Physics, asked by tillquote7095, 6 months ago

A centrifugal pump having outer diameter equal to two times the inner diameter and running at 1200 r.p.m. works against a total head of 75 m. The velocity of flow through the impeller is constant and equal to 3m/s. The vanes are set back at an angle of 30degrees at outlet. If the outer diameter of the impeller is 600 mm and width at outlet is 50 mm, determine (i) vane angle at inlet, (b) work done per second by impeller, (c) manometric efficiency.

Answers

Answered by vaishnaviv83
6

Answer:

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Answered by dreamrob
3

Given :

Speed of centrifugal pump N = 1200rpm

Total head H = 75m

Velocity of the flow through the impeller is constant = 3m/s

The vane are set back at an angle ∅ = 30°

The centrifugal pump outer diameter equal to two times of the inner diameter

D₂ = 2D₁

The outer diameter D₂ = 600mm = 0.6m

The inner diameter D₁ = D₂ / 2 = 0.6 / 2 =0.3m

Width at outlet B₂ = 50mm = 0.05m

To find :

a) vane angle at inlet

b) work done per second by impeller

c) manometric efficiency.

Solution :

a) u₁ = πD₁N / 60

u₁ = π × 0.3 × 1200 / 60

u₁ = 18.85m/s

u₂ = πD₂N / 60

u₂ = π × 0.6 × 1200 / 60

u₂ = 37.7m/s

Discharge Q = πD₂B₂Vf₂

Q = π × 0.6 × 0.05 × 3

Q = 0.283 m³/s

Vane angle at inlet

tanθ = Vf₁ / u₁

tanθ = 3 / 18.85

θ = 9°2' 34.27''

b) Work done by impeller on water per sec

Work  done = W×Vw₂×u₂/g

W = ρgQ × Vw₂×u₂/g

tan∅ = Vf₂ /(u₂ - Vw₂)

tan30° = 3 / (37.7 - Vw₂)

Vw₂ = 32.504m/s

W = 1000 × 9.8 × 0.283 × 32.504 × 37.7 /9.81

W = 346788.426Nm/s

c) η = gH / Vw₂u₂

η = 9.81 × 75 / 32.504 × 37.7

η = 0.6004 = 60.05%

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