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| playground:playground [2015/05/26 11:36] – team4 | playground:playground [2015/06/12 14:14] (current) – team4 | ||
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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: \\ | + | FEEDER: |
| - | The heat transfer rate Equation \ref{eq: | + | |
| - | $q$ - 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 |
| - | $U$< | + | |
| - | $A$< | + | |
| - | $ΔT$ - Difference of temperature \\ | + | |
| - | \begin{equation} | + | TO INCLUDE THIS IDEA IN 8.2 |
| - | \begin{aligned} | + | |
| - | \dot{q} = U_0\cdot A_0\cdot \Delta T_{Total} \\ | + | |
| - | | + | |
| - | | + | |
| - | \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 | ||
| - | Unknown coefficient is $U$< | + | 8.2 |
| + | ¿¿¿As | ||
| - | \begin{equation} | + | Water quality is very important for fish health. The control of the water is based on temperature, |
| - | \begin{aligned} | + | |
| - | | + | |
| - | | + | |
| - | \end{aligned} | + | |
| - | \end{equation} | + | |
| - | $U$< | ||
| - | $x$ - thicknes of the material \\ | ||
| - | $h$< | ||
| - | $h$< | ||
| - | $A$< | ||
| - | $A$< | ||
| - | $A$< | ||
| - | $k$< | ||
| - | First box it is resistance of water and the second box is resistance of air.\\ | ||
| - | To find $h$< | ||
| - | The average natural convection heat transfer coefficient can be expressed by the following general Equation \ref{eq: | ||
| - | \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: | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | $N$< | ||
| - | $h$ - film coefficient of heat transfer\\ | ||
| - | $L$ - length [m]\\ | ||
| - | $k$ - thermal conductivity [W/m·k]\\ | ||
| - | $a$ - constant [(table)] \\ | ||
| - | $ρ$ - density [kg/ | ||
| - | $g$ - gravitational acceleration 9.81 [m/ | ||
| - | $β$ - volumetric coefficient of expansion of the fluid (Equation \ref{eq: | ||
| - | $ΔT$ - positive temperature difference between the wall and bulk fluid in [K]\\ | ||
| - | $μ$ - viscosity [kg/m·s]\\ | ||
| - | $c$< | ||
| - | $m$ - constant [(table)] \\ | ||
| - | $N$< | ||
| - | $N$< | ||
| - | \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 represents our case of heat flow through wall Figure {{ref> | ||
| - | < | ||
| - | <figure flabel199> | ||
| - | {{ : | ||
| - | < | ||
| - | </ | ||
| - | </ | ||
| - | The next calculations represent heat loss inside ($h$< | ||
| - | We take the general formula for natural heat transfer Equation \ref{eq: | ||
| - | \begin{equation} | ||
| - | \begin{aligned} | ||
| - | | ||
| - | | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | Characteristics of water at our specific temperature we take from a website [(watertp)] and we will interpolate where we need. | ||
| - | $L$ = 0.77 m \\ | ||
| - | $ρ$< | ||
| - | $ΔT$ = 25-15 = 10 [ºC] \\ | ||
| - | $c$< | ||
| - | $μ$< | ||
| - | $k$< | ||
| - | For $β$ Equation \ref{eq: | ||
| - | \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: | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | We take the general formula for natural heat transfer Equation \ref{eq: | ||
| - | \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}} \\ | ||
| - | \label{eq: | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | Characteristics of air at our specific temperature we take from a website [(airp)] and we will interpolate where we need. | ||
| - | $L$ = 0.77 m \\ | ||
| - | $ρ$< | ||
| - | $ΔT$ = 25-15 = 10 [ºC] \\ | ||
| - | $c$< | ||
| - | $μ$< | ||
| - | $k$< | ||
| - | For $β$ Equation \ref{eq: | ||
| - | \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: | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | We take the general formula for natural heat transfer Equation \ref{eq: | ||
| - | \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 | ||
| - | {\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: | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | We return to the Equation \ref{eq: | ||
| - | \begin{equation} | ||
| - | \begin{aligned} | ||
| - | | ||
| - | | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | $A$< | ||
| - | $A$< | ||
| - | $k$< | ||
| - | $A$< | ||
| - | $A$< | ||
| - | \begin{equation} | ||
| - | \begin{aligned} | ||
| - | | ||
| - | | ||
| - | | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | Now we know $h$< | + | - The reasons why we decided Raspberry Pi B + against BEAGLEBONE Black are:\\ |
| + | - | ||
| + | - - 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 | ||
| + | - - 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\\ | ||
| + | - - 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.\\ | ||
| + | - 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 | ||
| + | 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} | ||
| - | | ||
| - | | ||
| - | U_0 = & 2.053 \ {\frac{W}{m^2 \cdot k}} \\ | ||
| - | | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | From the heat transfer Equation \ref{eq: | ||
| - | |||
| - | \begin{equation} | ||
| - | \begin{aligned} | ||
| - | \dot{q}_{wall}=& | ||
| - | \dot{q}_{wall}=& | ||
| - | \dot{q}_{wall}=& | ||
| - | \label{eq: | ||
| - | \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 {{ref> | ||
| - | |||
| - | <figure flabel198> | ||
| - | {{ : | ||
| - | < | ||
| - | </ | ||
| - | |||
| - | 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] \\ | ||
| - | $c$< | ||
| - | $μ$< | ||
| - | $k$< | ||
| - | For $β$ Equation \ref{eq: | ||
| - | |||
| - | \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: | ||
| - | \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 | ||
| - | {\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: | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | |||
| - | To know the heat loss in watts we need again to know the coefficient $U$< | ||
| - | |||
| - | \begin{equation} | ||
| - | \begin{aligned} | ||
| - | & {\frac{1}{U_0}}= | ||
| - | & {\frac{1}{U_0}}= | ||
| - | & {\frac{1}{U_0}}= | ||
| - | & U_0= 3.695 \ {\frac{W}{m^2 \cdot k}} \\ | ||
| - | | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | |||
| - | From the heat transfer Equation \ref{eq: | ||
| - | |||
| - | \begin{equation} | ||
| - | \begin{aligned} | ||
| - | \dot{q}_{ev}=& | ||
| - | \dot{q}_{ev}=& | ||
| - | \dot{q}_{ev}=& | ||
| - | \label{eq: | ||
| - | \end{aligned} | ||
| - | \end{equation} | ||
| - | |||
| - | And we finish with Equation \ref{eq: | ||
| - | |||
| - | \begin{equation} | ||
| - | \begin{aligned} | ||
| - | | ||
| - | | ||
| - | | ||
| - | | ||
| - | \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. | ||
| + | Table {{ref> | ||
| + | <table tlabe100> | ||
| + | < | ||
| + | ^ Resource Name ^ Cost [€] ^ Type ^ Intials ^ Allocation [%] ^ (Std) Rate [€/h] or [€/unit]^ Overtime [€/ | ||
| + | ^ Arick | 1 500.00| Work | | ||
| + | ^ Francisco | ||
| + | ^ Jan | 1 500.00| Work | | ||
| + | ^ Katoo | 1 500.00| Work | | ||
| + | ^ Rasmus | ||
| + | ^ Viorel | ||
| + | ^Raspberry Pi | 31.95| | ||
| + | ^Arduino | ||
| + | ^USB cable Raspberry Pi| 2.25| Material | ||
| + | ^USB cable Arduino | 2.85| Material | ||
| + | ^Power supply Raspberry Pi| 12.24| | ||
| + | ^Jumper wires | 12.00| | ||
| + | ^Memory card | ||
| + | ^Webcam | ||
| + | ^Wi-fi dongle | ||
| + | ^Arduino 4 channel relay module| | ||
| + | ^Water heater | ||
| + | ^Water pump | ISEP| Material | ||
| + | ^Light led | ||
| + | ^Automatic fish feeder| | ||
| + | ^Step motor | ISEP| Material | ||
| + | ^Temperature sensor| | ||
| + | ^Water flow sensor | 12.24| | ||
| + | ^Depth sensor | ||
| + | ^Infrared sensor | ||
| + | </ | ||
