# Renewable Energy 1
* [name=Tina] [time=Thu, Mar 4, 2025 1:30 PM]
>[Textbook](https://shahrsazionline.com/wp-content/uploads/2016/01/Renewable_Energy_Resourceswww.shahrsazionline.com_.pdf)
>[Renewable Energy 2](/JPp6kG0NSeSEPt2ir6izHA)
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:::spoiler **Outline**
[TOC]
:::

---
## 2023/02/23
### Renewable vs Non-renewable
- Renewable(再生) energy resources : an energy resources that is replaceednie
- EX :
- Renewable Example : Solar, Wind, etc.
- Non-Renewable Example : 火力, 核能, etc.

### Sustainability
- Sustainability(永續) energy resources : 廣泛是指不會改變地球的生態系，不會明顯的影響直到下一代...

### Global Energy Challenge
- Humanity's Top 3 Problems : Energy, Water, Food
- 80%+ of supply in 2000 from fossil-fuel
- Principle of Renewable Energy (2013 ARPA-E)
- How much is available?
- How should we use?
- What is the cost?

### Energay Consumption Tracking
- Growing population / developing / urbanization --> growing consumption
- Major fossil energy source (~85% of our supply) have problems
- What is energy
- York's Pee :+100:
- as ability to do work
- Energy Flow (of resource consumption)
- World : Renewables ~17% in 2010

### Practice working (unit & equivalent)
- 1 kW-hr(1度電:electric_plug: ) = 1 Watt = 1 joule/sec
- 懶得算(1kW-hr=3.6MJ):school:
- Energy Conversions :
- 1 Quad(quadrillion) = 10^15 BTU (British thermal unit) = 1.056 EJ(10^18J)

- Calories
- energy needed to increase the temperature of 1 gram of water by 1°C

### Energy Cost
- Transmission & Maintenance cost of renewable energy is larger than fossil energy
- coal : 118 yrs / oil : 46 yrs / gas : 59 yrs

### General conclusion
- Fossil energy : many & cheap
- Renewables wll NOT play a big role

### Taipower
- 發購電量 = 台電發電量 - 電廠內用電 + 購自民用發電量
- 108年，再生能源佔比6%

### Own energy usage
- How much extra energy? Estimate...
### Carbon Footprint
- Greenhouse Effect
- Lulu Carbon Vincent

---
## 2023/03/02
### Ex : Step 1, Examine the light bulb (60W) :bulb:
- ${E_{bulb_e}=4days*60W*24\frac{hr}{day}}=576Whr_{e}$
- ${E_{bulb}=5.76kWhr_{e}*\frac{3.6MJ}{kWhr_{e}}=20.736MJ}$
- In Taiwan : Energy source breakdown for electric power
In 2009 : Oil : 1.9%; Natural gas : 38.2%; Coal : 37.3%; Renewable : 6%; Nuclear : 13.4%
Avg. price$ ~${2.5217\frac{NTD}{kWhr}}$ for Residential & ~${2.4492\frac{NTD}{kWhr}}$ for Industrial
- Greenhouse gases emissions (www.moeaboe.gov.tw)
In 2009 : 電力排放係數 CO2 : 0.509${\frac{kg}{kWhr}}$; SO2 : ...; NOx : ...
=>${CO2_{bulb}=5.76kWhr_{e}*0.509\frac{kg}{kWhr}=2.9318kg=2.09*10^{-3}tonnes}$
- 
Notice : ~1/3 of the energy actually gets to you!
- :::success
60W light bulb => ${E_{bulb}=20.736MJ_e*\frac{3MJ}{MJ_e}=62.2MJ}$
:::
| Energy from | Avg(2019) | % |
|:-----------:|:---------:|:----:|
| Oil | 1.2MJ | ~2% |
| Coal | 23.2MJ | ~37% |
| Natial gas | 23.8MJ | ~38% |
| Nuclear | 8.3MJ | ~13% |
| Renewable | 3.7MJ | ~6% |
- BBL : Barrel of oil (原)石油桶 => ${1J=1.63*10^{-10}}$ BBL oil
- ${1.2MJ=(1.2*10^{6}J)*(1.63*10^{-10}\frac{BBL}{J})=1.956*10^{-4}}$ BBL
- ${1BBL=42gallons=158.9liter}$
- ${1.956*10^{-4}BBL*(42\frac{gallons}{BBL})=0.0082Gallons=0.0311LiterOil}$
- :::info
Table B.6 (P.741) : Crude Oil ${=\frac{35MJ}{Liter}}$ => ${\frac{1.2}{35}=0.0343}$ Liter Oil
:::
- Coal : ${23.2MJ}$
- :::warning
${(23.2*10^{6}J)*(\frac{3.41*10^{-8}kgCoal}{J})=0.79kg}$ Coal
:::
- Nat. gas : MCF = "M" for one thousand Cubic Feet :feet:
- ${1MCF=1000}$ cubic feet
- :::danger
${23.8MJ=(23.8*10^{6}J)*(9.48*10^{-10}\frac{MCF}{J})=2.26*10^{-2}MCF}$ Nat. gas
:::

### Ex : Step 2, The trip to the airport
- Assume your car gets ${20mpg(=8.5kpl)}$ and you need to go ${120km}$
- ${1mpg}$ (mile per gallon) ${=0.425kpl}$ (kilometer per liter) => ${\frac{120km}{8.5\frac{km}{liter}}=14.1}$ Liter petrol
- :::success
Table B.6 : Petrol : ${34\frac{MJ}{Liter}}$ => ${14.1Liter*\frac{34MJ}{Liter}=479.4MJ}$
> Carpool to airport ! :airplane:
:::
- From www.epa.gov : ${19.4\frac{lbsCO2}{Gallon}=2.327\frac{kgCO2}{Liter}}$ OR From 台灣環保署 ${=2.36\frac{kgCO2}{Liter}}$
=> ${14.1Liter*2.327=32.81kgCO2}$ (petrol) peko
- :::info
- For patrol : ${(\frac{328kg}{479.4MJ})_{CO_2}=0.0684\frac{kg}{MJ}CO_2}$
- For electricity : ${(\frac{0.509kg}{1kWhr})_{CO_2}=\frac{0.509kg}{3.6MJ}=0.1414\frac{kg}{MJ}CO_2}$
- Which is better for environment?
- All electric vehicle or High MPG vehicle?
:::

### Ex : Step 3, The airplane trip
- Assumptions :
- 95% capacity for thr flight, 2200km : what is carbon footprint? :feet}}$ km
- ${2200*128=281.6\frac{kgCO2}{airpassenger}}$
- Boeing 767-400ER : seats 245 people
- Consumes ~${0.0173\frac{gallons fuel}{passenger mile}}$ = ${\frac{0.0173*3.785 liter(fuel)}{1.6 passenger km}}$ = ${0.0409 \frac{\frac{liter(fuel)}{passenger km}}{95\% capacity}=0.0431\frac{liter(fuel)}{passengerkm}}$
- Your contribution for flight : ${2200km*0.0431\frac{liter}{passengerkm}=94.82}$ liter of aviation fuel
- Aviation fuel : Energy content (能量含量) of ~33 or 34 ${\frac{MJ}{liter}}$
${E_{airtrip}=94.82*34=3224MJ}$
- :::warning
- Total E = ${E_{bulb}+E_{airtrip}=62.2+479.4+3224=3.7656*10^{3}MJ}$
- Total ${CO_{2} = 2.93_{bulb}+32.8_{petrol}+281.6_{air} = 317.33 kgCO_2}$
:::
- Compensation? (${1ton=907kg}$)
- 1 tree removes ${0.2\frac{tons}{yr}}$ of CO2 ${=0.2*907\frac{kg}{yr}CO_2=181.4\frac{kg}{yr}CO_2}$
- How many trees need to plant to remove this CO2?
- :::danger
${\frac{317.33kg}{181.4kg/yr}=1.7493=2}$ trees-yr
:::

### Heat flow
* Convection(熱傳導)：與密度有關，熱能由高到低的溫度流動
* Radiation
* 原子的震動傳導熱能
* 密度愈低之材料熱很難被傳遞出去

---
## 2023/03/09
### Not in PPT calculation :D
- For general consideration, it's often useful to consider ++heat flow ${q}$ across unit area++ of surface. Then ++across a surface of area ${A}$++
- ${q=\frac{\Delta T}{r}}$, ${P=qA=\frac{\Delta T}{r/A}=\frac{\Delta T}{R}}$
- So, ${R=\frac{r}{A}}$ (unit of ${R}$ is ${KW^{-1}}$, ${K}$ is Kelvin)
${r=RA}$ (unit of ${r}$ is ${m^{2}KW^{-1}}$), ${r}$ is the "thermal resistivity" of unit area
- A common expression for the ++heat flow per unit area++ is
- ${q=h\Delta T}$, is the "heat transfer coefficient" (${Wm^{-2}K^{-1}}$)
- ${h=\frac{1}{r}=U}$, is sometimes called "thermal conductance", and denoted by symbol ${U}$.
- So, ${h=U=\frac{1}{r}}$ and ${UA=\frac{1}{R}}$
- Consider the heat flow ${P}$ by conduction through a slab of material, area ${A}$, thickness ${Δx}$ surface temoerature difference ${\Delta T}$,
- ${P=-kA\frac{\Delta T}{\Delta x}}$
${k}$ is the "thermal conductivity" (unit ${Wm^{-1}k^{-1}}$) ,
- By comparison of ${P=-kA\frac{\Delta T}{\Delta x}}$ and ${P_{ij}=\frac{\Delta T}{R_{ij}}}$,
the ++thermal resistance of conduction++ is ${R_{n}=\frac{\Delta x}{kA}}$,
the ++thermal resitivity of unit area++ is ${r_{n}=R_{n}A=\frac{\Delta x}{k}}$

### Ex 3.3 (RER 2nd)
- ++Case 1++ : Free convertion alone
- For the top : (using Table B.1 and eq. 3.25)
- ${A=\frac{g\beta x^{3}\Delta T}{kv}=(0.58*10^{8}m^{-3}K^{-1})(0.22m)^{3}(373K-293K)=4.9*10^{7}}$ (no unit)
=> ${k}$ is thermal ~~diffusivity~~ conductivity
=> ${T_{B_{(water)}}=100°C}$, ${T_{air}=20°C}$, ${\overline{T}=60°C}$ mean temp.
=> ${\frac{A}{x^{3}\Delta T}=\frac{g\beta}{kv}=0.5810^{8}m^{-3}K^{-1}}$ for ${\overline{T}=60°C}$ from *Table B.1*
- From *Table C.2* :
${N=0.14\mathcal{A}^{0.33}=(0.14)(4.9*10^{7})^{0.33}=48.4}$ for ${A>10^{5}}$
${P_{Top}=qA_{Top}}$ , ${A}$ is Area
${q=\frac{P}{A}=\frac{k(T_{S}-T_{F})}{\delta}=\frac{X}{\delta}\frac{k(T_{S}-T_{F})}{X}=N\frac{k(T_{S}-T_{F})}{X}}$ (eg. 3.17)
- ${T_{s}}$ : surface temp. & ${T_{F}}$ : temp across the fictitious boundary
${P_{Top}=qA_{Top}=\frac{A_{Top}kN\Delta T}{X}=\pi(\frac{0.22m}{2})^{2}(0.027Wm^{-1}K^{-1})(48.4)\frac{80K}{0.22m}=18W}$
${k}$ is thermal conductivity
> From *Table B.1* ${k=0.0288\approx0.027 Wm^{-1}K^{-1}}$
- ++For the slides++ : ${X=0.11m}$
${A_{Side}=\frac{g\beta x^{3}\Delta T}{kv}=6.1*10^{6}}$
- From *Table C2*, ${10^{4}<6.1*10^{6}<10^{9}}$, so, ${N=0.56\mathcal{A}^{0.25}=27.8}$
${P_{side}=\frac{A_{side}kN\Delta T}{X}=\frac{2\pi(\frac{0.22m}{2})(0.11m)(0.027)(27.8)(80K)}{0.11m}=41W}$
- So, Total power : ${P_{Tot}=P_{Top}+P_{side}=59W}$
${59*10^{-3}kW\cdot 1hr=0.059kwhr*3.6\frac{MJ}{kWh}=0.21\frac{MJ}{hr}}$
- ++Case 2++ : Forced plus free convection (if it's exposed to a breeze of ${3{m/s}}$)
- For the top : ${\mathcal{R}=\frac{ux}{v}}$ (eq. 3.21)
${\mathcal{R}}$ is Reynold's number, ${v}$ is kinematic viscosity ${=1.9*10^{-5}\frac{m^{2}}{s}}$ for ${60°C}$
${\mathcal{R}=\frac{(3\frac{m}{s})(0.22m)}{1.9*10^{-5}\frac{m^{2}}{s}}=3.5*10^{4}}$
- From *Tabl C3* : ${\mathcal{R}<5*10^{5}}$, ${N=0.664\mathcal{R}^{0.5}\mathcal{P}^{0.33}}$, ${\mathcal{P}}$ is Prandtl number
${P_{Top}=\frac{A_{Top}kN\Delta T}{x}=42W}$
- For the sides, ${\mathcal{R}=3.5*10^{4}}$
${N=0.26\mathcal{R}^{0.6}\mathcal{P}^{0.3}=124}$ => ${P_{sides}=\frac{A_{side}KN\Delta T}{x}=93W}$
${P_{forced}=P_{Top}+P_{sides}=135W}$
- The Total is ${P_{Tot}=P_{forced}+P_{free}=135W+59W=194W}$
${194*10^{-3}kW\cdot (1hr)=0.194kWh*3.6\frac{MJ}{kWh}=0.7\frac{MJ}{hr}}$
- **About ${3}$ times the energy per unit time of the sheltered cooking!!**

---
## 2023/03/16
### Solar Resource
> Ch2 (${3^{rd}) , Ch4 (2^{nd})}$

- Solar Radiation : ${-23.5 <= δ <= 23.5}$
- δ : declination 磁偏角
- ${δ_{0} = 23.5}$
- W ≡ hour angle : at P that earth has retated since solar noon
W = ${15^{o}(hr)(t_{solar} - 12 hr)= \frac{＃MinutesPastMidnight-720min}{4min/deg}}$
- ${t_{solar}}$ : actual solar time ≠${t_{zone}}$ (time on your clock) unless directly in center of time zone (on Local Standard Meridian)
- ${t_{solar} = LST + \frac{4min}{day}(LSM-Ψ)+ET}$
- ${LST}$：Local Stndard Tiime (subtract 1 hr for Daylight Standard Time)
- ${LSM}$ (標準子午線精度)：Longitude of Standard Meridian (子午線) in degress for the time zone.
- Earth : 24 time zones ${\frac{360°}{24}=15°}$
- Centered at LSM
- ${0, 15^{o}W, 30^{o}W,...,, 60^{o}W, 75^{o}W,...}$
- Same time zone +- 7.5 from LSM
- ${ET : Equation of Time }$
- true differences between solar time and measured solar time depending on seasonal cycles.
- Let ${n=}$ day of the year, ${n=1}$ on Jun.1
- ${ET~=9.87sin2B - 7.53cosB - 1.5sinB}$ (minutes)；${B=360^{o}(\frac{n-81}{365})}$

### Example
- If it is 9:00 am in Taipei on 7/21, what is ${t_{solar}}$?
- ${7/21, n=202}$, Taipei : ${Ψ=121.6°, \phi=25°N}$, ${ET=-6.05min}$
${LSM : 120°E (112.5°<=\Phi<=127.5°)}$, ${B=360°(\frac{202-81}{365})=119.3°}$
${t_{solar}=9AM+\frac{4min}{day} (120^{o}-121.6^{o})-6.05min = 8:47.55 AM}$
- ${w=15°(hr)(t_{solar}-12hr)=\frac{(＃MinutesPastMidnight-720min)}{4min/deg}=\frac{8* 60+47.55-720}{4}=-48.113°}$ from solar noon
- ${w}$ : negative in morning & positive in afternoon

### Example
- Calculate degree incidence (${\theta}$) of beam radiation at 9AM on 7/21 in Taipei with solar panel turned 20°E of S & tilted 40° to horizontal
- ${r_{s}=-20°,\beta=40°}$
From (4.8) : ${cos\theta=(A-B)sin\delta+[Csinw+(D+E)cosw]\cdot cos\delta}$
${A=sin\phi cos\beta=sin(25°)cos(40°)=0.3237}$
${B=cos\phi sin\beta cosr_{s}=cos(25°)sin(40°)cos(-20°)=0.5474}$
${C=sin\beta sinr_{s}=sin(40°)sin(-20°)=-0.2198}$
${D=cos\phi cos\beta=cos(25°)cos(40°)=0.6943}$
${E=sin\phi sin\beta cosr_{s}=sin(25°)sin(40°)cos(-20°)=0.2553}$
${\delta =23.5°\cdot sin\frac{360}{365}(202-81)=20.44°}$
- Matlab => ${cos\theta =0.6693}$, ${\theta =cos^{-1}(0.6693)=0.8375rad=47.985°}$

### Clear Sky
- ${G^{*}_{b}=Aexp(-km)}$
${m}$ : air mass ratio (空氣質量比) ${=\frac{1}{cos\theta_{z}}=\frac{1}{sin\alpha_{s}}}$
${A=1160+75sin[\frac{360}{365}(n-275)]\frac{w}{m^{2}}}$
${k=0.174+0.0035sin[\frac{360}{365}(n-100)]}$, ${n=}$ day number
- From (4.11), ${cos\theta_{z}=sin\phi sin\delta+cos\phi cos\delta cosw}$
${cos\theta_{z}=sin(25°)sin(2094°)+cos(25°)cos(20.44°)cos(-48.1°)=0.7146}$
${m=\frac{1}{0.7146}=1.3994}$
- ${\theta=47.985°(Taipei),\beta=40°}$
${G_{bc}=G^{*}_{b}cos\theta =(813.5)\cdot cos(47.985°)=544.5\frac{w}{m^{2}}}$

---
## 2023/03/23
### Example (Cont'd)
- ${G_{B}^{*}=813.5\frac{W}{m²}}$
1) ${G_{BC}=544.5\frac{W}{m²}}$
2) Diffuse radiation (approx.)
${G_{DC}=C\cdot G_{B}^{*}\cdot (\frac{1+cos\beta}{2})}$, ${C=0.095+0.04sin[\frac{360}{365}(n-100)]}$, ${n}$ = day
3) Reflected radiance (ignored except near snow / water often)
${G_{RC}=\rho\cdot G_{B}^{*}(sin\alpha_{s})\cdot(\frac{1-cos\beta}{2})}$
${\rho}$ given is the reflection coefficient (${<5}$% almost always)
4) Total irradiance on collector is always : ${G_{C}=G_{BC}+G_{DC}+G_{RC}}$
- Tracking Systems
1) 2-axis : always keeps ${\theta=0}$ by adjusting ${\beta=90-\alpha_{s}}$
--> follows ${r_{s}}$ (rotates ${E,W}$ by clock & formula we taught)
--> ${G_{BC}=G_{B}^{*}cos\theta=G_{B}^{*}}$
2) 1-axis : almost always tilt panel ${\beta=\phi}$-latitude
--> Rotate ${E-W}$ to track ${r_{s}}$
--> Then "${\beta}$ effective" = ${90° - \alpha_{s} + \delta}$
3) ${G_{BC}=G_{B}^{*}cos\delta}$
${G_{DC}=C\cdot G_{B}^{*}[\frac{1+cos\beta_{effective}}{2}]}$
${G_{RC}=\rho\cdot [\frac{1-cos\beta_{eff}}{2}]\cdot G_{B}^{*}\cdot (sin\alpha_{s}+C)}$

### Example


## 2023/03/30
### Example

- Caculate Resistor & Capacitance
1. Conduction Losses
- ${R_{b}=\frac{\Delta x}{kA}=\frac{0.1m}{(0.03\frac{W}{m°K})(1m^{2})}=3.33\frac{°K}{W}}$
${\Delta x=10}$ cm insulation, ${A=1m^{2}}$, ${k=0.03\frac{W}{m°K}}$
${P_{b}=\frac{T_{f}-T_{a}}{R_{b}}}$
- ${R_{g}=\frac{\Delta x}{kA}=\frac{5*10^{-3}m}{(1.1 W/m°K)(1m ^{2})}=0.00455m°K}$(very small~=0)
2. Radiative Losses
${R_{r,pg} =>2}$ flat
- Table C.5 -> eqn C.18, ${P_{12}=\frac{\sigma A_{1}(T^{4}_{1}-T^{4}_{2})}{\frac{1}{\varepsilon_{1}}+\frac{1}{\varepsilon_{2}}-1}}$ heat flow from body 1 -> 2
Let ${T_{1}=T_{f}}$, ${T_{2}=T_{g}}$, ${\varepsilon_{1}=\varepsilon_{2}=0.9}$ given, ${A_{1}=A_{2}=A=1m^{2}}$
${P_{fg}=P_{pg}=\frac{\sigma A(T^{4}_{f}-T^{4}_{g})}{\frac{1}{\varepsilon_{p}}+\frac{1}{\varepsilon_{g}}-1}}$, p-plate or "absorber" -> robber bag
${R_{r,pg}=\frac{T_f-T_g}{R_{r,fg}}=\frac{(\frac{1}{\varepsilon_{p}}+\frac{1}{\varepsilon_{g}}-1)(T_f-T_g)}{\sigma A(T_f^4+T_g^4)}=\frac{(\frac{1}{\varepsilon_{p}}+\frac{1}{\varepsilon_{g}}-1)}{\sigma A(T_f^2+T_g^2)(T_f+T_g)}}$
Guess ${T_{f}=70°C=343°K}$, ${T_{g}=\frac{1}{2}(T_{a}+T_{f})=45°=318°K}$
${R_{r,pg}=0.1491\frac{°K}{W}=0.15\frac{°K}{W}}$
- Table C.5 -> eqn C.17, ${P_{rad}=\varepsilon A\sigma(T_{g}^{4}-T_{sky}^{4})}$, ${A=1m^{2}}$
|||
|---|---|

${R_{r,gsky}=\frac{T_{g}-T_{sky}}{P_{rad}}=\frac{T_{g}-T_{a}}{P_{rad}}=\frac{T_{g}-T_{a}}{\varepsilon A\sigma(T_{g}^{4}-T_{sky}^{4})}=\frac{318-293}{(0.9)(1m^{2})(5.67*10^{8})(318^4-287^4)}=0.1424\frac{°K}{W}}$
${\sigma=}$ stefan-Boltzmann constant ${=5.67*10^{8}\frac{W}{m^{2}°K^{4}}}$
3. Convection Losses
- Mixed convection (Eqn C.15)
${h=a+bu}$, ${a=5.7\frac{W}{m^{2}°K}}$, ${b=3.8\frac{W}{m^{2}°K/(m/s)}}$, ${u=}$ wind speed
${R_{v,ga}=\frac{1}{h*A}=\frac{1}{[5.7+3.8(5)](1m^{2})}}$
- Free convection between glass & bag (example 3.2)
${\overset{°}{A}=\frac{g\beta}{Kv}\Delta TX^{3}}$, ${N=0.062\overset{°}{A}^{0.33}}$ if ${\overset{°}{A}>10^{5}}$, ${r_{r,pg}=\frac{X}{NK}}$ => ${R_{v,pg}=\frac{r_{v,pg}}{A_{eff}}}$
${\beta=\frac{1}{Avg.Temp}=\frac{1}{\frac{1}{2}(T_f+T_g)}=\frac{1}{\frac{1}{2}(343+318)°K}=\frac{1}{330°K}}$ where ${330°K}$ is the Avg.Temp of air between glass & bag => ${T_{avg}=330°K}$ => ${57°C}$, air
Table B.1 @57°C air
| Sign | Description | Value |
|:-----:|:--------------------:|:-----------------------:|
| ${v}$ | velocity | ${1.8*10^{-5}m²s^{-1}}$ |
| ${K}$ | Thermal Diffusivity | ${2.6*10^{-5}m²s^{-1}}$ |
| ${k}$ | Thermal Conductivity | ${0.028Wm^{-1}K^{-1}}$ |
| ${X}$ | | ${3cm=0.03m(given)}$ |
| ${g}$ | | ${9.81{m}s^{-2}}$ |
- ${N=0.062\overset{°}{A}^{0.33}=2.09}$ for ${\overset{°}{A}=4.288*10^4}$
- ${R_{v,pg}=\frac{X}{NkA}=\frac{0.03m}{(2.09)(0.028)(1m²)}=0.51\frac{°K}{W}}$
4. Thermal Capacitance
- ${C_{f}=mC=(97.7kg)(4190J(kg)^{-1}K^{-1})=4.09*10^{5}\frac{J}{kg}}$
${0.1m^{3}=100}$ liters --> ${m=\rho V=97.7kg}$
@70°C, ${\rho=0.977*10^{3}\frac{kg}{m²}}$ => ${4190J(kg)^{-1}K^{-1}}$ heat capacity
5. Current source
- ${\alpha\tau GA=(0.9)(0.9)(750\frac{W}{m^2})(1m^2)=608W}$
- 
${R_{pa}=R_{pg}+R_g+R_{ga}=0.15\frac{°K}{W}}$
1) Ignore ${R_{b}}$ because ${R_{b}}$ > ${R_{pa}}$
2) ${\frac{1}{R_L}=\frac{1}{R_{pa}}+\frac{1}{R_b}}$ => ${R_L=0.144\frac{°K}{W}=R_{pa}=0.15\frac{°K}{W}}$
- 
${P_c(t)=C_f\frac{dT_f}{dt}}$, ${P_L=\frac{T_f-T_a}{R_L}}$, ${P_s=P_c+P_L=C_f\frac{dT_f}{dt}+\frac{T_f-T_a}{R_L}}$
${\frac{dT_f}{dt}=(-\frac{1}{R_LC_f})T_f+(\frac{P_s}{C_f}+\frac{T_a}{R_LC_f})}$, ${T_f(t)=(P_sP_L+T_a)[1-e^{-\frac{t}{R_LC_f}}]+T_f(0)e^{-\frac{t}{R_LC_f}}}$
${\lambda=R_LC_f=(0.15)(4.09*10^5)=61400sec=17hrs}$
=> ${T_{f}(t)={[608(0.15)+293]}(1-e^{-\frac{t}{61400}})+T_f(0)e^{-\frac{t}{61400}}°K}$
${=(91.2+293)(1-exp\frac{-t}{17hrs})+T_f(0)exp\frac{-t}{17hrs}}$
${T_f(t=0)=Tf(0)given}$, ${T_f(t\to\infty)=91.2+293=384.2°K=111°C}$

---
## :memo: ( ö)/" Hey york

晚上要不要吃 ______ :chicken:
好像不錯欸
我的貢獻就是幫大家想晚餐
請選擇
(a) Yes (b) Yes (c) Yes (d) Yes (e) Both of all (f) Nope (g) 奶???
(a) 彥廷走路買 (b) 彥廷騎腳踏車買 (c) 彥廷坐經理的車買回來 (d) 彥廷左手牽右手一起買

Wendy：你是在大聲什麼啦 
Terry : 這我婆 

York：哥哥姊姊吃冰嗎 

:::spoiler **下方視頻連結以解鎖更多清涼的瑞光<3<3<3**
:::