Tutorial for Economic model

1. Introduction

Welcome to our comprehensive tutorial on running economic models. The economic model (economic_f.py) can be used to perform economic assessments of the treatment train or of individual units to identify economic hotspots. The economic assessment will help researchers, engineers, and decision-makers improve the system’s performance and select the most economically feasible integration of technologies.

In this tutorial, we provide step-by-step instructions on how to use economic models and analyse the results obtained using Example 1 as a case study (see Figure 1).

**Figure 1**. Process flow diagram of example 1.

2. Running Economic model

Table 1 gives an overview of the main inputs and outputs of the economic model (economic_f.py).

Table 1. Overview of Inputs and Outputs of economic models.
Input	Output
Selling price for products [€/ton] or [€/m³]	Operating cost (OPEX) [€/year]
Prices for energy [€/KWh], input chemicals [€/m³], cooling water [€/m³]	Investment cost (CAPEX) [€]
Operating hours, lifetime	Revenues from selling products [€/year]
Interest rate, Inflation rate
Equipment cost [€]
Assumptions on CAPEX and OPEX calculations

2.1. Import economic model functions

First, import the required functions from economic_f.py.

from desalsim.economic_f import econom
from desalsim.economic_f import revenue

2.2. Set constant values

# Operating hours per year
hr = 24 * 300  # hours/year
# Plant's lifetime
lf = 20  # years
# Interest rate
r = 0.06  # (-)
# Inflation rate
inf = 0.02  # (-)

Import constants.py for more constant parameters like molecular weights (MW) etc.

import desalsim.constants

2.3. Set input data

The market prices for utilities (electricity, chemicals, steam, cooling water, etc.) need to be set. Use updated market values.

# Market prices
# Electricity price
el_pr = 0.253  # euro/kwh
# Steam price
s_pr = 0  # euro/kwh
# Chemicals prices
hcl_pr = 5.78  # euro/l 1M solution
naoh_pr = 5.2  # euro/L 1M NaOH solution
antisc_pr = 0.002  # euro/ml
# Cooling water price
cw_pr = 0.000118  # euro/kg

The market prices for focused products need to be set. Use updated market values.

# Market prices
# Chemicals prices
hcl_pr = 5.78  # euro/l 1M solution
naoh_pr = 5.2  # euro/L 1M NaOH solution
# Industrial water
w_pr = 0.001  # euro/kg
# Salts prices
mgoh2_pr = 1  # euro/kg
caoh2_pr = 0.125  # euro/kg

2.4. Import results from technical model

Before importing the results from example_1.py, ensure that the script has been executed successfully. This can be done either by running the script directly or by making sure the necessary variables are defined and ready for import.

from example_1 import E_el_all, E_th_all, Qchem_all, Mhcl_need, Qnaoh_need, Q_w_in

Important

Ensure that example_1.py has been executed so that these variables are properly defined. If these variables do not exist or the script has not been run, you will encounter an error.

After importing, you can create lists with the results as shown below:

el_conc=E_el_all
s_conc=E_th_all
chem1_conc=Qchem_all
chem1_pr=[0.002,0,constants.naoh_pr_s,0]
chem2_conc=[0,0,Mhcl_need,0]
chem2_pr=[0.002,0,constants.hcl_pr_s,0]
wat_conc=[0,0, Qnaoh_need, Q_w_in]
cw_conc=[0,0, 0, 0]

2.5. Equipment cost

You can set equipment cost if it is known. In case of preliminary study, where equipment cost is still unknown, scale-up function (scaleup.py) can be used.

\[\frac{\text{Cost of purchased equipment (Plant A)}}{\text{Cost of purchased equipment (Plant B)}} = \left( \frac{\text{Capacity of Plant A}}{\text{Capacity of Plant B}} \right)^m\]

Where m is six-tenths factor rule (m=0.6 or m=0.8) depends on the nature of technology.

In this example, equipment costs from pilot units are used as reference. The equipment costs are included in the constantc.py so the user can call the function in case of scale-up and unknown equipment costs.

# Units: nf, mfpfr, edbm, ed
Mf_basic_sc = [2290.86, 595.03, 229.01, 253]

After setting the reference scenario, the equipment cost of the scaled-up unit can be calculated as below:

# Import equipment cost for reference scenario from constants function
eq_c = [constants.eq_c_nf, constants.eq_c_ed, constants.eq_c_mfpfr, constants.eq_c_edbm]
# Capacity of reference scenario
Mf_basic_sc = [constants.Mf_basic_sc[0], constants.Mf_basic_sc[8], constants.Mf_basic_sc[3], constants.Mf_basic_sc[6]]

# Capacity of evaluated treatment train
Mf_sce = [Qsw, Qed_in, Qin_mfpfr, Q_in_edbm]

# Calculation of the new equipment cost
for i in range(len(eq_c)):
    if Mf_basic_sc[i] != Mf_sce[i]:
        eq_c[i] = scaleup.scaleup_eq(eq_c[i], Mf_basic_sc[i], Mf_sce[i], tec_names[i])

2.6. Calculate Capital costs (CAPEX)

The CAPEX consists of fixed-capital investment and working capital, and the former includes hardware costs, costs of buildings, process, and auxiliary, land, working capital, and other indirect costs.

Table 2 gives an overview of the main assumptions made to calculate the CAPEX.

CAPEX
Installation: 25% of purchased equipment cost
Buildings, process, and auxiliary: 20% of purchased equipment cost
Land: 6% of purchased equipment cost
Indirect costs: 15% of direct cost
Working capital: 20% of total investment cost

Note

Note that the assumptions listed in Table 2 can be modified to suit different case studies.

# Set assumptions for Capital investment cost
inst_percent = 0.25  # % of purchased equipment cost
buildings_percent = 0.2  # of purchased equipment cost
land_percent = 0.06  # of purchased equipment cost
indirect_c_percent = 0.15  # of direct cost
working_c_percent = 0.2  # of total investment cost

Based on those assumptions the normalized factor is calculated and all the assumptions are assigned to the capex_assumptions list.

capex_assumptions = [inst_percent, buildings_percent, land_percent, indirect_c_percent, working_c_percent]

Then the CAPEX is calculated:

# Create list with capital cost for each unit. This will help us to identify the hotspots in capital investment.
capex_list = []
# Initialize value
CAPEX = 0
# Calculate investment cost (CAPEX)
for i in range(len(eq_c)):
    total_econom = econom(eq_c[i], el_conc[i], s_conc[i], chem1_conc[i], chem1_pr[i], chem2_conc[i], chem2_pr[i], cw_conc[i], wat_conc[i])
    total_econom.capex_calc(capex_assumptions)
    CAPEX = total_econom.t_capital_inv
    capex_list.append(total_econom.t_capital_inv)
print("Total investment cost (CAPEX) of system is " + str(round(CAPEX)) + " Euro")

2.6.1 Calculate annual capital costs

For the calculation of the annualized CAPEX, the amortization factor (α) is used:

\[\text{Annualized CAPEX} = \text{CAPEX} \times a\]

where a is the amortisation factor.

Amortisation factor is calculated as:

\[a = \frac{r \cdot (1+r)^{lf}}{(1+r)^{lf} - 1}\]

where r is discount rate, lf is plant lifetime (year).

# Calculate amortisation factor
lf = 20  # years
r = 0.06  # interest rate
a = (r * (1 + r) ** lf) / ((1 + r) ** lf - 1)

2.7. Calculate Operating costs (OPEX)

The OPEX refers to expenditure directly generated by manufacturing operation or connected to the equipment of a technical unit. Table 3 gives an overview of the costs that constitute OPEX (Peters, Timmerhaus and West, 2003). In this study, the utilities in this system are mainly energy, chemicals, and water costs. The calculation of yearly electrical (Cel) and thermal (Cth) energy costs follows equations:

\[C_{el} = E_{\text{tot, el}} \cdot t_{\text{operation}} \cdot P_{\text{el}}\]

\[C_{th} = E_{\text{tot, th}} \cdot t_{\text{operation}} \cdot P_{\text{st}}\]

\[C_{e} = C_{el} + C_{th}\]

Where:

E_el and E_th are the total energy consumption per operating hour (in kWh/hr), t_operation is the total operation time in one year (in hr), P_el and P_steam are the prices of electricity and steam, respectively (in €/kWh).

The calculation of chemicals and water costs is similar to the energy cost, multiplying the amount of consumption every year by their price. The other costs are calculated in opex_calc class (see economic.py).

Table 3 gives an overview of the main assumptions made to calculate the OPEX.

Annual OPEX
Maintenance: 3% of the fixed-capital investment
Operating Supplies: 5% of maintenance
Operating Labor: 15% of annual OPEX
Direct supervisory and clerical labor: 15% of operating labor
Laboratory charges: 15% of operating labor
Patents and royalties: 3% of annual OPEX
Fixed charges: 5% of annual OPEX
Plant overhead costs: 5% of annual OPEX

Note

Note that the assumptions listed in Table 3 can be modified to suit different case studies.

# Set assumptions for Operating cost (OPEX)
main_c_percent = 0.03  # % of the fixed-capital investment
oper_sup_c_percent = 0.05  # % of maintenance
oper_lab_c_percent = 0.15  # % of annual OPEX
super_c_percent = 0.15  # % of operating labor
lab_c_percent = 0.15  # % of operating labor
pat_c_percent = 0.03  # % of annual OPEX
fix_char_percent = 0.05  # % of annual OPEX
over_c_percent = 0.05  # % of annual OPEX

Based on those assumptions the normalized factor is calculated and all the assumptions are assigned to the economic assumptions list.

# Calculate normalized factor
norm_factor = (1 - oper_lab_c_percent - oper_lab_c_percent * super_c_percent - oper_lab_c_percent * lab_c_percent - pat_c_percent - fix_char_percent - over_c_percent)

# Create a list for economic assumptions
economic_assumptions = [main_c_percent, oper_sup_c_percent, oper_lab_c_percent, super_c_percent, lab_c_percent, pat_c_percent, fix_char_percent, over_c_percent, norm_factor]

Then the OPEX is calculated:

# Initialize values
OPEX = 0

# Create list with operating for each unit. This will help us to identify the hotspots in operating costs.
opex_list = []

# Calculate OPEX
for i in range(len(eq_c)):
    total_econom = econom(eq_c[i], el_conc[i], s_conc[i], chem1_conc[i], chem1_pr[i], chem2_conc[i], chem2_pr[i], cw_conc[i], wat_conc[i])
    total_econom.opex_calc(hr, el_pr, s_pr, cw_pr, w_pr, economic_assumptions)
    OPEX = total_econom.opex
    capex_list.append(total_econom.t_capital_inv)
print("Total operating cost (OPEX) is " + str(OPEX) + " Euro/year")

2.8. Calculate Revenues

The total amount of Revenues from selling products is used to evaluate the economic performance of the treatment train.

\[\text{revenues} = \sum_{i=1}^{n} Q_{\text{product}_i} \times \text{Selling price of product}_i\]

First, the updated market prices of the recovered products need to be set.

# Quantity of recovered products
prd = [Qw_tot, M_MgOH2_1, Q_b_out, Q_a_out]

# Specify products
prd_name = ["Water", "Mg(OH)2", "NaOH", "HCl"]

# Market prices
hcl_pr = 5.78  # euro/l 1M solution
w_pr = 0.001  # euro/kg
mgoh2_pr = 1.0  # euro/kg
naoh_pr = 7.2  # euro/L 1M NaOH solution

After the set of input parameters, the Revenues of the treatment train are calculated:

# Initialize lists
reve_t = 0
reve_list = []
# Revenue calculation
for i in range(len(prd)):
    rev_calc = revenue(prd[i], prd_name[i])
    rev_calc.rev(hr, w_pr, nacl_pr, mgoh2_pr, na2so4_pr, naoh_pr, hcl_pr)
    print("Revenues from selling product " + prd_name[i] + " are " + str(round(rev_calc.rev_prd, 2)) + " Euro/year")
    reve_t = reve_t + rev_calc.rev_prd
    reve_list.append(rev_calc.rev_prd)

3. Results evaluation

Running the economic model for Example 1, the following results are obtained.

Total operating cost (OPEX) is 3714921 Euro/year

Total investment cost (CAPEX) of the system is 6017026 Euro

Annual capital investment cost of the system is 524592 Euro/year

Revenues from selling product Water are 0.0 Euro/year

Revenues from selling product Mg(OH)2 are 2891274 Euro/year

Revenues from selling product NaOH are 3943229426 Euro/year

Revenues from selling product HCl are 3274716868 Euro/year