Question

A turbine pump is designed to generate water head of 60 m. The impeller speed of the pump is 2000 rpm and the manometric efficiency is 70%. Neglecting the impeller blade curvature, the impeller diameter of the pump, in cm, is . (rounded off to two decimal places)

Hint:

Neglecting blade curvature means the exit whirl velocity equals the impeller tip speed, so the ideal head is u2 squared over g; link that to the manometric efficiency and the diameter formula.

Answer 1

Correct Answer: 27.69

Step 1: Apply the assumption of no blade curvature.
When blade curvature is neglected, the pump behaves as if the water leaves the impeller tip carrying the full peripheral (tip) speed as its whirl velocity, so \( V_{w2} = u_2 \). Euler's pump equation for the ideal (theoretical) head then simplifies to \[ H_{th} = \dfrac{u_2 V_{w2}}{g} = \dfrac{u_2^2}{g} \]

Step 2: Bring in the manometric efficiency.
Manometric efficiency compares the actual delivered head to this ideal head: \( \eta_{man} = \dfrac{H_m}{H_{th}} \), so the ideal head is \[ H_{th} = \dfrac{H_m}{\eta_{man}} = \dfrac{60}{0.70} = 85.71\ \text{m} \]

Step 3: Solve for the impeller tip speed.
From Step 1, \( u_2 = \sqrt{g H_{th}} = \sqrt{9.81 \times 85.71} \approx \sqrt{840.86} \approx 29.00\ \text{m/s} \)

Step 4: Convert tip speed to impeller diameter.
The tip speed of a rotating impeller relates to its diameter and rotational speed by \( u_2 = \dfrac{\pi D N}{60} \), where \( N \) is in rpm. Rearranging, \[ D = \dfrac{60\,u_2}{\pi N} = \dfrac{60 \times 29.00}{\pi \times 2000} \approx 0.2769\ \text{m} = 27.69\ \text{cm} \]

Final Answer:
The impeller diameter of the pump is about 27.69 cm. \[ \boxed{27.69\ \text{cm}} \]