SULFUR DIOXIDE CONVERTER

Example 23.5-1

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restart;

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Q:=(T1,X)->H[OUT](X)-H[IN](T1);   This is the energy equation that best approximates the SO2 converter.  It was derived from a general energy equation in the Solution page of the website.  H[OUT](X) and H[IN](X) are the flow rate of enthalpy in and out of the converter.  They are functions of conversion within the reactor.  (The exit temperature, T2, in this example, is treated as a function of the conversion, X).

>	H[IN]:=T1->W[1,SO2]*H[1,SO2](T1)+W[1,O2]*H[1,O2](T1)+W[1,N2]*H[1,N2](T1);   The total enthalpy entering the converter.

>	H[OUT]:=X->W[2,SO2](X)*H[2,SO2](X)+W[2,SO3](X)*H[2,SO3](X)+W[2,O2](X)*H[2,O2](X)+W[2,N2](X)*H[2,N2](X);   The total enthalpy exiting the converter.

>	W[1,SO2]:=7.80*lbmol/hr; W[1,O2]:=10.80*lbmol/hr; W[1,N2]:=81.40*lbmol/hr;  Given entering molar flow rates.

>	W[2,SO2]:=X->W[1,SO2]*(1-X); W[2,SO3]:=X->X*W[1,SO2]; W[2,O2]:=X->W[1,O2]-0.5*X*W[1,SO2]; W[2,N2]:=X->W[1,N2]; W[2,TOT]:=X->W[2,SO2](X)+W[2,SO3](X)+W[2,O2](X)+W[2,N2](X);    These are the exit molar flow rates as functions of conversion.

>	K[p]:=X->(W[2,SO3](X)/W[2,TOT](X))/((W[2,SO2](X)/W[2,TOT](X))*(W[2,O2](X)/W[2,TOT](X))^(1/2));   This is the expression for the equilibrium constant, Kp.  It is given on page 741.

The plot below uses data taken from Bird, Stewart, & Lightfoot on page 741.  The data points were fitted with an exponential curve.  The value of r² is 0.9973, which indicates that the exponential fit is appropriate.  Extrapolation of this curve may not be justified, so we will only trust it in the range: 1080 R < T < 1620 R.

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T2:=X->-ln(K[p](X)/(10^10))/0.0129;    This equation rearranges the fit from the Excel graph to be explicit in T2, the exit temperature (which is in Rankine.  We do not use the units here because later this term appears inside an integral, and Maple doesn't seem to like units inside integrals.  We correct for this omission of units later).

>	H[1,SO2]:=T1->H[0,SO2]+R*int(A_SO2+B_SO2*T+D_SO2*T^(-2),T=T0..T1)*Rankine;   Enthalpy of SO2 at entrance temperature, T1

>	H[1,O2]:=T1->H[0,O2]+R*int(A_O2+B_O2*T+D_O2*T^(-2),T=T0..T1)*Rankine;     Enthalpy of O2 at entrance temperature, T1

>	H[1,N2]:=T1->H[0,N2]+R*int(A_N2+B_N2*T+D_N2*T^(-2),T=T0..T1)*Rankine;     Enthalpy of N2 at entrance temperature, T1

>	H[2,SO2]:=X->H[0,SO2]+R*int(A_SO2+B_SO2*T+D_SO2*T^(-2),T=T0..T2(X))*Rankine;     Enthalpy of SO2 at exit temperature, T2

>	H[2,SO3]:=X->H[0,SO3]+R*int(A_SO3+B_SO3*T+D_SO3*T^(-2),T=T0..T2(X))*Rankine;     Enthalpy of SO3 at exit temperature, T2

>	H[2,O2]:=X->H[0,O2]+R*int(A_O2+B_O2*T+D_O2*T^(-2),T=T0..T2(X))*Rankine;                 Enthalpy of O2 at exit temperature, T2

>	H[2,N2]:=X->H[0,N2]+R*int(A_N2+B_N2*T+D_N2*T^(-2),T=T0..T2(X))*Rankine;                 Enthalpy of N2 at exit temperature, T2

>	A_O2:=3.639: B_O2:=.506e-3: D_O2:=-0.227e-5:         Cp constants for O2

>	A_SO2:=5.699: B_SO2:=.801e-3: D_SO2:=-1.015e5:        Cp constants for SO2

>	A_SO3:=8.06: B_SO3:=1.056e-3: D_SO3:=-2.028e5:        Cp constants for SO3

>	A_N2:=3.280: B_N2:=.593e-3: D_N2:=0.040e5:        Cp constants for N2

>	T0:=536.67; This is the enthalpy reference temperature, in Rankine.  Zero energy is defined to be the energy of O2 and N2 at 536.67 Rankine.

>	R:=1.986*Btu/(lbmol*Rankine);  Gas Constant

>	H[0,SO2]:=-70960*(cal/mol)*0.0039657*(Btu/cal)*453.593*(mol/lbmol);   The given energy of SO2 and SO3 at 536.67 Rankine H[0,SO3]:=-94450*(cal/mol)*0.0039657*(Btu/cal)*453.593*(mol/lbmol);

>	H[0,O2]:=0; H[0,N2]:=0;    As mentioned earlier, zero energy is defined as O2 and N2 at 536.67 Rankine.

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Q(1283.67,0.95);  The purpose of this problem is to find the amount of heat that has to be removed in order to achieve 95% conversion in the reactor, when the reactants enter at 440 degrees Celsius (1283.67 Rankine).  Following is the answer (the negative sign means that this is the amount of heat that must be removed from the system.

>	Q_for_plot:=(T1,X)->Q(T1,X)*(hr/Btu);    This is the same as Q above, except it has no units so that it can be used in plots.

Below is a plot of Q as a function of conversion. The 6 lines represent operation of the converter at various feed entrance temperatures.  The first obvious trend is that for a given conversion, more heat must be removed as the entrance temperature is increased.  The second trend is that regardless of entrance temperature, the conversion is increased by removing more and more heat.  This explains why in industry, sulfur dioxide converters are run with several passes and interstage cooling between passes.  A very important feature of this plot to note is the scale of the Q axis.  The amount of heat that must be removed increases drastically as conversions in the upper 90% are reached.  4E5 Btu/hr is equivalent to about 420,000 kJ/hr of heat removed.  This would be a very large job for any single heat exchanger.  Again, this is why interstage cooling is used in industry.

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plot([eval(Q_for_plot(T1,X),T1=1083),eval(Q_for_plot(T1,X),T1=1183),eval(Q_for_plot(T1,X),T1=1283), eval(Q_for_plot(T1,X),T1=1383),eval(Q_for_plot(T1,X),T1=1483),eval(Q_for_plot(T1,X),T1=1583)],X=0.73..1.0,color=[black,violet,blue,green,orange,red],legend=["T1=1083 Rankine","T1=1183 Rankine","T1=1283 Rankine","T1=1383 Rankine","T1=1483 Rankine","T1=1583 Rankine"],title="Heat Removed as a Function of Conversion (X)",axes=BOXED,labels=["X (Conversion)","Q (Btu/hr) (Negative Q Signifies Removed Heat)"],labeldirections=[HORIZONTAL,VERTICAL]);

The reason that the domain of the above plot is restricted to conversions between 0.73 and 1 is because the data given for Kp is only in the range of temperatures from 1080 to 1620 Rankine.  These temperatures correspond to conversions of about 0.73 and almost 1, as shown by the plot below.    

>	plot(T2(X),X=0.73..1.0,title="Exit Temperature as a Function of Conversion",labels=["Conversion (X)","Exit Temperature (Rankine)"],labeldirections=[HORIZONTAL,VERTICAL]);

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The above plot shows that higher temperatures lead to lower conversions.  This relationship makes sense because we know that the reaction is exothermic, and according to Le Chatelier's Principle, equilibriums shift to offset any changes in reaction conditions.  Therefore, hotter reactants will be less in favor of producing products + heat.