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