Nitric Acid

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1. Problem Statement

In nitric acid manufacturing plant an absorber column uses a reaction gas mixture and 42% weak acid,at

5531 kg/hr as absorbent. The gas mixture enters the absorption column at 650c and leaves at 100c.The

reaction gas mixture and tail gas composition are follows:

Reaction gas Tail gas

Kg/hr Kmol/hr Kg/hr Kmol/hr

Nitrogen +inerts 27482 981.5 27482 981.5

Oxygen (O2) 1501 46.9 257 8.04

Nitric oxide (NO) 117 3.9 10 0.33

Nitrogen dioxide (NO2) 2099 45.6 25 0.55

Nitrogen tetroxide (N2O4) 1302 14.2 6 0.06

The operation is carried out at a pressure 950 KPa. The feed plate temperature is 500c.60% wt acid is

withdrawn at 11784 kg/hr. (dissolved contents include 1.623 kg NO and 115.47 kg NO2).Make up water

required is 1532 kg/hr. The vapour and liquid densities are 10.1 and 1350 kg/m3 respectively. Design the

sieve tray column.

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2. Given Data

Operating Parameters Operating pressure 950 kpa Operating temperature 650c to 100c

Column Inputs Reaction Gas Feed Components Kg/hr Kmol/hr

N2+inerts 27482 981.5

O2 1501 46.9

NO 117 3.9

NO2 2099 45.6

N2O4 1302 14.2

Total 32501 1902

Weak acid condensate (at 50 ℃ ) Nitric acid (42% wt)

5531 Kg/hr

Make up water 1532 Kg/hr

Output Requirements Product acid (60% wt) 11784 Kg/hr

Dissolved NO

Dissolved NO2

1.623 Kg

115.47Kg

Tail Gas components Kg/hr Kmol/hr

N2+inerts 27482 981.5

O2 257 8.04

NO 10 0.33

NO2 25 0.55

N2O4 6 0.06

Total 27780 990.48

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3. Additional Data

3.1 Nitric acid absorption

The absorption column is required to absorb nitrous reaction gases thus producing 60% (wt.) nitric acid

product (dissolved contents include NO and NO2). This is achieved by the countercurrent absorption of

the nitrogen oxide components from the reaction gas into a water/weak acid media. The column

specification requires an operating pressure of 950 kPa and an absorption temperature in the range of

10°C to 65°C.

The design must consider three feed streams and two product streams. The three inlet feed streams are

reaction gases, weak nitric acid solution and make-up water. Two outlet streams flow from the column.

These are tail-gas stream and product acid. Absorption of nitrous oxides increases as the temperature is

reduced. This effect, together with the exothermic oxidation/absorption reactions, requires installation of

an internal cooling circuit.

Specific heat of the components

components Heat of formation (KJ/Kmol) Specific heat (KJ/KmolK)

H2O (V) - 241830 34.34

H2O (L) - 285840 75.4

NO 90370 29.75

NO2 33800 37.8

N2O4 9300 41.65

HNO3 -206570 110.0

O2 0 29.8

N2 0 29.16

3.2 Analysis of the process chemistry

The main aspects are listed below.

(a) Formation of nitric acid occurs primarily by the reaction of nitrogen dioxide with water, although

some acid forms from the tetroxide. The major reaction is shown below.

3��(�) + ��(�) → 2��(��) + ��(�)………………1

This is an exothermic reaction which is dependent on the partial Pressure (therefore total operating

pressure) of nitrogen dioxide.

(b) Additional reaction of nitric oxide and nitrogen tetroxide with water generates nitrous acid which

decomposes into nitric acid according to following reactions.

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2�� + �� → �� + �� …………… . .2

�� + �� → �� + �� ………… . .3

3�� → �� + 2�� +��……… . .4

(c) Secondary reactions concern the equilibrium of nitrogen dioxide with the tetroxide form, and the

oxidation of nitrogen monoxide (formed in the production of acid) to nitrogen dioxide.

2��(�) ↔ �� ……………… .5

2��(�) + ��(�) → 2�� …………… .6

Reaction 3 is relatively slow and is regarded as the rate-determining step in the entire process. Other key

reactions proceed virtually instantaneously. Both Reactions (2) and (3) are exothermic, and are dependent

on the various partial pressures (operating pressure) and temperature. Lower temperatures favor the

overall absorption chemistry. This factor, together with the exothermic nature of the reactions, determines

the need for a cooling circuit within the column.

(d) The dissolution of both nitrogen monoxide and nitrogen dioxide into aqueous nitric acid is inversely

proportional to their partial pressures. The amounts dissolved decrease with increasing temperature.

(e) In the mechanical considerations those factors of greatest significance are column diameter and plate

spacing. These factors determine the average residence time of the gases in the space between the plates,

and hence the degree of oxidation of nitrogen oxide (the rate-determining step).

(f) Tray design is also of critical importance. Tray area, weir height, and the size and spacing of tray

holes are all important variables. These factors and the gas velocity determine the tray efficiency. Tray

efficiency has a major effect on the overall column performance.

4. Process Design Methodology

Six component analysis model was used to design the absorption column. It accounts for N2, 02, H20, NO,

NO, and N204 within the reaction gas stream. The operation of the column influences a number of key

stages; these stages are described as follows,

Stage 1: The reaction gas of known composition enters the column at 65°C.

Stage 2: The number of moles of nitric acid formed as the gas passes through the liquid mixture on the

plate is calculated. This calculation includes a term that effectively describes the tray efficiency. The

moles of acid formed are given by: [Ref.1]

�� = �� + �� − �(��

� + 2��)

(�� + �)�� (�� + ��) × � × (�� + ��)

�� …… . . ��. 1

Where,

FHNO3 = moles of acid formed;

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PNO = partial pressure of nitrogen monoxide (bar);

PNOX = partial pressure of nitrogen peroxide (dioxide + tetroxide) (bar);

GNO = flow rates of NO. (KMol/sec)

GNOX = flow rates of nitrogen peroxide (dioxide +tetroxide) (KMol/sec)

K2 = dioxide/tetroxide equilibrium constant (Bodenstein); [Ref.2]

Where, log�� =��

�− 9.226 And �� =

��

K4= equilibrium constant between nitrogen monoxide and nitrogen tetroxide as presented empirically by

the Carberry equation; [2]

log�� = 7.412 − 20.28921� + 32.47322�� − 30.87��…… . . ��. 2

W = concentration of acid present on the plate

E = equilibrium partial pressure of nitrogen dioxide as determined by solving the following equation:

3�� + (2��/��) + � = 3�� +�� …… . . ��. 3 [Ref.1]

C = degree of approach to equilibrium (tray efficiency).

Stage 3: The component mole balance is recalculated according to the moles of nitric acid formed. The

gaseous concentrations calculated after the formation of nitric acid, are used to calculate the degree of

oxidation of nitrogen oxide.

Stage 4: the oxidation of nitrogen monoxide takes place in the void area prior to the first sieve tray. The

degree of oxidation (α) is determined using the Fig 1.

Bodenstein equation for calculate the velocity constant K1 is [Ref.4]

log�� =641

�− 0.725

Assumes ‘plug’ flow in the void area between plates calculate two dimensionless quantity Y and m,

where

� =��

�

� And � =

��

��

And from the nomograph the degree of oxidation (α) of nitrogen oxide is found out.

Where

PNO = partial pressure of NO (atm)

NO2=mole of oxygen.

NNO = moles of nitrogen monoxide.

t = residence time (sec)

T = temperature in kelvin

Stage 5: the equilibrium extends of dimerization (β) of nitrogen dioxide to nitrogen tetroxide is calculated

by using the equation [Ref.3]

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�� =2��(1 − �)�

�(1 − 0.5 ∗ ��)…… . . ��. 4

Where

GNO2 = flow rates of NO2. (KMol/sec)

P= total pressure (Mpa)

Kp can be calculated by the given equations [Ref.3]

ln�� = 7.0505−6198

�+ 1.75�� + 0.011� − 1.64 ∗ 10��

Stage 6: The gas composition is revised. Inlet gas composition at i+1 plate can be calculated using

following equation. [Ref.3]

�� = (1− �)��

�� + 0.5��……… ��. 5

�� = (1 − �)��

�� − (1.5��) + �� + 0.5�� ………��. 6

�� = ��

�� + �� − 1.5�� + ��

�� + 0.5��………��. 7

�� = ��

�� − 0.5�� + 0.5��………��. 8

�� = ��

�� − 0.5�� ……… . ��. 9

Stage 7: The computation proceeds to the next tray; this constitutes a return to Stage 2 whereby the

fraction of nitrogen dioxide convert into nitric acid and remains monoxide oxidized in the void space

above the tray. The steps proceed as indicated in Stages 2 to 6.

The calculation is terminated when the number of moles of nitric acid required from the ‘next tray’ is

found to be less than or equal to zero

This model is considered the following assumption,

● Oxidation of nitric oxide proceed in void space between trays and HNO3 formation due to absorption is

found in liquid pool.

● The tray temperatures are set by the known values of feed and output temperatures. The bottom tray

was given the value of 65°C corresponding to the inlet gas temperature. The weak-acid feed tray was set

at 50°C (the temperature of this feed condensate) and the top tray was set to 10 °C. Temperatures for all

other trays in the column were set by linear interpolation between these points.

● Gas and liquid are ideally mixed in the pool of liquid over the plates.

● There has no concentration and temperature gradient in the pool of liquid.

● HNO2 completely decomposes into HNO3 NO and H2O.

● Heat loss to the environment is negligible and reaction heat is exchanged with heat transfer coil.

.The nomograph for finding the degree of oxidation of NO is given below,

9

Fig 1

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5. Process design Computation

5.1 Overall Material Balance:

Overall material balance:

(reaction gas in + make up water in + weak acid in) – (tail gas out + strong acid withdrawn)

=32501+5531+1531-27780-11784

= 0

5.2 Tray by tray calculation

Operating pressure: 950 KPa.

Column diameter: 1.8 m

Hole diameter: 5 mm

Residence time: 1.4 sec

Liquid density: 1350 kg/m3

Vapor density: 10.1 Kg/m3

Tray efficiency: 50 %

Process water

1531 kg/hr

Weak acid (42%)

5531 Kg/hr

Acid withdrawn (60%)

11784 kg/hr

Inlet gas

32501 kg/hr

Tail gas

27780 kg/hr

Cooling water in

Cooling water out

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Calculation for stage 1:

Tray temp: 65℃

Concentration of nitric acid on the plate, w: 0.6

Value of equilibrium constant of NO and N2O4, K4: 1.8239

Value of equilibrium constant of NO2 and N2O4, K2: 0.4256

Value of K3: 4.2859

Value of KP: 2.1127

Value of velocity constant K1:14.8476

Value of degree of oxidation α: 0.9

Value of extend of dimerization β: 0.005

Equilibrium partial pressure of NO, E: 0.2259 bar

Value of dimensionless group Y: 1.71

Value of dimensionless group, m: 6.40

Inlet composition of stage 1

Component kg/hr kmol/hr kmol/sec mol frac

Partial pressure

KPa bar atm Mpa

N2 27482 981.5 0.272639 0.893367 848.6989988 8.378075 8.270558 0.848699

O2 1501 46.9 0.013028 0.042689 40.55423642 0.400338 0.3952 0.040554

NO 117 3.9 0.001083 0.00355 3.372313902 0.03329 0.032863 0.003372

NO2 2399 52.15217 0.014487 0.047469 45.09576952 0.44517 0.439458 0.045096

N2O4 1302 14.2 0.003944 0.012925 12.27868139 0.121211 0.119656 0.012279

total 32801 1098.652 0.305181 1 950 9.378085 9.257734 0.95

● By solving eq. 1 the moles of nitric acid formed in the 1st stage is 2.9 mol/sec.

● The revised composition for oxidation of nitric acid in the void space is,

Component Kmol/sec mol frac

Partial pressure

KPa MPa bar atm

N2 0.272639 0.901939 856.8416 0.856842 8.458456357 8.349908

O2 0.013028 0.043098 40.94332 0.040943 0.404178913 0.398992

NO 0.002533 0.008381 7.961687 0.007962 0.078595132 0.077587

NO2 0.010137 0.033534 31.85738 0.031857 0.314485441 0.31045

N2O4 0.003944 0.013049 12.39649 0.012396 0.122373999 0.120804

total 0.302281 1.000001 950.0005 0.950001 9.378089842 9.257739

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By using the value of dimensionless group Y and m, from nomograph the degree of oxidation of nitric

acid is to be found 0.9.

By solving eq. 4, the extend of dimerization of NO2 to N2O4 is found 0.005.

Again the composition of the gas phase is revised as eq.5 to eq. 9, which is considered as the inlet

composition of stage 2.

Calculation of stage 2:

Tray temp: 63.75℃




Value of K3: 14.15

Value of KP: 1.9610







Inlet composition of stage 2:

Component kmol/sec mol frac

Partial Pressure

KPa Mpa Bar atm

N2 0.272639 0.905353 860.0853 0.86008529 8.490477 8.381516986

O2 0.011888 0.039476 37.502 0.037501997 0.370207 0.365456343

NO 0.000253 0.000841 0.799183 0.000799183 0.007889 0.007788022

NO2 0.012355 0.041026 38.97477 0.038974765 0.384746 0.379808447

N2O4 0.004007 0.013304 12.63927 0.012639267 0.124771 0.123169452

total 0.301141 1.000001 950.0005 0.950000503 9.37809 9.257739251



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Calculation of stage 3:

Tray temp: 62.50℃




Value of K3: 30.89

Value of KP: 1.8193








Partial Pressure

KPa Mpa Bar atm

N2 0.272639 0.910219 864.7083 0.864708309 8.536114 8.426568224

O2 0.011888 0.039688 37.70357 0.037703573 0.372197 0.367420697

NO 0.001058 0.003533 3.356636 0.003356636 0.033136 0.032710367

NO2 0.00994 0.033184 31.52478 0.031524783 0.311202 0.307208493

N2O4 0.004007 0.013376 12.7072 0.012707204 0.125441 0.123831497

total 0.299531 1.000001 950.0005 0.950000506 9.37809 9.257739277

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Inlet gas composition of stage 3:


Partial Pressure

KPa Mpa Bar atm

N2 0.272639 0.932287 885.6725 0.885672476 8.743065 8.630863684

O2 0.011438 0.039112 37.15651 0.03715651 0.366797 0.362089573

NO 0.000159 0.000543 0.515702 0.000515702 0.005091 0.005025511

NO2 0.010785 0.036879 35.03533 0.035035333 0.345857 0.34141874

N2O4 0.004061 0.013886 13.19134 0.013191337 0.130221 0.128549365

total 0.299081 1 971.5714 0.971571358 9.59103 9.467946874




Partial Pressure

KPa Mpa Bar atm

N2 0.272639 0.916646 870.8135 0.870813548 8.596383 8.486063681

O2 0.011438 0.038456 36.53313 0.036533135 0.360643 0.356014796

NO 0.000984 0.003307 3.142115 0.003142115 0.031018 0.030619862

NO2 0.00831 0.027939 26.54235 0.02654235 0.262017 0.258654764

N2O4 0.004061 0.013653 12.97003 0.012970026 0.128036 0.126392692

total 0.297431 1.000001 950.0012 0.950001175 9.378096 9.257745796






In this way tray by tray calculation was done and the total no of trays was found 56.

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Table 1: Concentration of nitric acid in different stage

Tray no HNO3 formed (mol/sec)

Acid conc. in tray (weight fraction)

Tray no HNO3 formed (mol/sec)

Acid conc. in tray (weight fraction)

1 0.0029 0.6 29 0.00023 0.121847

2 0.00161 0.581263 30 0.00019 0.111265

3 0.00165 0.569715 31 0.00018 0.102296

4 0.00167 0.556899 32 0.00017 0.093604

5 0.00165 0.542783 33 0.00016 0.085212

6 0.00158 0.527549 34 0.00011 0.077148

7 0.00146 0.511592 35 0.00011 0.071507

8 0.0013 0.495486 36 0.00011 0.065787

9 0.00118 0.479894 37 0.0001 0.059984

10 0.00102 0.46459 38 0.00009 0.054636

11 0.00088 0.450372 39 0.000096 0.049763

12 0.00077 0.437289 40 0.000075 0.0445

13 0.00115 0.425163 41 0.00007 0.040342

14 0.00145 0.405746 42 0.00007 0.036423

15 0.00125 0.378738 43 0.00006 0.032467

16 0.0011 0.352829 44 0.000055 0.029047

17 0.00095 0.327684 45 0.000056 0.025887

18 0.00082 0.303935 46 0.00005 0.022645

19 0.000708 0.281713 47 0.000048 0.019729

20 0.000615 0.261087 48 0.000045 0.016912

21 0.00053 0.24197 49 0.00004 0.014253

22 0.00048 0.224514 50 0.000038 0.011876

23 0.00042 0.207853 51 0.000036 0.009606

24 0.00045 0.192558 52 0.000032 0.007444

25 0.0004 0.175373 53 0.000032 0.005513

26 0.00033 0.159354 54 0.00003 0.003574

27 0.0003 0.145575 55 0.000029 0.001748

28 0.00024 0.132581 56 -2.4E-05

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5.3 Heat duty calculation

The required flow rate through each set of cooling coils is calculated with the known value of heat duty.

Using the heat capacity data and enthalpy data, the energy balance is performed.

Heat duty = net enthalpy out-net enthalpy in

= �� − ��

Contributing compound Heat duty (KJ/hr.)

Nitrogen 981.5 ∗ 29.16 ∗ (10 − 65) = −1574130

Oxygen 8.04 ∗ 29.8 ∗ (10 − 65) = −13177.56

Nitric oxide 0.38 ∗ 29.75 ∗ (10 − 65) = −621.775

Nitrogen dioxide 3.06 ∗ 37.8 ∗ (10 − 65) = −6361.74

Nitrogen tetroxide 0.06 ∗ 41.65 ∗ (10 − 65) = −137.445

Nitric acid [{36.87 ∗ 110 ∗ (65 − 50)} + {75.36 ∗ (−206570)}

+ {75.35 ∗ 110 ∗ (65− 50)}] = −15379887

Water [{47.44 ∗ 75.4 ∗ (50 − 10)} + {178.22 ∗ 75.4 ∗ (65 − 50)}

+ {37.67 ∗ 285840}] = 11112239

Nitric oxide deformation {3.52 ∗ (−90370)} = −318102.4

Nitrogen dioxide deformation {42.54 ∗ (−33800)} = −1437852

Nitrogen tetroxide deformation {14.14 ∗ (−9300)} = −131502

Total −7749532

.

So the cooling coil must remove 7.75 *106 KJ heat in each hour.

Required flow rate of cooling water,

Consider, the design inlet temperature of the coil is 7℃

And the design outlet temperature is 25℃

So required flow rates of water in cooling coil is

= ℎ��/{��ℎ�� ∗ (�� − ��)}

= 7749532/(4.2*(25-7))

= 102507 kg/hr.

= 102.507 cubic meter/hr.

Nitric Acid

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