Answer:
a.) [tex]V_{PFR}=26.3dm^{3}[/tex] and [tex]V_{CSTR}=174.5dm^3[/tex]
b.) [tex]T_0=76.85^0C[/tex]
c.) See attached picture.
d.)
500 dm -> X=0.977
250dm-> X_1=0.967 and X_2=0.992
Explanation:
Hello,
a.) At first, it is possible to find the PFR volume as shown below, considering its design equation:
[tex]\frac{dX}{dV}=\frac{-r_A}{F_A_0}=\frac{kC_AC_B}{F_A_0}[/tex]
Thus, since this is a nonisothermal reactor, it is defined as:
[tex]\frac{dX}{dV}=[k_{(300K)}*exp(\frac{E}{R}(\frac{1}{300}-\frac{1}{T})]}\frac{C_AC_B}{F_A_0}[/tex]
Thus, by leaving the formula in terms of conversion and temperature, we obtain (equal molar feed):
[tex]\frac{dX}{dV}=[k_{(300K)}*exp(\frac{E}{R}(\frac{1}{300}-\frac{1}{T})]}\frac{C_A_0^2}{F_A_0}(1-X)^2[/tex]
Nonetheless, by applying the energy balance for ΔCp=0 under adiabatic conditions:
[tex]T=T_0-\frac{(\Delta _RH)X}{\Sigma \theta_iCp_i } =27^0C-\frac{(-41000+20000+15000)\frac{cal}{mol}*0.85}{(1*15+1*15+0*30)\frac{cal}{mol^0C} } =27^0C-(-170^0C)\\T=197^0C[/tex]
Hence, integrating the dX/dV, we obtain:
[tex]C_A_O=0.1\frac{kmol}{m^3}*\frac{1000mol}{1kmol}*(\frac{1m}{10dm} )^3 =0.1\frac{mol}{dm^3}\\F_A_0=0.1\frac{mol}{dm^3}*2\frac{dm^3}{s}=0.2\frac{mol}{s}[/tex]
[tex]\int\limits^{0.85}_0 {\frac{1}{(1-X)^2} } \, dX =[0.01dm^3/(mol*s)*exp(\frac{10000cal/mol}{ 1.9872cal/mol*K}(\frac{1}{300K}-\frac{1}{(197+273.15)K})]}\frac{(0.1mol/dm^3)^2}{0.2mol/s}\int\limits^V_0 {} \, dV \\\\5.667=0.216dm^{-3}V\\V=5.667/0.216dm^{-3}\\V_{PFR}=26.3dm^{3}[/tex]Now, for the CSTR, just the design equation is changed by:
[tex]V_{CSTR}=\frac{F_A_0X}{kC_AC_B}[/tex]
Thus:[tex]V_{CSTR}=\frac{0.2mol/s*0.85}{[0.01dm^3/(mol*s)*exp(\frac{10000cal/mol}{ 1.9872cal/mol*K}(\frac{1}{300K}-\frac{1}{(197+273.15)K})]*(0.1mol/dm^3)^2(1-0.85)^2} \\V_{CSTR}=\frac{0.17mol/s}{4.33dm^3/(mol*s)*0.01mol^2/dm^6(0.0225)} \\V_{CSTR}=174.5dm^3[/tex]
b.) In this case, we assume that the outlet temperature is 550K or 276.85°C, thus, one modifies the energy balance in terms of the inlet temperature as:
[tex]T_0=T+\frac{(\Delta _RH)X}{\Sigma \theta_iCp_i } =276.85^0C+\frac{(-41000+20000+15000)\frac{cal}{mol}*1}{(1*15+1*15+0*30)\frac{cal}{mol^0C} } \\T_0=276.85^0C-200^0C\\T_0=76.85^0C[/tex]
c.) See attached picture.
d.) In this case, for 500 dm³, we solve the following nonlinear equation (design equation):
[tex]500dm^3=\frac{F_A_0X}{k(T)C_A_0^2(1-X)^2}[/tex]
Considering that:
[tex]k(T)=[0.01dm^3/(mol*s)*exp(\frac{10000cal/mol}{1.9872cal/mol*K}(\frac{1}{300K}-\frac{1}{(T+273.15)K})][/tex]
Whereas T is defined in the enegy balance:
[tex]T=27-\frac{-6000X}{30}[/tex]
Thus, solving by using a solver:
X=0.977.
Finally, for two series reactors, for the first one, the outlet conversion (using the same nonlinear equation and solver is:
[tex]250dm^3=\frac{F_A_0X_1}{k(T)C_A_0^2(1-X_1)^2}\\X_1=0.967[/tex]
So, now, the initial flow, concentrations and temperature entering to the new reactor are:
[tex]F_A_1=0.0066mol/s\\\\C_A_1=0.0033mol/dm^3\\\\T_1=220.4^0C[/tex]
So the equation changes by:
[tex]250dm^3=\frac{F_A_1X_2}{k(T)C_A_1^2(1-X_2)^2}[/tex]
and:
[tex]T=208.2-\frac{-6000X_2}{30}[/tex]
Thus, we obtain for the same k(T):
[tex]X_2=0.992[/tex]
Best regards.
