GANADIPATHY TULSI’S JAIN ENGINEERING COLLEGE Chittoor – Cuddalore Road, Kaniyambadi (Post), Vellore – 632102.
Department of Mechanical Engineering 6404 THERMAL ENGINEERING ME – 6404
Unit: 1 GAS POWER POWER CYCLES CYCLES
Part-B (16 Marks) 1) Derive Derive an expression for the air standard standard efficiency efficiency of diesel diesel diesel cycle and then deduce deduce it for mean effectiv effectivee pressure pressure.(Nov – (Nov – 2010) 2010)
This cycle was introduced by Dr. R. Diesel in 1897. It differs from Otto cycle in that heat is supplied at constant constant pressure pressure instead of at constant volume. volume. This cycle comprises comprises of the following operations operations : (i) 1-2......Adiabatic compression. (ii) 2-3......Addition of heat at constant pressure. (iii) 3-4......Adiabatic 3-4......Adiabatic expansion. expansion. (iv) 4-1......Rejection of heat at constant volume. Point 1 represents represents that the cylinder cylinder is full of air. Let p 1, V1 and T1 be the corresponding pressure, volume and absolute temperature. Working Process: During this addition of heat let volume increases from V 2 to V3 and temperature T 2 to T3, corresponding corresponding to point 3. This This point (3)is called called the point of of cut-off. The air then expands expands adiabatically to the conditions p 4, V4 and T4 respectively corresponding to point 4. Finally, the air rejects the heat to the cold body at constant volume till the point 1 where it returns to its original state.
P-V and T-S diagram for Diesel Cycle
Consider 1 kg of air. T 2) Heat supplied at constant pressure = C p p (T 3 – T
Heat rejected at constant volume = C v(T 4 – T T 1) Work done = Heat supplied – supplied – heat heat rejected = C p(T 3 – T T 2) – C C v(T 4 – T T 1)
Efficiency = Work done / Heat supplied = Cp(T3 – T – T2) – C – Cv (T4 – T – T1)/ Cp(T3 – T – T2) – T1)/ γ (T3 – T – T2)} = 11- { (T (T4 – T Where here Cp/ Cv= γ
It may be observed that for efficiency of diesel cycle is different from that of the Otto cycle only in bracketed f actor. This factor is always greater than unity, because ρ > 1. Hence for a given compression ratio, the Otto cycle is more efficient. The net work for diesel cycle can be expressed in terms of pv as follows :
2) A six cylinder four stroke petrol engine has a swept volume of 300cubic cm per cylinder, a compression ratio of 10 and operates at a speed of 35000rpm. If the engine is required to develop an output of 73.5kw at this speed, calculated the cycle efficiency, the necessary Rate of heat addition,the mean effective pressure , maximum temperature of the cycle and efficiency ratio. The pressure and temperature before isentropic compression are 1.0bar = 0.72 and γ = 1.4 (nov – 2010) and 15° C respectively, take
Given data : r 10 vs 300cm3 300 10 6 m3 n 35000 rpm w 73.5Kw p1 1.0bar T1 150 C 288 K cv 0.72 γ 1.4
solution : compression ratio, r
10
vs vc vc
300 10 6 vc vc
300 10
6
vc
vc 33.33 10
1 6
air standard efficiency,η 1
1
1 γ
r 1 1
101.41 0.602 60.2% heat supplied, Qs Qs
work output η
73.5 0.602
122.09Kw
work output, w pm vs
N
60
Z
73.5 103 pm 300 10 6
35000
6
60
pm 70000 KN / m 0.7bar 2
V 1 T1 V 2
T2
γ 1
γ 1
V T2 T1 1 V 2
288 101.41 723.4 K
3) A four stroke petrol , four cylinder petrol engine of 250 mm bore and 375 mm stroke works on Otto cycle. The clearance volume is 0.01052 .The initial pressure and temperature are 1bar and 47° C. if the maximum pressure is limited to 25 bar, find the i. ii.
the air standard efficiency of the cycle mean effective pressure. Take = 1.005 kJ/kg°k and γ = 1.4
Given data: cylinder diameter, d 250 mm 0.250m stroke length, l 375 mm 0.375m 3 clearancevolume, vc 0.01052m
p1 1bar 100 KN / m
2
T1 470 C 320 K
p3 25 bar 2500 KN / m
2
solution : strokevolume, vs
2
π d
4
l
π
4
vs 0.01839 m compression ratio , r
vc vs vc
2
0.250 0.375 3
0.01052 0.01839 0.01052
air standard efficiency ,η 1
1 γ 1
r η 53.78%
1
w.k .t v1 v2
2.74
2.74
1 1.4 1
6.888
(Nov- 2011)
consider process1 2 ( adiabatic compression);
v p2 p1 1 v2
γ
100 (2.74)1.4 410.0631KN / m2 pressureratio, k
p3 p2
2500 410.063
6.096
1.4 1 6.096 1 2.74 100 2.74 1.4 1 2.74 1 pm 993.43 Kpa
4) In an air standard dual cycle, the pressure and temperature at the beginning of compression are 1 bar and 57° C respectively. The heat supplied in the cycle is 1250 Kj.kg, two third of this being added at constant volume and rest at constant pressure. If the compression ratio is 16, determine the maximum pressure, temperature in the cycle, thermal efficiency and mean effective pressure. (Nov – 2011)
Given data : p1 1bar T1 47 0 C 320 k r 16 Qs 1250 KJ / kg Qs1 2 / 3Qs 833.33 KJ / Kg Qs2 1 / 3Qs 416.67 KJ / Kg solution : specific volume, v1
rT 1 p1
287 320 1 10
5
0.9184m3 / kg
v2 0.0574 m3 / kg
1 2 isentropic compression processes p2 (r )γ p1
161.4 1 48.5bar T (r )
γ 1
T 1
161.41 320 970.06 K 2 3 constant volumeheat addition process
cv (T3 T 2 ) 0.718(T 3 970.06) 2130.69 K w.k .t
T 3 p2 T 2
p3
2130.69
48.5
970.06 106.53bar
3 4 constant pressureheat addition process Qs2 C (T4 T3 )
416.67 1.005(T 4 2130.69) T4 2545.29 K
v4
T 4 T 3
v3
2545.29 2130.69
0.0574
0.0686m3 / Kg expansion ratio, r e
v4 v1
0.0686 2130.69
0.0747
4 5 isentropic expansion process T re
γ 1
T 4
(0.0747)1.41 2545.29 901.71K heat rejected fromthecycle, Qr Cv (T5 T1 )
0.718(901.71 320) 417.67 KJ / Kg workdone W Qs Qr 1250 417.67 832.33 KJ / Kg cycleefficiency ,η
w Qs
832.33 1250
66.59%
meaneffective pressure , pm
w v v2
832.33 0.9184 0.0574
9.67bar
5)In the engine working on dual cycle , the temperature and pressure at the beginning of the cycle are 90° C and 1 bar respectively. The compression ratio is 9 . The maximum pressure is limited to 68 bar and heat supplied per Kg of air is 1750KJ. determine : i. ii. iii.
Pressure and temperature at all salient points Air standard efficiency Mean effective pressure. (may – 2012)
p1 1bar T1 90 C 0
p3 p4 68bar r 9 Qs 1750 KJ / Kg solution :
1 2 : isentropic comp. process p2 r p1 9.1.4 1 γ
21.67bar T2 r
γ 1
T 1 90.4 363
874 K 2 3 constant volume heat addition process p3 T p 2 68 874 21.67 2743K 3 4 : constant pressureheat addition process Qs cv (T3 T2 ) c p (T4 T3 )
T3
1750 0.718(2743 874) 1.005( T 4 2743) T4 3149 K v
RT 1 p1
v3 v2
287 363 1 105
v1 r
1.04181m3 / Kg
1.04181
0.11576m3 / Kg
9
T 4 3149 v3 0.11576 T 2743 3
v4
0.132894m3 / Kg cut off ratio, ρ
v4 v3
pressure ratio , K
0.13289 0.11576
p3 p2
efficiency of thecycle
1.148
68 21.67
3.138
γ r 1 (k 1) γ ( ρ 1) 1 3.138 1.148 1.4 1 1 1.41 9 (3.138 1) 3.138 1.4(1.148 1)
η
1
1
k ρ 1 γ
58.19% net work of the cycle, wnet η Qs
0.5819 1750 1018.33KJ / Kg mean effective pressure, pm
wnet v1 v 2
1018.33 1.0418 1 0.11576
10.98bar
6) a. Consider an air standard cycle in which the air enters the compression at 1 bar and 20 °C. The pressure of air leaving the compressor is 3.5 bar and the temperature at turbine inlet is 600° C .determine per Kg of air. (May – 2012) i. ii. iii. iv. v.
Efficiency of the cycle Heat supplied to air Work available at the shaft Heat rejected in the cooler and Temperature of air leaving the turbine
Given data: p1 1bar T1 20 C T 3 600 P2 3.5
solution : consider the process1 2 adiabatic compression
p 2 T1 p1
γ 1
T2
p2 p1
γ 1 γ
T2
3.5
γ
T 1
1.41
1 419 K
1.4
293
consider the process 3 4 adiabatic exp ansion
p 4 T3 p3
γ 1
T4
p4 p3
T4
1 3.5
γ
γ 1 γ
T 3
1.41 1.4
873 674.3 K
air standard efficiency,η 1
1 γ 1
r p
1
γ
1 1.4 1
0.30
3 1.4 η 30% heat supplied Qs c p (T3 T2 ) 1.005(873 419) 456.2 KJ / Kg heat rejected Qr c p (T4 T1 ) 1.005(610.3 293) 318.8 KJ / Kg compressor work, wc c p (T2 T1 ) 1.005(419 293) 126.63 KJ
similarly for exapander , We c p (T3 T4 ) 1.005(873 610) 264.31 KJ work output w We wc 264.31 126.63 137.68 temperature of leaving theturbine 674.3K
b. The efficiency of an Otto cycle is 60% and γ= 1.5. What is the compression ratio
Solution: Efficiency of Otto cycle, η = 60% Ratio of specific heats, γ = 1.5 Compression ratio, r = ? Efficiency of Otto cycle is given
by,
7) a. A spark ignition engine working on ideal Otto cycle has the compression ratio 6. The initial pressure and temperature of air are 1 bar and 37 ° C. the maximum pressure in the cycle is 30bar. For unit mass flow, calculated i. ii.
P , v and T at various salient points of the cycle and The ratio of heat supplied to the heat rejected. Assume γ = 1.4 and R = 8.314KJ/k mol K (Nov – 2012)
Given data : r 6 p1 1bar 100 KN / m
2
T1 37 0 C 37 273 310 k p3 30 bar solution : consider process1 2( adiabatic process) : γ
V 1 p1 V 2
p2
γ
V p2 1 p1 V 2 p2 61.4 100 1228.6 KN / m 2 V 1 T1 V 2
T2
γ 1
γ
V T2 1 T 1 V 2 61.41 310 T2 634.78 K consider process 2 3( constant volume process); p3 p2
T 3 T 2
T3
p3 p2
T 2
3000 100
634.78
T3 19043.4 K consider process 3 4( adiabaticprocess) : p4 p3
p4
v3
γ
v4 v3 v4
γ
1.4
1 2 p3 3000 244.18 KN / m 6
v 3 T3 v4
T 4
v T4 3 v4
γ 1
γ 1
0.4
1 T3 19043.4 9300 K 6 heat supplied : Qs mCv (T3 T2 ) 1 0.718 (19043.3634.78) Qs 13217.39 KJ / Kg heat rejected mCv (T4 T1 ) 1 0.718 (9300 310) 6454.82 KJ / Kg Qs Qr
13217.39 6454.82
2.048
8) An air standard dual cycle has a compression ratio of 18, and compression begins at 1 bar 40° C. The maximum pressure is 85 bar. The heat transferred to air at constant pressure is equal to that at constant volume. Estimate: i. ii. iii.
The pressure and temperatures at the cardinal points of the cycle. The cycle efficiency and Mean effective pressure of the cycle (Nov – 2012)
Given data : r 18 p1 1bar T1 400 C 313k p3 85bar Qs1 Qs2 C p 1.005 KJ / KgK Cv 0.718 KJ / KgK solution : spceific volume, V1
RT 1 P1
287 313 1 10
5
0.92701m3 / Kg
V2 0.049906 m3 / Kg
1 2 isentropiccompression process p2 r p1 18 γ
T2 r
γ 1
1.4
1 57.19bar
T1 181.4 1 313 994.61K
2 3 cons tan volumeheat addition process
p3 85 T 994 1478.26 K 2 p 48.5 2 Qs Cv ( T3 T2 )
T3 1
0.718(1478 994.61) 347.26 KJ / Kg 3 4 constant pressure heat addition Qs1 Qs2 C p (T4 T3 )
347.26 1.005(T 4 1478.26) T4 1832.79 K v4
T 4 T 3
v
1832.79 1478.26
expansion ratio, r e
0.04996 0.061919 m 3 / Kg
v4 v1
0.061919 0.89831
0.06892
4 5 isentropic exp ansion process p5 reγ p4
0.068921.4 85 2.009bar T5 reγ T4 0.06892
1.4 1
1
cut off ratio, ρ
v4 v3
1832.79 628.7 K
0.06892 0.049906
1.380
p3 85 1.486 p2 57.19
pressure ratio, K thecycleefficiency
1 η
1
γ 1 r (1.486 1) 1.4 1.486 (1.380 1) 1.486 1.6380.1.4 1
67.83%
net heat supplied tocycle Qs Qs1 Qs2 311.612 311.612 692.521KJ / Kg network donetocycle, W Qs η 692.52 0.6783 469.73 KJ / Kg the mean efficitive pr essure pm
W v1 v 2
413.45 0.8931 0.049906
5.57bar
9) In an oil engine working on dual cycle the heat supplied at constant pressure is twice that of heat supplied at constant volume. The compression and expansion ratios are 8 and 5.3. The pressure and temperature at the beginning of cycle are 0.93 bar and 27 °C. find the efficiency of the cycle and mean effective pressure. Take = 1.005 KJ/kgK and = 0.718 KJ/kgK. Given data:
(May – 2013)
P1 0.93bar T1 27 C 0
r 8 Qs2 2 Qs1 k
v5 v4
v1
v4
5.3
C p 1.005KJ / KgK Cv 0.718KJ / KgK solution : specificvolumes, v1 v2
RT 1 p
v1 r
287 300 0.93105
0.926 8
0.926m3 / Kg
0.11572m3 / Kg
v3 v2 0.11572m / Kg 3
1 2 compression process p2 r p1 8 0.93 17.093bar γ
γ 1
T2 r
1.4
T1 81.41 300 689K
10) Air standard cycle consists of the following process. o
(a).Isentropic compression from 15 c and 1 bar to 5 bar. (b).2500KJ/Kg of heat is added at constant volume. (c).Isentropic expansion to initial volume. (d).Heat rejection at constant volume. Calculate the ideal efficiency, mean effective pressure and peak pressure.
(Nov 2013)
