A tank with a volume 20 m3 contains a mixture of carbon diox

A tank with a volume 20 m3 contains a mixture of carbon dioxide and water vapor at 18 C and 100 kPa with a relative humidity () of 80 % .

Part A

Determine the partial pressure of dry CO2

Part B

Determine the specific humidity () of the mixture.

Part C

Determine the dew point.

Part D

Determine the amount of vapor.

Part E

Determine the mass of CO2.

Part F

What-if scenario: What would the amount of vapor be if the mixture was composed of dry air and water vapor?

Solution

a)

Relative Humidity, RH = Pv / Psat

At 18 deg C, sat. vapor pressure Psat = 2.0647 kPa

0.8 = Pv / 2.0647

Pv = 1.6518 kPa

P = P_co2 + Pv

100 = P_co2 + 1.6518

P_co2 = 98.3482 kPa

b)

Specific humidity w = m_vap / m_co2

P_vap * V = m_vap * R_vap * T

1.6518*10^3 * 20 = m_vap*461.5*(18+273)

m_vap = 0.246 kg

P_co2 * V = m_co2 * R_co2 * T

98.3482*10^3 * 20 = m_co2 * 188.9 * (18+273)

m_co2 = 35.782 kg

w = 0.246 / 35.782

w = 0.00687 kg vapor per kg of co2

w = 6.87 grams vapor per kg of co2

c)

At Pv = 1.6518 kPa, we get sat. temperature = 14.48 deg C which is the dew point.

d)

m_vap = 0.246 kg

e)

m_co2 = 35.782 kg