A biopharmaceutical processing plant uses three ultrafiltration units—Unit A, Unit B, and … — GMAT

A biopharmaceutical processing plant uses three ultrafiltration units—Unit A, Unit B, and Unit C—to extract an active therapeutic protein from a $3,000\text{-liter}$ batch of liquid culture media. The liquid media contains $5\%$ active protein solute by volume. The operational specifications for each unit are as follows:

- Unit A: Processes raw media at a rate of $150\text{ liters/hour}$ with a protein solute recovery efficiency of $80\%$ .
- Unit B: Processes raw media at a rate of $200\text{ liters/hour}$ with a protein solute recovery efficiency of $85\%$ .
- Unit C: Processes raw media at a rate of $400\text{ liters/hour}$ with a protein solute recovery efficiency of $70\%$ .

The batch is processed in two sequential stages:
- Stage 1: Units A and B operate simultaneously for exactly $6\text{ hours}$ .
- Stage 2: Unit A is shut down, and Units B and C operate simultaneously to process all remaining liquid media from the $3,000\text{-liter}$ batch.

Match each operational outcome on the left with its corresponding calculated value on the right.

Total volume of raw culture media processed by Unit B across both stages $1,500\text{ liters}$
Duration of Stage 2 required to process the remaining batch $1.5\text{ hours}$
Overall solute recovery percentage for the entire $3,000\text{-liter}$ batch $80.5\%$
Total volume of solute recovered by Unit A during the operation $36.0\text{ liters}$

Cevap

The correct pairings match the operational outcomes as follows: Total volume processed by Unit B matches

1,500\text{ liters}

; Duration of Stage 2 matches

1.5\text{ hours}

; Overall solute recovery percentage matches

80.5\%

; Total solute recovered by Unit A matches

36.0\text{ liters}

Each calculation requires tracking both raw liquid flow rates and solute concentration extraction efficiencies across two operational phases. Unit A operates only in Stage 1 (

6\text{ hours}

), processing

900\text{ liters}

of media containing

45\text{ liters}

of solute and recovering

80\%

, which equals

36.0\text{ liters}

. The combined intake of Units A and B in Stage 1 is

2,100\text{ liters}

, leaving

900\text{ liters}

for Stage 2. Units B and C process the remaining volume at a combined rate of

600\text{ L/hr}

, requiring

1.5\text{ hours}

. Over both stages, Unit B processes

1,200 + 300 = 1,500\text{ liters}

. Total solute recovered by all units equals

36.0 + 51.0 + 12.75 + 21.0 = 120.75\text{ liters}

out of

150\text{ liters}

total solute, which gives an overall recovery percentage of

80.5\%

Adım Adım Çözüm

Calculate Stage 1 processing volumes and solute recovery for Units A and B.

In Stage 1 (

6\text{ hours}

): Unit A processes

150 \times 6 = 900\text{ L}

and recovers

900 \times 0.05 \times 0.80 = 36.0\text{ L}

of solute. Unit B processes

200 \times 6 = 1,200\text{ L}

and recovers

1,200 \times 0.05 \times 0.85 = 51.0\text{ L}

of solute. Total raw media processed in Stage 1 =

2,100\text{ liters}

Establishes baseline volumes completed before Stage 2 begins.

Determine the remaining volume and the duration of Stage 2.

Remaining volume =

3,000 - 2,100 = 900\text{ liters}

. Combined processing rate of Units B and C =

200 + 400 = 600\text{ L/hr}

. Stage 2 duration =

900 / 600 = 1.5\text{ hours}

Quantifies the time required for Stage 2 based on joint processing rates.

Calculate Stage 2 processing volumes and solute recovery for Units B and C.

In Stage 2 (

1.5\text{ hours}

): Unit B processes

200 \times 1.5 = 300\text{ L}

and recovers

300 \times 0.05 \times 0.85 = 12.75\text{ L}

of solute. Unit C processes

400 \times 1.5 = 600\text{ L}

and recovers

600 \times 0.05 \times 0.70 = 21.0\text{ L}

of solute.

Determines Unit B's second-stage contribution and Unit C's total output.

Calculate total volume processed by Unit B and cumulative solute recovery efficiency.

Total raw media processed by Unit B =

1,200 + 300 = 1,500\text{ liters}

. Total solute recovered by all units =

36.0 + 51.0 + 12.75 + 21.0 = 120.75\text{ liters}

. Total solute initially present in the batch =

3,000 \times 0.05 = 150\text{ liters}

. Overall recovery efficiency =

(120.75 / 150) \times 100\% = 80.5\%

Synthesizes multi-stage outputs to obtain total system efficiency metrics.

Anahtar Kavram

Sequential multi-stage work rates, solute mass-balance calculations, and weighted percentage recovery efficiency.

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