1. ENGR 2213 Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma

2. Gas Power Cycles Otto Cycle Diesel Cycle Brayton Cycle - Ideal cycle for spark-ignition engines - Ideal cycle for compression-ignition engines - Ideal cycle for gas-turbine engines

3. Gas Power Cycles Air-Standard Assumptions 1. The working fluid is air which continuously circulates in a closed loop and always behaves as an ideal gas. 2. All the processes which make up the cycle are internally reversible. 3. The combustion process is replaced by a heat addition process from an external source. 4. The exhaust process is replaced by a heat rejection process which restores the air to its initial state. ► Air has constant specific heats which are evaluated at room temperature (25 ºC or 77 ºF). Cold

4. Gas Power Cycles BDC TDC Displacement volume V min Stroke V max Bore

5. Gas Power Cycles Compression Ratio Mean Effective Pressure (MEP) Compression ratio is a volume ratio and should not be confused with the pressure ratio. W = MEP A P Stroke = MEP Displacement volume

6. Otto Cycles Nikolaus A. Otto (1876) – four-stroke engine Beau de Rochas (1862)

7. Otto Cycles Two-stroke vs. Four-stroke Advantages Disadvantages 1. Simple 2. Inexpensive 3. High power-to-weight and power-to-volume ratios 1. Less efficiency - Incomplete expulsion of the exhausted gases - Partial expulsion of the air-fuel mixture with the exhausted gases

8. Otto Cycles First Law q – w = Δu q in = u 3 – u 2 q out = u 4 – u 1 1 2 3 4 T S Q in Q out = c v (T 3 – T 2 ) = c v (T 4 – T 1 )

9. Otto Cycles The increase in thermal efficiency is not as pronounced at high compression ratios. At high compression ratios, autoignition may occur. Autoignition in gasoline engines hurts performance and cause engine damage.

10. Example 1 An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 100 kPa and 17 ºC, and 800 kJ/kg of heat is transferred to air during the constant-volume heat addition process. Accounting for the variation of specific heats of air with temperature, determine (a) the maximum temperature and pressure which occur during the cycle, (b) the net work output, (c) the thermal efficiency of this cycle, and (d) the mean effective pressure for the cycle.

11. Example 1 (continued) Process 1-2: Isentropic compression Table A-17 u 1 = 206.91 kJ/kg v r1 = 676.1 State 1: air at p 1 = 100 kPa and T 1 = 290 K Table A-17 T 2 = 652.4 K u 2 = 475.11 kJ/kg (a)

12. Example 1 (continued) Table A-17 T 3 = 1575.1 K v r3 = 6.108 = 1799.7 kPa Process 2-3: constant-volume heat addition q in = u 3 – u 2 u 3 = q in + u 2 = 800 + 475.11 = 1275.11 kJ/kg

13. Example 1 (continued) = 4347 kPa (b) Process 3-4: Isentropic expansion Table A-17 T 4 = 795.6 K u 4 = 588.74 kJ/kg q out = u 4 – u 1 = 588.74 – 206.91 = 381.83 kJ/kg

14. Example 1 (continued) w net = q net = q in – q out = 800 – 381.83 = 418.17 kJ/kg (c) = 1 – r 1-k = 1 – (8) 1-1.4 = 0.565 (d) = 574.4 kPa

15. Diesel Cycles Rudolph Diesel (1890)

16. Diesel Cycles Gasoline Engines Diesel Engines 1. Spark ignition (SI) Compression ignition (CI) 2. Constant-volume Constant-pressure heat heat addition addition 3. Low compression-ratio High compression-ratio 4. High fuel cost Low fuel cost 5. Low efficiency High Efficiency

17. Diesel Cycles First Law q – w = Δu q in = w + Δu q out = u 4 – u 1 = pΔv + Δu = c v (T 4 – T 1 ) Q in Q out 1 2 3 4 T S p = const v = const = h 3 – h 2 = c p (T 3 – T 2 )

18. Diesel Cycles Cutoff Ratio

19. Diesel Cycles ► η D increases as r c decreases ► η D → η O as r c → 1 ► η D > η O

20. Example 2 An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 2. At the beginning of the compression process, air is at 100 kPa, 27 ºC and 0.0018 m 3. Utilizing the cold-air-standard assumptions, determine (a) the temperature and pressure of air at the end of each process, (b) the net work output, (c) the thermal efficiency of this cycle, and (d) the mean effective pressure for the cycle.

21. Example 2 (continued) State 1: air at p 1 = 100 kPa, T 1 = 300 K and V 1 = 0.0018 m 3 (a) V 3 = r c V 2 = 2 (0.0001) = 0.0002 m 3 V 4 = V 1 = 0.0018 m 3 Process 1-2: Isentropic compression

22. Example 2 (continued) = 1906.6 K Process 3-4: Isentropic expansion Process 2-3: Constant-pressure heat addition p 3 = p 2 = 5719.8 kPa = 791.7 K

23. Example 2 (continued) = 0.0021(1.005)(1906.6 – 593.3) = 2.77 kJ Q in = m(h 3 – h 2 ) = mc p (T 3 – T 2 ) (b) Q out = m(u 4 – u 1 ) = mc v (T 4 – T 1 ) = 0.0021(0.718)(791.7 – 300) = 0.74 kJ Table A-2 c p = 1.005 kJ/kg·K c v = 0.718 kJ/kg·K

24. Example 2 (continued) W net = Q net = Q in – Q out = 2.77 – 0.74 = 2.03 kJ (c) (d) = 1194.1 kPa