A hydraulic structure is submerged or partially submerged in any body of water, which disrupts the natural flow of water. They can be used to divert, disrupt, or completely stop the flow. It also includes structures used for the storage of water such as tanks or large means of provision or conveyance of water. A common example of such is the dam. Table 1 shows common hydraulic structures and their estimated life span.
|S/No||Types of Hydraulic Structures||Estimated life span (years)|
|1||Earth dams, gravity dams||150|
|2||Loose rock dams||60|
|4||Timber dams (crib)||25|
|5||Wells||40 – 50|
|6||Pumps||15 – 25|
The planning of a water resources development project involves a number of alternative designs. The best alternative design is generally the one that is the most economical. To make the assessment possible, a Bill of Quantities (BOQ) for every project component needs to be estimated. Also, the capital and maintenance cost of each alternative needs to be worked out. The different alternatives are examined with regard to the economy and the most favourable overall design is chosen.
The following steps are commonly adapted in the process:
- Identify and clearly define each alternative design in physical terms.
- Prepare the preliminary estimate of each cost of each alternative.
- Determine the benefits likely to accrue from each alternative and convert all tangible benefits into money units.
- Compare the costs and benefits. Usually, the total capital costs (investments) are converted into equivalent annual costs for comparison with annual benefits.
- Determine the benefit/cost ratio (B/C – ratio) of the various alternatives.
- Considering the tangible benefits and other merits/demerits of each alternative design, including other factors like social, political and ecological, etc., select the best design.
Estimation of benefits cost ratio (B/C – ratio)
The benefit-cost ratio is the ratio of the annual benefits to the annual cost.
Mathematically, B/C – ratio = Total annual benefits/ Total annual costs
Estimation of the B/C – ratio first requires the conversion of all capital costs into annual costs such that they can be compared with annual benefits. A project that gives a B/C – ratio greater than unity (1) is considered economically viable (1.5 is usually specified for irrigation projects while 1.1 is ideal for flood control projects).
Estimation of annual costs
This is the total cost of a project and is usually divided into two categories viz:
Capital costs: This is the expenditure done at the time of construction. It is also called initial cost.
Recurring costs: This is the expenditure during the life of the project in maintenance and operation. It is usually expressed as the annual recurring cost.
The capital cost that is received during the lifetime of the project with a certain minimum alternative rate of return (i.e. interest rate) is usually converted to an equivalent annual recovery cost using the capital recovery factor (CRF).
CRF = [i (1 + i)n]/ [(1 + i)n – 1]
i = interest rate per annum
n = estimate of the life of the project in years
Equivalent annual cost = CRF x capital investment (or cost)
This means that the total initial investment will be recovered in n-years along with interest if the equivalent annual recovery is achieved every year throughout the structure’s life.
The total annual cost of the project = equivalent annual recovery cost + the annual recurring (maintenance cost).
Estimated annual benefits
These are the benefits that are likely to accrue by the construction of the project. It is easy to estimate for single purpose projects but difficult for multi-purpose projects. For example, for an irrigation project, the benefits can be estimated if we know the irrigated area, the type of crop, and the yield of the crop. For a hydropower project, the benefits may be estimated from the power generated and the rate of a unit of energy served by the populace.
Determine whether the hydropower project with the following particulars is economically viable
Capital cost = N 10,000,000
Annual maintenance cost = N 115,000
Interest rate, i = 7% per annum
The useful life of the project, n = 120 years
Power generated = 300 kW
Power rate = N 0.5 per kWh
CRF = [i (1 + i)n]/ [(1 + i)n – 1] = [0.07 (1 + 0.07)120]/ [(1 + 0.07)120 – 1] = 235.0451868/ 3356.788383 = 0.07002
Equivalent annual recovery cost = 0.07002 x 10,000,000 = N 700, 200
Total annual cost = 700, 200 + 115, 000 = N 815, 200
Annual benefits = (300 x 365 x 24) x 0.5 = N 1,314, 000
B/C – ratio = Annual benefits/ Annual costs = 1314000/815, 200 = 1.612 ˃ 1 (the project is economically viable).
In some water resources development projects, the benefits increase with an increase in the size of the project. However, as the size of the project grows, so does the cost. A stage is reached when an increase in size may not give the minimum attractive return. The size of the project is prudently fixed at this stage. This is necessary because the aim of the water resources project is often to maximize the net benefits rather than achieve the maximum benefits/cost ratio. To make a decision in this type of situation, economic analysis is carried out and three curves are usually plotted:
- Benefits versus costs curve
- Benefits-cost ratio versus cost curve
- Net benefits versus cost curve
Thus, the decision on the size of the project can be made based on net benefit versus benefit-cost ratio.
Using the data in the table below, plot the necessary curves for economic analysis to choose the most viable project.
From the graph in Figure 2, the benefit-cost ratio is maximum at point A. If the aim of the project is to achieve the maximum rate of return, the size of the project with cost corresponding to point A (Project A) is chosen. The maximum net benefits occur at point B. If the aim of the project is to achieve the maximum net benefits, the size of the project with cost corresponding to point B (Project C) is chosen. Among the two alternatives, project C is more viable because it has the maximum net benefits and the B/C – ratio is also greater than unity (1).
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