Discussion
Air enters a dryer at a humidity ratio of \( 0.010 \) and leaves at \( 0.022 \) kg per kg dry air. To evaporate \( 60 \) kg of water the dry air required is
Answer: D
Correct answer: 5000 kg
Each kilogram of dry air picks up the difference between the outlet and inlet humidity ratios:
\[ \Delta W = 0.022 - 0.010 = 0.012\ \mathrm{kg\ water\ per\ kg\ dry\ air} \]
\[ m_{air} = \frac{60}{0.012} = 5000\ \mathrm{kg\ dry\ air} \]
The distractor \( 2727 \) divides by the outlet humidity alone, ignoring the moisture the air already carried in. The calculation shows how large the air flow must be: five tonnes of air to remove sixty kilograms of water, which is why fan power is a significant part of a dryer's energy budget.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development