1. Isothermal heat transfer: hot reservoir to working fluid at $T_h$.
2. Adiabatic expansion of working fluid to $T_c$.
3. Isothermal heat transfer: working fluid to cold reservoir at $T_c$.
4. Adiabatic compression of working fluid to $T_h$.
## Carnot efficiency $\eta_c = 1-\frac{T_c}{T_h}$
# Otto Cycle
::: {.columns}
::: {.column}
* Gas power cycle (uses a gas as the working fluid)
* Spark ignition engines (gasoline)
* The *working fluid* is not fixed (intake, exhaust)
* **Four strokes**:
1. *Intake*
2. *Compression*
3. *Expansion*
4. *Exhaust*
:::
::: {.column}
:::
:::
## Otto Cycle: steps
::: {.columns}
::: {.column}
* Steps
- **0-1**: Intake, const P
- **1-2**: $W_{in}$: Adiabatic compression
- **2-3**: $Q_{in}$: Heat addition, combustion
- fast: constant V
- **3-4**: $W_{out}$: Adiabatic expansion
- **4-1**: $Q_{out}$: Heat rejection, valve open
- fast: constant V
- **1-0**: Exhaust
* How do these correspond to the 4 strokes?
* Note, the area under the PV diagram is the work; the area under the TS diagram is the heat.
:::
::: {.column}
:::
:::
## Otto cycle: net work
::: {.columns}
::: {.column}
Net work is area under curve 3-4 minus area under curve 1-2
***Question: how to best compute the net work?***
::: fragment
Write the net work in terms of the internal energies at points 1, 2, 3, 4.
::: incremental
* The compression and expansion steps are adiabatic.
- $W_{in} = u_2-u_1$
- $W_{out} = u_3-u_4$
- $W_{net} = W_{out}-W_{in} = (u_3-u_4) - (u_2-u_1)$
- there is a simplification here for combustion...
:::
:::
:::
::: {.column}
:::
:::
## Adiabatic compression/expansion
* Ideal gas compression and expansion [relations](https://en.wikipedia.org/wiki/Adiabatic_process)
* Use $γ = c_p/c_v$
* See [this link](https://en.wikipedia.org/wiki/Heat_capacity_ratio) for a list of values.
* $γ = 1.4$ for air
$$PV^γ = \mbox{const.}$$
$$TV^{γ-1} = \mbox{const.}$$
$$TP^{(1-γ)/γ} = \mbox{const.}$$
## Otto cycle: [efficiency](https://en.wikipedia.org/wiki/Thermal_efficiency)
* Assume air as a working fluid
* Assume constant heat capacity
$$η = \frac{W_{net}}{Q_{in}}$$
$$W_{net} = (u_3-u_4) - (u_2-u_1)$$
$$W_{net} = c_v(T_3-T_4) - c_v(T_2-T_1)$$
$$Q_{in} = u_3 - u_2 = c_v(T_3-T_2)$$
$$η = \frac{T_3 - T_4 - T_2 + T_1}{T_3-T_2} = 1 - \frac{T_4 - T_1}{T_3 - T_2}$$
Factor out $T_1$ from the numerator and $T_2$ from the denominator:
$$η = 1 - \left(\frac{T_1}{T_2}\right)\left(\frac{T_4/T_1 - 1}{T_3/T_2 - 1}\right)$$
## Otto cycle: efficiency
$$η = 1 - \left(\frac{T_1}{T_2}\right)\left(\frac{T_4/T_1 - 1}{T_3/T_2 - 1}\right)$$
Now, the compression and expansion steps are adiabatic, so $TV^{γ-1}=\mbox{const}$. Hence,
::: {.columns}
::: {.column}
$$\frac{T_4}{T_3} = \left(\frac{V_3}{V_4}\right)^{γ-1}$$
But $V_4 = V_1$ and $V_3 = V_2$, so
$$\frac{T_4}{T_3} = \left(\frac{V_3}{V_4}\right)^{γ-1} = \left(\frac{V_2}{V_1}\right)^{γ-1} = \frac{T_1}{T_2}$$
Hence, $T_4/T_1 = T_3/T_2$, giving
$$η = 1 - \frac{T_1}{T_2} = 1 - \left(\frac{V_2}{V_1}\right)^{γ - 1}$$
$$η = 1 - \frac{1}{r^{γ-1}}$$
::: ::: {.column}
*How does this efficiency compare to Carnot?*
::: ::: # Diesel cycle
## Diesel cycle: thermodynamics
Key difference with the Otto cycle is the constant pressure heat addition rather than constant volume.
## Diesel cycle: efficiency
Question: do you expect the efficiency of the Diesel cycle to be higher or lower than for the Otto cycle?
::: fragment
$$η = 1 - \frac{1}{r^{γ-1}}\left(\frac{α^γ-1}{γ(α-1)}\right),$$
where $r=V_1/V_2$ is the compression ratio and $α=V_3/V_2=T_3/T_2$ is the cut-off ratio. $\alpha$ can be written in terms of the high and low temperatures and the compression ratio: $\alpha=T_3/(T_1r^{\gamma-1})$.
:::
# Brayton cycle: gas turbines
## Brayton cycle: thermodynamics
* Gas turbine combustors
* Like the Otto cycle, but heat transfer steps are constant $P$ instead of constant $V.$
* Same efficiency as the Otto cycle, but often written in terms of the pressure ratio.
# Conversion efficiencies