- 64 Views
- Uploaded on
- Presentation posted in: General

Pulping and Bleaching PSE 476/Chem E 471

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Pulping and BleachingPSE 476/Chem E 471

Lecture #12

The H Factor

PSE 476: Lecture 12

- Train question
- Derivation of “H” factor equation
- Example “H” factor calulation
- Validity of “H” factor

PSE 476: Lecture 12

- Consider you are the engineer on this train. Your windows are painted shut and you need to travel from Seattle to exactly the station in Portland. How do you do it????????
- If all you have is your speedometer and your trusty watch, you could record your speed at specified time intervals and plot your progress. This is basically what is done in a pulp mill to determine when to stop the pulping reaction. This is accomplished through the use of the “H” factor.

PSE 476: Lecture 12

(1)

dx

= k f (composition) = rate of lignin removal

dt

(2)

x(t) = ƒ k dt

In order to determine when to stop a kraft cook, it is necessary

to know the extent of the reaction which is based on the rate of

lignin removal. This can be expressed as:

f (composition) = [lignin]a[OH-]b[HS-]c[phase of the moon]d

Within a given cook, there is assumed to be a unique

relationship between the extent of the reaction and the

composition so the top relationship can be integrated to give:

Cannot solve equation because k is

dependent on temperature

PSE 476: Lecture 12

Ea = activation energy(32 kcal/mole (kraft))

T = absolute temperature

R = gas constant

A = constant

-Ea/RT

(3)

k = A e

Taking the log of both sides:

(4)

ln k = ln A - (Ea/RT)

At 100°C, the above equation becomes:

(5)

ln k100 = ln A - (Ea/R373)

Subtracting equation 5 from 4 gives:

ln(k/k100) = -(Ea/RT) + (Ea/373R)

PSE 476: Lecture 12

(5)

ln(k/k100) = -(Ea/RT) + (Ea/373R)

Substituting in the appropriate values

(6)

ln kr = -(16,113/T) + 43.2

kr is called the relative rate constant or a comparison of the

rate constant at a temperature to that at 100°C

(43.2-16,113/T)

(7)

kr = e

From this equation, it is possible to see that kr is significantly

affected by temperature (see figure on page7)

PSE 476: Lecture 12

PSE 476: Lecture 12

(-16,113/T)

(43.2)

kr = e

e

(-16,113/T)

k = Ae

(43.2)

(10)

Where p = A/e

k = (p)(kr)

Equation 7 can be rewritten

(8)

Equation 4 becomes:

(9)

Combining these 2 equations leads to:

PSE 476: Lecture 12

Equation 10 can be substituted into equation 2 leaving:

(11)

ƒ k dt = p ƒ kr dt

(p is a constant)

The expression ƒ kr dt is referred to as the “H” factor:

(12)

H = ƒ kr dt

By combining equations 2, 11, and 12, it can be seen that the

extent to which a pulping reaction has proceeded is a function

of the H factor.

x(t) = (p)(H) or x = f(H)

PSE 476: Lecture 12

In order to solve for the factor, temperature readings are taken every 0.25 hours (or sooner - every minute) of the cook and relative rate constants kr determined. The kr is plotted versus time. The area under the curve is equivalent to the H factor (Figure slide 11). Sample calculations for the determination of the H factor can be found in slide 12. The accuracy of this method can be seen in slide 13 for the determination of endpoint at 3 different temperatures. It needs to be stressed that this equation only estimates the effects of time and temperature and assumes constant effective alkali, sulfidity, liquor/wood, wood species, etc. All these factors and more can change the rate,

PSE 476: Lecture 12

H factor equal to

Area under this

Curve

PSE 476: Lecture 12

PSE 476: Lecture 12

PSE 476: Lecture 12

As mentioned previously, H factor is used to determine the time required at a given EA and sulfidity to reach a desired kappa. This figure shows the effect that changing the active alkali has on the H factor required to reach a kappa.

PSE 476: Lecture 12