Skip to main content

Flow through partially and fully submerged orifice

Short answer

For fully submerged orifice, Q = Cd a √(2g(H1-H2)) where H1, H2 are upstream and downstream heads. Partially submerged flow sums free part Q1 = (2/3)Cd b √(2g)[H^(3/2) - H1^(3/2)] and submerged part Q2 = Cd b (H2 - H) √(2gH), where H is the level difference, H1 top edge, H2 bottom edge.

Derive an expression for flow through partially and fully submerged orifice.

A submerged orifice discharges below the surface of the liquid on the downstream side. It may be fully or partially submerged.

Fully submerged orifice

Let the upstream and downstream liquid surfaces be at heights H1H_1 and H2H_2 above the centre of the orifice, and H=H1−H2H = H_1 - H_2. Apply Bernoulli between a point in the upstream tank (negligible velocity) and the vena contracta, where the pressure is the downstream hydrostatic pressure γH2\gamma H_2:

H1+0=H2+Vc22g  ⇒  Vc=2g (H1−H2)=2gHH_1 + 0 = H_2 + \frac{V_c^2}{2g}\;\Rightarrow\; V_c = \sqrt{2g\,(H_1 - H_2)} = \sqrt{2gH}

With orifice area aa and coefficient of contraction CcC_c, Q=CcaVcQ = C_c a V_c. Including the velocity coefficient CvC_v gives Cd=CcCvC_d = C_cC_v:

Q=Cd a2g (H1−H2)Q = C_d\,a\sqrt{2g\,(H_1 - H_2)}

The discharge depends only on the head difference.

Partially submerged orifice

Here the downstream level lies between the top and bottom edges of the orifice, and the flow is the sum of two parts.

Let bb be the width of the (rectangular) orifice, H1H_1 and H2H_2 the depths from the upstream surface to the top and bottom edges, and HH the difference between the upstream and downstream levels (H1<H<H2H_1 < H < H_2).

  • Free part (top edge to the downstream level): the pressure outside is atmospheric, so it behaves like a free rectangular orifice with the head varying from H1H_1 to HH:
Q1=23Cdb2g[H3/2−H13/2]Q_1 = \frac{2}{3}C_d b\sqrt{2g}\left[H^{3/2} - H_1^{3/2}\right]
  • Submerged part (downstream level to bottom edge): height (H2−H)(H_2 - H) under the constant head difference HH:
Q2=Cd b (H2−H)2gHQ_2 = C_d\,b\,(H_2 - H)\sqrt{2gH}

Total discharge:

Q=23Cdb2g[H3/2−H13/2]+Cdb (H2−H)2gHQ = \frac{2}{3}C_d b\sqrt{2g}\left[H^{3/2} - H_1^{3/2}\right] + C_d b\,(H_2 - H)\sqrt{2gH}
 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ upstream level
   |    |    |
   |    |    H1
   |    |    v      ____  top edge
   |    H          |    |  free part
   H2   |          |    |
   |    v ~~~~~~~~~|~~~~|~~~~~~~ downstream level
   |               |    |  submerged part
   v               |____|  bottom edge

✓ Answer checked by Claude Opus 5.5 (high effort)

All Fluid Mechanics chapter 6 questions →