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Equivalent length of pipeline losses

Short answer

Sum of loss coefficients expressed in 200 mm pipe velocity heads is 51.04. Equivalent length of 200 mm pipe with f=0.02 is L_e = (51.04 × 0.2)/0.02 = 510.4 m.

A pipe line system consists of the following sources of head losses. (i) Entrance loss in 200 mm diameter (ii) Friction loss in 500 m of 200 mm diameter pipe with f = 0.02 (iii) Sudden expansion from 200 mm to 250 mm diameter (iv) Exit loss from 250 mm diameter pipe. Obtain the equivalent length of 200 mm diameter pipe with f = 0.02.

Idea. Express every loss in terms of the velocity head in the 200 mm pipe, V122g\dfrac{V_1^2}{2g}, and find the length of 200 mm pipe (f=0.02f = 0.02) that has the same total loss: fLeD1=∑Kf\dfrac{L_e}{D_1} = \sum K.

Areas: A1A2=(200250)2=0.64\dfrac{A_1}{A_2} = \left(\dfrac{200}{250}\right)^2 = 0.64, so V2=0.64V1V_2 = 0.64V_1.

LossExpressionIn terms of V12/2gV_1^2/2g
(i) Entrance0.5V122g0.5\dfrac{V_1^2}{2g}0.5
(ii) Friction, 500 m of 200 mmfLDV122g=0.02(5000.2)V122gf\dfrac{L}{D}\dfrac{V_1^2}{2g} = 0.02\left(\dfrac{500}{0.2}\right)\dfrac{V_1^2}{2g}50
(iii) Sudden expansion(V1−V2)22g=(1−A1A2)2V122g\dfrac{(V_1 - V_2)^2}{2g} = \left(1 - \dfrac{A_1}{A_2}\right)^2\dfrac{V_1^2}{2g}(1−0.64)2=0.1296(1 - 0.64)^2 = 0.1296
(iv) Exit from 250 mm pipeV222g=(0.64)2V122g\dfrac{V_2^2}{2g} = (0.64)^2\dfrac{V_1^2}{2g}0.4096
Total51.0392

Equivalent length of 200 mm pipe:

fLeD1=51.0392  ⇒  Le=51.0392×0.20.02=510.39 mf\frac{L_e}{D_1} = 51.0392 \;\Rightarrow\; L_e = \frac{51.0392\times 0.2}{0.02} = 510.39\ \text{m}

Answer: Equivalent length ≈510.4\approx 510.4 m of 200 mm pipe (with f=0.02f = 0.02). The minor losses (entrance, expansion, exit) alone are equivalent to about 10.4 m of pipe.

✓ Answer checked by Claude Opus 5.5 (high effort)

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