Illustrative Example

Illustrative Example – p.1/26

Recap

We derived the following equations: Split fractions:

ξ_{k} = \frac{α_{k/n}ξ_{n}}{1 + (α_{k/n} − 1)ξ_{n}}

Simplified bubble-point equation: \frac{P_{k}^{0}}{P} = \frac{α_{k/n}}{ᾱ}

Simplified dew-point equation: ∑_{i=1}^{n} \frac{y_{i}}{α_{i/n}} = \frac{P_{n}^{0}}{P}

Vapor Fraction and Split Fraction:

ξ_{n} = \frac{1}{\frac{1}{K_{n}ϕ} − \frac{1}{K_{n}} + 1}

Recap

We derived the following equations: Split fractions:

ξ_{k} = \frac{α_{k/n}ξ_{n}}{1 + (α_{k/n} − 1)ξ_{n}}

Simplified bubble-point equation: \frac{P_{k}^{0}}{P} = \frac{α_{k/n}}{ᾱ}

Simplified dew-point equation: ∑_{i=1}^{n} \frac{y_{i}}{α_{i/n}} = \frac{P_{n}^{0}}{P}

Vapor Fraction and Split Fraction:

ξ_{n} = \frac{1}{\frac{1}{K_{n}ϕ} − \frac{1}{K_{n}} + 1}

Recap

We derived the following equations: Split fractions:

ξ_{k} = \frac{α_{k/n}ξ_{n}}{1 + (α_{k/n} − 1)ξ_{n}}

Simplified bubble-point equation: \frac{P_{k}^{0}}{P} = \frac{α_{k/n}}{ᾱ}

Simplified dew-point equation: ∑_{i=1}^{n} \frac{y_{i}}{α_{i/n}} = \frac{P_{n}^{0}}{P}

Vapor Fraction and Split Fraction:

ξ_{n} = \frac{1}{\frac{1}{K_{n}ϕ} − \frac{1}{K_{n}} + 1}

Recap

We derived the following equations: Split fractions:

ξ_{k} = \frac{α_{k/n}ξ_{n}}{1 + (α_{k/n} − 1)ξ_{n}}

Simplified bubble-point equation: \frac{P_{k}^{0}}{P} = \frac{α_{k/n}}{ᾱ}

Simplified dew-point equation: ∑_{i=1}^{n} \frac{y_{i}}{α_{i/n}} = \frac{P_{n}^{0}}{P}

Vapor Fraction and Split Fraction:

ξ_{n} = \frac{1}{\frac{1}{K_{n}ϕ} − \frac{1}{K_{n}} + 1}

Recap

We derived the following equations: Split fractions:

ξ_{k} = \frac{α_{k/n}ξ_{n}}{1 + (α_{k/n} − 1)ξ_{n}}

Simplified bubble-point equation: \frac{P_{k}^{0}}{P} = \frac{α_{k/n}}{ᾱ}

Simplified dew-point equation: ∑_{i=1}^{n} \frac{y_{i}}{α_{i/n}} = \frac{P_{n}^{0}}{P}

Vapor Fraction and Split Fraction:

ξ_{n} = \frac{1}{\frac{1}{K_{n}ϕ} − \frac{1}{K_{n}} + 1}

Recap

We derived the following equations: Split fractions:

ξ_{k} = \frac{α_{k/n}ξ_{n}}{1 + (α_{k/n} − 1)ξ_{n}}

Simplified bubble-point equation: \frac{P_{k}^{0}}{P} = \frac{α_{k/n}}{ᾱ}

Simplified dew-point equation: ∑_{i=1}^{n} \frac{y_{i}}{α_{i/n}} = \frac{P_{n}^{0}}{P}

Vapor Fraction and Split Fraction:

ξ_{n} = \frac{1}{\frac{1}{K_{n}ϕ} − \frac{1}{K_{n}} + 1}

In particular, how do we use these equations in the following scenarios?

T_{V} , P_{V} specified  Isothermal flash
V/F = 0, P_{L} specified Bubble point temperature
V/F = 1, P_{V} specified Dew point temperature
T_{L} specified, V/F = 0 Bubble point pressure
T_{V} specified, V/F = 0 Dew point pressure
Q = 0, P_{V} specified  Adiabatic flash
Q, P_{V} specified  Nonadiabatic flash
V/F, P_{V} specified  Percent vaporization flash

In particular, how do we use these equations in the following scenarios?

T_{V} , P_{V} specified  Isothermal flash
V/F = 0, P_{L} specified Bubble point temperature
V/F = 1, P_{V} specified Dew point temperature
T_{L} specified, V/F = 0 Bubble point pressure
T_{V} specified, V/F = 0 Dew point pressure
Q = 0, P_{V} specified  Adiabatic flash
Q, P_{V} specified  Nonadiabatic flash
V/F, P_{V} specified  Percent vaporization flash

Illustrative Example

Consider the mixture with components, flowrates, boiling points and Antoine coefficients given in the following table:

Comp., k f_{k} (kmol/hr) B. pt. (K) A_{k}  B_{k}  C_{k}
Benzene, 1  30  353  15.9008 2778.51 -52.34
Toluene, 2  50  383  16.0137 3096.52 -53.67
O-xylene, 3  20  418  16.1156 3395.57 -59.4

where the Antoine equation is given by:

ln(P_{k}^{0}) = A_{k} − \frac{B_{k}}{(C_{k} + T)}

where P_{k}^{0} is in mm Hg and T is in K. Question 1: At what pressure will the stream come out as a saturated liquid at 390 K?

Illustrative Example – p.4/26

Illustrative Example

Consider the mixture with components, flowrates, boiling points and Antoine coefficients given in the following table:

Comp., k f_{k} (kmol/hr) B. pt. (K) A_{k}  B_{k}  C_{k}
Benzene, 1  30  353  15.9008 2778.51 -52.34
Toluene, 2  50  383  16.0137 3096.52 -53.67
O-xylene, 3  20  418  16.1156 3395.57 -59.4

where the Antoine equation is given by:

ln(P_{k}^{0}) = A_{k} − \frac{B_{k}}{(C_{k} + T)}

where P_{k}^{0} is in mm Hg and T is in K. Question 1: At what pressure will the stream come out as a saturated liquid at 390 K?

Illustrative Example – p.4/26

Illustrative Example

Consider the mixture with components, flowrates, boiling points and Antoine coefficients given in the following table:

Comp., k f_{k} (kmol/hr) B. pt. (K) A_{k}  B_{k}  C_{k}
Benzene, 1  30  353  15.9008 2778.51 -52.34
Toluene, 2  50  383  16.0137 3096.52 -53.67
O-xylene, 3  20  418  16.1156 3395.57 -59.4

where the Antoine equation is given by:

ln(P_{k}^{0}) = A_{k} − \frac{B_{k}}{(C_{k} + T)}

where P_{k}^{0} is in mm Hg and T is in K. Question 1: At what pressure will the stream come out as a saturated liquid at 390 K?

Illustrative Example – p.4/26

Solution

x_{k} = z_{k} Thus, x_{1} = 0.30, x_{2} = 0.50, x_{3} = 0.40

Compute vapor pressures from Antoine equation:

\begin{matrix}ln(P_{1}^{0}) = A_{1} − \frac{B_{1}}{(C_{1} + T)} \\ ⇒ P_{1}^{0} = exp [A_{1} − \frac{B_{1}}{(C_{1} + T)}] \end{matrix}

Plugging in the values of A_{1}, B_{1}, C_{1} we get P_{1}^{0} = 2059.49 mm Hg Similarly, we get P_{2}^{0} = 904.07 mm Hg and P_{3}^{0} = 344.96 mm Hg.

Choose toluene as the key component (as it is the most abundant).

Illustrative Example – p.5/26

x_{k} = z_{k} Thus, x_{1} = 0.30, x_{2} = 0.50, x_{3} = 0.40

Compute vapor pressures from Antoine equation:

\begin{matrix}ln(P_{1}^{0}) = A_{1} − \frac{B_{1}}{(C_{1} + T)} \\ ⇒ P_{1}^{0} = exp [A_{1} − \frac{B_{1}}{(C_{1} + T)}] \end{matrix}

Plugging in the values of A_{1}, B_{1}, C_{1} we get P_{1}^{0} = 2059.49 mm Hg Similarly, we get P_{2}^{0} = 904.07 mm Hg and P_{3}^{0} = 344.96 mm Hg.

Choose toluene as the key component (as it is the most abundant).

Illustrative Example – p.5/26

x_{k} = z_{k} Thus, x_{1} = 0.30, x_{2} = 0.50, x_{3} = 0.40

Compute vapor pressures from Antoine equation:

\begin{matrix}ln(P_{1}^{0}) = A_{1} − \frac{B_{1}}{(C_{1} + T)} \\ ⇒ P_{1}^{0} = exp [A_{1} − \frac{B_{1}}{(C_{1} + T)}] \end{matrix}

Plugging in the values of A_{1}, B_{1}, C_{1} we get P_{1}^{0} = 2059.49 mm Hg Similarly, we get P_{2}^{0} = 904.07 mm Hg and P_{3}^{0} = 344.96 mm Hg.

Choose toluene as the key component (as it is the most abundant).

Illustrative Example – p.5/26

x_{k} = z_{k} Thus, x_{1} = 0.30, x_{2} = 0.50, x_{3} = 0.40

Compute vapor pressures from Antoine equation:

\begin{matrix}ln(P_{1}^{0}) = A_{1} − \frac{B_{1}}{(C_{1} + T)} \\ ⇒ P_{1}^{0} = exp [A_{1} − \frac{B_{1}}{(C_{1} + T)}] \end{matrix}

Plugging in the values of A_{1}, B_{1}, C_{1} we get P_{1}^{0} = 2059.49 mm Hg Similarly, we get P_{2}^{0} = 904.07 mm Hg and P_{3}^{0} = 344.96 mm Hg.

Choose toluene as the key component (as it is the most abundant).

Illustrative Example – p.5/26

x_{k} = z_{k} Thus, x_{1} = 0.30, x_{2} = 0.50, x_{3} = 0.40

Compute vapor pressures from Antoine equation:

\begin{matrix}ln(P_{1}^{0}) = A_{1} − \frac{B_{1}}{(C_{1} + T)} \\ ⇒ P_{1}^{0} = exp [A_{1} − \frac{B_{1}}{(C_{1} + T)}] \end{matrix}

Plugging in the values of A_{1}, B_{1}, C_{1} we get P_{1}^{0} = 2059.49 mm Hg Similarly, we get P_{2}^{0} = 904.07 mm Hg and P_{3}^{0} = 344.96 mm Hg.

Choose toluene as the key component (as it is the most abundant).

Illustrative Example – p.5/26

x_{k} = z_{k} Thus, x_{1} = 0.30, x_{2} = 0.50, x_{3} = 0.40

Compute vapor pressures from Antoine equation:

\begin{matrix}ln(P_{1}^{0}) = A_{1} − \frac{B_{1}}{(C_{1} + T)} \\ ⇒ P_{1}^{0} = exp [A_{1} − \frac{B_{1}}{(C_{1} + T)}] \end{matrix}

Plugging in the values of A_{1}, B_{1}, C_{1} we get P_{1}^{0} = 2059.49 mm Hg Similarly, we get P_{2}^{0} = 904.07 mm Hg and P_{3}^{0} = 344.96 mm Hg.

Choose toluene as the key component (as it is the most abundant).

Illustrative Example – p.5/26

x_{k} = z_{k} Thus, x_{1} = 0.30, x_{2} = 0.50, x_{3} = 0.40

Compute vapor pressures from Antoine equation:

\begin{matrix}ln(P_{1}^{0}) = A_{1} − \frac{B_{1}}{(C_{1} + T)} \\ ⇒ P_{1}^{0} = exp [A_{1} − \frac{B_{1}}{(C_{1} + T)}] \end{matrix}

Plugging in the values of A_{1}, B_{1}, C_{1} we get P_{1}^{0} = 2059.49 mm Hg Similarly, we get P_{2}^{0} = 904.07 mm Hg and P_{3}^{0} = 344.96 mm Hg.

Choose toluene as the key component (as it is the most abundant).

Illustrative Example – p.5/26

Average relative volatility is given by:

\begin{aligned}ᾱ & = & α_{1/2}x_{1} + α_{2/2}x_{2} + α_{3/2}x_{3} \\ & = & \frac{P_{1}^{0}}{P_{2}^{0}} x_{1} + \frac{P_{2}^{0}}{P_{2}^{0}} x_{2} + \frac{P_{3}^{0}}{P_{2}^{0}} x_{3} \end{aligned}

Substituting the values of vapor pressures and mole fractions, we get α_{1/2} = 2.28, α_{2/2} = 1, α_{3/2} = 0.38 and ᾱ = 2.28 × 0.30 + 1 × 0.50 + 0.38 × 0.20 = 1.26 The simplified bubble point equation (in terms of the key

component) is given by \frac{P_{2}^{0}}{P} = \frac{1}{ᾱ}

Thus, the total pressure is given by P = \frac{P_{2}^{0}}{ᾱ}

Substituting the values of ᾱ and P_{2}^{0}, we get: P = 1139.13 mm Hg

What would be the answer if you chose benzene or o-xylene

as the key component?

Illustrative Example – p.6/26

\begin{aligned}ᾱ & = & α_{1/2}x_{1} + α_{2/2}x_{2} + α_{3/2}x_{3} \\ & = & \frac{P_{1}^{0}}{P_{2}^{0}} x_{1} + \frac{P_{2}^{0}}{P_{2}^{0}} x_{2} + \frac{P_{3}^{0}}{P_{2}^{0}} x_{3} \end{aligned}

Substituting the values of vapor pressures and mole fractions, we get α_{1/2} = 2.28, α_{2/2} = 1, α_{3/2} = 0.38 and ᾱ = 2.28 × 0.30 + 1 × 0.50 + 0.38 × 0.20 = 1.26 The simplified bubble point equation (in terms of the key

component) is given by \frac{P_{2}^{0}}{P} = \frac{1}{ᾱ}

Thus, the total pressure is given by P = \frac{P_{2}^{0}}{ᾱ}

Substituting the values of ᾱ and P_{2}^{0}, we get: P = 1139.13 mm Hg

What would be the answer if you chose benzene or o-xylene

as the key component?

Illustrative Example – p.6/26

\begin{aligned}ᾱ & = & α_{1/2}x_{1} + α_{2/2}x_{2} + α_{3/2}x_{3} \\ & = & \frac{P_{1}^{0}}{P_{2}^{0}} x_{1} + \frac{P_{2}^{0}}{P_{2}^{0}} x_{2} + \frac{P_{3}^{0}}{P_{2}^{0}} x_{3} \end{aligned}

Substituting the values of vapor pressures and mole fractions, we get α_{1/2} = 2.28, α_{2/2} = 1, α_{3/2} = 0.38 and ᾱ = 2.28 × 0.30 + 1 × 0.50 + 0.38 × 0.20 = 1.26 The simplified bubble point equation (in terms of the key

component) is given by \frac{P_{2}^{0}}{P} = \frac{1}{ᾱ}

Thus, the total pressure is given by P = \frac{P_{2}^{0}}{ᾱ}

Substituting the values of ᾱ and P_{2}^{0}, we get: P = 1139.13 mm Hg

What would be the answer if you chose benzene or o-xylene

as the key component?

Illustrative Example – p.6/26

\begin{aligned}ᾱ & = & α_{1/2}x_{1} + α_{2/2}x_{2} + α_{3/2}x_{3} \\ & = & \frac{P_{1}^{0}}{P_{2}^{0}} x_{1} + \frac{P_{2}^{0}}{P_{2}^{0}} x_{2} + \frac{P_{3}^{0}}{P_{2}^{0}} x_{3} \end{aligned}

Substituting the values of vapor pressures and mole fractions, we get α_{1/2} = 2.28, α_{2/2} = 1, α_{3/2} = 0.38 and ᾱ = 2.28 × 0.30 + 1 × 0.50 + 0.38 × 0.20 = 1.26 The simplified bubble point equation (in terms of the key

component) is given by \frac{P_{2}^{0}}{P} = \frac{1}{ᾱ}

Thus, the total pressure is given by P = \frac{P_{2}^{0}}{ᾱ}

Substituting the values of ᾱ and P_{2}^{0}, we get: P = 1139.13 mm Hg

What would be the answer if you chose benzene or o-xylene

as the key component?

Illustrative Example – p.6/26

\begin{aligned}ᾱ & = & α_{1/2}x_{1} + α_{2/2}x_{2} + α_{3/2}x_{3} \\ & = & \frac{P_{1}^{0}}{P_{2}^{0}} x_{1} + \frac{P_{2}^{0}}{P_{2}^{0}} x_{2} + \frac{P_{3}^{0}}{P_{2}^{0}} x_{3} \end{aligned}

Substituting the values of vapor pressures and mole fractions, we get α_{1/2} = 2.28, α_{2/2} = 1, α_{3/2} = 0.38 and ᾱ = 2.28 × 0.30 + 1 × 0.50 + 0.38 × 0.20 = 1.26 The simplified bubble point equation (in terms of the key

component) is given by \frac{P_{2}^{0}}{P} = \frac{1}{ᾱ}

Thus, the total pressure is given by P = \frac{P_{2}^{0}}{ᾱ}

Substituting the values of ᾱ and P_{2}^{0}, we get: P = 1139.13 mm Hg

What would be the answer if you chose benzene or o-xylene

as the key component?

Illustrative Example – p.6/26

\begin{aligned}ᾱ & = & α_{1/2}x_{1} + α_{2/2}x_{2} + α_{3/2}x_{3} \\ & = & \frac{P_{1}^{0}}{P_{2}^{0}} x_{1} + \frac{P_{2}^{0}}{P_{2}^{0}} x_{2} + \frac{P_{3}^{0}}{P_{2}^{0}} x_{3} \end{aligned}

Substituting the values of vapor pressures and mole fractions, we get α_{1/2} = 2.28, α_{2/2} = 1, α_{3/2} = 0.38 and ᾱ = 2.28 × 0.30 + 1 × 0.50 + 0.38 × 0.20 = 1.26 The simplified bubble point equation (in terms of the key

component) is given by \frac{P_{2}^{0}}{P} = \frac{1}{ᾱ}

Thus, the total pressure is given by P = \frac{P_{2}^{0}}{ᾱ}

Substituting the values of ᾱ and P_{2}^{0}, we get: P = 1139.13 mm Hg

What would be the answer if you chose benzene or o-xylene

as the key component?

Illustrative Example – p.6/26

\begin{aligned}ᾱ & = & α_{1/2}x_{1} + α_{2/2}x_{2} + α_{3/2}x_{3} \\ & = & \frac{P_{1}^{0}}{P_{2}^{0}} x_{1} + \frac{P_{2}^{0}}{P_{2}^{0}} x_{2} + \frac{P_{3}^{0}}{P_{2}^{0}} x_{3} \end{aligned}

Substituting the values of vapor pressures and mole fractions, we get α_{1/2} = 2.28, α_{2/2} = 1, α_{3/2} = 0.38 and ᾱ = 2.28 × 0.30 + 1 × 0.50 + 0.38 × 0.20 = 1.26 The simplified bubble point equation (in terms of the key

component) is given by \frac{P_{2}^{0}}{P} = \frac{1}{ᾱ}

Thus, the total pressure is given by P = \frac{P_{2}^{0}}{ᾱ}

Substituting the values of ᾱ and P_{2}^{0}, we get: P = 1139.13 mm Hg

What would be the answer if you chose benzene or o-xylene

as the key component?

Illustrative Example – p.6/26

\begin{aligned}ᾱ & = & α_{1/2}x_{1} + α_{2/2}x_{2} + α_{3/2}x_{3} \\ & = & \frac{P_{1}^{0}}{P_{2}^{0}} x_{1} + \frac{P_{2}^{0}}{P_{2}^{0}} x_{2} + \frac{P_{3}^{0}}{P_{2}^{0}} x_{3} \end{aligned}

Substituting the values of vapor pressures and mole fractions, we get α_{1/2} = 2.28, α_{2/2} = 1, α_{3/2} = 0.38 and ᾱ = 2.28 × 0.30 + 1 × 0.50 + 0.38 × 0.20 = 1.26 The simplified bubble point equation (in terms of the key

component) is given by \frac{P_{2}^{0}}{P} = \frac{1}{ᾱ}

Thus, the total pressure is given by P = \frac{P_{2}^{0}}{ᾱ}

Substituting the values of ᾱ and P_{2}^{0}, we get: P = 1139.13 mm Hg

What would be the answer if you chose benzene or o-xylene

as the key component?

Illustrative Example – p.6/26

Question 2: At what pressure will the stream come out as a saturated vapor at 390 K?

y_{k} = z_{k} Thus, y_{1} = 0.30, y_{2} = 0.50, y_{3} = 0.40

For calculating the pressure, we will use the dew point equation.

P = \frac{P_{n}^{0}}{(∑ \frac{y_{k}}{α_{k/n}} )}

We know y_{i} from the feed composition and we already computed the values of the vapor pressure of toluene as well as the relative volatilities of benzene and o-xylene w.r.t. toluene in the previous slides.

Illustrative Example – p.7/26

y_{k} = z_{k} Thus, y_{1} = 0.30, y_{2} = 0.50, y_{3} = 0.40

For calculating the pressure, we will use the dew point equation.

P = \frac{P_{n}^{0}}{(∑ \frac{y_{k}}{α_{k/n}} )}

We know y_{i} from the feed composition and we already computed the values of the vapor pressure of toluene as well as the relative volatilities of benzene and o-xylene w.r.t. toluene in the previous slides.

Illustrative Example – p.7/26

y_{k} = z_{k} Thus, y_{1} = 0.30, y_{2} = 0.50, y_{3} = 0.40

For calculating the pressure, we will use the dew point equation.

P = \frac{P_{n}^{0}}{(∑ \frac{y_{k}}{α_{k/n}} )}

We know y_{i} from the feed composition and we already computed the values of the vapor pressure of toluene as well as the relative volatilities of benzene and o-xylene w.r.t. toluene in the previous slides.

Illustrative Example – p.7/26

y_{k} = z_{k} Thus, y_{1} = 0.30, y_{2} = 0.50, y_{3} = 0.40

For calculating the pressure, we will use the dew point equation.

P = \frac{P_{n}^{0}}{(∑ \frac{y_{k}}{α_{k/n}} )}

We know y_{i} from the feed composition and we already computed the values of the vapor pressure of toluene as well as the relative volatilities of benzene and o-xylene w.r.t. toluene in the previous slides.

Illustrative Example – p.7/26

y_{k} = z_{k} Thus, y_{1} = 0.30, y_{2} = 0.50, y_{3} = 0.40

For calculating the pressure, we will use the dew point equation.

P = \frac{P_{n}^{0}}{(∑ \frac{y_{k}}{α_{k/n}} )}

We know y_{i} from the feed composition and we already computed the values of the vapor pressure of toluene as well as the relative volatilities of benzene and o-xylene w.r.t. toluene in the previous slides.

Illustrative Example – p.7/26

Substituting the values of y_{1}, y_{2}, y_{3} from above and the values of P_{2}^{0} and α_{1/2}, α_{2/2}, α_{3/2} computed earlier, we get

\begin{aligned}P & = & \frac{904.07}{(\frac{0.3}{2.28} + \frac{0.5}{1} + \frac{0.2}{0.38})} \\ & = & 780.79 mm Hg \end{aligned}

Note that at the same temperature, if the mixture comes out at as saturated vapor, the flash pressure is lower than when it comes out as saturated liquid.

\begin{aligned}P & = & \frac{904.07}{(\frac{0.3}{2.28} + \frac{0.5}{1} + \frac{0.2}{0.38})} \\ & = & 780.79 mm Hg \end{aligned}

Note that at the same temperature, if the mixture comes out at as saturated vapor, the flash pressure is lower than when it comes out as saturated liquid.

\begin{aligned}P & = & \frac{904.07}{(\frac{0.3}{2.28} + \frac{0.5}{1} + \frac{0.2}{0.38})} \\ & = & 780.79 mm Hg \end{aligned}

Note that at the same temperature, if the mixture comes out at as saturated vapor, the flash pressure is lower than when it comes out as saturated liquid.

\begin{aligned}P & = & \frac{904.07}{(\frac{0.3}{2.28} + \frac{0.5}{1} + \frac{0.2}{0.38})} \\ & = & 780.79 mm Hg \end{aligned}

Note that at the same temperature, if the mixture comes out at as saturated vapor, the flash pressure is lower than when it comes out as saturated liquid.

Question 3: If the flash pressure is 1000 mm Hg and the stream come out as a saturated liquid, what is the flash temperature?

We use the bubble point equation in the following form:

P_{n}^{0} = \frac{P}{ᾱ}

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and ᾱ at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.9/26

We use the bubble point equation in the following form:

P_{n}^{0} = \frac{P}{ᾱ}

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and ᾱ at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.9/26

We use the bubble point equation in the following form:

P_{n}^{0} = \frac{P}{ᾱ}

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and ᾱ at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.9/26

We use the bubble point equation in the following form:

P_{n}^{0} = \frac{P}{ᾱ}

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and ᾱ at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.9/26

We use the bubble point equation in the following form:

P_{n}^{0} = \frac{P}{ᾱ}

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and ᾱ at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.9/26

We use the bubble point equation in the following form:

P_{n}^{0} = \frac{P}{ᾱ}

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and ᾱ at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.9/26

We use the bubble point equation in the following form:

P_{n}^{0} = \frac{P}{ᾱ}

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and ᾱ at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.9/26

Guess T = 380 K Then, using the Antoine equation: P_{1}^{0} = 1599.45 mm Hg, P_{2}^{0} = 681.82 mm Hg, P_{3}^{0} = 250.38 mm Hg Thus,

ᾱ = \frac{1599.45}{681.82} × 0.30 + \frac{681.82}{681.82} × 0.50 + \frac{250.38}{681.82} × 0.20 = 1.28

LHS = P_{2}^{0} = 681.82

RHS = \frac{P}{ᾱ} = \frac{1000}{1.28} = 781.25

Since LHS ≠ RHS guess another temperature.

Guess T = 380 K Then, using the Antoine equation: P_{1}^{0} = 1599.45 mm Hg, P_{2}^{0} = 681.82 mm Hg, P_{3}^{0} = 250.38 mm Hg Thus,

ᾱ = \frac{1599.45}{681.82} × 0.30 + \frac{681.82}{681.82} × 0.50 + \frac{250.38}{681.82} × 0.20 = 1.28

LHS = P_{2}^{0} = 681.82

RHS = \frac{P}{ᾱ} = \frac{1000}{1.28} = 781.25

Since LHS ≠ RHS guess another temperature.

Guess T = 380 K Then, using the Antoine equation: P_{1}^{0} = 1599.45 mm Hg, P_{2}^{0} = 681.82 mm Hg, P_{3}^{0} = 250.38 mm Hg Thus,

ᾱ = \frac{1599.45}{681.82} × 0.30 + \frac{681.82}{681.82} × 0.50 + \frac{250.38}{681.82} × 0.20 = 1.28

LHS = P_{2}^{0} = 681.82

RHS = \frac{P}{ᾱ} = \frac{1000}{1.28} = 781.25

Since LHS ≠ RHS guess another temperature.

Guess T = 380 K Then, using the Antoine equation: P_{1}^{0} = 1599.45 mm Hg, P_{2}^{0} = 681.82 mm Hg, P_{3}^{0} = 250.38 mm Hg Thus,

ᾱ = \frac{1599.45}{681.82} × 0.30 + \frac{681.82}{681.82} × 0.50 + \frac{250.38}{681.82} × 0.20 = 1.28

LHS = P_{2}^{0} = 681.82

RHS = \frac{P}{ᾱ} = \frac{1000}{1.28} = 781.25

Since LHS ≠ RHS guess another temperature.

Guess T = 380 K Then, using the Antoine equation: P_{1}^{0} = 1599.45 mm Hg, P_{2}^{0} = 681.82 mm Hg, P_{3}^{0} = 250.38 mm Hg Thus,

ᾱ = \frac{1599.45}{681.82} × 0.30 + \frac{681.82}{681.82} × 0.50 + \frac{250.38}{681.82} × 0.20 = 1.28

LHS = P_{2}^{0} = 681.82

RHS = \frac{P}{ᾱ} = \frac{1000}{1.28} = 781.25

Since LHS ≠ RHS guess another temperature.

Guess T = 380 K Then, using the Antoine equation: P_{1}^{0} = 1599.45 mm Hg, P_{2}^{0} = 681.82 mm Hg, P_{3}^{0} = 250.38 mm Hg Thus,

ᾱ = \frac{1599.45}{681.82} × 0.30 + \frac{681.82}{681.82} × 0.50 + \frac{250.38}{681.82} × 0.20 = 1.28

LHS = P_{2}^{0} = 681.82

RHS = \frac{P}{ᾱ} = \frac{1000}{1.28} = 781.25

Since LHS ≠ RHS guess another temperature.

Guess T = 380 K Then, using the Antoine equation: P_{1}^{0} = 1599.45 mm Hg, P_{2}^{0} = 681.82 mm Hg, P_{3}^{0} = 250.38 mm Hg Thus,

ᾱ = \frac{1599.45}{681.82} × 0.30 + \frac{681.82}{681.82} × 0.50 + \frac{250.38}{681.82} × 0.20 = 1.28

LHS = P_{2}^{0} = 681.82

RHS = \frac{P}{ᾱ} = \frac{1000}{1.28} = 781.25

Since LHS ≠ RHS guess another temperature.

Guess T = 380 K Then, using the Antoine equation: P_{1}^{0} = 1599.45 mm Hg, P_{2}^{0} = 681.82 mm Hg, P_{3}^{0} = 250.38 mm Hg Thus,

ᾱ = \frac{1599.45}{681.82} × 0.30 + \frac{681.82}{681.82} × 0.50 + \frac{250.38}{681.82} × 0.20 = 1.28

LHS = P_{2}^{0} = 681.82

RHS = \frac{P}{ᾱ} = \frac{1000}{1.28} = 781.25

Since LHS ≠ RHS guess another temperature.

Guess T = 380 K Then, using the Antoine equation: P_{1}^{0} = 1599.45 mm Hg, P_{2}^{0} = 681.82 mm Hg, P_{3}^{0} = 250.38 mm Hg Thus,

ᾱ = \frac{1599.45}{681.82} × 0.30 + \frac{681.82}{681.82} × 0.50 + \frac{250.38}{681.82} × 0.20 = 1.28

LHS = P_{2}^{0} = 681.82

RHS = \frac{P}{ᾱ} = \frac{1000}{1.28} = 781.25

Since LHS ≠ RHS guess another temperature.

Guess T = 385 K Then, using the Antoine equation: P_{1}^{0} = 1818.41 mm Hg, P_{2}^{0} = 786.80 mm Hg, P_{3}^{0} = 294.61 mm Hg Thus, ᾱ = 1.27 LHS = P_{2}^{0} = 786.80

RHS = \frac{P}{ᾱ} = \frac{1000}{1.27} = 787.40

Since LHS ≈ RHS the guessed temperature of 385 K is correct. Note that while P_{k}^{0} varied substantially, ᾱ did not vary much when a new temperature was guessed.

Guess T = 385 K Then, using the Antoine equation: P_{1}^{0} = 1818.41 mm Hg, P_{2}^{0} = 786.80 mm Hg, P_{3}^{0} = 294.61 mm Hg Thus, ᾱ = 1.27 LHS = P_{2}^{0} = 786.80

RHS = \frac{P}{ᾱ} = \frac{1000}{1.27} = 787.40

Since LHS ≈ RHS the guessed temperature of 385 K is correct. Note that while P_{k}^{0} varied substantially, ᾱ did not vary much when a new temperature was guessed.

Guess T = 385 K Then, using the Antoine equation: P_{1}^{0} = 1818.41 mm Hg, P_{2}^{0} = 786.80 mm Hg, P_{3}^{0} = 294.61 mm Hg Thus, ᾱ = 1.27 LHS = P_{2}^{0} = 786.80

RHS = \frac{P}{ᾱ} = \frac{1000}{1.27} = 787.40

Since LHS ≈ RHS the guessed temperature of 385 K is correct. Note that while P_{k}^{0} varied substantially, ᾱ did not vary much when a new temperature was guessed.

Guess T = 385 K Then, using the Antoine equation: P_{1}^{0} = 1818.41 mm Hg, P_{2}^{0} = 786.80 mm Hg, P_{3}^{0} = 294.61 mm Hg Thus, ᾱ = 1.27 LHS = P_{2}^{0} = 786.80

RHS = \frac{P}{ᾱ} = \frac{1000}{1.27} = 787.40

Since LHS ≈ RHS the guessed temperature of 385 K is correct. Note that while P_{k}^{0} varied substantially, ᾱ did not vary much when a new temperature was guessed.

Guess T = 385 K Then, using the Antoine equation: P_{1}^{0} = 1818.41 mm Hg, P_{2}^{0} = 786.80 mm Hg, P_{3}^{0} = 294.61 mm Hg Thus, ᾱ = 1.27 LHS = P_{2}^{0} = 786.80

RHS = \frac{P}{ᾱ} = \frac{1000}{1.27} = 787.40

Since LHS ≈ RHS the guessed temperature of 385 K is correct. Note that while P_{k}^{0} varied substantially, ᾱ did not vary much when a new temperature was guessed.

Guess T = 385 K Then, using the Antoine equation: P_{1}^{0} = 1818.41 mm Hg, P_{2}^{0} = 786.80 mm Hg, P_{3}^{0} = 294.61 mm Hg Thus, ᾱ = 1.27 LHS = P_{2}^{0} = 786.80

RHS = \frac{P}{ᾱ} = \frac{1000}{1.27} = 787.40

Since LHS ≈ RHS the guessed temperature of 385 K is correct. Note that while P_{k}^{0} varied substantially, ᾱ did not vary much when a new temperature was guessed.

Guess T = 385 K Then, using the Antoine equation: P_{1}^{0} = 1818.41 mm Hg, P_{2}^{0} = 786.80 mm Hg, P_{3}^{0} = 294.61 mm Hg Thus, ᾱ = 1.27 LHS = P_{2}^{0} = 786.80

RHS = \frac{P}{ᾱ} = \frac{1000}{1.27} = 787.40

Since LHS ≈ RHS the guessed temperature of 385 K is correct. Note that while P_{k}^{0} varied substantially, ᾱ did not vary much when a new temperature was guessed.

Guess T = 385 K Then, using the Antoine equation: P_{1}^{0} = 1818.41 mm Hg, P_{2}^{0} = 786.80 mm Hg, P_{3}^{0} = 294.61 mm Hg Thus, ᾱ = 1.27 LHS = P_{2}^{0} = 786.80

RHS = \frac{P}{ᾱ} = \frac{1000}{1.27} = 787.40

Since LHS ≈ RHS the guessed temperature of 385 K is correct. Note that while P_{k}^{0} varied substantially, ᾱ did not vary much when a new temperature was guessed.

Question 4: If the flash pressure is 1000 mm Hg and the stream come out as a saturated vapor, what is the flash temperature?

We use the dew point equation in the following form:

P_{n}^{0} = P (∑ \frac{y_{k}}{α_{k/n}} )

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and α_{k/n} at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.12/26

We use the dew point equation in the following form:

P_{n}^{0} = P (∑ \frac{y_{k}}{α_{k/n}} )

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and α_{k/n} at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.12/26

We use the dew point equation in the following form:

P_{n}^{0} = P (∑ \frac{y_{k}}{α_{k/n}} )

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and α_{k/n} at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.12/26

We use the dew point equation in the following form:

P_{n}^{0} = P (∑ \frac{y_{k}}{α_{k/n}} )

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and α_{k/n} at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.12/26

We use the dew point equation in the following form:

P_{n}^{0} = P (∑ \frac{y_{k}}{α_{k/n}} )

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and α_{k/n} at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.12/26

We use the dew point equation in the following form:

P_{n}^{0} = P (∑ \frac{y_{k}}{α_{k/n}} )

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and α_{k/n} at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.12/26

We use the dew point equation in the following form:

P_{n}^{0} = P (∑ \frac{y_{k}}{α_{k/n}} )

and the following iterative algorithm

1. Guess a temperature.

2. Compute P_{n}^{0} and α_{k/n} at this temperature.

3. Check if LHS = RHS. If the two sides of the equation are not equal, guess another temperature and go to step 2.

Illustrative Example – p.12/26

Guess T = 395 K Then, using the Antoine equation: P_{1}^{0} = 2324.07 mm Hg, P_{2}^{0} = 1034.07 mm Hg, P_{3}^{0} = 402.01 mm Hg Thus, α_{1/2} = 2.25, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1034.07

RHS = P (∑ \frac{y_{k}}{α_{k/n}} ) = 1000 (\frac{0.3}{2.25} + \frac{0.5}{1} + \frac{0.2}{0.39})

= 1146.15

Since LHS ≠ RHS guess another temperature.

Guess T = 395 K Then, using the Antoine equation: P_{1}^{0} = 2324.07 mm Hg, P_{2}^{0} = 1034.07 mm Hg, P_{3}^{0} = 402.01 mm Hg Thus, α_{1/2} = 2.25, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1034.07

RHS = P (∑ \frac{y_{k}}{α_{k/n}} ) = 1000 (\frac{0.3}{2.25} + \frac{0.5}{1} + \frac{0.2}{0.39})

= 1146.15

Since LHS ≠ RHS guess another temperature.

Guess T = 395 K Then, using the Antoine equation: P_{1}^{0} = 2324.07 mm Hg, P_{2}^{0} = 1034.07 mm Hg, P_{3}^{0} = 402.01 mm Hg Thus, α_{1/2} = 2.25, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1034.07

RHS = P (∑ \frac{y_{k}}{α_{k/n}} ) = 1000 (\frac{0.3}{2.25} + \frac{0.5}{1} + \frac{0.2}{0.39})

= 1146.15

Since LHS ≠ RHS guess another temperature.

Guess T = 395 K Then, using the Antoine equation: P_{1}^{0} = 2324.07 mm Hg, P_{2}^{0} = 1034.07 mm Hg, P_{3}^{0} = 402.01 mm Hg Thus, α_{1/2} = 2.25, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1034.07

RHS = P (∑ \frac{y_{k}}{α_{k/n}} ) = 1000 (\frac{0.3}{2.25} + \frac{0.5}{1} + \frac{0.2}{0.39})

= 1146.15

Since LHS ≠ RHS guess another temperature.

Guess T = 395 K Then, using the Antoine equation: P_{1}^{0} = 2324.07 mm Hg, P_{2}^{0} = 1034.07 mm Hg, P_{3}^{0} = 402.01 mm Hg Thus, α_{1/2} = 2.25, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1034.07

RHS = P (∑ \frac{y_{k}}{α_{k/n}} ) = 1000 (\frac{0.3}{2.25} + \frac{0.5}{1} + \frac{0.2}{0.39})

= 1146.15

Since LHS ≠ RHS guess another temperature.

Guess T = 395 K Then, using the Antoine equation: P_{1}^{0} = 2324.07 mm Hg, P_{2}^{0} = 1034.07 mm Hg, P_{3}^{0} = 402.01 mm Hg Thus, α_{1/2} = 2.25, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1034.07

RHS = P (∑ \frac{y_{k}}{α_{k/n}} ) = 1000 (\frac{0.3}{2.25} + \frac{0.5}{1} + \frac{0.2}{0.39})

= 1146.15

Since LHS ≠ RHS guess another temperature.

Guess T = 395 K Then, using the Antoine equation: P_{1}^{0} = 2324.07 mm Hg, P_{2}^{0} = 1034.07 mm Hg, P_{3}^{0} = 402.01 mm Hg Thus, α_{1/2} = 2.25, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1034.07

RHS = P (∑ \frac{y_{k}}{α_{k/n}} ) = 1000 (\frac{0.3}{2.25} + \frac{0.5}{1} + \frac{0.2}{0.39})

= 1146.15

Since LHS ≠ RHS guess another temperature.

Guess T = 395 K Then, using the Antoine equation: P_{1}^{0} = 2324.07 mm Hg, P_{2}^{0} = 1034.07 mm Hg, P_{3}^{0} = 402.01 mm Hg Thus, α_{1/2} = 2.25, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1034.07

RHS = P (∑ \frac{y_{k}}{α_{k/n}} ) = 1000 (\frac{0.3}{2.25} + \frac{0.5}{1} + \frac{0.2}{0.39})

= 1146.15

Since LHS ≠ RHS guess another temperature.

Guess T = 399 K Then, using the Antoine equation: P_{1}^{0} = 2553.57 mm Hg, P_{2}^{0} = 1149.24 mm Hg, P_{3}^{0} = 452.91 mm Hg Thus, α_{1/2} = 2.22, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1149.24

RHS = 1000 (\frac{0.3}{2.22} + \frac{0.5}{1} + \frac{0.2}{0.39}) = 1147.96

Since LHS ≈ RHS the guessed temperature of 399 K is correct. Note that while P_{k}^{0} varied substantially, α_{k/n} did not vary much when a new temperature was guessed.

Illustrative Example – p.14/26

Guess T = 399 K Then, using the Antoine equation: P_{1}^{0} = 2553.57 mm Hg, P_{2}^{0} = 1149.24 mm Hg, P_{3}^{0} = 452.91 mm Hg Thus, α_{1/2} = 2.22, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1149.24

RHS = 1000 (\frac{0.3}{2.22} + \frac{0.5}{1} + \frac{0.2}{0.39}) = 1147.96

Since LHS ≈ RHS the guessed temperature of 399 K is correct. Note that while P_{k}^{0} varied substantially, α_{k/n} did not vary much when a new temperature was guessed.

Illustrative Example – p.14/26

Guess T = 399 K Then, using the Antoine equation: P_{1}^{0} = 2553.57 mm Hg, P_{2}^{0} = 1149.24 mm Hg, P_{3}^{0} = 452.91 mm Hg Thus, α_{1/2} = 2.22, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1149.24

RHS = 1000 (\frac{0.3}{2.22} + \frac{0.5}{1} + \frac{0.2}{0.39}) = 1147.96

Since LHS ≈ RHS the guessed temperature of 399 K is correct. Note that while P_{k}^{0} varied substantially, α_{k/n} did not vary much when a new temperature was guessed.

Illustrative Example – p.14/26

Guess T = 399 K Then, using the Antoine equation: P_{1}^{0} = 2553.57 mm Hg, P_{2}^{0} = 1149.24 mm Hg, P_{3}^{0} = 452.91 mm Hg Thus, α_{1/2} = 2.22, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1149.24

RHS = 1000 (\frac{0.3}{2.22} + \frac{0.5}{1} + \frac{0.2}{0.39}) = 1147.96

Since LHS ≈ RHS the guessed temperature of 399 K is correct. Note that while P_{k}^{0} varied substantially, α_{k/n} did not vary much when a new temperature was guessed.

Illustrative Example – p.14/26

Guess T = 399 K Then, using the Antoine equation: P_{1}^{0} = 2553.57 mm Hg, P_{2}^{0} = 1149.24 mm Hg, P_{3}^{0} = 452.91 mm Hg Thus, α_{1/2} = 2.22, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1149.24

RHS = 1000 (\frac{0.3}{2.22} + \frac{0.5}{1} + \frac{0.2}{0.39}) = 1147.96

Since LHS ≈ RHS the guessed temperature of 399 K is correct. Note that while P_{k}^{0} varied substantially, α_{k/n} did not vary much when a new temperature was guessed.

Illustrative Example – p.14/26

Guess T = 399 K Then, using the Antoine equation: P_{1}^{0} = 2553.57 mm Hg, P_{2}^{0} = 1149.24 mm Hg, P_{3}^{0} = 452.91 mm Hg Thus, α_{1/2} = 2.22, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1149.24

RHS = 1000 (\frac{0.3}{2.22} + \frac{0.5}{1} + \frac{0.2}{0.39}) = 1147.96

Since LHS ≈ RHS the guessed temperature of 399 K is correct. Note that while P_{k}^{0} varied substantially, α_{k/n} did not vary much when a new temperature was guessed.

Illustrative Example – p.14/26

Guess T = 399 K Then, using the Antoine equation: P_{1}^{0} = 2553.57 mm Hg, P_{2}^{0} = 1149.24 mm Hg, P_{3}^{0} = 452.91 mm Hg Thus, α_{1/2} = 2.22, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1149.24

RHS = 1000 (\frac{0.3}{2.22} + \frac{0.5}{1} + \frac{0.2}{0.39}) = 1147.96

Since LHS ≈ RHS the guessed temperature of 399 K is correct. Note that while P_{k}^{0} varied substantially, α_{k/n} did not vary much when a new temperature was guessed.

Illustrative Example – p.14/26

Guess T = 399 K Then, using the Antoine equation: P_{1}^{0} = 2553.57 mm Hg, P_{2}^{0} = 1149.24 mm Hg, P_{3}^{0} = 452.91 mm Hg Thus, α_{1/2} = 2.22, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1149.24

RHS = 1000 (\frac{0.3}{2.22} + \frac{0.5}{1} + \frac{0.2}{0.39}) = 1147.96

Since LHS ≈ RHS the guessed temperature of 399 K is correct. Note that while P_{k}^{0} varied substantially, α_{k/n} did not vary much when a new temperature was guessed.

Illustrative Example – p.14/26

Guess T = 399 K Then, using the Antoine equation: P_{1}^{0} = 2553.57 mm Hg, P_{2}^{0} = 1149.24 mm Hg, P_{3}^{0} = 452.91 mm Hg Thus, α_{1/2} = 2.22, α_{2/2} = 1, α_{3/2} = 0.39 LHS = P_{2}^{0} = 1149.24

RHS = 1000 (\frac{0.3}{2.22} + \frac{0.5}{1} + \frac{0.2}{0.39}) = 1147.96

Since LHS ≈ RHS the guessed temperature of 399 K is correct. Note that while P_{k}^{0} varied substantially, α_{k/n} did not vary much when a new temperature was guessed.

Illustrative Example – p.14/26

Question 5: If the flash pressure is 900 mm Hg and the flash temperature is 390 K what are the compositions of the vapor and liquid exiting the flash unit? What fraction of the feed is vaporized?

Note that in this problem, the composition of the exiting streams need to be computed.

Note that based on the answers of questions 1-4, we know that at the temperature and pressure given, part of the mixture is vapor and part of the mixture is liquid.

Question 5: If the flash pressure is 900 mm Hg and the flash temperature is 390 K what are the compositions of the vapor and liquid exiting the flash unit? What fraction of the feed is vaporized?

Note that in this problem, the composition of the exiting streams need to be computed.

Note that based on the answers of questions 1-4, we know that at the temperature and pressure given, part of the mixture is vapor and part of the mixture is liquid.

Question 5: If the flash pressure is 900 mm Hg and the flash temperature is 390 K what are the compositions of the vapor and liquid exiting the flash unit? What fraction of the feed is vaporized?

Note that in this problem, the composition of the exiting streams need to be computed.

Note that based on the answers of questions 1-4, we know that at the temperature and pressure given, part of the mixture is vapor and part of the mixture is liquid.

We will use the following iterative algorithm

1. Choose the key component n and guess its split fraction, ξ_{n}.

2. Compute α_{k/n} at the flash temperature and compute the split fractions of the non-key components using the equation:

ξ_{k} = \frac{α_{k/n}ξ_{n}}{1 + (α_{k/n} − 1)ξ_{n}}

3. Compute the liquid and vapor flowrates from the split fractions.

4. Compute x from the liquid flowrates and check the bubble point equation. If the bubble point equation is not satisfied, guess another split fraction and go to step 2.

Illustrative Example – p.16/26

We will use the following iterative algorithm

1. Choose the key component n and guess its split fraction, ξ_{n}.

2. Compute α_{k/n} at the flash temperature and compute the split fractions of the non-key components using the equation:

ξ_{k} = \frac{α_{k/n}ξ_{n}}{1 + (α_{k/n} − 1)ξ_{n}}

3. Compute the liquid and vapor flowrates from the split fractions.

4. Compute x from the liquid flowrates and check the bubble point equation. If the bubble point equation is not satisfied, guess another split fraction and go to step 2.

Illustrative Example – p.16/26

We will use the following iterative algorithm

1. Choose the key component n and guess its split fraction, ξ_{n}.

2. Compute α_{k/n} at the flash temperature and compute the split fractions of the non-key components using the equation:

ξ_{k} = \frac{α_{k/n}ξ_{n}}{1 + (α_{k/n} − 1)ξ_{n}}

3. Compute the liquid and vapor flowrates from the split fractions.

4. Compute x from the liquid flowrates and check the bubble point equation. If the bubble point equation is not satisfied, guess another split fraction and go to step 2.

Illustrative Example – p.16/26

We will use the following iterative algorithm

1. Choose the key component n and guess its split fraction, ξ_{n}.

2. Compute α_{k/n} at the flash temperature and compute the split fractions of the non-key components using the equation:

ξ_{k} = \frac{α_{k/n}ξ_{n}}{1 + (α_{k/n} − 1)ξ_{n}}

3. Compute the liquid and vapor flowrates from the split fractions.

4. Compute x from the liquid flowrates and check the bubble point equation. If the bubble point equation is not satisfied, guess another split fraction and go to step 2.

Illustrative Example – p.16/26

We will use the following iterative algorithm

1. Choose the key component n and guess its split fraction, ξ_{n}.

2. Compute α_{k/n} at the flash temperature and compute the split fractions of the non-key components using the equation:

ξ_{k} = \frac{α_{k/n}ξ_{n}}{1 + (α_{k/n} − 1)ξ_{n}}

3. Compute the liquid and vapor flowrates from the split fractions.

4. Compute x from the liquid flowrates and check the bubble point equation. If the bubble point equation is not satisfied, guess another split fraction and go to step 2.

We will use the following iterative algorithm

1. Choose the key component n and guess its split fraction, ξ_{n}.

2. Compute α_{k/n} at the flash temperature and compute the split fractions of the non-key components using the equation:

ξ_{k} = \frac{α_{k/n}ξ_{n}}{1 + (α_{k/n} − 1)ξ_{n}}

3. Compute the liquid and vapor flowrates from the split fractions.

4. Compute x from the liquid flowrates and check the bubble point equation. If the bubble point equation is not satisfied, guess another split fraction and go to step 2.

We will use the following iterative algorithm

1. Choose the key component n and guess its split fraction, ξ_{n}.

2. Compute α_{k/n} at the flash temperature and compute the split fractions of the non-key components using the equation:

ξ_{k} = \frac{α_{k/n}ξ_{n}}{1 + (α_{k/n} − 1)ξ_{n}}

3. Compute the liquid and vapor flowrates from the split fractions.

4. Compute x from the liquid flowrates and check the bubble point equation. If the bubble point equation is not satisfied, guess another split fraction and go to step 2.

Choose toluene as the key component and guess ξ_{2} = 0.90.

Compute α_{1/2} and α_{3/2} at the given temperature. P_{1}^{0} = 2059.49 mm Hg, P_{2}^{0} = 904.07 mm Hg, P_{3}^{0} = 344.96 mm Hg Thus, α_{1/2} = 2.28 and α_{3/2} = 0.38

The split fractions ξ_{1} and ξ_{3} are:

ξ_{1} = \frac{2.28 × 0.9}{1 + (2.28 − 1)0.9}
= 0.95
ξ_{3} = \frac{0.38 × 0.9}{1 + (0.38 − 1)0.9}
= 0.77

Choose toluene as the key component and guess ξ_{2} = 0.90.

Compute α_{1/2} and α_{3/2} at the given temperature. P_{1}^{0} = 2059.49 mm Hg, P_{2}^{0} = 904.07 mm Hg, P_{3}^{0} = 344.96 mm Hg Thus, α_{1/2} = 2.28 and α_{3/2} = 0.38

The split fractions ξ_{1} and ξ_{3} are:

ξ_{1} = \frac{2.28 × 0.9}{1 + (2.28 − 1)0.9}
= 0.95
ξ_{3} = \frac{0.38 × 0.9}{1 + (0.38 − 1)0.9}
= 0.77

Choose toluene as the key component and guess ξ_{2} = 0.90.

Compute α_{1/2} and α_{3/2} at the given temperature. P_{1}^{0} = 2059.49 mm Hg, P_{2}^{0} = 904.07 mm Hg, P_{3}^{0} = 344.96 mm Hg Thus, α_{1/2} = 2.28 and α_{3/2} = 0.38

The split fractions ξ_{1} and ξ_{3} are:

ξ_{1} = \frac{2.28 × 0.9}{1 + (2.28 − 1)0.9}
= 0.95
ξ_{3} = \frac{0.38 × 0.9}{1 + (0.38 − 1)0.9}
= 0.77

Choose toluene as the key component and guess ξ_{2} = 0.90.

Compute α_{1/2} and α_{3/2} at the given temperature. P_{1}^{0} = 2059.49 mm Hg, P_{2}^{0} = 904.07 mm Hg, P_{3}^{0} = 344.96 mm Hg Thus, α_{1/2} = 2.28 and α_{3/2} = 0.38

The split fractions ξ_{1} and ξ_{3} are:

ξ_{1} = \frac{2.28 × 0.9}{1 + (2.28 − 1)0.9}
= 0.95
ξ_{3} = \frac{0.38 × 0.9}{1 + (0.38 − 1)0.9}
= 0.77

Choose toluene as the key component and guess ξ_{2} = 0.90.

Compute α_{1/2} and α_{3/2} at the given temperature. P_{1}^{0} = 2059.49 mm Hg, P_{2}^{0} = 904.07 mm Hg, P_{3}^{0} = 344.96 mm Hg Thus, α_{1/2} = 2.28 and α_{3/2} = 0.38

The split fractions ξ_{1} and ξ_{3} are:

ξ_{1} = \frac{2.28 × 0.9}{1 + (2.28 − 1)0.9}
= 0.95
ξ_{3} = \frac{0.38 × 0.9}{1 + (0.38 − 1)0.9}
= 0.77

Choose toluene as the key component and guess ξ_{2} = 0.90.

Compute α_{1/2} and α_{3/2} at the given temperature. P_{1}^{0} = 2059.49 mm Hg, P_{2}^{0} = 904.07 mm Hg, P_{3}^{0} = 344.96 mm Hg Thus, α_{1/2} = 2.28 and α_{3/2} = 0.38

The split fractions ξ_{1} and ξ_{3} are:

ξ_{1} = \frac{2.28 × 0.9}{1 + (2.28 − 1)0.9}
= 0.95
ξ_{3} = \frac{0.38 × 0.9}{1 + (0.38 − 1)0.9}
= 0.77

Given the split fractions, the vapor and liquid flowrates can be computed.

v_{1} = ξ_{1}f_{1} = 0.95 × 30 = 28.5 kmol/hr v_{2} = ξ_{2}f_{2} = 0.90 × 50 = 45.0 kmol/hr v_{3} = ξ_{1}f_{1} = 0.77 × 20 = 15.4 kmol/hr l_{1} = 30 − 28.5 = 1.5 kmol/hr l_{2} = 50 − 45.0 = 5.0 kmol/hr l_{3} = 20 − 15.4 = 4.6 kmol/hr

The total liquid flow = 1.5 + 5.0 + 4.6 = 11.1 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{1.5}{11.1} = 0.14, x_{2} = \frac{5}{11.1} = 0.45, x_{3} = \frac{4.6}{11.1} = 0.41

v_{1} = ξ_{1}f_{1} = 0.95 × 30 = 28.5 kmol/hr v_{2} = ξ_{2}f_{2} = 0.90 × 50 = 45.0 kmol/hr v_{3} = ξ_{1}f_{1} = 0.77 × 20 = 15.4 kmol/hr l_{1} = 30 − 28.5 = 1.5 kmol/hr l_{2} = 50 − 45.0 = 5.0 kmol/hr l_{3} = 20 − 15.4 = 4.6 kmol/hr

The total liquid flow = 1.5 + 5.0 + 4.6 = 11.1 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{1.5}{11.1} = 0.14, x_{2} = \frac{5}{11.1} = 0.45, x_{3} = \frac{4.6}{11.1} = 0.41

v_{1} = ξ_{1}f_{1} = 0.95 × 30 = 28.5 kmol/hr v_{2} = ξ_{2}f_{2} = 0.90 × 50 = 45.0 kmol/hr v_{3} = ξ_{1}f_{1} = 0.77 × 20 = 15.4 kmol/hr l_{1} = 30 − 28.5 = 1.5 kmol/hr l_{2} = 50 − 45.0 = 5.0 kmol/hr l_{3} = 20 − 15.4 = 4.6 kmol/hr

The total liquid flow = 1.5 + 5.0 + 4.6 = 11.1 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{1.5}{11.1} = 0.14, x_{2} = \frac{5}{11.1} = 0.45, x_{3} = \frac{4.6}{11.1} = 0.41

v_{1} = ξ_{1}f_{1} = 0.95 × 30 = 28.5 kmol/hr v_{2} = ξ_{2}f_{2} = 0.90 × 50 = 45.0 kmol/hr v_{3} = ξ_{1}f_{1} = 0.77 × 20 = 15.4 kmol/hr l_{1} = 30 − 28.5 = 1.5 kmol/hr l_{2} = 50 − 45.0 = 5.0 kmol/hr l_{3} = 20 − 15.4 = 4.6 kmol/hr

The total liquid flow = 1.5 + 5.0 + 4.6 = 11.1 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{1.5}{11.1} = 0.14, x_{2} = \frac{5}{11.1} = 0.45, x_{3} = \frac{4.6}{11.1} = 0.41

v_{1} = ξ_{1}f_{1} = 0.95 × 30 = 28.5 kmol/hr v_{2} = ξ_{2}f_{2} = 0.90 × 50 = 45.0 kmol/hr v_{3} = ξ_{1}f_{1} = 0.77 × 20 = 15.4 kmol/hr l_{1} = 30 − 28.5 = 1.5 kmol/hr l_{2} = 50 − 45.0 = 5.0 kmol/hr l_{3} = 20 − 15.4 = 4.6 kmol/hr

The total liquid flow = 1.5 + 5.0 + 4.6 = 11.1 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{1.5}{11.1} = 0.14, x_{2} = \frac{5}{11.1} = 0.45, x_{3} = \frac{4.6}{11.1} = 0.41

v_{1} = ξ_{1}f_{1} = 0.95 × 30 = 28.5 kmol/hr v_{2} = ξ_{2}f_{2} = 0.90 × 50 = 45.0 kmol/hr v_{3} = ξ_{1}f_{1} = 0.77 × 20 = 15.4 kmol/hr l_{1} = 30 − 28.5 = 1.5 kmol/hr l_{2} = 50 − 45.0 = 5.0 kmol/hr l_{3} = 20 − 15.4 = 4.6 kmol/hr

The total liquid flow = 1.5 + 5.0 + 4.6 = 11.1 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{1.5}{11.1} = 0.14, x_{2} = \frac{5}{11.1} = 0.45, x_{3} = \frac{4.6}{11.1} = 0.41

v_{1} = ξ_{1}f_{1} = 0.95 × 30 = 28.5 kmol/hr v_{2} = ξ_{2}f_{2} = 0.90 × 50 = 45.0 kmol/hr v_{3} = ξ_{1}f_{1} = 0.77 × 20 = 15.4 kmol/hr l_{1} = 30 − 28.5 = 1.5 kmol/hr l_{2} = 50 − 45.0 = 5.0 kmol/hr l_{3} = 20 − 15.4 = 4.6 kmol/hr

The total liquid flow = 1.5 + 5.0 + 4.6 = 11.1 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{1.5}{11.1} = 0.14, x_{2} = \frac{5}{11.1} = 0.45, x_{3} = \frac{4.6}{11.1} = 0.41

v_{1} = ξ_{1}f_{1} = 0.95 × 30 = 28.5 kmol/hr v_{2} = ξ_{2}f_{2} = 0.90 × 50 = 45.0 kmol/hr v_{3} = ξ_{1}f_{1} = 0.77 × 20 = 15.4 kmol/hr l_{1} = 30 − 28.5 = 1.5 kmol/hr l_{2} = 50 − 45.0 = 5.0 kmol/hr l_{3} = 20 − 15.4 = 4.6 kmol/hr

The total liquid flow = 1.5 + 5.0 + 4.6 = 11.1 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{1.5}{11.1} = 0.14, x_{2} = \frac{5}{11.1} = 0.45, x_{3} = \frac{4.6}{11.1} = 0.41

v_{1} = ξ_{1}f_{1} = 0.95 × 30 = 28.5 kmol/hr v_{2} = ξ_{2}f_{2} = 0.90 × 50 = 45.0 kmol/hr v_{3} = ξ_{1}f_{1} = 0.77 × 20 = 15.4 kmol/hr l_{1} = 30 − 28.5 = 1.5 kmol/hr l_{2} = 50 − 45.0 = 5.0 kmol/hr l_{3} = 20 − 15.4 = 4.6 kmol/hr

The total liquid flow = 1.5 + 5.0 + 4.6 = 11.1 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{1.5}{11.1} = 0.14, x_{2} = \frac{5}{11.1} = 0.45, x_{3} = \frac{4.6}{11.1} = 0.41

v_{1} = ξ_{1}f_{1} = 0.95 × 30 = 28.5 kmol/hr v_{2} = ξ_{2}f_{2} = 0.90 × 50 = 45.0 kmol/hr v_{3} = ξ_{1}f_{1} = 0.77 × 20 = 15.4 kmol/hr l_{1} = 30 − 28.5 = 1.5 kmol/hr l_{2} = 50 − 45.0 = 5.0 kmol/hr l_{3} = 20 − 15.4 = 4.6 kmol/hr

The total liquid flow = 1.5 + 5.0 + 4.6 = 11.1 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{1.5}{11.1} = 0.14, x_{2} = \frac{5}{11.1} = 0.45, x_{3} = \frac{4.6}{11.1} = 0.41

Check the bubble point equation P = \frac{ᾱ}{α_{2/2}} P_{2}^{0}

ᾱ = 0.14 × 2.28 + 0.45 × 1 + 0.41 × 0.38 = 0.925 LHS = 900 mm Hg RHS = 0.925 × 904.07 = 836.26 mm Hg

Since LHS ≠ RHS the guessed split fraction of toluene (ξ_{2} = 0.9) is incorrect and it is necessary to make another guess. ᾱ = 0.14 × 2.28 + 0.45 × 1 + 0.41 × 0.38 = 0.925 LHS = 900 mm Hg RHS = 0.925 × 904.07 = 836.26 mm Hg

Since LHS ≠ RHS the guessed split fraction of toluene (ξ_{2} = 0.9) is incorrect and it is necessary to make another guess. ᾱ = 0.14 × 2.28 + 0.45 × 1 + 0.41 × 0.38 = 0.925 LHS = 900 mm Hg RHS = 0.925 × 904.07 = 836.26 mm Hg

Since LHS ≠ RHS the guessed split fraction of toluene (ξ_{2} = 0.9) is incorrect and it is necessary to make another guess. ᾱ = 0.14 × 2.28 + 0.45 × 1 + 0.41 × 0.38 = 0.925 LHS = 900 mm Hg RHS = 0.925 × 904.07 = 836.26 mm Hg

Since LHS ≠ RHS the guessed split fraction of toluene (ξ_{2} = 0.9) is incorrect and it is necessary to make another guess. ᾱ = 0.14 × 2.28 + 0.45 × 1 + 0.41 × 0.38 = 0.925 LHS = 900 mm Hg RHS = 0.925 × 904.07 = 836.26 mm Hg

Since LHS ≠ RHS the guessed split fraction of toluene (ξ_{2} = 0.9) is incorrect and it is necessary to make another guess. ᾱ = 0.14 × 2.28 + 0.45 × 1 + 0.41 × 0.38 = 0.925 LHS = 900 mm Hg RHS = 0.925 × 904.07 = 836.26 mm Hg

Since LHS ≠ RHS the guessed split fraction of toluene (ξ_{2} = 0.9) is incorrect and it is necessary to make another guess. ᾱ = 0.14 × 2.28 + 0.45 × 1 + 0.41 × 0.38 = 0.925 LHS = 900 mm Hg RHS = 0.925 × 904.07 = 836.26 mm Hg

Since LHS ≠ RHS the guessed split fraction of toluene (ξ_{2} = 0.9) is incorrect and it is necessary to make another guess.

Choose toluene as the key component and guess ξ_{2} = 0.65.

Compute α_{1/2} and α_{3/2} at the given temperature. Thus, α_{1/2} = 2.28 and α_{3/2} = 0.38

The split fractions ξ_{1} and ξ_{3} are:

ξ_{1} = \frac{2.28 × 0.65}{1 + (2.28 − 1)0.65}
= 0.81
ξ_{3} = \frac{0.38 × 0.65}{1 + (0.38 − 1)0.65}
= 0.41

Choose toluene as the key component and guess ξ_{2} = 0.65.

Compute α_{1/2} and α_{3/2} at the given temperature. Thus, α_{1/2} = 2.28 and α_{3/2} = 0.38

The split fractions ξ_{1} and ξ_{3} are:

ξ_{1} = \frac{2.28 × 0.65}{1 + (2.28 − 1)0.65}
= 0.81
ξ_{3} = \frac{0.38 × 0.65}{1 + (0.38 − 1)0.65}
= 0.41

Choose toluene as the key component and guess ξ_{2} = 0.65.

Compute α_{1/2} and α_{3/2} at the given temperature. Thus, α_{1/2} = 2.28 and α_{3/2} = 0.38

The split fractions ξ_{1} and ξ_{3} are:

ξ_{1} = \frac{2.28 × 0.65}{1 + (2.28 − 1)0.65}
= 0.81
ξ_{3} = \frac{0.38 × 0.65}{1 + (0.38 − 1)0.65}
= 0.41

Choose toluene as the key component and guess ξ_{2} = 0.65.

Compute α_{1/2} and α_{3/2} at the given temperature. Thus, α_{1/2} = 2.28 and α_{3/2} = 0.38

The split fractions ξ_{1} and ξ_{3} are:

ξ_{1} = \frac{2.28 × 0.65}{1 + (2.28 − 1)0.65}
= 0.81
ξ_{3} = \frac{0.38 × 0.65}{1 + (0.38 − 1)0.65}
= 0.41

Choose toluene as the key component and guess ξ_{2} = 0.65.

Compute α_{1/2} and α_{3/2} at the given temperature. Thus, α_{1/2} = 2.28 and α_{3/2} = 0.38

The split fractions ξ_{1} and ξ_{3} are:

ξ_{1} = \frac{2.28 × 0.65}{1 + (2.28 − 1)0.65}
= 0.81
ξ_{3} = \frac{0.38 × 0.65}{1 + (0.38 − 1)0.65}
= 0.41

v_{1} = ξ_{1}f_{1} = 0.81 × 30 = 24.3 kmol/hr v_{2} = ξ_{2}f_{2} = 0.65 × 50 = 32.5 kmol/hr v_{3} = ξ_{1}f_{1} = 0.41 × 20 = 8.2 kmol/hr l_{1} = 30 − 24.3 = 5.7 kmol/hr l_{2} = 50 − 32.5 = 17.5 kmol/hr l_{3} = 20 − 8.2 = 11.8 kmol/hr

The total liquid flow = 5.7 + 17.5 + 11.8 = 35 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{5.7}{35} = 0.16, x_{2} = \frac{17.5}{35} = 0.50, x_{3} = \frac{11.8}{35} = 0.34

v_{1} = ξ_{1}f_{1} = 0.81 × 30 = 24.3 kmol/hr v_{2} = ξ_{2}f_{2} = 0.65 × 50 = 32.5 kmol/hr v_{3} = ξ_{1}f_{1} = 0.41 × 20 = 8.2 kmol/hr l_{1} = 30 − 24.3 = 5.7 kmol/hr l_{2} = 50 − 32.5 = 17.5 kmol/hr l_{3} = 20 − 8.2 = 11.8 kmol/hr

The total liquid flow = 5.7 + 17.5 + 11.8 = 35 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{5.7}{35} = 0.16, x_{2} = \frac{17.5}{35} = 0.50, x_{3} = \frac{11.8}{35} = 0.34

v_{1} = ξ_{1}f_{1} = 0.81 × 30 = 24.3 kmol/hr v_{2} = ξ_{2}f_{2} = 0.65 × 50 = 32.5 kmol/hr v_{3} = ξ_{1}f_{1} = 0.41 × 20 = 8.2 kmol/hr l_{1} = 30 − 24.3 = 5.7 kmol/hr l_{2} = 50 − 32.5 = 17.5 kmol/hr l_{3} = 20 − 8.2 = 11.8 kmol/hr

The total liquid flow = 5.7 + 17.5 + 11.8 = 35 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{5.7}{35} = 0.16, x_{2} = \frac{17.5}{35} = 0.50, x_{3} = \frac{11.8}{35} = 0.34

v_{1} = ξ_{1}f_{1} = 0.81 × 30 = 24.3 kmol/hr v_{2} = ξ_{2}f_{2} = 0.65 × 50 = 32.5 kmol/hr v_{3} = ξ_{1}f_{1} = 0.41 × 20 = 8.2 kmol/hr l_{1} = 30 − 24.3 = 5.7 kmol/hr l_{2} = 50 − 32.5 = 17.5 kmol/hr l_{3} = 20 − 8.2 = 11.8 kmol/hr

The total liquid flow = 5.7 + 17.5 + 11.8 = 35 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{5.7}{35} = 0.16, x_{2} = \frac{17.5}{35} = 0.50, x_{3} = \frac{11.8}{35} = 0.34

v_{1} = ξ_{1}f_{1} = 0.81 × 30 = 24.3 kmol/hr v_{2} = ξ_{2}f_{2} = 0.65 × 50 = 32.5 kmol/hr v_{3} = ξ_{1}f_{1} = 0.41 × 20 = 8.2 kmol/hr l_{1} = 30 − 24.3 = 5.7 kmol/hr l_{2} = 50 − 32.5 = 17.5 kmol/hr l_{3} = 20 − 8.2 = 11.8 kmol/hr

The total liquid flow = 5.7 + 17.5 + 11.8 = 35 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{5.7}{35} = 0.16, x_{2} = \frac{17.5}{35} = 0.50, x_{3} = \frac{11.8}{35} = 0.34

v_{1} = ξ_{1}f_{1} = 0.81 × 30 = 24.3 kmol/hr v_{2} = ξ_{2}f_{2} = 0.65 × 50 = 32.5 kmol/hr v_{3} = ξ_{1}f_{1} = 0.41 × 20 = 8.2 kmol/hr l_{1} = 30 − 24.3 = 5.7 kmol/hr l_{2} = 50 − 32.5 = 17.5 kmol/hr l_{3} = 20 − 8.2 = 11.8 kmol/hr

The total liquid flow = 5.7 + 17.5 + 11.8 = 35 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{5.7}{35} = 0.16, x_{2} = \frac{17.5}{35} = 0.50, x_{3} = \frac{11.8}{35} = 0.34

v_{1} = ξ_{1}f_{1} = 0.81 × 30 = 24.3 kmol/hr v_{2} = ξ_{2}f_{2} = 0.65 × 50 = 32.5 kmol/hr v_{3} = ξ_{1}f_{1} = 0.41 × 20 = 8.2 kmol/hr l_{1} = 30 − 24.3 = 5.7 kmol/hr l_{2} = 50 − 32.5 = 17.5 kmol/hr l_{3} = 20 − 8.2 = 11.8 kmol/hr

The total liquid flow = 5.7 + 17.5 + 11.8 = 35 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{5.7}{35} = 0.16, x_{2} = \frac{17.5}{35} = 0.50, x_{3} = \frac{11.8}{35} = 0.34

v_{1} = ξ_{1}f_{1} = 0.81 × 30 = 24.3 kmol/hr v_{2} = ξ_{2}f_{2} = 0.65 × 50 = 32.5 kmol/hr v_{3} = ξ_{1}f_{1} = 0.41 × 20 = 8.2 kmol/hr l_{1} = 30 − 24.3 = 5.7 kmol/hr l_{2} = 50 − 32.5 = 17.5 kmol/hr l_{3} = 20 − 8.2 = 11.8 kmol/hr

The total liquid flow = 5.7 + 17.5 + 11.8 = 35 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{5.7}{35} = 0.16, x_{2} = \frac{17.5}{35} = 0.50, x_{3} = \frac{11.8}{35} = 0.34

v_{1} = ξ_{1}f_{1} = 0.81 × 30 = 24.3 kmol/hr v_{2} = ξ_{2}f_{2} = 0.65 × 50 = 32.5 kmol/hr v_{3} = ξ_{1}f_{1} = 0.41 × 20 = 8.2 kmol/hr l_{1} = 30 − 24.3 = 5.7 kmol/hr l_{2} = 50 − 32.5 = 17.5 kmol/hr l_{3} = 20 − 8.2 = 11.8 kmol/hr

The total liquid flow = 5.7 + 17.5 + 11.8 = 35 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{5.7}{35} = 0.16, x_{2} = \frac{17.5}{35} = 0.50, x_{3} = \frac{11.8}{35} = 0.34

v_{1} = ξ_{1}f_{1} = 0.81 × 30 = 24.3 kmol/hr v_{2} = ξ_{2}f_{2} = 0.65 × 50 = 32.5 kmol/hr v_{3} = ξ_{1}f_{1} = 0.41 × 20 = 8.2 kmol/hr l_{1} = 30 − 24.3 = 5.7 kmol/hr l_{2} = 50 − 32.5 = 17.5 kmol/hr l_{3} = 20 − 8.2 = 11.8 kmol/hr

The total liquid flow = 5.7 + 17.5 + 11.8 = 35 kmol/hr Thus, the mole fractions in the liquid stream are:

x_{1} = \frac{5.7}{35} = 0.16, x_{2} = \frac{17.5}{35} = 0.50, x_{3} = \frac{11.8}{35} = 0.34

ᾱ = 0.16 × 2.28 + 0.50 × 1 + 0.34 × 0.38 = 0.994 LHS = 900 mm Hg RHS = 0.994 × 904.07 = 898.6 mm Hg

Since LHS ≈ RHS the guessed split fraction of toluene (ξ_{2} = 0.65) is correct.

The vapor fraction can be computed as follows:

Total moles of vapor = Total feed - Total liquid
= 100 − 35
= 65 kmol/hr

Thus, V apor F raction = \frac{65}{100} = 0.65

ᾱ = 0.16 × 2.28 + 0.50 × 1 + 0.34 × 0.38 = 0.994 LHS = 900 mm Hg RHS = 0.994 × 904.07 = 898.6 mm Hg

Since LHS ≈ RHS the guessed split fraction of toluene (ξ_{2} = 0.65) is correct.

The vapor fraction can be computed as follows:

Total moles of vapor = Total feed - Total liquid
= 100 − 35
= 65 kmol/hr

Thus, V apor F raction = \frac{65}{100} = 0.65

ᾱ = 0.16 × 2.28 + 0.50 × 1 + 0.34 × 0.38 = 0.994 LHS = 900 mm Hg RHS = 0.994 × 904.07 = 898.6 mm Hg

Since LHS ≈ RHS the guessed split fraction of toluene (ξ_{2} = 0.65) is correct.

The vapor fraction can be computed as follows:

Total moles of vapor = Total feed - Total liquid
= 100 − 35
= 65 kmol/hr

Thus, V apor F raction = \frac{65}{100} = 0.65

ᾱ = 0.16 × 2.28 + 0.50 × 1 + 0.34 × 0.38 = 0.994 LHS = 900 mm Hg RHS = 0.994 × 904.07 = 898.6 mm Hg

Since LHS ≈ RHS the guessed split fraction of toluene (ξ_{2} = 0.65) is correct.

The vapor fraction can be computed as follows:

Total moles of vapor = Total feed - Total liquid
= 100 − 35
= 65 kmol/hr

Thus, V apor F raction = \frac{65}{100} = 0.65

ᾱ = 0.16 × 2.28 + 0.50 × 1 + 0.34 × 0.38 = 0.994 LHS = 900 mm Hg RHS = 0.994 × 904.07 = 898.6 mm Hg

Since LHS ≈ RHS the guessed split fraction of toluene (ξ_{2} = 0.65) is correct.

The vapor fraction can be computed as follows:

Total moles of vapor = Total feed - Total liquid
= 100 − 35
= 65 kmol/hr

Thus, V apor F raction = \frac{65}{100} = 0.65

ᾱ = 0.16 × 2.28 + 0.50 × 1 + 0.34 × 0.38 = 0.994 LHS = 900 mm Hg RHS = 0.994 × 904.07 = 898.6 mm Hg

Since LHS ≈ RHS the guessed split fraction of toluene (ξ_{2} = 0.65) is correct.

The vapor fraction can be computed as follows:

Total moles of vapor = Total feed - Total liquid
= 100 − 35
= 65 kmol/hr

Thus, V apor F raction = \frac{65}{100} = 0.65

ᾱ = 0.16 × 2.28 + 0.50 × 1 + 0.34 × 0.38 = 0.994 LHS = 900 mm Hg RHS = 0.994 × 904.07 = 898.6 mm Hg

Since LHS ≈ RHS the guessed split fraction of toluene (ξ_{2} = 0.65) is correct.

The vapor fraction can be computed as follows:

Total moles of vapor = Total feed - Total liquid
= 100 − 35
= 65 kmol/hr

Thus, V apor F raction = \frac{65}{100} = 0.65

ᾱ = 0.16 × 2.28 + 0.50 × 1 + 0.34 × 0.38 = 0.994 LHS = 900 mm Hg RHS = 0.994 × 904.07 = 898.6 mm Hg

Since LHS ≈ RHS the guessed split fraction of toluene (ξ_{2} = 0.65) is correct.

The vapor fraction can be computed as follows:

Total moles of vapor = Total feed - Total liquid
= 100 − 35
= 65 kmol/hr

Thus, V apor F raction = \frac{65}{100} = 0.65

ᾱ = 0.16 × 2.28 + 0.50 × 1 + 0.34 × 0.38 = 0.994 LHS = 900 mm Hg RHS = 0.994 × 904.07 = 898.6 mm Hg

Since LHS ≈ RHS the guessed split fraction of toluene (ξ_{2} = 0.65) is correct.

The vapor fraction can be computed as follows:

Total moles of vapor = Total feed - Total liquid
= 100 − 35
= 65 kmol/hr

Thus, V apor F raction = \frac{65}{100} = 0.65

ᾱ = 0.16 × 2.28 + 0.50 × 1 + 0.34 × 0.38 = 0.994 LHS = 900 mm Hg RHS = 0.994 × 904.07 = 898.6 mm Hg

Since LHS ≈ RHS the guessed split fraction of toluene (ξ_{2} = 0.65) is correct.

The vapor fraction can be computed as follows:

Total moles of vapor = Total feed - Total liquid
= 100 − 35
= 65 kmol/hr

Thus, V apor F raction = \frac{65}{100} = 0.65

ᾱ = 0.16 × 2.28 + 0.50 × 1 + 0.34 × 0.38 = 0.994 LHS = 900 mm Hg RHS = 0.994 × 904.07 = 898.6 mm Hg

Since LHS ≈ RHS the guessed split fraction of toluene (ξ_{2} = 0.65) is correct.

The vapor fraction can be computed as follows:

Total moles of vapor = Total feed - Total liquid
= 100 − 35
= 65 kmol/hr

Thus, V apor F raction = \frac{65}{100} = 0.65

ᾱ = 0.16 × 2.28 + 0.50 × 1 + 0.34 × 0.38 = 0.994 LHS = 900 mm Hg RHS = 0.994 × 904.07 = 898.6 mm Hg

Since LHS ≈ RHS the guessed split fraction of toluene (ξ_{2} = 0.65) is correct.

The vapor fraction can be computed as follows:

Total moles of vapor = Total feed - Total liquid
= 100 − 35
= 65 kmol/hr

Thus, V apor F raction = \frac{65}{100} = 0.65

Question 6: If the flash temperature is 390 K and 60% of the feed is vaporized, what are the compositions of the vapor and liquid exiting the flash unit? What is the flash pressure?

Guess the flash pressure to be P = 900 mm Hg.

Choose toluene as the key component and compute K_{n} at the given temperature and guessed pressure.

Thus, K_{2} = \frac{P_{2}^{0}}{P} = \frac{2059.49}{900} = 2.29.

Compute the key component split fraction ξ_{2} from the equation:

ξ_{n} = \frac{1}{\frac{1}{K_{n}ϕ} − \frac{1}{K_{n}} + 1}

Substituting ϕ = 0.60 and K_{n} = 2.29 we get ξ_{2} = 0.77 The rest of the solution is similar to that used for question 5.

Illustrative Example – p.23/26

Guess the flash pressure to be P = 900 mm Hg.

Choose toluene as the key component and compute K_{n} at the given temperature and guessed pressure.

Thus, K_{2} = \frac{P_{2}^{0}}{P} = \frac{2059.49}{900} = 2.29.

Compute the key component split fraction ξ_{2} from the equation:

ξ_{n} = \frac{1}{\frac{1}{K_{n}ϕ} − \frac{1}{K_{n}} + 1}

Substituting ϕ = 0.60 and K_{n} = 2.29 we get ξ_{2} = 0.77 The rest of the solution is similar to that used for question 5.

Illustrative Example – p.23/26

Guess the flash pressure to be P = 900 mm Hg.

Choose toluene as the key component and compute K_{n} at the given temperature and guessed pressure.

Thus, K_{2} = \frac{P_{2}^{0}}{P} = \frac{2059.49}{900} = 2.29.

Compute the key component split fraction ξ_{2} from the equation:

ξ_{n} = \frac{1}{\frac{1}{K_{n}ϕ} − \frac{1}{K_{n}} + 1}

Substituting ϕ = 0.60 and K_{n} = 2.29 we get ξ_{2} = 0.77 The rest of the solution is similar to that used for question 5.

Illustrative Example – p.23/26

Guess the flash pressure to be P = 900 mm Hg.

Choose toluene as the key component and compute K_{n} at the given temperature and guessed pressure.

Thus, K_{2} = \frac{P_{2}^{0}}{P} = \frac{2059.49}{900} = 2.29.

Compute the key component split fraction ξ_{2} from the equation:

ξ_{n} = \frac{1}{\frac{1}{K_{n}ϕ} − \frac{1}{K_{n}} + 1}

Substituting ϕ = 0.60 and K_{n} = 2.29 we get ξ_{2} = 0.77 The rest of the solution is similar to that used for question 5.

Illustrative Example – p.23/26

Guess the flash pressure to be P = 900 mm Hg.

Choose toluene as the key component and compute K_{n} at the given temperature and guessed pressure.

Thus, K_{2} = \frac{P_{2}^{0}}{P} = \frac{2059.49}{900} = 2.29.

Compute the key component split fraction ξ_{2} from the equation:

ξ_{n} = \frac{1}{\frac{1}{K_{n}ϕ} − \frac{1}{K_{n}} + 1}

Substituting ϕ = 0.60 and K_{n} = 2.29 we get ξ_{2} = 0.77 The rest of the solution is similar to that used for question 5.

Illustrative Example – p.23/26

Guess the flash pressure to be P = 900 mm Hg.

Choose toluene as the key component and compute K_{n} at the given temperature and guessed pressure.

Thus, K_{2} = \frac{P_{2}^{0}}{P} = \frac{2059.49}{900} = 2.29.

Compute the key component split fraction ξ_{2} from the equation:

ξ_{n} = \frac{1}{\frac{1}{K_{n}ϕ} − \frac{1}{K_{n}} + 1}

Substituting ϕ = 0.60 and K_{n} = 2.29 we get ξ_{2} = 0.77 The rest of the solution is similar to that used for question 5.

Illustrative Example – p.23/26

Compute α_{1/2} and α_{3/2} at the given temperatures and use these to compute split fractions ξ_{1} and ξ_{3} given ξ_{2}: α_{1/2} = 2.28 and α_{3/2} = 0.38 ξ_{1} = 0.88 and ξ_{3} = 0.56

Compute vapor and liquid flowrates

v_{1} = ξ_{1}f_{1} = 0.88 × 30 = 26.4 kmol/hr v_{2} = ξ_{2}f_{2} = 0.77 × 50 = 38.5 kmol/hr v_{3} = ξ_{3}f_{3} = 0.56 × 20 = 11.2 kmol/hr l_{1} = 30 − 26.4 = 3.6 kmol/hr l_{2} = 50 − 38.5 = 11.5 kmol/hr l_{3} = 20 − 11.2 = 8.8 kmol/hr

Compute liquid mole fractions and average relative volatility. x_{1} = 0.15, x_{2} = 0.48, x_{3} = 0.37 ᾱ = 2.28 × 0.15 + 1 × 0.48 + 0.38 × 0.37 = 0.963

Illustrative Example – p.24/26

Compute α_{1/2} and α_{3/2} at the given temperatures and use these to compute split fractions ξ_{1} and ξ_{3} given ξ_{2}: α_{1/2} = 2.28 and α_{3/2} = 0.38 ξ_{1} = 0.88 and ξ_{3} = 0.56

Compute vapor and liquid flowrates

v_{1} = ξ_{1}f_{1} = 0.88 × 30 = 26.4 kmol/hr v_{2} = ξ_{2}f_{2} = 0.77 × 50 = 38.5 kmol/hr v_{3} = ξ_{3}f_{3} = 0.56 × 20 = 11.2 kmol/hr l_{1} = 30 − 26.4 = 3.6 kmol/hr l_{2} = 50 − 38.5 = 11.5 kmol/hr l_{3} = 20 − 11.2 = 8.8 kmol/hr

Compute liquid mole fractions and average relative volatility. x_{1} = 0.15, x_{2} = 0.48, x_{3} = 0.37 ᾱ = 2.28 × 0.15 + 1 × 0.48 + 0.38 × 0.37 = 0.963

Illustrative Example – p.24/26

Compute α_{1/2} and α_{3/2} at the given temperatures and use these to compute split fractions ξ_{1} and ξ_{3} given ξ_{2}: α_{1/2} = 2.28 and α_{3/2} = 0.38 ξ_{1} = 0.88 and ξ_{3} = 0.56

Compute vapor and liquid flowrates

v_{1} = ξ_{1}f_{1} = 0.88 × 30 = 26.4 kmol/hr v_{2} = ξ_{2}f_{2} = 0.77 × 50 = 38.5 kmol/hr v_{3} = ξ_{3}f_{3} = 0.56 × 20 = 11.2 kmol/hr l_{1} = 30 − 26.4 = 3.6 kmol/hr l_{2} = 50 − 38.5 = 11.5 kmol/hr l_{3} = 20 − 11.2 = 8.8 kmol/hr

Compute liquid mole fractions and average relative volatility. x_{1} = 0.15, x_{2} = 0.48, x_{3} = 0.37 ᾱ = 2.28 × 0.15 + 1 × 0.48 + 0.38 × 0.37 = 0.963

Illustrative Example – p.24/26

Compute α_{1/2} and α_{3/2} at the given temperatures and use these to compute split fractions ξ_{1} and ξ_{3} given ξ_{2}: α_{1/2} = 2.28 and α_{3/2} = 0.38 ξ_{1} = 0.88 and ξ_{3} = 0.56

Compute vapor and liquid flowrates

v_{1} = ξ_{1}f_{1} = 0.88 × 30 = 26.4 kmol/hr v_{2} = ξ_{2}f_{2} = 0.77 × 50 = 38.5 kmol/hr v_{3} = ξ_{3}f_{3} = 0.56 × 20 = 11.2 kmol/hr l_{1} = 30 − 26.4 = 3.6 kmol/hr l_{2} = 50 − 38.5 = 11.5 kmol/hr l_{3} = 20 − 11.2 = 8.8 kmol/hr

Compute liquid mole fractions and average relative volatility. x_{1} = 0.15, x_{2} = 0.48, x_{3} = 0.37 ᾱ = 2.28 × 0.15 + 1 × 0.48 + 0.38 × 0.37 = 0.963

Illustrative Example – p.24/26

Compute α_{1/2} and α_{3/2} at the given temperatures and use these to compute split fractions ξ_{1} and ξ_{3} given ξ_{2}: α_{1/2} = 2.28 and α_{3/2} = 0.38 ξ_{1} = 0.88 and ξ_{3} = 0.56

Compute vapor and liquid flowrates

v_{1} = ξ_{1}f_{1} = 0.88 × 30 = 26.4 kmol/hr v_{2} = ξ_{2}f_{2} = 0.77 × 50 = 38.5 kmol/hr v_{3} = ξ_{3}f_{3} = 0.56 × 20 = 11.2 kmol/hr l_{1} = 30 − 26.4 = 3.6 kmol/hr l_{2} = 50 − 38.5 = 11.5 kmol/hr l_{3} = 20 − 11.2 = 8.8 kmol/hr

Compute liquid mole fractions and average relative volatility. x_{1} = 0.15, x_{2} = 0.48, x_{3} = 0.37 ᾱ = 2.28 × 0.15 + 1 × 0.48 + 0.38 × 0.37 = 0.963

Illustrative Example – p.24/26

Compute α_{1/2} and α_{3/2} at the given temperatures and use these to compute split fractions ξ_{1} and ξ_{3} given ξ_{2}: α_{1/2} = 2.28 and α_{3/2} = 0.38 ξ_{1} = 0.88 and ξ_{3} = 0.56

Compute vapor and liquid flowrates

v_{1} = ξ_{1}f_{1} = 0.88 × 30 = 26.4 kmol/hr v_{2} = ξ_{2}f_{2} = 0.77 × 50 = 38.5 kmol/hr v_{3} = ξ_{3}f_{3} = 0.56 × 20 = 11.2 kmol/hr l_{1} = 30 − 26.4 = 3.6 kmol/hr l_{2} = 50 − 38.5 = 11.5 kmol/hr l_{3} = 20 − 11.2 = 8.8 kmol/hr

Compute liquid mole fractions and average relative volatility. x_{1} = 0.15, x_{2} = 0.48, x_{3} = 0.37 ᾱ = 2.28 × 0.15 + 1 × 0.48 + 0.38 × 0.37 = 0.963

Illustrative Example – p.24/26

Check bubble point equation P = ᾱP_{2}^{0}. LHS = 900 mm Hg (guessed value) RHS = 0.963 × 904.07 = 870.62 mm Hg

LHS ≠ RHS and thus the guessed pressure is wrong.

Guess another pressure and iterate.

How would you solve the problem if the pressure was given but the temperature was not given.

Check bubble point equation P = ᾱP_{2}^{0}. LHS = 900 mm Hg (guessed value) RHS = 0.963 × 904.07 = 870.62 mm Hg

LHS ≠ RHS and thus the guessed pressure is wrong.

Guess another pressure and iterate.

How would you solve the problem if the pressure was given but the temperature was not given.

Check bubble point equation P = ᾱP_{2}^{0}. LHS = 900 mm Hg (guessed value) RHS = 0.963 × 904.07 = 870.62 mm Hg

LHS ≠ RHS and thus the guessed pressure is wrong.

Guess another pressure and iterate.

How would you solve the problem if the pressure was given but the temperature was not given.

Check bubble point equation P = ᾱP_{2}^{0}. LHS = 900 mm Hg (guessed value) RHS = 0.963 × 904.07 = 870.62 mm Hg

LHS ≠ RHS and thus the guessed pressure is wrong.

Guess another pressure and iterate.

How would you solve the problem if the pressure was given but the temperature was not given.

Check bubble point equation P = ᾱP_{2}^{0}. LHS = 900 mm Hg (guessed value) RHS = 0.963 × 904.07 = 870.62 mm Hg

LHS ≠ RHS and thus the guessed pressure is wrong.

Guess another pressure and iterate.

How would you solve the problem if the pressure was given but the temperature was not given.

Check bubble point equation P = ᾱP_{2}^{0}. LHS = 900 mm Hg (guessed value) RHS = 0.963 × 904.07 = 870.62 mm Hg

LHS ≠ RHS and thus the guessed pressure is wrong.

Guess another pressure and iterate.

How would you solve the problem if the pressure was given but the temperature was not given.

√ T_{V} , P_{V} specified  Isothermal flash
√ V/F = 0, P_{L} specified Bubble point temperature
√ V/F = 1, P_{V} specified Dew point temperature
√ T_{L} specified, V/F = 0 Bubble point pressure
√ T_{V} specified, V/F = 0 Dew point pressure
Q = 0, P_{V} specified  Adiabatic flash
Q, P_{V} specified  Nonadiabatic flash
√ V/F, P_{V} specified  Percent vaporization flash

Adiabatic and non-adiabatic flash will be covered in the "Energy Balance" lecture.

√ T_{V} , P_{V} specified  Isothermal flash
√ V/F = 0, P_{L} specified Bubble point temperature
√ V/F = 1, P_{V} specified Dew point temperature
√ T_{L} specified, V/F = 0 Bubble point pressure
√ T_{V} specified, V/F = 0 Dew point pressure
Q = 0, P_{V} specified  Adiabatic flash
Q, P_{V} specified  Nonadiabatic flash
√ V/F, P_{V} specified  Percent vaporization flash

Adiabatic and non-adiabatic flash will be covered in the "Energy Balance" lecture.