CENE 150
MATERIAL BALANCE EXAMPLE #1
Find a picture of a pipe emptying into a small stream or channel. Then show the Problem PROBLEM STATEMENT: A stream has a flow rate of 8 ft3/s. The SS concentration of the stream is 2 mg/L. A stormwater outfall averaging 10 gpm containing 175 mg/L SS enters the stream. Assuming complete mixing, determine the downstream SS concentration in the stream.
STATE: In order to solve this problem, we will use a MASS BALANCE. First, let’s visualize the problem.
DRAW: Begin drawing a stream; at the same time blinking highlight the words “a stream” in the problem statement. Show the stream flowing with an arrow showing the direction of flow; blinking highlight the words “a flow rate of 8 ft3/s.” Put “8 ft3/s” next to the arrow.
STATE: ft3/s means cubic feet of water per second; cubic feet is a volume, second is time, therefore ft3/s is a volumetric flow rate.
DRAW: Add little dots into the stream while blinking highlight the words “the SS concentration of the stream is 2 mg/L” – only add a few dots though. Show them flowing along with the stream. Put “2 mg/L SS” below the 8 ft3/s.
STATE: SS means suspended solids; mg/L means milligrams of suspended solids per L of stream water and is a measure of concentration. The concentration of the SS in the stream at all points is 2 mg/L.
DRAW: Add a pipe – it should come into the stream from the top, about halfway along the length of the stream while blinking highlight the words “a stormwater outfall….enters the stream” in the problem statement. Put “10 gpm” next to the flow of the pipe.
STATE: gpm means gallons of stormwater per minute; gallon is a volume, minute is time, therefore gpm is a volumetric flow rate.
DRAW: Add little dots to the pipe water – many more dots than in the stream; proportion them appropriately between the 2 mg/L of the stream and 175 mg/L of the pipe. Blinking highlight the words “containing 175 mg/L SS” and put “175 mg/L SS” below the 175 mg/L.
DRAW: On the part of the stream below the stormwater outfall, add a few more dots to indicate that the concentration of SS and state. >>>
STATE: Once the stormwater outfall enters the stream, it mixes with the stream water. Logic tells you that both the volumetric flow rate of the stream and the concentration of SS in the stream below the outfall must change due to the addition of the outfall: the total volume of the stream must increase and the concentration will be higher than 2 mg/L but lower than 175 mg/L. The problem asks you to determine the concentration of SS in the stream below the outfall.
DRAW: Blinking highlight the words “determine the downstream SS concentration in the stream.” Put a few black dots below the outfall with a bunch of question marks, then switch to a lot of black dots below the outfall with the question marks, then show a mid-range number of black dots below the outfall with the question marks.
STATE: To solve the problem, we use the concept of material balance. The first step is to change the above physical drawing of the problem to a simplified “block diagram” that represents all the data in the problem. Note that every statement in the problem is useful and needed.
STATE: The flowing stream is represented by an arrow.
DRAW: Draw an arrow.
STATE: We assign variables to the data: Q’s are used to denote flow rates; C’s are used to denote concentrations. Since this is the first flow we are presented with, we will subscript the variables with “1”.
DRAW: Put Q1= 8 ft/3s below the first arrow; put c1=2 mg/L below the 8 ft3/s.
REVISIONS START HERE:
STATE: “The stormwater outfall pipe is represented also by an arrow because it, too is flowing.”
DRAW: another arrow from the top to meet the end of the first arrow (this should be about in the middle of the page).
STATE: This stream will be stream “2”.
WRITE: Q2=10 pgm next to that arrow, and c2=175 mg/L below that.
STATE: “The mixing zone is denoted by a box; think of it as the region of the stream where the two streams completely mix, so that the flow out of the box is homogeneous.
The stream below the outfall is represented by an arrow leaving the box.”
DRAW: an arrow leaving the block.
STATE: “This stream will be denoted by subscript “3”. We don’t know it’s flow rate or it’s concentration, so we simply write Q3= and c3=.”
STATE: “We need to understand what mathematical equations are needed to perform a material balance. There are 2 equations: the overall mass balance (of the entire water stream) and a component balance (of an individual component within the stream). We perform this balance around the box in the block diagram. The first equation describes the mass of the water in the stream. Mass = volume times density. We don’t know the mass of water, we only know the volume, or, in this case, the volumetric flow rate. But the density of water is a known value, it’s 1 g/mL The equation “balancing” the streams into and out of the box are: INS = OUTS.”
WRITE: INS=OUTS”.
STATE: “Mathematically, we say Q1 times density plus Q2 times density equals Q3 times density because there are two streams coming in and only one stream leaving.”
WRITE: Q1*dens + Q2*dens = Q3*dens.
STATE: “Because we assume that the density of each water stream at normal environmental conditions is the same, the density term can be cancelled out, leaving the “mass” balance to become a “volume” balance. This is OK for the overall water balance. Note that this will NOT work for gas streams because volume and mass are dependent on temperature and pressure for gases. So the first mass balance equation becomes Q1 plus Q2 equals Q3.
WRITE: Q1 + Q2 = Q3.
STATE: “The second equation describes the component that we care about in the stream, the suspended solids. The mass balance is the same: INS = OUTS.”
WRITE: INS=OUTS.
STATE: “But we must use mass for this equation, we cannot use concentration or we will get the wrong answer – remember that logic tells us that the SS concentration won’t be less than 2 or greater than 175 – it will be somewhere in between. It can’t be 175 + 2 = 177!!!!! The correct equation is: (MASS SS in 1) + (MASS SS in 2) = (MASS SS in 3).”
WRITE: (MASS SS in 1) + (MASS SS in 2) = (MASS SS in 3).
STATE: “MASS is computed using flow rate and concentration”.
WRITE: MASS SS in 1 = Q1 * c1, or:
ft3/s water * mg/L SS * conversion factor for L/ft3 = mg/s SS
STATE: “The equation becomes: Q1 times c1 plus Q2 times c2 equals Q3 times c3.”
WRITE: Q1*c1 + Q2*c2 = Q3*c3.
STATE: “These two equations are INDEPENDENT of one another and can both be used to solve the problem. In the first equation, Q1 + Q2 = Q3, we have only one unknown (Q3), so we can immediately and directly solve for Q3. In the second equation, Q1*c1 + Q2*c2 = Q3*c3, we have two unknowns (Q3 and c3). Between both equations, we have two unknowns (Q3 and c3). Since we have 2 independent equations, we can solve them by solving equation 1 first for Q3 then using equation 2 to solve for c3, which is the answer requested in the problem statement. In order to actually do the math, we see that the UNITS of Q1 and Q2 are different. We must convert the units so that we can add them together. We must convert the 10 gpm to ft3/s. The conversion factors are 7.48 gal/ft3 and 60 s/min.”
WRITE: 10 gal/min * ft3/7.48 gal * min/60 s = 0.022 ft3
STATE: “You can see that the 10 gpm = 0.022 ft3/s, when compared to the stream flow rate of 8 ft3/s, is VERY small! Therefore, Q3 = Q1 + Q2 = 8 + 0.022 = 8.022 ft3/s.”
WRITE: Q3 = Q1 + Q2 = 8 + 0.022 = 8.022 ft3/s
STATE: “And then:”
WRITE: c3 = (Q1*c1 + Q2*c2)/Q3
STATE: “Substituting:”
WRITE:c3 = [(8 ft3/s)*(2mg/L) + (0.022 ft3/s)*(175mg/L)]/8.022 ft3/s = 2.47 mg/L
STATE: “NOTE that you didn’t need to convert ft3 to L to solve the problem!!!! The ft3/s, in both of the additive terms in the numerator AND in the term in the denominator, can be directly cancelled out, leaving the answer in mg/L!!!!!!”
STATE: “LOOK at the solution: the final concentration in the stream”
REDRAW: the first sketch and put the appropriate concentration of black dots in the stream).
STATE: “See how this makes sense: because the flow rate of the outfall pipe was so low compared to the stream, even a high concentration in the pipe did not significantly affect the stream – the concentration only rose from 2 mg/L to 2.47 mg/L.”