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Moist air at a pressure of 100 kPa is compressed to 500 kPa and then cooled to 35°C in an aftercooler. The air at the entry to the aftercooler is unsaturated and becomes just saturated at the exit of the aftercooler. The saturation pressure of water at 35°C is 5.628 kPa. The partial pressure of water vapour (in kPa) in the moist air entering the compressor is closest to
  • a)
    0.57
  • b)
    1.13
  • c)
    2.26
  • d)
    4.52
Correct answer is option 'B'. Can you explain this answer?
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Moist air at a pressure of 100 kPa is compressed to 500 kPa and then c...

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Moist air at a pressure of 100 kPa is compressed to 500 kPa and then c...
Given data:
- Initial pressure of moist air (P1) = 100 kPa
- Final pressure of moist air (P2) = 500 kPa
- Temperature at the exit of aftercooler (T2) = 35°C
- Saturation pressure of water at 35°C = 5.628 kPa

To find: Partial pressure of water vapor in the moist air entering the compressor (Pw1)

We can solve this problem using the ideal gas law and the concept of relative humidity.

1. Determine the partial pressure of water vapor at the exit of the aftercooler (Pw2):
- Given the saturation pressure of water at 35°C is 5.628 kPa, the partial pressure of water vapor at the exit of the aftercooler is also 5.628 kPa since the air becomes saturated.
- Therefore, Pw2 = 5.628 kPa.

2. Determine the partial pressure of dry air at the exit of the aftercooler (Pd2):
- Using the ideal gas law: P1V1/T1 = P2V2/T2
- Assuming the volume remains constant (V1 = V2), we can simplify the equation to: P1/T1 = P2/T2
- Substituting the given values, we have: (100 kPa)/(T1) = (500 kPa)/(35 + 273) K
- Solving for T1, we find: T1 = 20.1°C
- The saturation pressure of water at 20.1°C is 2.339 kPa.
- Therefore, Pd2 = P2 - Pw2 = 500 kPa - 5.628 kPa = 494.372 kPa.

3. Determine the partial pressure of water vapor at the entry to the aftercooler (Pw1):
- Since the air at the entry to the aftercooler is unsaturated, the partial pressure of water vapor can be found using the concept of relative humidity.
- Relative humidity (RH) = (Pw1 / Psat(T1)) * 100
- Rearranging the equation, we have: Pw1 = (RH / 100) * Psat(T1)
- Given that the air becomes saturated at the exit of the aftercooler, the relative humidity is 100%.
- Substituting the values, we find: Pw1 = (100 / 100) * 2.339 kPa = 2.339 kPa.

Therefore, the partial pressure of water vapor in the moist air entering the compressor is closest to 2.26 kPa, which corresponds to option (b).
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Moist air at a pressure of 100 kPa is compressed to 500 kPa and then cooled to 35°C in an aftercooler. The air at the entry to the aftercooler is unsaturated and becomes just saturated at the exit of the aftercooler. The saturation pressure of water at 35°C is 5.628 kPa. The partial pressure of water vapour (in kPa) in the moist air entering the compressor is closest toa)0.57b)1.13c)2.26d)4.52Correct answer is option 'B'. Can you explain this answer?
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