In a study of nitric oxide formation in laminar jet flames, the propane fuel is

Question

In a study of nitric oxide formation in laminar jet flames, the propane fuel is diluted with N_2 to suppress soot formation. The nozzle fluid is 60 percent N_2 by mass. The burner has a circular port. The fuel, nitrogen, and the air are all at 300 K and 1 atm. Compare the flame lengths for the following two cases with the undiluted base case (m_F = 5 middot 10^-6 kg/s). What is the physical significance of your results? Discuss. The total flowrate of the diluted flow (C_3H_8 + N_2) is 5 middot 10^-6 kg/s. The flowrate of the C_3H_8 in the diluted flow is 5 middot 10^-6kg/s.

Explanation / Answer

At 300 K and 1 atm we will have 1L propane will be

PV = nRT

1 x 1 = n x 0.08204 x 300

n = 0.0406 moles which is 1.79 g

1 Kg/s means 22 moles/sec.

Propane has 46.4 MJ/Kg or 43978 BTU/Kg ot 43978 BTU/s which is 1.58 x 108BTU/h

Flame length is = 0.00604*Q0.4776 (references to the API 521 standards)

(Assuming Q in BTU/hr and Length in feet)

So flame length with pure propane with flow of 1Kg/sec will be 49.78 feet.

A) When the flow is 5 x 10-6 Kg/sec is 1.13 x 10-4 moles/sec

MJ for 5 x 10-6 Kg/sec - 0.018 Kg/hx 46.4 Mj/Kg- 835200 J/hr - 791 BTU/hr

Flame length is = 0.00604*7910.4776 = 0.146 ft flame length

B) With 60% Nitrogen we will have propane 2 x 10-6Kg/sec

When the flow is 0.4 x 5 x 10-6 Kg/sec is 4.54 x 10-5 moles/sec

MJ for 2 x 10-6 Kg/sec - 0.0072 Kg/h x 46.4 Mj/Kg- 334080 J/hr - 316.6 BTU/hr

Flame length is = 0.00604*316.60.4776 = 0.0945 ft flame length

If you have a diluted flame since N2 has no BTU value flame length will be smaller

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