Question

(b) The daily design load curve of a residential area is given in the table below. The hydropower facility providing the power has an effective head of 15 m and a plant efficiency of 0.85. What is the minimum daily flow (m3/s) that could meet the demand? What is the minimum required storage capacity (within-day storage) to meet the peak load? If the daily flowrate were 320m3/s, what would be the minimum required storage capacity? Would there be some water released without passing through the turbines in this case, and if so, how much? Power kW 19,000 13,000 18,000 35,000 50,000 54,000 62,000 43,000 19,000 Hour 0300 0600 0900 1200 1500 1800 2100 2400 Hint: ΔΡ-γ(24h, where P is power of flowing water (kW), γ is specific weight of water at 9.8 kN/m3, Q is flow rate (m'/s), and h is energy head (m) You may have to repeat certain calculations for a number of times. You are expected to show the details and steps only once as an example, and you can then just provide the answers for the repeating on (12 marks)

Answer 1

our 152.0 9000 13000 800D 0300 O600 OSO0 144,0 280.1 SOOoD 12 oo yB217 y96.20 SuooD S00 62oo0 18 2100 2400 152. 06 3oo6 2. 3 4u.lu r) 254订752 CS') = 234.11 m3/S-Ars (i) ecapacit¥ is be Based On mininum through mbmes . Scanned with CamScanne CS

De-cienuuuattr Discha Houイ 0000ーーーク152.06 Ogoo う104, oy 205275 280.11 400.16 4321구 496.20 Osoo OS0o 200 so0 う ㅡㅡㅡㅡ→ ugi 048 --> S40324 3 44.14 う!S2.06 2100 2400 total Surplu Cale . 53 58, 636o 5359,284 r) Surplus, Deficena) 5353284 So provide Capacrty mavmum QH CSI Scanned with CamScanner

daily discharge released bypassing turbine = 320-294.11

                                                               =25.89m3/s (ans)