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Loss due to sudden enlargement and contraction

Short answer

Loss due to sudden enlargement is h_e = (V1 - V2)²/(2g) = (1 - A1/A2)² V1²/(2g). Loss due to sudden contraction is deduced by applying the enlargement formula at the vena contracta: h_c = (1/Cc - 1)² V2²/(2g), with Cc ≈ 0.62 giving k_c ≈ 0.375 (commonly ~0.5).

Derive an expression for the loss due to sudden enlargement in pipe flow and therefrom deduce the loss due to sudden contraction.

Loss due to sudden enlargement

  1                2
 _______          _________    Pipe area A1 -> A2 (A2 > A1)
        |  ~ ~ ~ |             Eddies form in the corner;
  V1,p1 |  eddy  |  V2,p2      the pressure on the annular face
 _______|  zone  |_________    equals p1.

Consider section 1 (just before the enlargement) and section 2 (downstream, where the flow has spread over the full area A2A_2). Assume the pipe is horizontal and neglect wall friction between 1 and 2. The pressure on the annular shoulder is taken as p1p_1 (eddy zone).

Momentum equation:

p1A1+p1(A2−A1)−p2A2=ρQ(V2−V1),Q=A2V2p_1A_1 + p_1(A_2 - A_1) - p_2A_2 = \rho Q (V_2 - V_1), \qquad Q = A_2V_2 p1−p2γ=V2(V2−V1)g⋯(1)\frac{p_1 - p_2}{\gamma} = \frac{V_2(V_2 - V_1)}{g} \quad \cdots (1)

Energy equation (head loss heh_e):

p1γ+V122g=p2γ+V222g+he⇒he=p1−p2γ+V12−V222g\frac{p_1}{\gamma} + \frac{V_1^2}{2g} = \frac{p_2}{\gamma} + \frac{V_2^2}{2g} + h_e \quad\Rightarrow\quad h_e = \frac{p_1-p_2}{\gamma} + \frac{V_1^2 - V_2^2}{2g}

Substituting (1):

he=V22−V1V2g+V12−V222g=V12−2V1V2+V222gh_e = \frac{V_2^2 - V_1V_2}{g} + \frac{V_1^2 - V_2^2}{2g} = \frac{V_1^2 - 2V_1V_2 + V_2^2}{2g} he=(V1−V2)22g=(1−A1A2)2V122g\boxed{h_e = \frac{(V_1 - V_2)^2}{2g} = \left(1 - \frac{A_1}{A_2}\right)^2\frac{V_1^2}{2g}}

Loss due to sudden contraction (deduced)

  1      c         2       Jet contracts to the vena contracta c
 ________                  (area Cc A2), then expands to fill
          \ ~~~ _          the smaller pipe A2.
  V1       \___|___ V2     Loss is only in the expansion c -> 2.
 ________/  Vc

In a sudden contraction from A1A_1 to A2A_2 the stream contracts to a vena contracta of area Ac=CcA2A_c = C_c A_2, then expands suddenly to fill the pipe A2A_2. The only significant loss is this expansion from cc to 2. Applying the enlargement result between cc and 2:

hc=(Vc−V2)22g,Vc=A2V2Ac=V2Cch_c = \frac{(V_c - V_2)^2}{2g}, \qquad V_c = \frac{A_2V_2}{A_c} = \frac{V_2}{C_c} hc=(1Cc−1)2V222g=kcV222g\boxed{h_c = \left(\frac{1}{C_c} - 1\right)^2\frac{V_2^2}{2g} = k_c\frac{V_2^2}{2g}}

With Cc≈0.62C_c \approx 0.62, kc=(1/0.62−1)2≈0.375k_c = (1/0.62 - 1)^2 \approx 0.375; a common value used is kc≈0.5k_c \approx 0.5 for a large area ratio.

✓ Answer checked by Claude Sonnet 5.5

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