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COMBUSTION POWER CYCLES Otto Cycle ( Basis of Comparison for Spark-Ignition Engines ) QA (V=C)
QR (V=C)
A. T D
L S=C
S=C
B. B D
A. PV and TS Diagram
3 P
3
T
S=C
S=C
V=C
V=C 4
2
Wnet
4 QT
2
V=C S=C S=C V=C
1 1
V
V2=V3
V1=V4
S
S1=S2
S3=S4
INTERNAL COMBUSTION ENGINE
ANALYSIS OF EACH STATE: P1 V1 T1 S1
< > < =
P2 V2 T2 S2
< = < <
P3 V3 T3 S3
> < > =
P4 V4 T4 S4
> = > >
P1 V1 T1 S1
B. PVT Relationship Process 1-2 : ( S=C ) k 1 V1 P2 K T2 P V T1 1 2
1
k 1
K 1
V1 V2 VD cVD VD C 1 V2 V2 cV D C Where : c = percent clearance
Isentropic Compression Ratio : rks Process 2-3 : ( V= C )
P3 T P 3; 3 P2 T2 P2
Presssure Ratio , rpv=c = ( Addition of heat) Compression
Process 3 – 4 ( S = C ) 1
k 1 k 1 K 1 V3 P4 K T4 V P T3 4 3
V4 = Isentropic expansion Ratio, V3
res=c
Process 4 – 1 ( V = C )
P4 T P 4; 4 P1 T1 P1
Pressure Ratio , rpv=c ( Rejection of Heat )
INTERNAL COMBUSTION ENGINE
C. Heat Added , QA = Σ + A Process 2 – 3 ( V = C ) QA = Q2-3 = mcvΔT = mcv(T3-T2)
D. Heat Rejected, QR = Σ – A Process 4 –1 : ( V = C ) QR = Q4-1 = - mcvΔT = - mcv(T4-T1)
E. Net Work, WNET WNET = / QA / - / QR / =
P2V2 P1V1 P4V4 P3V3 1 K 1 K
F. Thermal Efficiency, e
W NET T T1 x100% = 1 4 x100% QA T3 T2 1 x100% = (rK ) K 1
e=
G. Mean Effective Pressure, Pm
PM=
W NET W NET = V1 V2 VD
INTERNAL COMBUSTION ENGINE
Diesel Cycle – Basis of Comparison for Compression-Ignition Engines QA (P=C) QR (V=C)
TDC
x
L S=C
S=C
BDCC
PV & TS Diagram 3 P
2
P=C
3
T
S=C
S=C 4
2
P=C 4
Wnet QT V=C S=C S=C
1 1
V=C
1 V
S
ANALYSIS OF EACH STATE:
INTERNAL COMBUSTION ENGINE
P1 V1 T1 S1
< > < =
P2 V2 T2 S2
= < < <
P3 V3 T3 S3
> < > =
P4 V4 T4 S4
> = > >
P1 V1 T1 S1
A. PVT Relationship Process 1-2 : ( S=C ) P2 T2 P T1 1
k 1 K
1
V1 V 2
k 1
K 1
V1 V2 VD cVD VD C 1 V2 V2 cV D C Where : c = percent clearance
Isentropic Compression Ratio : rks Process 2-3 : ( P= C )
V3 T 3 V2 T2 Cut-off Ratio , rCP=C Process 3 – 4 ( S = C ) k 1 V3 P4 K T4 V P T3 4 3
1
k 1
K 1
V4 = Isentropic Expansion Ratio, V3
res=c
Process 4 – 1 ( V = C )
P4 T P 4; 4 P1 T1 P1
Pressure Ratio , rpv=c
B. Heat Added , QA = Σ + A
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Process 2 – 3 ( P = C ) QA = Q2-3 = mcpΔT = mcp(T3-T2)
C. Heat Rejected, QR = Σ – A Process 4 –1 : ( V = C ) QR = Q4-1 = - mcvΔT = - mcv(T4-T1)
D. Net Work, WNET WNET = / QA / - / QR / = / mcp(T3-T2) / - / - mcv(T4-T1)/
E. Thermal Efficiency, e
e=
C (T T1 ) W NET x100% = 1 V 4 x100% QA Cp (T3 T2 )
1 =1( rK ) K 1
(rC ) K 1 x100% K ( r 1 ) C
G. Mean Effective Pressure, Pm
PM=
W NET W NET = V1 V2 VD
INTERNAL COMBUSTION ENGINE
Dual Combustion Cycle (Sabathe Cycle) –basis of comparison spark and compression ignition engines.
P
P=C
TY S=C
IP
P=C
V=C
V=C
WNET
S=C
QA
V=C S=C S=C
V=C
V
SP
Thermal Efficiency, e e =1-
1 (rP ) K 1
K (rC rP ) 1 x100% ( rP 1) KrP (rC 1)
Where : Compression Rato ( rK) =
V1 V2
V4 V3 P3 Pressure Ratio ( rP) = P2 Cut-off Ratio ( rC ) =
INTERNAL COMBUSTION ENGINE
IP
Problems & Solutions : 1. An ideal diesel engine operates on 0.5 kg/s of air with a suction state of 100 Kpa and 45°C. The pressure at the end of compression is 3.25 Mpa and the cut-off is at 6% of the stroke from head-end dead center position. a) Compression Ratio b) Percent Clearance c) Network d) Thermal Efficiency e) Mean Effective Pressure Solution : A Solving for Compression Ratio P2 T2 P T1 1
k 1 K
1
V1 V 2
k 1
V1 V2 VD cVD VD C 1 V2 V2 cV D C
rks
1
rKS=C
K 1
1
K 1.4 = V1 P2 3250 12.02 V2 P1 100
B) Solving for Percent Clearance
V1 V2 VD cVD VD C 1 V2 V2 cV D C c 1 12.02 = c 12.02 C = C + 1 12.02 C - C = 1 rks
C = 0.0907 or 9.07 % C) Solving for NetWork WNET = / QA / - / QR / = / mcp(T3-T2) / - / - mcv(T4-T1)/ where : T4 = T1(rC)K c.o. c 0.06 0.0907 1.66 rC = = c 0.0907 so; T4 = 318(1.66)1.4 = 646.51 K and , T3 = T1(rkk-1)(rc) = 318 [ ( 12.02)1.4-1(1.66)] T3 = 1427.24 K Also,
T2 = T1(rk)k-1 = 318(12.02)1.4-1
INTERNAL COMBUSTION ENGINE
T2 = 859.78 K Therefore; WNET = / (0.5)(1.0062)(1427.24-859.78)/-/(0.5)(0.7186)(318-646.51)/ WNET = 167.455 KW d) Solving for the thermal efficiency e=
W NET W NET 167.455 KW x100 QA mair Cp air (T3 T2 ) (0.5)(1.0062)(1427.24 859.78) e = 58.655 % e) Solving for the Mean Effective Pressure
W NET W NET = V1 V2 VD V V1 rks 1 ; V2 V2 rKS C
PM=
but;
P1V1 = maRaT1 maRaT1 (0.5)(0.787)(318) m3 0 . 456 V1 = P1 (100) s
So,
m3 3 V2 = s 0.037 m 12.02 s 0.456
Therefore; PM=
167.455 400 KPa (0.456 0.037)
INTERNAL COMBUSTION ENGINE
2. A heat exchanger was install purposely to cool 0.50 kg of gas per second. Molecular weight is 28 and k=1.32. The gas is cooled from 150C to 80. Water is available at the rate of 0.30 kg/s and at a temperature of 12C. Calculate the exit temperature of the water in C. Ans. 46 Solution:
Gas constant, R
8.3143 kJ 0.297 28 kg K
C k 1.32 C C 1.32 C C C R 1.32 C C 0.297 C 0.928 P
V
P
V
P
V
V
V
V
C
P
1.32(0.928) 1.225
kJ kg K
HeatBalance : Heat gain by water Heat loss by gas
mC w
PW
T W mg C P T g g
0.3( 4.187)(t 12) 0.5(1.225)(150 80) t 46.1C 3. Determine the average Cp value in kJ/kg-K of a gas if 522 kJ of heat is necessary to raise the temperature from 300K to 800 K making the pressure constant. a. 1.044 b. 3.044
c. 2.044 d. 4.044
Solution: Q = mCpT 522 = 1 Cp (800-300) Cp = 1.044 kJ/kg-K
INTERNAL COMBUSTION ENGINE
4. What is the efficiency of an Otto cycle with compression ratio of 6:1. The gas used is air. a. 51.2 % c. 61.2 % b. 71.2 % d. 52 % Solution:
Eff 1
1
r
k 1 k
1
1 1.4 1
6
Eff 0.512 51.2% 5. A diesel generating set consumes 235 liters of fuel during 1 hour operation and produces 900 kw power. The density of fuel used is 0.8955 kg/l. Determine the specific fuel consumption of the diesel generating set in kg per kw-hr. Ans. 0.234 a. 0.234 b. 0.25 Solution:
c. 0.24 d. 0.26
Over-all spec. consumption = (235 li/900 kw-hr)(0.8955 kg/li) = 0.234 kg/kw-hr 6. The entropy of pure substance at a temperature of absolute zero is: a. One c. Zero b. Infinity d. Always Positive 7. A power plant operates on an ideal Brayton Cycle. The gas temperature at the turbine inlet is 1400K (1515.2 kJ/kg) and the gas temperature at the turbine exit is 800K (821.95 kJ/kg). Assume a turbine efficiency of 80%, what is the actual turbine work in kJ/kg? a. 554.6 b. 500
c. 555 d. 545
Solution: Turbine Work = 1515.2-821.95 = 693.25 kJ/kg Actual Turbine work = 693.25 x 0.80 = 554.6 kJ/kg 8. The ratio of the input work necessary to bring the pressure of a gas to a speed value in an isentropic process to the actual work input is known as: a. adiabatic efficiency of the compressor c. Q=+ b. Q=d. Isentropic Solution:
INTERNAL COMBUSTION ENGINE
Adiabatic efficiency of the compressor = (isentropic work/actual fluid work) 9. What is the area of the diagram from the relation of temperature and entropy plane? a. Work b. PM
c. Heat d. Energy
10. The first law of thermodynamics states about: a. Conservation of energy c. Zeroth Law b. Conservation of Mass d. Entropy 11. An engine operates on the air-standard OTTO cycle. The cycle work is 1000 KJ/kg . What is the compression ratio of the engine if the maximum cycle temperature is 3173 K and the temperature at the end of isentropic compression is 773 K. a. 6.85 c. 7.85 b. 8.85 d. 9.85 Solution: From: 1 x100% e=1(rK ) K 1 where : e=
W NET 1000 QA QA
Solving for QA : QA = mcv( T3 – T2 ) QA = (1) (0.716)(3173 –773 ) QA = 1718.4
KJ Kg
Then ; e=
1000 0.5819 58.19% 1718.4
thus; 0.5819 = 1 -
1 rk
1.4 1
; rk 8.85
12. The compression ratio of an OTTO cycle is 9. If the initial pressure is 150 Kpa, determinine the final pressure. a. 1 251.10 KPa b. 4 251.10 KPa
c. 2 251.10 KPa d. 3 251.10 KPa
Solution : k
P2 V1 k rk P1 V2
INTERNAL COMBUSTION ENGINE
P2 P1 ( rk ) k
P2 150(9)1.4 thus;
P2 3251.10 KPa 13. What is the final temperature after compression of a Diesel Cycle if the initial temperature is 32C and the clearance is 8%. a. 863.84 K c. 963.84 K b. 763.84 K d. 663.84 K Solution : k 1
T2 V1 (rk ) k 1 T1 V2 T2 T1 ( rk ) k 1 where : T1 32 273 305 K 1 C 1 0.08 C 0.08 rk 13.5 rk thus;
T2 305(13.5)1.4 1 T2 863.84 K 14. An Ideal Dual Combustion Cylce operates on 500 gram of air. At the beginning of the compression, the air is at 100 KPa, 45C. If rp 1.5, rc 1.65, and rk 10, determine the cycle efficiency. a. 53.88% c. 55.88% b. 54.88 % d. 56.88 % Solution : W e NET QA P
P=C
TY S=C
IP
P=C
V=C
V=C
WNET
S=C
QA
V=C S=C S=C
where :
V=C
W NET Q A Q R
V
SP
INTERNAL COMBUSTION ENGINE
IP
Q A Q AX Q AY Solving for Q AX and Q AY Q AX mCV (T3 T2 ) Q AY mC P (T4 T3 ) At point 1. mRT1 0.5 0.287 45 273 V1 0.456m 3 P1 100 At point 2. V 0.456 V2 1 0.0456m 3 rk 10 k 1
T2 V1 k 1 rk T1 V2 1.4 1 T2 45 27310 798.78K V P2 P1 1 V2 At point 3. P3 P2 rp
K
P1 rk 10010 k
1.4
2511.89 KPa
2511 .891.5 3767.84KPa
P 3767.84 T3 T2 3 798.78 1193 .17 K 2511 .89 P2 At point 4. V4 V3 rc 0.04561.65 0.075m 3
V T4 T3 4 V3
T3 rc 1198.171.65 1976.98K
At point 5.
V T5 T4 4 V5
k 1
0.075 T5 1976.98 0.456
Then;
1.4 1
960.375K
Q AX mc v T3 T2 0.5 0.7186 1198 .17 798.78 143.50 KJ
Q AY mc p T4 T3 0.51.0062 1976.98 1198 .17 391.82KJ Q A 143.50 391.82 535.32
QR mcv T3 T1 0.50 0.7186 960.375 318 230.81KJ W NET Q A QR 535.32 230.81 304.51KJ thus; e
304.51 56.88% 535.32
INTERNAL COMBUSTION ENGINE
15. The stoke of a petrol engine is 87.5 mm and the clearance is equal to 12.5 mm.A compression plate is now fitted which has the effect of reducing the clearance to 10 mm. Assuming the compression period to be the whole stroke, the pressure at the beginning of compression as 98.29 KPa and the law of compression PV 1.35 C , and the law of compression PV1.35=C, Calculate the pressure at the end of compression before and after the compression plate is fitted. a. 1628.10 KPa c. 1728.10 Kpa b. 1528.10 KPa d. 1828.10 KPa Solution : 1.35
P1V1
P2V2
1.35 1
98.29 87.5 12.5 thus; P2 1628.10 KPa
1.35
P2 12.5
1.35
16. The compression ratio of a petrol engine working on the constant volume cycle is 8.5. The pressure and temperature at the beginning of compression are 101.325 KPa and 40 C and the maximum pressure of the cycle is 3,141.08 KPa. Taking compression to follow the law PV1.35=C, Calculate the temperature at the end of compression. a. 388.97C c.355.65 C b. 288.97C d. 255.65 C Solution : k 1
T2 V1 T2 1.351 8.5 T1 V2 40 273 T2 661.97 K thus;
T2 388.97C
INTERNAL COMBUSTION ENGINE
17. When the piston is moving up in a two-stroke diesel engine, the scavenge ports are closed when the piston is 675 mm from the top of its stroke, the pressure and temperature of the air in the cylinder when being 13.80 KPag and 43C. The clearance is equal to 65 mm and the diameter of the cylinder is 650 mm. Calculate the mass of air compressed in the cylinder taking the atmospheric pressure as 101.325 KPa and R for air as 0.287 KJ/kg-K a. 0.21 kg b. 0.31 kg Solution :
c. 0.41 kg d. 0.51 kg
P1V1 = mRT1 Where : P1 13.8 101.325 115 .125 KPa T1 43 273 316 K 2 V1 0.65 0.675 0.065 0.2456m 2 4 KJ R 0.287 kg K Substitute
115.125 0.2456 m 0.287 316
thus; m= 0.31 kg 18. If the compression ratio and the cut-off ratio of a Diesel Engine are 15 and 3 respectively, what is the expansion ratio? a. 4 c. 6 b. 5 d. 7 Solution : V rk 1 15 V2
rc
V3 3 V2
re
V4 rk 15 V3 rc 3
re 5
INTERNAL COMBUSTION ENGINE
19. The air is compressed to 1/5 of its original volume in an Otto cycle . What is the final temperature of the air if the initial temperature is 25C a. 288.58C c. 255.66 C b. 388.58C d. 355.66 C Solution:
T2 V1 T1 V2
k 1
where: T1 25 273 298K V2
1 V1 5
V T2 1 298 1 V1 5 T2 561.58K
1.4 1
T2 288.58C
20. At the beginning of compression in an ideal dual combustion cycle, the working fluid is at 90F. If the pressure ration and compression ratio are 1.6 and 13 respectively, what is the temperature at the beginning of the constant pressure portion of combustion. a. 1886.64 F c. 1995.05 F b. 1495.05 F d. 1586.64 F Solution : k 1 1.4 1 T3 T1 rk rp 90 46013 1.6 2455.05 R T3 1995.05 F
INTERNAL COMBUSTION ENGINE
21. An ideal dual combustion cycle operates on ).60 kg of air. Find the cycle efficiency if the compression ratio is 12, pressure ratio is 1.6, cut-off ratio 1.5 and k = 1.3. a. 40.71 % c. 60.71 % b. 50.71 % d. 45.71 % Solution : e 1
c a
where:
A rk
k 1
12
1.31
2.11
k
C
rp rc 1 rp 1 rp k (rc 1)
e 1
1.61.5 1.3 1 1.6 1 1.61.5 1
1.04
1.04 50.71% 2.11
or
1 e =1(rP ) K 1
K (rC rP ) 1 x100% 50.71% (rP 1) KrP (rC 1)
22. What is the pressure ratio in an ideal dual combustion cycle if the pressure at the beginning and end of the constant volume portion of combustion are 2500 KPa and 4000 KPa respectively. a. 1.4 c. 1.6 b. 1.5 d. 1.7 Solution : P 4000 rp 3 1.6 P2 2500 23. In an ideal combustion cycle if the cut-off ratio is 1.5 and the volume at the beginning of the constant portion of combustion is 0.055 m3, what is the volume at the end of ombustion. a. 0.0625 m3 c. 0.0825 m3 3 b. 0.0725 m d. 0.0550 m3 Solution : V V4 rc 4 1.5; thus;V4 0.0825m 3 V3 0.055
INTERNAL COMBUSTION ENGINE
24. In an air standard OTTO cycle, the clearance volume is 18% of the displacement volume. Find the thermal efficiency. a. 0.52 c. 0.53 b. 0.55 d. 0.60 Solution : 1 e 1 A k 1 where : A rk solving for rk : 1 c 1 0.18 rk 6.556 c 0.18 then; A 6.556
1.4 1
2.12
thus; e 1
1 0.53 2.12
25. The compression ratio of an ideal OTTO cycle is 6:1. Initial conditions are 101.3 KPa and 20C. Find the pressure and temperature at the end of adiabatic compression. a. 1244.5 KPa,599.96C b. 1244.5 KPa,60 C
c. 1244.5 KPag, 60C d. 1244.5 KPa, 599.96 C
Solution : k
P2 V1 k rk P1 V2
P2 P1 rk
k
P2 101.3 6
1.4
P2 1244.5 KPa
T2 V1 T1 V2
T2 T1 rk
k 1
rk
k 1
k 1
T2 20 273 6
1.4 1
T2 599.96 K
INTERNAL COMBUSTION ENGINE
INTERNAL COMBUSTION ENGINE