Hardy-Cross Method & Program
 Qb
0
• For every node in a pipe network:
•
Iteration #
0
Eq . 1
b
Since a node pressure must be unique,
then net pressure loss head (losses-gains)
around any loop must be zero:

must be
h NetLoss
,i
 0 Eq . 2
Loop
• If we assume Qb0 to satisfy Eq. 1, the Eq. 2 will not be satisfied.
• So we have to correct Qb0 for DQloop, so that Eq. 2 is satisfied,
i.e.:
Express all h NetLoss
,i
 h loss , i ( Q ) as function
h
of Q . Since
 0 , then
0
loss , i
Loop
previous
h loss , i ( Q i  D Q loop )  h loss , i
next
previous
(Q ) 
dh loss , i
dQ
D Q loop ;

h loss , i  0 
next
D Q loop
Loop
L
Q
next
i
 Q i  D Q loop ...continu e until
convergenc
e
 D Q 
2
loop
loop 1
 tol
Hardy-Cross Method & Program (2)
...loops
...pipes
Example 1-13 (cont.)
Hardy-Cross subroutine
...guess Qloop(L1); Note l1
…while tolerance is not satisfied
Since assumed DQ>0
…correction (L1)
…new Qs (P1)=(PL)(L1)
…result (P1)
The results
after Hardy-Cross
iterations
Example 1-13
Loop 1
Loop 2
This 3rd loop is not independent (no new pipe in it)
Loops closed with Reference h=z=0
" matrix :
2 loops
0 

1 
 1 
1
 
 0
1
1
0 

 1
“Connection” matrix N
1

N  1
 0

3 pipes
Example 1-13 (cont.)
" Connection
T
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 DQ>0
…correction
…new Qs
…result
The results
after Hardy-Cross
iterations