of discharge of 0.95. The inlet pressure is 10 kPa (gauge), the vapour pressure of water is 4 kPa (abs) and the local atmospheric pressure is 96 kPa (abs).

Given:

For the Venturimeter

$D_1$=100 mm (Inlet diameter of Venturimeter)

$D_2$=50 mm (Throat of Venturimeter)

$C_d$=0.95 (Coefficient of Discharge)

$P_1$=10 kPa(gauge)(Inlet Pressure)

$P_vap$=4 kPa(abs)(Vapour Pressure of water)

$P_amb$=96 kPa(abs)(Local atmospheric or ambient Pressure)

To Find:

$Q ̇_max$ (Maximum dischage of water without cavitation)

Sol:

Now, the absolute pressure at the inlet of the venturimeter, $P_1=10+P_amb$=10+96=106 kPa

We know that as the velocity of fluid increases from inlet to throat due to converging nozzle, the pressure increases form inlet to the throat.

Thus as the discharge is increased a particular maximum value is reached, after which if the discharge is increased further the pressure falls below vapour pressure leading to the incidence of cavitation. This discharge value corresponds to the maximum discharge of water that can be carried without cavitation.

Hence to find $Q ̇_max⇒P_2=P_vap$=4 kPa(abs)

Applying Bernoulli’s Principle between the inlet section and the throat,

$\frac{P_1}{ρg}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{ρg}+\frac{v_2^2}{2g}+Z_2$

Where, Suffix 1 represents quantities at inlet while suffix 2 represents quantities at throat. All values are theoretical values.

Assuming the venturimeter to be horizontal ⇒$Z_1=Z_2$

∴$\frac{P_1}{ρg}+\frac{v_1^2}{2g}=\frac{P_2}{ρg}+\frac{v_2^2}{2g}$

∴$\frac{P_1}{ρg}-\frac{P_2}{ρg}=\frac{v_2^2}{2g}-\frac{v_1^2}{2g}$

Now, $Q ̇_th=v_1 A_1=v_2 A_2$

∴$\frac{P_1-P_2}{ρg}$=$\frac{Q_{th}^{2}}{2g}$ $[\frac{1}{A_2{^2}}-\frac{1}{A_1^{2}}]$

Now,

$A_1=\frac{π}{4} D_1^2=\frac{π}{4} 0.1^2=7.854∙10^{-3}$

and

$A_2=\frac{π}{4} D_2^2=\frac{π}{4}0.05^2=1.9635∙10^{-3}$

Substituting in the above equations,

∴$\frac{(106-4) 10^3}{9810}=\frac{Q ̇_th^2}{(2(9.81)} [\frac{1}{(1.9635∙10^{-3}){^2}} -\frac{1}{(7.854∙10^{-3}))^{2}} ]$

∴$10.3976=Q ̇_th^2 (12393.97)$

∴$Q ̇_th^2=8.3892∙10^(-4)$

∴$Q ̇_th=0.02896 m^3/s$

Now, we know that

$C_d=\frac{Q ̇_act}{Q ̇_th}$

Where, $Q ̇_act=Q ̇_max$

Substituting in the above equation,

0.95=$\frac{Q ̇_max}{0.02896}$

∴$Q ̇_max=0.02752 m^3/s =27.52 lts/s$

…this is the maximum discharge.