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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 Equation \ref{eq:two} [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:two} \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 \ref{eq:three} [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 \\ \label{eq:three} \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 (Equation \ref{eq:beta}) [1/K][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 [page]
$N$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 represents our case of heat flow through wall Figure 1 (section of water tank wall).
The next calculations represent heat loss inside ($h$i) and heat loss outside ($h$0): We take the general formula for natural heat transfer Equation \ref{eq:ght}.
\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 \\ \label{eq:ght} \end{aligned} \end{equation}
Characteristics of 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]
$k$water = 0.6004 [W/m·K]
For $β$ Equation \ref{eq:beta1} 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} \\ \label{eq:beta1} \end{aligned} \end{equation}
We take the general formula for natural heat transfer Equation \ref{eq:ght1} in order to know $h$i.
\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}} \\ \label{eq:ght1} \end{aligned} \end{equation}
Characteristics of 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]
$k$air = 0.0249 [W/m·k]
For $β$ Equation \ref{eq:beta2} 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} \\ \label{eq:beta2} \end{aligned} \end{equation}
We take the general formula for natural heat transfer Equation \ref{eq:ght2} in order to know $h$i.
\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}} \\ \label{eq:ght2} \end{aligned} \end{equation}
We return to the Equation \ref{eq:foruo}
\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:foruo} \end{aligned} \end{equation}
$A$i = 3.424 m^2
$A$o = 3.456 m^2
$k$PEHD thermal conductivity = 0.48 (http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html).
$A$Lm - log mean area
$A$Lm Equation \ref{eq:alm} 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 \\ \label{eq:alm} \end{aligned} \end{equation}
Now we know $h$i, $h$o and we can apply Equation \ref{eq:uof}
\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}} \\ \label{eq:uof} \end{aligned} \end{equation}
From the heat transfer Equation \ref{eq:qwall} we can determine $q$wall:
\begin{equation} \begin{aligned} \dot{q}_{wall}=& U_0 \cdot A_0 \cdot \Delta T_{Total}\\ \dot{q}_{wall}=& 2.053 \cdot 3.456 \cdot 10 \\ \dot{q}_{wall}=& 70.952 \ W \\ \label{eq:qwall} \end{aligned} \end{equation}
Until now we know lost of the heat through the wall in watts. The next calculation are for heat loss of water directly with air (top side water tank) as shown in Figure 2 .
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]
$k$air = 0.0257 [W/m·K]
For $β$ Equation \ref{eq:beta3} 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} \\ \label{eq:beta3} \end{aligned} \end{equation}
We replace in Equation \ref{eq:hw} 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}} \\ \label{eq:hw} \end{aligned} \end{equation}
To know the heat loss in watts we need again to know the coefficient $U$o. The Equation \ref{eq:uoev} is more simplified because in this case we do not have a material between water and air.
\begin{equation} \begin{aligned} & {\frac{1}{U_0}}= \boxed{\frac{1}{h_w \cdot {\frac{A_i}{A_o}}}}\\ & {\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}} \\ \label{eq:uoev} \end{aligned} \end{equation}
From the heat transfer Equation \ref{eq:qev}, we calculate heat lost through evaporation ($q$ev)
\begin{equation} \begin{aligned} \dot{q}_{ev}=& U_0 \cdot A_0 \cdot \Delta T_{Total}\\ \dot{q}_{ev}=& 3.695 \cdot 3.456 \cdot 10 \\ \dot{q}_{ev}=& 127.699 \ W \\ \label{eq:qev} \end{aligned} \end{equation}
And we finish with Equation \ref{eq:fuckingfinal} that results our lost heat in watts per seconds.
\begin{equation}
\begin{aligned}
\dot{q}_{total}=& \dot{q}_{wall} + \dot{q}_{ev} \\
\dot{q}_{total}=& 70.952 + 129.699 \\
\dot{q}_{total}=& 198.631 \ W \\
\label{eq:fuckingfinal}
\end{aligned}
\end{equation}
After this calculation we found one value, this value is the result after we apply a scenario. For safety, we will buy one 300 W heater.
