Fox and McDonald's Introduction to Fluid Mechanics
Fox and McDonald's Introduction to Fluid Mechanics
9th Edition
ISBN: 9781118912652
Author: Philip J. Pritchard, John W. Mitchell
Publisher: WILEY
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Chapter 10, Problem 1P

The geometry of a centrifugal water pump is r1 = 10 cm, r2 = 20 cm, b1 = b2 = 4 cm, β1 = 30°, β2 = 15°, and it runs at speed 1600 rpm. Estimate the discharge required for axial entry, the power generated in the water in watts, and the head produced.

Expert Solution & Answer
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To determine

The discharge required for axial entry, the power generated in the water in watts, and the head produced.

Explanation of Solution

Given:

Inlet impeller radius is r1=10cm.

Outlet impeller radius is r2=20cm.

Inlet blade angle is β1=30°.

Outlet blade angle is β2=15°.

Speed of the centrifugal water pump is N=1600rpm.

Inlet tangential velocity is Vt1=0

Impeller inlet width b1=4cm.

Impeller outlet width b2=4cm.

Calculation:

Calculate the angular speed of centrifugal water pump (ω).

  ω=2πN60=1600revmin×2πradrev×rev60s=167.46rad/s

Calculate the runner speed at inlet (U1).

  U1=ωr1=167.46radsec×10cm(1m100cm)=16.746m/sec

Calculate the runner speed at outlet (U2).

  U2=ωr2=167.46×20cm(1m100cm)=33.492m/s

Draw inlet and outlet velocity diagram as shown in Figure (1).

Fox and McDonald's Introduction to Fluid Mechanics, Chapter 10, Problem 1P

From Figure (1).

  cosβ=UVtVrbsinβ=VnVrbtanβ=VnUVt

At inlet, Vt1=0.

Calculate Vn1.

  tanβ=VnUVtVn1=U1tanβ1Vn1=16.74m/s×tan30°Vn1=9.66m/s

Using continuity equation calculate the discharge required at the entry (Q).

  Q=2πr1b1Vn=2×π×[10cm(1m100cm)]×[4cm(1m100cm)]×(9.66m/s)=0.2427m3/s

Thus, the discharge required at the entry (Q) is 0.2427m3/s_.

At Outlet,

Calculate tangential velocity at outlet (Vt2).

  Vt2=U2Q2πr2b2cot(β2)=33.492m/s0.2427m3/s2×3.14×[20cm×(1m100cm)]×4cm(1m100cm)cot(15°)=15.466m/s

Calculate the theoretical power generated in the water (W˙m).

  W˙m=(U2Vt2U1Vt1)m˙=(U2Vt2)ρQ=[(33.492m/s×15.466m/s)×1000kg/m3×0.2427m3/s]=125715.51W

Thus, the power generated in the water (W˙m) is 125715.51W_.

Calculate the theoretical head (H).

  H=1g(U2Vt2U1Vt1)H=1g(U2Vt2)=19.81m/s2×(33.492m/s×15.466m/s)=52.08m

Thus, the theoretical head (H) is 52.08m_.

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Chapter 10 Solutions

Fox and McDonald's Introduction to Fluid Mechanics

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