Calculation for the Output of Solar Still of an Individual Hour

Abstract

A study has been done over the 9-h cycle for the solar still and calculates their output for an individual hour. The values of solar intensity, coefficient of heat transfer, ambient temperature, depth of water, and the coefficient of heat transfer through the walls of base and sides of the water basin are measured from the approximate method for each hour and later check their performance. In the previous methods, we required the graphical constructions and also constant coefficients of heat transfer assume but this thing is not required in this proposed method and observed the coefficients of heat transfer value for all the earlier time interval in the equations of heat balance is must. Actually, observed the two alternatives method of calculation is obtained evaporation rate for each an hour. From the energy balance equations, we compared the results from one which is found by mathematical solution. By using the energy balance equations, results which are calculated are different from the results obtained from the 9-h cycle of the approximate method and having the ±1.5% error. Water depth is 5–15 cms, coefficient of heat transfer in wind is 5–40 W/m² K and the coefficient of heat transfer through side and base walls is 0–3 W/m² K; these are variable ranges which taken for the analysis.

Appendices

Appendix 1

• In a basin-type solar still, the energy balance and the individual heat transfer coefficient

In Fig. 10.7, thermal circuit diagram for a basin-type solar still is mentioned, and it is because of Dunkle in 1961 [1]. The coefficient of heat transfer of radiation, convection, and evaporation is given by the following equation between the water and glass cover “(Duffie and Beckman [7], Dunkle [1], Kreith and Kreider [8]).”

Fig. 10.7

Thermal circuit diagram of a solar still

$$h_{\text{cwg}} = 0.884\left[ {T_{\text{w}} - T_{\text{g}} + \frac{{\left( {P_{\text{w}} - P_{\text{g}} } \right)\left( {T_{\text{w}} + 273} \right)}}{{268.9 \times 10^{3} - P_{\text{w}} }}} \right]^{1/3}$$

(10.12)

$$h_{\text{rwg}} = 0.9\sigma \left[ {\left( {T_{\text{w}} + 273} \right)^{2} + \left( {T_{\text{g}} + 273} \right)^{2} } \right]\left[ {T_{\text{w}} + T_{\text{g}} + 546} \right]$$

(10.13)

$$h_{\text{ewg}} = 16.273 \times 10^{ - 3} h_{\text{cw}} \frac{{P_{\text{w}} - P_{\text{g}} }}{{T_{\text{w}} - T_{\text{g}} }}$$

(10.14)

between the atmosphere and the glass cover of solar still, the coefficient of radiative heat transfer is given by the following equation,

$$h_{{{\text{rsg}}}} = \varepsilon _{{{\text{eff}}}} \sigma \left[ {\left( {T_{{\text{g}}} + 273} \right)^{2} + \left( {T_{{\text{s}}} + 273} \right)^{2} } \right]\left[ {T_{{\text{s}}} + T_{{\text{g}}} + 546} \right]$$

(10.15)

coefficient of radiative heat transfer is located to surrounding temperature particularly for flat plate collectors which is given by,

$$h_{\text{rga}} = \varepsilon_{\text{eff}} \sigma \left[ {\left( {T_{\text{g}} + 273} \right)^{2} + \left( {T_{\text{s}} + 273} \right)^{2} } \right]\left[ {T_{\text{s}} + T_{\text{g}} + 546} \right]\frac{{\left( {T_{\text{g}} - T_{\text{s}} } \right)}}{{\left( {T_{\text{g}} - T_{\text{a}} } \right)}}$$

(10.16)

for small values of (T_g − T_a); this method gives a large value of h_rga. The sink temperature is higher than the actual sink temperature means (T_a ˃ T_s); due to this, the large value of h_rga is obtained. The sink temperature is used in the earlier work of Sharma and Mullick [5] in 1991 as a sky temperature (T_s) for thermal network which is the lowest temperature. Among all calculations, this method gives the better results related to the present problem; during the nighttime, this approach gives better results under such conditions. The wind heat transfer coefficient is given by the following relation, because the sink temperature is used like ambient temperature,

$$h_{\text{cgs}} = h_{\text{w}} \frac{{\left( {T_{\text{g}} - T_{\text{a}} } \right)}}{{\left( {T_{\text{g}} - T_{\text{s}} } \right)}}$$

(10.17)

from the Swinbank 1963, the ambient temperature in this work is given by,

$$T_{\text{s}} = 0.0522 \times T_{\text{a}}^{1.5}$$

(10.18)

• Equation formed for Latent heat and Water vapor Pressure as the temperature function.

Keenan and Keyes [9] in 1936 gives an empirical relation for partial pressure of water for the temperature limit of (10–150) °C is shown by the below equation,

$$p = 165960.72 \times 10^{{ - \left[ {X\left( {a + bX + cX^{3} } \right)} \right]/[T(1 + dX)]}}$$

(10.19)

in the above equation, the constants a, b, c, d, and X having the values of 3.2437, 5.8682 × 10⁻³, 1.1702 × 10⁻⁸, 2.1878 × 10⁻³, and 647.27, respectively; temperature (T) is measured in Kelvin.

In this current work for latent heat of vaporization, the relation given by Sharma and Mullick in 1991 [5] is as follows,

$$h_{\text{fg}} = 3044205.5 - 1679.1109T_{\text{w}} - 1.14258T_{\text{w}}^{2}$$

(10.20)

Appendix 2

• Calculate the data of solar insolation and ambient temperature

For 24-h period, for the mathematical solutions, it is easier to have ambient temperature and solar intensity in form of continuous functions. From the Fourier series with the 6 harmonics, the data of Delhi of June month for the clear day is shown by,

$$\begin{aligned} I & = 7.691\cos (6w - 48.247r) + 18.089\cos (5w - 183.169r) \\ & \quad + 22.051\cos (4w - 235.161r) + 18.917\cos (3w - 4.459r) \\ & \quad + 157.995\cos (2w - 20.85r) \\ & \quad + 475.165\cos (w - 188.961r) + 306.055 \\ \end{aligned}$$

(10.21)

$$\begin{aligned} & T_{a} = 0.157\cos (6w - 25.516r) + 0.298\cos (5w - 299.941r) \\ & \quad + 0.3\cos (4w - 155.915r) + 0.045\cos (3w - 77.879r) \\ & \quad + 1.107\cos (2w - 351.466r) + 6.666\cos (w - 230.105r) + 36.70 \\ \end{aligned}$$

(10.22)

where radius (r) = π/180 and w = 2π/(24 × 3600).

so, continuous functions are shown in Fig. 10.2.