Question 5.18: For the 100 mm diameter short tube acting as a mouthpiece in...
For the 100 mm diameter short tube acting as a mouthpiece in a tanks as shown in Fig. 5.40, (a) what flow of water at 24 ºC will occur under a head of 9.2 m? (b) What is the pressure head at vena contracta section c? (c) What maximum head can be used if the tube is to flow full at exit? (Take C_{d} = 0.82 and C_{v} = 1.0 Vapour pressure for water at 24 ºC is 3 kPa absolute, atmospheric pressure is 101 kPa).
Applying Bernoulli′s equation between the points a and b (the point b being at the exist plane) on a streamline (Fig. 5.40), with the horizontal plane through b as datum, we can write
\frac{p_{ atm }}{\rho g}+(h-z)+0+z=\frac{p_{ atm }}{\rho g}+\frac{V_{b}^{2}}{2 g}+0+\frac{V_{b}^{2}}{2 g}\left(\frac{1}{C_{c}}-1\right)^{2} (5.115)
where C_{c} is the coefficient of contraction (= C_{d} / C_{v} = 0.82/1 = 0.82). With the values given,
9.2=\frac{V_{b}^{2}}{2 g}\left[1+\left(\frac{1}{0.82}-1\right)^{2}\right]
which gives V_{b}=13.12 m/s
Then Q=C_{d} \times A \times V_{b}=0.82 \times(\pi / 4)(0.1)^{2} \times(13.12)=0.084 m ^{3} / s
(b) Applying Bernoulli′s equation between the points a and c on a streamline (the point c being at the vena contracta section), we get
\frac{p_{ atm }}{\rho g}+(h-z)+0+z=\frac{p_{c}}{\rho g}+\frac{V_{c}^{2}}{2 g}+0 (5.116)
Again from continuity,
A_{b} \times V_{b}=A_{c} \times V_{c}
where A_{b} \text { and } A_{c} are the areas at exist and vena contracta respectively
Hence, V_{c}=\frac{V_{b}}{A_{c} / A_{b}}=\frac{V_{b}}{C_{c}}=\frac{13.12}{0.82}=16 m/s
Substituting V_{c} in Eq. (5.116), we have
\frac{p_{ atm }}{\rho g}+9.2=\frac{p_{c}}{\rho g}+\frac{(16)^{2}}{2 g}
which gives P_{c} / \rho g=\left(\frac{p_{ atm }}{\rho g}-3.85\right) m of water
Therefore the pressure at vena contracta = 3.85 m of water vacuum.
(c) As the head causing flow through the short tube is increased, the velocity of flow at any section will increase and the pressure head at c will be reduced. For a steady flow with the tube full at exit, the pressure at c must not be less than the vapour pressure of the liquid at the working temperature. For any head h (the height of the liquid level in the tank above the centre line of the tube), we get from Eq. (5.116)
\frac{p_{ atm }}{\rho g}+h=\frac{p_{c}}{\rho g}+\frac{V_{c}^{2}}{2 g} (5.117)
again V_{c}=A_{b} V_{b} / A_{c}=V_{b} / C_{c}
Again, from Eq. (5.115),
V_{b}^{2}=\frac{2 g h}{\left(\frac{1}{C_{c}}-1\right)+1}
Therefore, \frac{V_{c}^{2}}{2 g}=\frac{h}{C_{c}^{2}\left[\left(\frac{1}{C_{c}}-1\right)^{2}+1\right]}
=\frac{h}{(0.82)^{2}\left[\left(\frac{1}{0.82}-1\right)^{2}+1\right]}=1.42 h
Substituting this value of V_{c} in Eq. (5.117), we get
\frac{p_{ atm }}{\rho g}+h=\frac{p_{c}}{\rho g}+1.42 h
or 0.42 h=\frac{p_{ atm }}{\rho g}-\frac{p_{c}}{\rho g}
For the maximum head h, p_{c}=p_{v}, the vapour pressure of water at the working temperature.
For the present case, \frac{p_{v}}{\rho g}=\frac{3 \times 10^{3}}{10^{3} \times 9.81}=0.306 m
while, \frac{p_{ atm }}{\rho g}=\frac{101 \times 10^{3}}{10^{3} \times 9.81}=10.296 m
Hence, h_{\max }=\frac{10.296-0.306}{0.42}=23.78 m