Question

Reservoir A renewable energy system (it can be solar or wind) does not produce power all the time. So when it is generating power some of the excess is used to pump water into a large reservoir that is 50 m above a lake. When the renewable system cannot produce power (no sunshine or wind) then the water can flow from the reservoir back through the turbine to the lake, producing power. How effective is this energy storage system? chin Lake Water Properties & Piping Parameters: H:= 0.00114a's = 0.011.poise P:= 1000 Ex= 0.1510ft = 0.046mm Q:= 1, sec = 15850 gpm mp := 90% Mr = 90% 4:= 100m D:= 35 cm = 13.78 in 22:= 50m Add 10% to losses to account for any minor losses (fittings, valves, etc.)

Answer 1

m .43 TX(0.35 First of all find the frictional head loss through the paping system g.e he flue; Given!- l = 0.001l pans, f-3000 kg/m3 E = 0.0 46 mm ,Q= 1 m3/s, Laivom, D=35cm ve en Axevasse = 10.4 m/s 10 Re: SvD - Inox 10.4 x 035 -- 3309090.9 .: fweren facter; f. 1.325 el drool for 500 < Re 5108 [en lieto + 5 346)]? ' and 10% < $ 5102 11325 D 350 mm 1.325 + Ton ( 1096 + 5139 I en , 01046 7 574 ,772 (3.7x8 Pen (3.554 xjö + 7:3262x76))" (330909019 1. 325 7-20. 047)2 - 947 = 0.01325 is he = {bv?= 0.0325 x 100 x Coogle = 20187 m 29D 2 89,81 x 0.35 Total be ad loss, hno hp + minor loss, which is 10% of total lass) 1 hn = h& torn hd - hn (1-0:1) = hf >> huo en hoe - 20.87 = 23. 188 m O19

.. 019 Now according to Bernouli's equation, the head developed by pumps with 90%. erheeney's F a rm = {* -z,) + mai mare to he]/09 Heres za +2,250-05 som, 130, V = (0.4 m/s, 7 meter borte 25,512 m P2 = P, o atmosphere pressure a zero gange -> Hp= (50+ 5.512 +23.18%) / 0.g = 7817. In una sony 87.44.m Now apply Bernoull's equation, the head developed in the turbine with 90%, efferency! Hy 5 (7,-2x) + ME-VES + ho - 62] *0.9 t o t = 550 + 5.512 +0 – 23.188] xooq = 32.324 x09 = 29.091 ms Mr Effere non of the systems = output head xico = 29.09! = 0.3326 = 33.26*. Latry in parte hered 87.444