A conical draft-tube having inlet and outlet diameters 0.8 m and 1.2 m discharges water at outlet with a velocity of 3 m/s. The total length of the draft-tube is 8 m and 2 m of the length of draft-tube is immersed in water. If the atmospheric pressure head is 10.3 m of water and loss of head due to friction in the draft-tube is equal to 0.25 times the velocity head at outlet of the tube, find : (i) Pressure head at inlet, and (ii) Efficiency of the draft-tube.

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Conical Draft-Tube Analysis

Problem Statement

A conical draft-tube having inlet and outlet diameters 0.8 m and 1.2 m discharges water at outlet with a velocity of 3 m/s. The total length of the draft-tube is 8 m and 2 m of the length of draft-tube is immersed in water. If the atmospheric pressure head is 10.3 m of water and loss of head due to friction in the draft-tube is equal to 0.25 times the velocity head at outlet of the tube, find : (i) Pressure head at inlet, and (ii) Efficiency of the draft-tube.

Given Data & Constants

  • Inlet diameter, \(D_1 = 0.8 \, \text{m}\)
  • Outlet diameter, \(D_2 = 1.2 \, \text{m}\)
  • Outlet velocity, \(V_2 = 3 \, \text{m/s}\)
  • Height of inlet above tailrace, \(H = 8 \, \text{m} - 2 \, \text{m} = 6 \, \text{m}\)
  • Atmospheric pressure head, \(H_{atm} = 10.3 \, \text{m}\)
  • Friction loss, \(h_f = 0.25 \times \frac{V_2^2}{2g}\)
  • Acceleration due to gravity, \(g = 9.81 \, \text{m/s}^2\)

Solution

1. Calculate Velocities and Head Losses

First, find the inlet velocity using the continuity equation.

$$ A_1 = \frac{\pi}{4}D_1^2 = \frac{\pi}{4}(0.8)^2 \approx 0.5027 \, \text{m}^2 $$ $$ A_2 = \frac{\pi}{4}D_2^2 = \frac{\pi}{4}(1.2)^2 \approx 1.131 \, \text{m}^2 $$ $$ V_1 = V_2 \frac{A_2}{A_1} = 3 \times \frac{1.131}{0.5027} \approx 6.75 \, \text{m/s} $$

Next, calculate the head loss due to friction.

$$ h_f = 0.25 \times \frac{V_2^2}{2g} = 0.25 \times \frac{3^2}{2 \times 9.81} \approx 0.115 \, \text{m} $$

(i) Pressure Head at Inlet

We apply Bernoulli's equation between the draft-tube inlet (1) and outlet (2). The pressure head at the outlet (\(P_2/\rho g\)) is the atmospheric pressure plus the head due to immersion.

$$ \frac{P_1}{\rho g} + \frac{V_1^2}{2g} + z_1 = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + z_2 + h_f $$ $$ \text{Here, } z_1 = 6 \text{ m}, z_2 = 0 \text{ (datum at outlet)}, \text{ and } \frac{P_2}{\rho g} = H_{atm} + (8-6) = 10.3 + 2 = 12.3 \text{ m (abs)} $$ $$ \frac{P_{1,abs}}{\rho g} + \frac{6.75^2}{2 \times 9.81} + 6 = 12.3 + \frac{3^2}{2 \times 9.81} + 0.115 $$ $$ \frac{P_{1,abs}}{\rho g} + 2.322 + 6 = 12.3 + 0.459 + 0.115 $$ $$ \frac{P_{1,abs}}{\rho g} + 8.322 = 12.874 $$ $$ \frac{P_{1,abs}}{\rho g} = 12.874 - 8.322 = 4.552 \, \text{m of water (absolute)} $$

(ii) Efficiency of the Draft-Tube (\(\eta_d\))

The efficiency is the ratio of the actual pressure head recovered to the ideal head that could be recovered (the change in kinetic energy head).

$$ \eta_d = \frac{\text{Actual Pressure Recovery}}{\text{Ideal Pressure Recovery}} = \frac{(\frac{V_1^2 - V_2^2}{2g}) - h_f}{(\frac{V_1^2 - V_2^2}{2g})} $$ $$ \text{Ideal Recovery} = \frac{6.75^2 - 3^2}{2 \times 9.81} = \frac{36.5625}{19.62} \approx 1.863 \, \text{m} $$ $$ \eta_d = \frac{1.863 - 0.459}{1.863} = \frac{1.404}{1.863} \approx 0.7536 $$
Final Results:

(i) Pressure head at inlet: \( \approx 4.55 \, \text{m} \) (absolute)

(ii) Efficiency of the draft-tube: \( \approx 75.4\% \)

Explanation of a Draft Tube

A draft tube is a diverging pipe fitted to the exit of a reaction turbine (like a Francis or Kaplan turbine). It has two main purposes:

  1. It allows the turbine to be placed safely above the tailwater level without losing any head.
  2. It converts the high kinetic energy of the water leaving the runner back into useful pressure energy. By gradually increasing the pipe's cross-sectional area, it slows the water down, causing a corresponding rise in pressure according to Bernoulli's principle.

Pressure at Inlet: The calculation shows the absolute pressure at the turbine outlet (draft tube inlet) is only 4.55 m, which is well below the atmospheric pressure of 10.3 m. This sub-atmospheric pressure effectively "sucks" the water through the runner, allowing the turbine to utilize the full head from the reservoir to the tailrace, even though it's physically located above the tailrace.

Efficiency: An efficiency of 75.4% means that the draft tube is converting over 75% of the available kinetic energy into pressure head, with the rest being lost to friction and the final exit velocity.

Related Posts

The figure below shows a pipe connecting a reservoir to a turbine which discharges water to the tailrace through another pipe. The head loss between the reservoir and the turbine is 8 times kinetic head in the pipe and that from the turbine to tailrace is 0.4 times the kinetic head in the pipe. The rate of flow is 1.36m3/s and the pipe diameter in both cases is 1m. Determine (a) the pressure at inlet and exit of turbine, and (b) the power generated by the turbine.

The figure below shows a pipe connecting a reservoir to a turbine which discharges water to the tailrace through another pipe. The head loss between the reservoir and the turbine is 8 times kinetic head in the pipe and that from the turbine to tailrace is 0.4 times the kinetic head in the pipe. The rate of flow is 1.36m3/s and the pipe diameter in both cases is 1m. Determine (a) the pressure at inlet and exit of turbine, and (b) the power generated by the turbine.

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