A centrifugal pump having outer diameter equal to two times the inner diameter and running at 1200 r.p.m. works against a total head of 75 m. The velocity of flow through the impeller is constant and equal to 3 m/s. The vanes are set back at an angle of 30° at outlet. If the outer diameter of the impeller is 600 mm and width at outlet is 50 mm, determine : (a) vane angle at inlet, (b) work done per second by impeller, (c) manometric efficiency.

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Centrifugal Pump Performance Analysis

Problem Statement

A centrifugal pump having outer diameter equal to two times the inner diameter and running at 1200 r.p.m. works against a total head of 75 m. The velocity of flow through the impeller is constant and equal to 3 m/s. The vanes are set back at an angle of 30° at outlet. If the outer diameter of the impeller is 600 mm and width at outlet is 50 mm, determine : (a) vane angle at inlet, (b) work done per second by impeller, (c) manometric efficiency.

Given Data & Constants

  • Speed, \(N = 1200 \, \text{r.p.m.}\)
  • Manometric Head, \(H_m = 75 \, \text{m}\)
  • Velocity of flow, \(V_{f1} = V_{f2} = 3 \, \text{m/s}\)
  • Outlet vane angle, \(\phi = 30^\circ\)
  • Outer diameter, \(D_2 = 600 \, \text{mm} = 0.6 \, \text{m}\)
  • Inner diameter, \(D_1 = D_2 / 2 = 300 \, \text{mm} = 0.3 \, \text{m}\)
  • Width at outlet, \(b_2 = 50 \, \text{mm} = 0.05 \, \text{m}\)
  • Density of water, \(\rho = 1000 \, \text{kg/m}^3\)
  • Acceleration due to gravity, \(g = 9.81 \, \text{m/s}^2\)

Solution

1. Calculate Tangential Velocities (\(u_1\) and \(u_2\))

$$ u_1 = \frac{\pi D_1 N}{60} = \frac{\pi \times 0.3 \times 1200}{60} \approx 18.85 \, \text{m/s} $$ $$ u_2 = \frac{\pi D_2 N}{60} = \frac{\pi \times 0.6 \times 1200}{60} \approx 37.70 \, \text{m/s} $$

(a) Vane Angle at Inlet (\(\theta\))

Assuming the water enters the impeller radially (a standard assumption for such problems unless stated otherwise), the whirl velocity at the inlet is zero (\(V_{w1} = 0\)).

$$ \tan(\theta) = \frac{V_{f1}}{u_1} $$ $$ \tan(\theta) = \frac{3 \, \text{m/s}}{18.85 \, \text{m/s}} \approx 0.15915 $$ $$ \theta = \arctan(0.15915) \approx 9.04^\circ $$

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

We use the outlet velocity triangle to find the whirl component of the velocity at the outlet.

$$ \tan(\phi) = \frac{V_{f2}}{u_2 - V_{w2}} $$ $$ u_2 - V_{w2} = \frac{V_{f2}}{\tan(\phi)} $$ $$ V_{w2} = u_2 - \frac{V_{f2}}{\tan(\phi)} = 37.70 - \frac{3}{\tan(30^\circ)} $$ $$ V_{w2} \approx 37.70 - 5.196 \approx 32.504 \, \text{m/s} $$

(b) Work Done per Second by Impeller (Power)

First, we find the volume flow rate (discharge, Q). Then we can calculate the power imparted to the water.

$$ Q = \text{Area at Outlet} \times V_{f2} = (\pi D_2 b_2) \times V_{f2} $$ $$ Q = (\pi \times 0.6 \times 0.05) \times 3 \approx 0.2827 \, \text{m}^3/\text{s} $$ $$ \text{Power} = \rho \times Q \times V_{w2} \times u_2 $$ $$ \text{Power} = 1000 \, \text{kg/m}^3 \times 0.2827 \, \text{m}^3/\text{s} \times 32.504 \, \text{m/s} \times 37.70 \, \text{m/s} $$ $$ \text{Power} \approx 346061 \, \text{W} \approx 346.06 \, \text{kW} $$

(c) Manometric Efficiency (\(\eta_{\text{mano}}\))

First, we calculate the theoretical head from the impeller (Euler Head). Then we compare it to the actual head achieved.

$$ \text{Euler Head, } H_e = \frac{V_{w2} u_2}{g} \quad (\text{since } V_{w1}=0) $$ $$ H_e = \frac{32.504 \times 37.70}{9.81} \approx 124.86 \, \text{m} $$ $$ \eta_{\text{mano}} = \frac{\text{Manometric Head}}{\text{Euler Head}} = \frac{H_m}{H_e} $$ $$ \eta_{\text{mano}} = \frac{75 \, \text{m}}{124.86 \, \text{m}} \approx 0.6006 $$
Final Results:

(a) Vane Angle at Inlet: \( \theta \approx 9.04^\circ \)

(b) Work Done per Second (Power): \( \approx 346.1 \, \text{kW} \)

(c) Manometric Efficiency: \( \eta_{\text{mano}} \approx 60.1\% \)

Explanation of Key Concepts

(a) Vane Angle at Inlet (\(\theta\)): This is the angle of the impeller blade at the point where water enters. It must be designed to accept the incoming flow smoothly to minimize shock losses. Assuming radial entry (\(V_{w1}=0\)) is a common design goal that simplifies this calculation.

(b) Work Done per Second (Power): This is the rate at which the impeller transfers energy to the water. It's calculated from the mass flow rate and the change in the water's angular momentum, represented by the product of whirl and tangential velocities (\(V_{w2}u_2\)).

(c) Manometric Efficiency (\(\eta_{\text{mano}}\)): This is a crucial performance metric. It compares the actual pressure head the pump delivers (\(H_m = 75\) m) to the theoretical head the impeller should produce based on its geometry and speed (Euler Head, \(H_e = 124.86\) m). The difference between these two values represents energy lost to internal hydraulic effects like friction and turbulence.

Related Posts

Find the rise in pressure in the impeller of a centrifugal pump through which water is flowing at the rate of 15 litres/s. The internal and external diameters of the impeller are 20 cm and 40 cm respectively. The widths of impeller at inlet and outlet are 1.6 cm and 0.8 cm. The pump is running at 1200 r.p.m. The water enters the impeller radially at inlet and impeller vane angle at outlet is 30°. Neglect losses through the impeller.

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