Page 229 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

184 Fundamentals of Water Treatment Unit Processes: Physical, Chemical, and Biological

PROBLEMS 8.6 Dissolved Air Concentration in Saturator Using
H(air)
8.1 History Let the context of the problem be a saturator, with air as
Problem the gas, and 700 kPa total pressure imposed.
Summarize the difference between the state-of-the-art of Given
flotation c. 1959 and in the 1990s (or other dates if you T ¼ 208C
wish). Cite one or two papers for each period.
H(air) ¼ H(O 2 ) X(O 2 ) þ H(N 2 ) X(N 2 )
8.2 System Description
¼ 43:49 mgO 2 =L=atm 0:209 þ 19:01 mg
Problem
N 2 =L=atm 0:78084
Summarize the flotation process, identifying key phases.
8.3 Dissolved Oxygen Concentration in Saturator by ¼ 23:93 mg air=L=atm
Henry’s Law ¼ 0:0002362 (kg air dissolved=m 3
The problem refers to a saturator; air is the gas, with 700
water=kPa air)
kPa total pressure imposed. Let f ¼ 1.0 for the purpose of
the calculation. Reference is Section 8.3.1. X(air) ¼ 1:00 mol air=mol air
P(total) ¼ 700 kPa
Given
Required
T ¼ 208C
3
H(O 2 ) ¼ 0.000428 (kg oxygen dissolved=m water=kPa Calculate the concentration of dissolved air at equilib-
rium by Henry’s law.
oxygen) at 208C
8.7 Condition for Gas Precipitation
X(O 2 ) ¼ 0.209 mol O 2 =mole air (Table B.7, Compos-
ition of Air) Given
P(total) ¼ 700 kPa (stated) Consider a flotation basin at sea level and T ¼ 208C,
. Atmospheric pressure, P(atm) ¼ 101.325 kPa (1.00
Required
atm)
Calculate the concentration of dissolved oxygen at equi- . Let nozzle depth below water surface, d(noz) ¼ 3.0 m
librium by Henry’s law.
(9.84 ft)
8.4 Dissolved Nitrogen Concentration in Saturator by . Let f 0.9 (saturator efficiency)
Henry’s law
Required
The problem refers to a saturator; air is the gas, with 700
Determine the minimum saturator pressure for gas pre-
kPa total pressure imposed. Let f ¼ 1.0 for the purpose of
cipitation in a flotation basin.
the calculation.
8.8 Separation Zone: Rise Velocity, v pb , and SOR
Given
Reference is Section 8.3.4.
T ¼ 208C Given
H(N 2 ) ¼ 19.01 (mg nitrogen dissolved=L water=atm B ¼ 10 bubbles=particle
nitrogen) d b ¼ 40 mm
3
¼ 0.0190 (kg nitrogen dissolved=m water=atm d p ¼ 50 mm
nitrogen) T ¼ 208C
3
3
3
¼ 0.0001875 (kg nitrogen dissolved=m water= r p ¼ 1.01 g=cm [ ¼ 1010 kg=m ]
kPa nitrogen)
Required
X(N 2 ) ¼ 0.78084 mol N 2 =mole air
Calculate the of rise velocity, v pb , of a particle–bubble
P(total) ¼ 700 kPa
agglomerate for conditions stated.
Required 8.9 Separation Zone: A=S ratio from Bubble-Particle
Calculate the concentration of dissolved N 2 at equilib- Ratio
rium by Henry’s law. Reference is Section 8.3.4.
8.5 Dissolved Air Concentration in Saturator by Henry’s Given
Law
Given
d p ¼ 50 mm r ¼ 0.10
Let T ¼ 208C, P(saturator) ¼ 700 kPa, which results in
3 N p ¼ 17000 #=mL C(saturator) ¼ 0.136 kg
C(O 2 ) ¼ 0.0628 kg dissolved O 2 =m water and C(N 2 ) ¼ 3
3 air=m water
0.10248 kg dissolved N 2 =m water. Let f ¼ 1.0 for the
3
r(floc) ¼ 1010 kg solids=m water Table CD8.3
purpose of the calculation.
Required
Calculate the concentration of ‘‘air’’ leaving the Required
saturator. Calculate the air-to-solids ratio, i.e., A=S.