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

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report [2016/06/14 12:47] – [Acknowledgement] team5report [2017/04/29 17:25] (current) – [7.4 Thermodynamics] team5
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 Parameters for the calculations:  Parameters for the calculations: 
   * H0 : Convection of heat transfer coefficient from cover surface to atmosphere   * H0 : Convection of heat transfer coefficient from cover surface to atmosphere
-    * Tc : Temperature of transparent cover +  * Tc : Temperature of transparent cover 
-    * Ta : Temperature of atmosphere +  * Ta : Temperature of atmosphere 
-  *  σ : Constant of Stefan-Boltzmann +  * σ : Constant of Stefan-Boltzmann 
-  *  ec : emissitivity factor for radiation from cover to atmosphere. For this the PMMA emissitivity is 0.84 +  * ec : Emissivity factor for radiation from cover to atmosphere. For this the PMMA emissivity is 0.84 
   * Hi : Convection of heat transfer coefficient from seawater container to cover surface   * Hi : Convection of heat transfer coefficient from seawater container to cover surface
   * Tb : Temperature of seawater in basin   * Tb : Temperature of seawater in basin
-  * Eb,c : Emissivity factor for radiation from seawater to cover+  * Eb,c: Emissivity factor for radiation from seawater to cover
   * D : Distillate outflow rate (kg/m/m²) from base   * D : Distillate outflow rate (kg/m/m²) from base
-  * λ : Enthalpy of condensation of water vapor is 2264 KJ/Kg+  * λ : Enthalpy of condensation of water vapor is 2264 kJ/kg
  
  
   - //Energy balance around the cover of the distiller and surrounding//   - //Energy balance around the cover of the distiller and surrounding//
  
-Q= Ho . Ac  . (Tc – To) + σ . e . Ac (Tc4 – To4) (1) +Q= Ho * Ac * (Tc – To) + σ * e * Ac (Tc4 – To4) (1)
  
 Convection Convection
-Ho . Ac  . (Tc – To) (2)+Ho * Ac * (Tc – To) (2)
  
 Radiation Radiation
-σ . e . Ac (Tc4 – To4) (3)+σ * e * Ac (Tc4 – To4) (3)
  
   - //Energy balance around the cover of the distiller and the inside of the system//   - //Energy balance around the cover of the distiller and the inside of the system//
  
-Q= Ho . Ac  . (Tb – Tc) + σ . e . Ac (Tb4 – Tc4) + D . λ + I . αc  (4)+Q= Ho * Ac * (Tb – Tc) + σ * e * Ac (Tb4 – Tc4) + D * λ + I * αc  (4)
  
 Convection Convection
-Hi . Ac . (Tb – Tc) (5)+Hi * Ac * (Tb – Tc) (5)
  
 Radiation Radiation
- σ . e . Ac (Tc4 – To4) (6)+ σ * e * Ac (Tc4 – To4) (6)
  
 Assumptions are taken for the calculations. The area of the seawater container, white base and cover stays equal. The cover and condensate film are at a temperature of Tc, the temperature Tb is uniform throughout the basin, liquid and vapor leakage is negligible and the cover condensate are opaque to thermal radiation from the seawater container. Over-all energy balance around 1m² of distiller is:  Assumptions are taken for the calculations. The area of the seawater container, white base and cover stays equal. The cover and condensate film are at a temperature of Tc, the temperature Tb is uniform throughout the basin, liquid and vapor leakage is negligible and the cover condensate are opaque to thermal radiation from the seawater container. Over-all energy balance around 1m² of distiller is: 
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 I(1-εr) = H0 (Tc – Ta ) + σ . ec. Ac (Tc4 – Tr4) +D (Tc – Ts) + B (Tb – Ts ) + Σ L (7)  I(1-εr) = H0 (Tc – Ta ) + σ . ec. Ac (Tc4 – Tr4) +D (Tc – Ts) + B (Tb – Ts ) + Σ L (7) 
  
-  * I : Solar radiation rate on horizontal surface [W/m2 ]+  * I : Solar radiation rate on horizontal surface [W/m²]
   * R: Loss of solar energy by reflection    * R: Loss of solar energy by reflection 
   * B: water outflow rate    * B: water outflow rate 
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 In this equation the solar radiation absorbed in the seawater container and on the distiller bottom per hour is equated to the sum of the heat transferred from the cover to the atmosphere by convection and radiation, the sensible heat carried out in the hot seawater and the warm condensate and heat losses, including the heat transferred through the bottom of the basin. The energy balance is written with the incoming saline water at the reference temperature and there is a further assumption that the condensate leaves the distiller substantially at the cover temperature. In this equation the solar radiation absorbed in the seawater container and on the distiller bottom per hour is equated to the sum of the heat transferred from the cover to the atmosphere by convection and radiation, the sensible heat carried out in the hot seawater and the warm condensate and heat losses, including the heat transferred through the bottom of the basin. The energy balance is written with the incoming saline water at the reference temperature and there is a further assumption that the condensate leaves the distiller substantially at the cover temperature.
- 
- 
    
 === - Practical concerns === === - Practical concerns ===
 The efficiency of the process relies on the mass of condensated water [kg/m³]. This can be calculated by  The efficiency of the process relies on the mass of condensated water [kg/m³]. This can be calculated by 
-∆m= - ∆es/(Rv . Tlowest)+∆m = - ∆es/(Rv * Tlowest)
 With es is the difference of water vapor pressure of the highest and the lowest temperature [bar]; With es is the difference of water vapor pressure of the highest and the lowest temperature [bar];
-Rv is the gas constant of water vapor [461.5 J (K Kg)-1];+Rv is the gas constant of water vapor [461.5 J (K.kg)-1];
 And the temperature at the lowest value [K] And the temperature at the lowest value [K]
  
-es = 6.1078 . 10 ^(7.5 . Ti/ (237.9+Ti) (8)+es = 6.1078 * 10 ^(7.5 * Ti/ (237.9+Ti) (8)
 with T [°C] with T [°C]
  
 The difference of water vapor pressure also can be calculated by the difference of Relative Humidity inside the system by:  The difference of water vapor pressure also can be calculated by the difference of Relative Humidity inside the system by: 
-RH = (e/es) . 100 (9)+RH = (e/es) * 100 (9)
 The e is the real vapor pressure inside the system The e is the real vapor pressure inside the system
  
 === - Water Pyramid Test Result === === - Water Pyramid Test Result ===
  
-Once the prototype was ready the team made some tests during 2h on the roof of the ISEP: +Once the prototype was ready the team made measured during 2h on the roof of ISEP: 
  
-  * Temperature on the Arduino Board: [45,89 ; 53,11] °C +  * Temperature on the Arduino Board: [45.89;53.11] °C 
-  * Pression on the Arduino Board: [1004,79 ; 1006,11] hPa +  * Pression on the Arduino Board: [1004.79;1006.11] hPa 
-  * Temperature inside the brine, on the inside wall and outside wall of the pyramid, Figure {{ref>flabel7313}} +  * Temperature inside the brine, on the inside wall and outside wall of the pyramid (Figure {{ref>flabel7313}}) 
-  * Humidity inside the pyramid+  * Relative Humidity inside the pyramid
  
 <WRAP centeralign> <WRAP centeralign>
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 </WRAP> </WRAP>
  
-  * Humidity inside the pyramid, figure {{ref>flabel7314}}+  * Humidity inside the pyramid (Figure {{ref>flabel7314}})
  
-This test was made to ensure the big difference between the temperature of the inside of the pyramid, the outside of the pyramid and the brine itself. In Fig. 46 a graph is shown which make clear that the Temperature from the inside of the pyramid is approximately 10 °C higher than the Brine temperature and approximately 18 °C higher is than the outside temperature. By this, the team can conclude that the process can work. Namely, the evaporation of the Brine will occur and the vapor can condense at the PMMA because the outside of the pyramid is colder than the inside of the pyramid. +This test was made to ensure the big difference between the temperature of the inside of the pyramid, the outside of the pyramid and the brine itself. In Figure {{ref>flabel7313}} a graph is shown which make clear that the Temperature from the inside of the pyramid is approximately 10 °C higher than the Brine temperature and approximately 18 °C higher is than the outside temperature. By this, the team can conclude that the process can work. Namely, the evaporation of the Brine will occur and the vapor can condense at the PMMA because the outside of the pyramid is colder than the inside of the pyramid. 
  
 <WRAP centeralign> <WRAP centeralign>
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 </WRAP> </WRAP>
  
-The Relative Humidity stays constant at approximately 60% in the inside of the pyramid, as seen in  {{ref>flabel7314}. The higher the Relative Humidity, the more brine evaporated. +The Relative Humidity stays constant at approximately 60 % inside of the pyramid (Figure {{ref>flabel7314}}). The higher the Relative Humidity, the more brine evaporated. 
  
 After this test, another test of 24 h was made. This for two reasons:  to make calculations and to compare the results with the experiment itself, and to get the difference between the minimum and maximum temperature of the inside-, outside of the system and the brine.  After this test, another test of 24 h was made. This for two reasons:  to make calculations and to compare the results with the experiment itself, and to get the difference between the minimum and maximum temperature of the inside-, outside of the system and the brine. 
  
-The results of the graph in Figure {{ref>flabel7771}} (A) shows the big difference during day and night. This dif-ference is important for the evaporation and condensation process. By day, the brine evap-orates. By night, the vapor condenses. This can also be proven by the Relative Humidity graph in Figure{{ref>flabel7771}} (B). During the evening/night the Relative Humidity increase up to almost 95 % and after this, after condensing, the Relative Humidity decreases. +The results of the graph in Figure {{ref>flabel7771}} (A) shows the big difference during day and night. This difference is important for the evaporation and condensation process. By day, the brine evaporates. By night, the vapor condenses. This can also be proven by the Relative Humidity graph in Figure {{ref>flabel7771}} (B). During the evening/night the Relative Humidity increase up to almost 95 % and after this, after condensing, the Relative Humidity decreases. 
  
 <WRAP centeralign> <WRAP centeralign>
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 With the results of the graphs it possible to calculate the percentage evaporated water.  With the results of the graphs it possible to calculate the percentage evaporated water. 
-  - Es1 = 122.21 = 6,1078 * 10 ^(7,5 * Tmaxbrine/ (237,9+Tmaxbrine)) +  - Es1 = 122.21 = 6.1078 * 10 ^(7.5 * Tmaxbrine/ (237.9+Tmaxbrine)) 
-  - Es2 = 20.41 = 6,1078 * 10 ^(7,5 * Tminbrine/ (237,9+Tminbrine)) +  - Es2 = 20.41 = 6.1078 * 10 ^(7.5 * Tminbrine/ (237.9+Tminbrine)) 
   - ∆Es = 101.80 = Es1-Es2   - ∆Es = 101.80 = Es1-Es2
-  - ∆m = 0.012 kg = Tmin inside/(461,5 * ∆Es) +  - ∆m = 0.012 kg = Tmin inside/(461.5 * ∆Es) 
-  - Water evaporated = 12.75 ml = ∆m *1000 ml/kg+  - Water evaporated = 12.75 ml = ∆m * 1000 ml/kg
   - % evaporated = 6.37 %   - % evaporated = 6.37 %
-In this experiment only 6.37 % evaporated. To increase the percentage evaporated, it is important to make the higher difference between the temperature . In future development it is necessary to take this in consideration. +In this experiment only 6.37 % evaporated. To increase the percentage evaporated, it is important to make the higher difference between the temperature. In future development it is necessary to take this in consideration. 
  
  

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