PROBLEM 1
Given:
source: Introduction to Carbon Capture and Sequestration, Pg. 290, Table 7.2.1
Assumption:
Membrane is operated at a steady state system/operation
Small stage cut which means the feed’s pressure is equal to the retentate’s pressure.
Calculations:
Formulas: source: Introduction to Carbon Capture and Sequestration, Pg. 293-300
Binary mixture mass balance:
Using these equations above, partial pressures of CO2 and N2 on each membrane’s sides could be related to flux per unit area through the membrane,
Separation factor (binary mixture):
Due to a very small stage cut we could make an assumption that the permeate pressure is zero. Thus, the separation factor becomes
Since the stage cut is very small, we could assume that:
Recalling the binary mixture mass balance equations and combining them with the equations above gives us result:
Rearranged into:
A.
Figure 1: % of in the permeate versus selectivity
MATLAB CODE:
xco2_feed = 0.12;
p_ratio = 5;
pFlue = 5;
j = 0;
xco2_permeate = [];
alpha = linspace(10,1000,500);
for a = alpha
formula = @(f_A) (pFlue.^2./p_ratio – pFlue.^2./(p_ratio.^2)).*f_A.^2 + (-(a+1).*pFlue./p_ratio +(1+(a-1).*xco2_feed).*pFlue).*f_A – a;
j = j+1;
f_A = fsolve(formula,50);
xco2_permeate(j) = f_A.*pFlue.*xco2_feed./(f_A.*pFlue./p_ratio + 1);
end
semilogx(alpha, xco2_permeate,’m–‘)
xlabel(‘alpha’)
ylabel(‘x_C_O_2_,_P’)
B.
Figure 2: % of in the permeate versus pressure ratio
MATLAB CODE:
xco2_feed = 0.12;
alpha = 25;
pFlue = 5;
j = 0;
xco2_permeate = [];
Rrange = linspace(10,10000,500);
for p_ratio = Rrange
formula = @(f_A) (pFlue.^2./p_ratio – pFlue.^2./(p_ratio.^2)).*f_A.^2 + (-(alpha+1).*pFlue./p_ratio +(1+(alpha-1).*xco2_feed).*pFlue).*f_A – alpha;
j = j+1;
f_A = fsolve(formula,50);
xco2_permeate(j) = f_A.*pFlue.*xco2_feed./(f_A.*pFlue./p_ratio + 1);
end
semilogx(Rrange, xco2_permeate,’m–‘)
xlabel(‘P_f_e_e_d/P_p_e_r_m_e_a_t_e’)
ylabel(‘x_C_O_2_,_P’)
C.
By observing Figure 1 in part A, we could see that as the selectivity increases the amount of percent carbon dioxide in the permeate increases as well. However, Figure 2 in part b shows that the amount of percent carbon dioxide in the permeate will start to get steady after the maximum has been reached which in this particular case is when the pressure ratio is about 1000.
Increasing the pressure ratio will lead us to a higher cost and energy requirement. Thus, the optimal operating range could be achieved by increasing the amount of percent carbon dioxide in the permeate and lowering the pressure ratio. We also need to consider the consequence of increasing the selectivity of the membrane since we would need to increase the surface area of the membrane as we increase its selectivity.
In conclusion for this particular case, the optimal operating range for this membrane is when the pressure ratio has not exceeded 15 (according to Merkel, T. C., et al.) while the selectivity is around 101 to 102 before it starts to get steady at a constant level.
PROBLEM 2
Given:
Problems
a
b
Calculations:
Using formula provided in the textbook (Introduction to Carbon Capture and Sequestration) page 335:
a. Plugging in known energy values in question part a, we got:
source : https://en.wikipedia.org/wiki/Boltzmann_constant
The function of Henry Coefficient in respect to Temperature was then plotted using MATLAB. (MATLAB code was included below the figure for part b).
Figure 3: The Henry Coefficient versus Temperature for Uc = -90 kJ/mole, Uw = -20 kJ/mol
b. Plugging in known energy values in question part b, we got:
source : https://en.wikipedia.org/wiki/Boltzmann_constant
The function of Henry Coefficient in respect to Temperature was then plotted using MATLAB. (MATLAB code was included below the figure for part b).
Figure 4: The Henry Coefficient versus Temperature for Uc = -25 kJ/mole, Uw = -20 kJ/mol
MATLAB CODE:
1. File 1: plotContribution.m
function h=plotContribution(Uc,Uw)
% Variables:
kb = 8.3144598 %J/K
Vc_VT = 1/2
Vw_VT = 1/10
T = linspace(260,310,100)
%Henry Coefficient Calculation Formula
Hc = @(T,kb,Vc_VT,Uc) ((1/(kb*T))*(Vc_VT)*exp((-Uc)/(kb*T)))
Hw = @(T,kb,Vw_VT,Uw) ((1/(kb*T))*(Vw_VT)*exp((-Uw)/(kb*T)))
% Constructing x and y data
cavity = []
window = []
for i= 1:length(T)
cavity(1,i) = Hc(T(i),kb,Vc_VT,Uc)
window(1,i) = Hw(T(i),kb,Vw_VT,Uw)
end
totalboth= cavity+window
x_rt = [298.15 298.15]
y_rt = [0 max(totalboth)]
x_z = [273.15 273.15]
y_z = [0 max(totalboth)]
% Plotting
h = figure
plot(T,cavity,’m–‘,T,window,’r–‘,T,totalboth,’g-‘,x_rt,y_rt,’b-‘,x_z,y_z,’b-‘)
xlabel(‘Temperature (Kelvin)’)
ylabel(‘The Henry Coefficient’)
j = length(T)/10
l = length(T)/4
cavity_ylabelpick = cavity(j)
cavity_xlabelpick= T(find(cavity==(cavity_ylabelpick)))
window_ylabelpick = window(j)
window_xlabelpick= T(find(window==(window_ylabelpick)))
totalboth_ylabelpick = totalboth(l)
totalboth_xlabelpick= T(find(totalboth==(totalboth_ylabelpick)))
text(cavity_xlabelpick,cavity_ylabelpick, ‘ Cavity Contribution (pink line) ‘)
text(window_xlabelpick,window_ylabelpick, ‘ Window Contribution (red line)’)
text(totalboth_xlabelpick,totalboth_ylabelpick, ‘ Total Henry Coefficient (green line)’)
text(273.15, max(totalboth), ‘ 0 degree Celcius’)
text(298.15, max(totalboth), ‘ Room Temperature’)
2. File 2: contributionFigure.m
clear
clc
plot_a = plotContribution(-90000,-20000) % part a
plot_b = plotContribution(-25000,-20000) % part b
c.
Model of Nanoporous Membranes
Source: Lecture slides Chapter 7 – Membranes II
The model of Nanoporous membrane shows that its window’s volume is smaller than its cavity volumes. Thus, by observing Henry coefficient formula and looking at both figure in part a and b, we find that the energy of the cavity dominates the total Henry coefficient at low temperature. The contribution of window to the Henry coefficient tends to be less than the contribution of cavity to the Henry coefficient at lower temperature. However, as the temperature increase the contribution of both the window and the cavity to the Henry coefficient starts to level off and becomes equal to zero.
As we increase the cavity adsorption energy, the difference between the contribution of cavity and the contribution of window will get even larger at low temperature. Increasing the cavity adsorption energy will increase the contribution of cavity and decrease the contribution of window at low temperature. It will also level off both contributions at a lower temperature. The Henry coefficient in figure part a (figure with a higher cavity adsorption energy) reaches zero at around 300 Kelvin, meanwhile in figure part b at around 300 Kelvin the Henry coefficient is still around 5.0 to 10.0.
PROBLEM 3
By controlling the volume of the cavity and the diameter of the window of the Nanoporous membrane (with an assumption that these changes would not impact the energies in both the cavity and the window), we can change the diffusion coefficient. By increasing the volume of the cavity, we increase the absorbed molecules’ entropy which leads to a decrease in the free energy. The decrease of free energy will increase the barrier for molecules to diffuse. In conclusion, increasing the volume of cavity will decrease the diffusion coefficient. Another way to change the diffusion coefficient is by changing the window diameter of the Nanoporous membrane. As the ratio between in figure question 2 part c increases the diffusion coefficient decreases.
First, figure (c) and figure (b) are compared. They both have the same ratio of , so the window diameter will not differentiate their diffusion coefficients. Yet, figure (c) has a smaller volume of cavity. The inversely proportional relationship between the volume of cavity and the diffusion coefficient makes figure (c) has a higher diffusion coefficient than figure (b).
Figure (a), figure (b), and figure (d) have the same volume of cavity, but they have different ratio of . As we can see from those figures, figure (a) has the highest value of , meanwhile figure (d) has the lowest value of
Based the principal that is stated in paragraph 1, the higher the ratio of , the higher the diffusion coefficient. Thus, we could conclude:
Lastly, figure (a) and figure (c) are compared. Figure (a) have both a higher volume of cavity and also a higher ratio. Thus, overall figure (a) has a higher total volume of cavity and window which allow the membrane to contain a larger number of molecules. Moreover, its larger windows will allow molecules to easily pass through each window. In conclusion:
Diffusion coefficient of Nanoporous membranes in decreasing manner:
PROBLEM 4
Given:
A natural gas well produces 1000 m3 / day (at 1atm and 20oC)
Using this membrane gives us 98 mol % purity of CH4 and the residual stream and 89 mol % CO2 in the permeate stream.
Assumption:
vacuum conditions on the permeate side
Calculation:
PROBLEM 5
There are numerous types of membrane with different properties and functions since different polymers construct membranes. Their high selectivity characteristic makes them good alternatives to perform a separation process. Moreover, the utilization of membrane technology in the separation process allows us to save up water and energy. Membrane requires a little amount of water to be fed into the separation process which will be removed once it enters vacuum pump1 and also requires low energy due to the absence of phase transformation in separation process by using a membrane. Although there is a small amount of additional energy needed to treat the flue gas before it enters the membrane, the overall energy required for separation using membrane is relatively smaller than the amount of energy required in absorption and adsorption separation process. Membranes are commonly produced from non-harmful material thus they would not threaten our environment2. A prediction made by The Institute of Sustainability Studies states that membrane will get into the commercial market in 20333. Global CCS Institute estimates that there will be a removal cost of 60 US dollars per tonne carbon dioxide with a target deployment before 2020. Unfortunately, using membrane will lead us into the trade-off between capital expenditure and operational expenditure as a consequence of the trade-off between the membrane area and the energy requirement.
In absorption, the water usage is relatively low due to the utilization of amine-based solvent such as monoethanolamine. The amine-based solvent is preferred to be used in CO2 absorption process because of its capability to absorb a more significant amount/ portion of CO2 from the flue gas. Unlike using a membrane, using a solvent in absorption process would reduce the capital expenditure and operational expenditure since the solvent could be reused for multiple runs of CO2 separation process by stripping off the CO2 that is contained in the solvent. However, when it comes to energy usage, absorption technology has energy losses (which one of them is thermodynamic losses) approximately around 10 to 20 percent in total4. There are also obstacles that are needed to be faced if we want to implement absorption technology to perform the CO2 separation. Using amine-based solvent would establish the possibility of equipment corrosion and solvent emission which is harmful to both human’s health and environment. We also need to deal with the degradation of the solvent while running the plant. The estimated cost of using amine-based solvent in absorption to separate CO2 is around 52 until 77 US dollar5.
Adsorption works quite similarly with absorption, but instead of using a solvent to separate the CO2, adsorption uses solids. By using solids, water usage in the adsorption process is relatively low. These solids are called as adsorbents which could also be regenerated like the solvent in absorption. Regeneration of adsorbents lead to a lower amount of capital expenditure and operational expenditure, however, stripping off the CO2 from the adsorbents requires heat which will increase the operational expenditure. It requires heat since we need to remove all of the moisture in order to regenerate the initial adsorbents’ CO2 loading capacity. Using solid adsorbent to commercialization costs a similar amount of money and time as using membranes, except for some specific adsorbents such as MOFs-based adsorbents which cost more money and consume a longer time. Adsorption’s impact on the environment depends on the types of adsorbents used. The current challenge we are facing upon implementing adsorption for CO2 separation at commercial scale is finding the ideal material which has high selectivity, has a large amount of CO2 loading capacity, low in cost (both operational cost and capital cost), and has no harmful impacts on the environment. The better the material, the better the performance of absorption column in adsorbing CO2 and the lower the operational expenditure and the capital expenditure. Even though zeolite is a suitable candidate to be used as adsorbents due to its pores which allow it to separate CO2, zeolite could have given a better performance under high temperature and high-pressure environment which would increase the operational cost of the adsorption column6. Another challenge which lowers the efficiency of adsorption technology is the energy cost required to transport the adsorbents in a huge quantity for a large-scale of separation operation. It is approximated that Temperature Swing Adsorption will cost around 50 to 80 US dollars per tonne of CO22.
SOURCES:
1. Went, Lecture 3 Unit 4
2. Rochelle, 2009 qtd in Yu et al (2012)
3. Merkel, T.C., Lin, H.Q., Wei, X.T., Baker, R. J. Membr. Sci. 2010,3 59, 126.
4. http://www.separationprocesses.com/Membrane/MT_Chp01c.htm
5. https://www.iass-potsdam.de/sites/default/files/files/working_paper_environmental_impacts_of_carbon_capture_technologies.pdf
6. Energy Fuels, 2005, 19 (3), pp 1153–1159