Temperature Calculation Using Polynomial Eqn

8/11/2019 Temperature Calculation Using Polynomial Eqn

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Estimating Temperatures in the primary reformer using K-Polynomials:

There are two major reactions taking place in the Primary reformer.

1. Reforming reaction : CH4 (g)+ H2O(g)CO(g)+ 3H2(g) (Equilibrium constantK1)

2.

Shift conversion : CO(g) + H2O(g)CO2 (g)+ H2 (g) (Equilibrium constantK2)

K1=()(

)

()()[P]2 and K2=

()()

()()[P]0

Where Pi is the partial pressure of component i and P is thetotal pressure.

From the Wet-Analysis of the outlet of primary reformer, we get the composition of the gas stream as :

Component Dry Analysis (mol%) Flow rates (Nm3

/hr) Wet Analysis (mol%)H2 66.51 59792.49 37.26

N2 0.82 737.18 0.46

CO 7.67 6895.33 4.3

CO2 10.86 9763.14 6.08

Ar 0.01 8.99 0.0056

CH4 14.13 12702.87 7.915

H2O 70585.76 44

Also, we know that,

Pi= xi * P where xi is the mole % of component i

Here, in the primary reformer, Ptotal= 33 kg/cm2 gauge.

And since the components in the outlet stream are in equilibrium, we take those compositions to calculate

the equilibrium partial pressures.

= 0.07195 x Ptotal

= 0.44 x Ptotal

= 0.043 x Ptotal


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= 0.0608 x Ptotal

= 0.3726 x ptotal

Kreforming= K1=()()

()()[P]2 =

()()

()()(

) = 74.87

And Kshift = K2=()()

()()[P]0 =

()()

()() = 1.197

K-Polynomials for calculating the Equilibrium Constant

K = exp (K1x InT + K2/T + K3 + T (K4+ T (K5+ K6x T)))

T = Temperature in Kelvin

Shift Reaction Ammonia Reforming Reaction

CO+H2OH2+CO2 3H2+N22NH3 CH4+ H2OCO+3H2

K1 -3.161655 x 10-1 -8.047466 8.611141

K2 5.016493 x 103 9.109037 x 103 -2.264800 x 104

K3 -4.104398 2.714340 x 101 -2.898262 x 101

K4

2.139622 x 10-3 4.730516 x 10-3 -4.980093 x 10-3

K5 -6.044361 x 10-7 -4.355303 x 10-7 3.977486 x 10-7

K6 7.582492 x 10-11 -6.422945 x 10-12 2.013355 x 10-11

Reforming reaction:

Given, K = (

( ( )))

lnK = K1lnT +

+ K3+ K4T + K5T

2+ K6T3

We have,

Kref= 74.87 ,


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And from the above table, for reforming reactions,

ln(74.87) = 8.611141lnT -

- 28.98262 - 0.004980093T + (3.977486x10-7)T2+ (2.013355 x 10-11)T3

Solving the above equation, we get,

Tref= 1039.037K = 766.0370C 7670C (as given in the data sheet)

Similarly, using the data given in the table for Shift reactions, we get,

Tshift= 766.7670C 767

0C

Thus, we can verify the operating values by theoretical methods.

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Temperature Calculation Using Polynomial Eqn