3.1.4. Step 4. Hydropower Output and Hydro-Pumped Storage

Hydropower plants supply the peaks of the load curve. The power produced annually from hydropower plants is highly dependent on the hydrological year. In a good year in terms of hydrological conditions, the annual production could reach 5 TWh. A special methodological approach is used for hydropower [35]. Before the integration of PV in the Greek power system, hydropower was used for peak demand supply. At that time peak demand occurred especially in summer period. Today, hydropower has a more complex role. The integration of PV has shifted the need for peak supply from summer to winter. Hydropower will continue to provide peak power supply whenever peak load demand occurs. Additionally, hydropower will balance the variability of rest RES power generation. Renewable power surplus and wind curtailment may occur in low demand periods or in windy ones. During low demand hours hydropower plants are switched off. During peak demand periods, if there is wind power surplus, hydropower plants may reduce their operation saving water for peak demand periods of low wind. So, wind power plants could save water in the hydro plants' reservoirs and hydro generation will not constrain wind power absorption. Good and bad hydrologic years occur in Greece, with an average annual hydro energy output in Greece of 5 TWh. This energy amount is split into the hours of peak demand using a simplified iterative approach, which finally defines the threshold of power demand for hydropower generation. The operation of hydroelectric power plants is defined by the "peak-shaving" method; it is defined by a minimum power demand threshold (*PT\_H*), when this load demand limit is surpassed hydroelectric power is produced. This limit is determined after an iterative procedure (goal seek function in MS Excel is used) for the whole year, so that the integral of hydro annual production will be equal with 5 TWh (typical hydraulic year). In this step, the remaining load demand after subtracting PV, lignite, natural gas, CHP, biomass and wind power absorbed is used (*PD-PV-CONV-W*). Then, hydropower output is defined for every hour of the year (Equations (4) and (5)) with respect to the nominal power of hydro plants in Greece. This approach is very close to the reality, since existing hydro-plants are operated today in

the way of seasonal storage and peak supply. Additionally, the PV's summer peak supply match well with the dry period of Greece from May to September.

$$P\_{H(i)} = \left\{ \begin{array}{c} 0, \text{ if } P\_{\text{D-PV-CONV-W}(i)} < P\_{T\_{-}H} \\\\ \text{Min}\{P\_{H\_{\text{nonW}}}, P\_{\text{D-PV-CONV-W}(i)} - P\_{T\_{-}H}\}, \text{ if } P\_{\text{D-PV-CONV-W}(i)} > P\_{T\_{-}H} \end{array} \right. \tag{4}$$
 
$$\sum\_{i=1}^{8760} P\_{H(i)} = 5 \text{ TWh} \tag{5}$$

By this approach, the power surplus (wind curtailment and PV surplus) *Psurplus* is minimized in most of the conservative scenarios. Obviously, for large-scale wind and PV integration, due to the technical constraints, power surplus occurs and could be transformed through hydro-pumped storage to useful peak demand supply. The aforementioned excess energy is stored in hydro-pumped storage units, by pumping water to the upper reservoir when there is a surplus of energy, while it is recovered through the hydro-turbines operation when other renewable sources are not available.

Due to the distribution of the curtailed power, it is not economically feasible to exploit 100% of this energy. Such a scenario would require enormous installed capacity of pumps and volume of the upper reservoirs which would be used only few hours per year. A sufficient degree of annual exploitation of excess system's power can be considered to be 70% [6]. On this basis, the required installed capacity of pumps *PP\_nom* is defined in order to achieve the target of 70% exploitation of the annual energy surplus. The hydro turbines capacity is considered to be equal with the pumps' nominal output. Therefore, the nominal power of the pumps is calculated (Equations (6) and (7)):

$$P\_{Pump(i)} = \dot{M} \dot{m} \left\{ P\_{P\_{\text{\\_nom}}} P\_{Surplus(i)} \right\} \tag{6}$$

$$\sum\_{i=1}^{8760} P\_{Pump(i)} = 70\% \cdot \sum\_{i=1}^{8760} P\_{surplus(i)}\tag{7}$$

where *Ppump* is the power used for pumping, *PP\_nom* is the nominal power of pumps and *Psurplus* is the available power surplus (wind curtailment and PV surplus). The volume of the upper reservoir is defined to ensure a 24-h operation of the turbines at their nominal capacity.
