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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 \ref{eq:one} [Christie J. Geankoplis (1993), Transport Process and Unit Operations, Third Edition, pg. 246]:

$q$ - total energy loss in time
$U$₀ - overall outside heat-transfer coefficient
$A$₀ - outside area
$ΔT$ - Difference of temperature

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

Unknown coefficient is $U$₀ and to find $U$₀ 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}} \label{eq:2} \end{aligned} \end{equation}

$U$₀ - heat-transfer coefficient
$x$ - thicknes of the material
$h$_i - heat-transfer coefficient inside
$h$_o - heat-transfer coefficient outside
$A$_Lm - logaritmic area of the film
$A$_i - inside area
$A$_o - outside area
$K$_PEHD - material polyethylene high-density

First box it is resistance of water and the second box is resistance of air.

To find h_i and h_o 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 (formula \ref{eq:beta})[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)}} \label{eq:beta} \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 Figure 1(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}