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DJA3032 CHAPTER 2
1.
CHAPTER 2: PISTON
ENGINE PROCESS ANALYSIS by MOHD SAHRIL BIN MOHD FOUZI, Grad. IEM (G 27763) DEPARTMENT OF MECHANICAL ENGINEERING © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
2.
General objective: To understand
the structure and various types of cycle engines. Specific objectives: At the end of this unit you should be able to: define the air standard cycle. define constant pressure (cp) and constant volume (cv). draw p-v diagram of Otto cycle, Diesel cycle and Combined cycle. explain Otto cycle, Diesel cycle and Combine/dual cycle. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
3.
Introduction : In this
unit we are to discuss the meaning of standard cycle, heat supplied at constant volume and heat supplied at constant pressure. An internal combustion engine can be classified into three different cycles and they are Otto cycle, Diesel cycle and Dual- combustion cycle. Pressure (P) 3 2 4 1 P3 P2 P4 P1 Volume ( v ) tconspv tan V1V2 Qin Qout V 4 1 2 3 P2 = P3 P1 V1= V4 V3V2 P4 PPressure, Volume tconspv tan Qin Qout Figure 2.1: P-V Diagram (Otto Cycle) Figure 2.2: P-V Diagram (Diesel Cycle) © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
4.
Qin P V 1 2 3 4 5 P3=p4 V2=V3 V1=V5 contsvp . Qin Qout Figure 2.3:
P-V Diagram (Dual/Combined Cycle) © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
5.
Air Standard Cycles The
air standard cycle is a cycle followed by a heat engine which uses air as the working medium. Since the air standard analysis is the simplest and most idealistic, such cycles are also called ideal cycles and the engine running on such cycles are called ideal engines. In order that the analysis is made as simple as possible, certain assumptions have to be made. These assumptions result in an analysis that is far from correct for most actual combustion engine processes, but the analysis is of considerable value for indicating the upper limit of performance. The analysis is also a simple means for indicating the relative effects of principal variables of the cycle and the relative size of the apparatus. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
6.
Assumptions: 1. The working
medium is a perfect gas with constant specific heats and molecular weight corresponding to values at room temperature. 2. No chemical reactions occur during the cycle. The heat addition and heat rejection processes are merely heat transfer processes. 3. The processes are reversible. 4. Losses by heat transfer from the apparatus to the atmosphere are assumed to be zero in this analysis. 5. The working medium at the end of the process (cycle) is unchanged and is at the same condition as at the beginning of the process (cycle). © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
7.
Figure 2.4: T-S
Diagram for Otto Cycle Figure 2.5: T-S Diagram for Diesel Cycle Figure 2.6: T-S Diagram for Combined/Dual Cycle T-S Diagram © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
8.
Compression ratio To give
direct comparison with an actual engine the ratio of specific volume, v1 / v2, is taken to be the same as the compression ratio of the actual engine, Compression ratio, rv = = 2 1 v v olumeclearencev olumeclearencevesweptvolum Pressure (P) 3 2 4 1 P3 P2 P4 P1 Volume ( v ) tconspv tan V1V2 Clearance volume Swept volume Minimum volume Maximum volume TDC BDC Swept volume Clearance volume Figure 2.7: P-V Diagram for Otto Cycle Figure 2.8: Animated 4 Stroke Engine © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
9.
Example 2.1 An Otto
cycle in a petrol engine with a cylinder bore of 55mm, a stroke of 80mm, and a clearance volume of 23.3 cm3 is given. Find the compression ratio of this engine. Solution: Cylinder bore, B = 55mm = 5.5 cm Stroke, S = 80mm = 8.0 cm Clearance Volume = 23.3 cm 3 Abstract the data from the question. Convert the data into centimetre unit. swept volume = x B² x S 4 = x (5.5)² x 8 = 190.07 cm 3 4 Compression ratio, rv = 2 1 v v olumeclearencev olumeclearencevesweptvolum 23.3 cm 3 = (190.07 + 23.3) cm 3 = 9.16 = © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
10.
Exercise 2.1 A petrol
engine with a cylinder bore of 73 mm, a stroke of 95 mm, and a clearance volume of 26.3 cm 3 is given. Find the compression ratio of this engine. [Ans: 16.12] Exercise 2.2 An air engine is operated with cylinder bore 65 mm with the stroke of 73 mm. The clearance volume for this engine is 1/10 of swept volume. Calculate the compression ratio for this air engine. [Ans: 11.00] Exercise 2.3 Given compression ratio for a petrol engine is 10.5.The cylinder bore and stroke length for this engine are 69 mm and 83 mm. Calculate the clearance volume for this engine. [Ans: 32.66 cm3 ] Exercise 2.4 An otto cycle engine with compression ratio 9.5 operating with clearrance volume 23.45 cm 3. The stroke length for this engine is 83 mm, find the cylinder bore for this engine. [Ans: 5.53 cm] Self-Exercise © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
11.
Otto Cycle Analysis Pressure
(P) 3 2 4 1 P3 P2 P4 P1 Volume ( v ) tconspv tan V1V2 Qin Qout Figure 2.9: P-V Diagram for Otto Cycle Otto Cycle Process: 1 to 2 is isentropic compression with compression ratio v1 / v2, or rv . 2 to 3 is reversible constant volume heating, the heat supplied Q1 3 to 4 is isentropic expansion, v4 /v3 is similar to v1 /v2. 4 to 1 is reversible constant volume cooling, the heat rejected Q2. Formula to find the thermal efficiency for Otto Cycle P1V1 P2V2= T1V1 1 = T2V2 1 Qin, Q1 = Cv(T3 ─ T2 ) Qout, Q2 = Cv(T4 ─ T1 ) Qout, Q2 = 1 - Qin, Q1 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
12.
Example 2.2 One petrol
engine is working at a constant volume, the compression ratio is 8.5:1. Pressure and temperature at a beginning compression process is 101 kN/m2 and 840 C. Temperature at the beginning of an expand process is14960 C. Calculate the temperature and pressure at the important points based on the Otto cycle. Solution: Pressure (P) 3 2 4 1 P3 P2 P4 P1 Volume ( v ) tconspv tan V1V2 Qin Qout P-V Diagram for Otto Cycle 3 4 2 1 v V V V V r = 8.5 *T1 = 84°C + 273K = 357 K P1 = 101 kN/m² *T3 = 1496°C + 273K = 1769 K * Temperature must be convert into Kelvin © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
13.
Point 2 (1
to 2 is isentropic compression ) P1V1 P2V2= P2 = V1 V2 [ ] x P1 P2 = 41. [8.5] x 101 kN / m² P2 = 2020.73 kN / m² T1V1 1 = T2V2 1 T2 = x T1 V1 V2 [ ] 1 T2 = 40. [8.5] x 357 K T2 = 840.30 K Point 3 (2 to 3 is reversible constant volume heating ) Constant volume P2V2 P3V3 T2 T3 = P3 = T3 T2 x P2 P3 = 1769 840.30 x 2020.73 kN / m² P3 = 4254.04 kN / m² © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
14.
Point 4 (3
to 4 is isentropic expansion ) P3V3 P4V4= P4 = V3 V4 [ ] x P3 P4 = 212.63 kN / m² P4 = 41. x 4254.04 kN / m²[ 1 8.5 ] T3V3 1 = T4V4 1 T4 = x T3 V3 V4 [ ] 1 T4 = 40. x 1769 K T4 = 751.55 K [ ]1 8.5 Qin, Q1 = Cv (T3 ─ T2 ) = 0.718 kJ/kg.K x ( 1769 K – 840.30 K) Qin, Q1 = 666.81 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 751.55 K – 357 K) Qout, Q2 = 283.29 kJ/kg © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
15.
Exercise 2.5 A four
cylinder engine operates in Otto cycle, the volume is constant and the compression ratio is 9:1, beginning pressure is 105 KN/m2 , temperature is 83 ° C and final temperature is 1520 °C. Draw a p-v diagram and find the temperature and pressure for each point. Lastly calculate the efficiency of Otto cycle. Exercise 2.6 One petrol engine is working at a constant volume, the compression ratio is 8.5:1. Pressure and temperature at a beginning compression process is 101 kN/m2 and 84 0C. Temperature at the beginning of an expand process is14960 C. Calculate the temperature and pressure at the important points based on the Otto cycle. Hence, calculate thermal efficiency for this engine. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
16.
Solution for Exercise
2.5 3 4 2 1 v V V V V r = 9 T1 = 83°C + 273K = 356 K P1 = 105 kN/m² T3 = 1520°C + 273K = 1793 K Pressure (P) 3 2 4 1 P3 P2 P4 P1 Volume ( v ) tconspv tan V1V2 Qin Qout P-V Diagram for Otto Cycle Point 2 (1 to 2 is isentropic compression ) P1V1 P2V2= P2 = V1 V2 [ ] x P1 P2 = 41. [9] x 105 kN / m² P2 = 2275.77 kN / m² T1V1 1 = T2V2 1 T2 = x T1 V1 V2 [ ] 1 T2 = 40. [9] x 356 K T2 = 857.33 K © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
17.
Point 3 (2
to 3 is reversible constant volume heating ) P3V3 P4V4= P4 = V3 V4 [ ] x P3 P4 = 41. x 4759.49 kN / m²[ 1 9 ] Constant volume P2V2 P3V3 T2 T3 = P3 = T3 T2 x P2 P3 = 1793 857.33 x 2275.77 kN / m² P3 = 4759.49 kN / m² Point 4 (3 to 4 is isentropic expansion ) P4 = 219.59 kN / m² T3V3 1 = T4V4 1 T4 = x T3 V3 V4 [ ] 1 T4 = 40. x 1793 K T4 = 744.53 K [ ]1 9 Solution for Exercise 2.5 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
18.
Solution for Exercise
2.5 Qin, Q1 = Cv (T3 ─ T2 ) = 0.718 kJ/kg.K x ( 1793 K – 857.33 K) Qin, Q1 = 671.81 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 744.53 K – 356 K) Qout, Q2 = 278.96 kJ/kg Thermal Efficiecy for Otto Cycle Qout, Q2 th, Otto = 1 - Qin, Q1 = 1 - 278.96 kJ.K/kg 671.81 kJ.K/kg = 0.585 @ 58.5 % © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
19.
Diesel Cycle Analysis V 4 1 2
3 P2 = P3 P1 V1= V4V3V2 P4 PPressure, Volume tconspv tan Qin Qout Figure 2.10: P-V Diagram for Diesel Cycle Diesel Cycle Process: 1 to 2 is isentropic compression with compression ratio v1 / v2, or rv . 2 to 3 is reversible constant pressure heating, the heat supplied Q1, v3 / v2, cut off ratio, rc 3 to 4 is isentropic expansion, 4 to 1 is reversible constant volume cooling, the heat rejected Q2. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
20.
Formula to find
the thermal efficiency for Diesel Cycle Point 2 (1 to 2 is isentropic compression ) P1V1 P2V2= Point 1 Get the data from the question. T1V1 1 = T2V2 1 To get pressure To get temperature (Kelvin) Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure P2V2 P3V3 T2 T3 = T3 T2 = rc V3 V2 = © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
21.
Point 4 (3
to 4 is isentropic expansion ) P3V3 P4V4= P4 = V3 V4 [ ] x P3 =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] V 4 1 2 3 P2 = P3 P1 V1= V4V3V2 P4 PPressure, Volume tconspv tan Qin Qout 1 2 Put 12 into =V3 V4 [ ] xrc[ ] 1 rv [ ] P4 = rc rv [ ] x P3 T3V3 T4V4= 1 1 T4 = V3 V4 [ ] x T3 1 3 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
22.
=V3 V4 [ ] xV3 V2 [
] V2 V1 [ ] =V3 V4 [ ] xrc[ ] 1 rv [ ] 4 Put 34 into T4 = rc rv [ ] x T3 1 Qin, Q1 = Cp(T3 ─ T2 ) Qout, Q2 = Cv(T4 ─ T1 ) Thermal Efficiency Qout, Q2 = 1 - Qin, Q1 th, Diesel V 4 1 2 3 P2 = P3 P1 V1= V4V3V2 P4 PPressure, Volume tconspv tan Qin Qout © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
23.
Example 2.3 Diesel engine
has an inlet temperature and a pressure at 15°C and 1 bar respectively. The compression ratio is 12/ 1 and the maximum cycle temperature is 1100°C. Calculate the air standard thermal efficiency based on the diesel cycle. = 12 T1 = 15°C + 273K = 288 K P1 = 1 bar T3 = 1100°C + 273K = 1373 K rv Solution © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
24.
Point 2 (1
to 2 is isentropic compression ) P1V1 P2V2= P2 = V1 V2 [ ] x P1 P2 = 41. [12] x 1 bar P2 = 32.42 bar T1V1 1 = T2V2 1 T2 = x T1 V1 V2 [ ] 1 T2 = 40. [12] x 288 K T2 = 778.15 K Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure P2V2 P3V3 T2 T3 = T3 T2 = rc V3 V2 = 1373 K 778.15 K = = 1.764 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
25.
Point 4 (3
to 4 is isentropic expansion ) P3V3 P4V4= P4 = V3 V4 [ ] x P3 =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] 1 2 Put 12 into =V3 V4 [ ] xrc[ ] 1 rv [ ] P4 = rc rv [ ] x P3 P4 = 1.764 12[ ] x 32.42 bar 41. T3V3 T4V4= 1 1 T4 = V3 V4 [ ] x T3 1 3 =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] =V3 V4 [ ] xrc[ ] 1 rv [ ] 4 Put 34 into T4 = rc rv [ ] x T3 1 T4 = 1.764 12[ ] x 1373 K 40. P4 = 2.21 bar T4 = 637.67 K © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
26.
Qin, Q1 =
Cp (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1373 K – 778.15 K) Qin, Q1 = 597.82 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 637.67 K – 288 K) Qout, Q2 = 251.06 kJ/kg Qout, Q2 th, Diesel = 1 - Qin, Q1 = 1 - 251.06 kJ/kg 597.82 kJ/kg = 0.580 @ 58 % Thermal Efficiecy for Diesel Cycle © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
27.
Exercise 2.7 One engine
operates in diesel cycle, the beginning temperature is 18 °C, and pressure is 1 bar. The compression ratio is 14:1 , maximun temperature is 1320°C. Given cp = 1.005 KJ/kg.K, R=287 KJ/kg.K. Calculate the temperature and rc and mechanical efficiency. Exercise 2.8 A diesel cycle’s engine running with the beginning of compression process temperature 17 °C and pressure 1.1 bar. The compression ratio for this engine is 13.5 and the temperature for beginning expansion process is 1290 °C. Calculate thermal efficiency for this diesel cycle engine. Exercise 2.9 An engine was operated with diesel cycle process with the beginning temperature and pressure at 16.8 °C and 1.05 bar. The compression ratio for this engine is 12.5 and cut of ratio for this engine is 1.75. Calculate thermal efficiency for this engine. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
28.
Solution for Exercise
2.7 = 14 T1 = 18°C + 273K = 291 K P1 = 1 bar T3 = 1320 °C + 273K = 1593 K rv P1V1 P2V2= P2 = V1 V2 [ ] x P1 P2 = 41. [14] x 1 bar P2 = 40.23 bar Point 2 (1 to 2 is isentropic compression ) T1V1 1 = T2V2 1 T2 = x T1 V1 V2 [ ] 1 T2 = 40. [14] x 291 K T2 = 836.27 K © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
29.
Point 3 (2
to 3 is reversible constant pressure heating ) Constant pressure P2V2 P3V3 T2 T3 = T3 T2 = rc V3 V2 = 1593 K 836.27 K = = 1.905 Solution for Exercise 2.7 P3V3 P4V4= P4 = V3 V4 [ ] x P3 =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] 1 2=V3 V4 [ ] xrc[ ] 1 rv [ ] Point 4 (3 to 4 is isentropic expansion ) Put 12 into P4 = rc rv [ ] x P3 P4 = 1.905 14[ ] x 40.23 bar 41. P4 = 2.47 bar © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
30.
T3V3 T4V4= 1 1 T4
= V3 V4 [ ] x T3 1 3 =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] =V3 V4 [ ] xrc[ ] 1 rv [ ] 4 Solution for Exercise 2.7 Put 34 into T4 = rc rv [ ] x T3 1 T4 = 1.905 14[ ] x 1593 K 40. T4 = 717.34 K Thermal Efficiecy for Diesel Cycle Qin, Q1 = Cp (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1593 K – 836.27 K) Qin, Q1 = 760.51 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 717.34 K – 291 K) Qout, Q2 = 306.11 kJ/kg Qout, Q2 th, Diesel = 1 - Qin, Q1 = 1 - 306.11 kJ/kg 760.51 kJ/kg = 0.597 @ 59.7% © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
31.
Solution for Exercise
2.8 = 13.5 T1 = 17°C + 273K = 290 K P1 = 1.1 bar T3 = 1290 °C + 273K = 1563 K rv P1V1 P2V2= P2 = V1 V2 [ ] x P1 P2 = 41. [13.5] x 1.1 bar P2 = 42.06 bar Point 2 (1 to 2 is isentropic compression ) T1V1 1 = T2V2 1 T2 = x T1 V1 V2 [ ] 1 T2 = 40. [13.5] x 291 K T2 = 821.36 K © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
32.
Point 3 (2
to 3 is reversible constant pressure heating ) Constant pressure P2V2 P3V3 T2 T3 = T3 T2 = rc V3 V2 = 1563 K 821.36 K = = 1.903 Solution for Exercise 2.8 P3V3 P4V4= P4 = V3 V4 [ ] x P3 =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] 1 2=V3 V4 [ ] xrc[ ] 1 rv [ ] Point 4 (3 to 4 is isentropic expansion ) Put 12 into P4 = rc rv [ ] x P3 P4 = 2.71 bar P4 = 1.903 13.5[ ] x 40.23 bar 41. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
33.
T3V3 T4V4= 1 1 T4
= V3 V4 [ ] x T3 1 3 =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] =V3 V4 [ ] xrc[ ] 1 rv [ ] 4 Solution for Exercise 2.8 Put 34 into T4 = rc rv [ ] x T3 1 T4 = 1.903 13.5[ ] x 1563 K 40. T4 = 713.84 K Thermal Efficiecy for Diesel Cycle Qin, Q1 = Cp (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1563 K – 821.36 K) Qin, Q1 = 745.35 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 713.84 K – 290 K) Qout, Q2 = 304.32 kJ/kg Qout, Q2 th, Diesel = 1 - Qin, Q1 = 1 - 304.32 kJ/kg 745.35 kJ/kg = 0.592 @ 59.2% © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
34.
Solution for Exercise
2.9 P1V1 P2V2= P2 = V1 V2 [ ] x P1 P2 = 41. [12.5] x 1.05 bar P2 = 36.05 bar Point 2 (1 to 2 is isentropic compression ) T1V1 1 = T2V2 1 T2 = x T1 V1 V2 [ ] 1 T2 = 40. [12.5] x 289.8 K T2 = 795.91 K = 12.5 T1 = 16.8°C + 273K = 289.8 K P1 = 1.05 bar rc rv = 1.75 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
35.
Point 3 (2
to 3 is reversible constant pressure heating ) Solution for Exercise 2.9 P3V3 P4V4= P4 = V3 V4 [ ] x P3 =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] 1 2=V3 V4 [ ] xrc[ ] 1 rv [ ] Point 4 (3 to 4 is isentropic expansion ) Put 12 into P4 = rc rv [ ] x P3 P4 = 2.30 bar P4 = 1.75 12.5[ ] x 36.05 bar 41. Constant pressure P2V2 P3V3 T2 T3 = T3 T2 = rc V3 V2 = 795.91 K = = 1.75T3 T3 = 1392.84 K © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
36.
T3V3 T4V4= 1 1 T4
= V3 V4 [ ] x T3 1 3 =V3 V4 [ ] xV3 V2 [ ] V2 V1 [ ] =V3 V4 [ ] xrc[ ] 1 rv [ ] 4 Solution for Exercise 2.9 Put 34 into T4 = rc rv [ ] x T3 1 T4 = 1.75 12.5[ ] x 1392.84 K 40. T4 = 634.38 K Thermal Efficiecy for Diesel Cycle Qin, Q1 = Cp (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1392.84 K – 795.91 K) Qin, Q1 = 599.91 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 634.38 K – 289.8 K) Qout, Q2 = 247.41 kJ/kg Qout, Q2 th, Diesel = 1 - Qin, Q1 = 1 - 247.41 kJ/kg 599.91 kJ/kg = 0.588 @ 58.8 % © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
37.
Dual/Combined Cycle Analysis QinP V 1 2 3
4 5 P3=P4 V2=V3 V1=V5 contsvp . Qin Qout Dual/Combined Cycle Process: 1 to 2 is isentropic compression with compression ratio v1 / v2, or rv , 2 to 3 is reversible constant volume heating, the heat supplied Q1 3 to 4 is reversible constant pressure heating, the heat supplied Q1, v4/ v3, cut off ratio, rc 4 to 5 is isentropic expansion, 5 to 1 is reversible constant volume cooling, the heat rejected Q2. Figure 2.11: P-V Diagram for Dual/Combined Cycle © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
38.
Formula to find
the thermal efficiency for Dual/Combined Cycle Point 2 (1 to 2 is isentropic compression ) P1V1 P2V2= Point 1 Get the data from the question. T1V1 1 = T2V2 1 To get pressure To get temperature (Kelvin) Point 3 (2 to 3 is reversible constant volume heating ) Constant volume P2V2 P3V3 T2 T3 = P3 = T3 T2 x P2 T3 = P3 P2 x T2 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
39.
Point 4 (3
to 4 is reversible constant pressure heating ) Constant pressure P3V3 P4V4 T3 T4 = T4 T3 = rc V4 V3 = Assume the heat added at constant volume is equal to the heat added at constant pressure; cv(T3 – T2 ) = cp (T4 – T3 ) T4 = cv(T3 – T2 ) + cp. (T3) cp © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
40.
Point 5 (4
to 5 is isentropic expansion ) P4V4 P5V5= P5 = V4 V5 [ ] x P4 =V4 V5 [ ] xV4 V3 [ ] V2 V1 [ ] 1 2 Put 12 into =V4 V5 [ ] xrc[ ] 1 rv [ ] P5 = rc rv [ ] x P4 QinP V 1 2 3 4 5 P3=P4 V2=V3 V1=V5 contsvp . Qin Qout © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
41.
T4V4 T5V5= 1 1 T5
= V4 V5 [ ] x T4 1 3 =V4 V5 [ ] xV4 V3 [ ] V2 V1 [ ] =V4 V5 [ ] xrc[ ] 1 rv [ ] 4 Put 34 into T5 = rc rv [ ] x T4 1 QinP V 1 2 3 4 5 P3=P4 V2=V3 V1=V5 contsvp . Qin Qout Qin, Q1 = Cv(T3 ─ T2 ) + Cp(T4 ─ T3 ) Qout, Q2 = Cv(T5 ─ T1 ) Thermal Efficiency Qout, Q2 = 1 - Qin, Q1 th, Dual © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
42.
Example 2.4 An oil
engine takes in air at 1.01 bar, 200C and the maximum cycle pressure is 69 bar. The compression ratio is 18/1. Draw the p-v diagram and calculate the air standard thermal efficiency based on the dual combustion cycle. Assume that the heat added at constant volume is equal to the heat added at constant pressure. Solution = 18 T1 = 20°C + 273K = 293 K P1 = 1.01 bar P3 = 69 bar rv © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
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Point 2 (1
to 2 is isentropic compression ) P1V1 P2V2= P2 = V1 V2 [ ] x P1 P2 = 41. [18] x 1.01 bar P2 = 57.77 bar T1V1 1 = T2V2 1 T2 = x T1 V1 V2 [ ] 1 T2 = 40. [18] x 293 K T2 = 931.06 K Point 3 (2 to 3 is reversible constant volume heating ) Constant volume P2V2 P3V3 T2 T3 = P3 = T3 T2 x P2 T3 = P3 P2 x T2 T3 = 69 bar 57.77 bar x 931.06 K T3 = 1112.05 K © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
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Assume the heat
added at constant volume is equal to the heat added at constant pressure; cv(T3 – T2 ) = cp (T4 – T3 ) T4 = cv(T3 – T2 ) + cp. (T3) cp T4 = 0.718 kJ/kg.K (1112.05 K – 931.06 K) + 1.005 kJ/kg.K(1112.05 K) 1.005 kJ/kg.K T4 = 1241.30 K Point 4 (3 to 4 is reversible constant pressure heating ) Constant pressure P3V3 P4V4 T3 T4 = T4 T3 = rc V4 V3 = rc = 1241.30 K 1112.05 K rc = 1.116 © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
45.
Point 5 (4
to 5 is isentropic expansion ) P4V4 P5V5= P5 = V4 V5 [ ] x P4 =V4 V5 [ ] xV4 V3 [ ] V2 V1 [ ] 1 2=V4 V5 [ ] xrc[ ] 1 rv [ ] Put 12 into P5 = rc rv [ ] x P4 P5 = 1.41 bar P5 = 1.116 18[ ] x 69 bar 41. T4V4 T5V5= 1 1 T5 = V4 V5 [ ] x T4 1 3 =V4 V5 [ ] xV4 V3 [ ] V2 V1 [ ] =V4 V5 [ ] xrc[ ] 1 rv [ ] 4 Put 34 into T5 = rc rv [ ] x T4 1 T5 = 1.116 18[ ] x 1241.30 K 40. T5 = 408.16 K © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
46.
Qin, Q1 =
0.718 kJ/kg.K(1112.05 K ─ 931.06 K ) + 1.005 kJ/kg.K (1241.30 K ─1112.05 K) Qin, Q1 = Cv(T3 ─ T2 ) + Cp(T4 ─ T3 ) Qin, Q1 = 259.85 kJ/kg Qout, Q2 = Cv (T5 ─ T1 ) Qout, Q2 = 82.68 kJ/kg Qout, Q2 = 0.718 kJ/kg.K x (408.16 K– 293 K) Qout, Q2 th, Dual = 1 - Qin, Q1 = 1 - 82.68 kJ/kg 259.85 kJ/kg = 0.682 @ 68.2 % Thermal Efficiecy for Dual/Combined Cycle © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
47.
Self –Assessment Exercise 2.11 In
a dual combustion cycle, the maximum pressure is 64 bar. Calculate the thermal efficiency when the pressure and temperature at the start of the compression are 1.01 bar and 18 0 C respectively. The compression ratio is 17/1. Exercise 2.10 The pressure and temperature of air standard dual combustion cycle are given below, i. T1 = 290 K ii. P1 = 1.01 bar iii. T2 = 871.1K iv. T3 = 1087.5 K v. T4 = 1236.3 K vi. T5 = 429.3 K and ratio of compression is 16:1. Calculate the thermal efficiency for this dual cycle engine. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
48.
Exercise 2.12 In a
dual combustion cycle, the maximum pressure is 62 bar. Calculate the thermal efficiency when the pressure and temperature at the start of the compression are 1.01 bar and 16.7 °C respectively. The compression ratio is 17.7. Exercise 2.13 In a combine cycle, the maximum pressure is 67 bar. Calculate the thermal efficiency when the pressure and temperature at the start of the compression are 1.03 bar and 17 °C respectively. The compression ratio is 16.9. Exercise 2.14 An engine is operating in dual combustion cycle, the maximum pressure is 66 bar. Calculate the thermal efficiency for this engine when the pressure and temperature at the beginning of the compression process are 1.03 bar and 20.3 °C respectively. The compression ratio for this engine is 17.9. © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
49.
Answer for Exercise
2.11 T1 = 291 K P1 = 1.01 bar T2 = 903.80 K P2 = 53.33 bar T3 = 1084.68 K P3 = 64 bar T4 = 1213.91 K P4 = 64 bar rc = 1.119 T5 = 408.85 K P5 = 1.42 bar Qin = 259.75 kJ/kg Qout = 84.61 kJ/kg ɳth, dual = 0.674 @ 67.4 % Answer for Exercise 2.12 T1 = 289.7 K P1 = 1.01 bar T2 = 914.40 K P2 = 56.43 bar T3 = 1004.72 K P3 = 62 bar T4 = 1069.25 K P4 = 62 bar rc = 1.064 T5 = 347.30 K P5 = 1.21 bar Qin = 129.70 kJ/kg Qout = 41.35 kJ/kg ɳth, dual = 0.681 @ 68.1 % © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
50.
Answer for Exercise
2.13 T1 = 290 K P1 = 1.03 bar T2 = 898.57 K P2 = 53.94 bar T3 = 1116.22 K P3 = 67 bar T4 = 1271.71 K P4 = 67 bar rc = 1.139 T5 = 432.40 K P5 = 1.54 bar Qin = 312.54 kJ/kg Qout = 102.25 kJ/kg ɳth, dual = 0.673 @ 67.3 % Answer for Exercise 2.14 T1 = 293.3 K P1 = 1.03 bar T2 = 929.94 K P2 = 58.46 bar T3 = 1049.94 K P3 = 66 bar T4 = 1135.68 K P4 = 66 bar rc = 1.082 T5 = 369.62 K P5 = 1.30 bar Qin = 172.33 kJ/kg Qout = 54.79 kJ/kg ɳth, dual = 0.682 @ 68.2 % © MSF @ POLITEKNIK UNGKU OMAR (DJA3032) INTERNAL COMBUSTION ENGINE
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