piping systems. piping systems- example 1-1 or (a)or (b) type i (explicit) problem : given: l i, d...

Piping Systems

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Piping Systems- Example 1-1

or (a)

or (b)

Type I (explicit) problem :Given: Li, Di, Qi; Find Hi

Type II (implicit) problem :Given: Li, Di, Hi; Find Qi

Type III (implicit) problem :Given: Li, Hi Qi; Find Di

Example 1-1 (Continue)

May be used to find any variableif the others are given, i.e.

for any type of problem (I, II, III)

Example 1-2 with MathCAD

For inlet (0.78)& exit (1), p.18

For valve (55) &elbows (2*30),NOTE: K=CfT

Re & fT

No pumpGiven

Solution

Satisfying conservation of mass and energy equationswe may solve or “guess and check” any piping problem …!

Hardy-Cross Method & Program

• For every node in a pipe network:

• Since a node pressure must be unique, then net pressure loss head around any loop must be zero:

2.0 Eqhj

bemustf j

• If we assume Q

to satisfy Eq. 1, the Eq. 2 will not be satisfied.

• So we have to correct Qfor Qloop, so that Eq. 2 is satisfied,

i.e.:

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QQhQdQ

dhQhQQh

hQhh

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nexti

loopj

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2

0

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then,0 Since.Q offunction as)( all Express

1.00 EqQ

Iteration #

Hardy-Cross Method & Program (2)

Example 1-13 (cont.)H

ardy

-Cro

ss s

ubro

utin

e

...loops...pipes

...guess (LX1)

…correction (LX1)

…while tolerance is not satisfied

…new Qs (PX1)=(PXL)(LX1)…result (PX1)

The resultsafter Hardy-Cross

iterations

Example 1-13

Loop 1

Loop 2

This 3rd loop is not independent (no new pipe in it)

Example 1-13 (cont.)

T

N

110

011

10

11

01

:matrix "Connection"

“Con

nect

ion”

mat

rix

N

2 loops

3 pi

pes

Example 1-13 (cont.)

d’s

L’s

units & g

Pump and reservoirs

dhd/dQ derivative

roughnesses

Kin. viscosity

Example 1-13 (cont.)

Re & fT

Laminar & turbulent f

No minor losses

Example 1-13 (cont.)

Loss & deviceheads

Derivative of h(Q)

Qi guesses from conservation of mass

“Connection” matrix N2 loops

3 pipes

About constant

Example 1-13 (cont.)

The resultsafter Hardy-Cross

iterations

...loops...pipes

...guess

…correction

…while tolerance is not satisfied

…new Qs…result

Har

dy-C

ross

sub

rout

ine

Since assumed Q>0