Osmotically dehydrated cherries, finish-dried from 18% (dry basis) to 11.5% (wet basis) moisture content, final mass flow rate 5000 kg/h. Dryer efficiency 70%, latent heat 2345 kJ/kg, c_p(air)=1.005 kJ/kg·K, drying temperature 50°C, specific volume of ambient air at 25°C = 0.866 m³/kg. The air flow requirement, in m³/min, is: [note: marked "Marks to All" in the official key owing to an ambiguity/error in the original question]
Options:
(A) 477
(B) 587
(C) 625
(D) 702
Official Answer Key:
MARKS TO ALL
Step-by-Step Solution & Explanation
• **Official Status:** Marked as **MARKS TO ALL (MTA)** in the official GATE 2012 AG answer key due to unstated ambient relative humidity / wet-bulb exit assumptions for drying air.
• **Engineering Mass & Energy Balance:**
- Product feed: Dried from $M_{initial} = 0.18\text{ d.b.}$ to $M_{final} = 0.115\text{ w.b.}$
- Moisture content on dry basis for final product: $M_{final,db} = \frac{0.115}{1 - 0.115} = \frac{0.115}{0.885} = 0.12994\text{ kg water/kg dry solid}$.
- Bone dry solids in final 5000 kg/h product: $W_s = 5000 \times (1 - 0.115) = 4425\text{ kg/h}$.
- Water evaporated $m_w = 4425 \times (0.18 - 0.12994) = 221.5\text{ kg/h}$.
- Heat required $= \frac{m_w \cdot \lambda}{\eta} = \frac{221.5 \times 2345}{0.70} = 742,025\text{ kJ/h}$.
- Air heating from $25^\circ\text{C}$ to $50^\circ\text{C}$: $\Delta h = c_p \Delta T = 1.005 \times 25 = 25.125\text{ kJ/kg}$.
- Mass of air required $= \frac{742025}{25.125} = 29533\text{ kg/h}$.
- Volumetric air flow $= \frac{29533 \times 0.866}{60} = 426.3\text{ m}^3/\text{min}$ (differing psychrometric exit paths led to conflicting options, hence full credit was awarded).
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