GATE AG 2011 • Q54 MCQ • 2 Marks

Section 6: Agricultural Process Engineering

Evaporation and Drying

Subtopic: Psychrometry – properties of air-water vapour mixture

Question Statement:

A vegetable with 70% moisture (wet basis) is to be dried in a countercurrent dryer to give a product with 5% moisture (wet basis). The feed rate to the dryer is 0.15 kg s⁻¹. The drying medium consists of air at 100 °C containing water vapour with a partial pressure of 1.0 kPa. The air leaves the dryer at 40 °C and is 70% saturated. The system works under a total pressure of 101.3 kPa. The vapour pressure of water that saturates air at 40 °C is 7.4 kPa. Molecular weight of water and air may be taken as 18 and 29 respectively. The humidity (kg of water vapour / kg of dry air) of the outlet air is

Options:

  • (A) 0.0027
  • (B) 0.0172
  • (C) 0.0335
  • (D) 0.0483

Official Answer Key:

C

Step-by-Step Solution & Explanation

Outlet air is 70% saturated at 40 °C, so its water-vapour partial pressure pv = 0.70 × 7.4 kPa = 5.18 kPa. Humidity (mass ratio) = (Mw/Ma) × pv/(P_total − pv) = (18/29) × 5.18/(101.3 − 5.18) = 0.6207 × 5.18/96.12 = 0.6207 × 0.05389 = 0.0335 kg water/kg dry air, matching option (C).

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