Question 16.SP.7: Refer to Fig. 16.11. A draft tube leading from the discharge......

Original draft tube:

Head loss due to draft tube,

h_{f}=f \frac{L}{D} \frac{V^{2}}{2 g}, \quad \text { where } \quad V=\frac{Q}{A}=\frac{130}{\pi(1.5)^{2}}=18.4 \mathrm{ft} / \mathrm{sec}

Thus:                 h_{f}=0.03 \frac{40}{3} \frac{(18.4)^{2}}{2(32.2)}=2.10 \mathrm{ft}

Head loss at discharge,

Eq. (8.74):         h_{d}^{\prime}=\frac{V_{2}^{2}}{2 g}-\frac{V_{C}^{2}}{2 g}=\frac{18.4^{2}}{2(32.2)}-0=5.26 \mathrm{ft}

Thus total head loss h_{L}=h_{f}+h_{d}^{\prime}=2.10 \mathrm{ft}+5.26 \mathrm{ft}=7.36 \mathrm{ft}

Replacement draft tube:

Head loss due to draft tube,

Eq. (8.78):            \quad h^{\prime}=k^{\prime} \frac{\left(V_{1}-V_{2}\right)^{2}}{2 g}, \quad where \quad V_{1}=18.4 \mathrm{ft} / \mathrm{sec}

and                        V_{2}=\left(\frac{D_{1}}{D_{2}}\right)^{2} V_{1}=\left(\frac{3}{8}\right)^{2} 18.4=2.59 \mathrm{ft} / \mathrm{sec}

Obtain k^{\prime} from Fig. 8.20a (we shall assume the solid line applies):

\tan \alpha=\frac{(8-3)}{40}=\frac{5}{40}=0.125, \quad \alpha=7.1^{\circ}, \quad k^{\prime}=0.13

Thus:                    h^{\prime}=0.13 \frac{(18.4-2.6)^{2}}{2(32.2)}=0.50 \mathrm{ft}

Head loss at discharge,

Eq. (8.74):               \quad h_{d}^{\prime}=\frac{V_{2}^{2}}{2 g}-\frac{V_{C}^{2}}{2 g}=\frac{2.59^{2}}{2(32.2)}-0=0.10 \mathrm{ft}

Thus: Total head loss h_{L}=h^{\prime}+h_{d}^{\prime}=0.50 \mathrm{ft}+0.10 \mathrm{ft}=0.60 \mathrm{ft}

Additional effective head available with the replacement draft tube

=7.36-0.60=6.76 \mathrm{ft}

Note: 5.26-0.10=5.16 \mathrm{ft} of the additional effective head is attributable to the decrease in the head loss at discharge. An extra 6.76 \mathrm{ft} of effective head is quite significant if we are dealing with a low-head installation.

h_d^{\prime}=\frac{V^2}{2 g}-\frac{V_c^2}{2 g}     (8.74)

h^{\prime}=k^{\prime} \frac{\left(V_1-V_2\right)^2}{2 g}          (8.78)