## Cycles
### Carnot Efficiency
$\eta_{\text{ Carnot}} = \frac{T_{H}-T_{L}}{T_{H}} = 1 - \frac{T_{L}}{T_{H}}$
![[image-34.png|500]]
### Rankine Cycle
| | | Constant | Effect |
| --- | --------- | -------- | ------ |
| 1 | Pump | S | P↑ |
| 2 | Boiler | P | |
| 3 | Turbine | S | |
| 4 | Condenser | P | |
![[927RCFig1.gif]]
![[Thermodynamic-Cycles-PV-and-TS-diagram-2.png|500]]
### Vapor Compression Cycle (Refrigeration)
![[ref-cycle-diagram-510 x 510.webp|500]]
![[image-49-1536x864.png|500]]
### Brayton Cycle
![[Brayton-cycle-example-calculation.png|500]]
![[Thermodynamic-Cycles-03-min.png|500]]
### Otto Cycle (Spark Ignition)
![[38_sys.png|500]]
![[Thermodynamic-Cycles-02-min.png|500]]
### Diesel Cycle (Compression Ignition)
![[internal-combustion-engine-four-stroke-cycle-in-a-typical-diesel-engine-HRKT5X.jpg|500]]
![[Thermodynamic-Cycles-01-min.png|500]]
## Shaft Work
### Shaft Work Machines
| Device | Function | Ideal Model | Efficiency |
| -------------- | ------------------------------------------ | ------------------------------------------------------------------------------------------------------------------------------------ | ----------------------------------------------------------------------------------------------------- |
| **Turbine** | P↓, T↓, h↓: $W_\text{out}
lt;br>Extract work | Steady, Adiabatic, reversible<br>$\dot{W}_{\text{rev}} = \dot{m}(h_{\text{in}} - h_{\text{out,rev}})lt;br>$s _\text{in}=s_\text{out}$ | ${} \eta_t = \frac{\dot{W}_{\text{act}}}{\dot{W}_{\text{rev}}} = \frac{h_{1}-h_{2}}{h_{1}-h_{2r}} {}$ |
| **Compressor** | P↑, T↑, h↑: $W_\text{in}lt;br>Apply work | $\dot{W}_{\text{rev}} = \dot{m}(h_{\text{out,rev}} - h_{\text{in}})lt;br>$s _\text{in}=s_\text{out}$ | ${} \eta_c = \frac{\dot{W}_{\text{rev}}}{\dot{W}_{\text{act}}}=\frac{h_{2r}-h_{1}}{h_{2}-h_{1}}$ |
### Ideal Gas Shaft Work
1st Law:
$-\frac{\dot{W}}{\dot{m}}=h_\text{out}-h_\text{in} = c_{p}(T_{2}-T_{1})$
2nd Law:
$\frac{\dot{S}_\text{gen}}{\dot{m}}=s_\text{out}-s_\text{in}=c_{p}\ln \frac{v_{2}}{v_{1}}+c_{v}\ln \frac{P_{1}}{P_{2}}$
For reversible case, $s_{2}=s_{1}$
Then 2nd law becomes: $P_{1}v_{1}^{\gamma}=P_{2}v_{2}^{\gamma}$
$\gamma = \frac{c_{p}}{c_{v}}$
$Pv = RT$
Benchmarks:
$\eta^{+}=\frac{\dot{W}_\text{actual}}{\dot{W}_\text{rev}}=\frac{(h_{2}-h_{1})_\text{actual}}{(h_{2}-h_{1})_\text{rev}}$
$\eta^{-}=\frac{1}{\eta^{+}}$
$R = c_{p}-c_{v}$
$-\frac{\dot{W}}{\dot{m}}=\frac{c_{p}}{R}\left( P_{2}v_{2}-P_{1}V_{1}\right)=\frac{c_{p}}{c_{p}-c_{v}}(P_{2}v_{2}-P_{1}v_{1})=\frac{\gamma}{\gamma-1}(P_{2}v_{2}-P_{1}v_{1}) = \frac{\gamma}{\gamma-1}\left( P_{2}\left( \frac{P_{1}}{P_{2}} \right)^{1/\gamma}v_{1}-P_{1}v_{1} \right)$
$\boxed{-\frac{\dot{W}}{\dot{m}}=\frac{\gamma}{\gamma-1}P_{1}v_{1}\left( \left( \frac{P_{2}}{P_{1}} \right)^{(\gamma-1)/\gamma}-1 \right)}$
1. Using isentropic properties, solve for the reversible state ($h_{2R}$)
2. Calculate $\dot{m}$ using $\frac{\dot{W}}{\dot{m}}=\eta(h_{1}-h_{2R})$
3. Caculate $h_{2}$ using $\frac{\dot{W}}{\dot{m}}=h_{1}-h_{2}$
4. Determine final state using $h_{2}$ and $P_{2}$
5. Calculate entropy generation rate with $S_\text{gen}=\dot{m}(s_{2}-s_{1})$
### Incompressible Shaft Work
$h = u + Pv=cT+Pv$
$-\frac{\dot{W}}{\dot{m}}=h_\text{out}-h_\text{in}=c(T_\text{out}-T_\text{in})+v(P_\text{out}-P_\text{in})$
Reversible case ($s_\text{out}-s_\text{in}$)
$-\left( \frac{\dot{W}}{\dot{m}} \right)_\text{rev}=v(P_\text{out}-P_\text{in})$
$\dot{W}_\text{rev}=\dot{V}(P_\text{in}-P_\text{out})$
$\eta^{+}=\frac{\dot{W}_\text{actual}}{\dot{W}_\text{rev}}=\frac{c(T_\text{in}-T_\text{out})+v(P_\text{in}-P_\text{out})}{v(P_\text{in}-P_\text{out})}$
$\eta^{-}=\frac{1}{\eta^{+}}$
## Nozzles and Diffusers
### Nozzles and Diffusers
| Device | Function | Ideal Model | Efficiency |
| ------------ | --------------------------- | ------------------------------------------------------------------------------------------------------------------------ | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |
| **Diffuser** | P↑, v↓, h↑<br>KE → Enthalpy | $\left( h + \frac{v^{2}}{2} \right)_\text{in}=\left( h+\frac{v^{2}}{2} \right)_\text{out}lt;br>$s_\text{out}=s_\text{in}$ | ${} \eta _\text{d} = \frac{(h_\text{in}-h_\text{out})_\text{rev}}{(h_\text{in}-h_\text{out})_\text{act}} = \frac{(v^{2}_\text{in}-v^{2}_\text{out})_\text{rev}}{(v^{2}_\text{in}-v^{2}_\text{out})_\text{act}} {}$ |
| **Nozzle** | P↓, v↑, h↓<br>Enthalpy → KE | $\left( h + \frac{v^{2}}{2} \right)_\text{in}=\left( h+\frac{v^{2}}{2} \right)_\text{out}lt;br>$s_\text{out}=s_\text{in}$ | ${} \eta _\text{ n}= \frac{(h_\text{in}-h_\text{out})_\text{act}}{(h_\text{in}-h_\text{out})_\text{rev}} = \frac{(v^{2}_\text{in}-v^{2}_\text{out})_\text{act}}{(v^{2}_\text{in}-v^{2}_\text{out})_\text{rev}} {}$ |
| **Throttle** | P↓,T↓, V↑<br>Pressure drop | $h_{\text{in}} = h_{\text{out}}$ | N/A |
#### Typical Assumptions:
- Steady
- PE negligible
- Adiabatic
- No shaft work
#### 1st Law
$\cancel{ \frac{dE_{\text{cv}}}{dt} }=\cancel{ \dot{Q} }-\cancel{ \dot{W} }+\sum_{\text{in}}\dot{m}_{\text{in}}\left( h+\frac{v^{2}}{2}+\cancel{ gz } \right)_{\text{in}}-\sum_{\text{out}}\dot{m}_{\text{out}}\left( h+\frac{v^{2}}{2}+\cancel{ gz } \right)_{\text{out}}$
$\left( h+\frac{v^{2}}{2} \right)_\text{in}=\left( h+\frac{v^{2}}{2} \right)_\text{out}$
$h_\text{in}-h_\text{out}=\frac{v^{2}_\text{out}}{2}-\frac{v^{2}_\text{in}}{2}$
#### 2nd Law
$\cancel{ \frac{dS_{\text{cv}}}{dt} }=\cancel{ \frac{\dot{Q}}{T_{\text{s}}} }+\dot{S}_{\text{gen}}+\sum_{\text{in}} \dot{m}_{\text{in}}S_{\text{in}} - \sum_{\text{out}} \dot{m}_{\text{out}} S_{\text{out}}$
$s_\text{out}-s_\text{in}=\frac{\dot{S}_\text{gen}}{\dot{m}}$
**Reversible:** $s_\text{out}=s_\text{in}$
![[Pasted image 20250425092804.png|275]]
$h_{1}-h_{2} < h_{1} - h_{2R}$
- Nozzle is analogous to a positive shaft work machine
- KE increase is smaller for an irreversible nozzle
![[Pasted image 20250425092826.png|275]]
$h_{1}-h_{2} < h_{2R}-h_{1}$
- Diffuser is analogous to a negative shaft work machine
- Higher KE expenditure needed to get to the same pressure for an irreversible nozzle
#### Efficiency
For defining efficiency, we use the same inlet state and outlet pressure, but not necessarily inlet velocity
**Nozzle Efficiency**
$\eta _\text{ nozzle}= \frac{(h_\text{in}-h_\text{out})_\text{actual}}{(h_\text{in}-h_\text{out})_\text{rev}} = \frac{(v^{2}_\text{in}-v^{2}_\text{out})_\text{actual}}{(v^{2}_\text{in}-v^{2}_\text{out})_\text{rev}}$
**Diffuser Efficiency
$\eta _\text{ diff} = \frac{(h_\text{in}-h_\text{out})_\text{rev}}{(h_\text{in}-h_\text{out})_\text{actual}} = \frac{(v^{2}_\text{in}-v^{2}_\text{out})_\text{rev}}{(v^{2}_\text{in}-v^{2}_\text{out})_\text{actual}}$
**Diffuser Pressure Recovery Factor**
$C_{p} = \frac{P_\text{out}-P_\text{in}}{\frac{1}{2}\rho _\text{in}v^{2}_\text{in}}$