Question 16.CSGP.108: Saturated liquid butane (note: use generalized charts) enter......
Saturated liquid butane (note: use generalized charts) enters an insulated constant pressure combustion chamber at 25°C, and x times theoretical oxygen gas enters at the same pressure and temperature. The combustion products exit at 3400 K.
Assuming that the products are a chemical equilibrium gas mixture that includes CO, what is x?
Butane: T _1=25^{\circ} C \text {, sat. liq., } x _1=0, T _{ c }=425.2 \,K , P _{ c }=3.8\, MPa
T _{ r 1}=0.7 \text {, Figs. D.1 and D.2, } \quad P _{ r 1}=0.10, \quad P _1= P _{ r 1} P _{ c }=380 \,kPa
Fig D.2: \left(\overline{ h }_1^*-\overline{ h }_1\right)_{ f }=4.85 RT _{ c }
Oxygen: T _2=25^{\circ} C , X^ * theoretical air \text { Products: } T _3=3400 \,K
C _4 H _{10}+6.5 X O _2 \Rightarrow 4 CO _2+5 H _2 O +6.5( X -1) O _2
The equilibrium constant is
K =\frac{ y _{ CO }^2 y _{ O _2}}{ y _{ CO_2 }^2}\left\lgroup \frac{ P _1}{ P _{ o }} \right\rgroup^{2+1-2}=\left\lgroup \frac{ a }{2- a } \right\rgroup^2\left\lgroup \frac{6.5 X -6.5+ a }{6.5 X -2.5+ a } \right\rgroup\left\lgroup \frac{ P _1}{ P _{ o }} \right\rgroup
@ T _3=3400 \,K \text { Table A.11, } \ln ( K )=0.346, \quad K =1.4134
1.4134=\left\lgroup\frac{ a }{2- a }\right\rgroup^2\left\lgroup\frac{6.5 X -6.5+ a }{6.5 X -2.5+ a }\right\rgroup (3.76) Equation 1.
Need a second equation:
Energy eq.:
Products @ 3400 K:
H _{ P }= H _{ R } \quad \Rightarrow \quad 1924820=541299 a +741656.5 X Equation 2.
Two equations and two unknowns, solve for X and a.
a ≅ 0.87, X ≅ 1.96