h, t. t,$. kt- , J 'J. 1 . t. ? 1 ' ,' ' , ;t / .. .. .. -'

- . ; - ) . ..,:.s..c s.:....:v . ..j: y,. j az.u h . ,

lslam ic U niversity of G aza Instructors:''

Engr. KHALIL AL-ASTALHydraulics (ECIV 3322) '.=>.' Engr, Hus..tu AIIEDSecond Semester (7005-2006) (%. , . J

4. * Engr. EMAN ALSWAI I'YM idterm Exa m **.rx ';';G''1 j 1 hr cs min. A llowable T me:

' ' ' ' ' Student N alue. Studcnt N o

Ouestion l (zlpoints) 1-1

l . Power transmitted througll a pipe is maximum when % = h1/2. 'l'l1e methods of supplyillg w ater are:

kt ) q----- - è - -- '---ldf?. - - ') ) ..

ïj?z-l ,., ----x/? -sk ..

- -. . . - (: ) - -

k...,-. - .fL-.' ,- ,'.

- - -.

, --.

c)$ ',..i-,z-

,.- .!t), -!,. i:

. 3. 1l1 stllvillg tlis tri btlli()ll làetvvtlrlt s I7r()l)lclI) sa tlle Ik)iIk) vvill g ctlllditiol: s 1T,tlst b e s ati s li e d

sillltlltcïlzekltlsly'. '

xx- -

a) . 'J- :s.J. o .......r:. . .. .. at ettch j tl ncti ol)..

t) ) <.t.' ï-u -zz c:l . --'''- fk ) r t, tt (1 l 1 ! t) ( ) 1) .

4. the tlisavantages ol' branching systel'n itre '.

) - - - -lhneo.-or 4-m- .-- vrA' ,-..(.Aa pl petafrr- 'oqzzs.lt'b'%1- sf'-. J:'>' .11 ru '

t') /4??.p.,o/zlr--. (&? ..+zQ- czulo.. l<- unz.s.îrmga I m''-z

Hydraulics 4 . : . '

f ul .' 'p , . ' . . ) M idterm Exam

Ouestion J: (Jgoiftts) <A-For the three reservoir problem shown the piptz characteristi

cs are:D1 = 1000 mm, /? = 3000 m

./; =0. 015DJ = 500 lnm, 12 = 600 rn

, .Lb = 0.024,Lh = 6(10 kum, % = l 20t) m,

./à = 0.0211.* Jt = 1 40 m 2.% ez I 20 m and I (5 = .0.60 rnJ/s towatkls. thtt rtlsethit'c ir

s 'deteratilltt Q.)3 ard Q3.

Neglect lniner Iosses.

QY'x > .-1-A x

-

.-.i

W C iP' 5- D YL/

o $' ï.- t%

1q# rq

l -.s> $ s'ç.7îo/o

Q1 () :,7* 'Mo ...a y

zz x Ik Q >t J1 '

a: ç. 'j- -' z a

. lxo wj,,.- / ?> ' $Z3

KM lI i .

j q .g .yr.j --1 I

î')' zv,tà ';'' -. - x .t'.r . . J o ç

'h.d r'%' :1 .. A rk '

h q o .= . l ' 2 '-z v.j G

As Zdq Z .=- 'd r M X-z . j.. !

. u.!> . cs ! s- (v. .4.. n /n ttp :4

$ î tl . tz. 'tL.:w! j u y = ! . G. .i- G ,:47 $,. ) b)1 . 'L m

.L )

2

- z'z.s- Q!W = O - ?- -. -'2- =

. ttn uuunn uxamHydraulics . t , .

Oubwtîon JJ (g poims) jrFor the two tltnks shown'

L.4z = 800m L st. = 6001n (for each branch) Lco = 400mihv = 200mm Dco = 20()ln1n

(a) Find 1) using Darcy-svçjsbach Formula whm'e f = 0.02 fbr all pipes and themtiior losses from A to C = l 5114 .

(b) (Bonu:! ! !) Calculate the valve Coefûcient Kv

V cz t Cs. ...j9 . gk-i -o xa'' . 2 j! t-a v''- * Jc, ,vk (!./4,,1.:4 L-F L-A' Iz co. ,$. s-s ? . w c. ...z.

ox e n a( , ( s , j a...' . ' ' M l t.. o z.ov kjora o r-: L= ) 17 ea t z .#- C. . Z4 F = . ' a.. .2---% . .- ..=b co .. .- o .,. Sa to t:.

.a. A . $04 z-' 'î $ z.'1k = co . i !2- % X' ' .l-.J ' ' ' '.. m , ). w >w D ' ï I rw A . . . . .

. . . - - u.uo . o -a-

e- éc qp q/t- l cqncs A. 2. $u . -cL-.- l t . 2-3/ '- V- - - kq.(xx r

w ' ' '& @ & V . .& p ' n C6 PvGY u:tz <' ' K J -?- = =' ' O ? ,%$ ,S=oe <

. - - - - = W e v ... o.oz-yG=a o.

q o. oa.w G:.,=.

x; ps . u - z- . -q v/ = l , ç u; h x> . 2 ' 2 t --*-- a.fjf = '-Q--=b t u. q-ku ...W .).. -&e'- - .s-su . : w= ï , ,

I . % P-

L qa cj = . . . ' ! z. u ../....-.1 g s645 :* l q' k= A- .mfi = a 'z. o e..

u xc s. zlt%l y-.i. 1 ,s- . -' S' D -> 6V I -àT C i 1 'Z * . ' & hH&,u ; . .

- . ms....zl- ; ' ..'f- .

. $

...u . d *.(..' 7 kj -- - - . $ qacj- j )m <.- L $ ' î a, ... s . . j mwtvwcx. .

z + Xk

% =k . (:1) - k . x. %.) '')

. î$ % l à) = .-k Wx ,w G.-j.+...1.% . k'/ y jjwxmu c . .=su--Su ks = .. uty u $% . .o1 @ix t

==' tb =. co - l 3, 'ro v'q ,. . . ' - , 's...A .r = c (;

''-'-

..'z -) ,- tw .sàx'L w $ ' R. .xq> tît. .. = k

y z..o ..q . s ..aa. ,j! ,)Z .

k' vx'xx o o .= u I'Lfj(> ?<&.4. = z

z!y draulics -... , ; 1. ' : ' ) . . c q t 1 èdidler:q E x aIn

(tle.vtion 4: ( l 1 2t) int.s ) IV J ..0 . : x .Neglecfing iyiypl

-l-osses deterlnine:

'ue tlew ihrouzcraok pipe -....s t'L - q.- v.'h1. -(&' xj2. 7'h e pressure lkead at n

mg--q--d 17.-J,+ (k q/- (7 ymJ' x. . - -Use the following data: v.z-v.'.--w -'r

.@ Friction tàctor,.f=-0.02, for all pipes. Z ...>.:(y.'; t'/-'btyye (:.1.) $'q J /'-

. #, Prrssuty head qt pode A = Jtl-m . s Ur' 'x -- - -?. . . ' > ' . ... - '

*' Perform only one iteration of Hardy c' ross mei 0d' txstitifile inftially tha Q / = 50 L/s ' om 1! to 1t - .-s, . . . w ..e. ..- XL.---.-

plode .,!... B C D E F # H

Elevation (m) 30 25 20 20 25 20 20above datum

- - .

( -- -'-= ?.o l =rh ' >Kc f

.0 L/s

- . . 200n1 jcQin U- %- S - - J'--ol so- '- ' - t

x x D

dyf? .j. tjy /J 0- . -. - .--.y E îoQ œ h ' v 1-o() ..y. 6o j J '., . - , .. jpQ J s

,1 ï?-..s

rn

. - - . o- c t 2 0 L.Y .T .

'. s+.îM '

.

-

.l .4:. .#-L- QAk

=.& .>-c7---. , -sA -- s .c a m src 'x/riz.k..g:h . , . z; .- ..- - - . (. 4 () ë .--y/ s - .1>- ,p. - z a n -n- -k - - - -.z-.-. .x g zm . ,j . c a . a sa----lz $

- y'o 'z.'+ i & - , ;11 ' o '' . 1. s .-.o q

D=0.2 lnc ' .-.0 , -

. -0 y os j .- -$ qxf ç .- .t- 4 u - f i. 0 L /) ,#< - . . w ' (

.. .# () L/

- gs ,isf. $ x.- - j-?- ,gs -.t-o. o Q Lqtp a w S-

..I

S- F / l o .V

cn . o $

o x q> 'I.m '!e.,q'.:>-

!-! C. .Y o o

'zwcc ,1/1= .= c% . okoà' X = 2-, K: f' L-/.s X. )!

Coce lteo

-s.) cu.p au sz. u--s c) &19Rpnq = ï'M'CX Lo':E? - z.-?.. lqx L/1Ca r# =

% 7C z/

c' . C.e.

g @s)=

(-Q.2,,+M s$

k-t!i . t 5 (0

P o0.. c-.- szsu-aozjlkqo

. ..x.) .f % .+ -:> ) ..t -a . j s Taa- w o .z ,s(. '

t.C -k-z-') = G%'V7V-'M'as o. - -- - - sxV q j Cp-s ) .4./- , .xa .,vN - Yf 'eo

w . -- p- - - gy eyv-A-avW y.c3 V#>

j.D=-X

Q.4a- In 10 line discuss the master flow illput & outlnut with solution procedure

b-

W bat is the engineering application of siphota phenom ena?'-

wctf.. r.f .:o.ih ../ J ( ) / ; e (ya v &?> J

.- - .> - '.w fa , , . t,va y! x! ;v, a ;, r s, a e.al -

- .

...fêr>' szz

-e s o l,XJ. x ch .ze..qJ. . zw s

..a oy ; , ; x g,y c yJ, g yyo o w cy <Q.J

,r77, rt-c.2, 4 z

e ) ; /.,,,q,; s vo c.y y-o sas, s &,

*1> ua z - uo' ''-

> V'ater flow tbrough a sttdel pipe, the clistance from A to B = 300m

' ..4 e g.p zz , S- o07e dîstance frorn yk to S = 100rn the dianaeter of pipe 0.2rn Nvater,

,

. c. ,/, . vv yyrlevel diffcrence between the two tanks = 20m if lpqyjntà

.t'!l qçglévepressure nt I3. oint S = - 6ms use /'= 0.021a deternlitie i Y. sty.px>zzv co vq. . . . - < . : .;- . . ..v '- . c j -

a y . z a ; ( .x,FZj .. j ''-Qt.. 3.% l OJ T' ** 1 H' I y'i J, g o t? z

-. & zz--xy z.- s! s x / sr-' - - - +: - - . ;y - - y p y= g . .

. ; z) f Lxz wt fg .

g . . - ..o zy-zs )- , , (, y':)- .

B A r / rz (lot? .

1.0 +-5/ ' #:.!28 à7 2/y Gz = 5. - 's; = Lz,q

o . a s.$.!. $zr,--Y + z.A

''-

,, ' g' )--' -' ''' au'

h 6 t -3. v/.=$ 1/7

? J .

C ? o ; o '? ? pE Ya

Floc xxj.a ; G

..y k - . .4. - j

a .o zm o . û'2) ---- --.,.. . .u-. - ..

.j zyz x. lj. 6 z t ï. 61

( :) . v y y k) 2rs .... - jy.j-, :.. ;%

.() y-I ; .ozj , -!.to-ti t 7

za 'k. . t . y J .v x

JE''.: .. .re,'j 1t'a 1. ''wp.

' .:.Lr .( V.

' ( 'V tk!. 2. ..:% xkf1tr-. L . Fi. . $ j ' 1.yj.y). ..,26. .' z .)tf , ï'' k

tjy/v. q . hr - - -V 'N N' .îl;/'R û vx; lo, . . , # . 2 - '<p # y4....

,.4) $tj own ngure determirle tllc flow ratc into the Mgh levcl reservoirlt.'y * r aII / es = p.pJJ k. k.k ?

-

ztL , =.J

-F :! () . () I z ,.. - k. +i5 5m

(1 .

; o

f N% oiè0% + ei + O. xvp p '9

p znx.pz. s . vj. . (-;N

nHog zz y' ' .X. . ' 4 Fljtzvo p .. V-';zzp /z Lzzpzn Jh

tr P 4 r't ' o. %,ln . , pump 1;0. <-j-. m)' 3 -'kl )(' . i

1+ 1 8 10

Q. (; o ( 2 k't''lit?/ %') -t) .(b q 7. > . .. .y' ? . .... . -=- pht, ' 7 ) y k j' t

A cg.,,s,cl c? p ,*.!J sg jtvee.,t %. p :

f!b i -. y +.Qc/ s'o 4/, *S

- .- . . . . j'. -... .- . .J ' ?p r' jj k (' '?.

2 Q 9 3 l t) '! ,5' a 7 î/ j ? q - :). () . I j 1 t/ j,.'y'?

. '

) ,'%i . t..l z;.o o I ./ . A 2. 4

s t j t. ko ?. . f. pc ,

h .i ; .:? f'', -:,.j , m o t) .

/7 f'n t? l

ê-kr) -k

,?, - '! 7. .:7 ? j .. , ! .. . J.J

4Z;> i '

. tlseful H vdraulic Form ulas

Flow through pipes

L 1/2 g y t (22. Darcy-w eisbacll h = f (- ) or h = .-.'I D 2

.g 1 ,c2 o jg

64# 64Priction factor / = '- = for Iaminar flow

. p p' f) Re

(i.jl 6 5.lwv=t.v fopsmooth pipes with 4000 < Re < 10R * '

o .. . L . . . '

1 e / o é.51-

'

' ctlp rqok' formula =- 2log ---+.

é s.7 jto y. ' 1 325

SWjnis A in j' = '. .;;. 0 a

$ 3.713 /?

. ! 1.:5 l 85

e63 a 54 Z (? Z l?'Hazen -W illiants lzrs0.85Csgz Rg S or hy = 10.7 or hr = 6.814

.*7 x J l l66 vD (,s s, D (.ss,

i 2/3 1/2 a (?2 /. a zM'

anning Form ula V = - Rh .% or h = 10 3n L or hy = 6.35 n p'n J q. l 6 / 3 .

) ) 35D f

Chczy Form ula Pr = C A%1/2 5132 Izr :uu # .#7/3 çl/2strickler Form ula

t, h u. s

. . . 7t . '' : . ' ' .' r. , k . : ifiù. o r tzo ss es '..f 'qt .i1. . ;..

.

rz2 jzgEntrance of a Pipe h =K Exit of a Pipe h =

lg 2.:

Z22 Zj2 (jk- jz )2sudden contraction h ZZ,K sudden Expansion %e=K,e or hs.ew-----l-

g K 2g

(,r - ,,: ) d /, - ,K r: - p',Gradual Enlargement h

-e= K u, radual Contraction

pc geq - s & lg - 2g

, . u.y v aBends in Pi 1 çs h =K F; ; jupe nttjugs hv = Kv1 b b 2: jtg

Pow er Transm ission T brough Pipes Power available at the outle!. yim ciepcy ;y =Power )u: y Q H power supplied at the inlet

e -../w

.!.xv) . W ater Dktribution System sk ' ' '

geom etric- nP = .!$ ( 1 + r )increase model

Fire demand QF = 65 P (1 - 0.01 P ) Qj;. = 53 #- . Z 7z -)2àf

A fdischarge A = f-s = Darcy-W iesbachh H azen W illiam s h

tion j2 . 1. 2 ).q 'fcorrec j gj

' (? (?