Transcript Otto Cycle, ideal for spark ignition engines

MT 313 IC ENGINES
LECTURE NO: 04
(24 Feb, 2014)
Khurram
[email protected]
Yahoo Group Address: ICE14
Air Standard Cycle
• The air as the working fluid follows the perfect
gas law
𝑝𝑉 = 𝑚𝑅𝑇
• The working fluid is homogeneous throughout
and no chemical reaction takes place
• Specific heats of air do not vary with temperature
• The mass of air in the cycle remains fixed
• The exhaust process is replaced by an equivalent
heat rejection process
• The combustion process is replaced by an
equivalent heat addition process
• All processes are internally reversible
Air Standard Cycle
• Thermal efficiency
𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒
η𝑡ℎ =
𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑
𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑 − 𝐻𝑒𝑎𝑡 𝑅𝑒𝑗𝑒𝑐𝑡𝑒𝑑
=
𝐻𝑒𝑎𝑡 𝑆𝑢𝑝𝑝𝑙𝑖𝑒𝑑
𝑄𝑠 − 𝑄𝑟
=
𝑄𝑠
𝑞𝑠 − 𝑞𝑟
=
𝑞𝑠
• Thermal efficiency is also called air standard
efficiency ηa
Important Formulas
• Swept Volume
𝜋 2
𝑉𝑠 = 𝑑 𝐿
4
• Clearance Volume
𝑉 = 𝑉𝑐 + 𝑉𝑠
• Compression Ratio
𝑉𝑐 + 𝑉𝑠
𝑟=
𝑉𝑐
• Clearance ratio
𝑉𝑐
𝑐=
𝑉𝑠
Ideal cycles are simplified
Otto Cycle, ideal for spark ignition engines
OTTO CYCLE
• Process No 1-2 – Isentropic Expansion
1
pVγ = c
P1
2
P2
γ=
𝑐𝑝
𝑐𝑣
V1
V2
p– V diagram
•
OTTO
CYCLE
Process No 1-2 – Reversible Adiabatic or
Isentropic Expansion
1
T1
T2
2
T – S diagram
No Heat is added or rejected Q 1-2 = 0
S1, S2
OTTO CYCLE
• Process No 2-3 – Constant volume cooling process
2
P2
P3
3
V2
p– V diagram
Heat is rejected by air getting cooled from T2 to T3
•
OTTO
CYCLE
Process No 2-3 – Constant volume cooling process
T2
2
T3
S3 , S4
T – S diagram
S1, S2
Heat is rejected by air getting cooled from T2 to T3
OTTO CYCLE
• Process No 3-4 – Isentropic Compression
4
pVγ = c
P4
P3
γ=
𝑐𝑝
𝑐𝑣
3
V4
p– V diagram
No heat is added or rejected
V3
OTTO CYCLE
• Process No 3-4 – Reversible Adiabatic or Isentropic Expansion
T4
4
T3
3
S3 , S4
T – S diagram
No Heat is added or rejected Q 3-4 = 0
OTTO CYCLE
• Process No 4-1 – Constant volume heating process
1
P1
P4
4
V2
p– V diagram
Heat is absorbed by air getting heated from T4 to T1
•
OTTO
CYCLE
Process No 4-1 – Constant volume heating process
1
T1
T4
4
S3 , S4
S1, S2
T – S diagram
Heat is absorbed by air getting heated from T4 to T1
OTTO Cycle
• Process 1-2
No heat is added or rejected
𝑄1−2 = 0
• Process 2-3 Heat is rejected by air getting
cooled from temperature T2 to T3
𝑄1−2 = 𝑚𝑐𝑣 𝑇2 − 𝑇3
• Process 3-4 No heat is added or rejected
𝑄3−4 = 0
• Process 4-1 Heat is absorbed by air getting
heated from temperature T4 to T1
𝑄4−1 = 𝑚𝑐𝑣 𝑇1 − 𝑇4
OTTO Cycle
• Work Done = Heat absorbed – Heat rejected
• Work Done = 𝑄4−1 − 𝑄1−2
• Work Done = 𝑚𝑐𝑣 𝑇1 − 𝑇4 - 𝑚𝑐𝑣 𝑇2 − 𝑇3
•η=
•η=
𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒
𝐻𝑒𝑎𝑡 𝐴𝑏𝑠𝑜𝑟𝑏𝑒𝑑
𝑚𝑐𝑣 𝑇1 −𝑇4 − 𝑚𝑐𝑣 𝑇2 −𝑇3
𝑚𝑐𝑣 𝑇1 −𝑇4
OTTO Cycle
•η=
𝑚𝑐𝑣 𝑇1 −𝑇4 − 𝑚𝑐𝑣 𝑇2 −𝑇3
𝑚𝑐𝑣 𝑇1 −𝑇4
• η=1−
• η=1−
• η=1−
𝑚𝑐𝑣
𝑚𝑐𝑣
𝑇2 −𝑇3
𝑇1 −𝑇4
𝑇2 −𝑇3
𝑇1 −𝑇4
𝑇2
𝑇3
𝑇3
𝑇1
𝑇4
𝑇4
−1
−1
OTTO Cycle
• For reversible adiabatic expansion process
1-2
•
𝑇2
𝑇1
=
𝑣1 γ−1
[ ]
𝑣2
• where expansion ratio =
𝑣2
𝑣1
OTTO Cycle
• For reversible adiabatic expansion process
3-4
•
𝑇3
𝑇4
=
𝑣4 γ−1
[ ]
𝑣3
• where expansion ratio =
𝑣3
𝑣4
•
𝑣3
𝑣4
=
OTTO Cycle
𝑣2
𝑣1
• η=1−
• η=1 −
• η=1−
𝑇3
𝑇4
𝑇2
𝑇1
1
𝑟 γ−1
OTTO CYCLE
• Process No 1-2 – Isentropic Expansion
1
pVγ = c
P1
2
P2
γ=
𝑐𝑝
𝑐𝑣
V1
V2
p– V diagram
OTTO Cycle
2
𝑝𝑑𝑉
1
• 𝑊1−2 =
• 𝑝𝑉 γ = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
• 𝑊1−2 = 𝐶
• 𝑊1−2 =
• 𝑊1−2 =
2 𝑑𝑉
1 𝑉γ
𝑝1 𝑉1 −𝑝2 𝑉2
γ−1
𝑅 (𝑇1 −𝑇2 )
γ−1
Problem 1
• Calculate the air standard efficiency of a four
stock Otto cycle engine with the following
data
Piston diameter (bore)= 13.7 cm
Length of stock
= 13.0 cm
Clearance volume
= 14.6 %
• Diagram
Solution
• Swept Volume
• 𝑉𝑠 =
•
=
𝜋
4
𝜋
4
𝐷2𝐿
13.72 ∗ 13
•
= 1916 cm
• Clearance Volume
• 𝑉𝑐 =
•
𝜋
4
∗ 𝑉𝑠
= 297.7 cm3
Solution
• Compression ratio
• 𝑟𝑐 =
𝑉𝑠 +𝑉𝑐
𝑉𝑐
•
= 7.85
• Air Standard efficiency
• η=1−
•
1
𝑟
γ−1
= 56.2%
Problem 2
• In an Otto cycle the compression ratio is 6 .
The initial pressure and temperature of the air
are 1 bar and 100˚C. the maximum pressure in
the cycle is 35 bar. Calculate the parameter at
the salient points of the cycle. What is the ratio
of heat supplied to heat rejected
• How does air standard efficiency of the cycle
compares with that of a Carnot cycle working
within the same extreme temperature limits?
Explain the difference between the two values
Problem 2
• If the engine has a relative efficiency of 50 %
determine the fuel consumption per kWh.
Assume the fuel used has a calorific value of
42,000 kJ/kg
Problem 3
• An Otto cycle working on air has a
compression ratio of 6 and starting condition
are 40˚C and 1 bar. The peak pressure is 50
bar. Draw the cycle on p-v and T-S coordinates
if compression and expansion follow the law
pV1.25 = C. Calculate mean effective pressure
and heat added per kg of air.
Problem 4
• An Otto cycle has compression ratio of 8 and
initial conditions are 1 bar and 15˚C. Heat
added during constant volume process is 1045
kJ/kg. Find :
• Maximum cycle temperature
• Air standard efficiency
• Work done per kg of air
• Heat rejected
• Take cv = 0.7175 kJ/kg-K and γ = 1.4
Problem 5
• Find out the compression ratio in an Otto for
maximum work output
• An Otto cycle engine has the following data.
Calculate compression ratio, air standard efficiency
and specific fuel consumption.
Piston diameter
=
13.7
Length of stock
=
13 cm
Clearance volume
=
280 cm3
Relative efficiency
=
60 %
Lower calorific volume of petrol =
41900kJ/kg