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

Tips for Saving Water Amid Water Crisis in Pakistan

It is pretty clear that water reserves in Pakistan will be reduced to one half of what they were on 11 August 2014 in next 6 years. Since the government is not doing anything serious about it, here are some ways in which us the common people can contribute to help save water for our future generations and to prevent sever draught conditions post 2020 AD.

 

If we manage to follow these very simple guidelines, according to a rough estimate, each of us can save up to ~18 cubic meter of water a year (that's ~18,000 Liters a year), multiply it by our population, it will be 3,240,000,000,000 Liters of water a year, i.e. 3,240,000,000 cubic meter of water a year. In comparison capacity of world's largest earth filled dam (Terbela Dam) is 13,690,000,000 cubic meter. so we'll save water equivalent to Terbela Dam's capacity every 4 years.

 

• Water your lawn at night. It is common sense that water evaporates quickly during the day time because of the sun, so sprinkle when it’s more likely to stay in the soil.
• Do not use a dish washer. A common Pakistani person may not have even heard the term dish washer, but some of us have and some of us use them too. Most dish washers use 1,000+ Watt of electrical power, if you have to use a dish washer use a solar powered one. To wash the dishes, fill a ~6 Liter bottle with water and use it to rinse and soap the dishes, then use tap directly to wash them, as quickly as possible.
• Some of us are privileged enough to have a swimming pool in our homes, we should cover them to prevent contamination of water and evaporation accelerated by the Sun during the day time. Buckets, tubs lying in the open should be covered as well.
• Do not go to the car wash. Car washers in petrol stations or dedicated one's, both use ~200 Liters of water per car wash, you can wash your car at home by using ~4 buckets of water, ~25 liters each. Wash and rinse one fourth of your car with each bucket, starting from roof and ending at the rim/tire. You can wash your car up to 3 times a year at home, still using less water than washing station. If you have to use a car wash, use it no more than once a year.
• This one is especially for home owners, collect the rain water. Make an underground tank in your home, under the lawn or the back yard, save water in it, then use water filters to use that water for toilet use or to wash your car or water the plants/lawn, even drinking and cooking after boiling.
• While brushing teeth, shaving, soaping your body while bathing, turn of the tap. While bathing it is preferable to not to use the shower, use old school tub and mug combo. In winters especially, do not waste cold water while you wait for hot geezer's water, collect that in separate bucket and reuse as necessary, or use instant electric geezers near to the taps/showers (minimum piping). Check out your toilets for leaks. Pour some color in the flush tank, if there is a leak, the color will help you verify it. Fix it if necessary, replace the toilet if unfixable.
• Wash your carpets once a year. While washing clothes, use appropriate amount of washing powder so you don't have to rinse them again more than twice.

9,000 Sq. Ft. House Design For Sale

Isometric View

 Isometric View

Rear Isometric View


6x bedrooms, 7x bathrooms, 2x Living Room, 1x Kitchen, 1x Drawing & Dining, 1x Store, Porch for 3x cars, 1x Servant area, 2x terraces.

If you need the CAD files or want to buy the design for commercial (Building, and selling) use contact via this account or mail to fadoobaba@live.com

 Right Side View

 Left Side View

 Front View

Rear View

4,500 Sq. Ft. House Design for Sale

4x bedrooms, 5x bathrooms, 2x Living Room, 1x Kitchen, 1x Drawing & Dining, Porch for 4x cars, 1x Servant area, 2x terraces.

If you need the CAD files or want to buy the design for commercial (Building, and selling) use contact via this account or mail to fadoobaba@live.com

9000 Sq. Ft. House Design for Sale

7x bedrooms, 8x bathrooms, 2x Living Room, 2x Kitchen, 1x Drawing & Dining, Porch for 4x cars, 1x Servant area, 2x terraces, 2 lawns.

If you need the CAD files or want to buy the design for commercial (Building, and selling) use contact via this account or mail to fadoobaba@live.com

 Front View

 Right View

 Top View

Isometric View

Dinner Table Design for Sale

8.75 ft. in dia. 2.5 ft. high, chairs 12, 1.625 ft. diameter, 1.5 ft. high.

If you need the CAD files or want to buy the design for commercial (Building, and selling) use, contact via this account or mail to fadoobaba@live.com

Comparison between Down-Force and Drag Produced by a Legacy Spoiler VS a Spoiler with Tubercles (Humpback Whale Fin's Inspired)

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

Without Bumps
Air Speed in Km/h	Down Force in N	Drag in N
120	98.682	33.234
110	82.88	27.957
100	68.266	23.02
90	55.299	18.668
80	43.529	14.697
70	33.284	11.255
60	24.438	8.272
50	16.982	5.769
40	10.83	3.688
30	6.08	2.081
20	2.681	0.929
10	0.648	0.235

With Bumps
Air Speed in Km/h	Down Force in N	Drag in N
120	108.238	30.47
110	90.599	25.549
100	74.818	21.047
90	60.423	17.014
80	47.695	13.443
70	36.441	10.27
60	26.682	7.532
50	18.504	5.228
40	11.82	3.352
30	6.613	1.886
20	2.909	0.841
10	0.685	0.211

Comparison between Down Force and Drag

Air Speed in Km/h	Percentage Less Drag		Percentage More Down Force
120	8.32		8.83
110	8.61		8.51
100	8.57		8.76
90	8.86		8.48
80	8.53		8.73
70	8.75		8.66
60	8.95		8.41
50	9.38		8.23
40	9.11		8.38
30	9.37		8.06
20	9.47		7.84
10	10.21		5.4

It is clear that the spoiler with humpback whale's fin's inspired profile not only produce more down force at a particular velocity but also less drag.

Data for Spoiler without Humpback Whale's Fin's Inspired Bumps:

Wing Span: 100 cm

Chord Length: 17.5 cm

Air Velocity: 0-120 Km/h head on

Vertical Pitch: 22.5 Degree Downwards

Gravity Considered

Fluid: Dry Air at STP

Mesh Settings: Coarse (3/10)

Data for Spoiler with Humpback Whale's Fin's Bumps:

Wing Span: 100 cm

Chord Length Large: 17.5 cm

Chord Length Small: 15.75 cm

Air Velocity: 0-120 Km/h head on

Vertical Pitch: 22.5 Degree Downwards

Gravity Considered

Fluid: Dry Air at STP

Mesh Settings: Coarse (3/10)

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

This Plot Shows Air Velocity VS Drag, Down-Force by the Spoiler without Bumps

This Plot Shows Air Velocity VS Drag, Down-Force by the Spoiler with Bumps

As you can see from above two plots; the spoiler with the whale's fin like profile generates more down force and less drag.

This Plot Shows Air Velocity VS Down-Force Generated by the Spoilers

The green line represents the Down-Force generated by the spoiler with whale's fin's inspired design. It is around eight percent more at each velocity.

This Plot Shows Air Velocity VS Drag Generated by the Spoilers

The green line represents the Drag generated by the spoiler with whale's fin inspired design. It is around nine percent less at each velocity.

This Plot Shows Air velocity VS Down-Force to Drag Ratio

It is clear from this plot that Down-Force to Drag ratio is around sixteen percent more for whale's fin's inspired spoiler than the legacy one at each velocity.

This Plot Shows Air Flow Around the Spoiler without Bumps at 120 Km/h from the Right Side.

This Plot Shows Air Flow Around the Spoiler without Bumps at 120 Km/h.

This Plot Shows Air Flow Around the Spoiler with bumps at 120 Km/h.

This plot Shows Air Flow Around the Spoiler with bumps at 120 Km/h.

A simple stress analysis was carried out on both spoilers at 120 Km/h. FOS was greater than 1 for both cases.

Advantages of Spoilers:

The main benefit of installing a spoiler on a car is to help it maintain traction at very high speeds. Particularly at speeds around 90 Km/h. A car with a spoiler installed will be easier to handle at highway speeds. Rear spoilers such as the one's analysed in this study; push the back of the car down so the tires can grip the road better and increase stability. It also increases the braking ability of the car.

To build the prototypes and complete the study further, I need donations. To donate your part send an email to fadoobaba@live.com , tweet @fadoobaba, PM at https://www.facebook.com/ThreeDimensionalDesign orhttps://grabcad.com/fahad.rafi.butt or comment with your contact details and I will contact you!. Thank you for reading!

Do comment and share!