Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 14, Problem 67P
To determine

The net head produced by the centrifugal pump in cm.

The required brake horse power in W.

Explanation of the angle at which fluid impinges on the impeller blade a critical parameter in the design of centrifugal pump.

Expert Solution & Answer
Check Mark

Answer to Problem 67P

The net head produced by the centrifugal pump in cm is 0.2932cm.

The required brake horse power in W is 13676.9063W.

Explanation of Solution

Given information:

The number of rotations of centrifugal pump is 750rpm. The inlet radius is 12cm and the blade width is 18cm. The outlet radius is 24cm and the outlet blade width is 16.2cm. The volume flow rate is 0.573m3/s. Assuming 100% efficiency. The inlet angle of blade is 10° and the outlet angle of blade is 35°.

Expression for the angular velocity for the pump

  ω=2πn˙60  .......(I)

Here, the angular velocity of the centrifugal pump is ω and the number of rotation is n˙.

Expression for the normal velocity component at the outlet of the pump

  V2=V˙2πr2b2.......(II)

Here, the volume flow rate is V˙, the velocity at the outlet of the pump is V2, the outlet radius is r2 and the blade width at the outlet is b2.

Expression for the tangential velocity component at the outlet of the pump

  V=V2tanα2.......  (III)

Here, the angle at the outlet of the pump is α2 and the tangential velocity component at the outlet of the pump is V.

Expression for the normal velocity component at the inlet of the pump

  V1=V˙2πr1b1.......  (IV)

Here, the volume flow rate is V˙, the velocity at the inlet of the pump is V1, the inlet radius is r1 and the blade width at the inlet is b1.

Expression for the tangential velocity component at the inlet of the pump

  V=V1tanα1.......(V)

Here, the angle at the inlet of the pump is α1 and the tangential velocity component at the inlet of the pump is V.

Expression for the equivalent head

  Hwater=H(ρ airρ water).......  (VI)

Here, the equivalent head is Hwater, the density of air is ρair, the density of water is ρwater and the acceleration due to gravity is g.

Expression for the horse power

  P=ρwatergHwaterV˙.......(VII)

Here, the brake horse power is P.

Expression for the net head

  H=(ωg)(r2Vr1V)   ....... (VIII)

Here, the net head is H.

Calculation:

Substitute 750rpm for n˙ in Equation (I).

  ω=2π( 750rpm)60=4712.38898rpm( 1rad/s 1rpm )60=78.5398rad/s

Substitute 0.573m3/s for V˙, 24cm for r2 and 16.2cm for b2 in Equation (II).

  V2=0.573 m 3/s2π( 24cm)( 16.2cm)=0.573 m 3/s2π( 24cm( 1m 100cm ))( 16.2cm( 1m 100cm ))=2.34557m/s

Substitute 35° for α2 and 2.34557m/s for V2 in Equation (III).

  V=(2.34557m/s)tan(35°)=(2.34557m/s)(0.70002)=1.64238m/s

Substitute 0.573m3/s for V˙, 12cm for r1 and 18cm for b1 in Equation (IV).

  V1=0.573 m 3/s2π( 12cm)( 18cm)=0.573 m 3/s2π( 12cm( 1m 100cm ))( 18cm( 1m 100cm ))=4.222m/s

Substitute 10° for α1 and 4.222m/s for V1 in Equation (V).

  V=(4.222m/s)tan(10°)=(4.222m/s)(0.176326)=0.7444m/s

Substitute 24cm for r2, 12cm for r1, 78.5398rad/s for ω, 998kg/m3 for ρwater, 1.2kg/m3 for ρair, 0.7444m/s for V, 1.64238m/s for V and 9.81m/s2 for g in Equation (VI).

  Hwater=[( 78.5398 rad/s 9.81m/ s 2 )( ( 24cm )( 1.64238m/s ) ( 12cm )( 0.7444m/s ) )( 1.2 kg/ m 3 998 kg/ m 3 )]=[(8s/m)( ( 24cm( 1m 100cm )×( 1.64238m/s ) ) ( 12cm( 1m 100cm )×( 0.7444m/s ) ) )(0.0012024)]=2.932×103m( 100cm 1m)=0.2932cm

Substitute 24cm for r2, 12cm for r1, 78.5398rad/s for ω, 0.7444m/s for V, 1.64238m/s for V and 9.81m/s2 for g in Equation (VIII).

  H=[( 78.5398 rad/s 9.81m/ s 2 )( ( 24cm )( 1.64238m/s ) ( 12cm )( 0.7444m/s ) )]=[(8s/m)( ( 24cm×( 1.64238m/s ) ) ( 12cm×( 0.7444m/s ) ) )]=243.874cm( 1m 100cm)=2.43874m

Substitute 2.43874m for H, 9.81m/s2 for g, 998kg/m3 for ρwater, 0.573m3/s for V˙ in Equation (VII).

  P=(998kg/ m 3)(9.81m/ s 2)(2.43874m)(0.573 m 3/s)=(998kg/ m 3)(9.81m/ s 2)(2.43874m)(0.573 m 3/s)=13681.05kgm2/s3( 1W 1 kg m 2 / s 3 )=13681.05W

Substitute 0° for α1 and 4.222m/s for V1 in Equation (V).

  V=(4.222m/s)tan(0°)=(4.222m/s)(0)=0

Substitute 24cm for r2, 12cm for r1, 78.5398rad/s for ω, 998kg/m3 for ρwater, 1.2kg/m3 for ρair, 0 for V, 1.64238m/s for V and 9.81m/s2 for g in Equation (VI).

  Hwater=[( 78.5398 rad/s 9.81m/ s 2 )( ( 24cm )( 1.64238m/s ) ( 12cm )( 0 ) )( 1.2 kg/ m 3 998 kg/ m 3 )]=(8s/m)(24cm( 1m 100cm ))(1.64238m/s)(0.0012024)=3.75×103m( 100cm 1m)=0.375cm

Substitute 24cm for r2, 12cm for r1, 78.5398rad/s for ω, 0 for V, 1.64238m/s for V and 9.81m/s2 for g in Equation (VIII).

  H=[( 78.5398 rad/s 9.81m/ s 2 )( ( 24cm )( 1.64238m/s ) ( 12cm )( 0 ) )]=(8s/m)(24cm)(1.64238m/s)=315.49cm( 1m 100cm)=3.1549m

Substitute 3.1549m for H, 9.81m/s2 for g, 998kg/m3 for ρwater AND 0.573m3/s for V˙ in Equation (VII).

  P=(998kg/ m 3)(9.81m/ s 2)(3.1549m)(0.573 m 3/s)=17698.634kgm2/s3( 1W 1 kg m 2 / s 3 )=17698.634W

If the angle is 0°, then the horse power is 17698.634W and the net positive head is 0.375cm but when the angle is 10°, then the horse power is 13681.05W and the net positive head is 0.2932cm. There is a swirl at the inlet of the pump, the net head produced by the pump and required brake horsepower is decreased. The angle at which the fluid impinges has an important role in the design of a centrifugal pump.

Conclusion:

The net head produced by the centrifugal pump in cm is 0.2932cm.

The required brake horse power in W is 13681.05.

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