(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  ### 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*} &nbsp  * **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*} ::: &nbsp  * **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*} ::: &nbsp  * **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