Figure 9.46 represents the flow sheet for the recovery of acetone from air. All compositions are on a weight basis. Make a material balance and determine the quantities of the foll... Figure 9.46 represents the flow sheet for the recovery of acetone from air. All compositions are on a weight basis. Make a material balance and determine the quantities of the following streams: (a) Water added in the absorber (b) Acetone-free air leaving the absorber (c) Aqueous solution of acetone leaving the absorber (d) Distillate product (e) Bottom product. Read more

Steps to Solve

1. Define the streams and the known quantities

Let's start by defining the key streams and their compositions:

• Incoming air: 1400 kg/h with 95% air, 3% acetone, and 2% water.
• Water added: ( W ) kg/h.
• Aqueous solution leaving absorber: Acetone (19%), Water (81%).
• Distillate: 99% acetone, 1% water.
• Bottom product: 4% acetone, 96% water.

2. Calculate the composition of the incoming air

From the incoming air stream, we can find the quantities of acetone and water:

• Acetone in incoming air: $$ \text{Acetone}_{\text{in}} = 1400 , \text{kg/h} \times 0.03 = 42 , \text{kg/h} $$
• Water in incoming air: $$ \text{Water}_{\text{in}} = 1400 , \text{kg/h} \times 0.02 = 28 , \text{kg/h} $$

3. Material balance around the absorber

For the absorber, we apply the mass balance for acetone and water: Let ( x_a ) be the mass flow rate of the aqueous solution leaving the absorber.

• Total mass balance: $$ 1400 + W = x_a + \text{Acetone-free air}_{out} $$
• Material balance for acetone: $$ 42 = 0.19 x_a $$ Solving for ( x_a ): $$ x_a = \frac{42}{0.19} \approx 221.05 , \text{kg/h} $$

4. Determine the water added in the absorber

Insert the above value into the total mass balance equation: $$ 1400 + W = 221.05 + \text{Acetone-free air}_{out} $$

5. Calculate the acetone-free air leaving the absorber

First, calculate the total mass of the aqueous solution: $$ \text{Aqueous solution} = 0.81 x_a = 0.81 \cdot 221.05 \approx 179.45 , \text{kg/h} $$ Next, we find the acetone-free air leaving the absorber: Subtract the incoming mass and the aqueous solution from the total: $$ \text{Acetone-free air}{out} = 1400 + W - 221.05 $$ Thus, $$ \text{Acetone-free air}{out} \approx 1400 - 221.05 + 179.45 $$

6. Determine distillate and bottom products

Use the distillation process:

• For distillate: If ( D ) is the flow rate of the distillate: $$ D = x_d , \text{kg/h} $$ Using the distillate composition, we can find the relationship with the bottom products.

7. Solving for all unknowns

This approach will yield all necessary unknown values step by step for the defined streams.