The head of water over an orifice of area 0.1 m² is 5 m. Cd= 0.4 The actual discharge in m³ per second if (assume g 10m/s2) is

Concept:

Coefficient of discharge (C_d) is defined as:

\(C_d=\frac{Q_{actual}}{Q_{theoretical}}\)

Theoretical discharge (Q_theoretical) can be calculated by:

Qtheoretical = Area of orifice × velocity of flow

Velocity of flow (V) can be calculated by:

\(V = \sqrt{2gh}\)

where h is head of fluid over the orifice

Calculation:

Given:

h = 5 m, Area = 0.1 m², C_d = 0.4

Velocity of flow is:

\(V = \sqrt{2gh}\)

\(V = \sqrt{2×10×5}=10~m/s\)

Theoretical discharge (Qtheoretical) is:

Qtheoretical = Area of orifice × velocity of flow

Qtheoretical = 0.1 × 10 = 1 m³/s

We know that,

\(C_d=\frac{Q_{actual}}{Q_{theoretical}}\)

Actual discharge is:

Q_actual = C_d × Q_theoretical 

Q_actual = 0.4 × 1 = 0.4 m³/s