Designing RC Airplanes for Range



Designing RC Airplanes for RangeW. B. Garner 2017ObjectiveThe objective of this document is to develop the mathematics required to support the design an electric powered model airplane for maximum flight range. The model assumes steady state, constant level, and constant air speed flight. Winds are not considered.MethodologyDistance traveled is defined as:R= 0TV*dtWhere:T is the time interval of travelV is the instantaneous velocitydt is the incremental change in time intervalIt is assumed that V is invariant with time so that R becomes:R = V*TFor a finite energy source, in this case a battery, the time to exhaustion is given by:T =EbPb, HoursWhere:Eb is the battery energy content, Watt-HoursPb is the rate of energy withdrawal, or power, from the battery, WattsBut Pb is equal to the thrust power divided by the efficiency of the motor-propeller combination.Pb =PtEm*Ep = D*VEm*Ep , WattsWhere:Pt is the thrust power, WattsEm is the motor efficiencyEp is the propeller efficiencyD is the total drag, a forceVis the air speedSubstituting:R = Em*Ep*EbD In level flight the drag is given by:D = Cd*ρ*S*V2 V2=2*Wt*gρ*S*ClSubstituting:D = 2*Wt*g*CdCl Substituting in the formula for R:R = Em*Ep*Eb*Cl2*Wt*g*CdWhere:Cl is the airplane lift coefficientCd is the total drag coefficientWt is the total mass, including the battery, Kgρ is the air density, Kg/m^3S is the wing area, m^2g is the gravitational constant = 9.81 m/sec^2The energy content of a battery, Eb, can be defined as a constant, Kb, times its mass, WbEb = Kb*Wb*3600Kb = watt-hoursKg Where the number 3,600 converts Watt-Hours to Watt-secondsAssuming SI units the final formula is:R = Em*Ep*Kb*Wb*Cl*36002*Wt*g*Cd , metersTo maximize range, maximize the numerator and minimize the denominator. For example the efficiency of good quality motors, Em, is about 0.85 while a propeller matched to the cruise speed in diameter and pitch may have efficiency, Ep, of about 0.7. High density LiPo batteries have a density of about 180 Watt-hours/Kg.Inserting the values for Em, Ep, Kb and g;Then R = 33027*Wb*ClWt*Cd meters.This graph plots the resulting range as a function of Cl/Cd with the ratio of masses Wb/Wt as a parameter. The results are clear; make Cl/Cd as large as possible and Wb/Wt as large as possible. These results are idealized as they do not take into account variations in flight conditions or imperfections in implementation. It also assumes that the batteries match their name plate ratings and that they can be fully discharged. Choosing Cl and CdThe objective in choosing the values of Cl and Cd is to maximize the ratio Cl/Cd. Cd is generally divided into three additive components; Profile (Cdo), Induced (Cdi) and Parasitic (Cdp). Profile drag is caused by air flowing over the wing surface and generating friction and turbulence. Induced drag is caused by vortex sheets shedding off the wings, mostly the wing tips. Parasitic drag is drag caused by the fuselage, tail surfaces and any other protruding items such as landing gear. The total drag coefficient is:Cd = Cdo + Cdi + CdpA wing airfoil meant for such flying will have a nearly constant Cdo as a function of Cl (the polar plot). For our purposes assume that:Cdo = .02Cdp is a function of the fuselage shape, cross section area and smoothness, including such items as landing gear. Assume that the fuselage has been designed for low drag and has no landing gear in flight and has a drag coefficient, Cdf, associated with the fuselage cross-sectional area. Referring it to the wing area:Cdpf = Cdf*fus_area/wing_areaAssume that Cdf = 0.3 and fus_area /Wing_area = 0.02Cdpf = .006The tail also produces drag. Assuming an airfoil shape with a Cdo = .02, and a Tail_area/Wing_area = .2 the Tail Cdpt = .02*.2 + =.004Then Cd = .02 +.006+.004 + Cdi = .03 + CdiCdi = .318*Cl2.9*AR where AR is the aspect ratio. AR = wing_span2wing_area Cl/Cd = Cl.03+.35*Cl2AR The following graph plots Cl/Cd as a function of Cl with Aspect Ratio as a parameter.Aspect ratio makes little difference at low Cl where the induced drag is small and the profile and parasitic drag predominates. As the Cl increases, the induced drag increases, eventually dominating the drag where the curves bend over. If Cl is increased more it will reach stall value and the curves drop dramatically (not shown). Increasing AR increases Cl/Cd somewhat, but the rate of increase diminishes as well. The higher the AR the more difficult the mechanical design as the wing will be long and narrow, generating increased bending moments at the root.To proceed with the example design, assume that an AR = 14 and Cl =1, yielding a Cl/Cd = 18. The stall Cl is assumed to be 1.4. Assume that the battery weight is 50% of the total weight. The estimated theoretical range is therefore about 300 Km.At this point it is necessary to choose a total weight; choose Wt = 5 Kg. Make the wing area, S, a variable and estimate the required flight speed and wing dimensions based upon all of these choices.V = 2*Wt*gρ*S*Cl Wing_span = AR*S = 3.74S , mρ = 1.225 Kg/m^3 at sea level, g = 9.81 m/sec^2For Cl = 1, Wt = 5 Kg, Cl stall = 1.4V = 80S m/secS, m^2Wingspan, mMAC, cmV, m/secV, km/hrVstallKm/hrFlight HoursWing loading, Kg/m^2.21.67122072614.225.42.151914.151435.912.5.62.92111.541.5357.28.33.83.35241036308.46.2513.74278.9532279.45This table illustrates the kind of trades involved in selecting physical parameters & their related aerodynamic properties. Start with the smallest wing area, 0.2 m^2. The wingspan is reasonable but the Mean Aerodynamic Chord (MAC) is very narrow and will be hard to build. The wing loading is extremely high, making mechanical integrity a major challenge. The stall speed is very high, meaning launch must be done at high acceleration and the power system may have trouble meeting this requirement. Flight time is the shortest.The greatest wing area produces the greatest wing span and a reasonable MAC. The stall speed is high but doable. The wing loading is modest but the flight time is longest. The results favor the largest wing area designs.Another consideration is the battery volume and its effect on fuselage dimensions. All of these entries have the same battery capacity and weight. The energy content is 180W-hrs/Kg x .5 x 5Kg = 450 W-hrs. These batteries have a specific volume of about 3 cm^3/W-hr. So this implies a total volume of about 1350 cm^3, or a cube of about 11 cm per side. Batteries come in discrete units, rectangular in shape & in this case multiple batteries in parallel would be required. The actual volume will be substantially greater. This volume must be concentrated on the CG and will therefore require a fairly large cross-section body in that area, potentially increasing the drag.Power SystemThe power system consists of a propeller, motor, ESC and battery. The analysis will start with the propeller and work to the battery. First, select the desired cruising speed and calculate the required thrust and thrust power. Assume the configuration yielding a cruise speed of 10 m/sec or 36 km/hr.Start with calculating the thrust and thrust power required.Thrust = DragDrag = Cd*ρ*S*V^2Cd = .03+ .35*Cl^2/AR = .046, Cl =1, AR = 14Ρ=1.225, S = 0.8, V = 10Drag = 4.58 NewtonsThrust Power, Pt = Drag * V = 4.58*10 = 45.8WattsThe next step is to find a propeller capable of supplying both the thrust and thrust power required.The next graph plots generic propeller efficiencies as a function of the advance ratio, J, with the ratio of pitch to diameter, p/Dp, as a parameter.J = Vn*Dp Where n is the prop revolution rate, rps, Dp is the prop diameter, m, and V is the free air velocity, m/sec. J is a measure of the ratio of incoming air speed to the propeller tip speed.The important aspects of this chart are the conditions required for maximum propeller efficiency. Maximum efficiency occurs for p/Dp = 0.9 at a J of about 0.8. That is, by selecting a prop with a p/Dp of 0.9 and operating it at a J ratio = 0.8 will provide the “best” efficiency.n*Dp = VJThrust =Ct*ρ*n2*Dp4, NewtonsBy substitution:Dp2=ThrustCt*ρ*(VJ)2 Prop input power, Pin = Cp*ρ*n3*Dp5 , WattsWhat are needed are the coefficients Ct and Cp at the operating point J.This table supplies those values. Ct =0.069 and Cp =0 .078 when J = 0.8 and the efficiency is 0.714.?p/Dp = 0.9?JCtCpEfficiency00.1240.1180.0000.10.1270.1060.1200.20.1280.0990.2600.30.1280.0950.4050.40.1260.0940.5360.50.1180.0950.6200.60.1050.0940.6730.70.0890.0880.7050.80.0690.0780.7140.90.0480.0630.68110.0240.0450.5251.10.0000.0000.000Dp^2 = Thrustρ*Ct*(VJ)2 =0.3468 m^2, Thrust = 4.58, Ct = .069, V = 10, J = .8Dp = 0.589 mPitch = 0.9*.589 = .53 mn = V/(J*Dp) = 21.22 rps (1274 rpm)Prop input power, Pin = Cp*ρ*n^3*Dp^5Pin = 65 Watts, Cp =0.078Prop efficiency = Pt/Pin = 45.8/65 = 0.7To summarize the propeller properties:Diameter = 0.589 mPitch = 0.53 mn = 21.22 rps (1274 rpm)Pt = 45.8Pin = 65 WThe next step is to calculate the properties of a motor that meet the rps and power output requirements.Motor output power is given by:Pin = rpmKv Im where Kv is in units of rpm per volt, Im is the motor current producing the power.Im = Pin*Kvrpm Assuming the no-load current is negligible, the required voltage driving the motor is given by:Motor Voltage, Vm = rpmKv +Im*Rm , where Rm is the motor resistance.Battery Power, Pbattery = Im * VmThere is no singular combination of Kv and Im so some possibilities are presented in the following table.Pin/rpm =65/1274 =0.051 and let Rm = .03 Ohms (typical value)Kv, rpm/VoltIm, ampsRpm/Kv, VoltsV motor, voltsPbatteryEm30015.34.254.772.9040020.43.1853.878.8350025.52.553.3285.77The efficiency, Em, is optimistic as it does not account for no-load current losses.The choice of Kv will be determined in part by available motors as they tend to come in discrete values of Kv.The ESC determines the motor voltage via the throttle control channel. Battery voltage can and should be greater than the minimum calculated here as extra thrust is needed for maneuvering and takeoff and climb. Battery voltages come in discrete multiples of nominal 3.7 Volts. It is suggested to choose a battery voltage that is at least twice the cruise motor value. This allows the thrust to increase as the square of the voltage at the expense of an increase in power by the cube of the voltage. The next table indicates the approximate thrust increase ratio, the current and power for each of the possibilities assuming 11 Volt (3S) or 7.4 Volt (2S) batteries.Kv, rpm/VoltVbatteryThrust Increase ratioPower Increase ratioIm, AmpsPbattery, W30011.15.613869364007.43.87.477.55775007.4511127935The table clearly indicates that higher Kv results in excessive current and power draws unless the input power is limited in some fashion. Note how sensitive the results are to Kv as current is proportional to Kv^2 and power to Kv^3.At this point it is suggested to search for a motor with a Kv of about 300 as it will produce a reasonable maximum power and current and a modest thrust for climbing or maneuvering. Then rerun the numbers based on the actual motor characteristics, including no-load current effects.SummaryAll of the previous analysis was done to get to approximate values that could then be examined in detail and modified based on the findings. Many assumptions were made along the way, all of which need to be verified or modified. Here is a summary of the findings.ItemValue or descriptionRange300 km (optimistic)Total Weight5 KgBattery Weight2.5 KgBattery Energy Content450 W-HrsBattery Voltage2 S (7.4V ) or 3S (11.1V)Motor Current and VoltageCurrent = 15.3 Amps, Voltage = 4.7 VoltsMotor ParametersKv = 300, Rm =.03 OhmsPropeller SizeDiameter 0.589 m, pitch 0.53 mPropeller Cruise Power45.8 WattsBattery Power72 Watts (optimistic)Cruise Thrust4.58 NewtonsCruise Cl & CdCl = 1, Cd = .046Cruise air speed 36 km/hrStall air speed30 km/hrWing Aspect ratio14Wing Area & SpanArea = 0.8 m^2, Span = 3.35 mTime to Target8.4 hours ................
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