roldy
Sep28-11, 09:42 PM
1. The problem statement, all variables and given/known data
Consider air in chemical equilibrium at 0.1 atm and T=4500 K. The chemical species are O2, O, N2, N (Ignore NO). Calculate the enthalpy and internal energy per unit mass of the mixture. Neglect electronic excitation in your calculations.
K_{p,O_2}=12.19 atm
K_{p,N_2}=0.7899*10^{-4}atm
(\Delta H_f^o)_O=2.47*10^8 J/(kg*mol)
(\Delta H_f^o)_N=4.714*10^8 J/(kg*mol)
(\Theta_v)_{N_2}=3390 K
(\Theta_v)_{O_2}=2270 K
2. Relevant equations
(1)h=\Sigma ^n_{i=1}c_ih_i + \Sigma ^n_{i=1}c_i \Delta h_{f_i}
c_i=X_i\frac{\mu_i}{\mu}
X_i=\frac{P_i}{P}
R_{air}=\frac{R_u}{\mu_{air}}
\mu_{air}=X_{O_2}*\mu_{O_2} + X_O*\mu_O + X_N*\mu_N + X_{N_2}*\mu_{N_2}
Diatomic Gas:
h=e_{sens}+RT
e_{sens}=3/2RT + RT +\frac{\frac{\Theta_v}{T}}{e^{\frac{\Theta_v}{T}}-1}RT
Monatomic Gas:
h=5/2RT
=e_{sens}=3/2RT
3. The attempt at a solution
The weight of each species is as follows:
O=16kg/(kg*mol)
N=14kg(kg*mol)
O_2=32kg/(kg*mol)
N_2=28kg/(kg*mol)
I then found what the gas constant was for each species with Ru=8314 J/(kg*K).
R_O=519.625
R_N=593.8571
R_{O_2}=259.8125
R_{N_2}=296.9286
I then calculated the sensible internal enthalpy for each species.
e_{sense,O}=3/2(519.625)(4500)=3.5075*10^6
e_{sense,N}=3/2(593.8571)(4500)=4.0085*10^6
e_{sense,O_2}=3/2(259.8125)(4500) + (259.8125)(4500) +\frac{\frac{2270}{4500}}{e^{\frac{2270}{4500}}-1}(259.8125)(4500)=3.8218*10^6
e_{sense,N_2}=3/2(296.9286)(4500) + (296.9286)(4500) +\frac{\frac{3390}{4500}}{e^{\frac{3390}{4500}}-1}(296.9286)(4500)=4.2359*10^6
Next the enthalpy per unit mass of each species was calculated.
h_O=5/2(519.625)(4500)=5.8458*10^6
h_N=5/2(593.8571)(4500)=6.6809*10^6
h_{O_2}=3.8218*10^6+259.8125(4500)=4.9910*10^6
h_{N_2}=4.2359*10^6+296.9286(4500)=5.5721*10^6
This is where I get stuck. I believe I use equation (1) in some way but I do not see how since I am not given the partial pressures to find c_i. I'm not sure how the equilibrium constants work into this problem.
Consider air in chemical equilibrium at 0.1 atm and T=4500 K. The chemical species are O2, O, N2, N (Ignore NO). Calculate the enthalpy and internal energy per unit mass of the mixture. Neglect electronic excitation in your calculations.
K_{p,O_2}=12.19 atm
K_{p,N_2}=0.7899*10^{-4}atm
(\Delta H_f^o)_O=2.47*10^8 J/(kg*mol)
(\Delta H_f^o)_N=4.714*10^8 J/(kg*mol)
(\Theta_v)_{N_2}=3390 K
(\Theta_v)_{O_2}=2270 K
2. Relevant equations
(1)h=\Sigma ^n_{i=1}c_ih_i + \Sigma ^n_{i=1}c_i \Delta h_{f_i}
c_i=X_i\frac{\mu_i}{\mu}
X_i=\frac{P_i}{P}
R_{air}=\frac{R_u}{\mu_{air}}
\mu_{air}=X_{O_2}*\mu_{O_2} + X_O*\mu_O + X_N*\mu_N + X_{N_2}*\mu_{N_2}
Diatomic Gas:
h=e_{sens}+RT
e_{sens}=3/2RT + RT +\frac{\frac{\Theta_v}{T}}{e^{\frac{\Theta_v}{T}}-1}RT
Monatomic Gas:
h=5/2RT
=e_{sens}=3/2RT
3. The attempt at a solution
The weight of each species is as follows:
O=16kg/(kg*mol)
N=14kg(kg*mol)
O_2=32kg/(kg*mol)
N_2=28kg/(kg*mol)
I then found what the gas constant was for each species with Ru=8314 J/(kg*K).
R_O=519.625
R_N=593.8571
R_{O_2}=259.8125
R_{N_2}=296.9286
I then calculated the sensible internal enthalpy for each species.
e_{sense,O}=3/2(519.625)(4500)=3.5075*10^6
e_{sense,N}=3/2(593.8571)(4500)=4.0085*10^6
e_{sense,O_2}=3/2(259.8125)(4500) + (259.8125)(4500) +\frac{\frac{2270}{4500}}{e^{\frac{2270}{4500}}-1}(259.8125)(4500)=3.8218*10^6
e_{sense,N_2}=3/2(296.9286)(4500) + (296.9286)(4500) +\frac{\frac{3390}{4500}}{e^{\frac{3390}{4500}}-1}(296.9286)(4500)=4.2359*10^6
Next the enthalpy per unit mass of each species was calculated.
h_O=5/2(519.625)(4500)=5.8458*10^6
h_N=5/2(593.8571)(4500)=6.6809*10^6
h_{O_2}=3.8218*10^6+259.8125(4500)=4.9910*10^6
h_{N_2}=4.2359*10^6+296.9286(4500)=5.5721*10^6
This is where I get stuck. I believe I use equation (1) in some way but I do not see how since I am not given the partial pressures to find c_i. I'm not sure how the equilibrium constants work into this problem.