EPS@ISEP | The European Project Semester (EPS) at ISEP

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playground:playground [2015/05/20 20:12] – epsatisepplayground:playground [2015/06/12 14:14] (current) – team4
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-This is a reference to a book [(GangOfFour)]. 
  
-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: \\ +FEEDER:
-The heat transfer rate equation [Christie J. Geankoplis (1993), Transport Process and Unit Operations, Third Edition, pg. 246]:\\+
  
-q - total energy loss in time\\ +Our first idea was to create a (feeder) to feed the fish. The construction of the feeder was this: it was a plexi- glaas box, which had a spindle at the bottom. The food fell directly to the spindle because of the slope we had designed in the bottom of the box (picture…). The screw was inside a PVC pipe open at the top. The spindle protruding from the box so that the food fell into the tank, so now the pipe was open at the bottom (picture…). The operation was: we introduced the food at the top of the box, the food fell directly to the worm. A motor turned the splindle until the food fell into the tank.
-U<sub>0</sub> - overall outside heat-transfer coefficient\\ +
-A<sub>0</sub> - outside area \\ +
-ΔT - Difference of temperature \\ +
-BLA BALA\ref{eq:cosinesimilarity}+
  
-\begin{equation}  +TO INCLUDE THIS IDEA IN 8.2
-\begin{aligned} +
-      \dot{q} = U_0\cdot A_0\cdot \Delta T_{Total}  +
-      \label{eq:eq:cosinesimilarity} +
-   \end{aligned} +
-\end{equation}+
  
 +another improvement we wanted to include was a feeder and control fish food. We explain the idea in the point... but this idea was not included because it exceeded our budget
  
 +8.2
 +¿¿¿As we have said due to lack of time and money in the budget, we did not include features that are important for the development of our aquaponic system. Water quality is very important for fish health.???
  
 +Water quality is very important for fish health. The control of the water is based on temperature, pH and dissolved oxygen. Our aquaponic system controls the temperature so in the future development of our  aquaponic system  it would be necessary to control the pH and dissolved oxygen. The pH is involved in the process of nitrification (conversion of fish waste in a less toxic and compound acceptable to plants), it is essential in the availability of plant nutrients (manganese, copper, zinc and boron) and must be appropriate to the species of fish that wants to grow. Dissolved oxygen has to be suitable (above 3 mg / L) for a good nitrification.
  
-Unknown coefficient is U<sub>0</sub> and to find U<sub>0</sub> 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} 
  
-U<sub>0</sub> - heat-transfer coefficient\\ 
-x - thicknes of the material \\ 
-A<sub>i</sub> - inside area \\ 
-A<sub>o</sub> - outside area \\ 
-K<sub>PEHD</sub> - 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<sub>Nu</sub> - 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/m<sup>3</sup>]\\ 
-g - gravitational acceleration 9.81 [m/s<sup>2</sup>]\\ 
-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<sub>p</sub> - heat capacity [J/Kg·K]\\ 
-m - constant \\ 
-N<sub>Pr</sub> - the Prandtl number (dimensionless/adimensional number)\\ 
-N<sub>Gr</sub> - 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).\\ 
-{{ :heat_transfer_diagram.png?400 |}} 
  
-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<sub>water</sub> at 23.75 ºC = 997.4 [Kg/m<sup>3</sup>] \\ 
-ΔT = 25-15 = 10 [ºC] \\ 
-c<sub>p water</sub> at 23.75 ºC = 4181.5 [J/Kg·K] \\ 
-μ<sub>water</sub> at 23.75 ºC = 9.2·10<sup>-4</sup> [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<sub>air</sub> at 16.75 ºC = 1.219 [Kg/m<sup>3</sup>] \\ 
-ΔT = 25-15 = 10 [ºC] \\ 
-c<sub>p air</sub> at 16.75 ºC = 1004.5 [J/Kg·K] \\ 
-μ<sub>air</sub> at 16.75 ºC = 1.8·10<sup>-5</sup> [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<sub>i</sub> = 3.424 m^2 \\ 
-A<sub>o</sub> = 3.456 m^2 \\ 
-A<sub>Lm</sub> -   \\ 
-A<sub>Lm</sub> 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<sub>PEHD</sub> - 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} 
  
  
-{{ :playground:heat_transfer_diagram2.png?400 |}} 
-For water surface in direct contact with the air. 
-D = 1.1 m 
-L = 0.9·D = 0.99 \\ 
-ρ<sub>air</sub> at 20 ºC = 1.205 [Kg/m<sup>3</sup>] \\ 
-ΔT = 25-15 = 10 [ºC] \\ 
-c<sub>p air</sub> at 20 ºC = 1004.7 [J/Kg·K] \\ 
-μ<sub>air</sub> at 20 ºC = 1.8·10<sup>-5</sup> [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:\\ +- The reasons why we decided Raspberry Pi B + against BEAGLEBONE Black are:\\   
-\begin{equation} +-   
-\begin{aligned} +- - USB: B + Raspberry Pi has 4 X USB 2.0 CONNECTOR vs 1 X USB 2.0 CONNECTOR BLACK BONE beable. For our aquaponic system we  need to connect with Arduino, camera and wireless Ethernet so we need more than one USB.\\   
-    {\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 \\ +- - COMMUNITY: Raspberry Pi has sold over one million units and it have more coverage in the medial comunication and global exposure. Beagle Bone glack has a growing community but it's not enough. raspberry pi production generates 13 times more productivity than beaglebone black\\   
-    {\frac{h_o \cdot L}{K_{air}}} = & a\cdot (1.1 \cdot 10^{9})^m \implies  a=0.14 \ ; \ m={\frac{1}{4}} \\ +- - As for our aquaponic system is as valid as Raspberry Pi B + bone Beagle but  the price was decisive. As we can see from the chart the price of Raspberry Pi B + is € 31.95 compared to € 61.56 Beagle Bone Black. As our product has to have a lower price of 250 € we decided that was a good choice Raspeberry.\\ 
-    {\frac{h_w \cdot L}{K_{air}}} = & 0.14 \cdot (1.1 \cdot 10^{9})^{\frac{1}{4}} \\ +- To decide between a component or another, we have chosen a model of Arduino uno and other TI LaunchPad MSP430 . After we have made a comparison table 
-    {\frac{h_w \cdot 0.99}{0.0257}} = & 0.14 \cdot (1.1 \cdot 10^{9})^{\frac{1}{4}} \\ + We have chosen Arduino Uno because:  
-    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}} \\ +- - Arduino have a massively huge community support. This is not to be under-estimated.\\   
-\end{aligned} +- - In Arduino there is no limit in a future expansion because the boards called '​shields'​ can be stacked on top of the board to add features, in the The MSP430 that’s not possible.\\   
-\end{equation}+- - Arduino one Can run at 5V or 3.3V (or anywhere between 1.8V and 5V if you change oscillators) while The MSP430 caps out at 3.6V.\\   
 +- - The Arduino have IO pins, 20 while the MSP430 have 16. It might not seem like a huge difference, but some  projects where It's used 18 of those pins after using a pin expander to gain 8 more.\\
  
  
-\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} +Table {{ref>tlabe100}} of resources: Human and Material 
-\begin{aligned} +<table tlabe100> 
-       {\frac{1}{U_0}}= & \boxed{\frac{1}{3.730 \cdot {\frac{3.424}{3.456}}}} \\ +<caption>Table of resources: Human and Material</caption> 
-       {\frac{1}{U_0}} = & 0.271 \\ +^ Resource Name   ^ Cost [€] ^ Type ^  Intials ^ Allocation [%] ^ (Std) Rate [€/h] or [€/unit]^ Overtime [€/h]  ^ Cost/User ^ Accrue ^ Base ^ 
-       U_0 = & 3.695 \ {\frac{W}{m^2 \cdot K}} +^ Arick            | 1 500.00| Work  |         A|            100|     4|       0|         0|Prorated|Standard|            
-\end{aligned} +^ Francisco        | 1 500.00| Work  |         F|            100|     4|       0|         0|Prorated|Standard|  
-\end{equation} +^ Jan              | 1 500.00| Work  |         J|            100|     4|       0|         0|Prorated|Standard|  
- +^ Katoo            | 1 500.00| Work  |         K|            100|     4|       0|         0|Prorated|Standard|  
-From the heat transfer equation:\\ +^ Rasmus           | 1 500.00| Work  |         R|            100|            4|       0|         0|Prorated|Standard|  
-\begin{equation} +^ Viorel           | 1 500.00| Work  |         V|            100|     4|       0|         0|Prorated|Standard|  
-\begin{aligned} +^Raspberry Pi      |  31.95|  Material  |  RP| |  31.95|  |  |Prorated| | 
-      q_{ev}=& U_0 \cdot A_0 \cdot \Delta T_{Total}\\ +^Arduino           |  22.90|  Material  |  AR| |  22.90|  |  |Prorated| | 
-      q_{ev}=& 3.695 \cdot 3.456 \cdot 10 \\ +^USB cable Raspberry Pi|  2.25|  Material  |  UCRP| |  2.25|  |  |Prorated| | 
-      q_{ev}=& 127.699 \ W +^USB cable Arduino |  2.85|  Material  |  UCAR| |  2.85|  |  |Prorated| |  
-    \end{aligned} +^Power supply Raspberry Pi|  12.24|  Material  |  PSRP| |  12.24|  |  |Prorated| | 
-\end{equation} +^Jumper wires      |  12.00|  Material  |  JW| |  12.00|  |  |Prorated| | 
- +^Memory card       |  10.90|  Material  |  MC| |  10.90|  |  |Prorated| | 
-\begin{equation} +^Webcam            |  11.90|  Material  |  W| |  11.90|  |  |Prorated| | 
-\begin{aligned} +^Wi-fi dongle      |  9.80|  Material  |  WF| |  9.80|  |  |Prorated| | 
-      q_{total}=& q_{wall} + q_{ev} \\ +^Arduino 4 channel relay module|  12.90|  Material  |  A4CRM| |  12.90|  |  |Prorated| | 
-      q_{total}=& 70.952 + 129.699 \\ +^Water heater      |  38.61|  Material  |  WH| |  38.61|  |  |Prorated| | 
-      q_{total}=& 198.631 \ W +^Water pump        |  ISEP|  Material  |  WP| |  0|  |  |Prorated| | 
-    \end{aligned} +^Light led         |  7.40|  Material  |  LL| |  7.40|  |  |Prorated| | 
-\end{equation}+^Automatic fish feeder|  15.89|  Material  |  AFF| |  15.89|  |  |Prorated| | 
 +^Step motor        |  ISEP|  Material  |  SM| |  0|  |  |Prorated| | 
 +^Temperature sensor|  7.30|  Material  |  TS| |  7.30|  |  |Prorated| | 
 +^Water flow sensor |  12.24|  Material  |  WFS| |  12.24|  |  |Prorated| | 
 +^Depth sensor      |  12.24|  Material  |  DS| |  12.24|  |  |Prorated| | 
 +^Infrared sensor   |  8.40|  Material  |  IS| |  8.40|  |  |Prorated| | 
 +</table>

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