1. Predicting Engine Exhaust Plume Spectral Radiance & Transmittance Engineering Project MANE 6980 – Spring 2010 Wilson Braz

2. IR Radiation Introduction  Infrared (a.k.a. thermal) portion of electromagnetic spectrum spans approximately 0.5m to 1000m

3. IR Radiation Introduction (cont.)  All matter emits energy  Perfect emitters are called ‘blackbodies’  Max Planck, in 1900, was the first to derive the equation describing ‘spectral’ radiation emission from a blackbody  Planck unwittingly revolutionized physics with the introduction of the Planck constant h which describes the size of ‘quanta’ in quantum mechanics Planck’s Law 800K 700K 600K 500K 400K

4. Definitions  Radiance - energy flux per unit solid angle  Spectral – modifier that denotes units are given as a function of wavelength (or frequency) E.g. ‘spectral radiance’ = radiance per unit wavelength  Transmissivity – Fractional amount of energy pass  Absorptivity – Fractional amount of energy absorbed by a medium

5. Exhaust Plume Radiation  Exhaust plume is gaseous and opaque  Radiation is absorbed, emitted, and transmitted through media at different wavelengths  Molecular resonances cause different behaviors at varying wavelengths, so spectral analysis is of interest – CO2 and H2O predominant elements in IR of plume  Beer’s Law describes transmissivity  Kirchoff’s Law describes emissions and

6. Effects of Soot in plume  Combustion process is never 100% efficient A small portion of fuel does not completely combust, and carbon molecules coalesce into small particles  Carbon particles, or soot, emit and absorb too  Absorption varies significantly with size and particle density  Effects of soot is considered in this project

7. Plume IR problem break-down  The method of calculating plume emissions broken down into 2 major steps Gaseous spectral radiance and transmissivity calculations, dominated by CO2 and H2O Soot

8. Chemical Reactions of Combustion  Fuel (CH2) combines with Oxygen (O2) and results in water (H2O), carbon dioxide (CO2), and heat energy 2 CH2 + 3 O2 = 2 H2O + 2 CO2  Given mass flow of air and fuel, and the temperatures of plume, we can calculate the particle concentrations using ideal gas law

9. MODTRAN for Radiance and Transmittance due to CO2 and H2O  MODTRAN uses various techniques for calculating CO2 and H20 radiance.  Leverage these methods to obtain solutions for C02 and H2O ‘Standard Atmosphere’‘Plume (no soot)’

10. GE-T700 – 100%MC 100% Burn Efficiency (No soot)  MODTRAN Inputs (ppmv) H2O = 40874 CO2 = 39334 O3 = 0.0686 N2O = 0.0 CO = 176.7 CH4 = 4.284 O2 = 144863 NO = 0.0 SO2 = 0.0 NO2 = 148.54 NH3 = 0.001 HNO = 0.0  Temp = 533°K  Path = 1 meter  Soot = 0  Integrated Radiance = 0.0141 W/cm 2 sr (1 – 12

11. Modeling particulate  Several techniques have been devised for particulate modeling  Proper usage depends upon particle size parameter where D is particle diameter and m is wavelength in the particle fluid.

12. Turbine engine exhaust soot  For soot, is generally < 0.3 therefore Mie Scattering Theory is used.  Mie equation yield:

13. Plot of a/C = 5/ From Mie equation using properties of propane combustion

14. Concentration of Soot in turbine engines from literature  A study performed on the sooting properties of various jet fuels in jet turbines yielded very small soot concentrations.  These values may, or may not be indicative of actual soot concentration in turbo-shaft engines.  Results will be presented for increasing levels of soot concentration  Recommend correlating results with measurements of plume radiance and transmittance m 3 soot per m 3 plume

15. Results using published values for soot particle density N = 4.24x10 6 cm -3 N = 4x10 11 cm -3

16. Results  MODTRAN and additional procedure to calculate soot radiant emissions Published measured valuesPredicted values

17. Project TODO  Vary soot concentration to see effects of soot; plot results  Show attenuation plots Plots as a function of distance from emitting plume Soot may become more important  Discussion of Conclusions

18. Results with increasing particle number density N = 4x10 7 cm -3 N = 4x10 8 cm -3

19. Results with increasing particle number density (continued) N = 4x10 9 cm -3 N = 4x10 10 cm -3

20. Results with increasing particle number density (continued) N = 4x10 11 cm -3 N = 8x10 11 cm -3