Let's break the problem into steps: **Given Data:** - **Feed rate:** 83000 kg/h - **Feed composition:** 40% benzene, 60% toluene (by weight) - **Overhead (distillate) purity:** 97% benzene (by weight) - **Bottoms purity:** 98% toluene (by weight) - **Reflux ratio:** 3.5 mol reflux / 1 mol distillate - **Latent heat:** - Benzene = 7,360 cal/gmol - Toluene = 7,960 cal/gmol - **Relative volatility (α):** 2.5 - **Feed at 95°C and 1 atm (boiling point) for case (i)** - **Molecular weights:** - Benzene: 78 g/mol - Toluene: 92 g/mol --- ## **(a) Calculate the moles of overhead (distillate) and bottom product per hour.** ### **Step 1: Find mass of each component in the feed** \[ \text{Total feed} = 83,000~\text{kg/h} \] \[ \text{Benzene in feed} = 0.40 \times 83,000 = 33,200~\text{kg/h} \] \[ \text{Toluene in feed} = 0.60 \times 83,000 = 49,800~\text{kg/h} \] --- ### **Step 2: Let D = mass of distillate (overhead), B = mass of bottoms** Let \(D\) kg/h of distillate (97% benzene), \(B\) kg/h of bottoms (98% toluene). **Material balances:** **Overall mass:** \[ D + B = 83,000 \] **Benzene:** \[ 0.97D + 0.02B = 33,200 \] **Toluene:** \[ 0.03D + 0.98B = 49,800 \] Either benzene or toluene balance is sufficient, so use benzene. --- ### **Step 3: Solve for D and B** From overall: \[ B = 83,000 - D \] Substitute into benzene balance: \[ 0.97D + 0.02(83,000 - D) = 33,200 \] \[ 0.97D + 1,660 - 0.02D = 33,200 \] \[ (0.97 - 0.02)D = 33,200 - 1,660 \] \[ 0.95D = 31,540 \] \[ D = \frac{31,540}{0.95} = \boxed{33,200~\text{kg/h}} \] \[ B = 83,000 - 33,200 = \boxed{49,800~\text{kg/h}} \] --- ### **Step 4: Convert to moles per hour** #### **Distillate:** - 97% benzene, 3% toluene \[ \text{Benzene in D: } 0.97 \times 33,200 = 32,204~\text{kg} \] \[ \text{Toluene in D: } 0.03 \times 33,200 = 996~\text{kg} \] \[ \text{Moles benzene: } \frac{32,204}{78} = 412.87~\text{kmol} \] \[ \text{Moles toluene: } \frac{996}{92} = 10.83~\text{kmol} \] \[ \text{Total moles in D: } 412.87 + 10.83 = \boxed{423.7~\text{kmol/h}} \] #### **Bottoms:** - 98% toluene, 2% benzene \[ \text{Toluene in B: } 0.98 \times 49,800 = 48,804~\text{kg} \] \[ \text{Benzene in B: } 0.02 \times 49,800 = 996~\text{kg} \] \[ \text{Moles toluene: } \frac{48,804}{92} = 530.5~\text{kmol} \] \[ \text{Moles benzene: } \frac{996}{78} = 12.77~\text{kmol} \] \[ \text{Total moles in B: } 530.5 + 12.77 = \boxed{543.3~\text{kmol/h}} \] --- ### **Final answers (a):** \[ \boxed{ \begin{align*} \text{Moles of overhead product (distillate), } D &= 423.7~\text{kmol/h} \\ \text{Moles of bottom product, } B &= 543.3~\text{kmol/h} \end{align*} } \] --- ## **(b) Determine the number of ideal plates and the feed plate position** We'll use the **McCabe-Thiele method**. First, need to convert weight fractions to mole fractions for the key products and feed. ### **Step 1: Convert weight % to mole fractions** #### **Feed:** - \( w_B = 0.40, w_T = 0.60 \) - Assume basis: 100 kg feed \[ \text{Benzene: } \frac{40~\text{kg}}{78~\text{kg/kmol}} = 0.513~\text{kmol} \] \[ \text{Toluene: } \frac{60~\text{kg}}{92~\text{kg/kmol}} = 0.652~\text{kmol} \] \[ \text{Total moles: } 0.513 + 0.652 = 1.165~\text{kmol} \] \[ x_{B,F} = \frac{0.513}{1.165} = 0.440 \] \[ x_{T,F} = 1 - 0.440 = 0.560 \] #### **Distillate (overhead):** - \( w_B = 0.97, w_T = 0.03 \) \[ \text{Benzene: } \frac{97~\text{kg}}{78} = 1.244~\text{kmol} \] \[ \text{Toluene: } \frac{3~\text{kg}}{92} = 0.0326~\text{kmol} \] \[ \text{Total: } 1.2766~\text{kmol} \] \[ x_{B,D} = \frac{1.244}{1.2766} = 0.974 \] #### **Bottoms:** - \( w_T = 0.98, w_B = 0.02 \) \[ \text{Toluene: } \frac{98~\text{kg}}{92} = 1.065~\text{kmol} \] \[ \text{Benzene: } \frac{2~\text{kg}}{78} = 0.0256~\text{kmol} \] \[ \text{Total: } 1.0906~\text{kmol} \] \[ x_{B,W} = \frac{0.0256}{1.0906} = 0.0235 \] --- ### **Step 2: Operating Lines** #### **Reflux Ratio (R):** Given on molar basis: \( R = 3.5 \) \[ y_{D} = x_{B,D} = 0.974 \] The distillate operating line: \[ y = \frac{R}{R+1} x + \frac{x_D}{R+1} \] \[ y = \frac{3.5}{4.5} x + \frac{0.974}{4.5} \] \[ y = 0.778x + 0.216 \] --- ### **Step 3: q-line** - \(q\) = liquid fraction of feed entering column #### **Case (i): Feed is saturated liquid** \[ q = 1 \] \[ y = \frac{q}{q-1} x - \frac{x_F}{q-1} \] But for \(q = 1\), the q-line is **vertical** at \( x = x_F = 0.440 \). --- ### **Step 4: Stripping Section Operating Line** The bottoms operating line: \[ y = \frac{S}{S+1} x + \frac{x_W}{S+1} \] But typically use mass balance: \[ y = \frac{(q-1) x}{q} + \frac{x_F}{q} \] But for calculations, it's easier to use: At the intersection of the q-line and the rectifying/stripping operating lines. --- ### **Step 5: Number of Stages (McCabe-Thiele, graphical or algebraic)** Since we have relative volatility (\(\alpha = 2.5\)), we can use the **Fenske equation** (for minimum number of stages at total reflux): \[ N_{\text{min}} = \frac{\log \left( \frac{ \dfrac{x_D}{1-x_D} }{ \dfrac{x_W}{1-x_W} } \right) }{ \log \alpha } \] \[ \frac{x_D}{1-x_D} = \frac{0.974}{1-0.974} = \frac{0.974}{0.026} = 37.46 \] \[ \frac{x_W}{1-x_W} = \frac{0.0235}{1-0.0235} = \frac{0.0235}{0.9765} = 0.0241 \] \[ \frac{37.46}{0.0241} = 1554.8 \] \[ N_{\text{min}} = \frac{ \log 1554.8 }{ \log 2.5 } = \frac{3.192}{0.398} = 8.02 \] So **minimum number of stages** at total reflux: **8.0**. #### **Calculate the actual number of stages (using Underwood-Gilliland correlation):** First, calculate \(R_{min}\) via Underwood equation. ##### **Underwood equation:** For saturated liquid feed (\(q=1\)): \[ \sum \frac{q_i x_{i,F}}{\alpha_i - \theta} = 1 \] For binary with \(\alpha_1 = 2.5\) (benzene), \(\alpha_2 = 1\) (toluene): Let \(\theta\) be the Underwood root. Set up the equation: \[ \frac{0.44}{2.5 - \theta} + \frac{0.56}{1 - \theta} = 1 \] Let’s solve for \(\theta\) numerically. Let’s try \(\theta = 1\): \[ \frac{0.44}{2.5 - 1} = \frac{0.44}{1.5} = 0.293 \] \[ \frac{0.56}{1 - 1} = \infty \] So, can’t be 1 exactly. Let’s try \(\theta = 0.9\): \[ \frac{0.44}{2.5 - 0.9} = \frac{0.44}{1.6} = 0.275 \] \[ \frac{0.56}{1 - 0.9} = \frac{0.56}{0.1} = 5.6 \] \[ 0.275 + 5.6 = 5.875 \] Still much greater than 1. Try \(\theta = 0.99\): \[ \frac{0.44}{2.5-0.99} = \frac{0.44}{1.51} = 0.291 \] \[ \frac{0.56}{1-0.99} = \frac{0.56}{0.01} = 56 \] Still way more than 1. Try \(\theta = 0.5\): \[ \frac{0.44}{2.5-0.5} = \frac{0.44}{2.0} = 0.22 \] \[ \frac{0.56}{1-0.5} = \frac{0.56}{0.5} = 1.12 \] \[ 0.22 + 1.12 = 1.34 \] Try \(\theta = 0.7\): \[ \frac{0.44}{2.5-0.7} = \frac{0.44}{1.8} = 0.244 \] \[ \frac{0.56}{1-0.7} = \frac{0.56}{0.3} = 1.867 \] \[ 0.244 + 1.867 = 2.111