Problem 5 - Thermodynamics, the Otto cycle One mole of ideal diatomic gas undergoes a cyclic transformation in four steps, corresponding to the Otto cycle. The first step A → B is an isovolumetric transformation, with pressure increasing from PA = 0.83 atm to PA = 2PB at constant volume VA = 0.1 m³. The second step B C is an adiabatic expansion. The third step C D is an isovolumetric transformation, leading to a pressure PD < PA. The fourth step D → A is an adiabatic compression, closing the thermodynamic cycle. (a) Draw the cycle in the P-V diagram. Discuss where the temperature reaches its maximum value, Tí, and its minimum value, TL, along the Otto cycle. Find the values TA and TB. (b) Evaluate (symbolically) the entropy change during the first step, ASAB, and the third step, ASCD; show that, on the basis of entropy change considerations, necessarily TB/TA = Tc/TD. (c) Show that the efficiency of the Otto cycle is always inferior to the efficiency of the corresponding Carnot cycle operating between the temperatures Tн and TL (defined in part (a) above). Assume that the efficiency of this particular Otto cycle is eotto 0.5. Find the values of Tc, TD, and the efficiency of the corresponding Carnot cycle. (a) P↑ PB Pc+ PA- Pp BA A VA (b) AS Ag = da dQ T A C D VC T = PAVA 0.83x105x0./m² = 1x10³k nR 8.3 MAX TEMP State B; TH = T8 MIN TEMP→ state D; TL = TD PAVA = PBVB 2PB VA = PBVB TB = 2TA TB = 2x0.83×105 ⇒ ⇒ = 2×10³ TA Тв TA TB 8.3 >V ncvln TB TA ASCD = 1; da = ncv To = -nculn Te C T (c) The efficiency of the otto cycle: e otto The efficiency of the carnot to = ⇒ Since Scycle = DSAB + ASCD = 0, it must be TB 1- QL QH 1 - QDC QAB = 1- AEV (TC-TD) AC√(TB-TA) TH = 1 - TD TB cycle: ecarnot If, for this particular Otto cycle TC = = TB = TA then necessarily Tc TL , e carnot TH 11 = H ⇒ Cotto = 1 - TC = 0.5 TB 19/100 11 14/10 1/2 =2 Tell-ID = 1 - Te 11 - 1/2 = 1 - Tc 끝)-1- TB TB TB-TA Since Tc > To carnot > Cotto 1/2 and Tp = TA Tc = ± Tc = ± TB. Therefore, since T₁ = To and TH = TB 1- TL = 0.75 TH TB -+
Problem 5 - Thermodynamics, the Otto cycle One mole of ideal diatomic gas undergoes a cyclic transformation in four steps, corresponding to the Otto cycle. The first step A → B is an isovolumetric transformation, with pressure increasing from PA = 0.83 atm to PA = 2PB at constant volume VA = 0.1 m³. The second step B C is an adiabatic expansion. The third step C D is an isovolumetric transformation, leading to a pressure PD < PA. The fourth step D → A is an adiabatic compression, closing the thermodynamic cycle. (a) Draw the cycle in the P-V diagram. Discuss where the temperature reaches its maximum value, Tí, and its minimum value, TL, along the Otto cycle. Find the values TA and TB. (b) Evaluate (symbolically) the entropy change during the first step, ASAB, and the third step, ASCD; show that, on the basis of entropy change considerations, necessarily TB/TA = Tc/TD. (c) Show that the efficiency of the Otto cycle is always inferior to the efficiency of the corresponding Carnot cycle operating between the temperatures Tн and TL (defined in part (a) above). Assume that the efficiency of this particular Otto cycle is eotto 0.5. Find the values of Tc, TD, and the efficiency of the corresponding Carnot cycle. (a) P↑ PB Pc+ PA- Pp BA A VA (b) AS Ag = da dQ T A C D VC T = PAVA 0.83x105x0./m² = 1x10³k nR 8.3 MAX TEMP State B; TH = T8 MIN TEMP→ state D; TL = TD PAVA = PBVB 2PB VA = PBVB TB = 2TA TB = 2x0.83×105 ⇒ ⇒ = 2×10³ TA Тв TA TB 8.3 >V ncvln TB TA ASCD = 1; da = ncv To = -nculn Te C T (c) The efficiency of the otto cycle: e otto The efficiency of the carnot to = ⇒ Since Scycle = DSAB + ASCD = 0, it must be TB 1- QL QH 1 - QDC QAB = 1- AEV (TC-TD) AC√(TB-TA) TH = 1 - TD TB cycle: ecarnot If, for this particular Otto cycle TC = = TB = TA then necessarily Tc TL , e carnot TH 11 = H ⇒ Cotto = 1 - TC = 0.5 TB 19/100 11 14/10 1/2 =2 Tell-ID = 1 - Te 11 - 1/2 = 1 - Tc 끝)-1- TB TB TB-TA Since Tc > To carnot > Cotto 1/2 and Tp = TA Tc = ± Tc = ± TB. Therefore, since T₁ = To and TH = TB 1- TL = 0.75 TH TB -+
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Transcribed Image Text:Problem 5 - Thermodynamics, the Otto cycle
One mole of ideal diatomic gas undergoes a cyclic transformation in four steps, corresponding to the Otto
cycle. The first step A → B is an isovolumetric transformation, with pressure increasing from PA = 0.83
atm to PA = 2PB at constant volume VA = 0.1 m³. The second step B C is an adiabatic expansion. The
third step C D is an isovolumetric transformation, leading to a pressure PD < PA. The fourth step D →
A is an adiabatic compression, closing the thermodynamic cycle.
(a) Draw the cycle in the P-V diagram. Discuss where the temperature reaches its maximum value, Tí,
and its minimum value, TL, along the Otto cycle. Find the values TA and TB.
(b) Evaluate (symbolically) the entropy change during the first step, ASAB, and the third step, ASCD; show
that, on the basis of entropy change considerations, necessarily TB/TA = Tc/TD.
(c) Show that the efficiency of the Otto cycle is always inferior to the efficiency of the corresponding
Carnot cycle operating between the temperatures Tн and TL (defined in part (a) above). Assume that the
efficiency of this particular Otto cycle is eotto 0.5. Find the values of Tc, TD, and the efficiency of the
corresponding Carnot cycle.
(a) P↑
PB
Pc+
PA-
Pp
BA
A
VA
(b) AS Ag = da
dQ
T
A
C
D
VC
T = PAVA
0.83x105x0./m² = 1x10³k
nR
8.3
MAX TEMP State B; TH = T8
MIN TEMP→ state D; TL = TD
PAVA
=
PBVB 2PB VA = PBVB TB = 2TA TB = 2x0.83×105
⇒ ⇒
= 2×10³
TA
Тв
TA
TB
8.3
>V
ncvln TB
TA
ASCD = 1; da = ncv To = -nculn Te
C
T
(c) The efficiency of the otto cycle: e otto
The efficiency of the carnot
to
=
⇒ Since Scycle = DSAB + ASCD = 0, it must be TB
1- QL
QH
1 - QDC
QAB
= 1-
AEV (TC-TD)
AC√(TB-TA)
TH
= 1 - TD
TB
cycle: ecarnot
If, for this particular Otto cycle
TC = = TB = TA
then necessarily Tc
TL
, e carnot
TH
11
=
H
⇒ Cotto = 1 - TC = 0.5
TB
19/100
11
14/10
1/2
=2
Tell-ID
= 1 - Te 11 - 1/2
= 1 - Tc
끝)-1-
TB
TB
TB-TA
Since Tc > To carnot > Cotto 1/2
and Tp = TA Tc = ± Tc = ± TB. Therefore, since T₁ = To and TH = TB
1- TL = 0.75
TH
TB
-+
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