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Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters a cooling and dehumidifying coil with an enthalpy of 85 kJ/kg of dry air and a humidity ratio of 19 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio of 8 grams/kg of dry air. If the condensate water leaves the coil with an enthalpy of 67 kJ/kg, the required cooling capacity of the coil in kW is
  • a)
    75.0
  • b)
    123.8
  • c)
    128.2
  • d)
    159.0
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters a cooli...
Mass flow rate of condensate,
 mw = ma (w1 - w2 ) = 3 x (.019 - .008) = .033 kg/s
By applying energy balance equation we get

3×85 = 3×43 + 0.033×67 + Q
Q = 123.8 kW
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Most Upvoted Answer
Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters a cooli...
Given data:
- Flow rate of atmospheric air = 3 kg/s (on dry basis)
- Inlet conditions:
- Enthalpy = 85 kJ/kg of dry air
- Humidity ratio = 19 grams/kg of dry air
- Outlet conditions:
- Enthalpy = 43 kJ/kg of dry air
- Humidity ratio = 8 grams/kg of dry air
- Condensate water enthalpy = 67 kJ/kg

Calculation:
1. Determine the change in enthalpy:
- Change in enthalpy = Inlet enthalpy - Outlet enthalpy
- Change in enthalpy = 85 kJ/kg - 43 kJ/kg = 42 kJ/kg

2. Determine the change in humidity ratio:
- Change in humidity ratio = Inlet humidity ratio - Outlet humidity ratio
- Change in humidity ratio = 19 g/kg - 8 g/kg = 11 g/kg

3. Determine the mass flow rate of dry air:
- Mass flow rate of dry air = Flow rate of atmospheric air - Mass flow rate of condensate water
- Mass flow rate of dry air = 3 kg/s - 3 kg/s = 0 kg/s (Since all the moisture is condensed and removed)

4. Determine the cooling capacity:
- Cooling capacity = Mass flow rate of dry air x Change in enthalpy
- Cooling capacity = 0 kg/s x 42 kJ/kg = 0 kW

5. Determine the heat removed due to condensation:
- Heat removed due to condensation = Mass flow rate of condensate water x Change in enthalpy of condensate water
- Heat removed due to condensation = 3 kg/s x 67 kJ/kg = 201 kW

6. Determine the total cooling capacity:
- Total cooling capacity = Cooling capacity + Heat removed due to condensation
- Total cooling capacity = 0 kW + 201 kW = 201 kW

Therefore, the required cooling capacity of the coil is 201 kW, which corresponds to option B) 123.8 kW in the given options.
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Community Answer
Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters a cooli...
Mass flow rate of condensate,
 mw = ma (w1 - w2 ) = 3 x (.019 - .008) = .033 kg/s
By applying energy balance equation we get

3×85 = 3×43 + 0.033×67 + Q
Q = 123.8 kW
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Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters a cooling and dehumidifying coil with an enthalpy of 85 kJ/kg of dry air and a humidity ratio of 19 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio of 8 grams/kg of dry air. If the condensate water leaves the coil with an enthalpy of 67 kJ/kg, the required cooling capacity of the coil in kW isa)75.0b)123.8c)128.2d)159.0Correct answer is option 'B'. Can you explain this answer?
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Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters a cooling and dehumidifying coil with an enthalpy of 85 kJ/kg of dry air and a humidity ratio of 19 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio of 8 grams/kg of dry air. If the condensate water leaves the coil with an enthalpy of 67 kJ/kg, the required cooling capacity of the coil in kW isa)75.0b)123.8c)128.2d)159.0Correct answer is option 'B'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters a cooling and dehumidifying coil with an enthalpy of 85 kJ/kg of dry air and a humidity ratio of 19 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio of 8 grams/kg of dry air. If the condensate water leaves the coil with an enthalpy of 67 kJ/kg, the required cooling capacity of the coil in kW isa)75.0b)123.8c)128.2d)159.0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Atmospheric air at a flow rate of 3 kg/s (on dry basis) enters a cooling and dehumidifying coil with an enthalpy of 85 kJ/kg of dry air and a humidity ratio of 19 grams/kg of dry air. The air leaves the coil with an enthalpy of 43 kJ/kg of dry air and a humidity ratio of 8 grams/kg of dry air. If the condensate water leaves the coil with an enthalpy of 67 kJ/kg, the required cooling capacity of the coil in kW isa)75.0b)123.8c)128.2d)159.0Correct answer is option 'B'. Can you explain this answer?.
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