2300 KASCH PARK RD 2018-01-01 MF Import

r � ; i� '. t Derivation for ar OriFice Outlet 262500 i= + 50T + 625 — 40Oo = 0 '� OUTFLOW This is a quadiatic equotion which moy be redu�ed to � I � lli = co ��2 yh (OriGce formola) the form: � � � Assume Ihol Ihc storm Ihal Glis Ihe basin lo the peak ax' + bz + c = O � ,, volume causes Ihe woter level lo rise at a constont rafe. 262500 (h = K �) where x is T,a = 1,b = 50 and c = 625 — 40 Qo � i� � The general solution is: I ',I � Qi = ca � . � � � �et K; = ca \ 2 fl K, x = — b t 2a6 4oc i ,I Qi = K.. 1"� I �I T 262500 � ' Vo - 60 f„ Qi dt T _ _ 50 t �50)= — 4 (1� �625 — 40 Qo � I I� Ao�: � �� ' i _ iversion foctor of b�l sec/min is required bemuse Qi is in cfs and t is m m�n�.�tes. � 26250 � i * 2500 — 2500 + -- Vo = 60 Y;, „ t"' dt T = — 50 t Qo__ � , Vo = 60 K,j�h) T"' 2 I �•:. Vo =� 40 K: l�i: (T) 65� I II Assume the maximum oufflow oaurs only�t the end of 'f = — 25 t Qo � � I ihe peak storm wch Ihaf po == K, T'" i Vo = 40 Oo T Derivafion for a Constant Rate Outlet i • INFLOW - A hosin with a constant outFlow device,such as a pump, � is simpler to derive.The constaot outlet rate implies that I � . Qn = CIA Ra�ional Forc�ula the total out(low is merely the rale multiplied by the � LetC = 100% durafion o( the sform. � i A = 1 Acre �75 Vo = 60 Qo T � 1 = � � � T + 25 Vs = Vn — Vo j � 0500 T Qn = (100%)i + 25 ��) Vs = 1' + 25 - 6QQoT i , . �� dVs (T + 25)�lOSOq� — 10500T�1) � ; ' � Qn = T + 5 5 dT (T �+ 75�: —60 Qr . I I�. dVs 262500 _60Qo Vn = C�n X T x 60sec/min dT � T' + 50T �- 625 ! . 1050�T 262500 � yn = T + ZS T' + 50T t 625 — 60 Qo � i STCRAGE 4375 I • � Vs = `/n — Vo T = -- 25 + J Qo 10500 T GRAPH. n both outlet situalions ih d�mtion of fhe I�� : Vs = — 40 Oo T � .�—'--S-' T + 2 5 s torm to f l l t h e 6'iisin to ds ma�umumsr .,he foun d as a � Since Qo is a fixed maximum outFlow that will nn!y iunction of ihr nlaXirt7um.��d_(low. The peok storage � li� . occur for the peak storm,it is necessary to iind the time volume can then 6e found h�� substitulinq the duralion � ' �` (rom the instant fhe sbrage 6egins until the insPnnt the 6me into the srorage volume equolion, reducing Ihe i �� •, peak storage is attained.This mn be done by toking the s�orage voli�mc cquation to an equalion with only one • � I ' first derivalive of the sbwge volume equcFion ond seb indzpendenl variable.TFis r+akes it possible to draw a I.L� I. ting if equal to zerr.. graph of storoge volume as o(unclion of the maximum � � ; outflcw rate.The included graph shows that as the oul• � ` 10500 T lel gets laryer,the�equired starage volume decreases. j dVs d T t 25 d(wtl QoT) The idecd basin will fall ot o point on its respedive wrve. ' ' � � � dT dT dT A 6asin ttial is oversized will fall al a poinf obove its re- dVs (T + 25)(10500) — (10500 T 1 zpective c,�rve,and as o result,it will not reacS its max�• � ; � �( �-40 Oo mum aut(low rale during the slorm,An ina�equot�basin � dT �T t 25)' will fall ct o�ioint below its curve,and will rise obove the ' � design depth, producinq more fhan Ihe design out(luw � I � dVs 262500 and ossibl c�usin }loodin It should be ointed oul . - dT = T' + SOT + 625—40 Qo = O P Y 9 9� P i . that the mnsfoal rote system allowi the lea�� slorage � � for a given size ouqeL 7his is due fo the(od thot ii func- � "1Qymr Requ•rcy norm equa�im de.doped by Hubbeil,Roth 6 ��ons ai fhc maximum m�e Ihraughout the slorm,while a f � Qork. Inc. Iw Ooklond Caunh.Mi<hlym (Conlino�d on p. 12� , I�' �� WnTER 6 SEWAGF WORKS, Decr.mber, 1973 — q� � �