Give step-by-step solution with explanation and final answer:
a subcooled mixture
(
q
=
1
.
3

5
5
%
MeOH
)
enters fractionating column with a partial
/
total condenser and partial reboiler set

Question

Give step-by-step solution with explanation and final answer: 
a subcooled mixture 
(
q
=
1
.
3
 
5
5
%
 ﻿MeOH
)
 ﻿enters fractionating column with a partial
/
total condenser and partial reboiler set up the distillate 
(
9
5
 ﻿MeOH
)
 ﻿shall be collected from the total condenser bottom will have 
0
.
9
 
%
 ﻿MEOH. using mccabe thiele method determine feed plate. reflux is 
0
.
8
2
6
9

VIPSolutions · Accepted Answer

Let's solve your distillation problem step by step using the **McCabe-Thiele method**.

---

### **Given Data**

- **Feed (F):** Subcooled mixture  
  $ q = 1.355 $ (q-line slope > 1, subcooled liquid)
- **Feed composition:** $ x_F = 0.355 $ MeOH (by mass or mole fraction, let's assume mole fraction)
- **Distillate composition:** $ x_D = 0.95 $ MeOH
- **Bottoms composition:** $ x_B = 0.009 $ MeOH
- **Reflux ratio:** $ R = 0.8269 $
- **Column type:** Fractionating column with total condenser and partial reboiler

---

### **Step 1: Draw/Describe the McCabe-Thiele Diagram**

#### **1.1. Plot the Equilibrium Curve**
- For MeOH-water, you would use T-x-y or P-x-y data. For simplicity, let's assume an **ideal solution** (or you can look up the appropriate equilibrium line for MeOH-water).

#### **1.2. Operating Lines**
- **Rectifying Section** (above the feed):
  $$
  y = \frac{R}{R+1}x + \frac{x_D}{R+1}
  $$

- **Stripping Section** (below the feed):
  $$
  y = \frac{S}{S-1}x - \frac{x_B}{S-1}
  $$
  But we often use mass balances to relate the two sections.

#### **1.3. q-Line**
- **q** is the thermal condition of the feed:
  $$
  q = \text{fraction of feed that is liquid}
  $$
  $ q > 1 $ means the feed is subcooled liquid.

The q-line equation:
  $$
  y = q/(q-1)x - x_F/(q-1)
  $$
  Or, **slope** of q-line is $ \frac{q}{q-1} $.

For $ q = 1.355 $:
  $$
  \text{slope} = \frac{q}{q-1} = \frac{1.355}{1.355-1} = \frac{1.355}{0.355} \approx 3.82
  $$
  This is a steep line (as expected for subcooled liquid).

---

### **Step 2: Calculate the Operating Lines**

#### **2.1. Rectifying Section Operating Line**

$$
y = \frac{R}{R+1}x + \frac{x_D}{R+1}
$$
Plugging in the given values:
- $ R = 0.8269 $
- $ x_D = 0.95 $

$$
y = \frac{0.8269}{1.8269}x + \frac{0.95}{1.8269}
$$
$$
y = 0.4527x + 0.5200
$$

#### **2.2. Stripping Section Operating Line**

The equation is:
$$
y = \frac{S}{S-1}x - \frac{x_B}{S-1}
$$

But more commonly, for total condenser and partial reboiler:

$$
y = \frac{R}{R+1}x + \frac{x_D}{R+1}
$$
for above feed

Below feed, the stripping line passes through $(x_B, x_B)$ and the intersection with the q-line.

---

### **Step 3: Locate the Feed Stage (Plate)**

The intersection of the **q-line** and the **operating lines** determines the feed stage.

#### **3.1. Find Intersection Point (x_q, y_q)**

- **q-line** passes through $(x_F, x_F)$ with slope $ \frac{q}{q-1} $
- **Rectifying line equation:** $ y = 0.4527x + 0.5200 $
- **q-line equation:** $ y - x_F = m(x - x_F) $, with $ m = 3.82 $

So,
$$
y = m(x - x_F) + x_F
$$
Set equal to rectifying line:
$$
0.4527x + 0.5200 = m(x - x_F) + x_F
$$
$$
0.4527x + 0.5200 = 3.82(x - 0.355) + 0.355
$$
$$
0.4527x + 0.5200 = 3.82x - 1.356 + 0.355
$$
$$
0.4527x + 0.5200 = 3.82x - 1.001
$$
$$
0.5200 + 1.001 = 3.82x - 0.4527x
$$
$$
1.521 = 3.3673x
$$
$$
x = \frac{1.521}{3.3673} \approx 0.4517
$$

Now, plug $ x $ back into rectifying line to get $ y $:
$$
y = 0.4527 \cdot 0.4517 + 0.5200
$$
$$
= 0.2045 + 0.5200 = 0.7245
$$
So, intersection at $ (x_q, y_q) = (0.452, 0.7245) $

---

### **Step 4: Step Off Stages (Graphical, or Estimate Number of Stages to Feed)**

Since you asked for the **feed plate** (stage where the feed is introduced), here's the standard procedure:

1. **Start at the distillate composition** ($ x_D = 0.95 $), at point (0.95, 0.95) on the 45-degree line.
2. **Step horizontally** to the equilibrium curve (use equilibrium data for MeOH-water).
3. **Step vertically** to the rectifying operating line.
4. **Repeat** steps 2 and 3, each pair is one theoretical stage, **until you reach the intersection point ($ x_q, y_q $)**. This is the feed plate.
5. Below this, you step between equilibrium curve and stripping line down to $ x_B $.

Since you do not have the equilibrium data here, I'll outline the process and estimate the feed stage.

#### **Estimate via Lever Rule (Stage Location)**

The feed plate is the stage where the operating line switches from rectifying to stripping (i.e., from above to below the q-line intersection).

- **Number of stages above feed:** Number of steps from $ x_D $ to $ x_q $.
- **Number of stages below feed:** From $ x_q $ to $ x_B $.

To get an exact stage number, you need the equilibrium curve. But, based on the compositions:

- Feed is at $ x_F = 0.355 $
- Intersection is at $ x = 0.4517 $ (q-line and rectifying line)
- Distillate at $ x_D = 0.95 $

So, the intersection is above the feed composition ($ x_q > x_F $), which is expected for subcooled feed.

---

### **Final Answer: Feed Plate Location**

**The feed plate is the stage where the operating lines (rectifying and stripping) meet, i.e., at the intersection of the q-line and the rectifying line, which is found at $ x = 0.452 $.**

**Step-by-step:**
1. Draw the equilibrium curve for MeOH-water.
2. Plot the operating lines:
   - Rectifying: $ y = 0.4527x + 0.5200 $
   - Stripping: determined by mass balance an

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