Download - Hardycross method
Hardycross Method
Muhammad Nouman
UET,Peshawar,Pakistan
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= KQ1.85 where K = (10.68L)/(C1.85*d4.87)
For any pipe in a closed loop
Q = Q1 + ∆ where Q = actual flow; Q1 = assumed flow and ∆ = required flow correction
H= KQx (i) x is an exponent whose value is generally 1.85
From (i) H = K(Q1 + ∆)x By Binomial theorm = K[Q1x + (xQ1
x-1∆)/1! + {x(x-1)Qx-2∆2}/2! + ……..]
As ∆ is very small as compared to Q, we can neglect ∆2 etc. Therefore, H = KQ1x + KxQ1
x-1∆ (ii)
For a closed loop ∑ H = 0 => ∑ Qx = 0 => ∑ k Q1x = - ∆∑ x Q1
x-1) => ∆ = - ∑ k Q1x / (∑ x Q1
x-1)
As H = K*Q x
∑ KQ1 x/ Q1 = H/Q 1
Therefore, ∆ = - ∑ H/ [x* ∑ (H)/ Q1]
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 = V1 + V2 or Q = Q1 + Q2 )
(iii) Compute the head losses in each pipe by Hazen William Equation H = 10.68 * (Q/C)1.85 * L/D4.87
(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)/ Q1] and apply it to all pipes.
(vii) Repeat the procedure with corrected values of flow and continue till the correction become very small
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 m3/h 2.4 m3/h C (219 m, 700m 0.5m3/h 825 m B 189 m Public water main
Example
Solution: Computation Table
Pipe Sect.
Flow (m3/h)
Length (m)
Pipe Dia mm
Head Loss (m/100 m
Flow Vel m/s
Head Loss (m)
Elev. of hydr. Grade (m)
Ground level elev (m)
Press Head (m)
Rem.
AB 2.9 700 32 3.3 0.85 23 A 260B 237
189 46 O.K
BC 0.5 825 19 1.6 0.5 13 B 237C 224
219 5 Just O.K
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
Solution
Circuit Pipe L (m) D (m) Q (m3/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