Design a packed tower to reduce
9
5
%
of NH
3

(
g
)
present in
2
8
3
.
1
7
m
^
3
/
min
(
1
0
,
0
0
0
acfm
)
of exhaust at a
2
5
%
mixing ratio. Emissions T
,
P conditions
=

4
6
C

Question

Design a packed tower to reduce 
9
5
%
 ﻿of NH
3
 
(
g
)
 ﻿present in 
2
8
3
.
1
7
 ﻿m
^
3
/
min 
(
1
0
,
0
0
0
 ﻿acfm
)
 ﻿of exhaust at a 
2
5
%
 ﻿mixing ratio. Emissions T
,
P conditions 
=
 
4
6
C
,
 
7
6
0
 ﻿mm Hg Base design on the following parameters: 
•
 ﻿Average scrubber temperature 
=
 
3
0
C 
•
 ﻿Pressure in scrubber 
=
 
1
 ﻿atm 
•
 ﻿Pure water is used as scrubber liquid 
•
 ﻿Water rate is 
1
.
5
 ﻿times minimum 
•
 ﻿Packing is 
1
.
0
 ﻿in ceramic Raschig rings. ky
.
�
�
=
1
5
lbmol
/
hr
.
ft
^
3
.
delta
�
�
 ﻿and kx
.
�
�
=
 
6
0
lbmol
/
hr
.
ft
^
3
.
deltax

VIPSolutions · Accepted Answer

Let's **design a packed tower** to absorb **95% of NH₃** from an air stream, given the conditions and parameters you provided.

---

## 1. **Given Data**

- **Air flow:**  
  $ Q_{air} = 283.17 $ m³/min = **10,000 acfm** (actual cubic feet per minute)
- **NH₃ mixing ratio:** 25% (by mole)
- **Fraction to remove:** 95%
- **Emission conditions:**  
  $ T = 46^\circ C $, $ P = 760 $ mmHg (1 atm)
- **Scrubber conditions:**  
  $ T = 30^\circ C $, $ P = 1 $ atm
- **Scrubbing liquid:** Pure water
- **Liquid rate:** 1.5 × minimum
- **Packing:** 1.0" ceramic Raschig rings
- **Gas-phase mass transfer coefficient, $ k_y a $:** 15 lbmol/hr·ft³·Δy
- **Liquid-phase mass transfer coefficient, $ k_x a $:** 60 lbmol/hr·ft³·Δx

---

## 2. **Mole Balance on NH₃**

### **a. Calculate Inlet NH₃ Flow Rate**

#### **Find total molar flow in gas:**

At 1 atm, 46°C:

- $ T = 46 + 273.15 = 319.15 $ K
- $ P = 1 $ atm = 101325 Pa

**1 mole of ideal gas at 1 atm, 46°C:**
$$
V_m = \frac{RT}{P}
$$
Where:
- $ R = 0.08206 $ L·atm/mol·K = 0.08206 × 10⁻³ m³·atm/mol·K

$$
V_m = 0.08206 \times 319.15 = 26.19 \text{ L/mol} = 0.02619 \text{ m}^3/\text{mol}
$$

#### **Gas flow in mol/min:**
$$
Q_{air} = 283.17 \text{ m}^3/\text{min}
$$
$$
n_{total} = \frac{Q_{air}}{V_m} = \frac{283.17}{0.02619} = 10,813.7 \text{ mol/min}
$$

But the air is 25% NH₃ by mole, so this is the total of NH₃ + air.

#### **Breakdown:**

Let $ n_{total} $ be the total mol/min.
Let $ y_{NH3, in} = 0.25 $.

$$
n_{NH3, in} = 0.25 \times n_{total} = 0.25 \times 10,813.7 = 2,703.4 \text{ mol NH}_3/\text{min}
$$
$$
n_{inert, in} = 0.75 \times 10,813.7 = 8,110.3 \text{ mol inert}/\text{min}
$$

### **b. Calculate NH₃ Removal**

- **95% removal:** Only 5% remains in exit gas.
$$
n_{NH3, out} = 0.05 \times n_{NH3, in} = 0.05 \times 2,703.4 = 135.2 \text{ mol NH}_3/\text{min}
$$
- **NH₃ removed:**  
  $ 2,703.4 - 135.2 = 2,568.2 $ mol NH₃/min

---

## 3. **Mole Fractions In and Out**

### **Inlet (to absorber):**
$$
y_1 = \frac{n_{NH3, in}}{n_{total, in}} = \frac{2,703.4}{10,813.7} = 0.25
$$

### **Outlet (from absorber):**
$$
y_2 = \frac{n_{NH3, out}}{n_{inert, in} + n_{NH3, out}}
= \frac{135.2}{8,110.3 + 135.2} = \frac{135.2}{8,245.5} = 0.0164
$$

---

## 4. **Liquid Rate (Water) Calculation**

### **a. Minimum Liquid Rate**

The minimum water flow is the amount needed to absorb all but the outlet NH₃ (i.e., absorb 95%).

- **Assume fresh water in ($ x_1 = 0 $),**
- **Saturated water out (with $ x_2 $),**

#### **NH3 absorbed:** $ 2,568.2 $ mol/min

$$
L_{min} = \frac{NH_3\ absorbed}{x_2 - x_1}
$$

#### **Find $ x_2 $:** (mole fraction NH₃ in water leaving)

**Assume water leaves at equilibrium with exit gas.**

#### **NH₃ equilibrium in water:**

At the bottom, gas phase has $ y_2 = 0.0164 $.

For NH₃, Henry's Law at 30°C:
- $ H = 62,000 $ atm (approx. for NH₃ at 30°C)
- But NH₃ is very soluble, so a better approach is to use the empirical solubility or consult tables.

For now, let's approximate using the equilibrium:

$$
y^* = m x
$$
Where $ m = \frac{H}{P} $.

But for NH₃, because it dissolves and reacts with water (forming NH₄OH), the equilibrium is complex. For dilute solutions and low partial pressure, an approximate **distribution coefficient** at 30°C can be estimated from data:

- At 25°C, the distribution coefficient ($ m $) is about 1000 (dimensionless; see Perry's Chem Eng Handbook). At 30°C, it decreases, let's estimate $ m = 800 $.

So:
$$
y^* = 800 \cdot x
$$
$$
x = \frac{y_2}{800} = \frac{0.0164}{800} = 2.05 \times 10^{-5}
$$

#### **Minimum water flow:**

$$
L_{min} = \frac{NH_3\ absorbed}{x_2} = \frac{2,568.2}{2.05 \times 10^{-5}} = 1.25 \times 10^8 \text{ mol/min}
$$
But this is unreasonably large, because the outgoing water is far from saturation. Let's check with a more practical approach:

- **Use dilute solution:** For very low $ x_2 $, minimum $ L $ is very high. In practice, the water leaving is essentially pure.

**So, instead, let's use the operating line directly:**

---

## 5. **Overall Material Balance and Operating Line**

$$
L (x_2 - x_1) = G (y_1 - y_2)
$$
Where:
- $ G $: total molar gas flow (NH₃ + inert), mol/min
- $ L $: total molar liquid flow (mainly water), mol/min

- $ y_1 = 0.25 $, $ y_2 = 0.0164 $
- $ x_1 = 0 $ (pure water in)

### **a. Assume Water Out is Essentially Pure**

So,
$$
L_{min} = \frac{G (y_1 - y_2)}{x_2}
$$

But as shown, $ x_2 $ is extremely small, leading to unreasonably large $ L_{min} $.

**In practice, for very soluble gases, the minimum liquid flow is just enough to dissolve all the NH₃ removed, i.e.:**

$$
L_{min} = \frac{2,568.2 \text{ mol NH}_3/\text{min}}{(x_2 - x_1)}
$$
But $ x_2 $ is limited by the solubility of NH₃ at 30°C.

- **Solubility of NH₃ in water at 30°C:** About **30% by weight**, which is **1,764 g/L** (NH₃ MW = 17), thus:

$$
C_{NH3, max} = \frac{1,764}{17} = 103.8 \text{ mol/L}
$$
But this is at saturation; in practice, the water will

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