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HGL and EGL for pipes in series

Short answer

With d1 < d3 < d2, EGL is steepest in d1, flattest in d2. HGL lies below EGL by V^2/2g, so gap largest in d1, smallest in d2. At expansion to d2, EGL drops and HGL rises; at contraction to d3, EGL drops and HGL falls. Both lines end at the downstream reservoir surface.

Draw HGL and EGL diagram for the flow system shown in the figure considering all major and minor losses. [Figure: Reservoir A connected to Reservoir B by three pipes in series of diameters d1, d2 and d3, with d1 < d3 < d2]

Reservoir A (higher) feeds reservoir B through three pipes in series, with d1<d3<d2d_1 < d_3 < d_2. Both reservoir surfaces are at rest, so HGL and EGL start at the water surface of A and end at the water surface of B.

Rules used

  • EGL falls in the direction of flow: slope =fV2/(2gd)= f V^2/(2gd), so it is steepest in the smallest pipe 1 and flattest in pipe 2.
  • HGL is below the EGL by V2/2gV^2/2g; the gap is largest in pipe 1 (highest velocity) and smallest in pipe 2.
  • Sudden drops of the EGL: entrance loss 0.5V12/2g0.5V_1^2/2g; expansion from d1d_1 to d2d_2: (V1−V2)2/2g(V_1 - V_2)^2/2g; contraction from d2d_2 to d3d_3: 0.5(1−A3A2)V32/2g0.5\left(1-\dfrac{A_3}{A_2}\right)V_3^2/2g (approximately); exit loss into B: V32/2gV_3^2/2g.
  • At the exit, the whole velocity head is lost, so the EGL falls to the surface of B and both lines end at the water level of B.
 A __                                           
    |\ entrance                 
    | \___ EGL                              
    |     `-.__ (steep, d1)                  
    |  HGL     `-.   | expansion drop        
    |  (gap     `-.__|`-.__ (flat, d2)     
    |   V1²/2g)         |  `-.__|contraction
    |                    HGL     |`-.__ (d3)
    |==[d1]=======[d2 big]=======[d3]=====> 
                                       exit drop
                                          \__ B
  • The HGL rises after the expansion (velocity falls) and falls again after the contraction (velocity rises).
  • Total of all drops (friction + minor losses) equals the difference of the two reservoir levels, H=hf1+hf2+hf3+hentry+hexp+hcontr+hexitH = h_{f1} + h_{f2} + h_{f3} + h_{entry} + h_{exp} + h_{contr} + h_{exit}.

✓ Answer checked by Claude Sonnet 5.5

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