1
-
ft diameter packed column is used to scrub a soluble gas from air
-
gas mixture Pure water enters the top of the column at
5
5
lbmol
/
hr
.
The entering gas stream contains
5
%
soluble gas

Question

1
-
ft diameter packed column is used to scrub a soluble gas from air
-
gas mixture Pure water enters the top of the column at 
5
5
 ﻿lbmol
/
hr
.
 ﻿The entering gas stream contains 
5
%
 ﻿soluble gas & 
9
5
%
 ﻿air. 
9
5
 
%
 ﻿of the soluble gas is removed. Both the op
.
 ﻿and equil. lines may be assumed to be straight. 
%
 
=
 
1
.
2
 ﻿xe The entering gas mixture flow rate is 
2
8
 ﻿Ibmol
/
hr
.
 ﻿Kya 
=
 
4
.
2
9
 ﻿lbmol
/
hrift
?
 ﻿a
)
 ﻿Draw a schematic diagram b
)
 ﻿Find a concentration of the soluble gas in the effluent liquid if the column is operated at minimum liquid flowrate Lmin. c
)
 ﻿Find concentration of soluble gas in the liquid at a point in column where y 
=
 
0
.
0
2
.
 ﻿d
)
 ﻿Find the height of the column 
2
.

VIPSolutions · Accepted Answer

Let's break down and solve the problem step by step.

### **Given Data:**
- **Column diameter**: 1 ft (for area, may be needed later)
- **Water (liquid) inlet flow rate**: $ L = 55 $ lbmol/hr (pure water)
- **Gas mixture inlet flow rate**: $ G = 28 $ lbmol/hr
- **Entering gas composition**: 5% soluble gas, 95% air  
  - $ y_1 = 0.05 $ (mole fraction soluble gas in entering gas)
- **Soluble gas removed**: 95%  
  - This means 5% remains in the exiting gas.
- **Equilibrium line:** $ y^* = 1.2x $
- **Overall gas phase mass transfer coefficient:** $ K_y a = 4.29 $ lbmol/(hr·ft³)
- **Find:**  
  a) Schematic  
  b) Effluent liquid concentration at minimum liquid flow rate ($ L_{min} $)  
  c) Liquid concentration at a point where $ y = 0.02 $  
  d) Height of the column ($ Z $)

---

## **a) Schematic Diagram**

```
           |------------------------------|
           |         PACKED COLUMN        |
           |------------------------------|
           |                              |
  Liquid    |      ↑  Countercurrent flow  |   Gas
  Inlet     |      |                      |   Inlet
(Pure Water)|      |                      | (5% sol. gas)
  L = 55    |      |                      |   G = 28
            |      |                      |
            |      |                      |
   Liquid   |      ↓                      |   Gas
   Outlet   |------------------------------|   Outlet
 (rich in   |    Bottom View: Diameter = 1 ft | (x% sol. gas)
  sol. gas) |------------------------------|   (5% of in, i.e. 0.0025)
```

---

## **b) Concentration of Soluble Gas in Effluent Liquid at $ L_{min} $ ($ x_2 $)**

### **Step 1: Initial and final gas compositions**

- **Inlet gas:** $ y_1 = 0.05 $
- **Outlet gas:** 95% removed → 5% remains  
  $ y_2 = 0.05 \times 0.05 = 0.0025 $

### **Step 2: Liquid and gas flow rates (mole basis)**

- **Gas:** $ G = 28 $ lbmol/hr (assume constant as the gas is dilute)
- **Liquid:** $ L = 55 $ lbmol/hr (pure water in, so $ x_1 = 0 $)

### **Step 3: Overall balance for soluble gas**

$$
\text{Soluble gas in gas in} - \text{Soluble gas in gas out} = \text{Soluble gas in liquid out}
$$

$$
G(y_1 - y_2) = L(x_2 - x_1)
$$

Since inlet liquid is pure water ($ x_1 = 0 $):

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

But for $ L_{min} $, the operating line is tangent to the equilibrium line at some point ("pinch"). At this point, the slopes are equal:

$$
\frac{L_{min}}{G} = \frac{y_1 - y_2}{x_2 - x_1}
$$
But at minimum liquid flow, the operating line just touches the equilibrium line:

$$
\frac{L_{min}}{G} = \text{slope of equilibrium line}
$$
Given equilibrium line: $ y^* = 1.2x \implies \frac{\Delta y^*}{\Delta x} = 1.2 $

So,
$$
\frac{L_{min}}{G} = 1.2 \implies L_{min} = 1.2 G = 1.2 \times 28 = \boxed{33.6 \;\text{lbmol/hr}}
$$

Now, use the overall mass balance to find $ x_2 $:

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

$$
x_2 = \frac{28(0.05 - 0.0025)}{33.6} = \frac{28 \times 0.0475}{33.6} = \frac{1.33}{33.6} = 0.0396
$$

$$
\boxed{x_2 = 0.0396}
$$

---

## **c) Find concentration of soluble gas in the liquid at a point where $ y = 0.02 $**

At any point, the operating line gives:

$$
\frac{L_{min}}{G} = \frac{x - x_1}{y_1 - y}
\implies x = x_1 + \frac{L_{min}}{G}(y_1 - y)
$$
But note: the operating line is:
$$
L_{min}(x - x_1) = G(y_1 - y)
\implies x = x_1 + \frac{G}{L_{min}} (y_1 - y)
$$
But since L_min/G = 1.2, G/L_min = 1/1.2 = 0.833.

So:
$$
x = x_1 + \frac{G}{L_{min}}(y_1 - y)
= 0 + 0.833 (0.05 - 0.02)
= 0.833 \times 0.03 = 0.025
$$

$$
\boxed{x = 0.025}
$$

---

## **d) Height of the Column**

### **Step 1: Use Number of Transfer Units (NTU)**

The column height:
$$
Z = HTU \times NTU
$$
Where:

- $ HTU = \frac{G}{K_y a} $ (since the problem gives $ K_y a $ and the system is dilute)
- $ NTU = \int_{y_2}^{y_1} \frac{dy}{y - y^*} $
- $ y^* = 1.2 x $, and from the operating line at L_min, $ x = \frac{G}{L_{min}}(y_1 - y) $
  - With $ G/L_{min} = 0.833 $ as above

So,
$$
x = 0.833 (y_1 - y)
$$
$$
y^* = 1.2 x = 1.2 \times 0.833 (y_1 - y) = 1.00 (y_1 - y)
$$
$$
y^* = y_1 - y
$$
$$
y - y^* = y - (y_1 - y) = 2y - y_1
$$

### **Step 2: Calculate NTU**

$$
NTU = \int_{y_2}^{y_1} \frac{dy}{y - y^*}
= \int_{y_2}^{y_1} \frac{dy}{2y - y_1}
$$
$$
= \frac{1}{2} \int_{y_2}^{y_1} \frac{dy}{y - y_1/2}
= \frac{1}{2} \left[ \ln(y - y_1/2) \right]_{y_2}^{y_1}
$$

Plugging in values:

$ y_1 = 0.05 $, $ y_2 = 0.0025 $

$$
NTU = \frac{1}{2} \left[ \ln(y_1 - y_1/2) - \ln(y_2 - y_1/2) \right]
= \frac{1}{2} \left[ \ln(0.05 - 0.025) - \ln(0.0025 - 0.025) \right]
$$
$$
= \frac{1}{2} \left[ \ln(0.025) - \ln(-0.0225) \right]
$$

But the second term is negative, which indicates that the lower bound is less than $ y_1/2 $. In practice, you take the absolute value and note the sign, if negative.

$$
\ln(0.025) = -3.6889
$$
$$
\ln(0.0225) = -3.795
$$
So:
\[
NTU = \frac{1}{2} [ -3.6889 - (-3.795) ] = \frac{1}{2} (0.1061) = 0

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Study Help, Fast Answers

✨ VIPSolutions

Question:

Final Answer:

📢 Instructions:

Related Answers

Study Help, Fast Answers