A mixture of H_2 and O_2 (equivalence ratio = 1.0) is introduced into a combusto

Question

A mixture of H_2 and O_2 (equivalence ratio = 1.0) is introduced into a combustor at pressure of 1 bar and temperature of 298.15 K. After reaction, the products contain H_2 O, H_2 and O_2, pressure is 1 bar and temperature 1000 k. Assume that the mole of H_2 O formed is 0.92. Calculate the heat of reaction for this process. Calculate the final temperature if the above heat is used to further heat-up the products. Is this temperature the same as the adiabatic flame temperature? Explain. Evaluate the following sums for a multicomponent system with N species. sigma^N_i = 1 Y_i sigma^N_i = 1 x_i sigma^N_i = 1 rho_i sigma^N_i = 1 J_i sigma^N_i = 1 dot omega_i Here J_i and dot omega_1 represent, respectively, the diffusional flux and reaction (production/consumption) rate (kg/m^3/s) of species i. Consider an ethane (C_2 H_6)-air nonpremixed flame established on a circular port. The port diameter is 10 mm. Both air and ethane are at 300 K and pressure is 1 atm. Use the kinematic viscosity as v = 50.0 times 10^-6 m^2/s. Calculate the flame height assuming the initial velocity profile issuing the port is uniform and the velocity is 10 cm/s. Make other suitable assumptions as necessary. Will the flame height change if the initial velocity is nonuniform with an average value of 10 cm/s. Calculate the effect of changing O_2 mole fraction in air from 0.21 to 0.3 on the flame

Explanation / Answer

Solution:

Problem1:

(a) Enthalpy of the reaction: H2(g) + ½ O2(g) -------> H2O(g)    can be given by:

H (rxn) = x – [H0 (H2) + 1/2 H0 (O2)]                     Where x is the enthalpy of formation of H2O.

H (rxn) = x – [ 0 + 1/2x0]                              [Since, H0 (H2) = H0 (O2) = 0]

H (rxn) = x = -242 kJ/mol                  [Since, Enthalpy of formation of H2O(g)= -242 kJ/mol]

Therefore, enthalpy/heat(Q) of formation of 0.92mol H2O= -242x0.92= -222.64 kJ= -222.64x103 J

(b) 0.92 mol H2O = 18x0.92 g = 16.56 g H2O

We know, Q = ms(T2-T1) -------(1); m, s =mass and specific heat of water, T2, T1 final and initial temperature respectively.

Now, s = 2.02 J g¯1K¯1 for H2O(g)

Therefor using equation (1) we have:

222.64x103 = 16.56x2.02x(T2-1000)

Or, (T2-1000) = 6655.66

Or, T2 = 6655.66+1000 = 7655.66 K

The adiabatic flame temperature is the temperature that results from a complete combustion process that occurs without any work, heat transfer or changes in kinetic or potential energy. Therefore, in this case the high increase in the temperature (7655.66 K) at 1 bar constant pressure seems to be an adiabatic flame temperature without any work and/or heat transfer.

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