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    Introduction

    Chemical engineering is broad in scope and has implications that are far reaching. The theories that

    underlie reaction engineering are capable of modeling systems beyond industrial processes. One

    such system that we shall look at, which can be modeled through the use of chemical reactor designequations and some simplifying assumptions, is the digestive system. Specifically the properties of

    a hippopotamus digestive tract will be explored.

    Facts Sheet

    1 Hippopotamus means river horse in Greek.

    2. Male hippos can be up to 15 feet long, 5 feet high, and 8,000 pounds in weight, making them oneof the largest terrestrial mammals. Only elephants and some white rhinos are larger.

    3. At birth a hippopotamus weighs 30 60 kg and it takes 4 to 5 years to reach maturity and a

    weight of 3600 kg.4.Hippo lips are about 2 feet wide.

    5.Hippos can turn each ear in a different direction at the same time.

    6. A bull hippos bellow has been measured at 115 decibels.

    7.Hippos cannot float because their heavy muscles weigh them down and cause them to sink. Thisdoes not present a problem, since they can either paddle to stay afloat or simply walk along the river

    bottom.

    8. Feeding always occurs on land. However, courtship, birth and nursing takes place underwater.9. Adult hippos can stay underwater for five to six minutes. However, baby hippos can only stay

    underwater for twenty seconds.

    10.Few animals can open their mouths as wide as hippos can. They use this ability to scare awayother animals.

    11. Around the turn of the eighteenth century, hippo tusks were used to make artificial teeth.

    12.When hippos fight each other, they use their teeth, toss water at each other, accost each otherwith a variety of sounds and swing their massive heads together.

    13. Although all hippos are herbivores and would rather run away than fight, several hundred

    people a year do not survive a close hippo encounters. In addition, hippos are more deadly than

    many feared predators, including lions.14.An angry hippo can run much faster than a human; they have been clocked in shorting running

    dashes at 30 mph.

    15.Mother hippos often form nurseries for up to 40 young hippos on sandy beaches near the water.When traveling, females keep the youngsters in a single-file line behind them.

    16. Mother hippos have been known to kill lions and bite crocodiles in half.

    17. It is speculated that gases produced during digestion from fermentation are passed out through ahippos nostrils.

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    Background

    Background on the Hippopotamus.

    Within the family hippopotamide which is a subclass of artiodactyla (who are distinguishedas having a plane of symmetry passing through the third and fourth digit of their foot, such as

    hippos, giraffes, and camels) there exist two different species of hippos. One is named Choeropsis

    liberiensis, which is the common pygmy hippo. An adult pygmy hippo weighs around 250 kg and

    can be found mostly in forested areas. The pygmy hippo is the smaller and less social version oftheir relative the Hippopotamus amphibius, or more commonly the river hippopotamus, which from

    this point on will be referred to generically as a hippo. Hippos are known to live in herds of up to

    forty animals, and an adult hippo can weigh as much as 3600 kg. Hippos can be found in the lakes,rivers and streams of central and sub-Saharan Africa. Hippos are amphibious and are

    excellent swimmers. In fact calves are born in the water and must immediately go to the surface to

    breathe, and they even nurse underwater.Both species are similar in body type. They have barrel shaped bodies and broad square

    mouths. To protect and as a preventative measure against the sun and dehydration, they secrete a

    pinkish substance known as blood sweat. Historically this secretion was named as such because it

    was believed that it was the hippos blood. However, this has since been shown to be false. Ahippos nostrils, eyes, and ears are located on the top of their heads so that upon submersion these

    organs can remain above the water7.

    The river hippopotamus spends a majority of its time during the day in the water, andemerges in the evening for terrestrial foraging. Also, they eat aquatic plants, which they obtain by

    diving and swimming. A hippos terrestrial foraging is far reaching; they often travel a few hundred

    meters where they spend approximately five hours a night intensely grazing. This has led to thembeing blamed for crop damage in regions such as the Elephant Marsh in Africa 5. Hippos are strictly

    unselective grazers. Their blunt teeth, flat snout and wide straight lips enable them to graze close to

    the soil surfaces, which means besides eating grass, roots and fruit they digest a quantity of soil 4.

    Hippos do not ruminate, which means that they do not chew their cud. They have threechambered stomachs consisting of the parietal blind sac, the stomach, and the glandular stomach 4.

    Their stomachs are designed to efficiently derive nutrition from the lower-energy foods off which

    they exist. Within the complex structure of the hippos stomach microbial fermentation takes place,which is followed by catalytic digestion in the small and large intestine which are of similar size

    and structure. The microbial fermentation of ingested material before catalytic fermentation classify

    hippos (along with cows, sheep, goats, and kangaroos) as foregut fermenters, as opposed to hingutfermentors where catalytic fermentation proceeds microbial fermentation (as demonstrated by

    horses, rhinos, rabbits, and koalas) 6.

    The only natural predators of adult hippos are humans who seek them for their ivory tusks.However, hyenas, lions, leopards, and crocodiles predate calves. The river hippopotamus is no yet

    endangered, but due to habitat loss and hunting their populations are declining.

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    Background on modeling the digestive system as chemical reactors.

    As a first approximation in the analysis of the digestive system, the system shall be assumedto operate at steady state conditions. However, in the actual case digestive reaction rates may be

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    affected by changes in temperature, in pH, and in the composition of the microbial community.Also, to apply the design equations for ideal reactors, variations in volume due to the absorption of

    digestive products are assumed to be negligible 2.

    The rate of digestion in the complex stomach can be modeled as either an enzyme catalyzed

    digestion or an autocatalytic microbial fermentation. For the autocatalytic digestion microbes break

    down the feed and in the process produces more microbes.MPMA ++

    or more precisely

    besmore_microPmicrobes

    A +

    The rate law is of the following form where CA is the concentration of feed and CB is the

    concentration of microbes:

    )day*3m

    kg(,

    AC

    MK

    BC

    AkC

    AMr

    +

    =

    Letting X be the conversion of feed in the stomach, in kg converted per kg feed, the function thencan be expressed as follows:

    X)(1AO

    CM

    K

    X)

    M

    Y

    B

    X)((1

    k2AOCAMr +

    +

    =

    This function can then be written as:

    eXd

    )2CX-bX(ak'

    AMr

    ++

    =

    A plot of the reciprocal rate as a function of conversion is shown below:

    1

    -rA

    X

    If on the other hand the digestion takes place by the animals own enzymes.EPA*EEA ++

    The rate law would be of the form:

    ACKm

    AkC

    Ar

    +

    =

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    This process is referred to as a catalytic digestion. This rate law expressed in the terms ofconversion then would have the following form:

    X)b(11

    X)a(1

    X)(1AO

    CKm

    X)(1AO

    kC

    Ar

    +

    =

    +

    =

    The reciprocal of the reaction rate as a function of the conversion is of the form:

    1

    -rA

    X

    We now could ask the question, for a fixed stomach volume, V, what would be the most efficient

    system to convert the food into products when:

    a) the digestion is autocatalyticb) the digestion is catalytic

    c) catalytic followed by autocatalytic (hindgut fermentors)

    d) autocatalytic followed by catalytic (foregut fermentors)In most cases, where there is a complex stomach, digestion occurs as described by case (c) or (d).

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    Graph showing the performance of a CSTR with respect to an Autocatalytic Reaction

    Graph showing the performance of a PFR with respect to an Autocatalytic Reaction

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    Graph showing the performance of a PFR and a CSTR with respect to an Autocatalytic Reaction

    To answer the question of what would be the most efficient system to convert food intoproducts (as in vitamins and nutrients) the assumption that the reactor scheme with the fastest

    throughput time is desirable needs to be made. In the case of a purely autocatalytic reaction as long

    as the conversion is kept low a CSTR will be more efficient than a PFR. This difference inefficiency is easily seen when the performance of a CSTR and PFR are graphed against each other

    with the autocatalytic reaction. Ideally, in the digestive system a maximum reaction rate is

    desirable. Thus, conversion will be kept low and a CSTR will be desired. Physiologically this isalso true because within a stomach upon exiting the feed is acidified which inhibits the fermentation

    process.

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    Graph showing the performance of a CSTR with respect to a Catalytic Reaction

    Graph showing the performance of a PFR with respect to a Catalytic Reaction

    Graph showing the performance of a CSTR and a PFR with respect to a Catalytic Reaction

    However, for the case of a catalytic reaction when the performance of a CSTR and PFR iscompared, the PFR is always more efficient than the CSTR. Once again this is seen best by

    graphing the performance of a PFR against a CSTR with catalytic digestion 2.

    Combining the two cases of a purely catalytic reaction and a purely autocatalytic reaction it

    is easily seen which ideal reactor is desired. For any point within the digestive system when an

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    autocatalytic reaction is occurring a CSTR will be more efficient than a PFR, but when a catalyticreaction occurs a PFR is more efficient than a CSTR. Hence, in the case of the hippopotamus where

    autocatalytic digestion occurs within the stomach, which is followed by catalytic digestion within

    the intestines, a CSTR-PFR reactor scheme is desired. It should be noted at this point that in the

    following development of the reactor modeling catalytic digestion is assumed in order to provide a

    first approximation for the behavior of the hippos digestive system.

    Design equations and calculations

    The hippopotamus digestive system is modeled as a CSTR and a PFR in a series. The continuous-

    stirred tank reactor(or the backmixreactor) is used very commonly in industrial processing and is

    normally run at steady state. Ideally it is operated to obtain a perfect mixing and therefore modeled

    as having no spatial variations in concentrations, temperature, or reaction rate throughout the vessel.

    Theplug-flow rectoris, as the CSTR, run at steady state and consists of a long cylindrical pipe. The

    flow is considered turbulent enough so that one can assume that there is no radial variation in

    concentration.

    For the system above A represents the grass that makes up the bulk of the hippos normal diet. F is

    the mass flow of A and X represents the conversion of A into proteins, vitamins, minerals andeverything else that the hippo needs to survive. The conversion X is defined as:

    0

    1A

    A

    C

    CX =

    A simple mole balance over the CSTR gives the following expression:

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    1010 XFFF AAA =

    or in term of the reaction rate:

    110 VrFF AAA =

    These two equation can be combined into a familiar expression, a form of the design equation of a

    CSTR:

    1)(Eq.101

    A

    A

    r

    XFV

    =

    In a similar way a mole balance over the PFR gives us the following expressions:

    1020 XFFF AAA =

    )A

    Ar

    dV

    dF=

    which can be combined to the design equation of the PFR:

    AAr

    dV

    dXF =0

    or in this case the more useful form:

    2)(Eq.

    2

    1

    02 =X

    X A

    A

    r

    dXFV

    Finding correct values of X is of course impossible but fairly good estimates can be done. The

    overall conversion of all dry matter, i.e. X2, is according to studies of the hippos digestive system

    about 45%. We further assumed that about 75% of the total conversion occurred in the first part of

    the digestive system, i.e. the CSTR or the stomach, and 25% in the second part, i.e. the PFR or the

    intestines. The assumption is based on the volume ratio between the two parts. Studies made of the

    stomach contents of the hippo are not extensive enough to make any final conclusions but at least

    the protein contents in different parts of the stomach are in concord with the aforementionedassumption.

    In brief this means that when assuming that no conversion occurs before the CSTR and that 75% of

    the total conversion occurs in the CSTR the values of X will be as follows:

    45.034.00210=== XXX

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    The volumes of the stomach (CSTR) and the intestines (PFR) are about 0.46 m3 respectively 0.15

    m3 , the density of the grass is 306 kg / m3 and the flow rate is assumed to be 40 kg / day based on

    what is known about the hippos feeding habits and diet. This means that the volumetric flow rate is

    about 0.13 m3 / day.

    The space time (or the mean residence time), which is the time necessary to process one reactor

    volume of fluid based on entry conditions, is defined as:

    v

    V=

    combining this definition with the previously developed design equations for the CSTR (1) and

    PFR (2) yields: since

    3)(Eq.101

    A

    A

    rXC

    vV

    =

    4)(Eq.2

    1

    02 =

    X

    X A

    A

    r

    dXC

    v

    V

    In order to be able to solve these equations one needs to know how rA varies with X. Assuming

    that the digestive reactions are catalyzed only be the hippos own enzyme both in the CSTR and the

    PFR the Michaelis-Menton expression can be used:

    EPEAEA

    kk

    k ++

    ++

    21

    1

    )1(

    )1(

    0

    0maxmax

    1

    21

    2

    XCK

    XCvr

    CK

    Cvr

    Ck

    kk

    CCkr

    AM

    A

    A

    AM

    A

    A

    A

    AE

    A

    +

    =

    +

    =

    +

    =

    +

    +

    +

    Since Vmax, KM and CA0 all are constants this expression can be rewritten as:

    5)(Eq.)1(''1

    )1('

    Xk

    XkrA +

    =

    When combining equation 5 with equations 3 and 4 two equations which can be solved are finally

    obtained:

    6)(Eq.)1('

    ))1(''1(

    1

    1101

    Xk

    XkXC

    r

    XC

    v

    VAO

    A

    A

    +

    =

    =

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    ( ) 7)(Eq.'

    ''

    1

    1ln'

    1

    '

    '')1ln(

    '

    1

    '

    '')1ln(

    '

    1

    '

    '')1ln(

    '

    1

    '

    ''

    )1('

    1

    )1('

    )1(''1

    12

    2

    1

    0

    11220

    2

    1

    0

    2

    1

    0

    2

    1

    0

    2

    1

    0

    2

    +

    =

    ++

    =

    +=

    +

    =

    +=

    =

    XXk

    k

    X

    X

    kC

    Xk

    kX

    kX

    k

    kX

    kC

    Xk

    kX

    kCdX

    k

    k

    XkC

    dXXk

    XkC

    r

    dXC

    v

    V

    A

    A

    X

    X

    A

    X

    X

    A

    X

    X

    A

    X

    X A

    A

    Solving equations 6 and 7 for k and k yields:

    0740.1'49.16'' ACkk ==

    Equations 5,6 and 7 can now be written as:

    5)(Eq.)1(49.161

    )1(74.1AOA

    CX

    Xr

    +

    =

    6)(Eq.)1(74.1

    ))1(49.161(

    1

    111

    X

    XX

    v

    V

    +

    =

    ( ) 7)(Eq.48.91

    1ln74.1

    112

    2

    12XX

    X

    X

    v

    V+

    =

    Using equation 5 rA/CA0 can be plotted against X. As X reaches full conversion, i.e. nothing of A is

    left, the reaction rate goes towards zero.

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    A more common way of showing this is however to plot CA0 / -rA against X (compare with the

    charts in the background section).

    Finally by plotting rA against CA one can see how the reaction rate increases with respect to the

    concentration.

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    Problems

    1. One day a very competitive baby hippo follows his mother and tries to keep up with her rate of

    ingestion of food. However, the conversion of the ingested food is only 40% of the mothers. How

    old is the baby hippo?

    References

    1. Fogler, H Scott.Elements of Chemical Reaction Engineering. 3rd Ed,

    Prentice Hall PTR, Upper Saddle River, NJ, USA, 19992. Deborah L. Penry, and Peter A. Jumars. Modeling Animal Guts as Chemical Reactors.

    American Naturalist, 1987, Vol. 129, pp 69-96

    3. Boris D. Abaturov, Fekadu Kassaye, German V. Kutznetsov, Magomed-Rasul D.

    Magomedov and Dmitry A. Petelin. Nutritional estimates of populations of some wild free-

    ranging African ungulates in grassland (Nechisar national park, Ethiopia) in dry season.,

    Ecography, 1995, Vol. 18,pp 164-172

    4. E.T. Clemens, and G.M.O. Malioy. The digestive physiology of three East African

    herbivores: the elephant, rhinoceros and hippopotamus,Journal of Zoology, 1982, Vol.

    198, pp 141-156

    5. Francis X. Mkanda, and Brighton Kumchedwa, Relationship between crop damage by

    hippopotamus ( Hippopotamus amphibius L.) and farmer complaints in the Elephant Marsh,Journal of African Zoology, 1997, Vol. 111, pp 27-38

    6. Eric D. Carlson, and Alice P. Gast. Animal Guts as Ideal Reactors, Chemical Engineering

    Education, Winter 1998, pp 24-29

    7. Animal Diversity Web Home Page:http://www.oit.itd.umich.edu

    The University of Michigans Museum of Zoology, Mi, USA, 1997

    http://www.oit.itd.umich.edu/http://www.oit.itd.umich.edu/http://www.oit.itd.umich.edu/