EV350 – Useful Equations Sheet
Mass Balance:
1. Read the problem carefully and identify all given information.
2. Draw a conceptual diagram of your system. Label all flows (Q), concentrations (C), and volumes (V).
Flows will be in units of volume/time and concentrations will be in units of mass/volume. Place a control
volume around your system (dotted line below).
Q1
C1
3.
Q2
C2
V, k, Cmix
Q3
C3
Apply the Mass Balance Equation to your problem.
Accumulation Rate = (Mass Flow Rate In) – (Mass Flow Rate Out) +/- (Transformation)
4.
5.
𝑑𝑑𝑑𝑑
= � 𝑄𝑄𝑖𝑖𝑖𝑖 𝐶𝐶𝑖𝑖𝑖𝑖 − � 𝑄𝑄𝑜𝑜𝑜𝑜𝑜𝑜 𝐶𝐶𝑜𝑜𝑜𝑜𝑜𝑜 ± 𝑟𝑟𝑉𝑉
𝑑𝑑𝑑𝑑
(where 𝑟𝑟 = 𝑘𝑘𝐶𝐶𝑚𝑚𝑚𝑚𝑚𝑚 for 1st order reactions)
𝑑𝑑𝑑𝑑
= −𝑘𝑘𝑘𝑘
𝑑𝑑𝑑𝑑
𝐶𝐶𝑡𝑡 = 𝐶𝐶𝑜𝑜 𝑒𝑒 −𝑘𝑘𝑘𝑘
[EQN 2-12]
[EQN 2-9]
Modified
[EQN 2-11]
[EQN 2-14]
Apply simplifying assumptions:
a. If the system is operating at equilibrium, and there is no net change of mass with respect to time
within the control volume, assume the system is at steady state:
dM/dt = 0
(Accumulation of mass = 0)
b.
Assuming the system is completely mixed, then:
(from example diagram above: Cmix = C2 = C3)
Cmix = Cout
c.
Assuming the system is conservative with respect to flow, then:
(from example diagram above: Q1 = Q2 + Q3)
∑Qin = ∑Qout
Solve! UNIT WARNING: Track units throughout calculation and apply unit conversions as needed.
Environmental Chemistry:
Modified [EQN 5-21]
(
Solubility:
K sp = [ A b + ] a [ B a − ]b ⇒ given : Aa Bb ( s ) ⇔ aA b + + bB a −
Acid/Base:
𝐾𝐾𝑠𝑠 = 𝐾𝐾𝑠𝑠𝑠𝑠
pH = -log [H+]
[EQN 5-29]
[H+] = 10-pH
𝐶𝐶𝐴𝐴 𝑉𝑉𝐴𝐴 = 𝐶𝐶𝐵𝐵 𝑉𝑉𝐵𝐵
Neutralization:
Flocculation/Sedimentation:
𝑄𝑄 =
𝑉𝑉
𝑡𝑡𝑜𝑜
[EQN 2-27]
vo =
Stoke’s Law:
g ( ρ s − ρ )d 2
vs =
18µ
Filtration:
va =
Freundlich Isotherm:
𝑞𝑞𝑒𝑒 =
Q
As
𝐾𝐾𝑠𝑠 = 10−𝑝𝑝𝐾𝐾𝑠𝑠
[EQN 6-47]
[OH-] = 10-(14-pH)
Q
As
Based on
[Fig. 6-32]
WL =
)
[EQN 5-19]
pKw = pH + pOH = 14
Q
WOR
[p. 314]
P = 100
vs
vo
[EQN 6-33]
[EQN 6-45]
Disinfection:
𝑁𝑁
𝑙𝑙𝑙𝑙𝑙𝑙 � 𝑁𝑁𝑜𝑜 � = 𝑘𝑘′𝐶𝐶𝐶𝐶
[EQN 6-57, (integrated)]
𝑡𝑡
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 (𝑚𝑚𝑚𝑚) (𝐶𝐶𝑜𝑜 − 𝐶𝐶𝑒𝑒 )𝑉𝑉
𝑥𝑥
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
=
=
=
= 𝐾𝐾𝐹𝐹 𝐶𝐶𝑒𝑒 𝑛𝑛
𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 (𝑔𝑔)
𝑚𝑚
[Class
As of: 8 APR 2021
BOD: 𝐵𝐵𝐵𝐵𝐵𝐵𝑡𝑡 = 𝐿𝐿𝑜𝑜 (1 − 𝑒𝑒 −𝑘𝑘𝑘𝑘 )
DO Sag Curve:
Dt =
𝐿𝐿𝑡𝑡 = 𝐿𝐿𝑜𝑜 𝑒𝑒 −𝑘𝑘𝑘𝑘
[EQN 7-4]
DOt = DOSAT − Dt
𝑘𝑘 𝑇𝑇 = 𝑘𝑘20
(𝜃𝜃)𝑇𝑇−20
WW Microbiology:
LogP = LogP0 + nLog 2
Activated Sludge:
𝑉𝑉
𝑄𝑄
𝜃𝜃 =
𝐹𝐹 𝑄𝑄𝑆𝑆𝑜𝑜
=
𝑀𝑀
𝑉𝑉𝑋𝑋
Modified
[EQN 8-18]



[EQN 8-26]
Atmospheric Stability: Neutral: Γ𝑎𝑎 =
[p. 626]
Unstable: Γ𝑎𝑎 =
Stable:
Gaussian Dispersion Model:
𝑉𝑉𝑋𝑋
𝑄𝑄𝑤𝑤 𝑋𝑋𝑟𝑟
Γ𝑎𝑎 =
Δ𝑇𝑇
Δ𝑧𝑧
Δ𝑇𝑇
Δ𝑧𝑧
Δ𝑇𝑇
Δ𝑧𝑧
S=
[EQN 8-19]
Modified
[EQN 8-39]
= 𝑧𝑧2−𝑧𝑧1 = Γ𝑑𝑑 =
2
1
𝑇𝑇2 −𝑇𝑇1
= 𝑧𝑧
< Γ𝑑𝑑
2 −𝑧𝑧1
𝑇𝑇2 −𝑇𝑇1
= 𝑧𝑧
[p. 403]
[p. 420]
−1.00𝑜𝑜 𝐶𝐶
100 m
[EQN 8-21]
𝜃𝜃𝑐𝑐 𝑌𝑌(𝑆𝑆𝑜𝑜 − 𝑆𝑆)
𝜃𝜃(1 + 𝑘𝑘𝑑𝑑 𝜃𝜃𝑐𝑐 )
Modified
[EQN 8-23]
Yobs =
Y
1 + k dθ c
𝑉𝑉𝑠𝑠𝑠𝑠 =
𝑀𝑀𝑠𝑠
(𝜌𝜌)(𝑆𝑆𝑠𝑠𝑠𝑠 )(𝑃𝑃𝑠𝑠 )
  22.414 L  
T2
  101.325kPa  6
 
10


mol
273.15
K
P




2



𝑇𝑇 −𝑇𝑇
[p. 403]
K s (1 + k dθ c )
θ c (µm − kd ) − 1
𝑋𝑋 =
Modified [EQN 8-37]
𝑸𝑸(𝑺𝑺𝒐𝒐 − 𝑺𝑺)
=
− 𝟏𝟏. 𝟒𝟒𝟒𝟒(𝑷𝑷𝒙𝒙 )
𝒇𝒇
 1
ppmv = C p 
 MW p

[EQN 7-46]
[EQN 8-2]
𝜃𝜃𝑐𝑐 =
Px = Yobs Q(S o − S )
𝑴𝑴𝑶𝑶𝟐𝟐
Air Conversion**:
k 
k − kd
1
ln  r 1 − D a r
k r − k d  k d 
k d La
Rate adjustments: For k and kd: θ = 1.135 (T between 4 – 20 oC)
θ = 1.056 (T between 20 – 30 oC)
For kr: θ = 1.024 (for all temperatures)
[EQN 7-6]
Sludge Production/O2:
tc =
[EQN 7-41]
Modified [EQN 7-3]
( when t = a: 𝐷𝐷𝐷𝐷𝑎𝑎 = 𝐷𝐷𝐷𝐷𝑆𝑆𝑆𝑆𝑆𝑆 − 𝐷𝐷𝑎𝑎 )
Modified [EQN 7-31]
k d La
(e − k d t − e − k r t ) + Da (e − k r t )
kr − kd
𝐿𝐿𝑡𝑡 = 𝐿𝐿𝑎𝑎 𝑒𝑒 −𝑘𝑘𝑑𝑑 𝑡𝑡
[EQN 7-3]
[EQN 8-36]
[EQN 8-61]
[variant of EQN 9-7]
= −0.01𝑜𝑜 𝐶𝐶/𝑚𝑚
> Γ𝑑𝑑
2 −𝑧𝑧1
Modified
[EQN 9-25]
H = h + ∆H
[EQN 9-26]
∆H =
vs d
u


 T s − Ta
−2
1.5 +  2.68 x10 ( P)
 Ts


s y = ax 0.894
Air Adsorption:
Conversion Factors:
Z −δ
tB = t
vf
[EQN 9-58]
1 mg/L = 1 ppm (in water)
1 m3/s = 22.8245 MGD
1 d = 86,400 s
1Pa = 1 kg/(ms2)
**NOTE: Cp units are g/L
vf =
[EQN 9-28]
(
Q g 1 + bC g*
aρ s ρ g Ac
)
 
d 

 
[EQN 9-27]
s z = cx d + f
[EQN 9-29]
[where
Qg = mass/time]
[EQN 9-60]
1 L = 0.2642 gal 1 μg = 10-3 mg = 10-6 g = 10-9 kg = 10-12 Mg
1 m3 = 1000 L
Degrees K = Degrees C + 273.15
g = 9.81 m/s2
1 m3 = 35.31 ft3
1 ppm = 1,000 ppb
As of: 8 APR 2021