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.

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The figure below shows a pump P pumping 90 lps of water from a tank. (a) What will be the pressure at points B and C when the pump delivers 14.5KW of power to the flow? Assume the losses in the system to be negligible. (b) What will be the pressure at C when the loss in the in the inlet to the pump is negligible and between the pump and the point C, a loss equal to 2 times the velocity head at B takes place.

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