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Journal of Aerospace Technology and Management ISSN: 1984-9648 [email protected] Instituto de Aeronáutica e Espaço Brasil

Buchireddy Sri, Karthik Reddy; Aneesh, Poondla; Bhanu, Kiran; Natarajan, M Design Analysis of Solar-Powered Unmanned Aerial Journal of Aerospace Technology and Management, vol. 8, núm. 4, octubre-diciembre, 2016, pp. 397-407 Instituto de Aeronáutica e Espaço São Paulo, Brasil

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Design Analysis of Solar-Powered Karthik Reddy Buchireddy Sri1, Poondla Aneesh1, Kiran Bhanu 1, M Natarajan1

ABSTRACT: One of the main problems in micro unmanned aerial is endurance or flight time since the general Introduction domain uses conventional fuel, which is a pollutant, has a limited life and is costly. Then, there is a huge demand In today’s world, there are more than 11,000 unmanned for using an unlimited non-exhaustible source of energy. is one of the unlimited available which aerial vehicles (UAVs) in service (or planned for future services) can be used to increase the endurance of unmanned aerial by the Military for various purposes (Kumar et al. 2015).These vehicle without adding significant mass or increasing the size vehicles can fly remotely as well as autonomously. In spite of of the fuel system. This paper aims to encourage research on their usage in various applications, they lack in performance renewable energy sources for aviation considering the basic challenges for a solar-: geographical area of because of power restrictions, which means they either have to operation, energy collection and storage, payload and design land to refuel or to depend on another UAV to complete the task. parameters. Hence, a plane is designed for 2 kg, including Nowadays the ability to fly without using conventional fossil payload, and is analyzed in various aspects. Besides, the design is optimized starting from airfoil to complete structure fuels is primarily focused in recent years, both in application for better performance. point of view and scientific field since the major concerns are an increase in global warming and a decrease in natural resources. KEYWORDS: , Unmanned aerial vehicle, Since then, the use of electric aircraft has been widespread but Renewable energy, Endurance, Conceptual design. here the crucial issue is their high power consumption when compared with their limited capability, which leads to an endurance that can rarely exceed half an hour. Increasing the size of the battery or incorporating more batteries increases the weight of the plane, which affects the flight time of the UAV as it is inversely proportional. So, now, the ability to fly an aircraft to increase the endurance limit has been a key issue in both UAVs and civilian aviation. One of the possibilities to increase the flight time is by using unlimited solar energy through solar cells. These cells, by connecting them with an electronic circuit, can provide sufficient power for motor and the electronics and, if in excess, it can also be stored in the battery; here the battery can be used as a buffer when flying in darkness or under clouds. Hence, the possible solution to enhance the endurance is by using solar- powered aircraft driven by electric-based propulsion systems in which the power is supplied continuously throughout the

1.VIT University – School of Mechanical Engineering Mechanical Department – Vellore/Tamil Nadu – India. Author for correspondence: Karthik Reddy Buchireddy Sri | VIT University – School of Mechanical Engineering Mechanical Department | Near Katpadi Road | Vellore – Tamil Nadu 632014 – India | Email: [email protected] Received: 04/17/2016 | Accepted: 06/21/2016

J. Aerosp. Technol. Manag., São José dos Campos, Vol.8, No 4, pp.397-407, Oct.-Dec., 2016 Reddy BSK, Aneesh P, Bhanu K, Natarajan M 398

day by the unlimited solar energy, which can eliminate fuel (2 kg) is kept constant (Table 1). There was no range requirement and also solve the limited energy storage capability problem. since the goal was simply to remain flying, not to reach any This paper studies the complete design of a solar-powered destination. UAV, which is mostly designed for low-altitude long-endurance Table 1. Mission specifications. (LALE) applications. Parameter SI units The main principle is to make use of available unlimited solar energy by converting it into through solar Gross weight 2 kg cells. When sunlight strikes the solar cell, the cell creates Payload 0.5 kg electrons and holes as charge carriers and, when a circuit is Altitude 30 – 50 m –3 made, the free electrons pass through a certain load in order to Average air density 1.22 kg∙m recombine with holes and, in this way, the current is generated. Clearness factor 0.7 (1 = clear sky) Here, by arranging solar cells in series on the top of the wing Take-off distance None (hand toss launch) and then wrapping the entire wing with transparent material for the safety of solar cells during flight, cells are arranged Estimation of the Number of Solar Cells in series to get the required voltage in order to safely charge Required and its Arrangement on the Wing the 3S battery; from there, the battery power is supplied For this design, 3S battery has been chosen for two reasons: to the motor for throttling during constant level flight. In this using more lithium polymer (LiPo) cells in series (> 3S) requires study, the aircraft was assumed to be a , which means it more voltage to initiate the charging process, leading to an also stores the energy in the battery during the period. increase in solar cells and wingspan, which, in turn, increases In this way, both flying the plane by using solar energy alone the weight of the model. On the other hand, if less LiPo cells and storing the energy in the battery in order to extend the in series (2S) are used, it will be difficult to supply the required flight time can be achieved. power to achieve a good climb rate. As for charging the 3S battery, it requires a constant DESIGN METHODOLOGY safe charging voltage of about 12.4 V. This voltage has to be The design process of the entire aircraft is composed supplied by the solar cells; those selected for this design were of two phases: conceptual and preliminary. In conceptual power C-60 photovoltaic. The Sun power cells were more design, the basic configuration arrangement, size, weight, efficient than most silicon-based ones. The 22% efficiency of and performance are calculated. Then, designing the CAD the C-60 (Table 2)was significantly better than typical silicon modeling of individual parts in order to meet the required solar cells with efficiencies around 15% (Kederoglu et al. 2012). targets and optimizing it are done in preliminary design. According to Table 2, each solar cell gives out 0.57 V, which The design requirements are used to guide and evaluate the means 22 solar cells are required to meet the targeted voltage; development of the overall aircraft configuration. There are here, two more cells are added for safe side. These 24 cells two simple balances: should be connected in series to achieve the required voltage. • Weight balance: the lift force has to be equal to the Since the number of cells is fixed, a minimum area of the weight of all the elements constituting the . wing is calculated. Two possible combinations in arranging • Energy balance: the energy that is collected during a Table 2. Specifications of sun power C-60 solar cell. day from the has to be equal to or higher than the electrical energy needed by the airplane Parameter SI unit during its level flight. Mass of Solar cell 0.008 kg Length and width 0.125 × 0.125 m Mission Specifications Area of single solar panel 0.0150 m2 One of the main aims of this paper is to check whether the Efficiency of solar cell 22% power available from the Sun is required to power an aircraft, which, Rated voltage 0.57 V in turn, helps in increasing the endurance of the aircraft from a Rated current 5.37 A couple of minutes to hours. Here the gross weight of the plane Source: sunpowercorp.com

J. Aerosp. Technol. Manag., São José dos Campos, Vol.8, No 4, pp.397-407, Oct.-Dec., 2016 Design Analysis of Solar-Powered Unmanned Aerial Vehicle 399

these cells in series (Type 1 and Type 2) over the wing can be all parameters. The ends of the wing are tapered from a root seen in Fig. 1. chord of 30 cm to a tip one of 20 cm over a length of 50 cm. The aspect ratio of the entire wing is 9.36. (a) Airfoil Selection After finalizing the wing design, the airfoil is selected based on the following requirements: high lift coefficient; low drag (b) coefficient; less camber for cell placement. Here, the 4 airfoils are analyzed, which are taken from previous designs (Bruscoli 2011). The approach to find the airfoil lift and drag is shown in Fig. 3, where the input parameters are aspect ratio (AR) and wingspan (b). The airfoils can be seen in Figs. 4. Figure 1. (a) Single row of cells — high-aspect-ratio wing For these 4 airfoils, XFLR analysis has been done to select (Type 1); (b) Two rows of cells — medium-aspect-ratio wing (Type 2). the airfoil with the best performance characteristics. The analysis is done for 30-cm chord length and by keeping the 4-degree The problems in Fig. 1a are: angle of attack for all airfoils. • The wing is two long, which requires a tough structure, In Fig 5, various graphs have been plotted by comparing increasing the weight and cost. the 4 airfoils for the 4-degree angle of attack (α), and the • Controlling will be difficult as it does not have ailerons. important parameters lift (CL) and drag (CD) coefficients are • While connecting the circuit with tapping wire, the 2 ends shown in Table 3. Hence, according to this table, Airfoil-4 of the wire are too far, resulting in loss of power efficiency. WE3.55 is chosen for its best performance as it has high

In Fig. 1b, all these problems are solved as the solar cells CL/CD ratio, which is the same airfoil designed for Sky-Sailor are placed only on the mid wing; the end wings are for stability (Noth 2008). with a polyhedral angle of about 7° on both sides, which also includes ailerons for easy control. In Fig. 2, it is observed that, Input AR,b in Type 1, more tapping wire is used in order to close the circuit, which means more loss while transferring the current. In Calculate li , Type 2, this problem is nullified as we can close the circuit with Reynolds number Input airfoil the usage of very less wire. As each solar cell has 0.125 m in length, for 12 solar cells Create 2-D li curve, in a row, a wingspan (middle portion) of 1.5 m is required. drag polar pitch in moment curve However, apart from this length, extra space is required for the soldering purpose. Besides, at the ends, at least 2 – 3 cm should 3-D sweep taper analysis be left because there will be abnormal forces at the junction. Then, a central wingspan of 1.63 m was chosen by considering No Optimum airfoil Type 1 Yes

Create 3-D + – li and drag Type 2

+ – Output airfoil Wing’s urface Solar cell Tapping wire li and drag input

Figure 2. Arrangement and connection of solar cells on the wing. Figure 3. Flowchart to find airfoil lift and drag.

J. Aerosp. Technol. Manag., São José dos Campos, Vol.8, No 4, pp.397-407, Oct.-Dec., 2016 Reddy BSK, Aneesh P, Bhanu K, Natarajan M 400

Circuit Diagram (a) From the circuit diagram shown in Fig. 6, the energy received 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 from the Sun is collected by the solar panel, which converts it into electricity and then there is the maximum power point (b) tracker (MPPT), which helps in tracking the maximum power 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 of any solar cell and provides it at all times. In MPPT, there are 3 terminals: one is connected to the solar panel; the other, to load (c) (motor); and the last one, to the battery. Th en, in level fl ight, 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 the MPPT sends power directly to the motor from the solar cells and, when gliding, as the motor does not require power, (d) the battery starts charging. If an excess of power is required, 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 during climbing or when the solar intensity is low, the battery Figure 4. (a) Aquila 9.3%; (b) Medium S9000 (9%); supplies the required power, which is stored into the motor. (c) S9037 (9%); (d) WE3.55. Th us, from the battery, the energy goes to the motor, which rotates the and, between them, there is an electronic speed control (ESC) regulating the speed. Besides the present 1.2 1.0 battery, there is an additional small pencil battery to control

Cl 0.8 the avionics; the reason to separate it is, even if there is any Aquila 9.3% smoothed 0.6 S9000 (9%) problem in the main circuit or else, any damage in the cells still 0.4 S9037 (9%) 0.2 WE3.55 controlling the plane can be done as the control system’s battery 0.0 is not connected to the main circuit, which helps in safe landing. 0.00 0.02 0.04 0.06 0.08 0.10 0.12 Cd

1.2 1.2 Servo 1 Battery 2 1.0 1.0 BEC Receiver (aileron) 0.8 0.8 l l Sun

C Servo 2 0.6 C 0.6 0.4 0.4 (elevator) 0.2 0.2 Solar cells MPPT Battery 1 0.0 0.0 Servo 3 0.0 5.0 10.0 15.0 0.0 0.2 0.4 0.6 0.8 (rudder) α X ESC 80 tr1 –0.02 70 60

–0.04 d 50 m / C Motor –0.06 l 40 C C –0.08 30 BEC: Battery eliminator circuit. 20 –0.10 10 Figure 6. Circuit diagrams for entire plane. 0.0 5.0 10.0 15.0 0.0 5.0 10.0 15.0 α α pOWER REquIRED fOR LEvEL fLIGHT Figure 5. Comparison of airfoil performance. At steady level fl ight, the lift force generated by the wing table 3. Comparison of the 4 selected airfoils. exactly compensates for the weight, and the propeller thrust, for the drag force (Simons 1994). Medium Aquila S9037 parameter S9000 WE3.55 9.3% (9%) ρ 2 (1) (9%) W=mg=L= CLSV (1) ρ C 0.83 0.75 0.83 0.91 2 L where: m is the mass of the aircraft (kg);L is the lift force; C 0.013 0.01127 0.116 0.0115 D 2 –3 ρ is theT= airD= densityCDSV (kg∙m ); S is the surface area of the wing (2) C /C 64 66.5 71 78.5 2 –1 L D (m ); V is theρ cruise velocity of the aircraft (m∙s ). 2 J. Aerosp. Technol. Manag., São José dos Campos, Vol.8, No 4, pp.397-407, Oct.-Dec., 2016

CDi= 2 ; CD = cd + CDi (3) 𝐶𝐶𝐿𝐿 π∗e∗AR

Preq = T×V (4)

δ =23.45 sin (360 ) (5) 284+n 365

-1 ωs = cos (tanϕ × tanδ) (6)

2 Ho= Isc × 3600(1 + 0.033(cos ))(ωs.sinϕ.sinδ+cosϕ.cosδ.sinωs) kJ/m ∙day (7) 24 360n π 365

-1 Smax= cos (−tanϕ.tanδ) (8) 2 15

= a + b (9) H S Ho (Smax) ω = (15h − 180) (10)

Ig= (c + d.cosω) ] × H (11) π cosω−cosws πws 24 s s [ sinw − 180 cosw

λ = (12) 𝐶𝐶𝑡𝑡 𝐶𝐶𝑟𝑟

Vh = (13) Sh∗Lh S∗c Design Analysis of Solar-Powered Unmanned Aerial Vehicle 401 222 W=mg=L= W=mg=L= CCLLLSVSV (1)(1) (1) ρρρ 222 W=mg=L=The W=mg=L= W=mg=L= axial2 2thrust2 CC LCisSVL LSVcalculatedSV as: efficiencies (1) involved(1) (1) must be included: propeller efficiency ρρρ of 80%, motor efficiency of 80%, ESC efficiency of 70% ρ 222 2 T= T= D= D=CCDDSVSV (2) and the (2)(2) (2)battery discharge efficiency of 90%, so that it is ρρρ 2 2 22 obtained a total of 40.38%. Here, extra 2 W are added to W=mg=L= T= T= T= 22D= 2D= D= CCLCDCSVDSVDSVSV (1) (2) (2) (2) where: D ρisρρ theρ drag force. other losses. Hence, the total power required from the The induced2222 drag is obtained as: batteries is: (Power required for cruise/Battery to thrust C CDiDi== 222 ;; ; CC CDD = = c cdd d + + C CDiDi (3)(3) (3) 𝐿𝐿𝐿𝐿𝐿𝐿 𝐶𝐶𝐶𝐶𝐶𝐶 2 efficiency) + Pother= 27.11 W. T= D= CDSV (2) C C πDiCπ∗∗Di=e∗Diee=∗∗=AR∗AR ; ; C ;C DCD =D = =c c dc d+d + +C C DiCDi Di (3) (3) (3) (3) ρ 222 The other circuit was from the battery to solar cells in 𝐶𝐶𝐶𝐶𝐿𝐿𝐶𝐶𝐿𝐿𝐿𝐿 which solar encapsulation was 90%; solar cells, 22%; cam- π2π∗∗ee∗∗ARAR P Preqreqreq = =π T∗ Te×∗×VARV (4)(4) (4) where: cd is the coefficient of drag of airfoil. ber efficiency, 90%; MPPT efficiency, 90%; and battery From P P reqPreq req Bandel= = =T T ×T×V× VetV al . (2015), e means Oswald’s efficiency, charging efficiency, (4) (4) (4) 90%, totaling 14.435%. In order to get this CDi= 2 ; CD = cd + CDi (3) equal𝐶𝐶 to𝐿𝐿 1.78 × (1 − 0.045 × AR0.68) − 0.64. The power required power, the required is: Total power required

δ δ =23.45 for=23.45 πlevel∗e∗ sinAR sinflight (360 (360 is: )) ) from batteries/ (5)(5) (5) (Sunlight to battery efficiency × Area of Solar 284284284++nn -2 -2 (4) cells) = 451.23 W∙m . Hence, 451.23 W∙m of solar irradi- δ Pδ δ=23.45req =23.45 =23.45 = T× sinV sin sin (360 (360 (360 365365 365 ) ) ) (4) (5) (5) (5) 284284++nn ance is required in order to have a level flight. Now, whether 284+n --11-1 365365 ω ωss s= = cosThe cos (tanrequired(tan(tanϕϕ × × tanδ) parameterstanδ) 365 for calculating the level flight are this much (6)(6) (6) amount of solar irradiance is available from the shown in Table 4. As the mass of the plane was initially fixed Sun has to be determined.2 -1-1-1 W=mg=L= CLSV (1) ω ω sω s= s= =cos cos cos(tan(tan(tanϕϕ ϕ× × ×tanδ) tanδ) tanδ) (6) (6) (6) δ =23.45as 2 kg, sinthe total(360 weight of) the plane becomes 19.6 N. From the (5) ρ 284+n 2 above two equations, the cruise velocity of the plane 22 2 that can Power Available from Solar Energy H Hoo=o= IIscIscsc × × 3 3600(1600(1 + + 0.033( 0.033(coscoscos ))))()(ω(ωss.sin.sins.sinϕϕ .sinδ+cos.sinδ+cos.sinδ+cos365 ϕϕ.cosδ.sinω.cosδ.sinω.cosδ.sinωss))s ) kJ/mkJ/m kJ/m∙∙day∙day (7)(7) (7) be obtained as the lift coefficient for the airfoil is 0.91, and the Here, the power available from solar energy is calculated 242424 360n360n360n 2 -1 2 2 2 T= D= CDSV (2) H πH H oπ =o o= = I scI Isc sc× × × 3 3 600(13 600(1 600(1 + + +0.033( 0.033( 0.033( ωssurface =cos365365cos 365cos area(tan))() ofω)ϕ((ω sω×the.sins.sins .sintanδ) wingϕϕ.sinδ+cosϕ.sinδ+cos.sinδ+cos can be ϕcalculated ϕ .cosδ.sinωϕ .cosδ.sinω .cosδ.sinω as s the) s )s kJ/m) kJ/m chordkJ/m ∙ day∙ equalday∙ day for April (6) 1st, (7) (7) 2016(7) since April and May represent the hottest ρ 242424 360n360n360n 2 2 to 30 cm and the length of 2.63 m, which is 0.8 m . Hence, the seasons W=mg=L= in which2 maximum CLSV solar radiation occurs ( n = 92, as (1) πππ 365365365--11-1 S Smaxmax== coscos cos (−tan(−tan(−tanϕϕ.tanδ).tanδ).tanδ) –2 (8)(8) (8) ρ st 2 velocity of the plane at cruise is 7.5585 m∙s . The total drag is the numeric sequence starts with 1 forW=mg=L= January 1 ).C LSV (1) 222 2 2 ρ the sum of 3 drags:-1-1-1 one caused by the wing cd, which is around Then, the latitude (ϕ) of VIT Vellore, India, for the H o = I sc × 3 600(1 + 0.033( cos S S maxS1515max15max==)=)(ω cos coss .sincos(−tanϕ(−tan(−tan.sinδ+cosϕϕ.tanδ)ϕ.tanδ).tanδ)ϕ .cosδ.sinω s ) kJ/m ∙ day C Di (7) = (8) (8) (8)2 ;2 CD = cd + CDi (3) 2 24 0.2479360n 2 N;22 the other is the induced drag, which accounts for W=mg=L= calculation, 𝐶𝐶 Cis 𝐿𝐿 L 12.9692°.SV 2 From this, the declination angle ( δ ) (1) T= ρD= CDSV (2) π 0.7855365 15 N1515 due to very high lift; and the last one is due to the body is calculatedπ∗e (Eq.∗ARρ 5), which represents the angular displacement 2 2 == a a + + b b (9)(9) (9) T= D= CDSV (2) (fuselage),HH whichSSS is taken 1.7 times the airfoil drag. Then, the of the Sun’s rays2 at the north (or south) of the Equator and is -1 Preq = T×V ρ (4) S max = H H o o o cos == (−tan=a a a+ S+S +Smaxb maxb ϕb.tanδ) (8) (9) (9) (9) 2 total drag for(( (the entire)) ) plane is 1.4646 N. From the total drag T=the angleD= C madeDSV by the Sun and the Earth with 2 its projection (2) 2 HHH SSS force and the cruise velocity, one can find the power required on theCDi equatorial=ρ ;plane. CD = cd + CDi (3) 15 ooo 2 HHH ((S(SmaxSmaxmax))) 𝐿𝐿 ω ω = = (15h (15h − − 180) 180) (10)(10) (10)2 𝐶𝐶 for a cruise: Preq = T × V = 10.13 W. CDi= ; CD = cd + CDi (3) π∗e∗AR 2 In order to get this much power, more power should δ =23.45 sin (360 ) 𝐶𝐶 𝐿𝐿 (5) (5) = ω ω a ω= += =(15h b(15h (15h − − −180) 180) 180) (9) (10) (10) (10) 284+n π∗e∗AR beH extracted fromS batteries as there are electronics in the CDi= 2 ; CD = cd + CDi (3) 𝐿𝐿 Preq = T×V 365 (4) o 𝐶𝐶 II gIg=g= (c(c(c ++circuit +H d.cosω)d.cosω) d.cosω) for which(Smax there) will be]] ] ××losses × HH H during conversion and By (11)(11) (11)using latitude and declination angle, the hour angle π∗e∗AR ππ coscoscosωω− −coscoscoswwsss - 1 Preq = T×V (4) πππww sss ( ω s)ω cans = becos found,(tan whichϕ × tanδ) is the angle through which the Earth (6) [ [ I 24[ 24gI 24=Ig =g = ωT(cable(c (c= + + (15h + d.cosω) 4.d.cosω) d.cosω) Required −sinsin sin 180)wwss−s− parameters 180 180180 coscos cos w w ss s for ] ] × ] calculating × × H H H level flight. (10) (11) (11) (11) πππ coscos cos ω ω −ω − cos −cos cos w w sw s s mustPreq = turn T× Vto bring the meridian at a point directly in line with (4) ππwπwsw s s δ =23.45 sin (360 ) (5) Parameterssss sss SI units [24[[2424 sin sin sinwww−− −180 180 180coscoscoswww the Sun’s rays. It is equivalent284+n to 15° per hour. From Jashnani λ λ = = (12)(12) (12) δ =23.45 sin (360 ) (5) et al. (2013): 365 2 Gross𝐶𝐶𝐶𝐶𝐶𝐶𝑡𝑡𝑡𝑡𝑡𝑡 weight Ho= 2 Ikgsc × 3600(1 + 0.033(cos ))(ωs.sinϕ.sinδ+cosϕ.cosδ.sinωs) kJ/m284∙day+n (7) Ig= (c + d.cosω) ] × H (11) λ 𝑟𝑟𝑟𝑟λ 𝑟𝑟= = 24 –2 360n (12) (12) π λ𝐶𝐶𝐶𝐶Gravity 𝐶𝐶=cos ω − cos w s 9.81 m ∙s δ =23.45 sin (360(12)-1 ) 365 (5) 𝑡𝑡𝑡𝑡πw ωs = cos (tanϕ × tanδ) (6) (6) 𝐶𝐶𝐶𝐶𝑡𝑡𝐶𝐶 s π 365 284+n [24 sinws− 180 cosws Airfoil lift𝐶𝐶 𝐶𝐶𝑟𝑟coefficient𝐶𝐶𝑟𝑟𝑟𝑟 0.91 -1 365 ωs = cos (tanϕ × tanδ) (6) V Vhh h = = (13)(13) (13) -1 Airfoil chord 0.3 m Smax= cos (−tanϕ.tanδ) (8) SSShhh∗∗L∗LLhhh Thus, daily average irradiance (H ) is: λ = 2 - 1(12)2 o Surface V V Vh h= h=area = of wing 0.7389 m ω s = cos (tan (13) ϕ(13) (13) × tanδ) 2 (6) S𝑡𝑡SS∗∗c∗cc Ho= Isc × 3600(1 + 0.033(cos ))(ωs.sinϕ.sinδ+cosϕ.cosδ.sinωs) kJ/m ∙day (7) 𝐶𝐶 hh hh 15 Airfoil dragSSh coefficientS∗L∗∗LhL 240.0115 360n 𝐶𝐶𝑟𝑟 2 SS∗Sc∗∗cc Ho= Isc × 3600(1 + 0.033(cos ))(ωs.sinϕ.sinδ+cosϕ.cosδ.sinωs) kJ/m ∙day (7) Aspect ratio π9.36 365 24= a + b 360n (7) (9) 2 Oswald’s efficiency Ho= Isc × 30.773600(1 + 0.033(cos ))(ωπs.sinϕ.sinδ+cosϕ.cosδ.sinω365s) kJ/m ∙day (7) Vh = (13)H -1 S 24 360n Smax= cos (−tanϕ.tanδ) (8) Sh∗Lh Ho (Smax) π 365 2 -1 S∗c J. Aerosp. Technol. Manag., São José dos Campos, Vol.8, Smax No =4, pp.397-407, cos (−tan Oct.-Dec.,ϕ.tanδ) 2016 (8) 15 ω = (15h − 180) 2 (10) -1 Smax= cos (−tanϕ.tanδ) 15 (8) 2 = a + b (9) H S 15 = a + b (9) Ho (Smax) Ig= (c + d.cosω) ] × HH S (11) π cosω−cosws = a + b π w s H o ( S max ) (9) 24 ω = (15h −sin 180)ws− cos w s (10) [H S 180 Ho Smax ( ) ω = (15h − 180) (10) λ = (12) ω = (15h − 180) 𝑡𝑡 (10) Ig= (c + d.cosω) 𝐶𝐶 ] × H (11) π 𝐶𝐶𝑟𝑟cosω−cosws πw Igs= (c + d.cosω) ] × H (11) [24 sinws− 180 cosws π cosω−cosws πws Vh = (13) Ig= (c + d.cosω) ] ×[ 24H sin w s − 180 cos w s (11) π cosωS−hcos∗Lhw s λ = πw s (12) 24 s S𝑡𝑡 ∗c s [ sinw −𝐶𝐶 180 cosw 𝐶𝐶 𝑟𝑟 λ = (12) 𝐶𝐶𝑡𝑡 λ = 𝐶𝐶 𝑟𝑟 (12) 𝐶𝐶 V𝑡𝑡 h = (13) 𝐶𝐶𝑟𝑟 Sh∗Lh Vh = (13) S∗c Sh∗Lh Vh = S ∗ c (13) Sh∗Lh S∗c

22 2 W=mg=L= W=mg=L= W=mg=L= C CLCLSVSVLSV (1) (1) (1) ρρρ 222 222 W=mg=L= W=mg=L=W=mg=L= CCCLLSVLSVSV (1) (1)(1) ρρρ 22 2 T= T= T= D= D= D=CCDCDSVSVDSV (2) (2) (2) 222ρρρ 222 22 2 T= T=T= D= D=D=CCCDDDSVSVSV (2) (2)(2) ρρρ C C DiCDi=Di== ; ; C ;C DCD = D= =c cd dc + d+ +C C DiCDi Di (3) (3) (3) 222 22 2 𝐶𝐶𝐶𝐶𝐿𝐿𝐿𝐿𝐶𝐶𝐿𝐿 2 W=mg=L= CLSV (1) ππ∗π∗ee∗ρ∗eARAR∗AR C CCDiDiDi=== ; ; ;C CCDDD = == c ccdd d+ ++ C CCDiDiDi (3) (3)(3) 222 2 𝐶𝐶𝐶𝐶𝐶𝐶𝐿𝐿𝐿𝐿𝐿𝐿 2 P P req PreqW=mg=L=req = = =T T ×T×V×V V C L SV (4) (4) (4) (1) πππ∗∗e∗e∗e∗AR∗ARAR 2 ρ T= D= CDSV (2) 2 ρ P PPreqreqreq = == T TT×××VVV (4) (4)(4) 2 δ δ =23.45 δ=23.45 =23.45 sin sin sin (360 (360 (360 2 )) ) (5) (5) (5) T= D= CDSV (2) ρ 284284284++n+nn CDi= ; CD = cd + 365C365365Di (3) 2 2 δ δδ =23.45 =23.45=23.45𝐶𝐶𝐿𝐿 sin sinsin (360 (360(360 )) ) (5) (5)(5) Reddy BSK,-1-1-1 Aneesh284284284 P,+++ nBhanunn K, Natarajan M 402 ωπ ω ∗sω se= ∗s= AR=cos cos cos(tan(tan(tanϕϕ ϕ× × ×tanδ) tanδ) tanδ) (6) (6) (6) Di D d Di C = 2 365365365 ; C = c + C (3) 𝐶𝐶𝐿𝐿 Preq = T×V (4) -1--11 ω ωωs ss= == cos coscos(tan(tan(tanϕπϕϕ ∗× ×e× ∗tanδ) ARtanδ)tanδ) (6) (6)(6) where: I is the extra-terrestrial irradiance. 22 2 From the graph shown in Fig. 7 and the data in Table 6, H H oHo==o=IIscscI ×sc × ×3 3 600(13600(1600(1 + + +0.033( 0.033( 0.033(coscoscos sc))))()(ω)ω(sω .sins.sins.sinϕϕ.sinδ+cosϕ.sinδ+cos.sinδ+cosϕϕ.cosδ.sinωϕ.cosδ.sinω.cosδ.sinωs)s) skJ/m )kJ/m kJ/m∙day∙day∙day (7) (7) (7) 24 24 24 The 360n 360n day 360n length Preq =or T ×maximumV sunshine hours can be found it is observed that, by(4) using alone, one is able to π π π δ =23.45365 sin365365 (360 ) (5) using: 222 fly the plane for up to 5 h. This flight time can be increased H HHoo=o== IIscIscsc × ×× 3 33600(1600(1600(1 + ++ 0.033( 0.033(0.033(coscoscos )))))()(ω(ωωs.sinss.sin.sinϕϕϕ.sinδ+cos284.sinδ+cos.sinδ+cos+n ϕϕϕ.cosδ.sinω.cosδ.sinω.cosδ.sinωs)ss) )kJ/m kJ/mkJ/m∙day∙∙dayday (7) (7)(7) as, during gliding, some amount of energy is stored in the 242424 360n360n360n -1-1-1 365 S S maxSmaxmax=== cos cos cos(−tan(−tan (−tanϕϕ.tanδ)ϕ .tanδ).tanδ) (8) (8) (8) (8) π ππ 365 365365 δ =23.45 sin (360 ) battery. (5) 222 -1 284+n ωs = cos 1515(tan15 ϕ × tanδ) T able 6. (6)Estimated global solar radiation on April 1st, 2016 -1--11 365 S SSmaxmaxmax==which= cos coscos was(−tan(−tan(−tan for 12:13.ϕϕϕ.tanδ).tanδ).tanδ) Then, monthly average daily global versus power (8) (8)(8) required for level flight.

radiation222 (H) can be-1 calculated for an altitude of 30 m and ω==s =a a + acos+ +b b b(tanϕ × tanδ) (9) (9) (9) (6) Estimated 151515 Duration of Global Power for estimatedH Husing:H SSS 2 radiation Ho= Isc × 3600(1 + 0.033(cos ))(ωs.sinϕ.sinδ+cosϕ.cosδ.sinωs) kJ/m ∙day the day (7) radiation level flight HHoHo o ((SS(maxmaxSmax))) N = 0.7 24 360n (h) (W m–2) sky (W m–2) === a aa + ++ b bb (9) (9)(9) ∙ –2 ∙ π 365 (9) (W∙m ) HHH SSS 2 H o = I sc × 3 600(1 + 0.033( ω ω cos ω= = =(15h (15h (15h )− )− (−180) ω180) s180).sin ϕ .sinδ+cos ϕ .cosδ.sinω s ) kJ/m ∙ day (10) (10) (10) (7) HHHooo (((SSSmaxmaxmax))) 1 – 5 0 0 451.23 24 -1 360n Smax= cos (−tanϕ.tanδ) (8) π where: a =365 −0.309 + 0.539.cosϕ − 0.0693E + 0.29 (S/Smax); 2 6 10.27 7.19 451.23 b = ω ωω1.527 = == (15h (15h(15h − 1.027.cos − −− 180) 180)180) ϕ + 0.0926 E − 0.359 ( S / S ); S is the (10) (10)(10) 15 max I Ig g=I=g= (c(c(c + + +d.cosω) d.cosω) d.cosω) -1 ]] × ]× ×H H H 7 (11) (11) (11) 185.72 130.0 451.23 average sunshineSmax= hourcos per(−tan day;ϕ .tanδ)E is the elevation of the location (8) πππ coscoscosωω−ω−cos−coscoswwsws s above mean sea2 level (200 πmπwwπ —sws s Vellore). 24 2424= a + b s s s s s s 8 (9) 392.16 274.51 451.23 [[[ 15 sin sin sinwww−− − 180180 180coscoscoswww I IgIg=g== (c(c(c + ++H Atd.cosω) d.cosω)d.cosω) each hour Sof the day (h),] the] ]× ×× H instantaneous HH hour angle (11) (11)(11) 9 602.52 421.76 451.23 πππ Ho Smaxcoscoscosωωω−−−coscoscoswwwsss for a particular( hour) canπππwwws sbes estimated as (in degrees): [ [ 24 [ 24 24 sin λ sin λsin =λ =w ww= sas− s − −+ 180 180180b cos cos cos w w w s ss 10 (12) (12) (12) (9)784.69 549.28 451.23 H 𝐶𝐶𝐶𝐶𝑡𝑡𝑡𝑡𝐶𝐶𝑡𝑡 S ω = (15h − 180) (10) (10) Ho 𝐶𝐶𝐶𝐶𝑟𝑟𝑟𝑟𝐶𝐶𝑟𝑟 (Smax) 11 908.41 635.89 451.23 λ λλ = == (12) (12)(12) 𝐶𝐶𝐶𝐶𝐶𝐶𝑡𝑡𝑡𝑡𝑡𝑡 12 952.22 666.55 451.23 The hourly ω =global (15h solar− 180) irradiance can be achieved (in (10) V V 𝐶𝐶hV𝐶𝐶h𝐶𝐶𝑟𝑟 =𝑟𝑟 𝑟𝑟=h = (13) (13) (13) -2 -1 13 908.41 635.89 451.23 Ig= (ckJ∙m + d.cosω)∙day ) by using:SShhS∗∗hLL∗hhL h ] × H (11) s π cosωSS∗−∗Scccos∗c w πws 14 784.69 549.28 451.23 24 V VVhh h= ==sinws− 180 cos w s (13) (13)(13) [ Ig= (c +SS Shd.cosω)hh∗∗L∗LLhhh ] × H (11) 15 (11)602.52 421.76 451.23 π cosω−cosws SSS∗∗c∗cc πws [ 24 λ = sin w s − 180 cos w s 16 (12) 392.16 274.51 451.23 𝐶𝐶𝑡𝑡 17 185.72 130.0 451.23 where: c =𝐶𝐶𝑟𝑟 0.409 + 0.516 × sin (ωs − 60) λ = (12) and d = 0.6609 − 0.4767 × sin (ωs − 60). 𝑡𝑡 18 10.27 7.19 451.23 The calculated solar 𝐶𝐶parametric values are shown in Table 5 Vh = 𝐶𝐶 𝑟𝑟 19 –(13) 24 0 0 451.23 and, by calculatingSh∗Lh the instantaneous hour angle for a particular hour, global solarS∗c irradiance can be found for each solar hour; Vh = (13) by taking the clear sky factor (N ) as 0.7, the readings in Table Sh∗Lh sky 1,200 Estimated global radiation ]

6 are obtained for the entire S∗c day. –2 Actual global radiation 1,000 Level ight power Table 5. Calculated solar parametric values. 800 Parameter SI units 600

Declination angle 4.4139° 400 200 Hour angle 91.0186°

Solar irradiance [W · m Solar irradiance 0 Daily average irradiance 37,537 kJ∙m-2∙day-1 0 2 4 6 8 10 12 14 16 18 20 22 24 Duration of the day [h] Day length 12.13 h Figure 7. Duration of the day (April 1st, 2016) versus Monthly average daily radiation -2 -1 24,266 kJ∙m ∙day hourly global solar radiation — level flight power.

J. Aerosp. Technol. Manag., São José dos Campos, Vol.8, No 4, pp.397-407, Oct.-Dec., 2016

2 W=mg=L= CLSV (1)

ρ 2 2 W=mg=L= CLSV (1) ρ 2 T= D= CDSV (2) Design2 Analysis of Solar-Powered Unmanned Aerial Vehicle ρ 403 2 2 T= D= CDSV (2) ρ PRELIMINARY DESIGN The spars2 will handle the tension and compression forces CDi= 2 ; CD = cd + CDi (3) Design𝐶𝐶𝐿𝐿 of the Wing on the wing and are attached to the fuselage so they can transfer π∗e∗AR the forces to it; here, the forces will be spread over the entire CDi= ; CD = cd + CDi (3) Wing Configuration 2 construction.𝐶𝐶𝐿𝐿 The ribs (made of lightweight balsa wood) are Preq = T×V (4) With respect to the angle made with the fuselage, from Fig. 8, intendedπ∗e∗AR to give the wing the desired airfoil shape and also to the central part of the wing is perpendicular to the fuselage stabilize the construction; they provide rigidity and strength

(straight) because it contains the solar cells and, in order to P thereq = wing T× Vby transferring the forces to the spars. The skin (4) to get maximum efficiency, these cells should be parallel to is also a part of the construction to resist the forces on the δ =23.45 sin (360 ) (5) the ground; this284 kind+n of wing has high rigidity but poor roll wing, which covers only the top surface with 1 mm of balsa stability. Hence, in365 order to have a good counter roll and lift, sheet in order to make a non-conductive flat surface for the the ends of the main wing are extended by keeping a positive δ =23.45integration sin (360of solar cells. ) (5) -1 284+n ωs = cosdihedral(tan angleϕ × oftanδ) 7°, which makes an obtuse angle with the The ends (6) of the wing are tapered from a root chord of 365 main wing. 30 cm to a tip chord of 20 cm over a length of 50 cm. This Regarding the position of the wing on the fuselage, a high part of- 1the wings includes ailerons in order to have stability ωs = cos (tanϕ × tanδ) (6) wing is selected, which means it is situated over the fuselage; and control over the plane; In these parts, two carbon fiber 2 Ho= Isc × 3600(1 + 0.033(cos this)) kind(ωs.sin of designϕ.sinδ+cos helps in ϕhaving.cosδ.sinω high grounds) kJ/m clearance,∙day less spars are kept (7) in order to give support and help in integrating 24 360n interference drag, better field performance at low speeds and the tapered parts to the main wing, since, at the junction π 365 2 maximum space for having Ha payload.o= Isc × 3600(1 + 0.033(cos of two)) (partsωs.sin of ϕthe.sinδ+cos wing, largeϕ.cosδ.sinω forces will act.s) kJ/mIn Fig.∙ day12, all (7) 24 360n -1 dimensions are in mm. Smax= cos (−tanϕ .tanδ) 1,630 π 365 (8) 2 Design of the Tail 15 200 300 -1 Smax= Thecos airfoil(−tan selectedϕ.tanδ) for the horizontal and vertical tail (8) o 2 500 7 is the NACA 0009 (Fig. 9) because it has a symmetrical = a + b 15 (9) H S profile in nature and a maximum lift coefficient of 0.8. For Figure 8. A draft model of the entire wing assembly. the horizontal tail, the mean chord length is 0.205 m and, for the Ho (Smax) vertical= a one,+ b the mean chord length is 0.25 m (AirfoilTools.com (9) H S Taper Ratio 2016). ω = (15h − 180) H o (10)(Smax ) It is the ratio (λ) between the tip and the root chord of the wings. A trapezoidal shaped (0 < λ < 1) is selected for the ω = (15h − 180) (10) ends of the main wing. Wing tapering is used to improve lift Ig= (c + d.cosω) ] × H (11) distribution characteristics, which will also result in better lateral π cosω−cosws control since the massπws moment of inertia of the wings around [24 sinws− 180 cosws the longitudinal axis will be less. Here, λ = 0.66. Ig= (c +Figure d.cosω) 9. NACA 0009 airfoil shape.] × H (11) π cosω−cosws πws (12)24 s s λ = [ Horizontal (12) sin Tailw − 180 cosw 𝐶𝐶𝑡𝑡 The dominant factors which control the aircraft pitch stability 𝐶𝐶𝑟𝑟 where: Ct is the tip chord; C r is the root chord. are gravity λ = center (GC) position and horizontal tail size and (12) position. The𝐶𝐶𝑡𝑡 effectiveness of horizontal tail is measured by Structure Vh = horizontal (13) tail𝐶𝐶𝑟𝑟 volume coefficient (V ): h In theSh ∗centralLh part of the wing, the construction consists S∗c of spars, ribs, and the skin. As the central part of the wing is Vh = (13) (13) 0.0163 m, which has solar cells integrated into it, the main Sh∗Lh priority is to achieve very small deflection. Then, 3 spars have S∗c been included, which are made of a carbon fiber (zoltac) material; where: Sh is the horizontal tail area; Lh is the distance from two of them are 6 × 4 mm and the middle one is 4 × 2 mm. the tail’s aerodynamic center (more of which later) to the

J. Aerosp. Technol. Manag., São José dos Campos, Vol.8, No 4, pp.397-407, Oct.-Dec., 2016 Reddy BSK, Aneesh P, Bhanu K, Natarajan M 404

aircraft’s GC; S is the wing area; c is the mean aerodynamic chord. a Reynolds number of 150,000 and by using SST k-ω viscous After several iterations, the horizontal stabilizer parameters model. The boundary conditions are given in Table 10. are shown in Table 7. 250 Table 7. Horizontal tail configuration. 300

Horizontal stabilizer parameter SI units

Area (S ) 0.14 m2 700 h 200

Moment arm (Lh) 0.96 m Figure 10. A draft model of the entire tail assembly. Volume coefficient (Vh) 0.64 Table 9. Comparison of L/D for different trails. Span (bh) 0.7 m

Mean chord (ch) 0.205 m Trail number Design modification L/D

Taper ratio (λh) 1.0 1 Basic cuboidal 15.244

Aspect ratio (ARh) 3.41 2 Tapered front and back 15.498

Elevator ratio (Sr/Sh) 0.3 3 Lightly streamlined 15.72 4 Final design 15.81 Vertical Tail Table 10. Boundary conditions. The role of vertical tail is to control the yaw movement Type of Edge Boundary condition by providing yaw damping. The effectiveness of vertical tail is boundary measured by the vertical tail coefficient (V ): v Inlet Velocity 8 m∙s–1 (x direction) Outlet Pressure Gauge pressure = 0 Pa Vv = (14) (14) Sv∗Lv Wall Stationary Velocity = 0 m∙s–1 S∗b where: Sv is the vertical tail area; Lv is the distance from the vertical tail’s aerodynamic center to the aircraft’s GC; b is the wingspan. (a) 2 Pcrit = (0.5After) ρV severalSh C Lmaxiterations, the vertical stabilizer parameters (15) are shown in Table 8. In Fig. 10, all dimensions are in mm.

Fuselage (b) This part is designed after a number of iterations in order to improve the lift-to-drag ratio. The trails shown in Table 9 are analyzed in fluent software as shown in Fig. 11, respectively, for

Table 8. Vertical tail configuration.

(c) Vertical stabilizer parameters SI units

2 Area (Sv) 0.075 m

Moment arm (Lv) 0.85 m

Volume coefficient (Vv) 0.0328 d) Span (bv) 0.3 m

Mean chord (cv) 0.25 m Aspect ratio (AR ) 2.4 v Taper ratio (λl) 1.0 Figure 11. (a) Basic cuboidal (Trail 1); (b) Tapered ends Rudder ratio (Sr/Sv) 0.3 (Trail 2); (c) Streamlined (Trail 3); (d) Final design (Trail 4).

J. Aerosp. Technol. Manag., São José dos Campos, Vol.8, No 4, pp.397-407, Oct.-Dec., 2016

Design Analysis of Solar-Powered Unmanned Aerial Vehicle 405

The final design is selected, having a lift force of 18.06 N and A software has been used to find the deflection and stress a drag one of 1.142 N. The primary function of the fuselage is using finite element method. In order to validate it, we took a to carry the electronics and payload. Hence, a length of 0.65 m problem which can be solvable by the analytical approach and and 0.82 m in width and height are obtained for the design. The compared it with the software results. propulsion system components can be mounted in the front of For the validation, a problem (Bansal 2010) has been solved the fuselage, and the electronics and payload may be shifted back both manually and in the software; the deviation was found or fronted in such a way to adjust the center of gravity position to be less than 5%. Then, for further complex analyses, the of the plane, which is at a distance of 9 cm from the leading software is used to find the results. The input properties for balsa edge of the wing. The tail boom is connected to the back of the wood (MakeitFrom.com 2016) and carbon fiber (GoodWinds fuselage, which is integrated to the main wing through carbon Composites 2016) have been taken for all the analyses. fiber rods. In Fig. 12, all dimensions are in meters. Finally, all In the wing deformation, the mid wing excluding the tapered parts are assembled forming the complete plane, as shown in parts is analyzed since mostly the lift force is acted only on Fig. 13, in which all the dimensions are in mm. the middle section of the wing. Then, a load of 4 kg is applied to the entire wing, taking a factor of safety equal to 2 and keeping 0.650 the central position of about 10 cm constant since that portion 0.082 is fixed to the fuselage, as shown in Fig 14. A deflection of only 1.4 cm is observed. Figure 12. A draft model of the fuselage. The tail boom has to resist to bending loads, keeping the empennage surfaces at the proper aerodynamic incidence with 500 the wing. If the tail boom flexed too much, control effectiveness o 7 would be reduced. Bending load is the greatest on the tail boom 1,630 82 2,630 when the empennage Vv = surfaces operate at a high lift coefficient (14) Sv∗Lv and the aircraft is flying at a high speed. The bending load on 300 900 S∗b the tail boom could be treated as a point load acting at the 670 1,580 end of the tail boom and its critical value can be calculated as:

2 700 330 Pcrit = (0.5) ρV Sh CLmax (15) (15) Figure 13. Entire plane draft model. The critical bending load characteristics for the tail boom Structural Analysis are shown in Table 11. A bending load that we got from the The two primary structures for the design are the wing spar calculation of 4.4 N is applied on one end of the tail, keeping and the tail boom. Both of these are analyzed in detail so they the other end constant. A deflection of 0.8 cm is observed. could be lightweight but strong enough to support the design The fuselage has to carry the payload and electronics, which loads. The wing spar has to carry 3 primary loads: the wing can weight more than 1 kg. Hence, a force of 30 N, which is bending load produced by lift, the transverse shear load caused Table 11. Critical bending load characteristics for tail boom. by wing bending, and the torque load produced by the pitching moment of the wing airfoil. The tail boom has to carry the ben- Parameter SI unit ding loads produced by the horizontal and vertical stabilizers. Velocity (V) 8 m∙s–1 The wing spar and tail boom are made of carbon fiber material Air density (ρ) 1.227 kg∙m–3 which can support very high loads before they fail. For aerodynamic Horizontal stabilizer area (S ) 0.14 m2 surfaces, even a small amount of deformation could potentially h Max tail coefficient (C ) 0.8 upset the stability or control of the airplane. Calculating the Lmax Critical bending load (P ) 4.4 N structural deformation using an analytical method would have crit been impractical. Instead, the potential approach is to use finite Tail boom (L) 0.6 m element analysis to determine how the structures are deformed. Modulus of elasticity (E) 131 Gpa

J. Aerosp. Technol. Manag., São José dos Campos, Vol.8, No 4, pp.397-407, Oct.-Dec., 2016

Reddy BSK, Aneesh P, Bhanu K, Natarajan M 406

3 kg (more than the entire plane weight), is applied to the front on carbon rods is negligible when compared to its permissible section by keeping the end section fixed, as shown in Fig 15. A stress. The red portion is only observed on the carbon fiber rods, deformation of 0.279 cm is observed. whereas the balsa wood portion is blue in a color which is the Concerning the stress analysis, the fuselage is analyzed to minimum stress. Hence, all the parts are under the safer limit find out whether the stress is below the permissible range or even under a factor of safety equal to 2. not for a load of 30 N. Besides, according to the analysis, as shown in Fig. 16, a von Mises stress of 4.4 MPa is observed, which is within the permissible value of 9.6 MPa. RESULTS AND CONCLUSION In the middle section of the wing, more stress is acted on the carbon fiber rods than that of balsa ribs and skin for a force of In this paper, the suitable airfoil has been chosen by 40 N, as shown in Fig. 17. Nevertheless, the stress which is acted analyzing different airfoils from previous designs for a high

A: static structural Total deformation Type: total deformation Unit: m Time: 1 04-Apr-2016 12:57 p.m.

Figure 14. Deflection of mid-wing.

B: static structural Total deformation Type: total deformation Unit: m Time: 1 07-Apr-2016 7:02 p.m.

Figure 15. Deflection of the fuselage.

B: static structural Equivalent stress Type: equivalent (von Mises) stress Unit: MPa Time: 1 07-Apr-2016 11:36 p.m.

Figure 16. Von Mises stress analysis of the fuselage.

A: static structural Equivalent stress Type: equivalent (von Mises) stress Unit: Pa Time: 1 05-Apr-2016 2:24 p.m.

Figure 17. An enlarged view of stress region near the middle portion of the mid-wing.

J. Aerosp. Technol. Manag., São José dos Campos, Vol.8, No 4, pp.397-407, Oct.-Dec., 2016 Design Analysis of Solar-Powered Unmanned Aerial Vehicle 407

lift-to-drag ratio. It was presented a way to calculate a number Arivazhagan N, for granting permissions to use the required of solar cells required for giving enough power to the motor equipment during the course of this research. We would also lift the 2-kg plane, which can extend for heavier ones. It like to show our gratitude to Team Vimaanas, VIT University, was performed detailed energy and power analysis to check which provided insight and expertise that greatly assisted the whether the design would be feasible or not. Then, all the research. components were designed, considering various factors for better performance, and analyzed for deflection and stress to keep the design under the safer limit. This gives a basic AUTHOR’S CONTRIBUTION foundation in order to fabricate a solar-powered UAV. Aneesh P was responsible for conceptual study and design. Data acquisition, analysis and drafting were carried Acknowledgements out by Reddy BSK. Bhanu K and Natarajan M performed solar cells experimentation and interpretation of data. We wish to express our profound gratitude and indebtedness All authors discussed the results and commented on the to Prof. M. Natarajan, for his inspiring guidance, and Prof. manuscript.

REFERENCES

AirfoilTools.com (2016) Airfoil tools; [accessed 2016 Aug 11]. Kederoglu OG, Yanbastioglu S, Saraç B, Kocaoğlan MH (2012) http://airfoiltools.com/index Solar powered aircraft: PV Power System Definition & Solution

Bandel A (2015) Aviation studies design build and fly minor. Development Project; [accessed 2016 Aug 11]. http://www. Amsterdam: University of Applied Sciences. academia.edu/7515954/Solar_Powered_Aircraft_Project

Bansal RK (2010) A textbook of the strength of materials. Kumar G, Sepat S, Bansal S (2015) Review paper of solar-powered Bangalore: Laxmi Publications. UAV. International Journal of Scientific & Engineering Research

Bruscoli S (2011) Airfoil optimization for a solar powered aircraft 6(2):41-44. (PhD thesis). Pisa: Università degli Studi di Pisa. MakeitFrom.com (2016) Balsa; [accessed 2016 Aug 11]. http:// GoodWinds Composites (2016) Carbon rods; [accessed 2016 www.makeitfrom.com/material-properties/Balsa/ Aug 11]. Available: http://gwcomposites.com/carbon-rods/ Noth A (2008) Design of solar powered for continuous Jashnani S, Nada TR, Ishfaq M, Khamker A, Shaholia P (2013) flight (PhD thesis). Zürich: ETH Zürich. Sizing and preliminary hardware testing of solar-powered UAV. The Egyptian Journal of Remote Sensing and Space Sciences Simons M (1994) aerodynamics. 3rd edition. 16(2):189-198. doi: 10.1016/j.ejrs.2013.05.002 Hemel Hempstead: Argus.

J. Aerosp. Technol. Manag., São José dos Campos, Vol.8, No 4, pp.397-407, Oct.-Dec., 2016