The diameter of an impeller of a centrifugal pump at inlet and outlet are 300 mm and 600 mm respectively. The velocity of flow at outlet is 2.5 m/s and vanes are set back at an angle of 45° at outlet. Determine the minimum starting speed of the pump if the manometric efficiency is 75%.

Solution

1. Principle of Minimum Starting Speed

To start pumping, the centrifugal head generated by the stationary water inside the impeller (a forced vortex) must equal the manometric head (\(H_m\)) the pump is designed to work against under its normal operating conditions.

$$ H_m = \text{Centrifugal Head} = \frac{u_2^2 - u_1^2}{2g} $$

2. Express Manometric Head (\(H_m\)) in Terms of Operating Conditions

The manometric head is related to the Euler head (\(H_e\)) by the manometric efficiency. We assume radial inlet (\(V_{w1} = 0\)).

$$ H_m = \eta_{\text{mano}} \times H_e = \eta_{\text{mano}} \times \frac{V_{w2} u_2}{g} $$

From the outlet velocity triangle, we can express \(V_{w2}\) in terms of \(u_2\):

$$ \tan(\phi) = \frac{V_{f2}}{u_2 - V_{w2}} \implies V_{w2} = u_2 - \frac{V_{f2}}{\tan(\phi)} $$ $$ \text{Substituting this into the } H_m \text{ equation:} $$ $$ H_m = \frac{\eta_{\text{mano}}}{g} \left( u_2 - \frac{V_{f2}}{\tan(\phi)} \right) u_2 $$

3. Equate the Head Expressions and Solve for \(u_2\)

Now we set the two expressions for \(H_m\) equal to each other.

$$ \frac{u_2^2 - u_1^2}{2g} = \frac{\eta_{\text{mano}}}{g} \left( u_2^2 - \frac{u_2 V_{f2}}{\tan(\phi)} \right) $$

We know \(u_1 = u_2 \frac{D_1}{D_2} = u_2 \frac{0.3}{0.6} = 0.5 u_2\), so \(u_1^2 = 0.25 u_2^2\). Substitute this in and cancel \(g\) from both sides:

$$ \frac{u_2^2 - 0.25 u_2^2}{2} = \eta_{\text{mano}} \left( u_2^2 - \frac{u_2 V_{f2}}{\tan(\phi)} \right) $$ $$ \frac{0.75 u_2^2}{2} = 0.75 \left( u_2^2 - \frac{u_2 \times 2.5}{\tan(45^\circ)} \right) $$ $$ \text{Divide both sides by 0.75 and note that } \tan(45^\circ) = 1: $$ $$ \frac{u_2^2}{2} = u_2^2 - 2.5 u_2 $$ $$ 2.5 u_2 = u_2^2 - 0.5 u_2^2 = 0.5 u_2^2 $$ $$ \text{Since } u_2 \neq 0, \text{ we can divide by } u_2: $$ $$ 2.5 = 0.5 u_2 \implies u_2 = 5 \, \text{m/s} $$

4. Calculate the Minimum Starting Speed (N)

Now that we have the required tangential velocity at the outlet, we can find the rotational speed.

$$ u_2 = \frac{\pi D_2 N}{60} $$ $$ N = \frac{u_2 \times 60}{\pi D_2} = \frac{5 \times 60}{\pi \times 0.6} = \frac{300}{0.6\pi} $$ $$ N \approx 159.15 \, \text{r.p.m.} $$

Explanation of the Method

This problem is unique because it connects the pump's steady-state operating point to its start-up condition. The key insight is that the head the pump must overcome to start is the same as the manometric head it produces during normal operation.

1. We derive one expression for the manometric head based on the pump's running conditions (flow rate, vane angles, efficiency). This expression is a function of the unknown tangential speed, \(u_2\).
2. We derive a second expression for the head based on the start-up condition (a forced vortex), which is also a function of \(u_2\).
3. By equating these two expressions, we create a single equation with \(u_2\) as the only unknown. Solving this gives us the required tangential velocity.
4. Finally, the rotational speed (N) is directly calculated from this tangential velocity and the impeller diameter.