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Wet air enters a dryer at 98.5 kPa and 60°C. The partial pressure of water vapor in the entering air is 8.7 kPa. Calculate the kg-mol of water present per 100 kg-mol of dry air.

Answer :

Answer:

The mass of water present per 100 kg-mol of dry air is 3.17 kg-mol.

Explanation:

Given parameters:

The partial pressure of water vapor in the entering air = 8.7 kPa

The temperature of the entering air = 60°C

The pressure of entering air = 98.5 kPa

We are to calculate the number of kg-mol of water present per 100 kg-mol dry air

Solution: The mole fraction of water vapor in the air can be calculated as:

Y = Pv / PaPv = Partial pressure of water vapor in the entering air = 8.7 kPaPa = Total pressure of entering air = 98.5 kPaY = 8.7 / 98.5Y = 0.08817

Therefore, the mole fraction of water vapor in the air is 0.08817.

The total number of moles of air present can be calculated as:nA = PA * VA / RT... (i)

where, PA = Total pressure of entering air = 98.5 kPaVA = Volume of entering air = 1 kg-mol dry air

R = Universal gas constant = 8.314 J / mol K... (ii)

T = Temperature of entering air = 60 + 273 = 333 K

T = 333 K

Substituting the given values in equation (i)

nA = 98.5 × 1000 / (8.314 × 333) = 35.89 kg-mol

Therefore, the total number of moles of air present is 35.89 kg-mol. The mass of water present per 100 kg-mol of dry air can be calculated as:

nw = nA × Y

nw = 35.89 × 0.08817

nw = 3.17 kg-mol

Therefore, the mass of water present per 100 kg-mol of dry air is 3.17 kg-mol.

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