This is the statement of the problema I've invented (I studied McCabe-Thiele in spanish, so maybe there's somethings that are not well written, sorry for that):

*To the upper distillation column, it is fed onto the partial condenser 100 kmol/h (courrent B ) of a mixture that has 70% (in moles) of the compound we are interested in. To the propper column, it is also fed 100 kmol/h (courrent A) of the same mixture, with 40% (in moles) of the same compound. We obtain a Residue with 5% in moles of that compound, and a Distilate of 80% in moles. *

*Consider that V*_{R}/R = 1.5 · V_{R}/R)_{minimum }

*Consider too that q*_{B} = q_{A} = q_{L}_{D} = q_{D} = 1 and q_{R} = q_{VR} = 0

*Consider as compound 1 the compound of interest*

*Molar flows of distillate and residue*
*Feed equations*
*Net flows of the column*
*Molar flows of both phases in the column sectors*
*Operative equations*
**Minimum number of plates**
**Number of plates when used the specified value of V**_{R}/R
*Most efficient position for the feed A*

*The diagram we will use is:*

Okey so I proceede to explain the answers I've obtained:

**The molar flow of D and R**

By making a mass balance we obtain two equations with two unknown variables (D and R)

A + B = D + R --> 100 + 100 = D + R

A·x_{A} + B·x_{B} = D·y_{D} + R·x_{R} --> 100·0.4 + 100·0.7 = D·0.8 + R·0.05

From here we obtain D and R

D = 133.33 kmol/h

R = 66.67 kmol/h

2. **Feed equations**

In general, the equation used to determine the equation of the feed is the following:

y = q/(q-1) · x - x_{A}/(q-1)

As q = 1 for both A and B, the feed equations are:

A: y = 0.4

B: y = 0.7

3. **Net flows of the column**

Because we have a feed in the partial condenser, there is no net flow in the rectifying sector (I think)

So:

Stripping sector:

L_{m} - V_{m} = Δ_{R} = R = 66.67 kmol/h

z_{ΔR} = x_{R} = 0.05

Medium sector:

Mass balance: L_{n} + A = V_{n} + R

V_{n} - L_{n} = Δ_{T} = A - R = 33.33 kmol/h

z_{ΔT} = (A·x_{A} - R·x_{R})/Δ_{T} = 1.10

4. **Molar flows of both phases in the column**

We have to determine firstly V_{R}/R)_{min}

We know that: L_{m}/V_{m})_{max} = (y_{ΔR }- y_{A,eq})/(x_{ΔR} - x_{A,eq})

y_{ΔR} = x_{ΔR} = 0.05

y_{A,eq} is the point where the feed line cuts the equilibrium curve in the y x diagram, in this case: y_{A,eq }= 0.755

x_{A,eq} is the point where the feed line cuts the equilibrium curve in the y x diagram, in this case: x_{A,eq }= 0.4

We obtain: L_{m}/V_{m})_{max} = 2.014

By a mass balance (i'm not going to make it because it would make the resolution too long) we know that:

L_{m}/V_{m})_{max} = (1+V_{R}/R]_{min})/V_{R}/R)_{min} so we obtain:

V_{R}/R)_{min} = 0.986

As we know the relation between V_{R}/R) = V_{R}/R)_{min} · 1.15 we obtain:

V_{R}/R) = 1.47887

Now we can answer the question;

Stripping section:

L_{m }= V_{m} + R

We know that V_{R} = V_{m} and that V_{R}/R = V_{m}/R = 1.47887

so we can obtain V_{m} and consequently L_{m}:

L_{m} = 165.26 kmol/h

V_{m} = 98.59 kmol/h

We can calculate now the medium section:

L_{n} + V_{m} + A = L_{m} + V_{n}

Balance to the vapour: V_{m }+ (1-q_{A})·A = V_{n} --> V_{n} = 98.59 kmol/h

Balance to the liquid: L_{n} + A·q_{A} = L_{m} --> L_{n} = 65.26

5**. Operation lines:**

Stripping operation line: y_{m }= L_{m}/V_{m}· x_{m} - R·x_{R}/V_{m}

y_{m} = 1.6762·x_{m} - 0.0338

Medium section operation line: y_{n} = L_{n}/V_{n}·x_{n} + A·x_{A}/V_{n} - R·x_{R}/V_{n}

y_{n} = 0.6619·x_{n }+ 0.3719

6. **Minimum number of plates**

3.6 - partial boiler - partial condenser --> 1.6 (see final image)

7. **Number of plates**

4.8 - partial boiler - partial condenser --> 2.8 (see final image)

8. **Most efficient position of A**

third plate counting from the bottom of the column

What do you think about it???