In the upstream oil & gas industry, produced water is a by-product of well production. Hydrocarbon wells initially produce less water but in late field life, the water content increases. Produced water can contain oil carryover and a host of salts with TDS ranging anywhere from 2,000 mg/L to 40,000 mg/L for which evaporation ponds are used to concentrate by evaporating the associated water. The energy requirement consists of pumping concentrate to the pond and in some cases aeration is provided to enhance the rate of evaporation. The ponds are lined with synthetic liner material to prevent seepage of water into the soil. In case of any corrosive compounds in the water, the number of layers is increased. Landscape and topography play a role in setting up evaporation ponds and it is necessary to have a flat terrain to avoid any overflow of the contents. Evaporation ponds must also ensure that the amount of water entering is minimized and avoid any flooding. As part of waste disposal, the ponds maybe designed to accumulate sludge over the life time of the operating wells or can be periodically removed. The below figure depicts an evaporation pond. The following article focuses on estimating the rate of evaporation, water surface temperature and rate of heat transfer to the water in an evaporation pond.

1. Page 1 of 3
Evaporation Pond Process Design in Oil & Gas Industry
Jayanthi Vijay Sarathy, M.E, CEng, MIChemE, Chartered Chemical Engineer, IChemE, UK
In the upstream oil & gas industry, produced
water is a by-product of well production.
Hydrocarbon wells initially produce less water
but in late field life, the water content increases.
Produced water can contain oil carryover and a
host of salts with TDS ranging anywhere from
2,000 mg/L to 40,000 mg/L for which
evaporation ponds are used to concentrate by
evaporating the associated water.
The energy requirement consists of pumping
concentrate to the pond and in some cases
aeration is provided to enhance the rate of
evaporation. The ponds are lined with synthetic
liner material to prevent seepage of water into the
soil. In case of any corrosive compounds in the
water, the number of layers is increased.
Landscape and topography play a role in setting
up evaporation ponds and it is necessary to have
a flat terrain to avoid any overflow of the
contents.
Evaporation ponds must also ensure that the
amount of water entering is minimized and avoid
any flooding. As part of waste disposal, the ponds
maybe designed to accumulate sludge over the
life time of the operating wells or can be
periodically removed. The below figure depicts an
evaporation pond.
Figure 1. Evaporation Pond [2]
The following article focuses on estimating the
rate of evaporation, water surface temperature
and rate of heat transfer to the water in an
evaporation pond.
Methodology
Similarities exist between mass, momentum &
heat transfer phenomenon. Therefore, the
empirical correlations for heat transfer are also
applicable for mass transfer. Schmidt number
plays a similar role to Prandtl number in
convection heat transfer. The heat transfer to the
water from the air supplies the energy required to
evaporate the water,
𝑞 = 𝑚ℎ𝑓𝑔 = ℎ𝐴[𝑡∞ − 𝑡𝑠] = ℎ𝑚𝐴ℎ𝑓𝑔[𝜌𝑠 − 𝜌∞] (1)
Where,
h = Convective heat transfer coefficient [W/m2.k]
hm = Convective Mass Transfer Coefficient [m2/s]
A = Surface Area [m2]
m = Evaporation Rate [kg/s]
ts,s = Surface temperature & vapour density [K, kg/m3]
t, = Air Temperature & vapour density [K, kg/m3]
The energy balance can be arranged as,
𝑠
− 
=
ℎ
ℎ𝑚
[
𝑡−𝑡s
ℎ𝑓𝑔
] (2)
The heat transfer coefficient, h can be estimated
based on Nusselt Number (Nu) as,
𝑁𝑢 = 0.664𝑅𝑒
1
2
⁄
𝑃𝑟
1
3
⁄
, For Laminar Flow (3)
𝑁𝑢 = 0.037𝑅𝑒
4
5
⁄
𝑃𝑟
1
3
⁄
, For Turbulent Flow (4)
The mass transfer coefficient, hm can be calculated
using Sherwood Number (Sh),
𝑆ℎ = 0.664𝑅𝑒
1
2
⁄
𝑆𝑐
1
3
⁄
, For Laminar Flow (5)
𝑆ℎ = 0.037𝑅𝑒
4
5
⁄
𝑆𝑐
1
3
⁄
, For Turbulent Flow (6)
Where,
h𝑚 =
𝑆ℎ×𝐷𝑣
𝐿
(7)
ℎ =
𝑁𝑢×𝑘
𝐿
(8)
Dividing both heat and mass transfer coefficients
and substituting in Eq. (2) yields,
ℎ
ℎ𝑚
=
𝑁𝑢×𝑘
𝑆ℎ×𝐷𝑣
= [
𝑃𝑟
𝑆𝑐
]
1
3
⁄ 𝑘
𝐷𝑣
(9)
𝑠
− 
= [
𝑃𝑟
𝑆𝑐
]
1
3
⁄ 𝑘
𝐷𝑣
[
𝑡−𝑡s
ℎ𝑓𝑔
] (10)

2. Page 2 of 3
The above expression is solved for s and hfg is
evaluated at surface temperature [ts]. Air
properties, k, Dv, Sc, Pr are evaluated at film
temperature [tf], as an average of t and ts.
𝑡𝑓 =
𝑡∞+𝑡𝑠
2
(11)
The solution is arrived beginning with a guess
value of surface temperature, ts in Eq. (10) &
iteratively solved until convergence. Relating
saturated vapour pressure [Ps] with moist air
temperature [T,
0C] using Arden Buck equation,
𝑃𝑠[𝑇∞ > 0℃] = 6.1121𝑒
[(18.678−
𝑇∞
234.5
)×(
𝑇∞
257.14+𝑇∞
)]
× 100 (12)
Where, Ps is saturated vapour pressure [Pa]
The vapour density [] for a given relative
humidity [RH] is calculated as,
∞
= [
𝑃𝑠×𝑀𝑊𝑤𝑎𝑡𝑒𝑟
8314.447×𝑇∞
] × [
𝑅𝐻%
100
] (13)
The mass diffusivity of moisture in air [Dv] is
estimated using Sherwood and Pigford, 1952
expression, valid for mass diffusivity of water
vapour in air up to 1,1000C
𝐷𝑣 = [
0.926
𝑃𝑎𝑚𝑏
] × [
𝑇2.5
𝑇+245
] ×
1
106 (14)
Where,
Dv = Mass diffusivity of moisture in air [m2/s]
Pamb = Atmospheric pressure [kPa]
T = Ambient Temperature [K]
The Schmidt Number (Sc) is estimated as,
𝑆𝑐 =
𝜇
𝜌𝑎𝑖𝑟×𝐷𝑣
(15)
Where,  = Dynamic Viscosity [kg/m.s]
air = Air density [kg/m3]
The Reynolds Number (Re) is estimated as,
𝑅𝑒 =
𝑢∞𝑎𝑖𝑟𝐿
𝜇
(16)
Where, L = Pond Length along direction of air [m]
For the range 00C to 800C, the surface
temperature [Ts] from curve fit data is,
T𝑠[𝐾] = [(19.45777 × 𝑙𝑛[𝜌𝑠]) + 100.4106] + 273.15 (17)
Case Study
Air at 250C & 101.325 kPa flows at 10 m/s along
the length of an evaporation pond of LW of
10m2m. The relative humidity is 60%. The rate
of heat transfer to water, rate of evaporation &
the water surface temperature is to be estimated.
Evaluating the saturated vapour pressure,
𝑃
𝑠 = 6.1121𝑒[(18.678−
25
234.5
)×(
25
257.14+25
)]
× 100 (18)
𝑃
𝑠 = 3,169 𝑃𝑎 (19)
The vapour density at 250C is estimated as,
∞
= [
3,169×18.02
8314.447×298.15
] × [
60
100
] = 0.013822
𝑘𝑔
𝑚3 (20)
Taking an initial guess of 150C, the tf is,
𝑡𝑓 =
15+25
2
= 20℃ = 293.15 𝐾 (21)
Evaluating air properties at tf = 200C, air =1.1975
kg/m3, =0.0000181 kg/m.s, k = 0.0257 W/m0C,
hfg = 2,465 kJ/kg, Pr = 0.7094,
𝐷𝑣 = [
0.926
101.325
] × [
293.152.5
293.15+245
] ×
1
106 = 2.5 × 10−5 𝑚2
𝑠
(22)
𝑆𝑐 =
0.0000181
1.1975×2.5×10−5 = ~0.6063 (23)
𝑠
= [
0.7094
0.6063
]
1
3
⁄
×
0.0257
2.5×10−5
[
298.15−288.15
2,465×1000
] + 0.01382 (24)
𝑠
= 0.01822 𝑘𝑔 𝑚3
⁄ (25)
Estimating the surface water temperature [Ts] for
s =0.0182 kg/m3,
T𝑠 = [(19.45777 × 𝑙𝑛0.01822) + 100.4106] + 273.15 (26)
T𝑠 = 295.62 𝐾 = ~22.5℃ (27)
Recalculating the air properties & iterating the
calculations, Ts = 200C, s = 0.01601 kg/m3, air =
1.1865 kg/m3,  = 0.00001826 kg/m.s, hfg = 2,454
kJ/kg and Dv = 2.5410-5 m2/s. The Reynolds
number & Sherwood number is estimated as,
𝑅𝑒 =
10×1.1865×10
0.00001826
= ~6,497,679 (28)
𝑆ℎ = 0.037 × 6,497,679
4
5
⁄
× 0.6058
1
3
⁄
= 8,828 (29)
The convective mass transfer coefficient is,
h𝑚 =
8,828×2.54×10−5
10
= 0.0224 𝑚/𝑠 (30)
The rate of evaporation [m] is,
𝑚 = 0.022426 × 20 × [0.01601 − 0.013822] (31)
𝑚 = ~0.000987
𝑘𝑔
𝑠
= 85.3
𝑘𝑔
𝑑𝑎𝑦
(32)
The Rate of heat transfer [q] is,
𝑞 = 𝑚ℎ𝑓𝑔 = 0.000987
𝑘𝑔
𝑠
× 2,454
𝑘𝐽
𝑘𝑔
= 2,422𝑊 (33)
References
1. “Heat Transfer”, 10th Ed, Holman JP.
2. https://www.fws.gov/ecological-
services/energy-development/oil-gas.html
3. 2005 ASHRAE Handbook - Fundamentals (SI),
Chapter 5

3. Page 3 of 3
Appendix A