Piping Systems
mass   mass  Energy  Energy  Conservati on Laws


; 


IN
OUT
IN
OUT

 
 
 
 for CV & steady state
 Ein,i  Eout, j  Eout,k  Ein,l if Ein , Eout on both side of Eq.
i
j
k
l
J
Energy
m 2  Energy
J

 e  h  head   2 ;
 H  Head   m
mass
N

 kg s  Weight
W pump
P V2
L V2
V2
V2
h  H  g  ; ; gz ;
 hp ; f
;K
; CfT
; etc.
 2
m
D 2
2
2
h
in,i
i
 hout , j  hout,k
j
k
VD QD
4 Q
 hin,l ; Re 



D


A

l
circle
VD
 shape 

  CfT
Power  Q  P  Q ( h)  m h ; f  f (Re,
) ; K  K 
D
size,Re


Piping Systems- Example 1-1
or (a)
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
or (b)
Example 1-1 (Continue)
May be used to find any variable
if 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
Given
No pump
Re & fT
Solution
Satisfying conservation of mass and energy equations
we may solve or “guess and check” any piping problem …!
Hardy-Cross Method & Program
0
Q
 b 0
• For every node in a pipe network:
•
Iteration #
Eq.1
b
Since a node pressure must be unique,
then net pressure loss head around any loop
must be zero:
mustbe
h
 f j  0 Eq.2
j
• If we assume Qb0 to satisfy Eq. 1, the Eq. 2 will not be satisfied.
• So we have to correct Qb0 for Qloop, so that Eq. 2 is satisfied,
i.e.:
Express all h f j  h f j (Q) as function of Q . Since
0
h
 f j  0 , then
j
previous
h next
(
Q


Q
)

h
(Q) 
fj
fj
next
i
Q
dh fprevious
j
dQ
Q ;  h next
 0  Qloop  Q
fj
j
 Qi  Qloop ...continu e until convergenc e
 Q 
L
loop1
2
loop
 tol
Hardy-Cross Method & Program (2)
...loops
...pipes
Example 1-13 (cont.)
Hardy-Cross subroutine
...guess (LX1)
…while tolerance is not satisfied
…correction (LX1)
…new Qs (PX1)=(PXL)(LX1)
…result (PX1)
The results
after Hardy-Cross
iterations
Example 1-13
Loop 1
Loop 2
This 3rd loop is not independent (no new pipe in it)
2 loops
T
 1 1 0 

 
 0 1  1
“Connection” matrix N
 1 0 


N   1 1 
 0  1


3 pipes
Example 1-13 (cont.)
" Connection " matrix :
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 & device
heads
Derivative of h(Q)
About constant
Qi guesses from conservation of mass
2 loops
3 pipes
“Connection” matrix N
...loops
...pipes
Example 1-13 (cont.)
Hardy-Cross subroutine
...guess
…while tolerance is not satisfied
Since assumed Q>0
…correction
…new Qs
…result
The results
after Hardy-Cross
iterations