A centrifugal pump is running at 1000 r.p.m. The outlet vane angle of the impeller is 30° and velocity of flow at outlet is 3 m/s. The pump is working against a total head of 30 m and the discharge through the pump is 0.3 m³/s. If the manometric efficiency of the pump is 75%, determine: (i) the diameter of the impeller, and (ii) the width of the impeller at outlet.

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Centrifugal Pump Design Calculation

Problem Statement

A centrifugal pump is running at 1000 r.p.m. The outlet vane angle of the impeller is 30° and velocity of flow at outlet is 3 m/s. The pump is working against a total head of 30 m and the discharge through the pump is 0.3 m³/s. If the manometric efficiency of the pump is 75%, determine: (i) the diameter of the impeller, and (ii) the width of the impeller at outlet.

Given Data & Constants

  • Speed, \(N = 1000 \, \text{r.p.m.}\)
  • Outlet vane angle, \(\phi = 30^\circ\)
  • Velocity of flow at outlet, \(V_{f2} = 3 \, \text{m/s}\)
  • Manometric Head, \(H_m = 30 \, \text{m}\)
  • Discharge, \(Q = 0.3 \, \text{m}^3/\text{s}\)
  • Manometric efficiency, \(\eta_{\text{mano}} = 75\% = 0.75\)
  • Acceleration due to gravity, \(g = 9.81 \, \text{m/s}^2\)

Solution

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

The manometric efficiency relates the actual head delivered to the theoretical head generated by the impeller.

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

2. Determine Tangential and Whirl Velocities (\(u_2, V_{w2}\))

We have two relationships for the outlet velocities, which we can solve simultaneously. The first is from the Euler Head equation (assuming radial inlet, \(V_{w1}=0\)).

$$ H_e = \frac{V_{w2} u_2}{g} \implies V_{w2} u_2 = H_e \times g $$ $$ V_{w2} u_2 = 40 \times 9.81 = 392.4 \quad \cdots(1) $$

The second relationship comes from the outlet velocity triangle.

$$ \tan(\phi) = \frac{V_{f2}}{u_2 - V_{w2}} \implies u_2 - V_{w2} = \frac{V_{f2}}{\tan(\phi)} $$ $$ u_2 - V_{w2} = \frac{3}{\tan(30^\circ)} \approx 5.196 \implies V_{w2} = u_2 - 5.196 \quad \cdots(2) $$

Substitute (2) into (1):

$$ (u_2 - 5.196) u_2 = 392.4 $$ $$ u_2^2 - 5.196 u_2 - 392.4 = 0 $$ $$ \text{Solving this quadratic equation gives } u_2 \approx 22.58 \, \text{m/s} $$

(i) Determine the Diameter of the Impeller (\(D_2\))

The tangential velocity \(u_2\) is directly related to the impeller diameter and rotational speed.

$$ u_2 = \frac{\pi D_2 N}{60} $$ $$ D_2 = \frac{u_2 \times 60}{\pi N} = \frac{22.58 \times 60}{\pi \times 1000} $$ $$ D_2 \approx 0.431 \, \text{m} $$

(ii) Determine the Width of the Impeller at Outlet (\(b_2\))

The discharge (flow rate) is a function of the outlet area and the velocity of flow.

$$ Q = \text{Area at Outlet} \times V_{f2} = (\pi D_2 b_2) \times V_{f2} $$ $$ b_2 = \frac{Q}{\pi D_2 V_{f2}} $$ $$ b_2 = \frac{0.3}{\pi \times 0.431 \times 3} $$ $$ b_2 \approx 0.0738 \, \text{m} $$
Final Results:

(i) Diameter of the impeller: \( D_2 \approx 431 \, \text{mm} \)

(ii) Width of the impeller at outlet: \( b_2 \approx 73.8 \, \text{mm} \)

Explanation of the Method

This problem is a "design" or "reverse" problem. Instead of being given the geometry to find the performance, we are given the required performance (Head, Discharge, Efficiency) and must determine the necessary geometry (Diameter, Width).

  1. Euler Head: We first determine the theoretical head the impeller must generate by using the given manometric efficiency and the required manometric head.
  2. Simultaneous Equations: The Euler head and the outlet velocity triangle geometry both provide relationships between the unknown tangential velocity (\(u_2\)) and whirl velocity (\(V_{w2}\)). Solving these two equations together is the key to unlocking the problem.
  3. Geometry Calculation: Once the required tangential velocity (\(u_2\)) is known, we can directly calculate the impeller diameter. With the diameter known, we can then use the given discharge rate to find the required width of the impeller outlet.

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