I write about Propulsion, Aerodynamics and Renewable Energy (Wind/Hydro Turbines).
Monday, 15 September 2014
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
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)*Cp
Velocity of Sound, ao=(1000*g*R*To).^(1/2)
Total to Static Temperature Ratio, tr=1+(((g-1)/2)*(Mo.^2))
Burner Exit Enthalpy to Ambient Enthalpy Ratio, tl=Tt4/To
Velocity Ratio, v9/ao=Mo*((tl/tr).^(1/2))
Thrust per Air Flow, F/mo=ao*((v9/ao)-Mo)
Fuel Air Ratio, f=((Cp*To)/hPR)*( tl-tr)
Thrust Specific Fuel Consumption, S=f/(F/mo)
Thermal Efficiency, hT=1-(1/tr)
Propulsive Efficiency, hP=2/(((tl/tr).^(1/2))+1)
Overall Efficiency hO=hT*hP
Total to Static Temperature Ratio for Maximum Thrust per Air Flow, tr (maxFmo)=t l.^(1/3)
Mach Number of Flight for Maximum Thrust per Air Flow , Mo (maxFmo)=((2/(g-1))*( tr (maxFmo)-1)).^(1/2)
Temperature at Station Tto (after Compression), Tto=tr*To
Nozzle Exhaust Temperature, T9=To*(tl/tr)
Total to Static Temperature Ratio, tr=1+(((g-1)/2)*(Mo.^2))
Burner Exit Enthalpy to Ambient Enthalpy Ratio, tl=Tt4/To
Velocity Ratio, v9/ao=Mo*((tl/tr).^(1/2))
Thrust per Air Flow, F/mo=ao*((v9/ao)-Mo)
Fuel Air Ratio, f=((Cp*To)/hPR)*( tl-tr)
Thrust Specific Fuel Consumption, S=f/(F/mo)
Thermal Efficiency, hT=1-(1/tr)
Propulsive Efficiency, hP=2/(((tl/tr).^(1/2))+1)
Overall Efficiency hO=hT*hP
Total to Static Temperature Ratio for Maximum Thrust per Air Flow, tr (maxFmo)=t l.^(1/3)
Mach Number of Flight for Maximum Thrust per Air Flow , Mo (maxFmo)=((2/(g-1))*( tr (maxFmo)-1)).^(1/2)
Temperature at Station Tto (after Compression), Tto=tr*To
Nozzle Exhaust Temperature, T9=To*(tl/tr)
Calculations
By using these equations, and by entering the following inputs:
For Mach Number of Flight= 1
To= 216.7 K, g=
1.4, Cp= 1.004 KJ/(Kg.K), Tt4= 1,600 K, Mo= 1,
hPR= 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 Tto and Tt2, 260.04 K
Temperature at Nozzle Entry, Station Tt4, 1,600 K
Temperature at Nozzle Exit, Station T9, 1,333.33 K
Station Temperatures for T-S Diagram
Ambient Temperature, Station T0, 216.7 K
Temperature after Compression, Station Tto and Tt2, 260.04 K
Temperature at Nozzle Entry, Station Tt4, 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 Tt4= 1,900 K, we get these results:
Station Temperatures for T-S Diagram
Ambient Temperature, Station To, 216.7 K
Temperature after Compression, Station Tt0 and Tt2, 260.04 K
Temperature at Nozzle Entry, Station Tt4, 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 Tt4= 2,200 K, we get these results:
Station Temperatures for T-S Diagram
Ambient Temperature, Station To, 216.7 K
Temperature after Compression, Station Tt0 and Tt2, 260.04 K
Temperature at Nozzle Entry, Station Tt4, 2,200 K
Temperature at Nozzle Exit, Station T9, 1,833.33 K
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 Tt4= 2,200 K, we get these results:
Station Temperatures for T-S Diagram
Ambient Temperature, Station To, 216.7 K
Temperature after Compression, Station Tt0 and Tt2, 260.04 K
Temperature at Nozzle Entry, Station Tt4, 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:
To= 216.7 K, g= 1.4, Cp= 1.004 KJ/(Kg.K), Tt4= 1,600 K, Mo= 2, hPR= 42,800 KJ/Kg
Station Temperatures for T-S Diagram
Ambient Temperature, Station To, 216.7 K
Temperature after Compression, Station Tt0 and Tt2, 390.06 K
Temperature at Nozzle Entry, Station Tt4, 1600 K
Temperature at Nozzle Exit, Station T9, 888.889 K
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:
To= 216.7 K, g= 1.4, Cp= 1.004 KJ/(Kg.K), Tt4= 1,600 K, Mo= 2, hPR= 42,800 KJ/Kg
Station Temperatures for T-S Diagram
Ambient Temperature, Station To, 216.7 K
Temperature after Compression, Station Tt0 and Tt2, 390.06 K
Temperature at Nozzle Entry, Station Tt4, 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 Tt4= 1,900 K, we get these results:
Station Temperatures for T-S Diagram
Ambient Temperature, Station To, 216.7 K
Temperature after Compression, Station Tt0 and Tt2, 390.06 K
Temperature at Nozzle Entry, Station Tt4, 1900 K
Temperature at Nozzle Exit, Station T9, 1055.56 K
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 Tt4= 1,900 K, we get these results:
Station Temperatures for T-S Diagram
Ambient Temperature, Station To, 216.7 K
Temperature after Compression, Station Tt0 and Tt2, 390.06 K
Temperature at Nozzle Entry, Station Tt4, 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 Tt4= 2,200 K, we get these results:
Station Temperatures for T-S Diagram
Ambient Temperature, Station To, 216.7 K
Temperature after Compression, Station Tt0 and Tt2, 390.06 K
Temperature at Nozzle Entry, Station Tt4, 2200 K
Temperature at Nozzle Exit, Station T9, 1222.22 K
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 Tt4= 2,200 K, we get these results:
Station Temperatures for T-S Diagram
Ambient Temperature, Station To, 216.7 K
Temperature after Compression, Station Tt0 and Tt2, 390.06 K
Temperature at Nozzle Entry, Station Tt4, 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
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
Wednesday, 13 August 2014
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.
Tuesday, 18 March 2014
9,000 Sq. Ft. House Design For Sale
Isometric View
Isometric ViewRear 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
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