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

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