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This is a reference to a book [1].

Table 1 illustrates a boxed table.

To be able to calculate how many watts we need for a heater we are forced to come with a scenario and the next formulas:
The heat transfer rate equation [Christie J. Geankoplis (1993), Transport Process and Unit Operations, Third Edition, pg. 246]:

q - total energy loss in time
U₀ - overall outside heat-transfer coefficient
A₀ - outside area
ΔT - Difference of temperature
BLA BALA\ref{eq:cosinesimilarity}

\begin{equation} \begin{aligned} \dot{q} = U_0\cdot A_0\cdot \Delta T_{Total} \label{eq:eq:cosinesimilarity} \end{aligned} \end{equation}

Unknown coefficient is U₀ and to find U₀ we have the next formula [pg. 227]: \begin{equation} \begin{aligned} {\frac{1}{U_0}=}\boxed{\frac{1}{h_i \cdot {\frac{A_i}{A_o}}}} + {\frac{x}{K_{PEHD} \cdot {\frac{A_{Lm}}{A_o}}}} + \boxed{\frac{1}{h_o}} \end{aligned} \end{equation}

U₀ - heat-transfer coefficient
x - thicknes of the material
A_i - inside area
A_o - outside area
K_PEHD - material

First box it is resistance of water and the second box is resistance of air.
To find hi and ho we will use formula of natural convection heat transfer.
The average natural convection heat transfer coefficient can be expressed by the following general equation [pg. 254].

\begin{equation} \begin{aligned} {N_{NU}=}{\frac{h \cdot L}{K}} = {a} \cdot \left( {\frac{L^3 \cdot \rho^2 \cdot g \cdot \beta \cdot \Delta T}{\mu^2}} \cdot {\frac{c_p \cdot \mu}{k}} \right)^m = {a} \cdot \left(N_{Gr} \cdot N_{Pr} \right)^m \end{aligned} \end{equation}

N_Nu - Nusselt number (dimensionless/adimensional number)
h - film coefficient of heat transfer
L - length [m]
K - thermal conductivity [W/m·k]
a - constant
ρ - density [Kg/m³]
g - gravitational acceleration 9.81 [m/s²]
β - volumetric coefficient of expansion of the fluid [pg. 253]
ΔT - positive temperature difference between the wall and bulk fluid in [K]
μ - viscosity [Kg/m·s]
c_p - heat capacity [J/Kg·K]
m - constant
N_Pr - the Prandtl number (dimensionless/adimensional number)
N_Gr - the Grashof number (dimensionless/adimensional number)

\begin{equation} \begin{aligned} {β}={\frac{\rho_b - \rho}{\rho(T-T_b)}} \end{aligned} \end{equation}

The water tank is uncovered and in our case is a cylinder with the next characteristics:

• radius 0.77 m
• height 0.55 m
• cylinder thickness 0.005 m
• material PEHD (polyethylene high-density)
• top uncovered
• bottom covered

For calculation of heat loss we need e scenario where conditions are a little bit exaggerated.
The scenario:
Water temperature is 25 ºC and outside of the tank, air temperature is 15 ºC.
The next diagram represent our case of heat flow through wall (section of cylinder).

The next calculations represent heat loss inside (hi) and heat loss outside (ho): We take the general formula for natural heat transfer. \begin{equation} \begin{aligned} {\frac{h_i \cdot L}{K_{water}}} = {a} \cdot \left( {\frac{L^3 \cdot \rho^2 \cdot g \cdot \beta \cdot \Delta T}{\mu^2}} \cdot {\frac{c_p \cdot \mu}{k}} \right)^m \end{aligned} \end{equation}

Characteristics for water at our specific temperature we take from a website (http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html) and we will interpolate where we need.

L = 0.77 m
ρ_water at 23.75 ºC = 997.4 [Kg/m³]
ΔT = 25-15 = 10 [ºC]
c_{p water} at 23.75 ºC = 4181.5 [J/Kg·K]
μ_water at 23.75 ºC = 9.2·10^-4 [Kg/m·s]
K = 0.6004 [W/m·k]
For β we calculate:

\begin{equation} \begin{aligned} {β} =& {\frac{\rho_{25} - \rho_{22.5}}{\rho_{22.5}(T_{22.5}-T_{25})}} = \\ {β} =& {\frac{997.1-997.7}{997.7-(22.5-25)}} = 2.4 \cdot 10^{-4} \end{aligned} \end{equation}

We replace in formula with the above data and results:
\begin{equation} \begin{aligned} {\frac{h_i \cdot L}{K_{water}}} = & {a} \cdot \left( {\frac{0.77^3 \cdot 997.4^2 \cdot 9.81 \cdot 2.4 \cdot 10^{-4} \cdot 10}{(9.2 \cdot 10^{-4})^2}} \cdot {\frac{4181.5 \cdot 9.2 \cdot 10^{-4}}{0.6004}} \right)^m \\ {\frac{h_i \cdot L}{K_{water}}} = & a\cdot (8.1 \cdot 10^{10})^m \implies a=0.13 \ ; \ m={\frac{1}{3}} \\ {\frac{h_i \cdot L}{K_{water}}} = & 0.13 \cdot (8.1\cdot 10^{10})^{\frac{1}{3}} \\ {\frac{h_i \cdot 0.77}{0.6004}} = & 0.13 \cdot (8.1\cdot 10^{10})^{\frac{1}{3}} \\ h_i = & {\frac{0.6004}{0.77}} \cdot 0.13 \cdot (8.1\cdot 10^{10})^{\frac{1}{3}} \\ h_i = & 438.648 \ {\frac {W}{m^2 \cdot K}} \\ \end{aligned} \end{equation}

Characteristics for air at our specific temperature we take from a website (http://www.engineeringtoolbox.com/air-properties-d_156.html) and we will interpolate where we need.

L = 0.77 m
ρ_air at 16.75 ºC = 1.219 [Kg/m³]
ΔT = 25-15 = 10 [ºC]
c_{p air} at 16.75 ºC = 1004.5 [J/Kg·K]
μ_air at 16.75 ºC = 1.8·10^-5 [Kg/m·s]
K = 0.0249 [W/m·k]
For β we calculate:

\begin{equation} \begin{aligned} {β}= & {\frac{\rho_{15} - \rho_{17.5}}{\rho_{17.5}(T_{17.5}-T_{15})}} \\ {β}= & {\frac{1.219-1.216}{1.216-(17.5-15)}} = 9.9 \cdot 10^{-4} \end{aligned} \end{equation}

We replace in formula with the above data and results:
\begin{equation} \begin{aligned} {\frac{h_o \cdot L}{K_{air}}} = & {a} \cdot \left( {\frac{0.77^3 \cdot 1.219^2 \cdot 9.81 \cdot 9.9 \cdot 10^{-4} \cdot 10}{(1.7 \cdot 10^{-5})^2}} \cdot {\frac{1004.6 \cdot 1.7 \cdot 10^{-5}}{0.0249}} \right)^m \\ {\frac{h_o \cdot L}{K_{air}}} = & a\cdot (1.5 \cdot 10^{8})^m \implies a=0.59 \ ; \ m={\frac{1}{4}} \\ {\frac{h_o \cdot L}{K_{air}}} = & 0.59 \cdot (1.5 \cdot 10^{8})^{\frac{1}{4}} \\ {\frac{h_o \cdot 0.77}{0.0255}} = & 0.59 \cdot (1.5 \cdot 10^{8})^{\frac{1}{4}} \\ h_o = & {\frac{0.0255}{0.77}} \cdot 0.59 \cdot (1.5 \cdot 10^{8})^{\frac{1}{4}} \\ h_o = & 2.107 \ {\frac {W}{m^2 \cdot K}} \\ \end{aligned} \end{equation}

We return to equation: \begin{equation} \begin{aligned} {\frac{1}{U_0}=}\boxed{\frac{1}{h_i \cdot {\frac{A_i}{A_o}}}} + {\frac{x}{K_{PEHD} \cdot {\frac{A_{Lm}}{A_o}}}} + \boxed{\frac{1}{h_o}} \end{aligned} \end{equation}

A_i = 3.424 m^2
A_o = 3.456 m^2
A_Lm -
A_Lm formula is:

\begin{equation} \begin{aligned} A_{L_m} = & {\frac{A_o-A_i}{ln{\frac{A_o}{A_i}}}} \\ A_{L_m} = & {\frac{0.55-0.545}{ln{\frac{0.55}{0.545}}}} \cdot 2 \cdot \pi = 3.44 \ m^2 \end{aligned} \end{equation}

K_PEHD - 0.48

\begin{equation} \begin{aligned} {\frac{1}{U_0}}= & \boxed{\frac{1}{438.648 \cdot {\frac{3.424}{3.456}}}} + {\frac{0.005}{0.48 \cdot {\frac{3.44}{3.456}}}} + \boxed{\frac{1}{2.107 \cdot 3.456}} \\ {\frac{1}{U_0}} = & 0.002 + 0.010 + 0.475 \\ U_0 = & 2.053 \ {\frac{W}{m^2 \cdot K}} \end{aligned} \end{equation}

From the heat transfer equation:
\begin{equation} \begin{aligned} q_{wall}=& U_0 \cdot A_0 \cdot \Delta T_{Total}\\ q_{wall}=& 2.053 \cdot 3.456 \cdot 10 \\ q_{wall}=& 70.952 \ W \end{aligned} \end{equation}

For water surface in direct contact with the air. D = 1.1 m L = 0.9·D = 0.99
ρ_air at 20 ºC = 1.205 [Kg/m³]
ΔT = 25-15 = 10 [ºC]
c_{p air} at 20 ºC = 1004.7 [J/Kg·K]
μ_air at 20 ºC = 1.8·10^-5 [Kg/m·s]
K = 0.0257 [W/m·k]
For β we calculate:

\begin{equation} \begin{aligned} {β}= & {\frac{\rho_{25} - \rho_{15}}{\rho_{15}(T_{25}-T_{15})}} \\ {β}= & {\frac{1.183-1.227}{1.227-(25-15)}} = 3.6 \cdot 10^{-3} \end{aligned} \end{equation}

We replace in formula with the above data and results:
\begin{equation} \begin{aligned} {\frac{h_w \cdot L}{K_{air}}} = & {a} \cdot \left( {\frac{0.99^3 \cdot 1.205^2 \cdot 9.81 \cdot 3.6 \cdot 10^{-3} \cdot 10}{(1.8 \cdot 10^{-5})^2}} \cdot {\frac{1004.7 \cdot 1.8 \cdot 10^{-5}}{0.0257}} \right)^m \\ {\frac{h_o \cdot L}{K_{air}}} = & a\cdot (1.1 \cdot 10^{9})^m \implies a=0.14 \ ; \ m={\frac{1}{4}} \\ {\frac{h_w \cdot L}{K_{air}}} = & 0.14 \cdot (1.1 \cdot 10^{9})^{\frac{1}{4}} \\ {\frac{h_w \cdot 0.99}{0.0257}} = & 0.14 \cdot (1.1 \cdot 10^{9})^{\frac{1}{4}} \\ h_w = & {\frac{0.0257}{0.99}} \cdot 0.14 \cdot (1.1 \cdot 10^{9})^{\frac{1}{4}} \\ h_w = & 3.730 \ {\frac {W}{m^2 \cdot K}} \\ \end{aligned} \end{equation}

\begin{equation} \begin{aligned} {\frac{1}{U_0}=}\boxed{\frac{1}{h_w \cdot {\frac{A_i}{A_o}}}} \end{aligned} \end{equation}

\begin{equation} \begin{aligned} {\frac{1}{U_0}}= & \boxed{\frac{1}{3.730 \cdot {\frac{3.424}{3.456}}}} \\ {\frac{1}{U_0}} = & 0.271 \\ U_0 = & 3.695 \ {\frac{W}{m^2 \cdot K}} \end{aligned} \end{equation}

From the heat transfer equation:
\begin{equation} \begin{aligned} q_{ev}=& U_0 \cdot A_0 \cdot \Delta T_{Total}\\ q_{ev}=& 3.695 \cdot 3.456 \cdot 10 \\ q_{ev}=& 127.699 \ W \end{aligned} \end{equation}

\begin{equation} \begin{aligned} q_{total}=& q_{wall} + q_{ev} \\ q_{total}=& 70.952 + 129.699 \\ q_{total}=& 198.631 \ W \end{aligned} \end{equation}