Gce Model Questions Vol1 2018 Agricultural Science Question 50

Practice objective / multiple choice question 50 from the 2018 Gce Model Questions Vol1 Agricultural Science examination.

Gce Model Questions Vol1 2018 Agricultural Science Objective / Multiple Choice Hard Difficulty
Question 50 GCE_MODEL_QUESTIONS_VOL1 • 2018 • AGRICULTURAL_SCIENCE • objective

A batch of cassava chips initially has a moisture content of 45 % (wet basis). The desired final moisture content is 12 % (wet basis). The drying follows the exponential model \(M(t) = M_{eq} + (M_{0} - M_{eq})e^{-kt}\), where \(M(t)\) is the moisture content at time \(t\) (hours), \(M_{eq}\) is the equilibrium moisture (assumed 10 % wb), \(M_{0}\) is the initial moisture, and \(k\) is the drying constant. After 5 hours of drying, the measured moisture content is 20 % wb.\n\n(a) Determine the drying constant \(k\).\n(b) Calculate the time required to reach the target moisture content of 12 % wb.\n(c) If the initial wet mass of the chips is 200 kg, compute the total amount of water removed during drying.

Answer Options

A. k ≈ 0.300 h⁻¹; time ≈ 13.2 h; water removed = 82 kg
B. k ≈ 0.210 h⁻¹; time ≈ 9.8 h; water removed = 68 kg
Correct Answer C. k ≈ 0.251 h⁻¹; time ≈ 11.4 h; water removed = 75 kg
D. k ≈ 0.180 h⁻¹; time ≈ 8.5 h; water removed = 60 kg
Correct Answer
C
Correct Option:
k ≈ 0.251 h⁻¹; time ≈ 11.4 h; water removed = 75 kg

About This Question

This is Gce Model Questions Vol1 2018 Agricultural Science Question 50. It is one of the objective questions from the 2018 Gce Model Questions Vol1 Agricultural Science examination.

Difficulty level: Hard .

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