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| playground:playground [2015/05/22 18:39] – team4 | playground:playground [2015/06/12 14:14] (current) – team4 | ||
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| - | $c$< | + | FEEDER: |
| - | $Q$ - energy lost or given [J] \\ | + | |
| - | $m$ - mass [g] \\ | + | |
| - | $ΔT$ - temperature difference [°C] \\ | + | |
| - | \begin{equation} \label{eq:1} | + | 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. |
| - | \begin{aligned} | + | |
| - | | + | |
| - | | + | |
| - | \end{equation} | + | |
| - | \begin{equation} \label{eq:2} | + | TO INCLUDE THIS IDEA IN 8.2 |
| - | \begin{aligned} | + | |
| - | | + | |
| - | \end{aligned} | + | |
| - | \end{equation} | + | |
| - | \begin{equation} \label{eq: | + | 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 |
| - | \begin{aligned} | + | |
| - | & | + | |
| - | & | + | 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, |
| - | | + | |
| - | this references | + | |
| - | For better results we take it from the power we need through the beginning. We have a 300 W heater in our budget and that means P=300 W | ||
| - | \begin{equation} | ||
| - | \begin{aligned} | ||
| - | & | ||
| - | \boxed{P\cdot t=Q} \\ \\ | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | \begin{equation} | ||
| - | \begin{aligned} | ||
| - | & | ||
| - | & 700~000\cdot 4.18\cdot ΔT=1~080~000 \\ | ||
| - | & 2~926~000\cdot ΔT=1~080~000 \\ | ||
| - | & ΔT=\frac{1~080~000}{2~926~000} \\ | ||
| - | & ΔT=0.37\, | ||
| - | \phantom{\hspace{35cm}} %%< | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | In this calculation ' | ||
| - | _____________________________________________________________________________________________________________ | ||
| - | 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\\ | ||
| - | U< | ||
| - | A< | ||
| - | ΔT - Difference of temperature \\ | ||
| - | \begin{equation} | ||
| - | \begin{aligned} | ||
| - | \dot{q} = U_0\cdot A_0\cdot \Delta T_{Total} | ||
| - | | ||
| - | \end{equation} | ||
| - | |||
| - | Unknown coefficient is U< | ||
| - | \begin{equation} | ||
| - | \begin{aligned} | ||
| - | | ||
| - | | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | U< | ||
| - | x - thicknes of the material \\ | ||
| - | A< | ||
| - | A< | ||
| - | K< | ||
| - | 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< | ||
| - | h - film coefficient of heat transfer\\ | ||
| - | L - length [m]\\ | ||
| - | K - thermal conductivity [W/m·k]\\ | ||
| - | a - constant \\ | ||
| - | ρ - density [Kg/ | ||
| - | g - gravitational acceleration 9.81 [m/ | ||
| - | 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< | ||
| - | m - constant \\ | ||
| - | N< | ||
| - | N< | ||
| - | \begin{equation} | ||
| - | \begin{aligned} | ||
| - | | ||
| - | \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} | ||
| - | | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | Characteristics for water at our specific temperature we take from a website (http:// | ||
| - | L = 0.77 m \\ | + | - The reasons why we decided Raspberry Pi B + against BEAGLEBONE Black are:\\ |
| - | ρ< | + | - |
| - | ΔT = 25-15 = 10 [ºC] \\ | + | - - 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.\\ |
| - | c< | + | - - 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\\ |
| - | μ< | + | - - 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.\\ |
| - | K = 0.6004 [W/m·k]\\ | + | - To decide between a component or another, we have chosen a model of Arduino uno and other TI LaunchPad MSP430 |
| - | For β we calculate: | + | We have chosen Arduino Uno because: |
| + | - | ||
| + | - - Arduino have a massively huge community support. This is not to be under-estimated.\\ | ||
| + | - - In Arduino there is no limit in a future expansion because the boards called ' | ||
| + | - - 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} | ||
| - | {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 | ||
| - | {\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} | ||
| - | + | Table {{ref> | |
| - | + | <table tlabe100> | |
| - | Characteristics for air at our specific temperature we take from a website (http:// | + | <caption>Table of resources: Human and Material</caption> |
| - | + | ^ Resource Name ^ Cost [€] ^ Type ^ Intials ^ Allocation | |
| - | L = 0.77 m \\ | + | ^ Arick | 1 500.00| Work | |
| - | ρ<sub>air</sub> at 16.75 ºC = 1.219 [Kg/ | + | ^ Francisco |
| - | ΔT = 25-15 = 10 [ºC] \\ | + | ^ Jan | 1 500.00| Work | |
| - | c< | + | ^ Katoo | 1 500.00| Work | K| 100| |
| - | μ< | + | ^ Rasmus |
| - | K = 0.0249 | + | ^ Viorel |
| - | For β we calculate: \\ | + | ^Raspberry Pi | 31.95| Material |
| - | + | ^Arduino | |
| - | \begin{equation} | + | ^USB cable Raspberry Pi| 2.25| Material |
| - | \begin{aligned} | + | ^USB cable Arduino | 2.85| Material |
| - | {β}= & {\frac{\rho_{15} - \rho_{17.5}}{\rho_{17.5}(T_{17.5}-T_{15})}} \\ | + | ^Power supply Raspberry Pi| 12.24| Material |
| - | {β}= & {\frac{1.219-1.216}{1.216-(17.5-15)}} = 9.9 \cdot 10^{-4} | + | ^Jumper wires | 12.00| Material |
| - | \end{aligned} | + | ^Memory card |
| - | \end{equation} | + | ^Webcam |
| - | + | ^Wi-fi dongle | |
| - | We replace in formula with the above data and results: | + | ^Arduino |
| - | \begin{equation} | + | ^Water heater |
| - | \begin{aligned} | + | ^Water pump | ISEP| Material |
| - | {\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 \\ | + | ^Light led |
| - | {\frac{h_o \cdot L}{K_{air}}} = & a\cdot (1.5 \cdot 10^{8})^m \implies | + | ^Automatic fish feeder| |
| - | | + | ^Step motor | ISEP| Material |
| - | {\frac{h_o \cdot 0.77}{0.0255}} = & 0.59 \cdot (1.5 \cdot 10^{8})^{\frac{1}{4}} \\ | + | ^Temperature sensor| |
| - | h_o = & {\frac{0.0255}{0.77}} \cdot 0.59 \cdot (1.5 \cdot 10^{8})^{\frac{1}{4}} \\ | + | ^Water flow sensor | 12.24| Material |
| - | h_o = & 2.107 \ {\frac {W}{m^2 \cdot K}} \\ | + | ^Depth sensor |
| - | \end{aligned} | + | ^Infrared sensor |
| - | \end{equation} | + | </ |
| - | + | ||
| - | We return to equation: | + | |
| - | \begin{equation} | + | |
| - | \begin{aligned} | + | |
| - | | + | |
| - | \end{aligned} | + | |
| - | \end{equation} | + | |
| - | + | ||
| - | A< | + | |
| - | A< | + | |
| - | A< | + | |
| - | A< | + | |
| - | + | ||
| - | \begin{equation} | + | |
| - | \begin{aligned} | + | |
| - | | + | |
| - | | + | |
| - | \end{aligned} | + | |
| - | \end{equation} | + | |
| - | + | ||
| - | K< | + | |
| - | + | ||
| - | \begin{equation} | + | |
| - | \begin{aligned} | + | |
| - | | + | |
| - | {\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}=& | + | |
| - | q_{wall}=& | + | |
| - | | + | |
| - | \end{aligned} | + | |
| - | \end{equation} | + | |
| - | + | ||
| - | + | ||
| - | {{ : | + | |
| - | For water surface in direct contact with the air. | + | |
| - | D = 1.1 m | + | |
| - | L = 0.9·D = 0.99 \\ | + | |
| - | ρ< | + | |
| - | ΔT = 25-15 = 10 [ºC] \\ | + | |
| - | c< | + | |
| - | μ< | + | |
| - | K = 0.0257 [W/ | + | |
| - | 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 | + | |
| - | | + | |
| - | {\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} | + | |
| - | | + | |
| - | \end{aligned} | + | |
| - | \end{equation} | + | |
| - | + | ||
| - | \begin{equation} | + | |
| - | \begin{aligned} | + | |
| - | | + | |
| - | {\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}=& | + | |
| - | q_{ev}=& | + | |
| - | | + | |
| - | \end{aligned} | + | |
| - | \end{equation} | + | |
| - | + | ||
| - | \begin{equation} | + | |
| - | \begin{aligned} | + | |
| - | | + | |
| - | q_{total}=& | + | |
| - | | + | |
| - | | + | |
| - | \end{equation} | + | |
