# F44091144 唐祥益 HW2
###### tags: `熱傳學`
## Problem 1
Applying conservatino of energy requirement in Cartesian CS,

$\dot{E_{st}}+\dot{E_g}-\dot{E_{out}}=\dot{E_{st}}$
where
$\dot{E_{st}}=\dfrac{\partial U_{sens}}{\partial t}=\rho c_p\dfrac{\partial T}{\partial t}\ dx\ dy\ dz$
$\begin{aligned}
\dot{E_{in}}-\dot{E_{out}}&=q_x+q_y+q_z-q_{x+dx}-q_{y+dy}-q_{z+dz}\\
&=-\dfrac{\partial q_x}{\partial x}dx-\dfrac{\partial q_y}{\partial y}dy-\dfrac{\partial q_z}{\partial z}dz
\end{aligned}$
In Cylindrical CS,

$\dot{E_{in}}-\dot{E_{out}}=-\dfrac{\partial q_r}{\partial r}dr-\dfrac{1}{r}\dfrac{\partial q_{\phi}}{\partial \phi}d\phi-\dfrac{\partial q_z}{\partial z}dz$
$\dot{E_{st}}=\rho c_p\dfrac{\partial r}{\partial t}\ r\ dr\ d\phi\ dz$
By Fourier's law we know that

$q_r=-k\dfrac{\partial T}{\partial r},\ q_{\phi}=-\dfrac{k}{r}\dfrac{\partial T}{\partial \phi},\ q_z=-k\dfrac{\partial T}{\partial z}$

Therefore, substitute the foregoing equations into the energy balance, we obtain

$\dfrac{1}{r}\dfrac{\partial}{\partial r}(kr\dfrac{\partial T}{\partial r})+\dfrac{1}{r^2}\dfrac{\partial}{\partial \phi}(k\dfrac{\partial T}{\partial \phi})+\dfrac{\partial}{\partial z}(k\dfrac{\partial T}{\partial z})+\dot{q}=\rho c_p\dfrac{\partial T}{\partial t}$
## Problem 2
- (a)

- (b)
Since $R_{total}=\dfrac{1}{UA'}, wherer A'=wL=3A$
Hence, $\begin{aligned}
UA&=(\dfrac{t/2}{K_cA}+R_c+\dfrac{t/2}{K_iA})^{-1}+(\dfrac{t}{K_gA})^{-1}+(\dfrac{t/2}{K_cA}+R_c+\dfrac{t/2}{K_iA})^{-1}
\end{aligned}$
Therefore, we substitute $UA$ into $R_{total}$
$\begin{aligned}
R_{total}&=\dfrac{1}{UA}\\
&=\dfrac{1}{(\dfrac{t/2}{K_cA}+R_c+\dfrac{t/2}{K_iA})^{-1}+(\dfrac{t}{K_gA})^{-1}+(\dfrac{t/2}{K_cA}+R_c+\dfrac{t/2}{K_iA})^{-1}}
\end{aligned}$
where A=wL/3
## Problem 3
1.
In this diagram, we can find that temparature do not change in the x-direction, hence we can assume that this slab is insulated

2.
Since the temperature gradient $dT/dx$ is negative, by Fourier's law, we know that heat flux is positive
3.
For this slab, the temperature distribution is symmetrical about the midplane, and this slab is uniform energy generation per unit volume.
4.
The temperature distribution is symmetrical about the midplane, hence the temperature at both surfaces are maitained at the same value. Since the first half plane's temperature gradient is negative, and its curve is much more deep, therefore we can assume that this plane has convection heating surface at x=0, and convection cooling surface at the other surface.
## Problem 4
- (a).
The long-fin model, that is$\ \ L\rightarrow\infty,\ \ \theta(L)\rightarrow 0$
And we also know that the ==Temperature Dirstribution== is

$e^{-mx}$ and $\ m^2\equiv hP/kA_c$

we have got, $T_b=200^{\circ}C,\ \ T_{\infty}=35^{\circ}C\quad \theta_b=T_b-T_{\infty}=165$
also $k=51.9\dfrac{W}{m-K},\ \ h=22\dfrac{W}{m^2-K}$
and the cross area $A=(12^{mm})^2=144\times10^{-6}\ m$,
perimeter $P=4\times (12^{mm})=48\times 10^{-3}\ m$

we evalute the $m^2=\dfrac{hP}{kA}=\dfrac{22\times 48\times 10^{-3}}{51.9\times144\times 10^{-6}}=141.297$
Hence, $m=11.8869$

Substituting into the temperature distribution $e^{-mx}$
$e^{-mx}=\theta / \theta _b=\dfrac{T-T_{\infty}}{165}\\
\Rightarrow e^{-11.8869x}=\dfrac{60-35}{165}\\
\Rightarrow x=0.1588\ m=158.75\ mm$
- (b)
when x=30 mm
$e^{-11.8869\times 30\times 10^{-3}}=\dfrac{T-35}{165}\\
\Rightarrow T=35+165\times 0.700\\
\Rightarrow T=150.5078^{\circ}C$
- C
In Convection model, we know the temperature distribution is,

$\dfrac{\theta}{\theta_b}=\dfrac{cosh\ m(L-x)+(h/mk)\ sinh\ m(L-x)}{cosh\ mL+(h/mk)\ sinh\ mL}$

with L=80 mm, find the temperature at the end of the fin (x=80 mm)

$\rightarrow \dfrac{1+(h/mk)\times 0}{cosh\ mL+(h/mk)\ sinh\ mL}\\
\Rightarrow \dfrac{\theta}{\theta_b}=\dfrac{1+0}{1.4873+22\times 1.1009/11.8869*51.9}\\
\Rightarrow \dfrac{T-35}{165}=0.6551\\
\Rightarrow T=143.09^{\circ}C$

## Problem 5
- effiviency

$\eta_f\equiv\dfrac{q_f}{q_{max}}=\dfrac{q_f}{hA_f\theta_b}\\
\rightarrow \dfrac{M\ \dfrac{sinh\ mL+(h/mk)\ cosh\ mL}{cosh\ mL+(h/mk)\ sinh\ mL}}{hPL\theta_b}
= \dfrac{\sqrt{hPkA}\ \dfrac{sinh\ mL+(h/mk)\ cosh\ mL}{cosh\ mL+(h/mk)\ sinh\ mL}}{hPL}$

we have, $h=22\ \dfrac{W}{m^2\cdot K},\ \ k=51.9\ \ \dfrac{W}{m^2\cdot K},\ \ P=48\times 10^{-3}\ m,\ \ A=144\times 10^{-6}\ m^2$

Substituting into $\eta_f$, we obtain

$\eta_f=\dfrac{(0.0888)\times \dfrac{(1.1009)+(0.0357)\times (1.4873)}{(1.4873)+(0.0357)\times (1.1009)}}{22\times 48\times 10^{-3}\times 80\times 10^{-3}}\\
\Rightarrow \eta_f=0.7945$

- effectiveness

In the above calculation, we got $q_f=11.0804$

$\epsilon_f=\dfrac{q_f}{hA\theta_b}\\
\Rightarrow \epsilon_f=\dfrac{11.0804}{22\times 144\times 10^{-6}\times 165}=21.1976$

- fin resistance

$R_{t,f}=\dfrac{\theta_b}{q_f}\\
\Rightarrow R=\dfrac{165}{11.0804}=14.8911$