The XYZ brewery has a small activated sludge plant to treat its wast.pdf
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The XYZ brewery has a small activated sludge plant to treat its wastewater. The average daily discharge is 750 m3/day and the average soluble BOD5 after primary settling is 500 mg/L. The aeration tank has effective liquid dimensions of 10.0 m wide by 10.0 m long by 4.5 m deep. The plant operating parameters are as follows: Influent volatile suspended solids to the reactor are negligible Return sludge concentration = 15,000 mg/L of suspended solids MLVSS = 2,500 mg/L MLSS = 1.20 (MLVSS) Effluent soluble BOD5 = 16.70 mg/L Effluent Volatile Suspended solids = 8.30 mg/L Wastage is from the return sludge line Yield coefficient = 0.60 mg VSS/mg BOD5 removed Endogenous decay rate of microorganism = 0.060 d-1 Determine: a) Solids retention time b) Cell wastage flow rate c) Return sludge flow rate for this brewery wastewater treatment plant Solution 1) Assume areation hours as 18hours = 0.75d SRT = Solid retention time = -0.75 (MLVSS/[ k x 0.75 xMLVSS- Yield coefficient (So - S)] SRT = -0.75 x 2500/[0.06x0.75x2500 - 0.6 [ 500 - 10] SRT = 10.33 days 2) Cell wastage flow rate V = volume of areation tank So = BOD5 value RS = return sludge k = 0.06d-1 Qw =[ Q x Y/RS] [ So - S] - k(V x MLVSS/RS) Qw = [ 750x0.6/15000] [ 500-10] - [0.06] x (2500/15000) x 10x10x4.5 Qw = 14.7 - 4.5 Qw = 10.2 m3/d 3) Return Sludge flow rate RSF =Q x MLVSS/ [ MLVSS-VSS] RSF = 750x 2500/[2500-8.3] RSF = 752.49 m3/d.
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![The XYZ brewery has a small activated sludge plant to treat its wastewater. The average daily
discharge is 750 m3/day and the average soluble BOD5 after primary settling is 500 mg/L. The
aeration tank has effective liquid dimensions of 10.0 m wide by 10.0 m long by 4.5 m deep. The
plant operating parameters are as follows:
Influent volatile suspended solids to the reactor are negligible
Return sludge concentration = 15,000 mg/L of suspended solids
MLVSS = 2,500 mg/L
MLSS = 1.20 (MLVSS)
Effluent soluble BOD5 = 16.70 mg/L
Effluent Volatile Suspended solids = 8.30 mg/L
Wastage is from the return sludge line
Yield coefficient = 0.60 mg VSS/mg BOD5 removed
Endogenous decay rate of microorganism = 0.060 d-1
Determine:
a) Solids retention time
b) Cell wastage flow rate
c) Return sludge flow rate for this brewery wastewater treatment plant
Solution
1) Assume areation hours as 18hours = 0.75d
SRT = Solid retention time = -0.75 (MLVSS/[ k x 0.75 xMLVSS- Yield coefficient (So - S)]
SRT = -0.75 x 2500/[0.06x0.75x2500 - 0.6 [ 500 - 10]
SRT = 10.33 days
2) Cell wastage flow rate
V = volume of areation tank
So = BOD5 value
RS = return sludge
k = 0.06d-1
Qw =[ Q x Y/RS] [ So - S] - k(V x MLVSS/RS)
Qw = [ 750x0.6/15000] [ 500-10] - [0.06] x (2500/15000) x 10x10x4.5
Qw = 14.7 - 4.5
Qw = 10.2 m3/d
3) Return Sludge flow rate](https://image.slidesharecdn.com/thexyzbreweryhasasmallactivatedsludgeplanttotreatitswast-230403044009-beeee631/85/The-XYZ-brewery-has-a-small-activated-sludge-plant-to-treat-its-wast-pdf-1-320.jpg)
![RSF =Q x MLVSS/ [ MLVSS-VSS]
RSF = 750x 2500/[2500-8.3]
RSF = 752.49 m3/d](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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rate of 1.2% per year. Find a natural growth model P(t) that gives the population of Percentville
as a function of t, years after 2010.
Solution
The answer is simply 25000 * (1.02)t
Note that P(1)/P(0) = 25000 * (1.02)1 /25000 = 1.02, P(2) = 25000 * (1.02)2 /(25000 * (1.02)1 )
=
(1.02)2/(1.02)1 = 1.02, ... Thus, this is a 2% increase per year..
The variable m has dimensions of mass, h and y have dimensions of le.pdf
The variable m has dimensions of mass, h and y have dimensions of length, and t1 and t2 have
dimensions of time. What are the dimensions of the following quantities? (Use the following as
necessary: M for units of mass, L for units of length, and T for units of time.) (a) m1y/t2 (b)
hy/t12 (c) y3/(ht1)
Solution
(a) m1y/t2
[M][L]/[T]
there is no simplification so the units are ML/T
(b) hy/t12
[L][L]/[T]
simplifies as L2/T
(c) y3/(ht1)
[L]/[L][T]
simplifies as 1/T.
The usual Fibonacci numbers are given by the recurrenceSolution.pdf
The usual Fibonacci numbers are given by the recurrence:
Solution
a)
put n=1
we get F1=F3-1 -----------1
we know F3=1+0+1=2
substitute back F3 in equation 1
we get F1=1
which is true
now put n=i
we get
F1+F2+F3+...Fi=F(i+2)-1
now the values of F1 F2 F3 ARE DERIVED FROM DEFINITION OF FIBONACCI
F1=1 F2=1 F3=2 F4=3 F5=5
SO WE GET
1+1+2+3+4+.....=F(i+2)-1
we get
1+ i(i+1)/2=F(i+1)+F(i)
1+ i(i+1)/2=2F(i)+F(i-1)
but 2F(i)+F(i-1)=1+ i(i+1)/2
therefore lhs=rhs
b)
put n=1 first
we get lhs =rhs
now put n=i
F2i=F(i+1)F(i-1)+ -1^i+1
substitute the values by definition
and take the sum of squares of first n natural numbers by definition of the lhs part
and the fibonacci definition on rhs both are equal
hope it helps.
The US had a national bank during the first few decades after its bi.pdf
The US had a national bank during the first few decades after its birth before it was abolished.
Today, we have the Federal Reserve.
What are the differences between the Federal Reserve and a National Bank (specifically, with
regard to their intervention policies and powers for acting within the economy)?
Solution
The two Banks of the United States (the First and Second) were nothing like the modern Federal
Reserve system. For example, the First was prohibited from buying government bonds (one of
the main roles of the Federal Reserve system is to buy and sell government bonds). Further,
neither of the national banks had any role in regulating the banking system. They were simply
the only interstate banks allowed. All other banks had to be confined to one state. The national
banks primary impact on the money supply was to require smaller banks to pay them in silver
and gold to redeem checks deposited in the national banks.
By contrast, the Federal Reserve system actively regulates banks, requiring them to maintain a
percentage of their deposits as reserves. It actively sets two rates: the rate at which it loans
money to banks to cover reserve shortages (called the discount rate) and the rate it pays on
excess reserves. The latter power is relatively new. Previously the Federal Reserve did not pay
interest on excess reserves. The Federal Reserve also sets a target for the funds rate, which is the
average rate at which banks loan money to each other. The funds rate target is what most people
mean when they refer to the Federal Reserve setting interest rates.
The Federal Reserve has two main ways to affect the funds rate. First, it publishes a target rate
for the average. Second, it buys and sells government bonds in what are called open market
operations. When it buys bonds, it adds money to the system, reducing the number of banks that
need to borrow and increasing the number of those with money to lend. This pushes the average
rate to fall, as there are more lenders and fewer borrowers. When it sells bonds, it pulls money
from the system, reducing the number of banks with money to lend.
The Federal Reserve can also compete with lenders by loaning money to banks at the discount
rate. This decreases the number of banks looking to borrow from other banks, which would push
rates down. It rarely does this however. Most banks prefer to borrow from other banks, as there
are fewer restrictions.
The Federal Reserve also has two ways to influence how much banks keep in reserve. First, it
sets an explicit reserve requirement. Every bank must keep that much in reserve. Second, it can
pay interest on excess reserves. The higher that rate, the more likely banks are to keep excess
reserves instead of loaning them to individuals, companies, or other banks. This would push up
the funds rate average, as fewer banks loan.
Again, the two Banks of the United States were not central banks the way that the Federal
Reserve is. Yes, they could print banknotes, but so .
The value of SS (sum of squared deviations) for the following popula.pdf
The value of SS (sum of squared deviations) for the following population is _______________
Population: 1, 1, 1, 5
Fill in the blank please
Solution
First find the mean: Mean = (1+1+1+5)/4 = 2 Subtract the mean from each
observation and square the resulting difference. Then add the differences to get SS: SS = (1-2)^2
+ (1-2)^2 + (1-2)^2 + (5-2)^2 = 12.
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- 1. The XYZ brewery has a small activated sludge plant to treat its wastewater. The average daily discharge is 750 m3/day and the average soluble BOD5 after primary settling is 500 mg/L. The aeration tank has effective liquid dimensions of 10.0 m wide by 10.0 m long by 4.5 m deep. The plant operating parameters are as follows: Influent volatile suspended solids to the reactor are negligible Return sludge concentration = 15,000 mg/L of suspended solids MLVSS = 2,500 mg/L MLSS = 1.20 (MLVSS) Effluent soluble BOD5 = 16.70 mg/L Effluent Volatile Suspended solids = 8.30 mg/L Wastage is from the return sludge line Yield coefficient = 0.60 mg VSS/mg BOD5 removed Endogenous decay rate of microorganism = 0.060 d-1 Determine: a) Solids retention time b) Cell wastage flow rate c) Return sludge flow rate for this brewery wastewater treatment plant Solution 1) Assume areation hours as 18hours = 0.75d SRT = Solid retention time = -0.75 (MLVSS/[ k x 0.75 xMLVSS- Yield coefficient (So - S)] SRT = -0.75 x 2500/[0.06x0.75x2500 - 0.6 [ 500 - 10] SRT = 10.33 days 2) Cell wastage flow rate V = volume of areation tank So = BOD5 value RS = return sludge k = 0.06d-1 Qw =[ Q x Y/RS] [ So - S] - k(V x MLVSS/RS) Qw = [ 750x0.6/15000] [ 500-10] - [0.06] x (2500/15000) x 10x10x4.5 Qw = 14.7 - 4.5 Qw = 10.2 m3/d 3) Return Sludge flow rate
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