A centrifugal pump is to discharge 0.12 m³/s at a speed of 1400 r.p.m. against a head of 30 m. The diameter and width of the impeller at outlet are 25 cm and 5 cm respectively. If the manometric efficiency is 75%, determine the vane angle at outlet.

Pump Outlet Vane Angle Calculation

Problem Statement

A centrifugal pump is to discharge 0.12 m³/s at a speed of 1400 r.p.m. against a head of 30 m. The diameter and width of the impeller at outlet are 25 cm and 5 cm respectively. If the manometric efficiency is 75%, determine the vane angle at outlet.

Given Data & Constants

  • Discharge, \(Q = 0.12 \, \text{m}^3/\text{s}\)
  • Speed, \(N = 1400 \, \text{r.p.m.}\)
  • Manometric Head, \(H_m = 30 \, \text{m}\)
  • Outlet diameter, \(D_2 = 25 \, \text{cm} = 0.25 \, \text{m}\)
  • Outlet width, \(b_2 = 5 \, \text{cm} = 0.05 \, \text{m}\)
  • Manometric efficiency, \(\eta_{\text{mano}} = 75\% = 0.75\)
  • Acceleration due to gravity, \(g = 9.81 \, \text{m/s}^2\)

Solution

1. Calculate Tangential Velocity at Outlet (\(u_2\))

$$ u_2 = \frac{\pi D_2 N}{60} = \frac{\pi \times 0.25 \times 1400}{60} \approx 18.326 \, \text{m/s} $$

2. Calculate Velocity of Flow at Outlet (\(V_{f2}\))

$$ \text{Area at Outlet, } A_2 = \pi D_2 b_2 = \pi \times 0.25 \times 0.05 \approx 0.03927 \, \text{m}^2 $$ $$ V_{f2} = \frac{Q}{A_2} = \frac{0.12}{0.03927} \approx 3.056 \, \text{m/s} $$

3. Calculate the Theoretical (Euler) Head (\(H_e\))

$$ \eta_{\text{mano}} = \frac{H_m}{H_e} \implies H_e = \frac{H_m}{\eta_{\text{mano}}} $$ $$ H_e = \frac{30 \, \text{m}}{0.75} = 40 \, \text{m} $$

4. Determine Whirl Velocity at Outlet (\(V_{w2}\))

Assuming radial entry (\(V_{w1}=0\)), the Euler head equation simplifies, allowing us to solve for \(V_{w2}\).

$$ H_e = \frac{V_{w2} u_2}{g} \implies V_{w2} = \frac{H_e \cdot g}{u_2} $$ $$ V_{w2} = \frac{40 \times 9.81}{18.326} \approx 21.412 \, \text{m/s} $$

5. Determine the Vane Angle at Outlet (\(\phi\))

The outlet vane angle is found from the geometry of the outlet velocity triangle.

$$ \tan(\phi) = \frac{V_{f2}}{u_2 - V_{w2}} $$ $$ \tan(\phi) = \frac{3.056}{18.326 - 21.412} = \frac{3.056}{-3.086} \approx -0.9903 $$ $$ \phi = \arctan(-0.9903) \approx -44.72^\circ $$
Final Result:

The vane angle at outlet is approximately \(44.7^\circ\).

(The negative sign indicates forward-curved vanes, where the angle relative to the tangent is obtuse, i.e., \(180^\circ - 44.7^\circ = 135.3^\circ\))

Explanation of the Result

The calculation for the outlet vane angle (\(\phi\)) yields a negative result. This is mathematically correct based on the given data but indicates an unusual pump design.

  • Velocity Triangle: The formula \( \tan(\phi) = V_{f2} / (u_2 - V_{w2}) \) is derived from the outlet velocity triangle. In our case, the whirl velocity (\(V_{w2} \approx 21.4\) m/s) is greater than the impeller's tangential velocity (\(u_2 \approx 18.3\) m/s).
  • Forward-Curved Vanes: This condition (\(V_{w2} > u_2\)) is characteristic of a pump with forward-curved vanes. Most centrifugal pumps use backward-curved vanes (\(\phi < 90^\circ\)) for more stable operation. Forward-curved vanes (\(\phi > 90^\circ\)) can produce a very high head but are often less stable and less efficient.

The calculated angle of -44.7° corresponds to an absolute angle of \(180^\circ - 44.7^\circ = 135.3^\circ\), which is consistent with a forward-curved design. While atypical, it is the correct answer derived from the problem's parameters.

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