CE 453 solution of midterm 2

CE 453 solution of midterm 2 - Midterm II -— Solution 5...

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Unformatted text preview: Midterm II -— Solution 5 Zuesn'cm I a) D=14 inches, h11=i : 7ft 3000 1000f; By using nomographa Q=2000 gpm from point B to point C The rate of discharge from elevated storage at point B 2000—1500: 500 gpm 5 + + h] ,ZB=10 n, Zc=20 fi, hL=21fi, and Vng i =150fi r g r 1” Hence, —‘=119fl —> R.=£=51.Spsi r 2.31 Question 2 l Soda Ash, meq/L CaU‘lCOflg 4.7 . 0 Mg(HCO3)3 0.5 . 0 MgSOa 1.4 . 1.4 Total . 1.4 The dOSages oflime=7.1*28+35=233.8 mg/L CaO The dosages of Soda Ash:l .4*53:71.2 mg/L NA2C03 c) Please read page 251-252 and Fig. 726 Question 3 600 ft —————I—:8+3.4:11.4 min 3.2flIsec*6OSec/m1n a) I, : Inlet time of Ike area1+ Inflow time = 8min+ t2 : Inlet time of the areaZ =10 min Sincet] >132, we use I, to find 1. Using I, : l 1.4 min, from the graph: I = 5.2 in/Iir Q: I *(ZCAJ) =5.2*(O.5 *1.5+ 0.45*4.2)=13.73cfs b) The pipe between MH—l and NIH-2 is only responsible for the area 1. So, to design this pipe we should calculate the quantity of flow entering the MH-l. I: Inlet time of the areal: 8min-MI: 5.5in/hr tin-5‘3 Q=CIA=O.5*5.5*1.5 :4.13cf.i:1851gpm —-—,,—>DElSin Mal; Q = 1 85 1 gpm} Numngrupgh v:3.2fi/s .—.——————_—————. Question 4 :1) Determine the following: - Length, width, and depth of the tank - Horizontal velocity through the tank Design criteria: - I > 30 min - 0.5ft/min <1»,1 <].Sft/min Q: 2.5*lofigpd:232fiJ/min Lett=45 min, tig—>V=Q*I=I232*45210440fi3 Leté=5¢=5WaA=nW=5W2 (i) LetH=10fl,A=%:10440 =104412‘1i 10 f () (i) and (ii) —>5W2 =1044» W =14.4ft, L =5W =72.2fi Now we have to check to see if the horizontal veloci range: vh=€$=L§§=L6filmin>L5fi/min The velocity is not acceptable, so we should change our design parameters: 14:15:55 L=4W—>A=L*W=4W2 So,4W2 :1044-—>W=16.2fi, L:4W:6448fi vh :£=%§ : 1.43fi/min < 1519/me (OK) 2‘ ._————-u——u_—~n.' '3) Determine the following: — the flow rate if the depth of flow equals 18 cm A the velocity of flow * 2 D:50cm—>A=” 4D =0.196m1 Q =0‘2m1/s'(fidllfl0w)—> V(fullflaw) =§=1 .GZm/s : 0.28aq = 0.0561113 fs d=18cm—>i:§=o_35M5L—L_, D 50 =O.84_>v=0.86m/s ...
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CE 453 solution of midterm 2 - Midterm II -— Solution 5...

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