FLUID MECHANICS FUNDAMENTALS+APPS
FLUID MECHANICS FUNDAMENTALS+APPS
4th Edition
ISBN: 2810022150991
Author: CENGEL
Publisher: MCG
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Chapter 2, Problem 85P

A frustum-shaped body is rotating at a constant angular speed of 200 rad/s in a container filled with SAE 10W oil at 20°C ( μ = 0.100 Pa s ), as shown in Fig. P2-85. If the thickness of the oil film on all sides is 1.2 mm, determine the power required to maintain this motion. Also determine the reduction in the required power input when the oil temperature rises to 80°C ( μ = 0.0078 Pa s ).

Chapter 2, Problem 85P, A frustum-shaped body is rotating at a constant angular speed of 200 rad/s in a container filled

Expert Solution & Answer
Check Mark
To determine

The torque required to rotate the inner cylinder at constant speed.

And the torque required for motion at 80°C viscosity.

Answer to Problem 85P

The torque required to rotate the inner cylinder at constant speed is 270.21W.

And the torque required for motion at 80°C viscosity is 249.14W.

Explanation of Solution

Given information:

The dynamic viscosity of oil at 20°C is (0.1Pas) (0.1Nsm2), dynamic viscosity of oil at 80°C is (0.0078Pas) (0.0078Nsm2), angular speed is 200rad/s length of the container is 12cm (12×102m) diameter at the top of container is 12cm (12×102m) at the bottom of container is 4cm (4×102m), thickness of the oil film is 1.2mm (1.2×103m).

Write the expression for shear stress on the surface of frustum.

  τw=μVh   ...... (I)

The wall shear is τw, dynamic viscosity is μ, tangential velocity is V, thickness of the oil film is h

Substitute rω for V in the Equation (I).

  τw=μrωh

Here, angular velocity of the container is ω and radius of frustum at the bottom is r.

Write the expression for shear force.

  dF=τwdA   ...... (II)

Here, area of the frustum is dA and shear force is dF.

Substitute μrωh for τw in the Equation (II).

  dF=(μrωh)dA   ...... (III)

Write the expression for torque developed.

  dT=rdF   ...... (IV)

Here, torque developed is dT.

Substitute (μrωh)dA for dF in the Equation (IV).

  dT=r( μrωh)dA=( μωh)r2dA   ...... (V)

Write the expression for torque by integrating Equation (V).

  dT=( μω h ) r 2dAT=μwhA r 2dA   ...... (VI)

Write the expression for shaft power developed.

  W˙sh=ωT   ...... (VII)

Here, developed shaft power is W˙sh.

Substitute μwhAr2dA in Equation (VII).

  W˙sh=(μω2h)Ar2dA   ...... (VIII)

Write the expression for shaft power for the top surface.

Substitute 2πrdr for dA with limits extending from r to D/2 in Equation (VIII).

  ( W ˙ sh)top=( μ ω 2 h)r D/2 r 2( 2πrdr)=( 2πμ ω 2 h)r D/2 r 3dr=( 2πμ ω 2 h)[ r 4 4]rD/2=( 2πμ ω 2 h)[ ( D/2 ) 44 r 44]

  ( W ˙ sh)top=(2πμω2h)[( D 4)64r44]   ...... (IX)

Substitute 0 for r in Equation (IX).

  ( W ˙ sh)top=( 2πμ ω 2 h)[( D 4 )64 0 44]=( 2πμ ω 2 h)[( D 4 )64]( W ˙ sh)top=πμω2D432h   ...... (X)

Here, diameter of frustum at the top is D and shaft power for the top surface is ( W ˙ sh)top.

Substitute 2πrdr for dA with limits extending from r to d/2 in Equation (VIII).

  ( W ˙ sh)bottom=( μ ω 2 h)r d/2 r 2( 2πrdr)=( 2πμ ω 2 h)r d/2 r 3dr=( 2πμ ω 2 h)[ r 4 4]rd/2=( 2πμ ω 2 h)[ ( d/2 ) 44 r 44]

  ( W ˙ sh)bottom=(2πμω2h)[( d 4)64r44]   ...... (XI)

Here, diameter of frustum at the bottom is d and shaft power for the bottom surface is ( W ˙ sh)bottom.

Substitute 0 for r in Equation (XI).

  ( W ˙ sh)bottom=( 2πμ ω 2 h)[( d 4 )64 0 44]=( 2πμ ω 2 h)[( d 4 )64]( W ˙ sh)bottom=πμω2d432h   ...... (XII)

Substitute 2πrdz for dA with limits extending from d/2 to D/2 in Equation (VIII).

  ( W ˙ sh)shide=( μ ω 2 h) d/2 D/2 r 2( 2πrdz)=( 2πμ ω 2 h) d/2 D/2 r 3dz   ...... (XIII)

Here, shaft power for the side surface is ( W ˙ sh)shide.

Write the expression to obtain value of r in term of z.

  r=d2+Dd2Lz

Differentiate r with respect to z.

  drdz=ddz(d2+ Dd 2Lz)drdz=0+Dd2L×1dr=( Dd 2L)dzdz=( 2L Dd)dr   ...... (XIV)

Substitute (2LDd)dr for dz in Equation (XIII).

  ( W ˙ sh)shide=( 2πμ ω 2 h) d/2 D/2 r 3( 2L Dd )dr=( 2πμ ω 2 h)( 2L Dd) d/2 D/2 r 3dr=( 4Lπμ ω 2 h( Dd ))[ r 4 4]d/2D/2=( 4Lπμ ω 2 h( Dd ))[ ( D/2 ) 24 ( d/2 ) 24]

  =( 4Lπμ ω 2 h( Dd ))[ ( D/2 ) 24 ( d/2 ) 24]( W ˙ sh)shide=( 4Lπμ ω 2 h( Dd ))( D 4 d 4 16)   ...... (XV)

Write the expression for total shaft power.

  ( W ˙ sh)total=( W ˙ sh)top+( W ˙ sh)bottom+( W ˙ sh)shide   ...... (XVI)

Substitute (πμω2D432h) for ( W ˙ sh)top, (πμω2d432h) for ( W ˙ sh)bottom, and ( W ˙ sh)shide for (4Lπμω2h( Dd))(D4d416) in Equation (XVI).

  ( W ˙ sh)total=( πμ ω 2 D 4 32h)+( πμ ω 2 d 4 32h)+( 4Lπμ ω 2 h( Dd ))( D 4 d 4 16)=( πμ ω 2 D 4 32h)[1+( d D )4+2L( 1 ( d D ) 4 )( Dd)]   ...... (XVII)

Write the expression for power required to maintain the motion at 80°C.

  ( W ˙ sh)totalat80°C=μat80°Cμat20°C( W ˙ sh)total   ...... (XVIII)

Here, viscosity at 80°C is μat80°C and viscosity at 20°C is μat20°C, and total shaft power at 80°C is ( W ˙ sh)totalat80°C.

Write the expression for reduction in power input required when the oil pressure rise at 80°C.

  Power=( W ˙ sh)totalat20°C( W ˙ sh)totalat20°C   ...... (XIX)

Here, total shaft power at 20°C is ( W ˙ sh)totalat20°C.

Calculation:

Substitute (0.1Nsm2) for μ, (200rad/s) for ω, 12×102m for D4, (1.2×103m) for h, (3×102m) for d4, (12×102m) for L in Equation (XVII).

  ( W ˙ sh)total=( π( 0.1N s m 2 ) ( 200 rad/s ) 2 ( 12× 10 2 m ) 4 32( 1.2× 10 3 m ))×[1+ ( ( 3× 10 2 m ) ( 12× 10 2 m ) ) 4+ 2( 12× 10 2 m )( 1 ( ( 3× 10 2 m ) ( 12× 10 2 m ) ) 4 ) ( ( 12× 10 2 m )( 3× 10 2 m ) )]=67.858[1+0.012+2.970](Nms)=270.21W

Substitute 0.0078(Pas) for μat80°C, 0.1(Pas) for μat20°C, and 270.21W for ( W ˙ sh)total in Equation (XVIII).

  ( W ˙ sh)totalat80°C=0.0078( Pas)0.1( Pas)(270.21W)=0.078×270.21W=21.075W

Substitute 270.21W for ( W ˙ sh)totalat20°C and 21.075W for ( W ˙ sh)totalat20°C in Equation (XIX).

  Power=270.21W21.075W=249.14W

Conclusion:

The torque required to rotate the inner cylinder at constant speed is 270.21W.

And the torque required for motion at 80°C viscosity is 249.14W.

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Chapter 2 Solutions

FLUID MECHANICS FUNDAMENTALS+APPS

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