CE 453 solution of midterm 2

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

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

## This note was uploaded on 12/14/2009 for the course CE 453 at USC.

### Page1 / 4

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online