Orifice Leakage Calculation

Saint Venant Formula

The velocity of fluid escaping from the reservoir through orifice is given by the following equation.

$$v = \sqrt{\left[2\left(\frac{\gamma}{\gamma-1}\right)\frac{p_{0}}{\rho_{0}}(1-r^{(\gamma-1)/\gamma})\right]}$$

where

$v$ is the flow velocity through orifice
$\gamma$ is the ratio of specific heats $C_p/C_v$
$p_0$ is the pressure in reservoir`
$\rho_0$ is the density in reservoir`
$r=p_0/p$ is the pressure ratio across orifice`

The density inside the reservoir $\rho_0$ can be obtained using the ideal gas law

$$\rho_0 = \frac{p_0}{R_gT_0}$$

where

$R_g$ is the specific gas constant
$T_0$ is the temperature in the reservoir in K`

The specific gas constant is obtained as

$$R_g = \frac{R_u}{MW}$$

where

$R_u$ is the universal gas constant (=8314 J/kmol.K)
$MW$ is the molecular mass of the fluid in kg/kmol or gm/mol`

The mass flow rate of the fluid escaping through the orifice can be obtained as follows:

(1) $$\dot{m}=Av\rho$$

where

$\dot{m}$ is the leakage mass flow rate
$A$ is the area of the orifice`
$\rho$ is the density of fluid as it just escapes the orifice`

Using isentropic process relationships we have

(2) $$\rho = \rho_0r^{1/\gamma}$$

Substituting the value of $v$ and $\rho$ in (1) we get

$$\dot{m} = A\rho_0\sqrt{\left[2\left(\frac{\gamma}{\gamma-1}\right)\frac{p_{0}}{\rho_{0}}r^{2/\gamma}(1-r^{(\gamma-1)/\gamma})\right]}$$

In actual practice, the flow will be less than what is derived above and this is addressed by introducing the coefficient of discharge term $C_d$. Introducing that in the above equation we get the final form of the equation which is also popular by the name Saint Venant Equation.

Important

Saint Venant Equation

$$\dot{m} = C_dA\rho_0\sqrt{\left[2\left(\frac{\gamma}{\gamma-1}\right)\frac{p_{0}}{\rho_{0}}r^{2/\gamma}(1-r^{(\gamma-1)/\gamma})\right]}$$