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BEST Example of Ammonia Material and Energy Balance

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BEST Example of Ammonia Material and Energy Balance Powered By Docstoc
					       MATERIAL BALANCE AND
        ENERGY BALANCE FOR
       AMMONIA PRODUCTION
              PLANT

                         BY: MUHAMMAD AIZAT
                                           AZMATUN
                                  NURUL ZAWANI
                                 SUHADA SYAZA
                                   NURWASIMAH




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                                      MATERIAL BALANCE




      97 Kmol CH4 are feeding into reactor.       n3 CO = 2 + 1 - 2
                                                  38.8=2 +72.75 - 2
      97 Kmol CH4( 1- 0.75)
                                                  2= 2 +72.75 - 38.8
      =24.25 Kmol CH4
                                                  2= 35.95
      n1 CH4 = 97 - 1
                                                  CH4 = 97 - 1
      n2 H2O = 291 - 1 - 2
                                                  = 97 - 72.75
      n3 CO = 2 + 1 - 2
                                                  =24.25 Kmol CH4
      n4 H2 = 31 + 2
      n5 CO2 = 2                                 H2O = 291 - 1 - 2
                                                  = 291- 72.75 -35.95
      To find 1                                  = 182.3 Kmol H2O
      n1 CH4 = 97 - 1                            CO = 2 + 1 - 2
      24.25 = 97 - 1                             = 2 + 72.75 -35.95
      1= 97 - 24.25                              =38.8 Kmol CO
      1= 72.75
                                                  H2 = 31 + 2
      To find 2                                  = 3(72.75) + 35.95
                                                  =254.2 Kmol H2
      1 Kmol CH4          1 Kmol CO
      97 Kmol CH4           97 Kmol CO            CO2 = 2
                                                  = 35.95 Kmol CO2
      So 97(0.4) = 38.8




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      n1 CH4 = 24.25 - 1- 2              CH4 = 24.25 - 1- 2
      n2 H2O = 182.3 - 1 + 2             = 24.25 - 10.48 - 10.13
      n3 CO = 38.8 + 1 - 22              = 3.64 Kmol CH4
      n4 H2 = 254.2 + 31
                                           H2O = 182.3 - 1 + 2
      n5 O2 = 30.4 - 32
                                           = 182.3 - 10.48 + 10.13
      n6 CO2 = 35.95 + 32                 = 192.08 Kmol H2O
      To find 2                           CO = 38.8 + 1 - 22
      Data given: O2 100% react            = 38.8 + 10.48 – 2(10.13)
      n5 O2 = 30.4 - 32                   = 29.02 Kmol CO
      = 30.4 Kmol O2 (1 - 1)               H2 = 254.2 + 31
      =0 Kmol O2                           = 254.2 + 3(10.48)
      0 = 30.4 - 32                       = 285.64 Kmol H2
      2=10.13                             O2 = 30.4 - 32
                                           = 30.4 – 3(10.13)
      To find 1
                                           = 0.01 Kmol O2
      Data given: fCH4 = 85%
                                           CO2 = 35.95 + 32
      24.25 Kmol CH4 (1 – 0.85)
                                           = 35.95 + 3(10.13)
      =3.64 Kmol CH4
                                           = 66.34. Kmol CO2
       n1 CH4 = 24.25 - 1- 2
      3.64 = 24.25 - 1- 10.13
      1 = 10.48




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        To find CO (Output of HTS)         To find CO (Output of LTS)
        29.02 Kmol CO (1-0.8)              5.804 Kmol CO (1-0.8)
        =5.804 Kmol CO                     = 1.1608 Kmol CO

        n1 CO = 29.02 - 1                 n1 CO = 5.804 - 1
        n2 H2O =192.08 - 1                n2 H2O =168.864 - 1
        n3 CO2 = 66.34 + 1                n3 CO2 = 89.556 + 1
        n4 H2 =285.64 + 1                 n4 H2 =308.856 + 1

        To find 1                         To find 1
        n1 CO = 29.02 - 1                 n1 CO = 5.804 - 1
        5.804 = 29.02 - 1                 1.1608 = 5.804 - 1
        1     = 29.02 -5.804              1     = 5.804 - 1.1608
        1     = 23.216                    1     = 4.6432

        CO = 29.02 - 1                    CO = 5.804 - 1
        = 29.02 - 23.216                   = 5.804 - 4.6432
        = 5.804 Kmol CO                    = 1.1608 Kmol CO

        H2O =192.08 - 1                   H2O = 168.864 - 1
        = 192.08 - 23.216                  =168.864 - 4.6432
        = 168.864 Kmol H2O                 =164.22 Kmol H2O

        CO2 = 66.34 + 1                   CO2 = 89.556 + 1
        = 66.34 + 23.216                   = 89.556 + 4.6432
        =89.556 Kmol CO2                   = 94.1992 Kmol CO2

        H2 = 285.64 + 1                   H2 =308.856 + 1
        = 285.64 +23.216                   =308.856 + 4.6432
        = 308.856 Kmol H2                  = 313.4992 Kmol H2




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                                           n1 CO = 94.1992 - 1
                                           n2 H2 =313.4992 - 31
                                           n3 CH4 = 3.64 + 1
                                           n4 H2O = 1

                                           To find 1

                                           1.1608 Kmol CO ( 1 – 1)
                                           = 0 Kmol CO


                                           n1 CO = 94.1992 - 1
                                           0 = 94.1992 - 1
                                           1= 94.1992

                                           CO = 94.1992 - 1
                                           =94.1992 - 94.1992
                                           = 0 Kmol CO

                                           H2 =313.4992 - 31
                                           =313.4992 – 3(94.1992)
                                           =30.901 Kmol H2

                                           CH4 = 3.64 + 1=
                                           3.64 + 94.1992
                                           =97.8392 Kmol CH4

                                           H2O = 1
                                           = 94.1992 Kmol H2O




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    n1 N2 = 113.83 - 1
    n2 H2 = 30.9016 - 31
    n3 NH3 = 21


    fH2 = 0.75

    30.9016 Kmol H2 X (1-0.75)
    = 7.7254

    So
    n2 N2 = 30.9016 - 31
    7.7254 = 30.9016 - 31
    31    = 130.9016 -7.7254
    1     = 7.7254

    N2 = 113.83 - 1
    = 113.83 - 7.7254
    =106.1046 Kmol N2

    H2 = 30.9016 - 31
    = 30.9016 – 3(7.7254)
    = 7.7254 Kmol H2

    NH3 = 21
    = 2 (7.7254) = 15.4508 Kmol NH3




Spreadsheet method for Synthesis Ammonia.



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Formula:

               A             B           C           D              E          F
 1                           1           2        3 (i)         3(ii)         4
 2      N2              113.83     38.1756           0    =0.85*C2     0.15*C2
 3                                                        =0.85*C3    0.15*C3
         H2         302.6176       75.6544           0
 4                                                        =0.85*C4    0.15*C4
         CH4            6.1508         3.64          0
 5                                                        =0.85*C5    0.15*C5
         NH3                 0 151.3088              0


Material Balance:

                    1           2 3 (i)         3(ii)           4
 N2            113.83     38.1756             0 32.44926 5.72634
 H2           30.9016     75.6544             0 64.30624 11.34816
 CH4          97.8392        3.64             0      3.094  0.546
 NH3                0     15.4508             0 13.13318 2.31762

                    1        2       3 (i)        3(ii)           4
 N2            113.83 38.1756        32.44926     32.44926 5.72634
 H2          302.6176 75.6544        64.30624     64.30624 11.34816
 CH4           6.1508     3.64            3.094        3.094  0.546
 NH3                0 151.3088       13.13318     128.6125 22.69632




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                                            ENERGY BALANCE




                                                 INPUT                                OUTPUT

Component         Phase            Molar flow        Heat capacity,     Molar flow        Heat capacity,
                                   rate, (mol)                          rate, (mol)
                                                         Ĥ (KJ/mol)                        Ĥ (KJ/mol)

    H2O           Liquid      291000 mol/h                  Ĥ1               -                  -

    H2O            Gas                  -                    -         182300 mol/h            Ĥ5

    CH4            Gas            97000 mol/h               Ĥ2         24250 mol/h             Ĥ6

     CO            Gas            2000 mol/h                Ĥ3         38800 mol/h             Ĥ7

     N2            Gas            1000 mol/h                Ĥ4          1000 mol/h             Ĥ8

    CO2            Gas                  -                    -         35950 mol/h             Ĥ9

     H2            Gas                  -                    -         254200 mol/h            Ĥ 10




Energy Balance Calculation



1. For Heat Capacity Ĥ (KJ/mol) at Input


All Capacity Ĥ (KJ/mol) at Input are based on references state at 25ºC



H2O(25 ºC) Ĥ1 = (Ĥf)      CH 4   = - 285.84 KJ/mol (from Table B.1)


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                            727
CH4 (727 ºC) Ĥ2 =       25
                                  (Cp )dt

                        727                        5.469  10 5 t 2 0.3661  10 8 t 3  11 .00  10 12 t 4
CH4 (727 ºC) Ĥ2 =            34 .31  10   3
                                                                                     
                        25                               2                  3                    4


                   = 38.222 KJ/mol CH4


From Table B.8


                        800  700 24.13  20.82
CO (727 ºC) Ĥ3 =                  
                        800  727   24.13  X


                  = 21.714 (KJ/mol) CO


                        800  700 23.86  20.59
N2 (727 ºC) Ĥ4 =                  
                        800  727   23.86  X




                  = 21.473 (KJ/mol) N2




2. For Heat Capacity Ĥ (KJ/mol) at Output


i. To find H2O (Ĥ5 Out)



   H2O ( l, 25º C)                 H2O ( V, 727 º C)



 a)   H20 ( l, 25º C)               H2O ( l, 100 º C)

            100

              75.4  10
                              3
                                   = 5.655 KJ/mol
             25




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b) H2O ( l,100 º C)        H2O (V, 100 º C)



             Ĥv= 40.656 KJ/mol



c) H2O (V, 100 º C)        H2O ( V, 864 º C)




                0.6880  10 5 t 2 0.7604  10 8 t 3  3.593  10 12 t 4
864

  33.46  10 
                 3
                                                    
100
                      2                  3                    4



= 29.22875 KJ/mol



 Ĥ (Total ) Ĥ5 = 5.655 KJ/mol + 40.656 KJ/mol + 29.22875 KJ/mol




Ĥ (Total ) Ĥ5    = 75.5397 KJ/mol H2O




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ii. To find CH4 (Ĥ6 Out)



(Ĥf) = -74.84 KJ/mol



                              864

                               
CH4 (864 ºC) Ĥ6 = (νĤf) + Cp(dt )t
                              727



                                    5.469 105 t 2 0.3661108 t 3  11.00 1012 t 4
                   864

                    34.3110 
                             3
(1)(74.84)                                                      
                   727
                                          2               3                 4



             Ĥ6 = -64.635 KJ/mol CH4




iii. To find (Ĥ7, Ĥ8, Ĥ9 at Output)



The following calculation value for Ĥ are obtained from table B.8



Example : Ĥ7 ( CO)



CO ( v, 727 º C)           CO ( v, 864 º C)



                               900  800 27.49  24.13
CO (864 ºC) Ĥ7 = (Ĥf) +                 
                               900  864   27.49  x



               X = -110.52 KJ/mol + 26.2804 KJ/mol CO



              Ĥ7 = -84.2396




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N2 (864 ºC) Ĥ8 = 26.06636 KJ/mol N2




CO2 (864 ºC) Ĥ9 = (Ĥf) + 38.866 KJ/mol CO2



                  = -393.5 KJ/mol + 38.866 KJ/mol CO2



             Ĥ9 = -354.6324KJ/mol CO2




H2 (864 ºC) Ĥ10 = (Ĥf) + 24.8213 KJ/mol H2




Component       Phase        INPUT                            OUTPUT

                             Molar flow      Heat capacity,   Molar flow     Heat capacity,
                             rate, (mol)                      rate, (mol)
                                               Ĥ (KJ/mol)                      Ĥ (KJ/mol)

    H2O           Liquid     291000 mol/h        - 285.84            -               -

    H2O            Gas               -                  -     182300 mol/h       75.5397

    CH4            Gas        97000 mol/h         38.222       24250 mol/h       -64.635

     CO            Gas         2000 mol/h         21.714       38800 mol/h       -84.2396

     N2            Gas         1000 mol/h         21.473       1000 mol/h        26.06636

    CO2            Gas               -                  -      35950 mol/h      -354.6324

     H2            Gas               -                  -     254200 mol/h       24.8213




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                        Component                      Input       Output

                                  H2O                -83179440        -

                                  H2O                    -       13770887.31

                                  CH4                3707534     -1567398.75

                                   CO                 43428      -3268496.48

                                   N2                 21473       26066.36

                                  CO2                    -       -12749034.78

                                   H2                    -       6309574.46

                                  Total              -79407005   2521598.12




H =       n ( out ) h ( out )
                                  n(in ) h (in )



     = 2521598.12 – (-79407005)



     = 8.19  107 KJ




  Q  H



    = 8.19  107 KJ




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COMPONENT                              INPUT                          OUTPUT

                         Number of                          Number of
                            Mol,           Heat Capacity,      Mol,      Heat Capacity,
                             n                    Ĥ             n              Ĥ
       CH4                 24250            -74.85 kJ/mol     36400      -64.015kJ/mol

       H20                  182500          33.32kJ/mol       19208       -204.14kJ/mol

       CO                   38800           27.49kJ/mol       29020      -105.888kJ/mol

        O2                  30400           27.616kJ/mol      30400       32.47kJ/mol

        H2                     -                 -           285640       29.04kJ/mol

       CO2                  35950           42.94kJ/mol       66340       -334.9kJ/mol

        N2                  113830          -25.99kJ/mol     113830       30.56kJ/mol



Reference states: CH4, H2O, CO, O2, H2, CO2, N2 (1atm, 864˚C)




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Energy Balance Calculation

In the following calculations, values of Ĥf come from Table B.1, formulas for Cp (T) come
from Table B.2. Value for each component Ĥ was getting from table B.8. Effect of any
pressure changes on enthalpies is neglected.

Specific Enthalpies of Selected gases involved



Ĥ1 (CH4)   =   Ĥ(900) = Ĥf = -74.85 kJ/mol


                               1000T2

Ĥ2 (CH4) = Ĥf      (CH4) +             34.31 x10-3 + 5.469x10-5 T + 0.3661x10-8 T2 –
                                   864



                                          11.00 x10-12 T3 dt
        = -74.85 kJ/mol + 10.835 kJ/mol

        = -64.015 kJ/mol



Ĥ3 (H2O) = Ĥ(900)
        = 33.32 kJ/mol



Ĥ4 (H20) = Ĥf     (H2O)     + Ĥ(1000)
         = -241.83 kJ/mol + 37.69 kJ/mol

         =-204.14 kJ/mol



Ĥ5 (CO) = Ĥ(900)
       = 27.49 kJ/mol
                             1000T2

Ĥ6 (CO) =Ĥf      (CO)   +     
                              864
                                         28.95 x10-3 + 0.4110 x10-5 T + 0.3548 x10-8 T2 –

                                          2.220 x10 -12 T3 dt
       = -110.52 kJ/mol + 4.63208 kJ/mol



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         = -105.888 kJ/mol



Ĥ7(O2 ) = Ĥ(864)
         =27.616 kJ/mol




Ĥ8 (O2) = Ĥf     (O2)   + Ĥ(1000)
         = 0 kJ/mol + 32.47 kJ/mol

         = 32.47 kJ/mol



Ĥ9(H2)      = Ĥf     (H2)   + Ĥ(1000)
          = 0 kJ/mol + 29.04

          = 29.04 kJ/mol



Ĥ10 (CO2) = Ĥ(900)
           = 42.94   kJ/mol



Ĥ11 (CO2) = Ĥf       (CO2)   + ĤCO2 (1000)
          = -393.5 + 48.60

          = -344.9 kJ/mol

Ĥ12 (N2) = Ĥ(864)
          = 25.99 kJ/mol



Ĥ13 (N2) = Ĥ(1000)
         = 30.56 kJ/mol




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Q= Htotal= Ʃ ṅout Ĥout    –   Ʃ ṅin Ĥin



= [ -2330146 + (-3921121.12) + (-3072869.76) + 987088 + 8294985.6 +

    (-22880666) + 3478644.8] - [(-1815112.5) + 6080900 + 1066612 +

     839526.4 + 1543693 + (-2958441.7)]

= (-19444084.48) – (4757177.2)

=-24201261.68 Kj




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For HTS:


  Component               nin (mol)        ∆Ĥin (kJ/Mol)    nout (mol)      ∆Ĥout (kJ/Mol)

     CO (g)                  29.02             ∆Ĥ1             5.804              ∆Ĥ7
     H2O (v)                192.08             ∆Ĥ2           168.864              ∆Ĥ8
     CO2 (g)                 66.34             ∆Ĥ3            89.556              ∆Ĥ9
      H2 (g)                285.64             ∆Ĥ4           308.856              ∆Ĥ10
     CH4 (g)                  3.64             ∆Ĥ5              3.64              ∆Ĥ11
      N2 (g)                113.83             ∆Ĥ6            113.83              ∆Ĥ12


Energy Balance Calculation
Calculation of Heat Capacity, Ĥ at input of HTS:
All Capacity Ĥ (KJ/mol) at Input of HTS are based on references state at 25ºC, 1 atm.
(The input temperature of HTS is 3760C from literature)

∆Ĥ1(CO) : 10000C – 3760C
For ∆Ĥ1,value of heat formation ∆Ĥf, are been used.
= - 74.85 kJ/mol


∆Ĥ2(H2O) : 10000C – 3760C
Temperature 1000 0C:
Ĥ = 37.69 kJ/mol (refer Table B.8)
For temperature 376 0C, interpolation method is been used to determine value of ∆Ĥ :

         =

x = 12.3516 kJ/mol
∆Ĥ2 = Ĥ376 – Ĥ1000
= 12.3516 - 12.3516
= - 25.3384 kJ/mol



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∆Ĥ3(CO2) : 10000C – 3760C
Temperature 1000 0C:
Ĥ = 48.60 kJ/mol (refer Table B.8)
For temperature 376 0C, interpolation method is been used to determine value of ∆Ĥ :

         =

x = 15.2052 kJ/mol
∆Ĥ3 = Ĥ376 – Ĥ1000
= 15.2052 - 48.60
= - 33.3948 kJ/mol


∆Ĥ4(H2) : 10000C – 3760C
Temperature 1000 0C:
Ĥ = 29.04 kJ/mol (refer Table B.8)
For temperature 376 0C, interpolation method is been used to determine value of ∆Ĥ :

         =

x = 10.1868 kJ/mol
∆Ĥ4 = Ĥ376 – Ĥ1000
= 10.1868 - 29.04
= - 18.8532 kJ/mol


∆Ĥ5(CH4) : 10000C – 3760C
There is no Ĥ value inside Table 8 for CH4, so integration of Heat Capacities, Cp are been
used to calculate Ĥ.

Ĥ5

     =

     = - 43.3489 kJ/mol


∆Ĥ6(N2) : 10000C – 3760C
Temperature 1000 0C:
Ĥ = 30.56 kJ/mol (refer Table B.8)



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For temperature 376 0C, interpolation method is been used to determine value of ∆Ĥ :

         =

x = 10.4228 kJ/mol
∆Ĥ6 = Ĥ376 – Ĥ1000
= 10.4228 - 30.56
= - 20.1372 kJ/mol


Calculation of Heat Capacity, Ĥ at output of HTS:


∆Ĥ7(CO) : 3760C - 4000C
Temperature 400 0C:
Ĥ = 11.25 kJ/mol (refer Table B.8)
For temperature 376 0C, interpolation method is been used to determine value of ∆Ĥ :

         =

x = 10.5108 kJ/mol
∆Ĥ7 = Ĥ400 – Ĥ376
= 11.25 - 10.5108
= 0.7392 kJ/mol


For the following calculation, ∆Ĥ value at temperature 376 are taken from previous
calculation (HTS input) except ∆Ĥ11.




∆Ĥ8(H2O) : 3760C - 4000C
∆Ĥ8 = Ĥ400 – Ĥ376
= 13.23 – 12.3516
= 0.8784 kJ/mol


∆Ĥ9(CO2) : : 3760C - 4000C
∆Ĥ9 = Ĥ400 – Ĥ376
= 16.35 – 15.2052
= 1.1448 kJ/mol




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∆Ĥ10(H2) : : 3760C - 4000C
∆Ĥ10 = Ĥ400 – Ĥ376
= 10.89 – 10.1868
= 0.7032 kJ/mol


∆Ĥ11(CH4) : 3760C - 4000C

Ĥ11

      =

      = 1.3305 kJ/mol


∆Ĥ12(N2) : 3760C - 4000C
∆Ĥ12 = Ĥ400 – Ĥ376
= 11.15 – 10.4228
= 0.7272 kJ/mol




Calculation Data


  Component               nin (mol)           ∆Ĥin (kJ/Mol)   nout (mol)    ∆Ĥout (kJ/Mol)

      CO (g)                 29.02                 - 74.85      5.804                 0.7392
      H2O (v)               192.08               - 25.3384    168.864                 0.8784
      CO2 (g)                66.34               -33.3948      89.556                 1.1448
       H2 (g)               285.64               -18.8532     308.856                 0.7032
      CH4 (g)                 3.64               -43.3489        3.64                 1.3305
       N2 (g)               113.83               -20.1372      113.83                 0.7272


             Component                     Input, (      in       Input, (      out
               CO (g)                         -2172.15               4.290317
              H2O (v)                         -4867.00               148.3301
              CO2 (g)                         -2215.41               102.5237



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                  H2 (g)                               -5385.23                217.1875
                 CH4 (g)                                -157.79                 4.84302
                  N2 (g)                               -2292.22                82.77718


           = (n1∆Ĥ1) + (n2∆Ĥ2) + (n3∆Ĥ3) + (n4∆Ĥ4) + (n5∆Ĥ5) + (n6∆Ĥ6)
           = (-2172.15) + (-4867.00)+( -2215.41)+( -5385.23)+( -157.79)+( -2292.22)
           = - 17089.8 kJ/mol

            = (n7∆Ĥ7) + (n8∆Ĥ8) + (n9∆Ĥ9) + (n10∆Ĥ10) + (n11∆Ĥ11) + (n12∆Ĥ12)
           = (4.290317)+( 148.3301)+( 102.5237)+( 217.1875)+( 4.84302)+( 82.77718)
           = 559.9518 kJ/mol



H    =      n ( out ) h ( out )
                                    n(in ) h (in )

      = (559.9518) - (- 17089.8)

      = 17649.75 kJ

Q     = H

      = 17649.75 kJ



For LTS:


    Component                          nin (mol)        ∆Ĥin (kJ/Mol)   nout (mol)    ∆Ĥout (kJ/Mol)

     CO (g)                               5.804             ∆Ĥ1           1.1608          ∆Ĥ7
     H2O (v)                            168.864             ∆Ĥ2           164.22          ∆Ĥ8
     CO2 (g)                             89.556             ∆Ĥ3          94.1992          ∆Ĥ9
      H2 (g)                            308.856             ∆Ĥ4         313.4992          ∆Ĥ10
     CH4 (g)                               3.64             ∆Ĥ5            3.64           ∆Ĥ11
      N2 (g)                             113.83             ∆Ĥ6           113.83          ∆Ĥ12


Energy Balance Calculation
Calculation of Heat Capacity, Ĥ at input of LTS:
All Capacity Ĥ (KJ/mol) at Input of HTS are based on references state at 25ºC, 1 atm.
Note that, values of Ĥ at input of LTS are equal with value of Ĥ at output of HTS.




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Calculation of Heat Capacity, Ĥ at output of LTS:
∆Ĥ7(CO) : 4000C – 2000C
Temperature 400 0C: Ĥ = 11.25 kJ/mol
Temperature 200 0C: Ĥ = 5.16 kJ/mol        (refer Table B.8)
∆Ĥ7 = Ĥ200 – Ĥ400
= 5.16 – 11.25
= - 6.09 kJ/mol


∆Ĥ8(H2O) : 4000C – 2000C
Temperature 400 0C: Ĥ = 13.23 kJ/mol
Temperature 200 0C: Ĥ = 6.01 kJ/mol        (refer Table B.8)
∆Ĥ8 = Ĥ200 – Ĥ400
= 6.01 – 13.23
= - 7.22 kJ/mol




∆Ĥ9(CO2) : 4000C – 2000C
Temperature 400 0C: Ĥ = 16.35 kJ/mol
Temperature 200 0C: Ĥ = 7.08 kJ/mol        (refer Table B.8)
∆Ĥ9= Ĥ200 – Ĥ400
= 7.08 – 16.35
= - 9.27 kJ/mol


∆Ĥ10(H2) : 4000C – 2000C
Temperature 400 0C: Ĥ = 10.89kJ/mol
Temperature 200 0C: Ĥ = 5.06 kJ/mol        (refer Table B.8)
∆Ĥ10 = Ĥ200 – Ĥ400
= 5.06 – 10.89
= - 5.83 kJ/mol


∆Ĥ11(CH4) : 4000C – 2000C

Ĥ11

      =


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    = -10.1457 kJ/mol


∆Ĥ12(H2) : 4000C – 2000C
Temperature 400 0C: Ĥ = 13.23 kJ/mol
Temperature 200 0C: Ĥ = 6.01 kJ/mol              (refer Table B.8)
∆Ĥ12 = Ĥ200 – Ĥ400
= 5.06 – 11.15
= - 6.09 kJ/mol




Calculation Data For LTS


  Component               nin (mol)           ∆Ĥin (kJ/Mol)          nout (mol)    ∆Ĥout (kJ/Mol)

     CO (g)                  5.804               0.7392                1.1608                 -6.09
     H2O (v)               168.864               0.8784                164.22                 -7.22
     CO2 (g)                89.556               1.1448               94.1992                 -9.27
      H2 (g)               308.856               0.7032              313.4992                 -5.53
     CH4 (g)                  3.64               1.3305                 3.64                 -10.15
      N2 (g)                113.83               0.7272                113.83                 -6.09


             Component                     Input, (     in               Input, (      out
               CO (g)                        4.290317                       -7.06927
              H2O (v)                        148.3301                       -1185.67
              CO2 (g)                        102.5237                       -873.227
               H2 (g)                        217.1875                       -1733.65
              CH4 (g)                         4.84302                        -36.946
               N2 (g)                        82.77718                       -693.225


          = (n1∆Ĥ1) + (n2∆Ĥ2) + (n3∆Ĥ3) + (n4∆Ĥ4) + (n5∆Ĥ5) + (n6∆Ĥ6)

          = (4.290317) + (148.3301)+( 102.5237)+( 217.1875)+( 4.84302)+( 82.77718)
          = 559.9518 kJ/mol

            = (n7∆Ĥ7) + (n8∆Ĥ8) + (n9∆Ĥ9) + (n10∆Ĥ10) + (n11∆Ĥ11) + (n12∆Ĥ12)
           = (-7.06927)+( -1185.67)+( -873.227)+( -1733.65)+( -36.946)+( -693.225)


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           = - 4529.7873 kJ/mol



H    =      n ( out ) h ( out )
                                    n(in ) h (in )

      = (- 4529.7873) - (559.9518)

      = -5089.7391 kJ

Q     = H

      = -5089.7391 kJ




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Since the component amounts of all streams are known from material balance, we may
proceed directly to the energy balance. We choose as references the elemental species that
form the reactants and products at 1 atm, 25°C and the nonreactive species at 1 atm, 25°C.
The inlet-outlet enthalpy table is shown below.

                                      Inlet-Outlet Enthalpy Table

                                    INPUT                                   OUTPUT
Component                  nin                   Ĥin                 nin             Ĥin
 CO                  -110.52 kJ/mol               -                   -               -
 H2                 313499.0 mol/h               Ĥ1             30901.0 mol/h        Ĥ4
CH4                   3640.0 mol/h               Ĥ2             97839.2 mol/h        Ĥ5
 H2O                         -                    -             94199.2 mol/h        Ĥ6
 N2                 113830.0 mol/h             Ĥ3             113830.0 mol/h         Ĥ7
                     Reference State: H2 (g), H2O (g), N2 (g) at 25°C, 1 atm



Ĥf (CO) = -110.52 kJ/mol




Calculation of Inlet Enthalpy

Ĥ2: Heat capacity of CH4


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T1 = 200°C T2 = 400°C
         T2

Ĥ2   C p (T )dT
         T1


        400
       
        200
              34.31x10-3 + 5.469x10-5 T + 0.3661x10-8 T2 + -11.00x10-12 T3 dT


    = [34.31x10-3 T + 5.469x10-5 T2/2 + 0.3661x10-8 T3/3 + 1-11.00x10-12 T4/4]

    = (13.724 – 6.862) + (4.3752 – 1.0938) + (0.07810 – 9.7624 x 10-3) + (-0.0704 +

        4.4 x 10-3)

    = 6.862 + 3.2814 + 0.06834 + (-0.066)

    = 10.15 kJ/mol



Calculation of Outlet Enthalpy

Ĥ5: Heat capacity of CH4

T1 = 400°C T2 = 300°C
         T2

Ĥ2   C p (T )dT
         T1


        300

       
        400
              34.31x10-3 + 5.469x10-5 T + 0.3661x10-8 T2 + -11.00x10-12 T3 dT


    = [34.31x10-3 T + 5.469x10-5 T2/2 + 0.3661x10-8 T3/3 + 1-11.00x10-12 T4/4]

    = (10.293 – 13.724) + (2.46105 – 4.3752) + (0.03295 – 0.078099) + (-0.022275 +

        0.0704)

    = (-3.431) + (-1.9142) + (-0.04515) + 0.04813

    = -5.34 kJ/mol

For Ĥ1, Ĥ3, Ĥ4, Ĥ6 and Ĥ7 value, it can take direct from table B.8.




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                                      Inlet-Outlet Enthalpy Table




                                    INPUT                                      OUTPUT
Component                    nin                 Ĥin                     nin               Ĥin
 CO                  -110.52 kJ/mol               -                       -                 -
 H2                 313499.0 mol/h            Ĥ1 = 5.06          30901.0 mol/h          Ĥ4 = 7.96
CH4                   3640.0 mol/h           Ĥ2 = 10.15          97839.2 mol/h          Ĥ5 = -5.34
 H2O                          -                   -              94199.2 mol/h          Ĥ6 = 9.57
 N2                 113830.0 mol/h         Ĥ3 = 5.13          113830.0 mol/h            Ĥ7 = 8.12
                     Reference State: H2 (g), H2O (g), N2 (g) at 25°C, 1 atm



Evaluate Htotal


When molecular species had been chosen as references for enthalpy calculation, the extents
of each reaction ( ξ1 and ξ2) would have been calculate. But when more than one reaction
occurs in a process, it is advised to choose element species as references and avoid these
complications.



               Input kJ/hr                                Output kJ/hr
                 -110.52                                   245971.96
              1586304.94                                  -522461.33
                 36946                                     901486.34
                583947.9                                   925111.6
          Total = 2207087.42                           Total = 1550108.57


Htotal = 1550108.57kJ/hr – 2207087.42kJ/hr

       = - 656978.85 kJ/mol

Reaction in reactor is exothermic

Q = Htotal

  = -6.57 x 103 kJ/mol




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  Component               nin (mol)        ∆Ĥin (kJ/Mol)     nout (mol)      ∆Ĥout (kJ/Mol)

      N2(g)                   0                 Ĥ1           106104.6              Ĥ5
      H2 (g)               30901.6              Ĥ2             7725.4              Ĥ6
     CH4 (g)               97839.2              Ĥ3            97839.2              Ĥ7
     H2O (v)               94199.2              Ĥ4            94199.2              Ĥ8
     NH3 (s)                  -                  -            15450.8              Ĥ9


Energy Balance Calculation
Calculation of Heat Capacity, Ĥ at input:
All Capacity Ĥ (KJ/mol) at Input are based on references state at 25ºC, 1 atm.


∆Ĥ1(N2) : 2500C – 4000C
For ∆Ĥ1,value of heat formation ∆Ĥf, are been used.
= 0 kJ/mol


∆Ĥ2(H2) : 2500C – 4000C
Temperature 400 0C:
Ĥ = 10.89 kJ/mol (refer Table B.8)
Temperature 250 0C, no value of ∆Ĥ for this temperature, so doing interpolation:


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         =

x = 6.51 kJ/mol
∆Ĥ2 = Ĥ400 - Ĥ250
= 10.89 - 6.51
= 4.38 kJ/mol


∆Ĥ3(CH4) : 2500C – 4000C
There is no Ĥ value inside table for this component, so calculate using integration of Heat
Capacities, Cp.



     =

     = -10.1457 kJ/mol


∆Ĥ4(H2O) : 2500C – 4000C
Temperature 400 0C:
Ĥ = 13.23 kJ/mol (refer Table B.8)
Temperature 250 0C, value of ∆Ĥ for this temperature, so doing interpolation:

         =

x = 7.79 kJ/mol
∆Ĥ4 = Ĥ400 - Ĥ250
= 13.23 – 7.79
= 5.44 kJ/mol




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Calculation of Heat Capacity, Ĥ at output:


∆Ĥ5(N2) : 4000C – 2700C
Temperature 400 0C:
Ĥ = 11.15 kJ/mol (refer Table B.8)
Temperature 270 0C, value of ∆Ĥ for this temperature, so doing interpolation:

         =

x = 7.277 kJ/mol
∆Ĥ5 = Ĥ270 – Ĥ400
= 7.277 – 11.15
= -3.873 kJ/mol


∆Ĥ6(H2) : 4000C – 2700C
Temperature 400 0C:
Ĥ = 10.89 kJ/mol (refer Table B.8)
Temperature 270 0C, value of ∆Ĥ for this temperature, so doing interpolation:

         =

x = 7.0897 kJ/mol
∆Ĥ6 = Ĥ270 – Ĥ400
= 7.0897 – 10.89
= -3.8003 kJ/mol


∆Ĥ7(CH4) : 4000C – 2700C
There is no Ĥ value inside table for this component, so calculate using integration of Heat
Capacities, Cp.

Ĥ7

     =

     = -6.84029


∆Ĥ8(H2O) : 4000C – 2700C




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Temperature 400 0C:
Ĥ = 13.23 kJ/mol (refer Table B.8)
For temperature 270 0C, interpolation method is been used to determine value of ∆Ĥ :

         =

x = 8.50195 kJ/mol
∆Ĥ8 = Ĥ270 – Ĥ400
= 8.50195 – 13.23
= -4.72805 kJ/mol


∆Ĥ9(NH3) : 4000C – 2700C
There is no Ĥ value inside Table 8 for NH3, so integration of Heat Capacities, Cp are been
used to calculate Ĥ.

Ĥ9

     =

     = -5.8874 kJ/mol


Calculation Data


  Component               nin (mol)            ∆Ĥin (kJ/Mol)   nout (mol)      ∆Ĥout (kJ/Mol)

      N2(g)                   0                       0        106104.6             -3.873
      H2 (g)               30901.6                  4.38         7725.4            -3.8003
     CH4 (g)               97839.2                -10.1457      97839.2           -6.84029
     H2O (v)               94199.2                  5.44        94199.2           -4.72805
     NH3 (s)                  -                       -         15450.8            -5.8874


             Component                     Input, (     in         Input, (     out
               N2(g)                              0                  -410943.12
               H2 (g)                        135349.008               -29358.84
              CH4 (g)                        -992647.17              -669248.50
              H2O (v)                         512443.65              -445378.53
              NH3 (s)                             -                   -90965.04


          = (n1∆Ĥ1) + (n2∆Ĥ2) + (n3∆Ĥ3) + (n4∆Ĥ4)
          = (0) + (30901.6(4.38)) + (97839.2(-10.1457)) + (94199.2(5.44))



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          = -344854.5154 kJ/mol



            = (n5∆Ĥ5) + (n6∆Ĥ6) + (n7∆Ĥ7) + (n8∆Ĥ8) + (n9∆Ĥ9)
           = (106104.6 (-3.873)) + (7725.4(-3.8003)) + (97839.2(-6.84029)) +
              (94199.2(- 4.72805)) + (15450.8(-5.8874))
           = -1645894.022 kJ/mol



H    =      n ( out ) h ( out )
                                    n(in ) h (in )

      = (-1645894.022) - (-344854.5154)

      = -1301039.507 kJ

Q     = H

      = -1301039.507 kJ




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s

               CH4 (g)                         -992647.17              -669248.50
               H2 O (v)                         512443.65              -445378.53
               NH3 (s)                              -                   -90965.04


           = (n1 ∆Ĥ1 ) + (n2 ∆Ĥ2 ) + (n3 ∆Ĥ3 ) + (n4 ∆Ĥ4 )

           = (0) + (30901.6(4.38)) + (97839.2(-10.1457)) + (94199.2(5.44))



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           = -344854.5154 kJ/mol



            = (n5 ∆Ĥ5 ) + (n6 ∆Ĥ6 ) + (n7 ∆Ĥ7 ) + (n8 ∆Ĥ8 ) + (n9 ∆Ĥ9 )
            = (106104.6 (-3.873)) + (7725.4(-3.8003)) + (97839.2(-6.84029)) +
               (94199.2(- 4.72805)) + (15450.8(-5.8874))

            = -1645894.022 kJ/mol



H =         n ( out ) h ( out )
                                    n(in ) h (in )

      = (-1645894.022) - (-344854.5154)

      = -1301039.507 kJ

Q     = H

      = -1301039.507 kJ




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DOCUMENT INFO
Description: Material Balance and Energy Balance Example for Ammonia Plant.