MATERIALS SCI + ENGR: INT W/ACCESS
10th Edition
ISBN: 9781119808084
Author: Callister
Publisher: WILEY
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Chapter 13, Problem 1FEQP
To determine
To find:
What happens to refractory ceramic brick when its porosity is increased?
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Estimate the required air flow rate for the new activated sludge plant at Pea Ridge (Problems
23-223-723-10, and 23-13). The flow rate is 8,450 m³/day, the concentration
of bCOD going into the system (So) is 137 mg/L, the concentration of bCOD leaving the
system (S) is 16.3 mg/L, and the mass of cells produced per day (Pxvss) is 277.4 kg/d. Use the
following assumptions to estimate the required air flow rate:
. Clean water correction, a = 0.50
. Salinity correction, B = 0.95
Fouling factor = 0.9
Wastewater temperature = 12°C
Atmospheric pressure = 101.325 kPa
.Elevation 500 m
.
Depth of aerator = 5.6 m
Operating DO 2.0 mg/L
Percent oxygen leaving aeration tank - 19%
■ Manufacturer's SOTR = 535 kg/d
Manufacturer's air flow rate at standard conditions 50 m³/d - aerator
Express your answer with the units of m³/d and round to the nearest integer.
Use Newton-Raphson method to solve the system
x²
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2x-y+0.5= 0
x² + 4y² 4 = 0
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with the starting value (xo,yo) = (2,0.25) and two iteration number.
Determine the required solids retention time (SRT) of a completely mixed
activated sludge aeration tank for a conventional activated sludge system
treating a design flow rate of 34,560 m³/d, where the effluent standards are
30.0 mg/L for BODs and 30.0 mg/L for total suspended solids (TSS). Assume
that the BOD5 of the effluent TSS is 70% of the TSS concentration. Assume the
BODs concentration leaving the primary clarifier is 128 mg/L that the MLVSS
concentration (X₂) is 2,500 mg/L. Assume the following values for the growth
constants:
Ks 100 mg/L BODS
⚫ Hm - 2.5 d 1
kd = 0.050 d 1
Y = 0.50 mg VSS/mg BODs removed
Express your answer in days and round to the nearest 0.1.
Chapter 13 Solutions
MATERIALS SCI + ENGR: INT W/ACCESS
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