Article pubs.acs.org/IECR
Energy Consumption in Spiral-Wound Seawater Reverse Osmosis at the Thermodynamic Limit Mingheng Li* Department of Chemical and Materials Engineering, California State Polytechnic University, Pomona, California 91768, United States ABSTRACT: Several theorems are presented and proved analytically for normalized specific energy consumption (NSEC) in single- or multistage spiral-wound seawater reverse osmosis (RO) desalination with or without brine energy recovery. It is shown that a spiral-wound RO unit employing an infinite number of RO stages and interstage booster pumps is essentially equivalent to a fully reversible RO process in terms of NSEC if the energy in the brine can be fully recovered. Without energy recovery, the minimum of NSEC decreases monotonically from 4 to 3.1462 and the corresponding optimal recovery increases monotonically from 50% to 68.22% as the number of RO stages increases from 1 to infinity. NSEC decreases monotonically as the number of stages or the energy recovery device (ERD) efficiency increases. An effective way to reduce NSEC in seawater RO is to use an ERD at low recoveries and a multistage design with interstage booster pumps at high recoveries. A combination of the two methods would bring NSEC toward its lowest theoretical limit, but the capital cost should also be taken into account. Brine recirculation does not reduce NSEC because of an increase in feed salinity.
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INTRODUCTION Specific energy consumption (SEC) is an important topic in reverse osmosis (RO) desalination. It is defined as pump energy consumption per unit amount of produced permeate. SEC has dimensions of pressure, and the quantity obtained when SEC is normalized by the initial transmembrane osmotic pressure, namely, the normalized specific energy consumption (NSEC), is dimensionless. The lowest NSEC occurs in an ideal, thermodynamically reversible RO process, where both the driving force (the difference between the transmembrane hydraulic pressure and the osmotic pressure) and the permeate flux are infinitesimal at any instantaneous time. For a unit amount of feedwater, if Δπ0 and Y denote the initial transmembrane osmotic pressure and water recovery, respectively, the transmembrane hydraulic pressure should be Δπ0/(1 − Y) when the permeate production is dY. Therefore, the NSEC corresponding to a final recovery Yo is determined as follows:1
pressure plus the pressure loss due to friction) would lead to the lowest energy cost.2,3 It should be noted that even though the driving force is close to zero at the end of the spiral-wound RO element, it is fairly large at the entrance (see Figure 1).
Y
NSEC =
Δπ ∫0 o 1 − Y0 dY
Yo·Δπ0
=−
ln(1 − Yo) Yo
(1) Figure 1. Schematic of the thermodynamic limit in spiral-wound seawater RO (solid line, ΔP; dashed line, Δπ).
Equation 1 provides the lower limit of NSEC posed by thermodynamics. It reveals that brine rejection is absolutely necessary because NSEC skyrockets as Yo approaches 1. Equation 1, even though useful, may be inadequate to describe an industrial RO process, which typically employs spiral-wound membrane elements. In such a configuration, retentate flows through feed spacer channels in the axial direction while clean water passes through the membrane into permeate channels and then flows in a spiral direction into a collection tube located in the center. Both hydraulic and osmotic pressures change with location, and therefore, the system behaves differently from a thermodynamically reversible RO. In spiral-wound RO, it has been shown that operation near the minimum level of applied pressure (i.e., the brine osmotic © 2014 American Chemical Society
Because of the distributed nature of the pressures in spiralwound RO, its NSEC at the thermodynamic limit (i.e., where the driving force is zero at the end of the RO) is not governed by eq 1. Detailed computational studies of thermodynamic restrictions on energy minimization in spiral-wound RO have Received: Revised: Accepted: Published: 3293
December 2, 2013 January 20, 2014 January 29, 2014 January 29, 2014 dx.doi.org/10.1021/ie404067s | Ind. Eng. Chem. Res. 2014, 53, 3293−3299
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Figure 2. Schematic of a multistage RO system with energy recovery.
been carried out.4 A commonly acknowledged fact is that a spiral-wound RO unit has an optimal recovery of 50% and that the corresponding minimum of NSEC is 4. The optimal solution occurs at the thermodynamic limit, where the driving force at the end of the RO unit is zero.4 Apparently, this NSEC is much higher than the one calculated by eq 1; the latter is only 1.4 when Yo = 50%. How to reduce the gap between spiralwound and fully reversible RO becomes an interesting problem. Recently, two important dimensionless parameters γ and κ have been introduced by the author to quantify the effect of membrane capacity demand ratio and retentate pressure drop ratio.5−7 The parameter γ is defined as ALpΔπ0/Q0, where A, Lp, and Q0 are the membrane area, hydraulic permeability, and feed rate, respectively. The parameter κ is defined as kQ02/Δπ0, where k is the pressure drop coefficient. It has been shown that for seawater RO (SWRO), κ is close to zero while γ is large, and the minimum of NSEC occurs near the thermodynamic limit.5−7 Analysis and optimization have been carried out for single- or multistage SWRO desalination with or without the use of an energy recovery device (ERD), mainly using numerical methods. However, for brackish water RO (BRWO), modern plants are not built with large enough γ. Moreover, κ is large and the pressure drop effect cannot be ignored. Different from SWRO, the minimum of NSEC in BWRO occurs far away from the thermodynamic limit.7−9 The model-based optimization results were validated in a BWRO desalination plant in Chino, California, and a 10% reduction in SEC and a significant saving in brine disposal cost were demonstrated.10 This work aims to provide a fundamental understanding of NSEC in seawater desalination employing spiral-wound RO elements. With the assumptions of negligible κ and adequately large γ (which are reasonable for seawater RO), it is possible to derive several analytical equations to demonstrate the gap between the NSEC in spiral-wound RO and the one in fully reversible RO and how this gap can be reduced.
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NSEC =
N[(1 − Yo)−1/ N − 1] + (1 − ηerd) Yo
(2)
where Yo is the system-level water recovery, N is the number of RO stages, and ηerd is the ef f iciency of the ERD. Proof. A generic RO system is shown in Figure 2. In such a system, pressurized feed is sent to the first RO stage. A fraction of the feed is recovered as permeate product, and the retentate is sent to a booster pump before going to the next stage for further water recovery. The reject after the last RO stage, or brine, which has a high pressure, exchanges energy with a feed stream to one of the RO stages to subsidize the total pump energy consumption. The NSEC of such an RO system, neglecting the pump efficiency, was derived in the author’s previous work:6
NSEC =
N − 1 Yj αj
∑j=1
+
1 αN
−
ηerd(1 − YN ) αN
j=1
1 − Π (1 − Yj) N
(3)
where αj is the transmembrane osmotic pressure/hydraulic pressure ratio at stage j (i.e., αj = Δπj/ΔPj) and Yj is the fractional recovery at stage j. Equation 3 was derived on the basis of the total pump energy consumption divided by the total permeate production and the transmembrane osmotic pressure at the feed location. It is valid even if the number of RO stages is 1 or no ERD is used. At the thermodynamic limit (where γj is sufficiently large), αj + Yj = 1 for 1 ≤ j ≤N.5,6 Therefore, the NSEC expression may be simplified as follows: N
NSEC =
1
∑ j = 1 α − N + (1 − ηerd) j
j=1
1 − Π αj N
(4)
The minimization of the NSEC is subject to a specified overall recovery (Yo):
RESULTS AND DISCUSSION
j=1
Yo = 1 − Π αj
Theorems on NSEC at the Thermodynamic Limit. In this subsection, several theorems for NSEC of spiral-wound seawater RO at the thermodynamic limit will be presented and proved. The NSEC is based on the RO unit (which may have several stages; each stage may have several pressure vessels in parallel, and each pressure vessel may contain several RO elements in series) instead of the entire desalination plant. The effects of pre- and post-treatment as well as capital cost will be discussed in a subsequent section. Theorem 1. The minimum of NSEC for a single- or multistage spiral-wound RO unit at the thermodynamic limit is
N
(5)
The optimal solution is determined by setting the partial derivatives of NSEC with respect to αj (j = 1, ..., N) equal to zero and solving all of the resulting equations. It can be shown that α1 = α2 = ... = αN = α and that α satisfies the following equation: α = (1 − Yo)1/ N
(6)
Therefore, the minimum of NSEC at the thermodynamic limit with energy recovery is 3294
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Industrial & Engineering Chemistry Research NSEC =
Article
N[(1 − Yo)−1/ N − 1] + (1 − ηerd)
NSEC = −
Yo
(7)
N[(1 − Yo)−1/ N − 1] + 1 Yo
(8)
Theorem 2. For a single- or multistage spiral-wound RO unit, the reduction in NSEC due to an ERD at the thermodynamic limit is ηerd/Yo. The reduction in NSEC increases as ηerd increases. The reduction in NSEC due to an ERD is more signif icant at low recoveries. Proof. On the basis of eqs 7 and 8, the difference between the NSEC values with and without the ERD, Δerd, is given by Δerd =
N[(1 − Yo)−1/ N − 1] + (1 − ηerd) Yo −1/ N
− =−
N[(1 − Yo)
− 1] + 1
Yo ηerd Yo
(9)
Therefore, the reduction in NSEC increases as ηerd increases. When ηerd = 100%, the reduction in NSEC is 1/Yo. Another way to prove this theorem is to calculate the energy that can be recovered from the brine stream. The brine flow rate is Q0(1 − Yo), and the pressure is Δπ0/(1 − Yo) at the thermodynamic limit. Therefore, Δerd = −
ηerd[Q 0(1 − Yo)][Δπ0/(1 − Yo)] Q 0YoΔπ0
=−
ΔN = NSECN + 1 − NSECN
ηerd Yo (10)
x
> 2x−1(x[1/(N + 1) + 1/(N − 1)]/2 − x1/ N ) > 0 for any x > 1 and N ≥ 2
(18)
Therefore, g(x) > g(1) = 0, from which it follows that
}
ΔN − 1 < ΔN
−ln(1 − Yo) + (1 − ηerd) Yo
(17)
g ′(x) = x[1/(N + 1)] − 1 − 2x[1/ N ] − 1 + x[1/(N − 1)] − 1
{
=
(16)
For g(x) = (N + 1)x1/(N+1) − 2Nx1/N + (N − 1)x1/(N−1), the derivative of g with respect to x is
− 1 + (1 − ηerd)x
e−x ln(1 − Yo) − 1 + (1 − ηerd)x 1 = lim Yo x → 0 x 1 = lim e−x ln(1 − Yo)[−ln(1 − Yo)] Yo x → 0 + (1 − ηerd)
(15)
shows that
+ (N − 1)(1 − Yo)−1/(N − 1)]
Yo
(1 − Yo) 1 lim Yo x → 0
−1
ΔN − ΔN − 1 = (NSECN + 1 − NSECN ) − (NSECN − NSECN − 1) 1 = [(N + 1)(1 − Yo)−1/(N + 1) − 2N (1 − Yo)−1/ N Yo
N[(1 − Yo)−1/ N − 1] + (1 − ηerd) −x
(14)
This implies that NSEC decreases monotonically as N increases. Moreover, consider the difference between ΔN and ΔN−1 at the same Yo:
(11)
N →∞
=
(N + 1)(1 − Yo)−1/(N + 1) − N (1 − Yo)−1/ N − 1 Yo
ΔN = NSECN + 1 − NSECN < 0
at the thermodynamic limit. If ηerd = 100%, the process is equivalent to an ideal reversible RO process. Proof. For an infinite number of RO stages, lim NSEC = lim
=
Therefore, f(x) < f(1) = 0. Setting x = (1 − Yo)
−ln(1 − Yo) + (1 − ηerd)
N →∞
(N + 1)[(1 − Yo)−1/(N + 1) − 1] − N[(1 − Yo)−1/ N − 1] Yo
f ′(x) = x[1/(N + 1)] − 1 − x[1/ N ] − 1 < 0 for any x > 1
It is apparent that Δerd ′ (Yo) = ηerd/Yo > 0, or that the absolute value of Δerd decreases monotonically as Yo increases, which means that an ERD is more effective at low recoveries. Theorem 3. For an inf inite number of RO stages, Yo
=
For f(x) = (N + 1)x1/(N+1) − Nx1/N − 1, the derivative of f with respect to x is
2
NSEC =
(13)
Equation 13 is exactly the same as eq 1,1 which shows the lowest NSEC an RO process can achieve without violating the laws of thermodynamics. Therefore, an infinite number of RO stages (using spiral-wound elements) with interstage pumps and a 100% ERD at the thermodynamic limit is essentially equivalent to the ideal reversible RO process in terms of NSEC. This conclusion clearly reveals the gap between NSEC in industrial operation and its theoretical limit posed by thermodynamics. Because most seawater RO employs a single-stage operation, its NSEC, even if equilibrium is reached at the end of the RO, is much higher than the one in the fully reversible RO. Theorem 4. For a spiral-wound seawater RO, the NSEC at the thermodynamic limit decreases monotonically as the number of RO stages N increases. However, the reduction in NSEC with increasing N is less signif icant as N becomes larger. Multistage designs work better at high recoveries. Proof. Let NSECN+1 and NSECN be the NSEC values when the number of RO stages is N + 1 or N, respectively. If ηerd is the same in both cases, the difference in NSECs, ΔN, is given by
Without energy recovery (i.e., ηerd = 0), eq 7 reduces to NSEC =
ln(1 − Yo) Yo
(19)
for any 0 < Yo < 1, which implies that the reduction in NSEC with increasing number of stages mainly occurs when N is small. As N becomes large, a further increase in N may not lead to a differentiable reduction in NSEC.
(12)
When ηerd = 100%, eq 12 simplifies to 3295
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Finally, consider the derivative of ΔN with respect to x, where x = (1 − Yo)−1 > 1:
∂f (x , n) >0 ∂n
⎡ f (x) ⎤′ Δ′N (x) = ⎢ ⎥ ⎣ 1 − x −1 ⎦
As a result, if f(x,n) = 0 is valid, then ∂f (x , n)/∂n dx =− 0 for n ≥ 1 and 0 < x < 1
(28)
This equation can be solved numerically. The solution is Yo = 0.6822, and the corresponding NSEC is 3.1462. Theorem 6. At the thermodynamic limit, when Yo = 0.6822, the NSEC of a single-stage RO system with 100% ERD is equivalent to that of an inf inite-stage RO without ERD. Below this recovery, a single-stage RO with 100% ERD has a lower NSEC. Above this recovery, an inf inite-stage RO has a lower NSEC. Proof. The difference between the NSEC values for these two cases is
(1 − N )(1 − Yo)(N + 1)/ N + (N + 1)(1 − Yo) − 1 = 0
For f(x,n) = (1 − n)x
(27)
This proves that when n changes from N to N + 1, x decreases, or Yo = 1 − x increases. Let YoN+1 and YoN be the optimal values of Yo when the number of stages is N + 1 or N, respectively. Then NSECN+1(YoN+1) < NSECN+1(YoN). Moreover, NSECN+1(YoN) < NSECN(YoN) on the basis of Theorem 4. Therefore, NSECN+1(YoN+1) < NSECN(YoN), that is, the optimal NSEC decreases monotonically as N increases. The lower and upper bounds of NSEC and Yo can be found as follows. For a single-stage RO without energy recovery, it is readily proved that Yo = 0.5 and that the minimum of NSEC is given by [Yo(1 − Yo)]−1 = 4, and these results are widely acknowledged in the RO desalination community.4 For an infinite number of stages without ERD, the minimum of NSEC can be obtained by taking the derivative of NSEC with respect to Yo and setting it equal to zero:
, where z > 1, the numerator in the above
(n+1)/n
(26)
(25)
Therefore, g(x) > g(1) = 0, from which it follows that 3296
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1 Yo(1 − Yo)
NSECbr =
This is exactly the same as the NSEC for a single-stage RO without ERD at the thermodynamic limit. In other words, the energy recovered from brine cancels the additional energy required for separation due to an increase in feed salinity. For multistage RO, where N ≥ 2, the function g( f) = (1 − f) + f(1 − Yo)1/N − (1 − f Yo)1/N can be considered. It can be shown that 1 g ′(f ) = (1 − Yo)1/ N − 1 + (1 − fYo)[1/ N ] − 1Yo (36) N and that
Figure 3. Schematic of a reverse osmosis water desalination process using brine recirculation.
g ″ (f ) = −
Therefore, RO with brine recirculation is equivalent to RO with an ERD, where the feed transmembrane osmotic pressure is Δπ0 Δπ0 = = (1 − f ) + f (1 − Yo) 1 − fYo
Δπ0br
(37)
It should also be noted that g(0) = g(1) = 0 and g′(0) < 0 < g′(1). Therefore, g( f) < 0 holds for any 0 < f < 1, so from the definition of g(f) it follows that
(31)
Yo Y (1 − f ) = o 1 + (1 − Yo)[f /(1 − f )] 1 − fYo
⎞ 1 ⎛⎜ 1 − 1⎟(1 − fYo)[1/ N ] − 2 Yo 2 ⎠ N⎝N
> 0 for any N ≥ 2, 0 < f < 1, and 0 < Yo < 1
The recovery on the system level (i.e., the product divided by the fresh feed) is: Yobr =
(35)
(1 − f ) + f (1 − Yo)1/ N < (1 − fYo)1/ N
(38)
Thus, (32)
(1 − Yo)−1/ N − 1
