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Conical draft tube inlet pressure and efficiency

Short answer

Using continuity, inlet velocity V1=5.88 m/s. Bernoulli equation with friction loss hf=0.092 m and outlet pressure head 1.5 m yields inlet pressure head -7.21 m gauge (3.09 m absolute). Efficiency ηd = (V1^2−V2^2)/2g − hf divided by V1^2/2g = 68.8%.

A conical draft tube having inlet and outlet diameters 1.5 m and 2.1 m discharges water at outlet with a velocity of 3 m/s. The total length of the draft tube is 7.5 m and 1.5 m of length of draft tube is immersed in water. If loss of head due to friction in draft tube is 0.2 × velocity head at the tube outlet, determine pressure head at inlet of the draft tube. Also find the efficiency of the draft tube.

Given: inlet diameter D1=1.5D_1 = 1.5 m, outlet diameter D2=2.1D_2 = 2.1 m, outlet velocity V2=3V_2 = 3 m/s, total length 7.5 m, length immersed in tailwater y=1.5y = 1.5 m, friction loss hf=0.2 V22/2gh_f = 0.2\,V_2^2/2g. Take atmospheric pressure head =10.3= 10.3 m of water.

  (1) inlet  D1 = 1.5 m
          /  \          ^
         /    \         | 6.0 m above TWL
 ~~~~~~~/      \~~~~~~~~|~~~~~ TWL
       /        \       | 1.5 m below TWL
  (2) outlet D2 = 2.1 m, V2 = 3 m/s

Height of inlet above tailwater Hs=7.5−1.5=6.0H_s = 7.5 - 1.5 = 6.0 m.

Velocity at inlet (continuity)

V1=V2(D2D1)2=3×(2.11.5)2=5.88 m/s\begin{aligned} V_1 &= V_2\left(\frac{D_2}{D_1}\right)^2 = 3 \times \left(\frac{2.1}{1.5}\right)^2 = 5.88\ \text{m/s} \end{aligned}

Velocity heads and loss

V122g=5.88219.62=1.762 mV222g=3219.62=0.459 mhf=0.2×0.459=0.092 m\begin{aligned} \frac{V_1^2}{2g} &= \frac{5.88^2}{19.62} = 1.762\ \text{m} \\ \frac{V_2^2}{2g} &= \frac{3^2}{19.62} = 0.459\ \text{m} \\ h_f &= 0.2 \times 0.459 = 0.092\ \text{m} \end{aligned}

Pressure head at inlet

Bernoulli between 1 and 2 (datum at outlet, gauge pressures; p2/ρg=1.5p_2/\rho g = 1.5 m):

p1ρg+V122g+7.5=p2ρg+V222g+hfp1ρg+1.762+7.5=1.5+0.459+0.092p1ρg=−7.21 m (gauge)\begin{aligned} \frac{p_1}{\rho g} + \frac{V_1^2}{2g} + 7.5 &= \frac{p_2}{\rho g} + \frac{V_2^2}{2g} + h_f \\ \frac{p_1}{\rho g} + 1.762 + 7.5 &= 1.5 + 0.459 + 0.092 \\ \frac{p_1}{\rho g} &= -7.21\ \text{m (gauge)} \end{aligned}

Absolute pressure head =10.3−7.21=3.09= 10.3 - 7.21 = 3.09 m of water (abs).

Efficiency of the draft tube

ηd=V12−V222g−hfV122g=1.762−0.459−0.0921.762=1.2111.762=0.688\begin{aligned} \eta_d &= \frac{\dfrac{V_1^2 - V_2^2}{2g} - h_f}{\dfrac{V_1^2}{2g}} = \frac{1.762 - 0.459 - 0.092}{1.762} \\ &= \frac{1.211}{1.762} = 0.688 \end{aligned}

Answer: pressure head at inlet =−7.21= -7.21 m of water (vacuum), i.e. 3.09 m of water absolute; efficiency of draft tube =68.8%= 68.8\%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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