Multiple steady states: the branch depends on which way you arrived

Type a small vessel model and one feed temperature. Over one interval there are three steady solutions, two of them stable, and the setting alone does not say which one you are on.

Live result

At a feed of 300.000 K three solutions exist, and arriving from below keeps the cold one.

Cold solution300.084 K
Middle solution384.644 K
Hot solution549.802 K

The middle solution is the boundary between the other two rather than a place anything sits.

Two quantities that meet, and one boundary

Try a record

Feed temperature300K

Arrival

A declaration of how this setting was reached, not an instruction.

Edit the vessel model

Rate factor number

The rate constant this reaction would reach at an unreachable temperature, written as a number and a power of ten.

Rate factor power of ten

The power of ten the number above is multiplied by.

Activation temperatureK

The activation energy divided by the gas constant, in kelvin.

Residence timemin

The mean time a parcel of feed spends in the vessel.

Adiabatic riseK

How far the temperature would rise on complete conversion with no cooling at all.

Heat removal number

The cooling duty relative to the feed's own heat capacity.

Coolant temperatureK

The coolant temperature, which must sit inside the range you declared below.

Vessel range fromK

It must be wide enough to hold both stable solutions.

Vessel range toK

Exact feed temperatureK

Six decimal places, so a jump setting can be typed exactly.

Algebraic root count: 3

Every solution

Cold

Vessel temperature: 300.084212 K
Conversion: 0.000337
Heat made: 0.168424 K
Heat taken: 0.168424 K
Stable: yes
An operating point: yes

Middle

Vessel temperature: 384.643663 K
Conversion: 0.338575
Heat made: 169.287325 K
Heat taken: 169.287325 K
Stable: no
An operating point: no

Hot

Vessel temperature: 549.801983 K
Conversion: 0.999208
Heat made: 499.603965 K
Heat taken: 499.603965 K
Stable: yes
An operating point: yes

Every solution
Solution	Vessel temperature	Conversion	Heat made	Heat taken	Stable	An operating point
Cold	300.084212 K	0.000337	0.168424 K	0.168424 K	yes	yes
Middle	384.643663 K	0.338575	169.287325 K	169.287325 K	no	no
Hot	549.801983 K	0.999208	499.603965 K	499.603965 K	yes	yes

The two jump settings

Upward jump feed temperature

Value: 382.432210 K

Vessel temperature at the upward fold

Value: 354.486605 K

Downward jump feed temperature

Value: 113.865542 K

Vessel temperature at the downward fold

Value: 436.185169 K

Width between jump settings

Value: 268.566668 K

Branch this arrival keeps

Value: Cold branch

The two jump settings
The two jump settings	Value
Upward jump feed temperature	382.432210 K
Vessel temperature at the upward fold	354.486605 K
Downward jump feed temperature	113.865542 K
Vessel temperature at the downward fold	436.185169 K
Width between jump settings	268.566668 K
Branch this arrival keeps	Cold branch

The feed temperature you are setting

FromK

ToK

In steps ofK

It does not check the values in between.

Across the values sampled, the stable branch this arrival keeps stays Cold branch from 100 K, the lowest value sampled to 380 K. The nearest value where it changes is 400 K, where it becomes Hot branch.

The answer	Cold branch
Values sampled	100 K to 400 K, 16 values, step 20 K
Holds from	100 K, lowest sampled
Holds to	380 K
Nearest value that changes it	400 K
The answer there	Hot branch

What an unchanged region does and does not prove is set out in a short lesson.

The sweep samples only the values you entered. Between samples the answer is not checked, so read the boundary as observed, not proven.

Here is the maths

The formula, in words

The reaction rate rises steeply with temperature, so the fraction converted rises from almost nothing to almost everything across a narrow band. The heat made follows it as an S shaped curve between zero and the adiabatic rise. The heat taken is a straight line in vessel temperature whose height falls as feed temperature rises. Every equality is a solution of the steady balance. The line can meet the curve once or three times. The middle equality has the balance sloping the wrong way, so a small departure grows rather than dies. The two feed temperatures where solutions appear or cease to exist come from exact tangency.

A worked example

At a feed of 300.000000 K, solving heat made minus heat taken equals zero gives roots at 300.084212 K, 384.643663 K and 549.801983 K. $Q_g=Q_r$ , where $Q_g$ is the heat made and $Q_r$ the heat taken.
At the cold root, conversion to nine decimals is 0.000336848, so heat made is $500 \times 0.000336848 = 0.168424$ K, equal to the reported heat taken.
At the middle root, conversion to eight decimals is 0.33857465, so heat made is $500 \times 0.33857465 = 169.287325$ K, equal to the reported heat taken.
At the hot root, conversion to eight decimals is 0.99920793, so heat made is $500 \times 0.99920793 = 499.603965$ K, equal to the reported heat taken.
The cold and hot roots are stable. The middle root is unstable and is not an operating point.
Exact tangency gives the upward jump at a feed of 382.432210 K and vessel temperature 354.486605 K, and the downward jump at a feed of 113.865542 K and vessel temperature 436.185169 K.
The interval between the jump settings is 382.432210 minus 113.865542: $382.432210-113.865542=268.566668$ K.

What the middle solution is and is not

It is a temperature at which the two quantities are equal. It is not a place a vessel sits, it is not a state a vessel passes through on the way to another, and there is no calculation here that places anything at it. The chart gives it a boundary line and no point. If the arrival declaration changes, the selection ring is drawn afresh at the other stable root in the same frame.

What this does not model

It solves the small model you typed and nothing else. It does not determine that a real vessel is well mixed, that its parameters have been identified, or that there is no unmodelled phase, side reaction, controller, delay or heat transfer limit.

Numbers and precision

This continuous model uses ordinary floating point, reports every temperature to six decimals, and treats a feed setting within 0.00000075 K of a derived fold as being at that fold. Both figures are the calculation's declared numeric contract.

Check it yourself

Set the sweep from 378 to 386 in steps of 1 with arrival from below. Nine values are checked. The answer holds through 382 and changes at 383, while the exact setting is 382.432210. Then set arrival from above and sweep from 110 to 118 in steps of 1. The answer holds through 113 and changes at 114, while the exact setting is 113.865542. The two settings are 268.566668 kelvin apart on the same model.

The calculation domain

Rate factor from 1e−12 to 1e30 per minute, activation temperature from 1,000 to 50,000 K, residence time from 0.001 to 100,000 minutes, adiabatic rise from 1 to 2,000 K, heat removal number from 0 to 20, and a searched range at least 100 K wide between 150 and 1,500 K.

Source anchors

University of Colorado Boulder LearnChemE, multiple steady states in a stirred tank with heat exchange. Hosted continuous stirred tank module on multiple steady state behaviour. Both published pages set out the multiple solution structure this page solves, and the second describes the separate arrival directions. Every parameter on this page comes from your own typing.

Calculation version

1.0.0

The whole record prints. Your browser can save it as a PDF instead of printing it.

It does not establish that a real vessel is well mixed, that its parameters are known, or that nothing outside this model is happening in it. This page will never turn the middle solution into a path, animate a passage through it, or present either jump setting as a command, an intervention or a safety measure. It describes where two quantities are equal in the model you typed. It determines neither a course of action nor safety.

Vessel equality record