Comparison between Lift and Drag Produced by a Legacy Wing VS a Wing with Tubercles (Humpback Whale Fin's Inspired)

* Link for Plots (now showing here for some reason) http://3dimensionaldesigningandmanufacturing.blogspot.com/2015/07/plots-for-comparison-between-lift-and.html

Following data was obtained from the CFD Simulations carried out in SolidWorks Flow Simulation Premium.

Project: Design of a Wing/Blade with Tubercles for Airplanes and/or Turbines

Without Tubercles
Air Speed in Km/h	Lift in N	Drag in N
150	46.307	14.775
140	39.942	12.917
130	33.432	11.057
120	28.807	9.498
110	24.234	7.928
100	20.593	6.625
90	15.836	5.352
80	12.482	4.205
70	9.411	3.243
60	7.272	2.406
50	4.873	1.680
40	3.130	1.082
30	1.763	0.612
20	0.810	0.279
10	0.231	0.072

 

 

With Tubercles
Air Speed in Km/h	Lift in N	Drag in N
150	50.616	11.360
140	48.131	10.008
130	37.190	8.505
120	30.988	7.309
110	24.784	6.079
100	20.892	5.094
90	17.225	4.146
80	13.412	3.287
70	9.955	2.507
60	7.444	1.849
50	4.955	1.286
40	2.991	0.828
30	1.652	0.468
20	0.725	0.212
10	0.214	0.057

 

Comparison between Lift and Drag

Air Speed in Km/h	Percentage Less Drag		Percentage More Lift
150	23.113		8.513
140	22.520		17.014
130	23.080		10.105
120	22.974		7.038
110	23.322		2.219
100	23.109		1.431
90	22.534		8.064
80	21.831		6.934
70	22.695		5.465
60	23.150		2.311
50	23.452		1.655
40	23.475		-7.523
30	23.529		-6.719
20	24.014		-11.72
10	20.833		-7.94

 

It is clear that the wing with tubercles not only produces more lift at a particular velocity but also less drag.

Data for the Wing without Tubercles:

Wing Span: 1.07 m

Chord Length: 0.229 m

Air Velocity: 0-150 Km/h head on

Vertical Pitch: 0 Degree

Gravity Considered

Fluid: Dry Air at STP

Mesh Settings: Coarse (3/8)

Data for the Wing with Tubercles:

Wing Span: 1.067 m

Chord Length Large: 0.229 m

Chord Length Small: 0.203 m

Air Velocity: 0-150 Km/h head on

Vertical Pitch: 0 Degree

Gravity Considered

Fluid: Dry Air at STP

Mesh Settings: Coarse (3/8)

Let's now take a look at visual representation of data.

This Plot Shows Air Velocity VS Drag, Lift by the Wing without Tubercles

This Plot Shows Air Velocity VS Drag, Lift by the Wing with Tubercles

As you can see from above two plots; the wing with tubercles generates more lift and less drag.

This Plot Shows Air Velocity VS Lift Generated by the Wings

The green line represents the Lift generated by the wing with tubercles. It is between two to six percent more at each velocity.

This Plot Shows Air Velocity VS Drag Generated by the Wings

The green line represents the Drag generated by the wing with tubercles. It is around twenty two percent less at each velocity.

This Plot Shows Air velocity VS Lift to Drag Ratio

It is clear from this plot that Lift to Drag ratio of the wing with tubercles is around thirty three percent more for the wing without tubercles at a velocity point.

 

This Plot Shows Air Flow around the Wings at 150 Km/h from the Right Side

This Plot Shows Air Flow around the Wings at 150 Km/h

The Need for Tubercles

In aviation there are four forces at play, Lift which over comes Weight and Thrust which overcomes Drag. For a cruise speed at a particular altitude, three of these forces are almost constant. Our goal is to minimize Thrust, Drag and Weight and maximize Lift, this is because Thrust costs in terms of fuel flow rate and Weight and Drag negatively impacts on the agility of the aircraft. Aerodynamically efficient Wings and/or Blades with "Tubercles" will not only increase Lift and but also decrease Drag. This all means that we will need less Thrust for a cruise speed than before, that results in savings in terms of fuel which will result in healthier environment.

 

Applications:

 

Canal Turbine Concept

It's a concept I am currently working on, so far I gave made a CAD model (renderings attached) of it in SolidWorks and analyzed it using its built in CFD module.

There are many advantages of canal turbines over wind turbines, prominent one's being:

 

Unidirectional flow

Water flows in one direction in a canal so we don't need pitch and yaw control surfaces. That simplifies the design process and reduces weight.

Constant flow rate

We (humans) control water flow rate through canals and it's almost same all year, so we don't have to worry about blade aero foil design to suit variable/abruptly variable flow rate, that makes design process further straight forward.

Large Electricity potential

Canals are 100s of km long, imagine the electricity potential in the canals. You can put these turbines in irrigation canals and it'll power nearby villages and all the irrigation equipment etc.

Higher Power/Discharge Ratio

Water is ~816 times dense (powerful) than air, so for the same discharge (flow) rate we get potentially 816 times more power. Which means more we can make designs that are lighter, smaller and easier to manage and maintain.

Easy maintenance

Fitted less than ~1 m deep inside the canal and can be retracted for maintenance at ground level, making maintenance very easy or better yet, we can maintain them while canals are being cleaned.

 

T-S Diagrams for Parametric Cycle Analysis of an Ideal Ramjet Engine

The text is available for reading at http://3dimensionaldesigningandmanufacturing.blogspot.com/2014/09/parametric-cycle-analysis-of-ideal.html, somehow, the diagrams are showing there.

1,600 K, Mach 1 

1,900 K Mach 1 

2,200 K Mach 1 

 1,900 K, Mach 2

 2,200 K, Mach 2

1,600 K, Mach 2

Parametric Cycle Analysis of an Ideal Ramjet Engine

Extract from my rejected thesis from my university days. Optimization of Thrust Per Unit Mass Flow of a Jet Engines by Optimizing the Overall Compression Ratio, to design Multi-diameter Inlets, because an inlet works best for a set specific of conditions. Hope it helps you in your scholarly work.

Ramjet engine

A ramjet engine is the simplest of all aircraft engines. It consists of an inlet or a diffuser, a combustor and a nozzle. Air is fed into the diffuser, it increases the static pressure and temperature of air by slowing it down. Then the air is fed into the combustor, where it is mixed with fuel and is ignited, resulting in increased energy. This corresponds to an increased temperature. The combustion occurs at a constant pressure for sub sonic flows. The nozzle then expand the gas to ambient pressure, with a decrease in temperature. This process results in increased in KE for gas.

Formulations

The formulations for the cycle analysis are as follows.

General Gas Constant, R=((g-1)/ g)*C_p

Velocity of Sound, a_o=(1000*g*R*T_o).^(1/2)
Total to Static Temperature Ratio, t_r=1+(((g-1)/2)*(M_o.^2))
Burner Exit Enthalpy to Ambient Enthalpy Ratio, t_l=T_t4/T_o
Velocity Ratio, v₉/a_o=M_o*((t_l/t_r).^(1/2))
Thrust per Air Flow, F/m_o=a_o*((v₉/a_o)-M_o)
Fuel Air Ratio, f=((C_p*T_o)/h_PR)*( t_l-t_r)
Thrust Specific Fuel Consumption, S=f/(F/m_o)
Thermal Efficiency, h_T=1-(1/t_r)
Propulsive Efficiency, h_P=2/(((t_l/t_r).^(1/2))+1)
Overall Efficiency h_O=h_T*h_P
Total to Static Temperature Ratio for Maximum Thrust per Air Flow, t_{r (maxFmo)}=t _l.^(1/3)
Mach Number of Flight for Maximum Thrust per Air Flow , M_{o (maxFmo)}=((2/(g-1))*( t_{r (maxFmo)}-1)).^(1/2)
Temperature at Station T_to (after Compression), T_to=t_r*T_o
Nozzle Exhaust Temperature, T₉=T_o*(t_l/t_r)

Calculations

By using these equations, and by entering the following inputs:

For Mach Number of Flight= 1

T_o= 216.7 K, g= 1.4, C_p= 1.004 KJ/(Kg.K), T_t4= 1,600 K, M_o= 1, h_PR= 42,800 KJ/Kg

We get the Following Results:

Station Temperatures for T-S Diagram

Ambient Temperature, Station T₀, 216.7 K
Temperature after Compression, Station T_to and T_t2, 260.04 K
Temperature at Nozzle Entry, Station T_t4, 1,600 K
Temperature at Nozzle Exit, Station T₉, 1,333.33 K

T-S Diagram generated from MATLAB

Results Based on the Data Entered

General Gas Constant= 0.286857 KJ/(Kg.K)

Velocity of Sound= 295.003 m/s

Total to Static Temperature Ratio= 1.2

Burner Exit Temperature to Ambient Temperature Ratio= 7.38348

Gas Exit Velocity to Velocity of Sound Ratio= 2.4805

Thrust per Unit Airflow= 436.753 N/(Kg/s)

Fuel to Air Ratio= 0.0314327

Thrust Specific Fuel Consumption= 71.9691 (mg/s)/N

Thermal Efficiency= 16.6667 %

Propulsive Efficiency= 57.4629 %

Overall Efficiency= 9.57716 %

For Optimum Mach Number of Flight

Total to Static Temperature Ratio for Maximum Thrust per Air Flow= 1.94724
Optimum Mach Number of Flight for Maximum Thrust per Air Flow= 2.17629

For all the same values as above, except T_t4= 1,900 K, we get these results:

Station Temperatures for T-S Diagram

Ambient Temperature, Station T_o, 216.7 K
Temperature after Compression, Station T_t0 and T_t2, 260.04 K
Temperature at Nozzle Entry, Station T_t4, 1,900 K
Temperature at Nozzle Exit, Station T₉, 1,583.33 K

T-S Diagram from MATLAB

Results Based on the Data Entered
General Gas Constant= 0.286857 KJ/(Kg.K)
Velocity of Sound= 295.003 m/s
Total to Static Temperature Ratio= 1.2
Burner Exit Temperature to Ambient Temperature Ratio= 8.76788
Gas Exit Velocity to Velocity of Sound Ratio= 2.70307
Thrust per Unit Airflow= 502.41 N/(Kg/s)
Fuel to Air Ratio= 0.0384701
Thrust Specific Fuel Consumption= 76.5712 (mg/s)/N
Thermal Efficiency= 16.6667 %
Propulsive Efficiency= 54.0093 %
Overall Efficiency= 9.00155 %

For Optimum Mach Number of Flight

Total to Static Temperature Ratio for Maximum Thrust per Air Flow= 2.06205
Optimum Mach Number of Flight for Maximum Thrust per Air Flow= 2.30439

For all the same values as above, except T_t4= 2,200 K, we get these results:

Station Temperatures for T-S Diagram

Ambient Temperature, Station T_o, 216.7 K
Temperature after Compression, Station T_t0 and T_t2, 260.04 K
Temperature at Nozzle Entry, Station T_t4, 2,200 K
Temperature at Nozzle Exit, Station T₉, 1,833.33 K

T-S Diagram from MATLAB

Results Based on the Data Entered


General Gas Constant= 0.286857 KJ/(Kg.K)
Velocity of Sound= 295.003 m/s
Total to Static Temperature Ratio= 1.2
Burner Exit Temperature to Ambient Temperature Ratio= 10.1523
Gas Exit Velocity to Velocity of Sound Ratio= 2.90865
Thrust per Unit Airflow= 563.057 N/(Kg/s)
Fuel to Air Ratio= 0.0455075
Thrust Specific Fuel Consumption= 80.8222 (mg/s)/N
Thermal Efficiency= 16.6667 %
Propulsive Efficiency= 51.1686 %
Overall Efficiency= 8.5281 %

For Optimum Mach Number of Flight

Total to Static Temperature Ratio for Maximum Thrust per Air Flow= 2.16532
Optimum Mach Number of Flight for Maximum Thrust per Air Flow= 2.41383

For Mach Number of Flight= 2

By using these equations, and by entering the following inputs:

T_o= 216.7 K, g= 1.4, C_p= 1.004 KJ/(Kg.K), T_t4= 1,600 K, M_o= 2, h_PR= 42,800 KJ/Kg

Station Temperatures for T-S Diagram

Ambient Temperature, Station T_o, 216.7 K
Temperature after Compression, Station T_t0 and T_t2, 390.06 K
Temperature at Nozzle Entry, Station T_t4, 1600 K
Temperature at Nozzle Exit, Station T₉, 888.889 K

T-S Diagram from MATLAB

Results Based on the Data Entered

General Gas Constant= 0.286857 KJ/(Kg.K)
Velocity of Sound= 295.003 m/s
Total to Static Temperature Ratio= 1.8
Burner Exit Temperature to Ambient Temperature Ratio= 7.38348
Gas Exit Velocity to Velocity of Sound Ratio= 4.05065
Thrust per Unit Airflow= 604.947 N/(Kg/s)
Fuel to Air Ratio= 0.0283827
Thrust Specific Fuel Consumption= 46.9177 (mg/s)/N
Thermal Efficiency= 44.4444 %
Propulsive Efficiency= 66.1086 %
Overall Efficiency= 29.3816 %

For Optimum Mach Number of Flight

Total to Static Temperature Ratio for Maximum Thrust per Air Flow= 1.94724
Optimum Mach Number of Flight for Maximum Thrust per Air Flow= 2.17629


For all the same values as above, except T_t4= 1,900 K, we get these results:

Station Temperatures for T-S Diagram

Ambient Temperature, Station T_o, 216.7 K
Temperature after Compression, Station T_t0 and T_t2, 390.06 K
Temperature at Nozzle Entry, Station T_t4, 1900 K
Temperature at Nozzle Exit, Station T₉, 1055.56 K

T-S Diagram from MATLAB

Results Based on the Data Entered

General Gas Constant= 0.286857 KJ/(Kg.K)
Velocity of Sound= 295.003 m/s
Total to Static Temperature Ratio= 1.8
Burner Exit Temperature to Ambient Temperature Ratio= 8.76788
Gas Exit Velocity to Velocity of Sound Ratio= 4.41409
Thrust per Unit Airflow= 712.163 N/(Kg/s)
Fuel to Air Ratio= 0.0354201
Thrust Specific Fuel Consumption= 49.7359 (mg/s)/N
Thermal Efficiency= 44.4444 %
Propulsive Efficiency= 62.3627 %
Overall Efficiency= 27.7168 %

For Optimum Mach Number of Flight

Total to Static Temperature Ratio for Maximum Thrust per Air Flow= 2.06205
Optimum Mach Number of Flight for Maximum Thrust per Air Flow= 2.30439

For all the same values as above, except T_t4= 2,200 K, we get these results:

Station Temperatures for T-S Diagram

Ambient Temperature, Station T_o, 216.7 K
Temperature after Compression, Station T_t0 and T_t2, 390.06 K
Temperature at Nozzle Entry, Station T_t4, 2200 K
Temperature at Nozzle Exit, Station T₉, 1222.22 K

T-S Diagram from MATLAB

Results Based on the Data Entered

General Gas Constant= 0.286857 KJ/(Kg.K)
Velocity of Sound= 295.003 m/s
Total to Static Temperature Ratio= 1.8
Burner Exit Temperature to Ambient Temperature Ratio= 10.1523
Gas Exit Velocity to Velocity of Sound Ratio= 4.7498
Thrust per Unit Airflow= 811.2 N/(Kg/s)
Fuel to Air Ratio= 0.0424575
Thrust Specific Fuel Consumption= 52.3391 (mg/s)/N
Thermal Efficiency= 44.4444 %
Propulsive Efficiency= 59.261 %
Overall Efficiency= 26.3382 %

For Optimum Mach Number of Flight

Total to Static Temperature Ratio for Maximum Thrust per Air Flow= 2.16532
Optimum Mach Number of Flight for Maximum Thrust per Air Flow= 2.41383

Conclusions

By looking at the results, we find the following results.

1.       The performance of any ramjet engine relies heavily on the stagnation temperature increase across the burner.
2.       To have efficient compression of the air, the ramjet requires high flight speeds.
3.       For ramjets, the static thrust is zero, they must be moving to develop thrust.

T-s diagrams available here, some how not showing here.
http://3dimensionaldesigningandmanufacturing.blogspot.com/2014/09/t-s-diagrams-for-parametric-cycle.html