Let's break down the problem **step by step** and address each part: --- # **Step 1: Problem Understanding and Given Data** - **Absorbent feed:** 100 kg/hr (basis) - **Absorbent entering extractor:** 25 mass% absorbent, rest is hexane/oil mixture - **Hexane needed for extraction:** 5 kg fresh hexane per 5 kg absorbent (1:1 ratio, but based on oil content in absorbent) - **Recycle:** 99% of hexane is recovered in the evaporator and recycled - **Product oil stream:** Hexane is lost with the oil and product, so pure hexane is added to the recycle stream - **Oil molar mass:** 150 g/mol --- # **Step 2: Flowchart (Part a)** ## **Flowchart Structure** ```plaintext +----------------------+ | Absorbent Feed | | (100 kg/hr) | +----------+-----------+ | v +----------------------+ | Extractor | +---+--------------+---+ | | Oil-rich Solid stream stream (to evaporator) (to product) | v +-------------------+ | Evaporator | +---+-----------+---+ | | Recycle Product stream hexane (oil + hexane) (to extractor) ``` **Label:** - Feed to extractor: Absorbent + hexane (fresh + recycle) - Output from extractor: Oil-rich stream, solid stream (to evaporator) - Output from evaporator: Recycle hexane (to extractor), product oil stream (oil + some hexane) --- # **Step 3: Degree of Freedom Analysis (Part a)** ### **Streams and Unknowns** - Feed: 100 kg/hr (basis, known) - Fresh hexane: unknown - Recycle hexane: unknown - Product oil stream: unknown - Product hexane stream: unknown - Recycle flow: unknown **Known relationships:** - Absorbent entering extractor: 25% absorbent by mass - Oil lost with absorbent: 1% of oil entering is lost (99% recovered) - 1:1 mass ratio for feed absorbent to fresh hexane **Number of unknowns is equal to the number of independent equations (material balances + composition constraints), so system is solvable.** --- # **Step 4: Material Balances and Calculations** ## **b. Find the mass% of oil product stream and mass flow rates (product and recycle)** ### **1. Absorbent Stream to Extractor** Let: - \( F_A = \) mass of absorbent feed \( = 100 \) kg/hr - \( F_H = \) mass of hexane in feed to extractor (fresh + recycle) - Mass fraction absorbent in extractor feed: 25% - \( F_A + F_H = \) total feed to extractor - \( \frac{F_A}{F_A + F_H} = 0.25 \Rightarrow F_H = 3F_A = 300 \) kg/hr ### **2. Fresh Hexane Calculation** - Absorbent entering extractor: 100 kg/hr - Fresh hexane needed: 5 kg per 5 kg absorbent \( \Rightarrow \) 100 kg hexane for 100 kg absorbent. - So, **Fresh hexane feed** = 100 kg/hr ### **3. Recycle Hexane Calculation** - Total hexane in extractor feed = 300 kg/hr - Fresh hexane = 100 kg/hr - **Recycle hexane** = \( 300 - 100 = 200 \) kg/hr ### **4. Oil Recovery** Let \( O_{in} \) = oil entering extractor (amount not given, so we’ll use symbols and relate to product stream). - 99% of oil entering is recovered in the product oil stream. - 1% leaves with absorbent (assume negligible for hexane/oil balance). ### **5. Product Oil Stream** - The product stream from evaporator contains oil and hexane (hexane lost with oil). - Hexane lost with oil = hexane in product stream. #### **Oil in Product Stream** Let \( O_p \) = oil in product stream \( = 0.99 O_{in} \) #### **Hexane in Product Stream** All hexane not recovered is lost here, so: - Hexane lost per cycle = 1% of total hexane in system (since 99% is recycled) - Hexane in product stream \( = 1\% \) of total hexane entering evaporator - But in this setup, the evaporator recovers 99% of hexane, so hexane in product stream is 1% of hexane entering evaporator. **But according to the problem, 99% of the oil is recovered, not hexane. Hexane is recycled except for what leaves with the oil, so the hexane in the product stream is whatever is not recycled.** ### **6. Mass Flow Rates** #### **Product Oil Stream Mass Flow Rate** Let’s assume all the oil in the absorbent feed is extracted, so the oil entering the extractor is all in the feed (not specified, so we may use a symbol or assign a value for demonstration). Let \( O_{in} \) = mass of oil in absorbent feed. So, mass of oil in product stream \( = 0.99 O_{in} \). #### **Hexane in Product Stream** Hexane in product stream = all hexane lost (since 99% is recycled): - Hexane lost per cycle = 1% of hexane entering evaporator = 1% × 300 kg = 3 kg/hr Thus, product oil stream contains: - Oil: \( 0.99 O_{in} \) - Hexane: 3 kg/hr **If we need to use a percentage for oil in the product stream, we need the oil mass. If not given, leave as symbol.** --- # **Step 5: Molar Flowrate of Product Stream (Part c)** - **Oil molar mass:** 150 g/mol = 0.15 kg/mol - **Oil in product:** \( 0.99 O_{in} \) kg/hr - **Hexane in product:** 3 kg/hr - **Hexane molar mass:** ~86 g/mol = 0.086 kg/mol Moles of oil per hour = \( \frac{0.99 O_{in}}{0.15} \) mol/hr Moles of hexane per hour = \( \frac{3}{0.086} \) mol/hr Total molar flowrate = sum of above. --- # **Summary Table** | Stream | Mass Flowrate (kg/hr) | Moles/hr | |---------------------|----------------------|-----------------------------------| | Absorbent Feed | 100 | -- | | Fresh Hexane | 100 | -- | | Recycle Hexane | 200 | -- | | Product Oil Stream | \( 0.99 O_{in} \) (oil), 3 (hexane) | \( \frac{0.99 O_{in}}{0.15} \) (oil), \( \frac{3}{0.086} \) (hexane) | --- # **Final Answers** ## **a. Flowchart** *(See flowchart above.)* ## **b. Mass % and Mass Flow Rates** - **Mass % of oil in product stream:** \( \frac{0.99 O_{in}}{0.99 O_{in} + 3} \times 100 \) % - **Mass flowrate of product stream:** \( 0.99 O_{in} + 3 \) kg/hr - **Mass flowrate of recycle stream:** 200 kg/hr ## **c. Molar Flowrate of Product Stream** - **Total molar flowrate:** \[ \text{Moles/hr} = \frac{0.99 O_{in}}{0.15} + \frac{3}{0.086} \] --- ## **Note:** If the mass of oil in the absorbent feed (\( O_{in} \)) is specified, you can substitute it directly to obtain numerical values. --- ### **Flowchart (Text Representation)** ```plaintext [Absorbent Feed 100 kg/hr] | +-------------------------+ | Extractor | +----------+--------------+ | [Solid Stream to Evaporator] | +-------------------------+ | Evaporator | +----+-------------+------+ | | [Recycle 200kg/hr] [Product Oil Stream: 0.99 O_in oil + 3 hexane kg/hr] ``` *(You can convert this into a diagram as needed.)* --- If you have a value for \( O_{in} \), I can provide concrete numbers!

# Step 1: Problem Understanding and Given Data In this problem, we need to analyze an extraction process where an absorbent material removes crude oil from a mixture. The extracted oil is then separated from the absorbent using hexane, which is recycled in the process. Here are the key components and their significance: - **Absorbent Feed (100 kg/hr)**: This is the mass flow rate of the absorbent material that is used to capture the crude oil. It acts as the medium for oil extraction. - **Absorbent Composition**: The absorbent feed consists of 25% absorbent material by mass. This means that in the total feed, 25% is the absorbent and the remaining 75% is a mixture of hexane and the oil. - **Hexane for Extraction**: A mass ratio of 1:1 is maintained between the amount of absorbent and fresh hexane required for effective extraction. This means for every kg of absorbent, 1 kg of fresh hexane is needed. - **Hexane Recycling**: After extraction, hexane is evaporated and recycled back into the extractor. However, a small percentage of the hexane is lost along with the oil during the process, necessitating the addition of pure hexane to maintain effective extraction. - **Oil Recovery**: The system is designed to recover 95% of the oil contained in the absorbent, which is critical for evaluating the process efficiency. - **Molar Mass of Oil**: The molar mass is given as 150 g/mol, which will be used for converting mass flow rates to molar flow rates later in the calculations. These components and their relationships form the basis for setting up material balances and flow rates throughout the extraction process. Understanding these relationships is crucial for determining the output product streams and performing the necessary calculations.