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Ideal Otto Cycle

An ideal Otto cycle with an air working fluid and a compression ratio of 10 is analyzed. The air is initially at 100 kPa and 25°C and undergoes: 1) Isentropic compression to 748.9 K 2) Constant volume heating of 1000 kJ/kg, raising the temperature to 2141.9 K 3) Isentropic expansion back to 852.7 K 4) Constant volume cooling, closing the cycle. The thermal efficiency is 60%, the mean effective pressure is 778.7 kPa, the highest temperature is 2141.9 K, and the temperature at the start of heat rejection is 852.7 K.

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Ralph Evidente
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692 views3 pages

Ideal Otto Cycle

An ideal Otto cycle with an air working fluid and a compression ratio of 10 is analyzed. The air is initially at 100 kPa and 25°C and undergoes: 1) Isentropic compression to 748.9 K 2) Constant volume heating of 1000 kJ/kg, raising the temperature to 2141.9 K 3) Isentropic expansion back to 852.7 K 4) Constant volume cooling, closing the cycle. The thermal efficiency is 60%, the mean effective pressure is 778.7 kPa, the highest temperature is 2141.9 K, and the temperature at the start of heat rejection is 852.7 K.

Uploaded by

Ralph Evidente
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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An ideal Otto cycle has a compression ratio of 10.

Air at 100 kPa and 25 oC


undergoes isentropic compression and 1000 kJ/kg of heat was transferred to air at
constant volume. Assume air behaves like an ideal gas and its properties remain
constant with CP = 1.005 kJ/kg-K and k = 1.4, calculate the (a) thermal efficiency,
, (b) mean effective pressure (kPa) , (c) highest temperature in the cycle and (d)
temperature at the beginning of heat rejection

Step 1 Step 2 (isentropic compression)


Step 2 Step 3 (addition of 1000 kJ/kg of heat at constant volume)
Step 3 Step 4 (isentropic expansion)
Step 4 Step 1 ( heat removal at constant volume)
(1) Air (at 100 kPa, 298.15 K) Compression (2)
s 1 = s2
k1
T1 V2
T2
= ( )
V1
V1 V max
r=10= =
V2 V min
V 1=10 V 2
1.41
298.15 V2
T2 = (
10 V 2
)

T2 = 748.9189 K
RT1 (CpCv) T 1
V1 = P1
= P1

Cp 1.005
*k = 1.4 = Cv = Cv

kJ
Cv = 0.71786 kg K

kJ
R = Cp Cv = 1.005 0.71786 = 0.28714 kg K

RT1 kJ
0.28714 (298.15 K ) m
3
V1 = P1
= kg K = 0.85610791 kg
(V max)
100 kPa
V 1=10 V 2
m3
V 2=0.085610791 (V min)
kg

(2)(3) addition of 1000 kJ/kg of heat at constant volume

q = CV(T3 - T2)
kJ kJ
1000 kg = 0.71786 kg K (T3 - 748.9189 K )
T3 = 2141.9482 K (highest temperature in the cycle)

th= 1 - r 1k
th = 1 - 1011.4
th = 0.60

W net
th = q

W net
0.60 = 1000 kJ
kg
kJ
W net =600
kg
W net
MEP (Mean Effective Pressure)= Vmax V min
kJ
600
kg
MEP = m3 m3
0.85610791 0.085610791
kg kg
MEP = 778.7180 kPa
k1
T4 V2
T3
= ( )
V1
1.41
T4 V2
2141.9482 K
= (
10 V 2
)

T4 = 852.7250 K (temperature at the beginning of heat rejection)

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