\%\).

14. A heat engine absorbs 3000 J and rejects 1800 J to the cold reservoir. Calculate the work done.
Solution: \(W = Q_{\text{hot}} – Q_{\text{cold}} = 3000 – 1800 = 1200\ \text{J}\).

15. Determine the efficiency of a Stirling engine operating between two reservoirs at 800 K and 400 K.
Solution: \(\eta = 1 – \frac{T_{\text{cold}}}{T_{\text{hot}}} = 1 – \frac{400}{800} = 0.5\) or \(50\%\).

16. A heat engine rejects 500 J to the cold reservoir while doing 300 J of work. Calculate the heat absorbed from the hot reservoir.
Solution: \(Q_{\text{hot}} = W + Q_{\text{cold}} = 300 + 500 = 800\ \text{J}\).

17. Determine the efficiency of an Ericsson cycle with hot and cold reservoirs at 1000 K and 250 K.
Solution: \(\eta = 1 – \frac{T_{\text{cold}}}{T_{\text{hot}}} = 1 – \frac{250}{1000} = 0.75\) or \(75\%\).

18. A heat engine does 700 J of work and has an efficiency of 70%. Calculate the heat rejected to the cold reservoir.
Solution: \(Q_{\text{cold}} = \frac{W}{\eta} – W = \frac{700}{0.7} – 700 \approx 1000 – 700 = 300\ \text{J}\).

19. Determine the efficiency of a Brayton cycle with pressure ratios of 6.5 and \(\gamma = 1.4\).
Solution: \(\eta = 1 – \frac{1}{r_p^{(\gamma – 1)/\gamma}} \approx 0.433\) or \(43.3\%\), where \(r_p\) is the pressure ratio.

20. Calculate the work done by a heat engine that absorbs 250 J and rejects 100 J to the cold reservoir.
Solution: \(W = Q_{\text{hot}} – Q_{\text{cold}} = 250 – 100 = 150\ \text{J}\).

21. A heat engine operates between 450 K and 150 K. Find the Carnot efficiency.
Solution: \(\eta = 1 – \frac{T_{\text{cold}}}{T_{\text{hot}}} = 1 – \frac{150}{450} \approx 0.667\) or \(66.7\%\).

22. Determine the efficiency of an Atkinson cycle with a compression ratio of 10.
Solution: \(\eta = 1 – \frac{1}{r^{(\gamma – 1)}} \approx 0.593\) or \(59.3\%\),

where \(r\) is the compression ratio and \(\gamma\) is the heat capacity ratio.

20. A heat engine absorbs 180 J and has an efficiency of 40%. Calculate the work done.
Solution: \(W = \eta \cdot Q_{\text{hot}} = 0.4 \cdot 180 = 72\ \text{J}\).

Note: For problems involving specific heat cycles (like Otto, Diesel, etc.), certain values like heat capacity ratio \(\gamma\) were used, which are typical for certain gases (e.g., \(\gamma = 1.4\) for air). These values might differ depending on the specific substances used in the engine.