An idealized centrifugal pump (blade outer radius of 50 mm) consumes 2 kW power while running at 3000 rpm. The entry of the liquid into the pump is axial and exit from the pump is radial with respect to impeller. If the losses are neglected, then the mass flow rate of the liquid through the pump is ------------------ kg/s (round off to two decimal places).

The correct answer is 8.11.

Step-by-Step Explanation:

1. Identify the given parameters:
- Power consumed by the pump, $P=2\text{kW}=2000\text{W}$
- Rotational speed, $N=3000\text{rpm}$
- Impeller blade outer radius, $R_2=50\text{mm}=0.05\text{m}$
- Entry is axial, meaning the inlet whirl velocity is zero: $V_{w1}=0$.
- Exit is radial with respect to the impeller, meaning the outlet blade angle is 90 degrees. For a radial-tipped impeller, the outlet whirl velocity equals the peripheral blade speed at the outlet: $V_{w2}=u_2$.

2. Calculate the peripheral velocity of the impeller at the outlet ($u_2$):
The angular velocity $\omega$ of the impeller is given by:
$\omega =\frac{2\pi N}{60}$
Substituting the value of $N$:
$\omega =\frac{2\times \pi \times 3000}{60}=100\pi \text{rad/s}$
Now, the outer peripheral velocity $u_2$ is:
$u_2=\omega \times R_2=100\pi \times 0.05=5\pi \approx 15.708\text{m/s}$

3. Relate power to mass flow rate ($\dot{m}$):
For an idealized pump with no losses, the power input to the impeller is given by Euler's turbomachinery equation:
$P=\dot{m}\left(V_{w2}{u}_2-V_{w1}{u}_1\right)$
Since the inlet flow is axial ($V_{w1}=0$) and the exit is radial with respect to the impeller ($V_{w2}=u_2$), the power equation simplifies to:
$P=\dot{m}{u}_2^2$

4. Calculate the mass flow rate ($\dot{m}$):
Substitute the values of $P$ and $u_2$ into the equation:
$2000=\dot{m}\times {\left(15.708\right)}^2$
$2000=\dot{m}\times 246.74$
$\dot{m}=\frac{2000}{246.74}\approx 8.106\text{kg/s}$

Rounding off to two decimal places, the mass flow rate of the liquid through the pump is 8.11 kg/s.