qustion solution
Solution to Tutorial 4
Given:
F = 1000 kg-mole/hr, xF = 0.20 (ethanol is MVC)
Feed is saturated liquid, thus q = 1.0 xD = 0.80, xB = 0.02 (maximum); R = 5/3
First, plot the equilibrium curve using the VLE data given. Note that you need to convert mole% ethanol into mole fraction. Then, apply the McCabe-Thiele method to find the number of theoretical (equilibrium) trays required for the separation. Step 1:
Since R and xD are known (5/3, and 0.80 respectively), plot the ROL equation - it passes through xD on the 45o diagonal, i.e. the point (0.80, 0.80).
The intercept is 0.80 / (5/3+1) = 0.30.
Step 2:
Plot the feed line using the q-line equation with q = 1.0 and xF = 0.20 - it passes through xF on the 45o diagonal.
In this case, with q = 1.0, the q-line is a vertical line at xF = 0.20. Then locate its intersection with ROL.
Step 3:
Plot the SOL, by joining the point xB = 0.02 on the 45o diagonal to the intersection point between the ROL and q-line.
Step 4:
Draw triangles starting from xD (between ROL and equilibrium curve) down to xB (between SOL and equilibrium curve). The change from ROL to SOL occurs when xF is reached.
[ Back on Top ]
From the graph plotted, number of triangles = 15. Thus, number of equilibrium stages required = 14 + 1 reboiler. Optimum feed plate location = stage #13.
The distillate and bottoms flowrates (D and B) can be obtained using material balances:
Total: F = D + B ; thus, B = 1000 - D
MVC: F xF = D xD + B xB
Substituting for B gives: ( 1000 ) ( 0.20 ) = ( D ) ( 0.80 ) + ( 1000 - D ) ( 0.02 )
Solving gives: D = 230.77 kg-mole/hr
And B = 769.23 kg-mole/hr [ Back on Top ]
Analysis of Design:
It is noted that the Rectifying Section has more trays (12) than the Stripping Section (3, including the feed tray but excluding the reboiler). This is because the relative volatility of ethanol decreases sharply with increase in ethanol concentration.
The relative volatility is very high
Given:
F = 1000 kg-mole/hr, xF = 0.20 (ethanol is MVC)
Feed is saturated liquid, thus q = 1.0 xD = 0.80, xB = 0.02 (maximum); R = 5/3
First, plot the equilibrium curve using the VLE data given. Note that you need to convert mole% ethanol into mole fraction. Then, apply the McCabe-Thiele method to find the number of theoretical (equilibrium) trays required for the separation. Step 1:
Since R and xD are known (5/3, and 0.80 respectively), plot the ROL equation - it passes through xD on the 45o diagonal, i.e. the point (0.80, 0.80).
The intercept is 0.80 / (5/3+1) = 0.30.
Step 2:
Plot the feed line using the q-line equation with q = 1.0 and xF = 0.20 - it passes through xF on the 45o diagonal.
In this case, with q = 1.0, the q-line is a vertical line at xF = 0.20. Then locate its intersection with ROL.
Step 3:
Plot the SOL, by joining the point xB = 0.02 on the 45o diagonal to the intersection point between the ROL and q-line.
Step 4:
Draw triangles starting from xD (between ROL and equilibrium curve) down to xB (between SOL and equilibrium curve). The change from ROL to SOL occurs when xF is reached.
[ Back on Top ]
From the graph plotted, number of triangles = 15. Thus, number of equilibrium stages required = 14 + 1 reboiler. Optimum feed plate location = stage #13.
The distillate and bottoms flowrates (D and B) can be obtained using material balances:
Total: F = D + B ; thus, B = 1000 - D
MVC: F xF = D xD + B xB
Substituting for B gives: ( 1000 ) ( 0.20 ) = ( D ) ( 0.80 ) + ( 1000 - D ) ( 0.02 )
Solving gives: D = 230.77 kg-mole/hr
And B = 769.23 kg-mole/hr [ Back on Top ]
Analysis of Design:
It is noted that the Rectifying Section has more trays (12) than the Stripping Section (3, including the feed tray but excluding the reboiler). This is because the relative volatility of ethanol decreases sharply with increase in ethanol concentration.
The relative volatility is very high