The Design and Development of a Solar Dehydrator for Fruits †

: An active solar dehydrator is designed for dehydrating fruits and vegetables. The solar collector transfers thermal energy to the drying air. The temperature inside the drying chamber is maintained between 45 ◦ C and 50 ◦ C by a feedback control system. The design is based on the December solar irradiance of Islamabad, which is found to be 572 watts per square meter. The energy required to dry one kilogram of apple and banana is 245 watts and 220 watts, respectively. The dried product from the solar dehydrator is better in terms of quality and taste compared to the product produced by ordinary open sun drying.


Introduction
Dehydration adds worth to fresh vegetables and fruits by eliminating water, reducing their weight, and increasing the time of usability.It also opens a window for the farmers to sell their surplus produce to local and international markets that would otherwise be wasted due to harsh weather conditions.Drying lessens the burden and quantity of the product, making it easy to keep and use.This will finally lower the cost of storing, packaging, and transportation.

Literature Review
The losses of fruit and vegetables are in the order of 40%, and under extremely hindering circumstances, they are as high as 80% [1].A major percentage of losses is either due to inaccurate or early drying of natural product chips (banana, mango, apple, etc.) [2].The indirect passive solar drying system was designed and built using convective heat flow principles [3].For the selected plants, a direct correlation between item temperature and drying rate was postulated.The proposed experimental relationship can depict the drying energy of the picked crops well [4].The cost of dehydrating chili per kilogram was reduced by 39% in the case of a double-pass solar drier [5].In the indirect solar dehydrator, the incoming air is heated using solar energy in a separate solar receiver and sent to the drying chamber.A consistent condition of evaporation is maintained with the flow of heat [6].

Design Methodology
Andress, Elizabeth L., and Judy A. [7] tabulated the dehydration temperature for fruits and vegetables.The current study is related to banana and apple.Accordingly, the moisture content of apple is 85.9%, and that of banana is 74%.Both dehydration temperature is to be maintained at 45 to 50 • C. Using these data, the moisture amount to be removed from the apple is calculated [8].In general, the total mass (m) of one standard-size apple is taken as 200 g, and the moisture content is 85.9% by mass (w = 0.859).On a dry basis, the mass of the apple is calculated to be m o = 107.58g.Hence, the mass of water present is =200 − 107.58 = 92.42g.The equilibrium dry mass of an apple with moisture content is 16% by mass [7].So, the mass of the apple on a dry basis is calculated as m = 124.79g.Hence, the water content present at equilibrium state = 124.79− 107.58 = 17.21 g and the mass of water to evaporate per 200 g of sample = 92.42− 17.21 = 75.21g Now, repeating the same process for banana, the mass (m) of one standard-size banana is taken as 120 g, the moisture content is 74 percent by mass, and the equilibrium moisture content in dried banana is 13% [7].Hence, the mass of banana on a dry basis = 77.92g, the water content present at equilibrium state = 77.92− 68.96 = 8.96 g, and the mass of water to evaporate per 120 g of sample = 51.03− 8.96 = 42.07 g

The Volumetric Air Flow Rate Required to Dry One Kilogram of Apple
Unsaturated air is passed over wet material to take away water from the material being dried.This water is evaporated a drying chamber (Figure 1).The volume of air (V) is cooled in the process of evaporating a mass of water from the fruit sample.
Eng. Proc.2023, 45, x 2 of 5 apple is taken as 200 g, and the moisture content is 85.9% by mass (w = 0.859).On a dry basis, the mass of the apple is calculated to be mo = 107.58g.Hence, the mass of water present is =200 − 107.58 = 92.42g.The equilibrium dry mass of an apple with moisture content is 16% by mass [7].So, the mass of the apple on a dry basis is calculated as m = 124.79g.Hence, the water content present at equilibrium state = 124.79− 107.58 = 17.21 g and the mass of water to evaporate per 200 g of sample = 92.42− 17.21 = 75.21g Now, repeating the same process for banana, the mass (m) of one standard-size banana is taken as 120 g, the moisture content is 74 percent by mass, and the equilibrium moisture content in dried banana is 13% [7].Hence, the mass of banana on a dry basis = 77.92g, the water content present at equilibrium state = 77.92− 68.96 = 8.96 g , and the mass of water to evaporate per 120 g of sample = 51.03− 8.96 = 42.07 g

The Volumetric Air Flow Rate Required to Dry One Kilogram of Apple
Unsaturated air is passed over wet material to take away water from the material being dried.This water is evaporated in a drying chamber (Figure 1).The volume of air (V) is cooled in the process of evaporating a mass of water from the fruit sample.
In one kilogram of apple, there are approximately five (5) apples.So, for the water content in one kilogram of apple sample, the mass of water removed per kg is mw = 5 (75.

The Heat Energy Required to Dry One Kilogram of Apple
The amount of energy transferred through the air is equal to the amount of energy absorbed by the water.The energy (Q) required to dry the sample is given by where mw is the mass of water to be removed, and LH is the latent heat of vaporization.So, the energy required to dry one kilogram of apple is To dry it in one hour, the heat rate is given Q = 849.76kJ/1 h = 849.76kJ/h = 245 W. (1) So, to dehydrate a kilogram of apples in one hour, an air flow rate of 61.57m 3 /h is required.

The Heat Energy Required to Dry One Kilogram of Apple
The amount of energy transferred through the air is equal to the amount of energy absorbed by the water.The energy (Q) required to dry the sample is given by where m w is the mass of water to be removed, and L H is the latent heat of vaporization.To dry it in one hour, the heat rate is given Q = 849.76kJ/1 h = 849.76kJ/h = 245 W. So, the energy required for apple is 245 W. Similarly, for banana, the energy required is 220 W.

Solar Collector Design
Based on the design methodology discussed by Twidell, J., and Weir, T., the solar collector is designed to meet energy requirements for one kilogram of apple when placed inside a dehydrating chamber.For the collector, the efficiency is η c = 60%.Total useful power is given by [8]: The solar irradiance on the first June is G hmax = 885 w/m 2 , and solar irradiance on the first December is G hmax = 572 w/m 2 [8].Using irradiance G = 572 w/m 2 and the power required for the apple (calculated above), the energy needed to dry is P u = 245 watts.So, the area required is A = 0.71 m 2 .Hence, the collector is designed for the month of December.The dimensions used for the collector are length 1.2 m and width 0.6 m (A = 0.72 m 2 ).When the blower is not operating, the flow is stagnated, and the maximum temperature of the plate is observed.The stagnation temperature of the plate in June is 105 • C, and in December, it is 94.6 • C.

Experimental Setup and Results
An active solar dehydrator is designed with the wood (Figure 2).The drying chamber maintains a temperature range of 45 to 50 • C, with two temperature sensors: one at the inlet and one inside the chamber.The sensor controls the speed of blowers and the exhaust fan's on/off mechanism.When the chamber temperature is 45 • C or higher, blowers operate at maximum speed, and speed decreases with temperature.The exhaust fan starts working at 50 In the open sun and dehydrator case, samples are weighed to determine moisture content, impacting color, taste, and overheating, leading to decolorization.Dusty environment also contributes to the end product as shown in Figure 3.While in the active solar dehydrator, the drying is faster.The product obtained from an active solar dehydrator is better in terms of quality and hygiene, as shown in Figure 4.While in the active solar dehydrator, the drying is faster.The product obtained from an active solar dehydrator is better in terms of quality and hygiene, as shown in Figure 4.While in the active solar dehydrator, the drying is faster.The product obtained from an active solar dehydrator is better in terms of quality and hygiene, as shown in Figure 4.

Conclusions
Almost 300 watts per kilogram is required to dry one kilogram of sample that has a moisture content of around 80%.In the active solar dehydrator, the drying is faster.The product is dried in 2 h in a solar dehydrator compared to the more than four hours needed for open sun drying.

Conclusions
Almost 300 watts per kilogram is required to dry one kilogram of sample that has a moisture content of around 80%.In the active solar dehydrator, the drying is faster.The product is dried in 2 h in a solar dehydrator compared to the more than four hours needed for open sun drying.

Figure 1 .
Figure 1.Active solar dehydrator: (a) Schematic diagram; (b) Actual system.In one kilogram of apple, there are approximately five (5) apples.So, for the water content in one kilogram of apple sample, the mass of water removed per kg is m w = 5 (75.2) = 376 g.So, the volume of air required to dry one kilogram of apple is V = (m w L H )/(ρ C p ∆T) = ((0.376)(2260))/((1.15)(1.0)(50− 38)) = 61.57m 3(1) So, the energy required to dry one kilogram of apple is Q = (0.376) (2260) = 849 kJ Eng.Proc.2023, 45, 48 3 of 4 The overall dimensions of the active solar dehydrator are length = 1.5118 m, width = 0.6096 m, and height = 1.21 m.The dimensions of the solar collector are 1.206 m long and 0.6069 m wide.The interior of the solar dehydrator is painted black to enhance the absorption of heat energy.

Figure 2 .
Figure 2. Reduction in moisture content with dehydration time.