Question

Problem 6.14 Humidity and Condensation Your answer is partially correct. Try again. Air at 70.0% relative humidity is cooled isobarically at 1 atm absolute from 85.0°C to 35.0°C. a. Estimate the dew point and the degrees of superheat of the air at 85.0°C. Dew point: 76.2 loc 吖: loc 8.8 Degrees of superheat:

Answer 1

Given: Relative humidity, h_r =70%=0.70

Initial temperature of air, T₁=85 ⁰C=358.15 K

Final temperature of air, T₂=35 ⁰C=308.15 K

Pressure, P=1 atm=760 mm Hg

(a) Dew point and Degree of superheat

Obtain the vapor pressure at 85 ⁰C from the Vapor Pressure of Water Table.

At 80⁰C, vapor pressure, p*=355.1 mm Hg and at 90⁰C, p*=525.76 mmHg. Interpolate the values

Thus vapor pressure at 85 ⁰C, p*(85 ⁰C)=440.43 mm Hg

Mole fraction of water, y_H2O, is

hrp(85°C no =

where h_r is the Relative humidity

P is the Pressure

$\Rightarrow y_{H_{2}O}=\frac{0.7*440.43\ mmHg}{760\ mmHg}$

ун,0-0.4056 mol H2O/ mol

Dew point temperature, T_dp, is

$y_{H_{2}O}P=p^{*}(T_{dp})$

where p*(T_dp) is the Vapor pressure at dew point temperature

$\Rightarrow p^{*}(T_{dp})=0.4056\ mol\ H_{2}O/mol*760\ mmHg$

$\Rightarrow p^{*}(T_{dp})=308.256\ mmHg$

Refer Vapor Pressure of Water Table and obtain the temperature corresponding to 308.256 mmHg.

At 70⁰C, p*=233.7 mmHg and at 80⁰C, p*=355.1 mmHg. Interpolate the values.

The vapor pressure is 308.256 mmHg at a temperature of 76.14⁰C

Dew point temperature, T_dp=76.14⁰C

Degree of superheat=T₁-T_dp=(85-76.14)⁰C

Degree of Superheat=8.86⁰C

(b) To calculate water condensed per m³ of feed gas

Basis: 1 m³ feed gas

Number of moles of gas in feed,n, is obtained using the gas law

where P', V', n' and T' is the Pressure, Volume, Moles and Temperature at standard conditions

V is the Volume of feed gas

$\Rightarrow n=\frac{VT'n'}{T_{1}V'}$

$\Rightarrow n=\frac{1\ m^{3}feed\ gas*273.15\ K*1\ mol}{358.15\ K*0.0224\ m^{3}(STP)}$

$\boldsymbol{n=34.029\ mol}$

At saturation conditions, the mole fraction of water, y, is

$y=\frac{p^{*}(35^{\circ}C)}{P}$

where p*(35⁰C) is the Vapor pressure of water at 35⁰C, p*(35⁰C)=42.175 mmHg

$\Rightarrow y=\frac{42.175\ mmHg}{760\ mmHg}$

$\boldsymbol{y=0.056\ mol\ H_{2}O/mol}$

Dry air balance is

$(1-y_{H_{2}O})n=n_{1}(1-y)$

$(1-0.4056)n=n_{1}(1-y)$

$\Rightarrow 0.5944\ mol\ dry\ air/mol*34.029\ mol=n_{1}(1-0.056\ mol\ H_{2}O/mol)$

$\Rightarrow 20.2268=0.944n_{1}$

$\boldsymbol{n_{1}=21.427\ mol}$

Total mole balance is

where n₂ is the Moles of water condensed

$\Rightarrow 34.029\ mol=21.427\ mol+n_{2}$

$\mathbf{n_{2}=12.602\ mol\ H_{2}O/m^{3}}$

Thus moles of water condensed is n₂=12.602 mol H₂O/m³ feed gas

(c) Pressure at which mist forms is obtained using Raoult's law

$y_{H_{2}O}P=p^{*}(85^{\circ}C)$

$\Rightarrow P=\frac{p^{*}(85^{\circ}C)}{y_{H_{2}O}}$

$\Rightarrow P=\frac{440.43\ mmHg}{0.4056\ mol\ H_{2}O/mol}$

$\Rightarrow P=1085.87\ mmHg$

$\mathbf{P=1.43\ atm}$

Thus the pressure at which mist forms is P=1.43 atm.