Conservation Principles in Engineering Mechanics

 2000, W. E. HaislerConservation of Mass1

ENGR 211

Principles of Engineering I

(Conservation Principles in Engineering Mechanics)

Conservation of Mass

• Multicomponent Systems
• Rate Equation
• Multicomponent Rate Expressions
• Multiple Inlets and Outlets
• Steady State
• Multiple Unit Processes
• Recycle and Bypass

We consider conservation of mass in the general conservation statement and write:

where

We can write a more compact mass conservation statement as

We can make the statement even more compact by defining the following:

and

so that conservation becomes

Multicomponent Systems

Conservation of mass applies to the total mass of a multicomponent system and for each chemical component (or species).

Consequently, for a system that contains n species we can write n species conservation statements that must sum to give total mass conservation.

For species i mass conservation, we write

We can write n of the above equations (i=1,…,n). Summing up the individual conservation statements for all n species gives:

The summation of the mass components (species) must equal the total mass. Thus, we write

, , etc

and the top equation becomes

For the case where there is no massenergy conversion, then there is no net generation of total mass so that

and the total mass conservation equation becomes

In summary, for a mass system with n components (species), we have n species conservation equations:

i = 1, …, n

and 1 total mass conservation equation:

Example: Ten pounds mass (lbm) of air is placed in a cylinder. Air consists of a mixture of 77% nitrogen (by mass) and 23% oxygen (by mass). If 5 lbm of air is removed from the cylinder and then 3 lbm of oxygen is added, what is the resultant composition of the gas mixture in the cylinder?

The system is considered to be the contents of the cylinder. The time period is whatever time is required to remove the air and add the pure Oxygen (O2).

At the beginning of the time period, we have 10 lbm of air

which is composed of 7.7 lbm of N2 (0.77 x 10) and

2.3 lbm of O2 (0.23 x 10).

At the end of the time period, we have an unknown amount of air and unknown amounts of N2 and O2 .

Applying conservation to the oxygen and nitrogen components, we can tabulate the following known and unknown information:

Species

lbmlbmlbmlbmlbmlbm

______

OxygenX-0.23(10)=3-0.23(5)+0-0

NitrogenY-0.77(10)=0-0.77(5)+0-0

______

TotalZ-10=3-5+0-0

Solving the O2 equation gives X== 4.15 lbm O2

Solving the N2 equation gives Y== 3.85 lbm N2

Solving the total mass equation gives Z== 8 lbm air

Consequently we have complete total mass and species mass balance (conservation).

Species

lbmlbmlbmlbm

______

Oxygen4.15-2.3=3-1.15

Nitrogen3.85-7.7=0-3.85

______

Total8-10=3-5

Notice that the total mass at the end (8 lbm) consists of a mixture of 4.15/8 = 52% oxygen and 48% nitrogen. It is no longer “standard” air.

Rate Equation

Rather than consider finite time periods (tbeg to tend), it is useful to consider rate equations and then integrate over the time period.

Suppose we have the rate of mass entering: . Then the total amount of mass entering is the integral of the mass rate from tbeg to tend:

.

We can similarly write the total system mass in rate form: and integrate it over the time period to obtain the obtain the mass accumulation in the system (total change over the time period):

With these ideas, consider the mass conservation statement written previously:

In the above equation, we now replace the accumulation term on the left and all the other terms by their integral forms to obtain:

Removing the integrals provides the mass conservation statement for a differential time period:

So the rate form of the mass conservation equation becomes (divide by dt):

In words: Accumulation rate = input rate - output rate

+ generation rate - consumption rate

We could have obtained the integral form in a different way. Start with the mass conservation statement written previously:

Write the terms on the right hand side in terms of rates:

where . Now divide the above equation by to obtain

Now: = accumulation rate

Substitute above into the mass conservation statement to get:

Multiplying the above by dt gives the mass conservation law for a differential period of time

The above equation can be integrated from tbeg to tend to obtain the integral form of the mass conservation law

Example: Consider a problem wherein mass is generated and

consumed at a certain rate during a time dependent process. The mass flow rates and generation and consumption rates are given by the following:

where a= a constant = 1 (hr-1). We want to determine the time required to accumulate 4 pounds of mass in the system.

Tabulating each term in the mass rate equation gives

lbm/hrlbm/hrlbm/hrlbm/hrlbm/hr

______

=+

Integrating the rate equation from tbeg to tend gives

The left side of the above equation is the accumulation of mass in the system from tbeg to tend. We let the beginning time be 0. The desidred accumulation is 4 pounds so we write

The above equation gives (recall a= 1 hr-1)

and solving for tend gives t = 1.61 hr

Multicomponent Rate Equations

Previously we wrote the conservation equations for multicompoent systems in integral form as

In summary, for a mass system with n components (species), we have n species conservation equations:

i = 1, …, n

and 1 total mass conservation equation:

We can write the above in rate form following the same ideas introduced earlier.

In summary, for a mass system with n components (species), we have n species conservation equations in rate form:

i = 1, …, n

and 1 total mass conservation equation in rate form:

Example: Silica gel (a solid) is used as a drying agent to remove water from processed food. The processed food must be below a certain water content before it can be shipped.

1)Determine the amount of silica get required per pound mass of wet food if the food is initially 30% water (by weight) and must be reduced to 20% water. It is assumed that 3.2 lbm of silica gel has the capacity to absorb 1 lbm of water.

2)Find the time required to absorb the water if the silica gel absorbs water at the rate where t is time in minutes, a = 1 min-1 and b=0.13 lbm/min.

Schematically, we have the following for the beginning and ending states.

In order to determine the process required on a per pound of food basis, we assume we have 1 pound of food to begin with. The mass of the dried food is unknown so we let be an unknownx. The wet food is considered to be the system. We set up the mass rate table for the beginning and ending states:

Species

lbmlbmlbmlbm

______

Dry Food0.80x-0.70(1)=0-0

Water0.20x-0.30(1)=0-?

______

Totalx-1=0-?

From the dry food equation, we see that x = 0.70/0.8 = 0.875 lbm (there will be 0.875 lb of food with 20% water content at the end). Substituting x into the water equation, we see that 0.125 lbm of water must be extracted by the silica gel.

We know that 3.2 lbm of silica gel has the capacity to absorb 1 lbm of water. So the amount of silica gel required to abosrb 0.125 lbm of water is:

lbm of silica gel =

= 0.4 lbm silica gel

To determine the time required to absorb 0.125 lb of water, we write the rate equation for water in the wet food system:

The gen/con term has been omitted. The water input term is zero. Multiply the above by dt and integrate from 0 to t (where is the time required to reduce water content in the food to 20%).

or

Note: the negative accumulation means the amount of water in the food decreases with time. The integral term represents the rate of absorption of water by the silica gel. We know that the silica gel absorbs water at the rate of . So we have the equation:

Solving for t from the above equation gives

Multiple Inlets and Outlets

Steady State

Steady state means that the state of system is steady; it does not change with time. Alternately, one can say that the accumulation of mass within the system is zero. Mathematically, steady state is described by

or

Note that steady state implies the following:

and for no net generation:

Example: The feed to a continuous flash distilliation consists of 25 lbm water, 10 lbm ethanol and 5 lbm methanol per hour.

The temperature and pressure of the systeam are 95 C and 1 atm. The total flow rate from the bottom of the column is 24 lbm per hour. The composition of the bottom stream is 83.3% water (by weight), 12.5 % ethanol and 4.2 % methanol. The system is assumed to be operating at steady state and no reactions are occurring in the system.

a)Set up the species accounting and total mass conservation equation for the system.

b)Make the proper assumptions, and solve for the unknows.

c)Determine the composition and mass flow rate of the distillate

Since the system is steady state and no reactions are occurring in the system, mass conservation reduces to: .

We let the input feed stream be labeled "F", the distillate stream be "D" and the bottom stream be "B." We can now write conservation of mass (rate form) in table form:

=

SpeciesStream FStream BStream D

(lbm/hr)(lbm/hr)(lbm/hr)

Water25=0.833(24)+?

Ethanol10=0.125(24)+?

Methanol5=0.042(24)+?

______

Total40=24+?

The distillate stream (D) can now be easily solved for:

Water = 5 (lbm/hr), Ethanol = 7 (lbm/hr), Methanol = 4 (lbm/hr) and Total = 16.

The composition on a mass basis of the distillate is easily calculated: Water = (5/16) 100% = 31.2%, Ethanol = 43.8% and Methanol = 25%.
Multiple Unit Processes

Multiple unit processes involve flow of mass from one system into another; often as a "chain" of systems. An example schematic is shown below.

In the above, A, B and C can be treated as individual systems. In addition, a system boundry can be drawn around the combination of A, B and C. See p. 50-52 for an example.

Recycle and Bypass Systems

In many processes, such as reactions and separation, some of the material that leaves the system is introduced back into the system as a recycle stream. Bypass streams are used to remove material that requires no further processing. A schematic of a recycle and bypass system is shown below.

Example: A membrane system is used to filter waster products from the blood stream.

The membrane can extract 30 mg/min of waste , without removing any blood. The waste concentration in the entering blood is 0.17 wgt %, and the flow rate of blood is 25 g/min. A recyle stream is used to help control the concentration of the waste before it enters the membrane.

a)If the recycle rate is twice that of the filtered blood rate, what is the composition at A?

b)If the composition at A is to be 0.1 % waste, what must be the recycle rate, assuming streams U, W and F remain the same?

We first label the individual streams U, F, W, A, B, and R as shown on the sketch. U=unfiltered blood, A=input stream to filter (A=U stream + R stream), B = output stream from filter, F=filtered blood, W = pure waste, R=recycle stream.

We can draw several “systems.” We will choose the membrane plus recycle stream as the first system.

We start with the rate equation

For steady state and no gen/con in the system, the above reduces to . We construct the mass balance table:

Note that the Recycle Stream is inside the system and does not enter into the mass in and out terms.

=

SpeciesStream UStream WStream F

(g/min)(g/min)(g/min)

Blood0.9983(25)=0+?

Waste0.0017(25)=0.030+?

______

Total25.0=0.030+?

From the above, we conclude that for stream F the filtered blood flow rate is 24.9575 g/min, the waste flow rate is 0.0125 g/min or 12.5 mg/min and the total flow rate is 24.9700 g/min. The filtered blood contains 0.05 % waste (0.125/24.97x100%)

To evaluate stream A we can consider the junction of streams U, R and A as a system:

We write the mass balance equation for the A, U and R streams:

Note: The factor of 2 in stream R comes from the problem statement which states the recycle rate (stream R) is twice that of the filtered blood rate (stream F).

=

SpeciesStream UStream RStream A

(g/min)(g/min)(g/min)

Blood24.9575+2(24.9575)=?

Waste0.0425+2(0.0125)=?

______

Total25.0+2(24.9700)=?

From the above, stream A consists of 74.8725 g/min blood, 0.0675 g/min of waste, total mass rate of 74.9400 g/min and the waste in stream A is 0.09% by mass.

We finally take the membrane itself as the system and construct the following table to evaluate stream B:

=

SpeciesStream AStream WStream B

(g/min)(g/min)(g/min)

Blood74.8725=0+?

Waste0.0675=0.030+?

______

Total74.9400=0.030+?

From the above, stream B consists of 74.8725 g/min blood, 0.0375 g/min of waste, total mass rate of 74.9100 g/min.

For part b) we wish to determine the recycle rate so that the composition in stream A is 0.10% waste (assuming that streams U, W, and F remain the same). Again, we assume steady state and no net generation. Schematically, we take the membrane as the system. The flow rates in streams A and B are unknown.

We construct our mass conservation table for the membrane:

=

SpeciesStream AStream WStream B

(g/min)(g/min)(g/min)

Blood0.999A=0+0.9995B

Waste0.001A=0.03+0.0005B

______

TotalA=0.03+B

Solving the Blood and Waste equations simultaneously, we obtain stream A Total = 59.97 g/min and stream B total = 59.94.

The recycle rate will have changed since stream B has changed. Again, we draw a schematic of system and take the juncuture of the B, F and R streams.

We write mass conservation for the B, F and R streams in table form:

=

SpeciesStream BStream FStream R

(g/min)(g/min)(g/min)

Blood59.91=24.9575+?

Waste0.03=0.0125+?

______

Total59.94=24.97+?

Solving the Blood and Waste equations simultaneously, we obtain stream A Total = 59.97 g/min and stream B total = 59.94.

Consequently, the recycle stream R flow rates are: Stream R blood=34.9525 g/min, stream R waste=0.0175 g/min and stream R total=34.97 g/min.