Engineering, 21.04.2020 16:30 camp591

A relatively nonvolatile hydrocarbon oil contains 4.0 mol % propane and is being stripped by direct superheated steam in a stripping tray tower to reduce the propane content to 0.2%. The temperature is held constant at 422 K by internal heating in the tower at 2.026 × 105 Pa pressure. A total of 11.42 kg mol of direct steam is used for 300 kg mol of total entering liquid. The vapor–liquid equilibria can be represented by y = 25x, where y is mole fraction propane in the steam and x is mole fraction propane in the oil. Steam can be considered as an inert gas and will not condense. Plot the operating and equilibrium lines and determine the number of theoretical trays needed.

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Answer from: Levantine2643

The number of theoretical trays needed is 20.

From the given data,

Equilibrium relation

y = 25x

$L_s=L_2(1-x_2)\\L_s=300(1-0.04)= 288Kmol$

Applying total balance equation

$G_sy_1+L_sx_2=G_sy_2+L_sx_1\\G_s(y_1-y_2)=L_s(x_1-x_2)\\x_1=\frac{0.002}{1-0.002}\\x_1=0.002\\y_1=0, y_2=?\\x_2=\frac{0.04}{1-0.04}; x_2=0.0417$

substituting the values into the equation;

$11.42(0-y_2)=288(0.002-0.0417)\\0-11.42y_2=-11.4048\\y_2=0.998$

The numbers of trays

$N=\frac{In[(\frac{(x_2-y_1)/m}{(x_1-y_1)/m}(1-A)+A }{In(1/A)}\\$

But $A=\frac{L_s}{mGs}$

$N=\frac{x_2-x_1}{(x_1-y_1)/m}=\frac{0.0417-0.002}{0.002}\\N=19.85$

The numbers of tray is approximately 20.

learn more about steam stripping and numbers of tray here

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A relatively nonvolatile hydrocarbon oil contains 4.0 mol % propane and is being stripped by direct

Answer from: lackin2056

Number of Trays = Six (6)

Explanation:

Given that: y' = 25x' , in terms of molecular ratio, we can write it as

$\frac{Y'}{1 + Y'} =25 \frac{X'}{1 + X'}$ ......... 1

after plotting this we get equilibrium curve as shown in the attached picture.

inlet concentration and outlet concentration of liquid phase is

x₂ = 4% = 0.04 (inlet)

so that can be converted into molar

$X_2 = \frac{x_2}{1-x_2} = \frac{0.04}{1-0.04} = 0.04167$

and

x₁ = 0.2% = 0.002

$X_1 = \frac{x_1}{1-x_1} = \frac{0.002}{1-0.002} = 2.004*10^{-3}$

Now we have to use the balance equation a

$\frac{G_s}{L_s} = \frac{X_2-X_1}{Y_2-Y_1}$ .............. a

here amount of solute is comparably lower than

Here we have

L = 300 kmol (total)

L_s = 300(1 - 0.04) = 288 kmol pure oil

G = G_s = 11.42 kmol

Y_1 = 0 , solvent free steam

substitute into the equation a

$\frac{11.42}{288} = \frac{0.04167 - 2*10^{-3}}{Y_2 - 0}$

Y₂ = 1.0003

Now plot the point A(X₁ , Y₁) and B(X₂ , Y₂) and join them to construct operating line AB.

Starting from point B, stretch horizontal line up to equilibrium curve and from there again go down to operating line as shown in the picture attached. This procedure give one count of tray and continue the same procedure up to end of operating.

at last count, the number of stage, gives 6.

∴ Number of trays = 6

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