Question 5.19: An external mouthpiece converges from the inlet up to the ve...
An external mouthpiece converges from the inlet up to the vena contracta to the shape of the jet and then it diverges gradually. The diameter at the vena contracta is 20 mm and the total head over the centre of the mouthpiece is 1.44 m of water above the atmospheric pressure. The head loss in flow through the converging passage and through the diverging passage may be taken as one per cent and five per cent respectively of the total head at the inlet to the mouthpiece. What is the maximum discharge that can be drawn through the outlet and what should be the corresponding diameter at the outlet. Assume that the pressure in the system may be permitted to fall to 8 m of water below the atmospheric pressure head, and the liquid conveyed is water.
In terms of meters of water, Total head available at the inlet to the mouthpiece h_{1} = 1.44 m above the atmospheric pressure
Loss of head in the converging passage =0.01 \times 1.44
= 0.0720 m
Total head available at the vena contracta
= 1.44 – 0.0144 = 1.4256 m above the atmospheric pressure
At vena contracta, we can write
\frac{p_{c}}{\rho g}+\frac{V_{c}^{2}}{2 g}=1.4256+\frac{p_{ atm }}{\rho g}
For a maximum velocity V_{c}, the pressure P_{c} will attain its lower limit which is 8 m below the atmospheric pressure. Therefore,
\frac{p_{ atm }}{\rho g}-8+\frac{V_{c}^{2}}{2 g}=1.4256+\frac{p_{ atm }}{\rho g}
which gives V_{c} = 13.6 m/s
Therefore, the maximum possible discharge becomes
Q_{\max }=13.6 \times \pi(0.02)^{2} / 4=0.0043 m ^{3} / s
Pressure at the exit is atmospheric. Application of Bernoulli′s equation between the vena contracta section and the exit section gives
\frac{p_{ atm }}{\rho g}+1.4256=\frac{p_{ atm }}{\rho g}+\frac{V_{2}^{2}}{2 g}+0.0720
Hence, v_{2}, the exit velocity = 5.15 m/s
Therefore, the diameter d_{2} at the exit is given by
\left(\pi d_{2}^{2} / 4\right) \times 5.15=0.0043
or d_{2} = 0.0326 m = 32.60 mm