Linear Theory Method

8/13/2019 Linear Theory Method

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INTRODUCTION

Linear graph theory used for pipe network analysis is to make the methodsystematic

A numerical method that uses linear graph theory is presented for the

steady-state analysis of flow and pressure in a pipe network including its

hydraulic components (pumps, valves, junctions, etc.)

The linear theory method is another looped network analysis method

presented by Wood and Charles (1972)

The entire network is analyses altogether like the Newton-Raphson (NR)

method

The NR method requires a good initial vector for fast convergence,

whereas the LTM oscillates near the true solution (Nielsen 1989)



Pipe network analysis methods consist mainly of

a constitutive relation;

the formulation of system equations; and

a solution algorithm

Formulation of the model uses either the loop or node

equations, or a combination of the two

This method is applied in practice to

Pipe discharge equations i.e. Q equation

The nodal head equations i.e. H equation



Pipe Discharge Equations

They are taken as the basic unknown in formulating the Q

equations

The application of the LTM for the solution of these

equations is described for different situation as

Networks with known pipe resistances

Networks with unknown pipe resistances

Networks with pumps

Networks with valves



Networks With Known Pipes Resistances

The nonlinear loop-head loss equation are as

( ∈ |tQx|n-1)Qx = 0, c = 1,…., C (A)

Where,

= known resistance constant of pipe x

tQx = assumed discharge in pipe x for the tth iteration

Qx = the unknown parameter

Equation (1) in linear form becomes

t′ ∈ Qx = 0, c = 1,…., C (B)



t′ = modified resistance constant of pipe x for tth iteration & is

given by

t′ = |tQx|n-1 (C)

For first iteration tQx taken as Unity, we have

1′ =

, x = 1 , … … , X (D)

Eq. (1) modified to

∈ Qx = 0, c = 1,…., C (E)

The simultaneous solution of the linear node-flow continuity

equation and the linearized loop-head loss Eq. (5) would give

the pipe discharges at the end of the first iteration i.e.

Qx(1), x = 1,…, X



From the second iteration onwards the value of Qx

should be taken as the square root of the product of the

assumed and obtained values of Qx in the previous

iteration. Thus

tQ

x =|(t-1)

Qxx

Q

x (t-1)|0.5, t = 2,3,…. (F)

The Qx values obtained from the first iteration are taken

as the assumed values for the second iteration for

evaluating 2′

Thus,

2Q

x = Q

x(1), x = 1,…, X (G)



If this procedure of taking the obtained values of the

previous iteration as the assumed values for the next

iteration is continued further, it is observed that the Qx

values start oscillating

To prevent oscillations ,the avg. of the obtained Qx values in

the previous two iterations should be taken as the assumed

value for the next iteration. Thus, from the third iterationonwards,

(H)



Instead of taking the average of the obtained values

in the previous two iterations, taking the average of

the assumed and obtained values in the previousiteration.

Therefore,

(I)



Example 1

Solve example by linear theory method based

on Q equations

Solution: in the given network of fig.1, the

node-flow continuity linear relationships for

nodes 3,…,6 give respectively



Fig.1 Two-source, four-demand-node looped network



From the two basic loop I and II and the pseudo loop III, we

have, respectively,

For the first iteration, 1Q1 = ….. = 1Q7 = 1 and therefore Eqs.

(5),….(7) becomes respectively,



Solving Eqs. (1),…, (4) and (8),…, (10) simultaneously, we get

Q1(1) = 0.16 m3/s

Q2(1) = 0.37 m3/s

Q3(1) = - 0.14 m3/s

Q4(1) = 0.08 m3/s

Q5(1) = 0.25 m3

/s

Q6(1) = 0.25 m3/s

Q7(1) = 0.15 m3/s

These values are used as 2Qx values to calculate 2R’x values for the seconditeration.

Thus, 2R’1 = 40(0.16)0.85 = 8.42 and 2R’2 = 20(0.37)0.85 = 8.59

Similarly, 2R’3 = 10|-0.14|0.85 = 1.88 and so on.



Thus, linearized form Eqs. (5),…, (7) for the second iteration becomes

Solving Eqs. (1),…, (4) and (11),…, (13) simultaneously, we get

Q1(2)

= 0.5402 m3/s

Q2(2) = 0.5464 m3/s

Q3(2) = 0.2402 m3/s

Q4(2)

= - 0.0872 m3/s

Q5(2) = 0.2593 m3/s

Q6(2) = 0.2407 m3/s

Q7(2) = 0.1407 m3/s



For the third iteration the average of the obtained values in the previous two iterations,

i.e. iteration 1 and 2, are taken

Thus,

3Q1 = [Q1(1) + Q1(2)]/2 = (0.16 + 0.5402)/2 = 0.3501 m3/s and so on.

The iterative procedure is continued and the first five iteration are shown in Table 1 Ex.1

The final solution with an accuracy of 0.0001 m3/s in pipe discharge, is obtained in 14

iteration as

Q1 = 0.3246 m3/s

Q2 = 0.4715 m3/s

Q3 = 0.0246 m3/s

Q4 = - 0.0219 m3/s

Q5 = 0.2496 m3/s

Q6 = 0.2504 m3/s

Q7 = 0.1504m3/s



The first five iteration in the alternative procedure of taking

the average of the assumed and obtained values in the

previous iteration [Eq. (I)], from third iteration onwards are

shown in Table 2 Ex.1. This procedure requires six iterations

to obtain the final solution.

From the two tables it can be observed that for any iteration,

the average of the assumed and obtained values in the

previous two iteration is closer to the final value than the

average of the obtained values in the previous two iterations.

Thus, the convergence is faster in Table 2 than in Table 1



Table 1 Example 1 First five iteration using Eq. (H)

Table 2 Example 1 First five iteration using Eq. (I)



Networks With Unknown Pipes Resistances

When Q-R equation are formulated and solved, the number of the

unknowns is X + XUR and the approach is implicit

When Q equation are formulated and solved, the number of the unknowns is

X and the approach is explicit

In the loop-head loss equation for the implicit approach, the terms indicating

the head loss through pipes of known resistance are linearized for pipe

discharge and those indicating head loss through pipes of unknown

resistance are linearized for pipe resistances

Thus,

tℎ = ( ∙ tQx

n-1) Qx (linearized for basic unknown Qx) (J)

tℎ = (tQxn) ∙ (linearized for basic unknown ) (K)



In the implicitly approach, the unknown pipe resistances are

evaluated along with the unknown pipe discharge for each iteration

When a pipe with unknown resistances is internal, it appears inmore than one loop-head loss equation in the formulated Q-R

equations and therefore the unknown resistances evaluated

implicitly

If a pipe with unknown resistance is external, the unknown

resistance appears in only one equation. Therefore this equation

can be set aside, and the unknown resistance can be evaluated in

the end

In the explicitly approach, the pipes with unknown resistances are

externalized and only Q equations are formulated and solved. The

unknown resistances are evaluated in the end



Example 2

Solve by the linear theory method based on pipe

discharge equations using the implicit and explicit

approach

Solution:

Implicit Approach

The network of fig. 2, has six pipes of which pipe 4 has

an unknown resistance. Thus, the number of unknowns

for Q-R equations is seven



Fig.2 Network with unknown pipe resistance



The seven Q-R equations are

For the first iteration, 1Q1 = ….= 1Q6 = 1. The term R4Q42 is

linearized for R4. Therefore, taking

1Q

4 = 1, R

4Q

4

2 becomes R4

for the first iteration. Thus, Eqs. (5),…, (7) in the linearized form

are



Equations (1),…, (4) and (8),…., (10) are solved

simultaneously. Equations (8),…, (10) are updated and the

first three iteration are shown in Table Ex 2. The final solution

is obtained in five iterations

Explicit Approach

For the explicit approach, Eqs. (1),…, (5) and (7) provide six

equations for the six unknowns Q1,…., Q6.

These six equation are continuously updated and solved

until satisfactory Q values are obtained. The value of R4 is

obtained in the end.



Table 3 Example 2 First three iterations for network with

unknown pipe resistance



For the explicit approach, to obtain only one equation involving

R3, pipe 3 is externalised, by considering the overlapping loop.

Therefore, one equation from Eqs. (5) and (6) is retained and theother is replaced by an equation for the overlapping loop.

Therefore, retaining Eq. (5) and (6) is replaced by

For the pseudo loop comprising pipes1, 4 and 5. Thus, Eqs.

(1),…, (4), (7) and (12) provide six Q equations for the six

unknowns Q1, …, Q6. These equations are then solved by the

LTM to obtain the final Q values and subsequently the value of R3



Networks With Pumps

The formulation of Q equations for networks with

pumps is quite simple

The head supplied by the pump hp is expressed in

terms of the discharge through it and therefore, in

terms of discharge in the pipe in which the pump is

located



Example 3

Solve by the linear theory method based on pipe dischargeequations

Solution: The network of fig. 3, has seven pipes and therefore

the basic Q unknowns are Q1,…., Q7.

From the node-flow continuity relationship at nodes 3, 4, 5 and6 we get, respectively



Fig.3 A Network with Pumped source node



Taking f = 0.02 for all pipes, the first-iteration R values of the

pipes are obtained as shown in Table 4 Ex 3. Using R values,

the head loss equations for loops I, II and III are respectively,

The head delivered by pump, hp in Eq. (7) can be expressedin terms of Q1



Incorporating the value of from Eq. (8). Eq. (7)

becomes

On simplification Eq. (9) becomes

Taking 1Q1 =…= 1Q7 =1 for the first iteration,

linearized of Eq. (5), (6) and (10) gives



Note that 1Q1 = 1 is used only for the linearization of

nonlinear terms and therefore 2.697Q1 in Eq.(10) remains

as 2.697Q1 .

Equations (1),…, (4) and (11),…., (13) are solved

simultaneously and the Q1(1),…., Q7(1) values are obtained.

Using the assumed Q values, the f and R values are

updated, the Eqs. (5), (6) and (10) are linearized and solved

simultaneously along with Eqs. (1),…, (4).

The iterative procedure is continued and the first three

iterations are shown in Table 4 Ex 3




pumped source node



Network With Valves

A check valve, the flow through it is initially assumed to

be in the permissible direction and Q equations are

formulated

When the direction of flow reverses, the discharge in

the pipe containing the valve is made zero in the node-

flow continuity relationship for the end node of this pipe

and the loop-head loss equation for the loop containing

this pipe is dropped



A flow control valve, FCV is initially assumed inoperative

and Q equations are formulated and solved

When the pipe discharged stabilise, the discharge in the

pipe with FCV, say Qij is compared with Qset value in the

FCV

Qij ≤ Qset – solution is acceptable

Qij > Qset – the discharge is fixed at Qset, the pipe with

FCV is removed and the outflow at upstream node is

increased by Qset while that at downstream node is

decreased by Qset



Example 4

Solve by the linear theory method based on pipe

discharge equations

Solution: The network of fig. 4, assuming the PVR is

in the operative mode the PRV is replaced by source

and sink nodes and a pseudo loop is formed.

Thus, Q equations for nodes 2,…, 6 and loops I and II

are respectively,



Fig.4 A Network with PRV



When the PRV becomes inoperative for another pressure setting,

loop II is replaced by loop III shown, Eq. (7) is replaced by

Note that the resistance constant of pipe 7 now includes the

resistance constant of the PRV also

Equations (1),…, (6) and (8) are solved by linear theory

Here, pipe 7 containing the PRV is dropped

Therefor, the term Q7 is dropped in Eq. (1),…, (6)

As loop II containing pipe 7 is now absent, Eq. (7) or (8) for loop II

is also absent

The required Q equation for basic unknowns Q1, ….., Q6 are

provided by Eq. (1),…, (6) in which Q7 is absent



Nodal Head Equations

The nodal heads are taken as the basic unknowns in the

formulation of H equations

The application of linear theory method for the solution of

these equations for different situations as

Networks with known pipe resistances

Networks with unknown pipe resistances

Networks with pumps

Networks with valves



Networks With Known Pipe Resistances

Pseudo loops are not necessary in the formulation of H

equation

These equations, formulated for single and multiple looped

networks are non-linear

From the node-flow continuity relationship at nodes of

known flow,

(L)

Wh H H f d Th f f th tth



Where, Hi = Hoj for source nodes. Therefore, for the tth

iteration, Eq. (L) can be written as

(M)

Rox = known resistance constant for pipe x,tHi , tH j = known or assumed nodal heads, for tth iteration, at

nodes i, j, respectively

Hi , H j = unknown nodal heads

Expressing Eq. (M) in the linearized form, we get

(N)

Wh



Where

tC’x = modified conductance of pipe x for the tth

iteration and is given by

(N) (N) (O)

Or

( P)

tQx, thx = discharge and head loss in pipe x for the tth

iteration, respectively



To begin the iterative procedure, it is necessary to initialise

the discharge and head loss in a pipe i.e. to select the values

of 1Qx, 1hx, x = 1,…, X.

Collins and Johnson (1975) suggest that 1Qx should be

chosen to correspond to the Reynolds number of 200,000

As Re = 4Q/ΠDv, we get,

(Q)

(R)

The discharge & dia. Are in m3/sec and meters, resp.



Isaacs and mills (1980a) suggest that an initial pipe

discharge may be set to the same value for all pipes

An initial pipe discharge ranging from 0.001 m3/s to 1

m3/s, when used in a test network of sixty pipes, had

no significant effect on the solution

The value of the initial head loss 1hx corresponding to

the assumed discharge 1Qx is then evaluated from

(S)



Eq. (P) now gives 1C’x, x = 1,…, X and thus the

nonlinear node-flow continuity equations are

linearized in the form of Eq. (N) for the first iteration

The linearized Eq. are solved to obtain H j values at

the end of iteration 1 i.e. H j(1) values

These H j(1) values are then used to determine C’x(1)

values from Eq. (O) or Eq. (P)

For the tth iteration,

(T)

A i il it i t b t t t l t k d h d li t k



A similarity exists between a structural network and a hydraulic network

such as a water distribution network, these are as:

The algebraic sum of the forces at a joint of a structural network mustbe zero to maintain equilibrium. Similarly, the algebraic sum of the

flows at a node of hydraulic network must be zero to maintain flow

continuity

The displacement of the ends of the members meeting at a joint of a

structural network must be the same. Similarly, the head at the ends of

the pipes meeting at a node of a hydraulic network must be the same

The force-displacement relationship specified by the geometric and

elastic properties of the members must be satisfied in a structural

network.



Eqs. (U) & (V), the discharge head loss relationship for a pipe

in a hydraulic network is similar to the force-displacement

relationship of a member in a structural network

Thus, similarity exists between structural and a hydraulic

network

For structural network, a finite-element method is well

developed to formulate and simultaneously solve Eq.(U).

Therefore, it can be also used for hydraulic network to

formulate and simultaneously solve Eq.(V).

The linear theory method based on linearized nodal-head

equation as the finite element method

E l 5



Example 5

Solve example by linear theory method based on (1)

H equations and (2) H-q equations

Solution:H equations –

In the network of fig. 5, the linearized H equations

written for the ode-flow continuity relationship at

nodes 3,…, 6 are respectively,



Fig.5 Two-source, four-demand-node looped network



As the pipe dia. Are not given, we assume the

discharge in the pipes to be 0.1 m3/s

The corresponding 1hx and 1Qx values are obtained

from Eqs. (S) & (P) respectively, and are shown in

Table 5 Ex. 5.

Therefore, for first iteration Eq. (9) becomes

Table 5 Example 5 First two iterations for



Table 5 Example 5 First two iterations for

two-source network



Solving Eq. (10) we get,

H3(1) = 95.66 m

H4(1) = 96.98 m

H5(1) = 94.71 m and

H6(1) = 94.72 m

The corresponding values of 1hx and 1Qx [using Eq.

(S)] and C’x(1) [using Eq.(P)] are obtained

The avg. of 1C’x and C’x(1) is taken as 2C’x for the

second iteration

Th f E (9) b



Therefore Eq. (9) now becomes

Solving Eq.(11) we get,

H3(2) = 95.48 m

H4(2) = 96.63 m

H5(2)

= 94.32 m and

H6(2) = 94.47 m

The first two iteration shown in Table 5 Ex 5

H i



H-q equations –

In the approach using H-q equations, the node-flow

continuity relationship is applied to the nodes of

unknown flow i.e. to nodes 1 and 2

Therefore, H-q equations for nodes 1 and 2 are,

respectively

E ti (12) (13) d (1) (4) i th



Equations (12), (13), and (1),…., (4) give the

required linearized H-q equations. These equations

are expresses in the matrix form as

Using the 1C’ values from Table 5 Ex 5 Eq (14) becomes



Using the 1C x values from Table 5 Ex 5, Eq. (14) becomes

Solving Eq.(15) we get,

q1= 1.84 m3/s

q2= - 0.84 m3/s

H3 = 95.66 m

H4 = 96.98 m

H5 = 94.71 m and

H6 = 94.72 m

At the different nodes either the flow is unknown or the head is



At the different nodes, either the flow is unknown or the head is

unknown, this is the usual case in practice where the heads are

known and the flows unknown at source nodes and flows are

known and heads unknown at demands nodes

For such networks, the formulation of linearized H-q equation in

the matrix form of Eq. (14)

The second matrix in Eq. (14) is column matrix for nodal heads

known at source nodes and unknown at demand nodes

The column matrix on the right hand side represents the known

flows at demand nodes and unknown flows at source nodes, the

preceding sign for the nodal flows being negative for outflows

and positive for inflows

Networks With Unknown Pipe Resistances



Networks With Unknown Pipe Resistances

When H-q-R equation are formulated, the

formulation is implicit and R values are updated at

every iteration

In the explicit approach, the R values are obtained

after satisfactory convergence is reached for the H

values

Implicit Approach



Implicit Approach –

The head loss in pipe is linearized for Rx and

therefore can be expressed as

(W)

(X)

(Y)

(Z)

EXAMPLE 6



EXAMPLE 6

Solve example by the linear theory method based on(1) Implicit and (2) Explicit Approach

Solution:

1. Implicit Approach –

In the network of fig. (6), basic unknowns are q1, H3,

H4, R4 and H5.



Fig.6 Network with unknown pipe resistance

The linearized H-q-R equation written for the



The linearized H-q-R equation written for the

node-flow-continuity relationship at nodes

1,…, 5 are, respectively

Eqs. (1),…, (5) can be expressed in the matrix form as



q ( ), , ( ) p

Initially, let the 1Qx values be 0.1m3/s. The corresponding 1hx

and 1C’x values are evaluated and shown in Table 6 Ex 6.Therefore, for the first iteration Eq. (6) becomes

Solving Eq (7) we get



Solving Eq. (7) we get,

q1(1) = 1.2025 m3/s

H3(1) = 98.81 m

H4(1) = 99.99 m

R4(1) = - 183.17 and

H5(1) = 91.54 m

Using these values, the second iteration is carried out from which we

get,

q1(2) = 1.200 m3/s

H3(2) = 98.685 m

H4(2) = 99.853 m

R4(2) = - 150.33 and

H5(2) = 90.551 m

Table 6 Example 6 First two iterations for network with



unknown pipe resistance

2 Explicit Approach –



2. Explicit Approach

Networks With Pumps



Networks With Pumps

It is preferable to use head discharge relationship

that is consistent with the head loss relationship for

the pipes

EXAMPLE 7



EXAMPLE 7

Solve by the linear theory method based on head

equation

Solution:

The linearized node-flow-continuity equations for

nodes3,…, 6 are



Fig.7 A Network with Pumped source node

Eqs. (1),…, (4) can be expressed in the matrix form as



The pump developed by the pump is given by

Since Qp = 1Q1 = 0.04712 m3/s, hp = 12.04 m. Substituting the values for thp

and 1C’x Eq. (5) becomes




p

pumped source node



Network With Valves



Network With Valves

H equations are linearized so that the discharge in pipe ij,connecting nodes i and j and containing a check valve with

permissible direction of flow from node i to node j is given by

C’ij (Hi – H j)

During the iterative procedure if (Hi – H j) becomes negative,

C’ij is made zero so that the discharge in pipe is zero

A flow control valve FCV is initially assumed to be inoperative

The H equations are formulated, linearized and solved and

HGL values of end nodes are obtained

If the HGL difference is not more than h*ij for the set discharge Qset,



ij g set,

the assumption is correct and the iteration continue

If the HGL difference is more than h*ij

, the FCV becomes operative,

this pipe is removed and outflow at upstream node is increased by

Qset, while the outflow at downstream node is decreased by Qset

For a network with a pressure reducing valve, it is initially assumed

that the PRV is in the operative mode and therefore it is replaced by

source and sink nodes and H equation are linearized and iteratively

solved

A check is made at the end of each iteration to see that the PRV has

remained in the operative mode

If the PRV becomes inoperative or the direction of flow reverse, the

PRV behaves like a check valve

EXAMPLE 8



EXAMPLE 8

Solved by the linear theory method based on H equations

Solution:

The network of fig. (8), initially it is assumed that that PRV is

in the operative mode

Using the known head of 100 m at node 1 and the set

pressure head of 80 m at d, the downstream end of the PRV,

the linearized H equations for nodes 2,…., 6 of the network

for the tth iteration are respectively,



Fig.8 A Network with PRV



• As the PRV is situated just downstream of node 3, C’d-6 = C’7

• For the first iteration, taking 1Q1 = … = 1Q7 = 0.1 m3/s, the 1C’x values

determined and solving Eq. (6) we get the values of H2(1),…., H6(1)

• The discharge in pipe 7 is evaluated and the pressure head at the

downstream end of the PRV is determined to check the PRV

At the pressure head setting of 89 m, the PRV becomes



inoperative during the iterative procedure

At this stage, C’d-6 is replaced by C’7 , corresponding to theincreased resistance from R7 to R7 + RPRV and the known

pressure head setting of 89 m is replaced by unknown

pressure head H3

Therefore Eq. (6) becomes



Relationship Between Newton-Raphson and Linear



Theory Methods

The newton-raphson method linearized the

nonlinear equations through partial differentiation

while the linear theory methods linearizes the nonlinear equations by merging a part of the non linear

term in the resistance or conductance of the pipes

Pipe – Discharge Equation

Nodal – Head Equation

Pipe - Discharge Equations



p g q

Considering a pipe with resistance constant R0

carrying discharge Q so that the head loss in the

pipe is R0Qn

Considering the 1st iteration, let the assumed

discharge in the pipe be 1Q and the correction be

ΔQ

R0 (1Q + ΔQ) n = R0Qn (1)

According to the Newton-Raphson method, we get



R0 (1Q n + n ∙ 1Q

n -1 ∙ ΔQNR) = R0Qn (2)

or

ΔQNR = Q n

−1Q n

n ∙ 1Q n −1 (3)

Where, ΔQNR = correction obtained according to Newton-

Raphson method

According to the linear theory method, we get

R0 (1Q n -1 ) Q(1) =

R0Qn (4)

Where, Q(1) is the obtained discharge in the pipe at the end of

iteration1 If ΔQLT represents the correction to 1Q according to the linear

theory, then Q(1) = 1Q + ΔQLT and thus Eq. (4) becomes

R0 (1Q n -1 ) (1Q + ΔQLT ) = R0Q

n (5)

ΔQ Q n −1Q

n(6)



ΔQLT = Q Q

1Q n −1

(6)

Thus, from Eq. (3) & (6), we get

ΔQLT = n ∙ ΔQNR (7)

As n = 1.852 or 2, it is evident that the absolute value of the

correction obtained by the linear theory method is lager than

that obtained by the Newton-Raphson method and therefore

oscillation occur

Discharge for the 2nd iteration in the liner theory method can

be taken as

2Q = 1Q +

ΔQLT (8)

Q Q

[Q Q] (9)



2Q = 1Q +

[Q(1) – 1Q] (9)

For Darcy-Weisbach and Manning head lossformulae, n = 2, so that Eq. (9) gives

2Q =1Q +Q(1)

2 (10)

Thus, for the tth iteration, the assumed discharge in

pipe x is given by

tQx =(t−1)Qx+Qx(t−1)

2 (11)

When the Hazen-Williams head loss formulae is



When the Hazen Williams head loss formulae is

used, n = 1.852 so that Eq. (9) lead to

tQx = 0.46(t-1)Qx + 0.54 Qx (t-1) (12)

However, because the difference between Eqs. (11)

& (12) is not much, Eq. (11) can be uniformly used

throughout the liner theory method irrespective of the

head loss formula

Nodal Head Equations



Consider a pipe with resistance constant R0 connecting

nodes i and j with head Hi and H j (Hi > H j) so that the

discharge Q in the pipe is given by

Q =Hi

− H j R0

1/n

(1)

Let the head Hi be known and fixed and the assumed head at

node j for the 1st iteration be 1H j with correction ΔH j so that

Hi − (1H j

+ ΔH j)

R0

1/n

=Hi −

Hj R0

1/n

(2)



According to linear theory method, we get



(6)

Since H j(1) = 1H j + ∆H j-LT from Eq. (6), we get

(7)

From which we get

(8)

Thus, from Eqs.(5) and (8), we get



∆H j-LT =

∙ ∆H j-NR (9)

thus, the modified H j(1) value i.e. H j(1)m value is given by

H j(1)m = 1H j+ n ∙ ∆H j (10)

H j(1)m = 1H j + n [H j(1) - 1H j] (11)

These H j(1)m values are used to calculate the modified values,

i.e. hx(1)m, Qx(1)m, and C’x(1)m values, so that

(12)

Instead of taking n = 1.852 or 2 depending upon the



head loss formula, if n is uniformly taken as 2, we

get

(13)

Similarly from Eq.(10), we get

(14)



(15)

Substituting the values of 1C’x and C’x(1)m from

Eqs.(13) & (14) respectively,

(16)

Expanding the term within brackets in Taylor’s series

and considering the first two terms, we get

(17)

(18)



Now the term on the right-hand side Eq.(18) represent the first

two terms of expansion in Taylor’s series of

i.e C’x(1). Therefore, we get

2C’x = C’x(1) (19)

For the tth iteration,

tC’x = C’x(t-1) (20)

Linear Theory Method

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