hydrology problem numerical, Exams of Fluid Mechanics

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J = Tuncrion ~— Reservoirs — Ap & a Pipe .- Ai®, & Zp = pitrometc head ab junction D Ze bp, Datum Condition 4 Canaition 2 Za = (pa pe Fe, aq = hPa snp. Ze Za = (PD, +p, + 28 Za =ChpD, +B Oa = Ber Oe” . . 7 5 8a =0 Zp 7Ze : we " a ®, = 26 Condition & Cooaieon © Zn t(hDat pe Ze = &pP)p+ &Pe F Ze 2 ayy! 8, + Ge = Be eres Zp 4 2p “Typ es st Peotolern. pe 4 “ Ze =, 3 Qe =? 3 Rye 2 ow Ze? > Barts Type Gar; Be® p Solution , Ven evdure- oo a noe ae, [ee oa) ‘ es Te Type Ase Calentate TA es ) Pa =, OK = —— WY) Zp = Za —&HD, Wi) Lenn pare. zea* Zp i) hfe = \te- 20 v) Ge = bie WY Oe = an @ Xe, “dh ACE ete AT? 1 vi) Wie = We Oe Ae YW) 2e= Zp- Ube and ad Aireck on, of Se