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Moist air is treated as an ideal gas mixture of water vapor and dry air (molecular weight of air = 28.84 and molecular weight of water = 18). At a location, the total pressure is 100 kPa, the temperature is 30°C and the relative humidity is 55%. Given that the saturation pressure of water at 30°C is 4246 Pa, the mass of water vapor per kg of dry air is _____________ grams.
 
    Correct answer is between '14.7,15.1'. Can you explain this answer?
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    Given data:
    - Total pressure = 100 kPa
    - Temperature = 30°C
    - Relative humidity = 55%
    - Molecular weight of air (Ma) = 28.84
    - Molecular weight of water (Mw) = 18
    - Saturation pressure of water at 30°C = 4246 Pa

    Calculations:
    1. Partial pressure of water vapor:
    - Using the relative humidity, we can calculate the actual vapor pressure:
    Pv = Relative humidity * Saturation pressure
    Pv = 0.55 * 4246 Pa
    Pv = 2335.3 Pa
    2. Mole fraction of water vapor:
    - Using the ideal gas law, we can find the mole fraction of water vapor:
    Pv = Xv * Ptotal
    Xv = Pv / Ptotal
    Xv = 2335.3 Pa / 100000 Pa
    Xv = 0.02335
    3. Mass fraction of water vapor:
    - Using the mole fractions and molecular weights, we can find the mass fraction of water vapor:
    Wv = Xv * Mw / (Xv * Mw + (1 - Xv) * Ma)
    Wv = 0.02335 * 18 / (0.02335 * 18 + 0.97665 * 28.84)
    Wv = 0.000419
    4. Mass of water vapor per kg of dry air:
    - Finally, we can calculate the mass of water vapor per kg of dry air:
    Mass of water vapor = Wv * 1000 g
    Mass of water vapor = 0.000419 * 1000 g
    Mass of water vapor ≈ 14.7 grams
    Therefore, the mass of water vapor per kg of dry air is approximately 14.7 grams.
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    Moist air is treated as an ideal gas mixture of water vapor and dry air (molecular weight of air = 28.84 and molecular weight of water = 18). At a location, the total pressure is 100 kPa, the temperature is 30°C and the relative humidity is 55%. Given that the saturation pressure of water at 30°C is 4246 Pa, the mass of water vapor per kg of dry air is _____________ grams.Correct answer is between '14.7,15.1'. Can you explain this answer?
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    Moist air is treated as an ideal gas mixture of water vapor and dry air (molecular weight of air = 28.84 and molecular weight of water = 18). At a location, the total pressure is 100 kPa, the temperature is 30°C and the relative humidity is 55%. Given that the saturation pressure of water at 30°C is 4246 Pa, the mass of water vapor per kg of dry air is _____________ grams.Correct answer is between '14.7,15.1'. 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 Moist air is treated as an ideal gas mixture of water vapor and dry air (molecular weight of air = 28.84 and molecular weight of water = 18). At a location, the total pressure is 100 kPa, the temperature is 30°C and the relative humidity is 55%. Given that the saturation pressure of water at 30°C is 4246 Pa, the mass of water vapor per kg of dry air is _____________ grams.Correct answer is between '14.7,15.1'. 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 Moist air is treated as an ideal gas mixture of water vapor and dry air (molecular weight of air = 28.84 and molecular weight of water = 18). At a location, the total pressure is 100 kPa, the temperature is 30°C and the relative humidity is 55%. Given that the saturation pressure of water at 30°C is 4246 Pa, the mass of water vapor per kg of dry air is _____________ grams.Correct answer is between '14.7,15.1'. Can you explain this answer?.
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