Cycle Wars: The Rise of Otto Cycles
DOFPro Team
Carnot (IE, IT, IE, IT) Review
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\[\eta\]
\[r\]
\[\eta\]
\[r_P\]
Otto (IE, IC, IE, IC)
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\[\eta=\frac{|Q_\mathrm{hot}|-|Q_\mathrm{cold}|}{|Q_\mathrm{hot}|}=1-\frac{|Q_\mathrm{cold}|}{|Q_\mathrm{hot}|}\]
\[Q_\mathrm{hot}=C_v\left(T_a-T_d\right)\]
\[Q_\mathrm{cold}=C_v\left(T_c-T_b\right)\]
\[\eta=1-\frac{T_b-T_c}{T_a-T_d}=1-\frac{T_c}{T_d}=1-r^{(1-\gamma)}\]
\[W_{\rm{net}}=-\left(Q_\mathrm{hot}+Q_\mathrm{cold}\right)=C_v\left[\left(T_b+T_d\right)-\left(T_a+T_c\right)\right]\]
Otto (IE, IC, IE, IC)(cont.)
\[\eta\]
\[r\]
\[\eta = 1 - \frac{T_\mathrm{c}}{T_\mathrm{d}} = 1 - r^{(1-\gamma)} \]
Diesel (IE, IB, IE, IC)
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\[\eta=1+\frac{Q_\mathrm{cold}}{Q_\mathrm{hot}}=1+\frac{C_v\left(T_c-T_b\right)}{C_p\left(T_a-T_d\right)}\]
\[=1+\frac{1}{\gamma}\frac{\left(T_b-T_c\right)}{\left(T_a-T_d\right)}\]
\[Q_\mathrm{hot}=Q_2 = C_p\left(T_a-T_d\right)\]
\[Q_\mathrm{cold}=Q_4 = C_p\left(T_c-T_b\right)\]
\[r_e \equiv \frac{V_b}{V_a}\]
\[r \equiv \frac{V_c}{V_d}\]
\[W_{\rm{net}}=-\left(Q_\mathrm{hot}+Q_\mathrm{cold}\right)=C_p\left[\left(T_d-T_a\right)+\left(T_b-T_c\right)\right]\]
Diesel (IE, IB, IE, IC)(cont.)
\[\eta\]
\[r\]
\[\eta = 1 - \frac{1}{\gamma}\left[\frac{\left(\frac{1}{r_e}\right)^{\gamma-1}-\left(\frac{r_e}{r}\right)\left(\frac{1}{r}\right)^{\gamma-1}}{1-\frac{r_e}{r}}\right]\]
\[= 1 - \frac{1}{\gamma}\left[\frac{\left(\frac{1}{r_e}\right)^{\gamma-1}-\left(\frac{1}{r}\right)^\gamma}{\frac{1}{r_e}-\frac{1}{r}}\right]\]
\[r \equiv \frac{V_c}{V_d}\]
\[r_e \equiv \frac{V_b}{V_a}\]
Graph is for \(r_e = 0.5 r\)
Brayton (IE, IB, IE, IB)
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\[\eta=1-\frac{|Q_\mathrm{cold}|}{|Q_\mathrm{hot}|}=1+\frac{\left(T_d-T_a\right)}{\left(T_c-T_b\right)}\]
\[=1-\frac{T_a}{T_b}=1-r_P^{\frac{1-\gamma}{\gamma}}\]
\[Q_\mathrm{hot}=Q_2 = C_p\left(T_c-T_b\right)\]
\[Q_\mathrm{cold}=Q_4 = C_p\left(T_d-T_a\right)\]
\[W_{\rm{net}}=-\left(Q_\mathrm{hot}+Q_\mathrm{cold}\right)=C_p\left[\left(T_b+T_d\right)-\left(T_a+T_c\right)\right]\]
Brayton (IE, IB, IE, IB)(cont.)
\[\eta\]
\[r\]
\[\eta = 1 - r_P^{\left(\frac{1-\gamma}{\gamma}\right)}\]
Brayton With Regenerator (IE, IB, IE, IB)
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\[W_{s_\mathrm{in}}\]
\[W_{s_\mathrm{out}}\]
\[W_{s_\mathrm{in}}\]
\[W_{s_\mathrm{out}}\]
\[Q_\mathrm{hot}=C_p\left(T_c-T_x\right)\]
\[W_{\mathrm{net}}=W_{\mathrm{turb}}+W_{\mathrm{comp}}\]
\[-W_{\mathrm{turb}}=- \Delta H = C_p\left(T_c-T_d\right)\]
With an ideal regenerator, \(T_d = T_x\).
\[Q_\mathrm{hot}=-W_{\mathrm{turb}}\]
\[\eta=\frac{|W_{\mathrm{turb}}+W_{\mathrm{comp}}|}{Q_\mathrm{hot}}=1-\frac{|W_{\mathrm{comp}}|}{|W_{\mathrm{turb}}|}\]
\[\eta=1+\frac{T_b-T_a}{T_c-T_d}=1-\frac{T_a}{T_c}\left(\frac{P_b}{P_a}\right)^{\frac{\gamma-1}{\gamma}}\]
\(\eta\)
\(r_p\)
\(T_a/T_c = 1/3\)
\(T_a/T_c = 1/4\)
\(T_a/T_c = 1/5\)
Turbojet (IE, IB, IE, IE IB)
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Raminagrobis, CC BY 4.0, via Wikimedia Commons
In ideal or actual cycle
\(\ \ \ \ |W_1|=|W_3|\) Just enough to drive
\(\ \ \ \ W_1=C_p\left(T_b-T_a\right)\)
\(\ \ \ \ Q_2=C_p\left(T_c-T_b\right)\)
\(\ \ \ \ W_3=C_p\left(T_d-T_c\right)\)
Turbojet (IE, IB, IE, IE IB)(cont.)
Raminagrobis, CC BY 4.0, via Wikimedia Commons
\(\ \ \implies C_p\left(T_b-T_a\right)=C_p\left(T_c-T_d\right)\)
\(\ \ \ \ T_d=T_c+\left(T_a-T_b\right)\)
In the nozzle (Step 4)
\(\ \ \ \ \Delta \hat{H} + \frac{\Delta (u^2)}{2}=0\)
\(\ \ \ \ d\hat{H}+udu=0\)
\(\ \ \ \ d\hat{H} = Td\hat{S} +\hat{V}dP\)
For isentropic \((\Delta S = 0)\)
\(\ \ \ \ \hat{V}dP+udu=0\)
\[\ \ \ \ \int_{u_d}^{u_e} udu=- \int_{\hat{V}_d}^{\hat{V}_e} \hat{V}dP\]
\[\text{ also }P\hat{V}^{\gamma}=\text{const.}\]
\[u_e^2-u_d^2=\frac{2\gamma P_d \hat{V}_d\ (\text{or }RT_d)}{\gamma-1}\left[1-\left(\frac{P_e}{P_d}\right)^{\frac{\gamma-1}{\gamma}}\right]\]
\[\frac{T_e}{T_d}=\left(\frac{\hat{V}_d}{\hat{V}_e}\right)^{\gamma-1}=\left(\frac{P_e}{P_d}\right)^{\frac{\gamma-1}{\gamma}}\]
Ericsson (IT, IB, IT, IB)
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\[Q_1=-W_1=RT_a \ln \frac{P_a}{P_b}\]
\[Q_3=-W_3=RT_c \ln \frac{P_c}{P_d}\]
\[Q_2=\Delta \hat{H}_2=C_p\left(T_c-T_b\right)\]
\[Q_4=\Delta \hat{H}_4=C_p\left(T_a-T_d\right)=-C_p\left(T_c-T_b\right)\]
\[W_2=- P_2 \Delta \hat{V}_2=-P_b\left(\hat{V}_c-\hat{V}_b\right)\]
\[W_4=-P_d\left(\hat{V}_a-\hat{V}_d\right)\]
Ericsson (IT, IB, IT, IB)(cont.)
After lots of algebra
\[W_{\mathrm{net}}=-R\left(T_3-T_1\right)\ln \frac{P_2}{P_4}\]
\[Q_\mathrm{hot}=C_p \left(T_1-T_3\right)+RT_3\ln \frac{P_2}{P_4}\]
\[\eta=\frac{R\left(T_3-T_1\right)\ln \frac{P_2}{P_4}}{C_p\left(T_1-T_3\right)+RT_3 \ln \frac{P_2}{P_4}}\]
Note that \(|Q_2|=|Q_4|\)
If we could use a perfect heat exchanger between Steps 2 and 4
\[\eta=\frac{R\left(T_3-T_1\right)\ln \frac{P_2}{P_4}}{RT_3 \ln \frac{P_2}{P_4}}=1-\frac{T_1}{T_3}\]
Stirling (IT, IC, IT, IC)
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Similar to the Ericsson cycle
\[W_{\mathrm{net}}=-R\left(T_3-T_1\right)\ln \frac{P_b}{P_a}\]
\[Q_\mathrm{hot}=C_v \left(T_3-T_1\right)+RT_3\ln \frac{P_b}{P_a}\]
\[Q_4=C_v \left(T_1-T_3\right)\]
For perfect HX between Steps 2 and 4
\[\eta=\frac{R\left(T_3-T_1\right)\ln \frac{P_b}{P_a}}{RT_3 \ln \frac{P_b}{P_a}}=1-\frac{T_1}{T_3}\]
Thanks for watching!
The previous video in the series is in the link in the upper left. The next video in the series is in the upper right. To learn more about Chemical and Thermal Processes, visit the website linked in the description.
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