Question 11.1: A PHES system (refer to Figure 11.2) equipped with a reversi......

A PHES system (refer to Figure 11.2) equipped with a reversible pump–turbine and motor–generator units is operating under the following conditions:

• Static head, that is, elevation difference between upper and lower reservoirs H = 600 m
• Volumetric water flow rate in pumping mode VpV_{p} = 90 m³/s
• Daily duration of pumping period τp\tau_{p} = 8.5 h
• Overall pumping efficiency ηp\eta_{p} = 0.83
• Volumetric water flow rate in generating mode VtV_{t} = 135 m³/s
• Daily duration of power generation period τgen\tau_{gen}= 6.5 h
• Overall power generation efficiency ηgen\eta_{gen} = 0.92

Calculate (a) the plant power output in generating mode, (b) the daily electricity generation, (c) the pumping power input, (d) the daily pumping energy demand, and (e) the turnaround efficiency of the energy storage.

Annotation 2023-05-21 204128
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1. Plant power output in generating mode

Pgen=grVtHhgen=9.81×1000×135×600×0.92=731MWP_{\mathrm{gen}}\,=\,g\mathbf{r}\,V_{t}H\mathbf{h}_{\mathrm{gen}}\,=\,9.8 1\times1 000\times1 35\times600\times0.92\,=731\,\mathrm{MW}

2. Daily electricity generation

Egen=tgenPgen=6.5×731=4751.8  MWh/dayE_{\mathrm{gen}}\,=\,\mathbf{t}_{gen}P_{\mathrm{gen}}\,=\,6.5\times731\,=\,4751.8\,\,\mathrm{MW\,h/day}\,

3. Pumping power input

Pp=grVρH/hp=9.81×1000×90×600/0.83=638.24  MWP_{p}\,=\,g\,\mathbf{r} V_{\rho}H\,/\,\mathbf{h}_{p}\,=\,9.81\times1 000\times90\times600\,/\,0.83\,=\,638.24\,\,\mathrm{MW}

4. Daily pumping energy demand

Ep=tpPp=8.5×638.24=5425 MWh/dE_{\mathrm{p}}\,=\,\mathbf{t}_{p}P_{\mathrm{p}}\,=\,8.5\times638.24\,=\,5425~\mathrm{MWh/d}

5. Turnaround efficiency of PHES system based on energy output EgenE_{gen} and energy input EpE_{p}

ηs=Egen/Ep=4751.8/5425=0.876=87.6%.\eta_{s}=E_{\mathrm{gen}}/E_{p}=4751.8/5425=0.876=87.6\%.

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