1. Materials Balances with Reactors
Generally the purpose of a reactor is to create or destroy some material
Examples
Water treatment reactors
Wastewater treatment reactors

2. Examples Wastewater treatment
Objective is to convert dissolved organic compounds to suspended solid
material that can the be removed from the waste stream
Organics are converted to microorganisms
Excess organisms are wasted from the system - called waste activated sludge

3. An activated sludge system has an
influent flow of 10 mgd and a
suspended solids concentration of
50 mg/L. The waste activated sludge
flow rate is 0.2 mgd at a solids
concentration of 1.2%. The effluent
has a solids concentration of 20 mg/L.
What is the rate of solids production
in the system (commonly called the
yield)
Consider a mass balance around the whole system
Mass Balance for Flow rate
0 = 10 – [0.2 + QE] + 0
QE = 9.8 mgd

4. 0 = (QICI) – [(QECE) + QWCW] + X
Mass balance on Solids
Rate of
Solids
accumulated
Solids In
Solids Out
Generation = - +
0 = (QICI) – [(QECE) + QWCW] + X
X = the rate of solids production ( the rate at which organics
are converted to microorganisms
Units
(Gallons . Mg)/(day. L)
Not appropriate units for X - convert to lb/day
1,000,000gal/day x 1 mg/L x 3.78 L/gal x 10-6 kg/mg x 2.2 lb/kg = 8.34 lb/day
0 = (10)(50)(8.34) – [(9.8)(20)(8.34) – (0.2)(12,000)(8.34)] + X
X = 17,438 lb/day

5. Solids Destruction Anaerobic Digestion Pathogen Destruction
Volatile Organic Destruction
Gas Production
CO2 and Methane

6. Example The influent flow to a digester is
0.1 m3/s and has a solids concentration
of 4%, of which 70% are volatile solids
The supernatant has 2% solids of
which 50% are volatile solids. The
digested solids have a solids
concentration of 6% (50% volatile).
Find the flow rate of the supernatant and digested sludge?
Three possible materials balances of concern:
Volatile solids
Total Solids
Sludge volume

7. Mass Balance on Volatile Solids
Rate of
Volatile Solids
accumulated In
Out
Generation = - +
0 = (Q0C0) – [(QGCG + QSCS + QDCD] - X
X = the rate of volatile solids consumption
Units
(m3/s) (mg/L) (1000 L/m3) (1g/1000mg) = g/s
0 = (0.1)(40,000)(0.7) – [0 + QS(20,000)(0.5) + QD(60,000)(0.5)] - X
0 = 2800 – 10,000 QS – 30,000 QD - X

8. Mass Balance on Total Solids
Rate of
Total Solids
accumulated
Total Solids In
Out
Generation = - +
0 = Q0C0 – [QGCG + QSCS + QDCD] - X
0 = (0.1)(40,000) – [0 + QS(20,000) + QD(60,000)] - X
Balance on Liquid Flows
0 = Q0 – [QS + QD + QG] - 0
Since there is very little liquid in the gas flow stream, QG
is negligible
0 = 0.1 – QS - QD

9. Now we have three equations, with three unknowns:
0 = 2800 – 10,000 QS – 30,000 QD - X
0 = (0.1)(40,000) – QS(20,000) - QD(60,000) - X
0 = 0.1 – QS - QD
Solving:
QD = 0.01 m3/s
QS = 0.09 m3/s
X = 1600 g volatile solids/s