Levenspiel plot: mixed flow is the rectangle and plug flow the area

Scale

The four fixed panels share one scale, 0.0 to 80.0 percent across and 0.00 to 11.00 m³ up, and it does not move. The live chart above rescales with what you type.

Basis

One reference reactant, at the conversions and rates you typed, with the volumes read between the first and last of them

Model version

1.0

The formula, in words

• The vertical axis is already a volume. $\frac{F_{A0}}{-r_A}$ is $\frac{\mathrm{kmol}}{\mathrm{h}} \div \frac{\mathrm{kmol}}{\mathrm{m}^3\cdot\mathrm{h}}$, which is $\mathrm{m}^3$. Every height on the plot is a volume, so every area and every rectangle on it is a volume too, and nothing on this page needs converting after that one division.
• Plug flow is the area, summed as trapezoids. Between two consecutive typed points the volume is $\frac{\Delta X_i(g_i+g_{i+1})}{2}$. The whole ideal plug flow volume is those contributions added up, and each one is printed in its own row so the total can be checked a line at a time.
• Mixed flow is a rectangle on the exit height. $\Delta X \times g_n$. One height, used everywhere, because an ideal mixed vessel has one state and the exit is it. Where the first typed conversion is not zero, the width is the change and not the conversion.
• The straight segments are the method, not a drawing choice. The trapezoid rule joins consecutive points with straight lines, so the shaded region is exactly the region that was summed. A smoothed curve would shade an area the arithmetic never computed, and this page draws no curve it did not add up.
• The ratio is $\frac{V_M}{V_P}$ at one target on one curve. It rises with the target when the height rises with conversion, it falls below one when the height falls, and it has no consistent direction at all on a curve that turns. It is a consequence of the points, and the page reports it without ranking anything.

A worked example

The reaction and the one transformation

A reference reactant is fed at 60.0 $\frac{\mathrm{kmol}}{\mathrm{h}}$. Its disappearance rate has been measured at five conversions and typed in. Every rate becomes a height by one division.

The rate falls as conversion rises, so the height rises. Nothing else about the reaction is known to the page and nothing else is needed.

Case A, the full span, 0.0% to 80.0%

Four trapezoids, each 0.200 wide in conversion.

Ideal plug flow volume is $3.70\ \mathrm{m}^3$. Ideal mixed flow volume is $0.800 \times 10.00 = 8.00 \mathrm{m}^3$. Rectangle minus area is $8.00 - 3.70 = 4.30 \mathrm{m}^3$. Volume ratio is $8.00 \div 3.70 = 2.16\times$.

43.24% of the plug flow volume buys the last twenty points of conversion, and it is the same mechanism that makes the rectangle large: the exit is where the rate is slowest, and both ideals pay for that, but only the mixed vessel pays for it across the whole span.

Case B, the same curve stopped at 60.0%

Drop the last point. Four points, three intervals, everything else identical.

Plug flow is $0.45 + 0.65 + 1.00 = 2.10 \mathrm{m}^3$. Mixed flow is $0.600 \times 6.00 = 3.60 \mathrm{m}^3$. The gap is $1.50\ \mathrm{m}^3$ and the ratio is $1.71\times$.

The ratio fell from $2.16\times$ to $1.71\times$ and not one typed figure changed except the target. The rectangle now stands on 6.00 instead of 10.00, and the interval that carried 43.24% of the volume is simply not being bought.

Case C, the same five rates in the other order

Conversions are unchanged. The rates are 6.0, 10.0, 15.0, 24.0 and 30.0, so the heights are 10.00, 6.00, 4.00, 2.50 and 2.00.

Ideal plug flow volume is $3.70\ \mathrm{m}^3$, identical to Case A to the last digit. Ideal mixed flow volume is $1.60\ \mathrm{m}^3$. Rectangle minus area is $-2.10 \mathrm{m}^3$ and the volume ratio is $0.43\times$. The mixed flow volume in Case A divided by this one is exactly 5.00.

Same feed, same span, same five rates, same four intervals. One ideal volume did not move and the other moved by a factor of five. The four trapezoids are the same four trapezoids read backwards, so their sum cannot change; the rectangle depends on one height and that height went from 10.00 to 2.00.

Case D, a rate that dips and recovers

The rates are 6.0, 15.0, 30.0, 15.0 and 6.0, so the heights are 10.00, 4.00, 2.00, 4.00 and 10.00.

Ideal plug flow volume is $4.00\ \mathrm{m}^3$. Ideal mixed flow volume is $8.00\ \mathrm{m}^3$ and the volume ratio is exactly $2.00\times$.

The ratio lands between Case A at $2.16\times$ and Case C at $0.43\times$, and it says nothing at all about where a stage boundary belongs, because the curve it was read from goes down and then up. The largest interval is the first 1.40 interval, not the second.

The partial inlet case

Take Case A's last four points only, so the first typed conversion is 20.0%. The heights are 2.50, 4.00, 6.00 and 10.00.

Plug flow is $0.65 + 1.00 + 1.60 = 3.25 \mathrm{m}^3$. Mixed flow is $0.600 \times 10.00 = 6.00 \mathrm{m}^3$ and the ratio is $1.85\times$.

The rectangle's width is the conversion change, 0.600, and not the conversion, 0.800. Using the conversion would give $8.00\ \mathrm{m}^3$ and would be a claim about a span the visitor did not type.

Check it yourself

Type 15.0 into all five rate rows. Every height becomes $4.00\ \mathrm{m}^3$, both ideal volumes become $3.20\ \mathrm{m}^3$, the ratio reads $1.00\times$ and the two shapes land exactly on top of each other, because a flat height means every slice costs the same and there is nothing for the area and the rectangle to disagree about. Now change the last row's rate on its own, from 15.0 to 30.0. The area moves a little, from 3.20 to 3.00, because only the one interval that touches that point changed. The rectangle moves in full, from 3.20 to 1.60, because it stands on that height and on nothing else, and the ratio reads $0.53\times$. One typed figure, two very different consequences, and that difference is the page.

What this does not model

Kinetics: there is no rate law, no order, no rate constant and no temperature anywhere on this page, and the typed points are the only description of the reaction it has. Heat, in any form, which is the mechanism most likely to make a real rate curve turn and the reason Case D is a scenario rather than a curiosity. Pressure drop along a bed or a tube, which can suppress conversion the plot has no way to show. Real mixing, which is neither of the two ideals, and the residence time methods that describe it. Transport to a catalyst surface or across a phase boundary, and the resistances in series that a measured rate already contains without saying so. Selectivity and side products, so a volume that fits the reference reactant can miss the thing you actually wanted. Catalyst decay, so a rate curve measured once is not a rate curve that lasts. Staging and sequence, which need the whole curve and the volumes already on the ground rather than one interval at a time. Every property a real design needs, none of which this page holds. Cost, safety, operability and control, none of which appear in an area comparison. And measurement error in the typed points, which the arithmetic cannot tell apart from a real feature of the curve.