Hardycross method

civil engineering,environmental

1. Hardycross Method
Muhammad Nouman
UET,Peshawar,Pakistan

2. Hardy Cross Method:-
•To analyze a given distribution system to determine the pressure and flow available in any section of the
system and to suggest improvement if needed a number of methods are used like Equivalent pipe, Circle
method, method of section and Hardy-cross method. Hardy-cross method is a popular method. According
to this method the sum of the loss of head for a closed network / loop is equal to zero. Also the sum of
inflow at a node / joint is equal to the out flow i.e. ∑inflow = ∑outflow or total head loss = 0
Assumption:-
Sum of the inflow at a node/point is equal to outflow.∑inflow = ∑outflow; ∑total flow = 0
Algebraic sum of the head losses in a closed loop is equal to zero. ∑head losses = 0
Clockwise flows are positive. Counter clockwise flows are negative.
• Derivation:-
According to Hazen William Equation H= 10.68 (Q/C)
1.85
(L / d)
4.87
H= KQ
1.85
where K = (10.68L)/(C
1.85
*d
4.87
)
For any pipe in a closed loop
Q = Q
1
+ ∆ where Q = actual flow; Q
1
= assumed flow and ∆ = required flow correction
H= KQ
x
(i) x is an exponent whose value is generally 1.85
From (i) H = K(Q
1
+ ∆)
x
By Binomial theorm = K[Q
1
x
+ (xQ
1
x-1
∆)/1! + {x(x-1)Q
x-2
∆
2
}/2! + ……..]
As ∆ is very small as compared to Q, we can neglect ∆
2
etc. Therefore, H = KQ
1
x
+ KxQ
1
x-1
∆ (ii)
For a closed loop ∑ H = 0 => ∑ Q
x
= 0 => ∑ k Q
1
x
= - ∆∑ x Q
1
x-1
) => ∆ = - ∑ k Q
1
x
/ (∑ x Q
1
x-1
)

3. As H = K*Q
x
∑ KQ
1
x
/ Q
1
= H/Q 1
Therefore, ∆ = - ∑ H/ [x* ∑ (H)/ Q
1
]
Above equation is used in Hardy Cross Method
Procedure : (i) Assume the diameter of each pipe in the loop.
(ii) Assume the flow in the pipe such that sum of the inflow = sum of the outflow at any junction or node
( V = V
1
+ V
2
or Q = Q
1
+ Q
2
)
(iii) Compute the head losses in each pipe by Hazen William Equation H = 10.68 * (Q/C)
1.85
* L/D
4.87
(iv) Taking clock wise flow as positive and anti clock wise as negative.
(v) Find sum of the ratio of head loss and discharge in each pipe without regard of sign ∑ ( H/Q1 )
(vi) Find the correction for each loop from ∆ = - ∑ H/ [x* ∑ (H)/ Q
1
] and apply it to all pipes.
(vii) Repeat the procedure with corrected values of flow and continue till the correction become very small

4. Example 2: For the branching pipe system shown below: At B and C, a
minimum pressure of 5 m. At A, maximum pressure required is 46 m and the
minimum is 36 m. Select a suitable diameter for AB and BC.



0.15 l/s
219 m A

2.9 m
3
/h 2.4 m
3
/h C (219 m,
700m
0.5m
3
/h
825 m
B 189 m

Public water main

Example

5. Solution: Computation Table
Pipe
Sect.
Flow
(m
3
/h
)
Lengt
h (m)
Pipe
Dia
mm
Head
Loss
(m/10
0 m
Flow
Vel
m/s
Head
Loss
(m)
Elev.
of
hydr.
Grade
(m)
Groun
d level
elev
(m)
Press
Head
(m)
Rem.
AB2.9700323.3
0.8523
A
260
B
237
18946O.K
BC0.5825191.60.513
B
237
C
224
2195
Just
O.K

6. Example
Example: Obtain the flow rates in the network shown below.

90 l/s

A 55 600 m B
45

35 600 m
254 mm
600 m C C
152 mm 15
15 60l/s
66600 600 m
E 600 m 5 D 152 mm
152 mm
254 mm
10
+ve
600
152 mm
ABDE is one loop above and BCD is the second loop.
Note that the clockwise water flows are positive while the anti-clockwise ones are negative.
Positive and negative flows give rise to positive and negative head losses respectively

7. Solution
Circuit Pipe L (m) D (m) Q (m
3
/s) hf (m) hf/Q Q
AB 600 0.254 + 0.055 2.72 49.45
I BD 600 0.152 + 0.01 1.42 142
DE 600 0.152 - 0.005 - 0.39 78 0.008
EA 600 0.152 - 0.035 -14.42 412
Total - 10.67 681.45



BC 600 0.254 + 0.045 1.88 41.8
II CD 600 0.152 - 0.015 - 3.01 200.67 0.004
DB 600 0.152 - 0.010 - 1.42 142
Total - 2.55 384.47

Sample Calculation: Using the Hazen Williams Equation in Step 2 :

hf for pipe AB = 10.67 x 135
– 1.85
x 0.254
-4.87
x 0.055
1.85
x 600 = 2.72