Universidade do Estado do Rio de Janeiro
Instituto de Química
Prof André Luís Alberton
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Fechamento de balanços materiais¶

Em muitos casos, o fechamento de balanços materiais leva à formulação do problema na forma de um sistema linear. Segue um exemplo.

Enunciado¶

Feche o balanço material para a espécie 𝐴,𝐵,𝐶 no fluxograma a seguir. São informadas:

  • as reações que ocorre nos reatores R1 e R2, com sua conversão associadas
  • as recuperações das espécies no separador S1 na corrente 7 e no separador S2 na corrente 9
  • a vazão de 𝐵 na corrente 1, sendo igual a 50 kmol/s
  • a vazão C na corrente 10, sendo igual a 1200 kmol/s
  • a razão das vazões de A/C na corrente 10, sendo igual a 0,05

  • a razão entre as vazões totais da corrente 5 pela corrente 6, sendo igual a 0,7.

    Fluxograma

Solução do exercício¶

Vamos aos balanços: $$ \begin{array}{ccc} \rm{ para \ o \ A} & \rm{para \ o \ B} & \rm{ para \ o \ C} \\ \hline & Mix & \\ N_{A,1} + N_{A,6}+ N_{A,9} = N_{A,2} & N_{B,1} + N_{B,6}+ N_{B,9} = N_{B,2} & N_{C,1} + N_{C,6}+ N_{C,9} = N_{C,2} \\ \hline & Reator & \\ N_{A,3} = N_{A,2} \cdot \left(1-XR1 \right) & N_{B,3} = N_{B,2} + 2 \cdot N_{A,2} \cdot XR1 & N_{C,3} = N_{C,2} \\ \hline & Separador \ S1 & \\ N_{A,7} = N_{A,3} \cdot rA7 & N_{B,7} = N_{B,3} \cdot rB7 & N_{C,7}=N_{C,3} \cdot rC7 \\ N_{A,4} = N_{A,3} \cdot \left(1-rA7\right) & N_{B,4} = N_{B,3} \cdot \left(1-rA7\right) & N_{C,4}=N_{C,3} \cdot \left(1-rA7\right) \\ \hline & Tê \ de \ divisao & \\ N_{A,5} = 0,7 \cdot N_{A,6} & N_{B,5} = 0,7 \cdot N_{B,6} & N_{C,5} = 0,7 \cdot N_{C,6} \\ N_{A,4} = N_{A,5} + N_{A,6} & N_{B,4} = N_{B,5} + N_{B,6} & N_{C,4} = N_{C,5} + N_{C,6} \\ \hline & Reator R2 & \\ N_{A,8} = N_{A,7} \cdot \left(1-XR2 \right) & N_{B,8} = N_{B,7} & N_{C,8} = N_{C,7} + N_{A,7} \cdot XR2 \\ \hline & Separador \ S2 & \\ N_{A,9} = N_{A,8} \cdot rA9 & N_{B,9} = N_{B,8} \cdot rB9 & N_{C,9}=N_{C,8} \cdot rC9 \\ N_{A,10} = N_{A,8} \cdot \left(1-rA9\right) & N_{B,10} = N_{B,8} \cdot \left(1-rB9\right) & N_{C,10}=N_{C,8} \cdot \left(1-rC9\right) \\ \hline & Informacoes \ adicionais & \\ N_{A,10} = 0,05 \cdot N_{C,10} & N_{B,1} = 5 & N_{C,10} = 1200 \\ \hline \end{array} $$

Podemos empilhar estas 30 equações na forma (primeiro todas as equações do A, depois todas as equações do B, e depois todas as equações do C):

$$ \begin{array}{c|ccc} & N_{A,1} - N_{A,2} + N_{A,6}+ N_{A,9} & = & 0 \\ & N_{A,2} \cdot \left(1-XR1 \right) - N_{A,3} & = & 0 \\ & N_{A,3} \cdot rA7 - N_{A,7} & = & 0\\ & N_{A,3} \cdot \left(1-rA7 \right) - N_{A,4} & = & 0\\ \rm{Eq \ A } & N_{A,5} - 0,7 \cdot N_{A,6} & = & 0 \\ & -N_{A,4} + N_{A,5} + N_{A,6} & = & 0 \\ & N_{A,7} \cdot \left(1-XR2 \right) - N_{A,8} & = & 0 \\ & N_{A,8} \cdot rA9 - N_{A,9} & = & 0 \\ & N_{A,8} \cdot \left(1- rA9 \right) - N_{A,10} & = & 0 \\ & N_{A,10} - 0,05 \cdot N_{C,10} & = & 0 \\ \hline & N_{B,1} - N_{B,2} + N_{B,6}+ N_{B,9} & = & 0 \\ & N_{A,2} \cdot XR1 + N_{B,2} - N_{A,3} & = & 0 \\ & N_{B,3} \cdot rB7 - N_{B,7} & = & 0\\ & N_{B,3} \cdot \left(1-rB7 \right) - N_{B,4} & = & 0\\ \rm{Eq \ B } & N_{B,5} - 0,7 \cdot N_{B,6} & = & 0 \\ & -N_{B,4} + N_{B,5} + N_{B,6} & = & 0 \\ & N_{B,7} - N_{B,8} & = & 0 \\ & N_{B,8} \cdot rB9 - N_{B,9} & = & 0 \\ & N_{B,8} \cdot \left(1- rB9 \right) - N_{B,10} & = & 0 \\ & N_{B,1} & = & 5 \\ \hline & N_{C,1} - N_{C,2} + N_{C,6}+ N_{C,9} & = & 0 \\ & N_{C,2} - N_{C,3} & = & 0 \\ & N_{C,3} \cdot rC7 - N_{C,7} & = & 0\\ & N_{C,3} \cdot \left(1-rC7 \right) - N_{C,4} & = & 0\\ \rm{Eq \ C } & N_{C,5} - 0,7 \cdot N_{C,6} & = & 0 \\ & -N_{C,4} + N_{C,5} + N_{C,6} & = & 0 \\ & N_{A,7}\cdot XR2 + N_{C,7} - N_{C,8} & = & 0 \\ & N_{C,8} \cdot rC9 - N_{C,9} & = & 0 \\ & N_{C,8} \cdot \left(1- rC9 \right) - N_{C,10} & = & 0 \\ & N_{C,10} & = & 1200 \\ \end{array} $$

Ou seja, temos um sistema com 30 equações. Apenas para visualizá-lo, vamos representar sua forma:

$$ \begin{array}{ccc} \begin{pmatrix}1&-1&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0.25&-1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0.92&0&0&0&-1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0.08&-1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&1&-0.7&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&1&-1&-1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0.58&-1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0.94&-1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0.06&0&-1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&-0.05\cr 0&0&0&0&0&0&0&0&0&0&1&-1&0&0&0&1&0&0&1&0&0&0&0&0&0&0&0&0&0&0\cr 0&1.5&0&0&0&0&0&0&0&0&0&1&-1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0.06&0&0&0&-1&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0.94&-1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-0.7&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0&1&-1&-1&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-1&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0.93&-1&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0.07&0&-1&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&0&1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-1&0&0&0&1&0&0&1&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-1&0&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&0&0&-1&0&0&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&-1&0&0&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-0.7&0&0&0&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-1&-1&0&0&0&0\cr 0&0&0&0&0&0&0.42&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&1&-1&0&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0.03&-1&0\cr 0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0.97&0&-1\cr 1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\cr \end{pmatrix} & \cdot \begin{pmatrix}\mathsf{\text{}}\cr \mathsf{\text{NA_1}}\cr \mathsf{\text{NA_2}}\cr \mathsf{\text{NA_3}}\cr \mathsf{\text{NA_4}}\cr \mathsf{\text{NA_5}}\cr \mathsf{\text{NA_6}}\cr \mathsf{\text{NA_7}}\cr \mathsf{\text{NA_8}}\cr \mathsf{\text{NA_9}}\cr \mathsf{\text{NA_10}}\cr \mathsf{\text{NB_1}}\cr \mathsf{\text{NB_2}}\cr \mathsf{\text{NB_3}}\cr \mathsf{\text{NB_4}}\cr \mathsf{\text{NB_5}}\cr \mathsf{\text{NB_6}}\cr \mathsf{\text{NB_7}}\cr \mathsf{\text{NB_8}}\cr \mathsf{\text{NB_9}}\cr \mathsf{\text{NB_10}}\cr \mathsf{\text{NC_1}}\cr \mathsf{\text{NC_2}}\cr \mathsf{\text{NC_3}}\cr \mathsf{\text{NC_4}}\cr \mathsf{\text{NC_5}}\cr \mathsf{\text{NC_6}}\cr \mathsf{\text{NC_7}}\cr \mathsf{\text{NC_8}}\cr \mathsf{\text{NC_9}}\cr \mathsf{\text{NC_10}}\cr \end{pmatrix} & = & \begin{pmatrix}0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 50\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 0\cr 1200\cr \end{pmatrix} \end{array} $$

Do ponto de vista de variáveis, no sistema acima, empilhanos primeiro todas as vazões de A, depois todas as vazões de B, depois todas as vazões de C. Ou seja, temos 30 variáveis, de 1 a 10 primeiras seriam as vazões de A, da 11 a 20 as vazões de B e de 21 a 30 as vazões de C:

$$ N_{A,1},N_{A,2},N_{A,3},...,N_{A,10} \ \rm{correspondem \ a } \ x_1,x_2,x_3,...,x_{10}, $$$$ N_{B,1},N_{B,2},N_{B,3},...,N_{B,10} \ \rm{correspondem \ a } \ x_{11},x_{12},x_{13},...,x_{20}, $$$$ N_{C,1},N_{C,2},N_{C,3},...,N_{C,10} \ \rm{correspondem \ a } \ x_{21},x_{21},x_{23},...,x_{30}, $$

Note que as equações, em termos das variáveis x1 a x30 ficam:

$$ \begin{array}{c|ccc} & x_{1} - x_{2} + x_{6}+ x_{9} & = & 0 \\ & x_{2} \cdot \left(1-XR1 \right) - x_{3} & = & 0 \\ & x_{3} \cdot rA7 - x_{7} & = & 0\\ & x_{3} \cdot \left(1-rA7 \right) - x_{4} & = & 0\\ \rm{Eq \ A } & x_{A,5} - 0,7 \cdot x_{6} & = & 0 \\ & -x_{4} + x_{A,5} + x_{6} & = & 0 \\ & x_{7} \cdot \left(1-XR2 \right) - x_{8} & = & 0 \\ & x_{8} \cdot rA9 - x_{A,9} & = & 0 \\ & x_{8} \cdot \left(1- rA9 \right) - x_{10} & = & 0 \\ & x_{10} - 0,05 \cdot x_{30} & = & 0 \\ \hline & x_{11} - x_{12} + x_{16}+ x_{19} & = & 0 \\ & x_{A,2} \cdot XR1 + x_{12} - x_{A,3} & = & 0 \\ & x_{13} \cdot rB7 - x_{17} & = & 0\\ & x_{13} \cdot \left(1-rB7 \right) - x_{14} & = & 0\\ \rm{Eq \ B } & x_{15} - 0,7 \cdot x_{16} & = & 0 \\ & -x_{14} + x_{15} + x_{16} & = & 0 \\ & x_{17} - x_{18} & = & 0 \\ & x_{18} \cdot rB9 - x_{19} & = & 0 \\ & x_{18} \cdot \left(1- rB9 \right) - x_{20} & = & 0 \\ & x_{11} & = & 5 \\ \hline & x_{21} - x_{22} + x_{26}+ x_{29} & = & 0 \\ & x_{22} - x_{23} & = & 0 \\ & x_{23} \cdot rC7 - x_{27} & = & 0\\ & x_{23} \cdot \left(1-rC7 \right) - x_{24} & = & 0\\ \rm{Eq \ C } & x_{25} - 0,7 \cdot x_{26} & = & 0 \\ & -x_{24} + x_{25} + x_{26} & = & 0 \\ & x_{7}\cdot XR2 + x_{27} - x_{28} & = & 0 \\ & x_{28} \cdot rC9 - x_{29} & = & 0 \\ & x_{28} \cdot \left(1- rC9 \right) - x_{30} & = & 0 \\ & x_{30} & = & 1200 \\ \end{array} $$

A seguir é apresentado o código em Scilab.

Scilab

clear;clc;close;

XR1 = 0.75; XR2 = 0.42
rA7 = 0.92; rB7 =0.06; rC7=1;
rA9 = 0.94; rB9 =0.93; rC9=0.03;

A = zeros(30,10)
y = zeros(30,1)
razao_5_6 = .7
// A*x =y
// x1 a x10 ==> NA1 a NA10
// x11 a x20 ==> NB1 a NB10
// x21 a x30 ==> NC1 a NC10

// Balanço para A
A(1,[1,2,6,9]) = [1,-1,1,1]
A(2,[2,3]) = [(1-XR1),-1]
A(3,[3,7]) = [rA7,-1]
A(4,[3,4]) = [1-rA7,-1]
A(5,[5,6]) = [1,-razao_5_6 ]
A(6,[4,5,6]) = [1,-1,-1]
A(7,[7,8]) = [1-XR2,-1]
A(8,[8,9]) = [rA9,-1]
A(9,[8,10]) = [1-rA9,-1]
A(10,[10,30]) = [1,-0.05]

// Balanços p/ B
A(11,[11,12,16,19]) = [1,-1,1,1]
A(12,[2,12,13]) = [2*XR1,1,-1] // 
A(13,[13,17]) = [rB7,-1]
A(14,[13,14]) = [1-rB7,-1]
A(15,[15,16]) = [1,-razao_5_6 ]
A(16,[14,15,16]) = [1,-1,-1]
A(17,[17,18]) = [1,-1]
A(18,[18,19]) = [rB9,-1]
A(19,[18,20]) = [1-rB9,-1]
A(20,11) = 1
y(20) = 5

// Balanços p/ C
A(21,[21,22,26,29]) = [1,-1,1,1]
A(22,[22,23]) = [1,-1] // 
A(23,[23,27]) = [rC7,-1]
A(24,[23,24]) = [1-rC7,-1]
A(25,[25,26]) = [1,-razao_5_6 ]
A(26,[24,25,26]) = [1,-1,-1]
A(27,[7,27,28]) = [XR2,1,-1]
A(28,[28,29]) = [rC9,-1]
A(29,[28,30]) = [1-rC9,-1]
A(30,1) = 1
y(30) = 1200

xsol = inv(A)*y

Em Python, o código a seguir resolve o problema como:

Python

In [113]:
import numpy as np

# Entrada de dados
XR1 = 0.75; XR2 = 0.42
rA7 = 0.92; rB7 =0.06; rC7=1;
rA9 = 0.94; rB9 =0.93; rC9=0.03;
razao_5_6 = .7

A = np.zeros((30,30))
y = np.zeros((30,1))

# Formulação do problema como A*x =y
# x1 a x10 ==> NA1 a NA10
# x11 a x20 ==> NB1 a NB10
# x21 a x30 ==> NC1 a NC10

# Os balanços serão escritos como [equacao,[posicoes_nao_nulas],[valores]]. Posteriormente, estes dados serão usados para montar as matrizes do sistema linear.
# Balanço para A
Acoef=[[0,  [1,2,6,9],[1,-1,1,1]],
       [1,  [2,3],    [1-XR1,-1]],
       [2,  [3,7],    [rA7,-1]],
       [3,  [3,4],    [1-rA7,-1]],
       [4,  [5,6],    [1,-razao_5_6]],
       [5,  [4,5,6],  [1,-1,-1]],
       [6,  [7,8]  ,  [1-XR2,-1]],
       [7,  [8,9]  ,  [rA9,-1]],
       [8,  [8,10] ,  [1-rA9,-1]],
       [9,  [10,30],  [1,-0.05]]]

# Balanço para B
Bcoef =[[10,  [11,12,16,19] ,    [1,-1,1,1]],
        [11,  [2,12,13]     ,    [2*XR1,1,-1]],
        [12,  [13,17]       ,    [rB7,-1]],
        [13,  [13,14]       ,    [1-rB7,-1]],
        [14,  [15,16]       ,    [1,-razao_5_6]],
        [15,  [14,15,16]    ,    [1,-1,-1]],
        [16,  [17,18]       ,    [1,-1]],
        [17,  [18,19]       ,    [rB9,-1]],
        [18,  [18,20]       ,    [1-rB9,-1]],
        [19,  [11]          ,    [1]]]
y[19] = 5

# Balanço para C
Ccoef = [[20,   [21,22,26,29],   [1,-1,1,1]],
         [21,   [22,23]      , [1,-1]],
         [22,   [23,27]      ,  [rC7,-1]],
         [23,   [23,24]      ,  [1-rC7,-1]],
         [24,   [25,26]      ,  [1,-razao_5_6]],
         [25,   [24,25,26]   ,  [1,-1,-1]],
         [26,   [7,27,28]    ,  [XR2,1,-1]],
         [27,   [28,29]      ,  [rC9,-1]],
         [28,   [28,30]      ,  [1-rC9,-1]],
         [29,   [1]          ,  [1]]]
y[29] = 1200

# Monta uma lista de coeficientes
M = Acoef + Bcoef + Ccoef
y = y.reshape(30) # redimensiona o y

# Monta o sistema linear
neq = len(M)
for i in range(30):
    eq = M[i][0]
    pos = M[i][1]
    val = M[i][2]
    m = len(pos)
    for j in range(m):
        A[i][pos[j]-1] = val[j]

# Resolve o sistema e linear e imprime o resultado.
x = np.linalg.solve(A,y)
print(x.reshape((30,1)))

[[ 1.20000000e+03]
 [ 1.39075725e+03]
 [ 3.47689311e+02]
 [ 2.78151449e+01]
 [ 1.14532950e+01]
 [ 1.63618499e+01]
 [ 3.19874166e+02]
 [ 1.85527017e+02]
 [ 1.74395396e+02]
 [ 1.11316210e+01]
 [ 5.00000000e+00]
 [ 3.25849981e+03]
 [ 5.34463568e+03]
 [ 5.02395754e+03]
 [ 2.06868840e+03]
 [ 2.95526914e+03]
 [ 3.20678141e+02]
 [ 3.20678141e+02]
 [ 2.98230671e+02]
 [ 2.24474699e+01]
 [ 8.82852699e+01]
 [ 9.51708087e+01]
 [ 9.51708087e+01]
 [-0.00000000e+00]
 [ 0.00000000e+00]
 [-0.00000000e+00]
 [ 9.51708087e+01]
 [ 2.29517959e+02]
 [ 6.88553876e+00]
 [ 2.22632420e+02]]

Os resultados estão apresentados a seguir (vazões em kmol/s):

$$ {\begin{pmatrix}\mathsf{\text{NA1}}&\mathsf{\text{1200}}&\mathsf{\text{NB1}}&\mathsf{\text{5}}&\mathsf{\text{NC1}}&\mathsf{\text{88.28527}}\cr \mathsf{\text{NA2}}&\mathsf{\text{1390.7572}}&\mathsf{\text{NB2}}&\mathsf{\text{3258.4998}}&\mathsf{\text{NC2}}&\mathsf{\text{95.170809}}\cr \mathsf{\text{NA3}}&\mathsf{\text{347.68931}}&\mathsf{\text{NB3}}&\mathsf{\text{5344.6357}}&\mathsf{\text{NC3}}&\mathsf{\text{95.170809}}\cr \mathsf{\text{NA4}}&\mathsf{\text{27.815145}}&\mathsf{\text{NB4}}&\mathsf{\text{5023.9575}}&\mathsf{\text{NC4}}&\mathsf{\text{0}}\cr \mathsf{\text{NA5}}&\mathsf{\text{11.453295}}&\mathsf{\text{NB5}}&\mathsf{\text{2068.6884}}&\mathsf{\text{NC5}}&\mathsf{\text{0}}\cr \mathsf{\text{NA6}}&\mathsf{\text{16.36185}}&\mathsf{\text{NB6}}&\mathsf{\text{2955.2691}}&\mathsf{\text{NC6}}&\mathsf{\text{0}}\cr \mathsf{\text{NA7}}&\mathsf{\text{319.87417}}&\mathsf{\text{NB7}}&\mathsf{\text{320.67814}}&\mathsf{\text{NC7}}&\mathsf{\text{95.170809}}\cr \mathsf{\text{NA8}}&\mathsf{\text{185.52702}}&\mathsf{\text{NB8}}&\mathsf{\text{320.67814}}&\mathsf{\text{NC8}}&\mathsf{\text{229.51796}}\cr \mathsf{\text{NA9}}&\mathsf{\text{174.3954}}&\mathsf{\text{NB9}}&\mathsf{\text{298.23067}}&\mathsf{\text{NC9}}&\mathsf{\text{6.8855388}}\cr \mathsf{\text{NA10}}&\mathsf{\text{11.131621}}&\mathsf{\text{NB10}}&\mathsf{\text{22.44747}}&\mathsf{\text{NC10}}&\mathsf{\text{222.63242}}\cr \end{pmatrix}}$ $$