A centrifugal pump delivers 0.3 m³/s against a head of 30 m at 1400 rpm. The external diameter of impeller is 0.5 m and the outlet width is 0.05 m. If the manometric efficiency is 80%, find the vane angle at outlet.

Step-by-Step

Learn more on how do we answer questions.

*Given data:*

Discharge Q=0.3 \mathrm{~m}^3 / \mathrm{s}

Speed of pump *N =1400 rpm*

Head H_m=30 \mathrm{~m}

External diameter of impeller D_2=0.5 \mathrm{~m}

Width at outlet B_2=0.05 \mathrm{~m}

Manometric efficiency \eta_{\text {mano }}=0.80

Tangential velocity of impeller at outlet is given by

U_2=\frac{\pi D_2 N}{60}=\frac{\pi \times 0.5 \times 1400}{60}=36.65 \mathrm{~m} / \mathrm{s}

The velocity of flow at the impeller outlet is

V_{f 2}=\frac{Q}{\pi D_2 B_2}

=\frac{0.3}{\pi \times 0.5 \times 0.05}=3.82 \mathrm{~m} / \mathrm{s}

Manometric efficiency is found from Eq. (20.10) as

\eta_{m a n o}=\frac{H_{m}}{\frac{V_{w2}U_{2}}{g} }=\frac{g H_{m}}{V_{w2}U_{2}} (20.10)

\eta_{\text {mano }}=\frac{g H_m}{V_{w 2} U_2}

or 0.80=\frac{9.81 \times 30}{V_{w 2} \times 36.65}

or V_{w 2}=10.04 \mathrm{~m} / \mathrm{s}

The outlet velocity triangle is shown in Fig. 20.8.

From the outlet velocity triangle, we have

\tan \beta_2=\frac{V_{f 2}}{U_2-V_{w 2}}

or \beta_2=\tan ^{-1} \frac{3.82}{36.65-10.04}=8.17^{\circ}

Question: 20.22

Given data:
Power absorbed by model [la...

Question: 20.21

Given data:
For pump 1,
Diameter of impeller [lat...

Question: 20.20

Given data:
Speed of pump ...

Question: 20.19

Given data:
Speed of pump ...

Question: 20.18

Given data:
Speed of pump N = 1000 ...

Question: 20.14

Given data:
Speed of pump ...

Question: 20.13

Given data:
External diameter of impeller ...

Question: 20.1

Given data:
Speed of pump ...

Question: 20.3

Given data:
Internal diameter of impeller [late...

Question: 20.4

Given data:
Speed of pump ...