GUS - The Drive Train Section of the Design Report

8/7/2019 GUS - The Drive Train Section of the Design Report

http://slidepdf.com/reader/full/gus-the-drive-train-section-of-the-design-report 1/19

1

THE DRIVE TRAIN POWER

REQUIREMENTS

By Augustine Owusu

Nomenclature:

α - Angle of descent

γ – Descent gradient

ω – Angular velocity

a – Acceleration

ar – radial/angular acceleration

CD – Drag coefficient

CL – Lift coefficient

g – Acceleration due to gravity

F - Thrust

L – Lift force

mb – mass of single fan blade

D – Drag force

S – Area of propeller blade or Wing area

r - Radius of fan

Re – Reynolds number

V – Speed of aircraft

Vfan – This is the tangential speed of the fan

Vsim - This is the tangential speed of the fan

W – Weight of Aircraft



2

Table of ContentsNOMENCLATURE ...............................................................................................................................1

OUTLINE OF STRATEGY ................................................................................................................... 3

OUTLINE OF REPORT ......................................................................................................................... 3

THEORY ................................................................................................................................................ 4

AN ATTEMPT TO CHOOSE A SET OF AEROFOILS ....................................................................... 6

CHOOSING A SUITABLE SOURCE OF POWER FOR THE AIRCRAFT ........................................ 9

Deriving an equation to calculate the Torque and RPM ....................................................................... 12

Calculating RPM for takeoff ................................................................................................................14

CONCLUSION ....................................................................................................................................17

APPENDIX ...........................................................................................................................................18



3

OUTLINE OF STRATEGY

1. Lump the aircraft into one mass and calculate energy requirement for a typical flight

(Use CL and CD to calculate the amount of Energy needed throughout the flight)2. Find the upper bound of power requirement

3. Use upper bound of power requirement to determine type of engines/ power sources

to use.

4. Look at how much fuel will be needed for a normal flight (based on efficiency of

power source and fuel burn)

5. Make considerations for fuel weight and weight of power source.

6. Determine the maximum flight time from based on the engine efficiency, engine size

and fuel consumption ratio.

OUTLINE OF THIS REPORT

This is a report about the procedure, which was followed in order to design the drive train. It

concludes with specifications relevant to manufacturing the components responsible for the

drive train.

This section of the report is in five parts:

Theory

Numerical calculations (Appendix)

Experiments to determine the best group of aerofoils to use or help group choose by

justifying why such aerofoils cannot be used.

Possible sources of drive power based on numerical calculations

Conclusion - This section gives a specification of the components of the drive trainbased on analytical calculations.



4

THEORY

This section starts with the aerodynamics of the aircraft as a whole. It then goes into further

Estimations to determine power requirements of the aircraft

Deductions of power requirements can be made from three flight conditions.

Gliding (Descent)

Flying straight and level in the air

Climbing

Gliding and climbing

While gliding, there needs not to be any thrust. Therefore F can be assumed to be zero (F=0)

The descent gradient Gamma (γ) can be expressed in terms of Lift and drag. In fact the

descent gradient is equal to the maximum lift-drag ratio.

(1)

Angle of descent

(2)

γ and α are positive when the aircraft is climbing.

Rate of climb or rate of descent (3)

For maximum rate of climb and minimum rate of descent, fly at 75% the speed of (L/D)max

Thrust horse power: (THP)

This is the effective amount of power an engine generates at full capacity taking into account

its efficiency.

THP = ηBHP (4)

Break horse power of engine (BHP)

This is the amount of power that an engine could generate at full capacity if it was 100%

efficient.



5

The general Engine performance equation

In unaccelerated level flight when thrust = Drag,

Power required = DV (5)

If a constant acceleration is made at level flight, them

F-D = (W/g)a (6)

When climbing, the weight of the aircraft (W) is being lifted at a rate of climb (vc)

So, Power required = Wvc (7)

Putting this all together gives the total power requirement of the Engine (I.e. Thrust Horse

Power)

(BHP has been converted from FPSR units into Watts, hence the 550)

Conversion:

BHP is denoted by P

THP = ηP = FV/550 = Pt (8)

And Pe = 550P (9)

Pe is when BHP is converted into lbf/s

Weight of aircraft (W) = Empty weight + payload (9a)



6

Minimum power to fly level in the air1

(10)

i.e.: power to fly level+ power to maintain airspeed

(11)

And also

(12)

Field performance

The shortness of take-off distance (TOD) depends on how quickly enough lift force can be

generated to be greater than the aircraft’s weight. This primarily depends on Weight to thrust

ratio (W/F) or Power to Thrust ratio (P/F).

(13)

(14)

AN ATTEMPT TO CHOOSE A SET OF AEROFOILS

An attempt was made to run a preliminary investigation into which group of aerofoils would

give optimal performance to meet the design requirements.

While choosing the dimensions for the aerofoil, different factors were considered as listed

below:

Lift (as high as possible)

Drag (As low as possible) – Can go for Corresponding drag of the stall angle, as low

speed requirement.

1According to Charles Renard (1847-1905) – The Design of the Aeroplane by Stinton, D



7

Aspect ratio (As high as possible)

Stall angle (as high as possible)

Strategy

Each factor is to be recorded for each aerofoil and the top 5 selected for each.

The aerofoil that wins top 5 for all factors is to be chosen

Choosing the shape of aerofoil

With the NACA dimensions for aerofoils, an experiment was conducted to choose the best

kind of aerofoil which will give the highest possible lift coefficient and lowest possible drag

coefficient.

Java-foil, (an Aerofoil aerodynamics simulation) software was used to compute the

aerodynamic parameters of each shape of aerofoil.

For the aerofoil shapes, thickness was first varied to optimise stall angle and lift coefficient.

The camber of the aerofoil shape with optimum thickness was then varied to create a new set

of shapes.

Finally, the camber location of the optimum shape was varied to get the best shape possible.

Results were as tabulated in Fig.1.

RESULTS

Fig.1



8

Graph comparing the magnitudes of Stall angle and highest lift for different aerofoil shapes

Fig.2

Conclusion

From the table in fig. 1 and from the graph in fig.2, it is clear that NACA 30440 is a superior

choice for optimal lift coefficient and stall angle. However, it was considered impractical to

model in solid-works due to the fact that, even thought Solidworks allows for the user to

import the coordinates of the shape and extrude it into a solid, it does NOT allow options to

edit (i.e. cut through the 3-dimensional extrusion model made from it).

Based on the above justification, the NACA convention was NOT used by the group for

aerofoils. Please refer to the “The Wing Calculations” section of the report for the aerofoil

used.

0

5

10

15

20

25

30

35

40

0 0 0 3

C

0 0 0 6

C

0 0 0 9

C

0 0 1 2

C

0 0 1 5

C

0 0 1 8

C

0 0 2 1

3 4 2 1

C

6 4 2 1

C

9 4 2 1

1 2 4 2 1

C

1 5 4 2 1

C

1 8 4 2 1

C

2 1 4 2 1

C

3 0 4 4 0

C

3 0 5 4 0

Stall angle /(⁰)

Highest lift

MAGNITUDE



9

CHOOSING A SUITABLE SOURCE OF POWER FOR THE AIRCRAFT

Considerations made while choosing engine

A bigger engine is needed to provide enough thrust if the takeoff distance is shorter, the

altitude is higher and the climate is hotter.

Range in Air- The distance that the aircraft is able to fly with the fuel carried. This depends

on the engine’s fuel consumption and cc or carrying capacity for fuel.

Possible power sources were

Electric

o Batteries

o Fuel cells

o Capacitors

Internal Combustion Engines

Fig.3 below compares the considered sources of power on a graph of Energy Density (How

long each Kg of power medium would generate power for) against Power Density(How much

power each Kg of power medium would generate)

Figure comparing different power sources

http://www.ftconferences.com/userfiles/file/Whittingham%20Stan%20White%20paper.pdf

Fig.3



10

Justification for Choosing Internal Combustion Engines

Even though gas turbines have higher performance than the energy sources listed above, they

were considered impractical for the design, due to the fact that this aircraft was going to bepowered by rotor fans based on the output Torque of the power source. Gas turbines give

direct thrust for aircrafts by letting out heated and compressed gas. Therefore this ruled out

any considerations for Gas turbines to power up the aircraft.

Referring to Fig.3. it can be seen that

o Capacitors have a high power density but low energy density,

o Pseudo-capacitors have a medium power density, and medium Energy Density

o Batteries have low power density, but high Energy density

o Fuel Cells performance is similar to batteries.

Internal Combustion Engines (ICEs) had Power Densities ranging from 100kW/kg to

1000kW/kg, matching Energy Densities ranging from 100 to 1000 kWh/Kg.

Therefore ICEs gave the best option of engine size to match the most output for power and

duration of power.

Based on the numerical calculations the minimum Break horse power of engine was decided

(as the upper bound of Break horse power required for a normal cruise flight – please refer

to table in appendix 1).

It was concluded that an engine with a minimum of 14622 BHP had to be chosen in order to

safely meet the flight requirements.

Based on the 14622 BHP power requirements, a decision was made to target a minimum

BHP requirement of 15000, in order to allow room for engine size and lower engine

efficiencies.



11

Challenges faced with implementing Break Horse Power into design

A major challenged faced during the investigation into suitable Engines to rotate the fan was

the fact that there were no internal Combustion Engines with a single Unit producing 15000

nor half the value (7500BHP) to have one to power up each fan on either side of the aircraft.

A consideration was made that 15000BHP could be divided among numerous (number

justified in fig.5) automotive Engines.

Considered locations to implement the Engine into the design

Original Image (Without annotation) by Dan Evans

Fig.4

Ideally, there must be a few engines as possible.

Looking at automotive engines, there is legislation governing the maximum BHP of Engines.

Currently, the boundaries are around 2000BHP on the Dodge Viper Engine.

This would render it impractical to use any existing factory batch automotive engines to

successfully fly the aircraft. However, a new engine with the required BHP could be

designed.

With reference to the groups work on the aircraft’s wing, there was not a possibility to

generate different RMPs to simulate Magnus effect. This would have allowed for

experimental values, with which the corresponding RPM of shaft could be known for the

required BHP. Therefore manual calculation and iteration were made to estimate the required

RPM of the fan’s shaft, which would help determine the Horse power of the engine required.



12

Deriving an equation to calculate the Torque and RPM

An analytical method was developed estimate the RPM requirement of the fan’s shaft in

order for the aircraft to be able to take off. This method is as follows:

Free body diagram of the wing:

Fig.5

Approximations can be made such that each rotor blade is modelled as a single aerofoil

Convention

If a convention is used where the rotor blades are named numbers 1 to 30 where the

parameter for each is subscripted its number. So rotor 1 is R1

For each rotor, we can calculate its contribution to the lift, thrust, drag and weight as it rotates

one full revolution with respect to discrete time (dt).

We can then say that (14)

And of course the frequency of shaft spin will be (15)

θ



13

From this we can calculate RPM and this will be a component of our following derived

equations.

Since we have 30 blades in the fan at equally spaced angles,

(16)

Equation for lift

The Contribution of lift for each one of the blades is a vertical component. Therefore it will

vary will vary with (dt) as the fan spins. The angular speed will therefore determine the lift.

We could make an assumption that, since the fan blades are identical, each will make

approximately the same magnitude of contribution to the. Therefore the lift force required canbe divided into 30 parts, and each part will be the vertical component of each fan blade for a

given time,

Now an equation showing the relationship between lift and angle of attack is needed.

From prove theory, radial acceleration For the sake of this method, if the circular path of rotation for the fan blades was stretched

out into one plane, then ar would be in linear motion.

This can be used to calculate the Lift contribution of each fan blade with respect to angle of attack (α) or angular velocity (ω) or frequency of revolution (f).

These values are arbitrary, dependent on the chosen ar

Summary of derived method

This method is based on the fact that we could pretend to stretch the circular motion of the

fan into a straight line while it is spinning in order to get approximate values of lift, drag and

trust.

Angularvelocityω



14

Fig.6

For the linear motion, each magnitude of a blade’s “swim force”

Swim force being the force that the blade exerts as a result of contact with air, which has the

direction normal to the circumference of the rotor fan;

This force only calculated the force normal to the path of the fan blades while they were

rotating

Knowing the mass of each fan blade = 6.408kg

Calculating RPM for takeoff

At takeoff, the vertical component of Lift must be greater than the aircraft’s weight. (i.e

L>W)

Takeoff weight of aircraft is known to be:

Starting Lift at 35316N, we can substitute lift and weight into the lift equation (as recollected

from equation (12))

(16a)

W = 35316 ; L= 35316 + Varying ; ρ= 1.29kgm3; Radius of fan (r) = 500mm = 0.5m

For the coefficient of Lift, D

Vsim – This is the effective velocity of air around the aerofoil. It is the sum of the tangential

velocity (Vfan) of the fan and the velocity of the aircraft.

Linear velocity (V)



15

So (16b)

The Vsim happens because the air being blown by the fan is travelling around the aerofoil at

approximately the fan speed; as well the aircraft is travelling at takeoff speed, which will

increase the speed of air travelling around the aerofoil as well.

To convert it back to Vfan will then be converted back to ω for the fan once it has been

determined.

Conversion was made between angular and linear velocity as (17)

Since the fan is spinning and blowing air over the aerofoil, the following method is

based on the assumption that the speed of air over the aerofoil is equal to the

tangential velocity of the fan.

A Java fluid dynamics Java application called (JAVAFOIL) was used to simulate the flow

pattern of air around a single fan blade.

Fig. 7 shows the flow field of air around a single fan blade

Fig.7



16

A maximum climbing angle of 10degrees was selected for the aircraft. This was due to the

fact that the aircraft rate of increase of the lift coefficient began to only increase

exponentially to a peak where it would start decreasing after stall. This is illustrated in the

Java foil graph in fig.7

For the takeoff of 10o

, each fan blade had a Lift coefficient of 0.9.

Now we can calculate air speed around aerofoil before takeoff

Inserting the deduced values of L, and CL into the Lift Equation (16a),

Vfan was calculated to get a value of: 50703m/s. This corresponds to an angular velocity of

101406 rads-1

.

This can be converted back into angular velocity by dividing the 0.5m radius of the fan, and

into RPM by multiplying the resultant by (

The resultant spin rate of the shaft needed has an upper-bound approximation of

989,000RPM (16483 revolutions per second). This was the lower bound value for the RPM

range.

NB: It was acknowledged that the lift contribution due to the takeoff speed would ideally be

taken into account. However, this speed (17 m/s) was negligible compared to the

Tangential velocity of the fan (50703m/s) and so it did not make any difference to the

approximation of RPM.

(18)

Polar moment of inertia (J): 412.7kgm2

Where torque = (19)

ar is 4885 rads-2

Torque required is therefore = 2016040Nm

So power required is 2.05 x 1011

Watts. This is an equivalent of 2.75x108

Thrust Horse

power.

Therefore an engine with average of 35% percent efficiency will have to have a BHP of

(2.75x108

/0.35) => i.e 7.86x108BHP

Pricing

BHP Engine



17

price

592 £51,000.0

1183 £600,000.0

730 £300,000.0

1001 £273,000.0

588 £165,000.0

Fig.8 (Data collected fromhttp://www.supercarworld.com/)

A graph was created from fig.8 to find a possible relationship between engine price and

Break horsepower.

Fig.9

Fig. 9 is a graph of engine price against break horsepower. The two quantities show a

logarithmic correlation.

CONCLUSION

An engine with a minimum of 7.86x108 Break Horse power (BHP) will make the aircraft fly.

This is clearly unrealistic a horsepower to find in any automotive engine. Currently, the legal

limit for automotive engines is 2000BHP.

£0.0

£100,000.0

£200,000.0

£300,000.0

£400,000.0

£500,000.0

£600,000.0

£700,000.0

0 200 400 600 800 1000 1200 1400

A graph of ENGINE PRICE against

Break Horse PowerEngine price

BHP

http://www.supercarworld.com/





18

Based on this a recommendation has been made in this report to manufacture a new engine to

meet the power requirements.

Fig. 9 shows a linear correlation between Engine price and its

Based on calculation data, the following were concluded about the aircraft for design

specification

The aircraft will have a 5.3 second run time before takeoff (based on the equations of

linear motion)

The engine will have a Break horse power of 7.86x108BHP

APPENDIX 1

Numerical calculations (done with equations from the “Theory” section).

Weight of

aircraft with

passengers

(plus 100kg

allowance)/Kg

Drag

force(D)

True air

speed

/(m/s)

(L/D)

ratio

Descent

gradient

Angle

of

descent

Rate of

climb

(vc)

Rate of

descent

(vd)

Engine

Power

requirement

(BHP)

3700 0.78813 15.00 3.51 0.29 17.31 3.03 -4.33 11225

3700 1.02312 16.00 3.30 0.31 18.45 3.44 -4.92 12759

3700 0.84388 17.00 3.55 0.29 17.13 3.40 -4.85 12585

3700 1.13943 18.00 3.42 0.30 17.80 3.74 -5.34 13854

3700 1.22132 19.00 3.46 0.29 17.58 3.90 -5.57 14449

3700 1.4344 20.00 3.60 0.28 16.90 3.94 -5.63 14622



19

REFERENCES

BOOKS

The design of the Aeroplane By Darrol Stinton

Aircraft Engine Technology By Ren Yin Tai AND Ikwan Jamaludin

Engineering Fundamentals of the Internal Combustion Engine, By Willard Pulkrabek

Websites:

www.supercarworld.com

http://www.bbc.co.uk/topgear/

http://www.mh-aerotools.de/airfoils/javafoil.htm

http://www.engineeringtoolbox.com/moment-inertia-torque-d_913.html

http://en.wikipedia.org/wiki/Equations_of_motion

Documents

GUS - The Drive Train Section of the Design Report