---
# System prepended metadata

title: '(Attempt 3) Goal 2: Water Layer - [`notebook`](water_surf_temp.ipynb)'

---

# (Attempt 3) Goal 2: Water Layer - [`notebook`](water_surf_temp.ipynb)

## To find

* Temperature of Water Surface $(T_w)$
* Total heat flux entering the house through the roof, $(q_t)$

## Nomenclature

* $S$ = Intensity of Solar Radiation (i.e. Solar Constant)
* $u_o$ = water velocity
* $v_o$ = wind velocity
* $\epsilon_w$ = emissivity of water surface
* $\sigma$ = Stefan-Boltzmann constant $(5.67*10^{-8}\ W/m^2K^4)$
* $T_r$ = inside room temperature
* $T_w$ = water surface temperature
* $T_{\infty}$ = outside air temperature
* $\tau_w$ = fraction of solar radiation absorbed by water
* $k_w$ = thermal conductivity of water
* $L_w$ = length of water path
* $h_w$ = convection coefficient of still water
* Medium:
  * A = water
  * B = air
* $h_r$ = radiative heat transfer coefficient
* $h_c$ = convective heat transfer coefficient

## Assumptions

1. Steady state
2. Water is still ($u_o = 0$) but gentle breeze is present ($v_o = 10\ km/h$)
3. Room is maintained at fixed ambient temperature
4. Surroundings are dry
<!-- 6. Length of water path = 5m -->

### Note

* Reynolds number in convection is for air 10km/h
* Characteristic length (for calculating Gr) is same as $L_w$

## Equations

Using energy balance:
$$
  q_t = q_c + q_r - q_e
$$

with,

* **Radiation:**

\begin{align*}
  q_r &= \tau_w\cdot S - h_r \cdot (T_{\infty} - T_w) \\
\\
  h_r &= \epsilon_w\cdot \sigma\cdot \frac{(\overline T_w)^4 - (\overline T_\infty - 12)^4}{\overline T_\infty - \overline T_w}
\end{align*}
<p>&nbsp</p>
<!------------------------->

* **Convection:**

\begin{align*}
  q_c &= h_c\cdot (T_{\infty} - T_w) \\
\\
  h_c &= 5.678 \cdot (1 + 0.85\cdot(v_o - u_o))
\end{align*}

<!------------------------------------------------------------>
:::spoiler old method (textbook based)
$$
  q_c = h_1\cdot (T_{\infty} - T_w)
\\
  h_1 = \frac{k_w}{L_w}\cdot [0.14\cdot(Gr\cdot Pr)^{1/3} + 0.644\cdot (Pr\cdot Re)^{1/3}]
$$

\begin{align*}
  Gr &= \frac{g\cdot \beta\cdot (T_w - T_a)\cdot L_w^3}{\nu^2}
\\
  \nu &= \frac{\mu}{\rho}
\\
  Re &= \frac{\rho \cdot v_o \cdot L_w}{\mu}
\end{align*}
:::
<p>&nbsp</p>
<!------------------------->

* **Evaporative:**

\begin{align*}
  q_e &= 0.013\cdot h_c\cdot (p(\overline{T}_w) - \gamma\cdot p(\overline{T}_\infty)) \\
\\
  p(T) &= R_1\cdot T + R_2
\end{align*}

<!------------------------------------------------------------>
:::spoiler old method (textbook based)
\begin{align*}
  q_e &= h_{fg} \cdot n_A^{''} \\
\\
  n^{''}_A &= h_m\cdot (p_{A, sat} - p_{A, \infty}) \\
\\
  p_{A, sat} &= R_1\cdot T_w + R_2
\\
  h_m &= \frac{h_1\cdot D_{AB}\cdot Le^{1/3}}{k}
\\
  Le &= \frac{\alpha}{D_{AB}}
\end{align*}
:::
<p>&nbsp</p>
<!------------------------->


* **Total:**

\begin{align*}
  q_t &= \frac{T_w - T_r}{R_{net}} \\
\\
  R_{net} &= \frac{1}{h_r} + \sum_{i=1}^{3} \frac{L_i}{k_i} + \frac{1}{h_{w}} \\
\\
  h_w &= \frac{k_w}{L_w}\cdot [0.14\cdot(Gr\cdot Pr)^{1/3} + 0.644\cdot (Pr\cdot Re)^{1/3}] \\
\\
  Gr &= \frac{g\cdot\beta\cdot(T_w-T_a)\cdot(L_w)^{3}}{\nu^2} \end{align*}

## Properties

Outside Air:
* Mild breeze $v_o = 2.78\ m/s$
* $T_\infty \in [300, 330] K$
* $T_f = 320K$
* $\beta = \frac{1}{T_f} = 0.0031\ K^{-1}$
* Table A.4, air ($T_f$):
  * $\nu = 18 \cdot 10^{-6}\ m^2/s$
  * $\alpha = 25 \cdot 10^{-6}\ m^2/s$
  * $Pr = 0.702$
  * $k = 27.7 \cdot 10^{-3}\ W/m\cdot K$
* $p_{A, \infty} \approx 0$
* S = 1366\ W/m^2$
* $R_1=325\ Pa/^\circ C$ and $R_2 = -5155\ Pa$ (from [page 310](../docs/papers/Experimental_validation_of_a_thermal_mod.pdf))
* $\gamma=0.27$ (approx average over a day)
* $r = 0.1$

Water layer:
* $L_w = 0.1\ m$ (approx thickness of water layer)
* Table A.6, water ($T_w$):
  * $\nu = 18 \cdot 10^{-6}\ m^2/s$
* Still water $u_o = 0$
* $\epsilon_w = 0.95$
* $\tau_w=0.2$

Roof:
* $t = 0.2\ m$ thick with,
  * Cement = $5\ cm$
  * Brick = $10\ cm$
  * Lime = $5\ cm$
* $K_i$, Conductivity of each layer,
  * Cement = $0.72\ W/m\cdot K$
  * Brick = $0.71\ W/m\cdot K$
  * Lime = $0.73\ W/m\cdot K$

Inside air:
* $T_r = 300K$ (Room Temperature)
* $h_r = 8.4\ W/m^2\cdot K$

## Solving

\begin{align*}
  h_c &= 5.678 \cdot(1 + 0.85 \cdot 2.78) \\
      &= 19.1 \ W/m^2\cdot K \\
\\
  q_c &= 19.1 \cdot (T_{\infty} - T_w) \\
\\
  q_e &= 0.013 \cdot 19.1 \cdot ((R_1\cdot \overline{T}_w + R_2) - \gamma\cdot(R_1\cdot \overline{T}_\infty + R_2)) \\
      &= 0.248 \cdot (R_1\cdot(\overline{T}_w - \gamma\cdot \overline{T}_\infty) + R_2 \cdot(1 - \gamma)) \\
      &= 0.248 \cdot (325\cdot(\overline{T}_w - 0.27\cdot \overline{T}_\infty) - 3763) \\
      &= 80.6\cdot \overline{T}_w - 21.76\cdot \overline{T}_\infty - 993 \\
\\
  R_0 &= 0.013 \cdot h_c \\
      &= 0.2483 \\
  h_r &= 5.3865\cdot 10^{−8}\cdot \frac{(T_w+273.15)^4 - (T_a+261.15)^4}{T_w-T_a} \\
  Gr  &= \frac{9.8\cdot\frac{1}{T_a}\cdot(T_w-T_a)\cdot(5)^{3}}{\nu^2} \\
      &= \frac{1225}{\nu^2} \cdot \left( \frac{T_w}{T_a}-1 \right) \\
  h_1 &= (\frac{300}{5})\cdot 0.14\cdot(Gr\cdot Pr)^{1/3}\\
      &= 8.4\cdot(Gr\cdot Pr)^{1/3}
\end{align*}

Resistance net,
\begin{align*}
  R_{net} &= \frac{1}{h_r} + \sum_{i=1}^{3} \frac{L_i}{k_i} + \frac{1}{h_w}
\\ \\
          &= \frac{1}{8.4} + \frac{0.05}{0.72} + \frac{0.10}{0.71} + \frac{0.05}{0.73} + \frac{1}{X}
\\ \\
          &= 0.398 + X\ m^2K/W
\end{align*}

## References

* Using **Eq. 15** on [page 308](../docs/papers/Experimental_validation_of_a_thermal_mod.pdf)
* Pg 411 textbook for evaporative cooling
