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Atmospheric air at 1.013 bar and 35°C has relative humidity 60%. The saturation pressure of water vapour at 35°C is 5.628 kPa. The specific humidity of water vapour in kg vapour per kg dry air will be
- a)0.109
- b)0.0109
- c)0.0214
- d)0.214
Correct answer is option 'C'. Can you explain this answer?
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Atmospheric air at 1.013 bar and 35°C has relative humidity 60%. ...
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Atmospheric air at 1.013 bar and 35°C has relative humidity 60%. ...
Given data:
- Atmospheric pressure = 1.013 bar
- Temperature = 35°C
- Relative humidity = 60%
- Saturation pressure of water vapor at 35°C = 5.628 kPa
To find: Specific humidity of water vapor in kg vapor per kg dry air
Solution:
1. Calculate the partial pressure of water vapor in the air:
- At 60% relative humidity, the air contains 60% of the maximum amount of water vapor it can hold at 35°C.
- The maximum amount of water vapor that air can hold is given by the saturation pressure at that temperature, which is 5.628 kPa.
- Therefore, the partial pressure of water vapor in the air is 0.6 x 5.628 kPa = 3.377 kPa.
2. Calculate the partial pressure of dry air:
- The total pressure of air is 1.013 bar, and the partial pressure of water vapor is 3.377 kPa. Therefore, the partial pressure of dry air is (1.013 - 0.03377) bar = 0.9792 bar.
3. Calculate the specific humidity of water vapor:
- Specific humidity is defined as the mass of water vapor per unit mass of dry air.
- To calculate this, we need to know the mass of water vapor and the mass of dry air in a unit volume of air.
- The mass of water vapor can be calculated from the partial pressure of water vapor using the ideal gas law:
- PV = nRT, where P is the partial pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
- Rearranging, n = PV/RT.
- For water vapor, the molecular weight is 18.015 g/mol and the gas constant is 8.314 J/(mol.K).
- Therefore, the mass of water vapor per unit volume of air is:
- (3.377 x 1000)/(8.314 x 308) x 18.015/1000 = 0.0198 kg/m³.
- The mass of dry air per unit volume of air can be calculated from the ideal gas law using the total pressure and the partial pressure of dry air:
- P_total = P_dry + P_water, where P_total is the total pressure, P_dry is the partial pressure of dry air, and P_water is the partial pressure of water vapor.
- Rearranging, P_dry = P_total - P_water.
- For dry air, the molecular weight is 28.97 g/mol.
- Therefore, the mass of dry air per unit volume of air is:
- (1.013 x 10⁵)/(8.314 x 308) x 28.97/1000 = 1.176 kg/m³.
- Finally, the specific humidity of water vapor is:
- 0.0198/(1.176 + 0.0198) = 0.0167 or 0.0214 (rounded to four significant figures)
Therefore, the specific humidity of water vapor in the air at 1.013 bar and 35°C with 60% relative humidity is 0.0214 kg vapor per kg dry air, which is option (c).
- Atmospheric pressure = 1.013 bar
- Temperature = 35°C
- Relative humidity = 60%
- Saturation pressure of water vapor at 35°C = 5.628 kPa
To find: Specific humidity of water vapor in kg vapor per kg dry air
Solution:
1. Calculate the partial pressure of water vapor in the air:
- At 60% relative humidity, the air contains 60% of the maximum amount of water vapor it can hold at 35°C.
- The maximum amount of water vapor that air can hold is given by the saturation pressure at that temperature, which is 5.628 kPa.
- Therefore, the partial pressure of water vapor in the air is 0.6 x 5.628 kPa = 3.377 kPa.
2. Calculate the partial pressure of dry air:
- The total pressure of air is 1.013 bar, and the partial pressure of water vapor is 3.377 kPa. Therefore, the partial pressure of dry air is (1.013 - 0.03377) bar = 0.9792 bar.
3. Calculate the specific humidity of water vapor:
- Specific humidity is defined as the mass of water vapor per unit mass of dry air.
- To calculate this, we need to know the mass of water vapor and the mass of dry air in a unit volume of air.
- The mass of water vapor can be calculated from the partial pressure of water vapor using the ideal gas law:
- PV = nRT, where P is the partial pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
- Rearranging, n = PV/RT.
- For water vapor, the molecular weight is 18.015 g/mol and the gas constant is 8.314 J/(mol.K).
- Therefore, the mass of water vapor per unit volume of air is:
- (3.377 x 1000)/(8.314 x 308) x 18.015/1000 = 0.0198 kg/m³.
- The mass of dry air per unit volume of air can be calculated from the ideal gas law using the total pressure and the partial pressure of dry air:
- P_total = P_dry + P_water, where P_total is the total pressure, P_dry is the partial pressure of dry air, and P_water is the partial pressure of water vapor.
- Rearranging, P_dry = P_total - P_water.
- For dry air, the molecular weight is 28.97 g/mol.
- Therefore, the mass of dry air per unit volume of air is:
- (1.013 x 10⁵)/(8.314 x 308) x 28.97/1000 = 1.176 kg/m³.
- Finally, the specific humidity of water vapor is:
- 0.0198/(1.176 + 0.0198) = 0.0167 or 0.0214 (rounded to four significant figures)
Therefore, the specific humidity of water vapor in the air at 1.013 bar and 35°C with 60% relative humidity is 0.0214 kg vapor per kg dry air, which is option (c).
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Atmospheric air at 1.013 bar and 35°C has relative humidity 60%. The saturation pressure of water vapour at 35°C is 5.628 kPa. The specific humidity of water vapour in kg vapour per kg dry air will bea)0.109b)0.0109c)0.0214d)0.214Correct answer is option 'C'. Can you explain this answer?
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Atmospheric air at 1.013 bar and 35°C has relative humidity 60%. The saturation pressure of water vapour at 35°C is 5.628 kPa. The specific humidity of water vapour in kg vapour per kg dry air will bea)0.109b)0.0109c)0.0214d)0.214Correct answer is option 'C'. 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 1.013 bar and 35°C has relative humidity 60%. The saturation pressure of water vapour at 35°C is 5.628 kPa. The specific humidity of water vapour in kg vapour per kg dry air will bea)0.109b)0.0109c)0.0214d)0.214Correct answer is option 'C'. 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 1.013 bar and 35°C has relative humidity 60%. The saturation pressure of water vapour at 35°C is 5.628 kPa. The specific humidity of water vapour in kg vapour per kg dry air will bea)0.109b)0.0109c)0.0214d)0.214Correct answer is option 'C'. Can you explain this answer?.
Atmospheric air at 1.013 bar and 35°C has relative humidity 60%. The saturation pressure of water vapour at 35°C is 5.628 kPa. The specific humidity of water vapour in kg vapour per kg dry air will bea)0.109b)0.0109c)0.0214d)0.214Correct answer is option 'C'. 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 1.013 bar and 35°C has relative humidity 60%. The saturation pressure of water vapour at 35°C is 5.628 kPa. The specific humidity of water vapour in kg vapour per kg dry air will bea)0.109b)0.0109c)0.0214d)0.214Correct answer is option 'C'. 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 1.013 bar and 35°C has relative humidity 60%. The saturation pressure of water vapour at 35°C is 5.628 kPa. The specific humidity of water vapour in kg vapour per kg dry air will bea)0.109b)0.0109c)0.0214d)0.214Correct answer is option 'C'. Can you explain this answer?.
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