5. An engine working on Otto cycle has a volume of 0.45 m3, pressure 1 bar and
temperature 30°C at the b&ginning of compression stroke. At the end of compression
stroke, the pressure is 11 bar. 210 kJ of heat is added at constant volume. Determine:
(i) Pressures, temperatures and volumes at salient points in the cycle. (ii) Percentage
clearance. (iii) Efficiency. (iv) Net work per cycle. (v) Mean effective pressure. (vi)
Ideal power developed by the engine if the number of working cycles per minute is 210.
Assume the cycle is reversible.
Answers
Answer:
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Explanation:
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Answer:
Question:
Elements of Mechanical Engineering [EXP-715]
An engine working on Otto cycle has a volume of 0.45 m^{3}m
3
, pressure 1 bar and temperature 30°C at the beginning of compression stroke. At the end of compression stroke, the pressure is 11 bar. 210 kJ of heat is added at constant volume. Determine :
(i) Pressures, temperatures and volumes at salient points in the cycle.
(ii) Percentage clearance.
(iii) Efficiency.
(iv) Mean effective pressure.
(v) Ideal power developed by the engine if the number of working cycles per minute is 210.
Assume the cycle is reversible.
Step-by-Step
Solution
Report Solution
Refer to Fig. 4.12.
Volume, V_{1}=0.45 m ^{3}V
1
=0.45m
3
Initial pressure, p_{1}=1 \text { bar }p
1
=1 bar
Initial temperature, T_{1}=30+273=303 KT
1
=30+273=303K
Pressure at the end of compression stroke, p_{2}=11 \text { bar }p
2
=11 bar
Heat added at constant volume = 210 kJ
Number of working cycles/min. = 210.
(i) Pressures, temperatures and
volumes at salient points :
For adiabatic compression 1-2
p_{1} V_{1}^{\gamma}=p_{2} V_{2}^{\gamma}p
1
V
1
γ
=p
2
V
2
γ
or \frac{p_{2}}{p_{1}}=\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=(r)^{\gamma}
p
1
p
2
=(
V
2
V
1
)
γ
=(r)
γ
or r=\left(\frac{p_{2}}{p_{1}}\right)^{\frac{1}{\gamma}}=\left(\frac{11}{1}\right)^{\frac{1}{1.4}}=(11)^{0.714}=5.5r=(
p
1
p
2
)
γ
1
=(
1
11
)
1.4
1
=(11)
0.714
=5.5
Also \frac{T_{2}}{T_{1}}=\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}=(r)^{\gamma-1}=(5.5)^{1.4-1}=1.977 \simeq 1.98
T
1
T
2
=(
V
2
V
1
)
γ−1
=(r)
γ−1
=(5.5)
1.4−1
=1.977≃1.98
\therefore \quad T_{2}=T_{1} \times 1.98=303 \times 1.98= 6 0 0 K∴T
2
=T
1
×1.98=303×1.98=600K
Applying gas laws to points 1 and 2
\frac{p_{1} V_{1}}{T_{1}}=\frac{p_{2} V_{2}}{T_{2}}
T
1
p
1
V
1
=
T
2
p
2
V
2
\therefore \quad V_{2}=\frac{T_{2}}{T_{1}} \times \frac{p_{1}}{p_{2}} \times V_{1}=\frac{600 \times 1 \times 0.45}{303 \times 11}= 0 . 0 8 1 ~ m ^{ 3 }∴V
2
=
T
1
T
2
×
p
2
p
1
×V
1
=
303×11
600×1×0.45
=0.081 m
3
The heat supplied during the process 2-3 is given by :
Q_{s}=m c_{v}\left(T_{3}-T_{2}\right)Q
s
=mc
v
(T
3
−T
2
)
\text { where } m=\frac{p_{1} V_{1}}{R T_{1}}=\frac{1 \times 10^{5} \times 0.45}{287 \times 303}=0.517 kg where m=
RT
1
p
1
V
1
=
287×303
1×10
5
×0.45
=0.517kg
\therefore \quad 210=0.517 \times 0.71\left(T_{3}-600\right)∴210=0.517×0.71(T
3
−600)
or T_{3}=\frac{210}{0.517 \times 0.71}+600=1172 KT
3
=
0.517×0.71
210
+600=1172K
For the constant volume process 2-3
\frac{p_{3}}{T_{3}}=\frac{p_{2}}{T_{2}}
T
3
p
3
=
T
2
p
2
\therefore \quad p_{3}=\frac{T_{3}}{T_{2}} \times p_{2}=\frac{1172}{600} \times 11= 2 1 . 4 8 \text { bar. }∴p
3
=
T
2
T
3
×p
2
=
600
1172
×11=21.48 bar.
V_{3}=V_{2}= 0 . 0 8 1 m ^{ 3 }V
3
=V
2
=0.081m
3
For the adiabatic (or isentropic) process 3-4
p_{3} V_{3}^{\gamma}=p_{4} V_{4}^{\gamma}p
3
V
3
γ
=p
4
V
4
γ
p_{4}=p_{3} \times\left(\frac{V_{3}}{V_{4}}\right)^{\gamma}=p_{3} \times\left(\frac{1}{r}\right)^{\gamma}p
4
=p
3
×(
V
4
V
3
)
γ
=p
3
×(
r
1
)
γ
=21.48 \times\left(\frac{1}{5.5}\right)^{1.4}=1.97 \text { bar. }=21.48×(
5.5
1
)
1.4
=1.97 bar.
Also \frac{T_{4}}{T_{3}}=\left(\frac{V_{3}}{V_{4}}\right)^{\gamma-1}=\left(\frac{1}{r}\right)^{\gamma-1}=\left(\frac{1}{5.5}\right)^{1.4-1}=0.505
T
3
T
4
=(
V
4
V
3
)
γ−1
=(
r
1
)
γ−1
=(
5.5
1
)
1.4−1
=0.505
\therefore \quad T_{4}=0.505, T_{3}=0.505 \times 1172= 5 9 1 . 8 K∴T
4
=0.505,T
3
=0.505×1172=591.8K.
V_{4}=V_{1}= 0 . 4 5 ~ m ^{3}V
4
=V
1
=0.45 m
3
.
(ii) Percentage clearance :
Percentage clearance
=\frac{V_{c}}{V_{s}}=\frac{V_{2}}{V_{1}-V_{2}} \times 100=\frac{0.081}{0.45-0.081} \times 100=
V
s
V
c
=
V
1
−V
2
V
2
×100=
0.45−0.081
0.081
×100
= 21.95%.
(iii) Efficiency :
The heat rejected per cycle is given by
Q_{r}=m c_{v}\left(T_{4}-T_{1}\right)Q
r
=mc
v
(T
4
−T
1
)
= 0.517 × 0.71(591.8 – 303) = 106 kJ
The air-standard efficiency of the cycle is given by
\eta_{\text {otto }}=\frac{Q_{s}-Q_{r}}{Q_{s}}=\frac{210-106}{210}=0.495 \text { or } 49.5 \%η
otto
=
Q
s
Q
s
−Q
r
=
210
210−106
=0.495 or 49.5%
\left[\begin{array}{l}\text { Alternatively: } \\\eta_{\text{otto }}=1-\frac{1}{(r)^{\gamma-1}}=1-\frac{1}{(5.5)^{1.4-1}}= 0 . 4 9 5 \text { or 49.5\% }\end{array}\right][
Alternatively:
η
otto
=1−
(r)
γ−1
1
=1−
(5.5)
1.4−1
1
=0.495 or 49.5%
]
(iv) Mean effective pressure, p _{ m }p
m
:
The mean effective pressure is given by
p_{m}=\frac{W \text { (work done) }}{V_{s}(\text { swept volume })}=\frac{Q_{s}-Q_{r}}{\left(V_{1}-V_{2}\right)}p
m
=
V
s
( swept volume )
W (work done)
=
(V
1
−V
2
)
Q
s
−Q
r
=\frac{(210-106) \times 10^{3}}{(0.45-0.081) \times 10^{5}}= 2 . 8 1 8 \text { bar. }=
(0.45−0.081)×10
5
(210−106)×10
3
=2.818 bar.
(v) Power developed, P :
Power developed, P = work done per second
= work done per cycle × number of cycles per second
= (210 – 106) × (210/60) = 364 kW.
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