Chapter.5 and Chapter.11

() = = () .

1 .

.

.

C .

r=0 , .

.

1/2. : - - - -

.

. = /

() .

Poiseulli .

, , , . , . 1. : : 20 * 0.5// = 10/ Q 2. P, L . 3. V, D .

* , .

, , : Re < 2000 : Re > 4000 : 2000

5.2 Derive the equations of the head loss for the problem 5.1 such as Darcy-Weisbach equation. : Darcy-Weisbach Darcy-Weisbach . , . Darcy-Weisbach . .

. () = = ()

.

. 1 .

, .

Osborne Reynolds Reynolds . : . .

Figure 3.4

Re < 2000 Laminar Flow 2000 < Re < 4000 Transition Region Re > 4000 Turbulent Flow Poiseuille . -> -> -> : - -> : / -> , ,

5.3 Derive the equations of the velocity from the head loss equation for the problem 5.2 such as Manning's, Chezy's, and Hagen-William's formula. 1) Manning's for turbulent regions of rough pipes, or for canals.

R = = / = perimeter/area n = = roughness coefficient : hydraulic radius ( for a circular pipe)

2) Chezy's , .

(5.6.1)

. () .

. () (secondary flow) () .

5.15 ()

5.15 , . f2 . . . . () () .

(3.10.1) .

.

(5.6.2)

, (3.11.2) . P () ()(wetted perimeter). ( ) . L sin = z1 - z2 (5.6.3) . (5.6.2), (5.6.3) (5.6.1) (5.6.4)

. R (hydraulic radius) . R . . (5.6.4) m N/N ft lb/lb ()(head loss due to friction) . () S (5.6.5) , V (5.6.6)

. C . Chezy . Chezy C () () . C .

3) Hagen-William's Hazen-Williams .

5.4 Explain the concept of Reynold's number and hydraulic radius. Reynolds . Reynolds (), , .

Reynolds Mach .

, .

. , () .

(transverse interchange of momentum) . , , Reynolds .

() () . 1. (). () . 2. () (). () .

2 () , Reynolds () () . , , , Reynolds .

, , . Reynolds R . (5.1.1)

Reynolds 5.1 . , . () , () () () () .Reynolds , . .

. Reynolds .

. . () .

. Reynolds .

5.1 Reynolds

Reynolds , .

() Reynolds(Reynolds upper critical number) . Reynolds Reynolds .

Reynolds Reynolds R 2,000 .

() Reynolds(Reynolds lower critical number) , . Reynolds 2,000~4,000 . . , 1.7~2.0 .

Reynolds () Reynolds . , () . () , () .

Reynolds . .

, () , , .

(inertial force), . () . Reynolds . Reynolds . () (fine scale) () () , (large scale) . , . , () . () .

. Reynolds . . .

() ()( 1 2 ) . Reynolds () .

hydraulic radius: , . (hydraulic radius) . . .

5.5 Explain the computational algorithm of the pipe network using the concept as shown in the section of 4.3-4.5 in the Brebbia's textbook. , . I D, C, L .

, k, j , i . i k, j , . Poiseuille ( ).

, . .

Hazen-Williams ( ). :

.

( ) .

, .

, +, - ( ). .

.

, , , .

. , .

, , .

, () ( , ), ( ) .

, , .

.

, j .

. .

5 . 5 .

.

, 5 , . 3 . 5 .

, ., . .

(ii) . (iii) .

(iv) .

.( . , .)

i k, j . - (k,k) (j,j) . - (k,j) (j,k) .

Fortran .

c c... initialize global matrix ... c npmax : total number of nodes = 6 c glob : global matrix after assembling c do i=1,npmax do j=1,npmax glob(i,j) = 0 enddo enddo c c... assemble each element matrix for overall element ... c nelmax : total nuber of elements : 9 c do nel=1, nelmax c c... read information about element connectivity ... c nodemax : total number of node in each elment : 2 c element1st node2nd node112213c read(nodeele(nel,ie), ie=1, nodemax) c... c... compute each element of element matrix at each node ... c ig : global nodal number of ieth node in element nel c jg : global nodal number of jeth node in element nel c do ie=1, nodemax do je=1, nodemax ig=nodeele(nel,ie) jg=nodeele(nel,je) c c... compute each element in k matrix ... c ek= rk(nel)*(-1)**(ie+je) c c... assemble each element in k matrix ... c glob(ig,jg) = glob(ig,jg)+ek c enddo enddo enddo

. , , , .

element1st node2nd node112213

5.6 Determine the discharge through the system as shown below by using the concept of the equivalent pipe where, =0.5, =300m, =600mm, = 2mm, =240 m, =1m, =0.3mm, , and H=6m.

5.21 , 5.21 .

5.7 Determine flow through each pipe and the pressure at B as shown below for a total flow of 12 cfs. Where, =3000ft, =1ft, =0.00 ft; =2000ft, =8in, =0.0001ft; =4000ft, =16in, =0.0008ft; =2.00 slugs/ , =0.00003 /s, =8psi, =100ft, =80ft.

, 2 (parallel-pipe) .

, .

, .

11.9 (11.5.1)

. , , .

. (1) . (2) , , .

, . .

.

. 1. 1 . 2. . 3. . 4. 3 , . 5. .

. , 1 (11.5.2) . .

5.8 Find the discharges for water at 20C with the following pipe data and reservoir elevations as shown below. Where

. . , . Darcy-Weisbach . J . .

. .

, . , . ,

.

,

0.029m3/s . .

5.9 Explain the algorithm of the Hardy Cross Method for the pipe network analysis. . .

, . .

1. 0 . () 2. . () 3. Darcy-Weisbach . , . ( : =, )

: : :

, 11.15 A G AG AFEDG .

. 11.15

. Hardy Cross (3) . , .

. , r (11.1.1) . r (Darcy-Weisbach ), -(loop-balancing) .

. , (11.7.1) . . . . , .

. (11.7.2) (11.7.1) . , . .

1. , . 2. . . (11.7.2) , . 3. 2 . . 4. (he) 2 3 . . r . . .

5.10 The distribution of flow through the network as shown below is desired for the inflows and outflows as given. For simplicity n has been given the value 2.0. Implement the Hardy Cross Method for this problem.

11.16(a) . 1 . . 2 . 15 + 11.06 = 26.06 , 35 + (-21.17) + (-11.06) = 2.77. 11.16(b) 11.16(c) 4 .

11.16 , 15 100 . , , Hardy Cross - .

.

11.16

[4-7] , Hardy Cross, (node balancing) .

Newton-Raphson .

[5,6] .

5.11 The program in Fig. 11.18 in the text book is used to solve the network problem displayed as below. The pump data are as follows: Implement the pipe network modeling for this problem by using thesuggested program of Hardy Cross Method.

. . .

, (pseudo elements) . . , . 3 ( 11.17) .

(11.8.1)

1 4 . , . (11.8. (11.7.2) . Newton( B) . . 8 - 3 . . 4

(11.8.2)

5 8 . (11.8.2) Newton . , - 0 . 11.18 BASIC . Hardy-Cross - , Hazen-Williams Darcy-Weisbach . , , . USC Sl .