Sludge wasting rate (Qw) from the solids residence time (Thetac = mcrt) calculation. Given the following information from the previous problem. The total design flow is 15,000 m3/day. Theoretical hydraulic detention time (Theta) = 8 hours. The NPDES limit is 25 mg/L BOD/30 mg/L TSS.
Assume that the waste strength is 170 mg/L BOD after primary clarification.
XA=MLSS = 2200 mg/L,
Xw = Xu = XR = 6,600 mg/L,
qc = 8 days.
Make sure you account for the solids in the discharge.
What volume of sludge (Qw=m3/day) is wasted each day from the secondary clarifiers?
Answers
Answer:
The volume of sludge wasted each day from the secondary classifiers is Qw = 208.33 m^3 / day
Explanation:
Check the file attached for a complete solution.
The volume of the aeration tank was first calculated, V = 5000 m^3 / day.
The value of V was consequently substituted into the formula for the wasted sludge flow. The value of the wasted sludge flow was calculated to be Qw = 208.33 m^3 / day.
Related Questions
Design a circuit that will tell whether a given month has 31 days in it. The month is specified by a 4-bit input, A3:0. For example, if the inputs are 0001, the month is January, and if the inputs are 1100, the month is December. The circuit output, Y, should be HIGH only when the month specified by the inputs has 31 days in it. However, if the input is not a valid month (such as 0 or 13) then the output is don’t care (can be either 0 or 1).
Answers
Answer:
see attachments
Explanation:
A Karnaugh map for the output is shown in the first attachment. The labeled and shaded squares represent the cases where Y = HIGH. The associated logic can be simplified to
Y = A3 xor A0
when the don't care at 1110 gives an output of HIGH.
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The second attachment shows a logic diagram using a 4:1 multiplexer to do the decoding. A simple XOR gate would serve as well. If AND-OR-INV logic is required, that would be ...
Y = Or(And(A3, Inv(A0)), And(A0, Inv(A3)))
The roof of a refrigerated truck compartment consists of a layer of foamed urethane insulation (t2 = 21 mm, ki = 0.026 W/m K) between aluminum alloy panels (t1 = 9 mm, kp = 180 W/m K). The length of the roof is L = 13 m. The net solar radiation into the roof is 107 W/m2. The temperature of the inner surface is Ts,i = -4 oC and the air temperature is T[infinity] = 29 oC. The convective heat transfer coefficient for the top surface of the truck is h = 47 W/m2 K. Find Ts,o in oC.
Answers
Answer:
Tso = 28.15°C
Explanation:
given data
t2 = 21 mm
ki = 0.026 W/m K
t1 = 9 mm
kp = 180 W/m K
length of the roof is L = 13 m
net solar radiation into the roof = 107 W/m²
temperature of the inner surface Ts,i = -4°C
air temperature is T[infinity] = 29°C
convective heat transfer coefficient h = 47 W/m² K
solution
As when energy on the outer surface at roof of a refrigerated truck that is balance as
Q = [tex]\frac{T \infty - T si }{\frac{1}{hA}+\frac{t1}{AKp}+\frac{t2}{AKi}+\frac{t1}{aKp}}[/tex] .....................1
Q = [tex]\frac{T \infty - Tso}{\frac{1}{hA}}[/tex] .....................2
now we compare both equation 1 and 2 and put here value
[tex]\frac{29-(-4)}{\frac{1}{47}+\frac{2\times0.009}{180}+\frac{0.021}{0.026}} = \frac{29-Tso}{\frac{1}{47}}[/tex]
solve it and we get
Tso = 28.153113
so Tso = 28.15°C
