This principle of conservation of mass is extremely useful. It means that if the amount of a pollutant somewhere (say, in a lake) increases, then that increase cannot be the result of some ``magical'' formation out of nowhere. The pollutant must have been either carried into the lake from elsewhere or produced via chemical reaction from other compounds that were already in the lake. And, if chemical reactions produced the mass increase in our pollutant, they must also have caused a corresponding decrease in the mass of some other compounds. Thus, conservation of mass allows us to compile a budget of the mass of our pollutant in the lake. This budget keeps track of the amounts of pollutant entering the lake, leaving the lake, and the amount formed or destroyed by chemical reaction. This budget can be balanced for a given time period, similar to the way you might balance your checkbook:
Note that each term of this equation has units of mass. This form of
balance is most useful when there is a clear beginning and end to the
balance period, so that 
is meaningful. For example, in a
checkbook balance 
is usually one month. In environmental
problems, however, it is usually more convenient to work with values
of mass flux---the rate at which mass enters or leaves a
system. To develop an equation in terms of mass flux, the mass
balance equation is divided by 
to produce an equation with
units of mass per unit time. Dividing
equation 12 by 
and moving the
first term on the right (mass at time t) to the left hand side yields
the following equation.
Note that each term in this equation has units of mass/time. The
left hand side of equation 13 is equal to 
. In the limit as 
, this becomes
, the rate of change of pollutant mass in the lake. We will
refer to 
as the accumulation rate of the pollutant. As
, the first two terms on the right side of
equation 13 become the mass flux into the lake and
the mass flux out of the lake. The last term of
equation 13 is the rate of chemical
production or loss. To stress the fact that each term in the new
equation refers to a flux or rate, we will use the symbol
to refer to a mass flux with units of mass/time. The
equation for mass balances is then
Equation 14 is the governing equation for the mass balances we will work with in this course.
In the remainder of this section, we will examine the importance of carefully defining the region over which the mass balance is applied and discuss the terms of equation 14. We will then present examples of the main types of situations for which mass balances are useful.
A mass balance is only meaningful in terms of a specific region of
space, which has boundaries across which the terms 
and
are determined. This region is called the
control volume. In our derivation of the mass balance equation, we
have referred to the mass of pollutant in a lake and the fluxes of
pollutant into and out of the lake---that is, we have used a lake as
our control volume. Theoretically, any volume of any shape and
location can be used. Realistically, however, certain control volumes
are more useful than others. The most important attribute of a
control volume is that it have boundaries over which you can calculate
and 
.
A well-mixed tank is an analogue for many control volumes used in environmental engineering. For example, in our lake example it might be reasonable to assume that pollutants dumped into the lake are rapidly mixed throughout the entire lake. In environmental engineering and chemical engineering, the term Continuously Stirred Tank Reactor, or CSTR is used for such a system. An example of a CSTR is shown in Figure 1. We will use a mass balance for a control volume which encloses the CSTR in Figure 1 as an example to describe the meaning of each term in equation 14.
The mass accumulation rate is, by definition, 
, or 
. The total mass in the CSTR cannot usually be measured.
For example, if the CSTR represented an entire lake, measuring the
total pollutant mass would require analyzing all of the water in the
lake. However, our assumption that the CSTR is well-mixed means that
this is not necessary. If the tank is well-mixed, then the
concentration of our pollutant is the same everywhere in the tank, and
we need only to measure the concentration in a sampling from the tank.
Using concentration units of (mass)/(volume), the total pollutant mass
in the tank is equal to 
, where V is the volume of the
CSTR. Thus, the accumulation rate is equal to
Here, we have made the assumption that the volume of the CSTR is
constant. This is usually a reasonable assumption for liquids,
although it may not always be valid for gases. However, 
will
always be equal to 
.
Mass balance problems can be divided into those that are in steady
state and those that are non-steady state. A steady-state situation
is one in which things do not change with time---the incoming
concentration and flow rate are constant, the outgoing flow rate is
constant, and therefore the concentration in the control volume is
constant. For steady-state systems, then, 
.
Non-steady-state conditions result whenever flows start or stop, or
when the concentration in an incoming stream changes. For
non-steady-state situations, 
is nonzero.
The example in Figure 1 includes one pipe entering the CSTR. We will again use concentration measured in mass/volume units to calculate the flux entering the CSTR through the pipe.
Often, we know the volumetric flow rate, Q, of each input stream.
For the example of Figure 1 the pipe has
a flow rate of 
, with corresponding pollutant
concentration of 
. The mass flux is
then given by
If it is not immediately clear how one knows that 
gives a
mass flux, consider the units of each term:
If the volumetric flow rate is not known, it may be calculated from
other parameters. For example, if the fluid velocity v and the
cross-sectional area A of the pipe are known, then 
.
Another way to describe the flux is in terms of a flux density
J times the area through which the flux occurs. J has units of
, and we will study it in more detail when we
cover diffusion in section 3. This
type of flux notation is most useful at interfaces where there is no
fluid flow, such as the interface between the air and water and the
surface of a lake.
The flux out of the CSTR is similarly equal to the product of volumetric flow rate in the exit pipe times the concentration in the exit pipe. Since the CSTR is well-mixed, the concentration in the liquid leaving the CSTR is equal to the concentration inside the CSTR. It is conventional to refer to the concentration within the CSTR simply as C. Thus,
The term 
refers to the net rate of
production of our pollutant from chemical reactions, in units of
mass/time. Thus, if other compounds react to form our pollutant,
will be greater than zero; if our pollutant reacts
to form some other compounds, resulting in a loss of the pollutant,
will be negative. Production or loss of a
compound by a chemical reaction is usually described in terms of
concentration, not mass. So it is necessary to multiply the chemical
rate of change of concentration by the volume of the CSTR to
obtain units of mass/time:
There are a number of possibilities for the form of 
, and
the resulting 
. The most common include:
, implying that 
also.
-order decay. The rate of loss of the pollutant
is constant. For a pollutant with 
-order decay, 
and 
.
-order decay. The rate of loss of the pollutant
is directly proportional to its concentration: 
. For
such a pollutant, 
.
is greater than zero. Examples of this type of situation
will be given in Part III of the course.
Reactor Analysis refers to the use of mass balances to analyze pollutant concentrations in a control volume which is a chemical reactor. Do not let the term ``reactor'' fool you, however. The reactor can be any control volume we want it to be. So the term reactor analysis is used to describe the application of the mass balance process to environmental situations also. Reactor analyses can be divided into two types: CSTRs (Continuously Stirred Tank Reactors) and PFRs, or Plug Flow Reactors. We have defined CSTRs already---they are simply well-mixed tanks which are used to model well-mixed environmental reservoirs. Plug Flow Reactors are essentially pipes, and they are used to model things like rivers, in which fluid is not mixed in the upstream-downstream direction.
In this section, we will present examples of the types of situations CSTRs are used to model. Plug Flow Reactors are described and used in examples in the following section. Example 2.1 demonstrates the use of CSTR analysis to determine the concentration of a substance resulting from the mixing of two or more influent flows. This type of calculation will be used again in the third part of this course to determine the initial BOD loading in a river downstream of a sewage outflow. Examples 2.2 through 2.4 refer to the tank in Figure 1 and demonstrate steady-state and non-steady-state situations with and without first-order chemical decay. Calculations completely analogous to those in examples 2.2, 2.3, and 2.4 can be used to determine the concentration of sewage pollutants exiting a treatment reactor, the rate of increase of pollutant concentrations within a lake resulting from a new pollutant source, and the period required for pollutant levels to decay from a lake or reactor once the source is removed.
.
. Mixing Problem
A sewage pipe from a wastewater treatment plant discharges
1.0 m
/s of effluent containing 5.0 mg/l of phosphorus compounds
(reported as mg P/l) into a river with an upstream flow rate of
25 m
/s and a phosphorus concentration of 0.010 mg P/l (see
Figure 2). What is the resulting concentration
of phosphorus in the river downstream of the sewage outflow, in units
of mg/l?
) and,
second, the downstream phosphorus concentration (
). But
first, we must select a control volume. To ensure that the input and
output fluxes cross the control volume boundaries, the control volume
must cross the river upstream and downstream of the sewage outlet and
must cross the sewage pipe. The selected control volume is shown in
Figure 2 as a dotted line. We must assume that
the control volume extends down the river far enough that the sewage
and the river water become well-mixed before leaving the control
volume. As long as that assumption is met, it makes absolutely no
difference to our analysis how far downstream the control volume
extends.
Before beginning our analysis, we should determine whether this is a steady-state or non-steady-state problem, and whether the chemical reaction term will be nonzero. Since the problem statement does not refer to time at all, and it seems reasonable to assume that both the river and sewage have been flowing for some time and will continue to flow, this is a steady-state problem. Sewage does participate in chemical and biological reactions. However, we are interested here in mixing---that is, in what concentration results right after the two flows mix. So we will assume that the mixing occurs instantly, without sufficient time for any reactions to occur.
? We will conduct a mass balance on the
total river-water mass. In this case, the ``concentration'' of river water
in mass/volume units is simply the density of the water, 
.
where the term 
has been set to zero because we are
ignoring chemical reaction. Since this is a steady-state problem,
. Therefore, as long as the density 
is constant,
, or 
.
? Again, we have steady-state with no
chemical formation or decay:
Plugging in, we find that
.
. Steady-state CSTR with 1
st-order
Decay
The CSTR shown in
Figure 1 is used to treat an
industrial waste, using a reaction which destroys the waste according
to first-order kinetics: 
, where
. The reactor volume is 500 m
, the
volumetric flow rate of the single inlet and exit is 50 m
/day, and
the inlet waste concentration is 100 mg/l. What is the
outlet concentration?
A single, constant outlet concentration is asked for and all problem
conditions are constant. Therefore, this is a steady-state problem.
The mass balance equation is
Solving for C, we find that
The numerical solution is
.
. Non-steady-state CSTR, Conservative
Substance
The CSTR shown in Figure 1 is used with
a conservative substance. The reactor is filled with clean water
before it is started. After starting, waste containing a 100 mg/l of
a pollutant is added at a flow rate of 50 m
/day. The volume of
the reactor is 500 m
. What is the concentration exiting the
reactor as a function of time after it is started?
Because of the extra term on the right (
), this equation
cannot be immediately solved in the way that
example 2.4 was solved. However, if we make a
change of variables, we can make the form of this equation similar to
that of example 2.4. Let 
. Since
is constant, 
.
Therefore, the last equation above is equivalent to
Rearranging and integrating,
which yields
If we now substitute 
for y, we obtain
The second equation is obtained using the observation that 
,
since the tank is started clean. Rearranging, we can obtain
This is the solution to the question posed in the problem statement.
Note what happens as 
: 
\
and 
. This is not surprising, since this is a
conservative substance. If we run the reactor for a long enough
period, the concentration in the reactor will eventually reach the
inlet concentration. Using the equation we have derived for C as a
function of time, we could determine how long it would take for
the concentration to reach, say, 90% of the inlet value.
.
. Non-steady-state CSTR with 1
-order
Decay
The manufacturing process that generates the waste in
example 2.2 has to be shut down, and, starting at t=0, the
concentration 
entering the CSTR is set to 0. What is the
outlet concentration as a function of time after the concentration is
set to 0? How long does it take the tank concentration to reach 10%
of its initial, steady-state value?
which yields
Since 
is equal to 
,
and, exponentiating both sides,
Plugging in the values from the problem, with 
equal to the
steady-state solution of 32 mg/l yields
? At the time when 
, we have
Taking the natural logarithm of both sides,
The Plug Flow Reactor (PFR) is used to model the chemical transformation of compounds as they are transported in ``pipes.'' The ``pipe'' may represent a river, a region between two mountain ranges through which air flows, or a variety of other conduits through which liquids or gases flow. Of course, it can even represent a pipe. A schematic diagram of a PFR is shown in Figure 3.
As fluid flows down the PFR, the fluid is mixed in the radial
direction, but mixing does not occur in the axial direction---each
plug of fluid is considered a separate entity as it flows down
the pipe. However, as the plug of fluid flows downstream, time
passes. Therefore, there is an implicit time dependence even in
steady-state PFR problems. However, because the velocity of the
fluid in the PFR is constant, time and downstream distance are
interchangeable: 
. We will use this observation together with
the mass balance formulations we have worked with already to determine
how pollutant concentrations vary during flow down a PFR.
To develop the equations which describe pollutant concentration in the plug of fluid as it flows down the PFR, we will conduct a mass balance on a control volume which encloses a section of the PFR of infinitesimally small thickness dx, as shown in Figure 4. Since the thickness is small, we can assume that the fluid in that region of the PFR is well-mixed. The mass balance equation for this control volume is
We have set 
equal to zero, indicating that this is a
steady-state problem. We are assuming here that conditions at a given
location in the PFR are constant. Concentrations can still vary along
the PFR, however.
Noting that the volume of our control volume is given by 
, dividing by dx, and rearranging, we obtain
In the limit as 
, the left hand side becomes the
derivative 
, so we obtain
As discussed earlier, 
can take a variety of forms,
depending on the type(s) of chemical reaction that are occurring.
. From
equation 20, this means that 
---there is no
variation of concentration along the pipe. (Of course, this result is
obvious, since if there is no reaction the fluid is just moving
along the pipe without changing in any way.)
. Plugging into
equation 20 for this case, we obtain
which can be integrated as follows:
Therefore, for a PFR of length l,
where the volume of the PFR, V, is equal to the length times time area.
Equation 26 describes the way in which concentration
decreases during passage down a PFR with loss via a first-order
reaction. Note that, since the time which passes during transport
down the PFR is equal to 
, equation 26 is
equivalent to
which is the solution to the differential equation which describes the
loss of a pollutant by first-order kinetics: 
.
That is, in a plug flow reactor time and distance are interchangeable,
and the concentration at any location in the PFR may be calculated simply
by determining the chemical decay during the time it took to reach
that location.
The CSTR and the PFR are fundamentally different. When a parcel of fluid enters the CSTR, it is immediately mixed throughout the entire volume of the CSTR. In contrast, each parcel of fluid entering the PFR remains separate during its passage through the reactor. This difference results in differing behavior. We will look at these differences for one special case: the continuous addition of a pollutant to each reactor, with destruction of the pollutant within the reactor according to first-order kinetics. The two reactors are shown in Figure 5.
We will assume that the incoming concentration (
), the flow
rate (Q), and the first-order reaction rate constant (k) are given
and are the same for both reactors. Then, we will consider two common
problems: (1) if we know the volume V (same for both reactors), what is the
resulting outlet concentration (
)? and (2) if we need a
specified outlet concentration, what volume of reactor is required?
Table 2 summarizes the results of this
comparison.
The results shown in Table 2 indicate
that, for equal reactor volumes, the plug flow reactor is more
efficient that the CSTR and, for equal outlet concentrations, a
smaller PFR is required. Why is this? The answer has to do with the
fundamental difference between the two reactors. In a PFR, each and
every molecule spends the same amount of time in the reactor; that
period is equal to 
. Since first-order decay occurs according to
, the concentration in each parcel of fluid
entering the reactor drops by this amount. In contrast, in a CSTR
there is no single amount of time that each small parcel of fluid spends in
the reactor. Some parcels may spend a long time mixing around inside
the CSTR; other parcels may, by chance, reach the exit in a relatively
short time. Since all these parcels are mixed together and result in
a single outlet concentration, an average value of 
\
results.
To see why that average value is higher than the corresponding value for a
PFR, consider what happens when 
is equal to 2,
approximately the value in the first example of
Table 2. Then,
. This is the value of 
\
that would result in the PFR. Let's assume that we can model the
mixing in the CSTR by splitting the fluid entering the CSTR
into two parcels. The first parcel remains in the CSTR only one
quarter of the time a parcel would take to pass through the PFR, while
the second parcel remains in the CSTR four times as long as it would
in the PFR. (So the average time spent in the CSTR by the two parcels
is the same as the time spent in the PFR---both are equal to 
.)
The concentration in the first parcel when it reaches the CSTR exit is
determined by its value of kt, which is 4 times larger than the
value for the PFR: 
. The concentration in the second
parcel is reduced less, because it spends a shorter time in the
reactor: 
. The actual concentration in the exit of the CSTR in this
situation would be the average of the concentrations in the two
parcels, so 
.
Thus, the resulting value of 
for the CSTR is higher
than that for the PFR (0.30 versus 0.14), even though the average
residence time is the same for both reactors. The reason for this is
illustrated in Figure 6, and results from the fact
that concentration decays exponentially with time for a first-order
reaction. Thus, the parcel that spends a shorter period of time in
the CSTR exits with a concentration that is increased significantly
relative to the PFR. However, the parcel that spends a longer period
in the CSTR exits with a concentration that is decreased only a small
amount (again, relative to the PFR).
.
. Required Volume in a PFR
Determine the
volume required for a PFR to obtain the same degree of pollutant
reduction as the CSTR in example 2.2. Assume that
the flow rate and first-order decay rate constant are unchanged
(
, 
).
. From
equation 26,
Solving for V, we obtain
As expected, this volume is smaller than the 500 m
required for
the CSTR in example 2.2.
A number of
terms are used to describe the average period spent in a given
reactor. The terms retention time, detention time, and
residence time are all used to refer to 
, the average
period spent in the reactor. This parameter has units of time. As
discussed above, for a plug flow reactor the retention time is
actually the time spent in the reactor. However, for a CSTR the
retention time is the average period spent in the reactor.
The reciprocal of the retention time, 
, has units of inverse
time---the same units as a first-order rate constant. This value is
sometimes referred to as the exchange rate.
.
. Retention Time in CSTR and PFR
Calculate the retention times in the CSTR of
example 2.2 and the PFR of
example 2.5.
For the PFR in example 2.5,
.
. Retention Times for the Great
Lakes
Calculate the retention times for Lake Michigan and Lake
Ontario using the data given in
Table 3.
. Note
that we are assuming that the lakes can be modeled as CSTRs (or PFRs,
but a CSTR makes more sense for a lake). This assumption is not far
off for retention times significantly longer than one year.
Modern society is dependent on the use of energy. Such use requires transformations in the form of energy and control of energy flows. For example, when coal is burned at a power plant, the chemical energy present in the coal is converted to heat, which is then converted in the plant's generators to electrical energy. Eventually, the electrical energy is converted back into heat for warmth or used to turn motors. However, energy flows and transformation are also the cause of environmental problems. Thermal heat energy from electrical power plants can result in increased temperature in rivers used for cooling water; ``greenhouse'' pollutants in the atmosphere alter the energy balance of the earth and may cause significant increases in global temperatures in the future; and many of our uses of energy are themselves associated with emissions of pollutants.
We can keep track of the movement of energy and changes in its form using energy balances, which are analogous to the mass balances we discussed in the previous section. We can do this because of the law of conservation of energy which states that energy can neither be produced nor destroyed. (Conservation of energy is sometimes referred to as the first law of thermodynamics.) As long as we consider all the possible forms of energy, there is no term in energy balances which is analogous to the chemical reaction term in mass balances. That is, we can treat energy as a conservative substance.
The forms of energy can be divided two types: internal energy and external energy. Energy which is a part of the molecular structure or organization of a given substance is internal. Energy which results from the location or motion of the substance is external. Examples of external energy include gravitational potential energy and kinetic energy. Gravitational potential energy is the energy gained when a mass is moved to a higher location above the earth. Kinetic energy is the energy which results from the movement of objects. When a rock thrown off of a cliff accelerates toward the ground, the sum of kinetic and potential energy is conserved (neglecting friction)---as it falls it loses potential energy, but increases in speed, gaining kinetic energy. Examples of some common forms of energy are given in Table 4.
Heat is a form of internal energy. It results from the random motions of atoms. Heat is thus really a form of kinetic energy, although it is considered separately. When you heat a pot of water, you are adding energy to the water. That energy is stored in the form of internal energy, and the change in internal energy of the water is given by
where C is the heat capacity or specific heat of the water, with
units of [energy]/([mass][temperature]). Heat capacity is a property of
a given material. For water, the heat capacity is 1 BTU/(
), or 4184 J/(
).
Chemical internal energy reflects the energy in the chemical bonds of a substance. This form of energy is composed of two parts:
In analogy with the mass balance equation (equation 14), we will use the following equation to conduct energy balances:
We will illustrate the use of this relationship with some examples.
.
. Heating water
A 40-gallon electric water heater is used to heat tap water
(temperature 50
F (10
C)). The heating level is set to
the maximum level while several people take consecutive showers. If,
at the maximum heating level, the heater uses 5 kW of electricity, and
the water use rate is a continuous 2 gallons/min, what will be the
temperature of the water exiting the heater? Assume that the system
is at steady-state and that the heater is 100% efficient; that is, it
is perfectly insulated and all of the energy used goes to heat the
water. 
Each term of this equation is an energy flux, and has the units of (energy/time). To solve, we need to use the same units in each term. We will use the definition of watts: watts are defined as Joules/s. In addition, we need to convert the water flow rate (gallons/min) to mass of water per unit time, using the density of water. Combining the first and third terms we obtain
which is a cold shower! (You may have foreseen this answer if you
have ever taken a shower after the hot water in the tank was used up
by previous showerers.)
.
. Heating water
From the previous example,
we see that if one wants a hot shower, it is necessary to wait until
the water in the tank can be reheated. How long would it take the
temperature to reach 130
F (54
C) if no hot water were
used during the heating period and the water temperature started at
20
C?
We will solve this for 
, given that 
is equal to 
C.
The two previous examples related to the controlled conversion and
transfer of energy for a beneficial use. However, the use of energy
for heat always results in some loss to the environment due to
imperfect insulation, resulting in higher energy use or less heating
than one would calculate. In addition, the second law of
thermodynamics states that it is impossible to convert heat energy to
work with 100% efficiency. Conversion of heat to work is essentially
what is done in the generator of an electric power plant, and as a
result a significant fraction of the energy released from fuel
combustion is lost during the conversion. Modern large power plants
convert fuel energy to electricity with an overall efficiency in the
30--35% range. The next example looks at what happens to the heat
energy that is not converted to electricity. Finally,
example 2.11 considers the implications of
another aspect of the burning of fossil fuels in power plants,
vehicles, and for heating. Whenever fossil fuels are burned, carbon
atoms in the fuel are converted to carbon dioxide (
) and
released into the atmosphere. As a result of this process, the
concentration in the atmosphere is increasing at a rate of
about 1 ppmv/year. Carbon dioxide contributes to the greenhouse
effect, which is considered in example 2.11.
.
. Thermal Pollution.
A 1000 MW
(
W) power plant is located next to a river and uses
cooling water from the river to remove its waste heat. What is the
resulting increase in river temperature? (The power plant has an
overall efficiency of 33%. Assuming that all of the waste heat from
the power plant is removed with cooling water and added to the adjacent
river. The river flow rate is 100 m
/s.
). The heat energy added
to the river is the amount that is not converted to electricity, or
(3000-1000=2000 MW). We can now write our energy balance over the
region of the river to which the heat is added. We will use 
\
to represent the temperature of the water upstream, and 
to
represent the temperature after heating:
Rearranging, we obtain
The remainder of this problem is basically a problem of unit
conversions. To obtain 
requires multiplication of the
given river volumetric flow rate by the density of water
(1000 kg/m
). We also use the heat capacity of water, 
C. Thus,
.
. Earth's Energy Balance and the
Greenhouse Effect.
The global average surface temperature of the
earth is determined by a balance between the energy added to the earth
by the sun and the energy radiated away by the earth to space.
Greenhouse gases, both natural and anthropogenic (or, human-affected),
affect this energy balance. In this example, we will calculate the
global average temperature without greenhouse gases and show the
effect which greenhouse gases have on the earth's energy
balance.
1 yr, it is reasonable to
assume that the system is in steady state, so our energy balance is
simply
The energy flux in is equal to the solar energy intercepted by the
earth. At the earth's distance from the sun, the sun radiates
342 W/m
. We will refer to this value as S. The earth intercepts
an amount of energy equal to S times the cross-sectional area of
the earth: 
. However, because the earth reflects
part of this energy back to space, 
is equal to only 70%
of this value:
The second term, 
, is equal to the energy radiated to
space by the earth. The energy emitted per unit surface area of the
earth is given by Boltzmann's Law:
where 
is Boltzmann's constant, equal to
. To obtain 
, we
multiply this value by the total surface area of the earth, 
. (We use the total surface area of the sphere here because
energy is radiated away from the earth during both day and night.)
We can now solve our energy balance by setting 
equal to
.
Plugging in the values for S and 
and taking the fourth
root yields an average temperature of T=255
K, or -18
C.
This is too cold! In fact, the globally averaged temperature at the
surface of the earth is much warmer: 287
K. The reason for the
difference is the presence of gases in the atmosphere that absorb the
infrared radiation emitted by the earth and prevent it from reaching
space. We neglected these gases in our energy balance. However, if
we denote the energy flux absorbed and retained by these gases by
, we can then correct our value for 
:
The reduction in 
which results from greenhouse gas
absorption is sufficient to cause the higher observed surface
temperature. Clearly, this is largely a natural phenomenon---surface
temperatures were well above 255
K long before people began
burning fossil fuels. The main natural greenhouse gas is water vapor.
However, increasing atmospheric concentrations of carbon dioxide and
other gases emitted by human activities are increasing the value of
. So far, this increase is approximately 2 W/m
\
averaged over the entire earth, and projections indicate that the
increase could be as high as 5 W/m
over the next 50 years.
According to our energy balance, this increase in 
is
expected to result in an increase in the globally averaged
temperature. (There is considerable uncertainty in the precise value
of the resulting increase, however, due to a number of complexities
that we have not considered.)
.
.
A pond is used to treat sewage wastewater before the liquid is
discharged into a river. The inflow to the pond is sewage at a flow
rate of 
and with a BOD concentration of
. The volume of the pond is 20,000 m
. The
purpose of the pond is to allow time for the decay of BOD to occur
before discharge into the environment. BOD decays in the pond with a
first-order rate constant equal to 0.25/day. What is the BOD
concentration at the outflow of the pond, in units of mg/l?
answer: 11 mg/l
.
.
For each of the following problems, would you use a steady-state or
non-steady-state mass balance to obtain a solution? For each
situation, also indicate whether the compound for which you would
conduct a mass balance is conservative or non-conservative. Give
an explanation for each of your answers. (You do not need to
actually solve these problems.)
) emissions from fossil fuel burning are mixed
throughout the atmosphere. Assume that these emissions
are mixed immediately throughout the entire atmosphere, and
that 
does not degrade chemically
If you know the total emission rate of carbon dioxide and the
volume of the atmosphere, what would be the rate of increase
of atmospheric carbon dioxide levels in ppm/yr?
-order decay rate constant of the perfume is very
slow relative to the amount of time it takes to mix fresh air
through the room.)
answer: (a) non-steady state
.
.
A mixture of two gas flows is used to calibrate an air pollution
measurement instrument. The calibration system is shown in
Figure 7.
If the calibration gas concentration 
is 4.90 ppmv, the
calibration gas flow rate 
is 0.010 l/min, and the total gas
flow rate 
is 1.000 l/min, what is the concentration of
calibration gas after mixing (
)? (Assume that the
concentration upstream of the mixing point is zero.)
answer: 49.0 ppbv
.
.
You are in an old spy movie, and have been locked into a small room
(volume 1000 ft
). You suddenly realize that a poison gas has just
started entering the room through a ventilation duct. Recognizing the
type of poison from its smell, you know that if the gas reaches a
concentration of 100 mg/m
, you will die instantly, but that you
are safe as long as the concentration is less than 100 mg/m
. If
the ventilation air flow rate in the room is 100 ft
/min and the
incoming gas concentration is 200 mg/m
, how long do you have to
escape?
answer: 6.9 minutes
.
.
Sewage waste is added to a stream
through a discharge pipe. The river flow rate upstream of the
discharge point is 
. The discharge occurs at a
flow of 
and has a BOD concentration of 50.0 mg/l.
Assuming that the upstream BOD concentration is negligible
(a) What is the BOD concentration just downstream of the discharge point?
(b) If the stream has a cross sectional area of 10 m
, what would
the BOD concentration be 50 km downstream? (BOD is removed with a
first-order decay rate constant equal to 0.20 day
)
answer: (a) 4.7 mg/l. (b) 4.2 mg/l
.
.
(a) Calculate the hydraulic residence times (the retention time) for Lake
Superior and for Lake Erie using the data in
Table 3.
(b) Assume that both lakes currently are polluted with the same
compound at a concentration which is 10 times the maximum acceptable
level. If all sources of the compound are removed, how long will it
take the concentration to reach acceptable levels in each lake?
Assume that the pollutant does not decay chemically.
(c) Comment on the significance of your answers.
answer: (a) Lake Superior: 179 years; Lake Erie: 3 years. (b) Lake Superior: 412 years; Lake Erie: 6 years.
.
.
How many watts of power would it take to
heat 1 liter of water (weighing 1.0 kg) by 10
C in 1.0 hour?
Assume that no heat losses occur, so that all of the energy expended
goes into heating the water.
answer: 12 Watts
