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3. In a steady-state methane-air flame at approximately atmospheric pressure, the temperature is raised from 68 °F to 3200 °F. The incoming air-gas mixture and the products of combustion may both be considered ideal gases with a molecular weight = 28 g/mol. The flame is a thin, flat region perpendicular to the gas flow. If the flow comes into the flame at a velocity of 2 ft/s, what is the pressure difference from one side of the flame to the other?

3. In a steady-state methane-air flame at approximately atmospheric pressure, the temperature is raised from 68 °F to 3200 °F. The incoming air-gas mixture and the products of combustion may both be considered ideal gases with a molecular weight = 28 g/mol. The flame is a thin, flat region perpendicular to the gas flow. If the flow comes into the flame at a velocity of 2 ft/s, what is the pressure difference from one side of the flame to the other?

Applications and Investigations in Earth Science (9th Edition)
Applications and Investigations in Earth Science (9th Edition)
9th Edition
ISBN:9780134746241
Author:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
Publisher:Edward J. Tarbuck, Frederick K. Lutgens, Dennis G. Tasa
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3. In a steady-state methane-air flame at approximately atmospheric pressure, the temperature is raised from 68 °F to 3200 °F. The incoming air-gas mixture and the products of combustion may both be considered ideal gases with a molecular weight = 28 g/mol. The flame is a thin, flat region perpendicular to the gas flow. If the flow comes into the flame at a velocity of 2 ft/s, what is the pressure difference from one side of the flame to the other? T2 m(Vin - Vout) + F = 0 m = PinAVin F = A(Pin-Pout) Vout Pout = Vin Pin m[Vin - Vout] + F = 0 Pout Pin T(Rº) = 68F + 460 = 528R° & 3200F + 460 = 3660Rº
Transcribed Image Text:3. In a steady-state methane-air flame at approximately atmospheric pressure, the temperature is raised from 68 °F to 3200 °F. The incoming air-gas mixture and the products of combustion may both be considered ideal gases with a molecular weight = 28 g/mol. The flame is a thin, flat region perpendicular to the gas flow. If the flow comes into the flame at a velocity of 2 ft/s, what is the pressure difference from one side of the flame to the other? T2 m(Vin - Vout) + F = 0 m = PinAVin F = A(Pin-Pout) Vout Pout = Vin Pin m[Vin - Vout] + F = 0 Pout Pin T(Rº) = 68F + 460 = 528R° & 3200F + 460 = 3660Rº
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