materials balances with one material. materials balances with one material a materials balance is...
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Materials Balances With One Material
Materials Balances with One MaterialA materials balance is based on the principle of conservation of mass, that is, mass is conserved in any system as long as nuclear reactions do do occur.
Rate at which mass is
accumulated in a system
=Rate of Mass flow into the
system
_Rate of Mass flow out of the system
+Rate of Generation or consumption
of mass in the system
Macroscopic Mass Balance (Black Box)
Inputs – InfluentsOutputs - effluents
Macroscopic Mass Balance (Black Box)
Simplifying assumption – Steady-State
Steady-State d/dt = 0
Nothing Varies with time
Macroscopic Mass Balance (Black Box)
Rate At which Mass is
Accumulated in a system
=Rate of Mass flow into the
system
Rate of Mass flow out of the system
+Rate of Generation or consumption
of mass in the system
_
Rate of Mass flow into the
system
Rate of Mass flow out of the system
=
X0 = X1
V dc/dt
0, S.S. 0
Splitting Flow Streams
01
2
X0
X1
X2
0, S.S. 0
Rate of Mass flow into the
system
Rate of Mass flow out of the system
=
X0 = X1+ X2
Rate At which Mass is
Accumulated in a system
=Rate of Mass flow into the
system
Rate of Mass flow out of the system
+Rate of Generation or consumption
of mass in the system
_
ExampleA city generates 102 tons/day of refuse, all of which goes to a transfer station.At the transfer station the refuse is split into four flow streams. Three of the Flow streams go to incinerators, the fourth goes to a landfill. If the capacity of the incinerators is 20, 50, and 22 tons/day, how much goes to the landfill?
0, S.S. 0
Rate of Mass flow into the
system
Rate of Mass flow out of the system
=
102 = 20 + 50 + 22 + M
M = 10 tons/day
Rate At which Mass is
Accumulated in a system
=Rate of Mass flow into the
system
Rate of Mass flow out of the system
+Rate of Generation or consumption
of mass in the system
_
Combining Flows into a Single Stream
Rate of Mass flow into the
system
Rate of Mass flow out of the system
=
X1 + X2 + X3 = X0
ExampleA trunk sewer has the capacity of 4.0 m3/s. If the flow to the sewer is exceeded, it will not be able to transmit the sewer through the pipe and a backflow occurs. Three neighborhoods contribute to the sewer, and their peak flows are 1.0, 0.5, and 2.7 m3/s. A builder wants to construct a new development that will contribute another 0.7 m3/s to the trunk line. Will it cause a backup?
Rate of Mass flow into the
system
Rate of Mass flow out of the system
=
1.0 + 0.5 + 2.7 + 0.7 = X0
X0 = 4.9 m3/s The sewer line would be overloaded. In fact, it already is overloaded
Another Example
A storm sewer carrying water to manhole 1 has a constant flow of 20,947 l/min. At manhole 1 a constant flow of 100 l/min comes in from a lateral. What is the flow to the next manhole down stream?
Rate At which Mass is
Accumulated in a system
=Rate of Mass flow into the
system
Rate of Mass flow out of the system
+Rate of Generation or consumption
of mass in the system
_
0, S.S. 0
0 = (QA + QB) – QC
0 = (20,947 +100) - QC
QC = 21,047 L/min
As the previous example showed, it is often convenient to draw system boundaries around any junction of three or more flows.
Sometimes the flows are contained in pipes of rivers, etc.,. Sometimes they are not. When they are not it is often convenient to visualize them as if they were.
Or example, consider a system where rain falls to earth where it can either percolate into the groundwater or runoff into a watercourse.
A system may contain any number of processes or junctions thatmay be treated as black boxes.
Consider the hydrologic system in which precipitation falls to the earth. Some of it percolates into the groundwater where it joins a groundwater reservoir. The water may then be pumped from the ground and used for irrigation. The irrigation water either goes back into the atmosphere by the process of evaporation or transpiration (evapotranspiration) or it percolates back into the groundwater.
Example40 inches of rain falls in an area each year. 50% of this percolates into the ground. A farmer irrigates his crops using well water. Of the extracted water, 80% is lost by evapotranspiration, the rest percolates back into the ground. How much groundwater could a farmer extract without depleting the groundwater if he has a 2000 acre farm?
40 in/yr(1/12 ft/in)x 2000 acres(43560ft2/acre) = 2.90 x 108 ft3/yr
Rate At which Mass is
Accumulated in a system
=Rate of Mass flow into the
system
Rate of Mass flow out of the system
+Rate of Generation or consumption
of mass in the system
_
0, S.S.0
0 = 2.9x108 ft3/yr – [QR + QN] + 0 - 0
0 = 2.9x108 ft3/yr – [QR + QN] + 0 - 0
QR = 0.5 QP = QN
So: 0 = 2.90 x 108 – 2 QR
QR = 1.45 x 108 ft3/yr = QN
Now do mass balance around second box to get QW
Materials Balances1) Draw a system diagram, include all inputs and outputs2) Add the available information, assign symbols3) Draw a dotted line representing the system boundary4) Decide what materials to balance5) Write the mass balance equation6) If only one variable is unknown, solve7) If there is more than one unknown, draw another materials banalce