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 2, Problem 123P

A 10-cm diameter cylindrical shaft rotates inside a 50-cm-long 10 .3-cm diameter bearing. The space between the shaft and the bearing is completely filled with oil whose viscosity at anticipated operating temperature is 0 .300 N s/m 2 . Determine the power required to overcame friction when the shaft rotates at a speed of (a) 600 rpm and (b) 1200 rpm .

Expert Solution
Check Mark
To determine

(a)

The power required to overcome the friction when the shaft rotates at a speed of 600rpm.

Answer to Problem 123P

The power required to overcome the friction when the shaft rotates at a speed of 600rpm is 247.55W.

Explanation of Solution

Given information:

The diameter of cylindrical shaft is 10cm, the length of shaft is 40cm, the diameter of bearing is 10.3cm, the viscosity of the oil filled between the shaft and the bearing is 0.3Ns/m2, and the speed of the shaft is 600rpm.

Write the expression for the gap between shaft and the bearing.

  l=DbDs2...... (I)

Here, tthe gap between the shaft and the bearing is l, the diameter of the bearing is Db, and the diameter of the shaft is Ds.

Write the expression for the torque generated by the shaft.

  T=μ4π2R3nLl...... (II)

Here, the torque is T, the radius of the shaft is R, the speed of the shaft is n, the length of the shaft is L, and the viscosity of the oil is μ.

Write the expression for the radius of the shaft.

  R=Ds2...... (III)

Write the expression for the angular velocity.

  ω=2πn60...... (IV)

Here, the angular velocity is ω.

Write the expression for the power.

  P=Tω...... (V)

Here, the power is P.

Calculation:

Substitute 10.3cm for Db and 10cm for Ds in Equation (I).

  l=10.3cm10cm2=0.15cm×( 1m 100cm)=1.5×103m

Substitute 10cm for Ds in Equation (III).

  R=10cm2=5cm×( 1m 100cm)=0.05m

Substitute 0.3Ns/m2 for μ, 0.05m for R, 600rpm for n, 40cm for L and 1.5×103m for l in Equation (II).

  T=(0.3Ns/ m 2)4π2× ( 0.05m )3×600rpm×40cm1.5× 10 3m=0.98696Ns×(600rpm× 1rps 60rpm)×(40cm× 1m 100cm)=3.94Nm

Substitute 600rpm for n in Equation (IV).

  ω=2π×600rpm60=3769.960rad/s=62.83rad/s

Substitute 62.83rad/s for ω and 3.94Nm for T in Equation (V).

  P=3.94Nm×62.83rad/s=247.55Nm/s×( 1W 1 Nm/s )=247.55W

Conclusion:

The power required to overcome the friction when the shaft rotates at a speed of 600rpm is 247.55W.

Expert Solution
Check Mark
To determine

(b)

The power required to overcome the friction when the shaft rotates at a speed of 1200rpm.

Answer to Problem 123P

The power required to overcome the friction when the shaft rotates at a speed of 1200rpm is 991.45W.

Explanation of Solution

Given information:

The speed of the shaft is 1200rpm.

Calculation:

Substitute 0.3Ns/m2 for μ, 0.05m for R, 1200rpm for n, 40cm for L and 1.5×103m for l in Equation (II).

  T=(0.3Ns/ m 2)4π2× ( 0.05m )3×1200rpm×40cm1.5× 10 3m=0.98696Ns×(1200rpm× 1rps 60rpm)×(40cm× 1m 100cm)=7.89Nm

Substitute 1200rpm for n in Equation (IV).

  ω=2π×1200rpm60=7539.860rad/s=125.66rad/s

Substitute 125.66rad/s for ω and 7.89Nm for T in Equation (V).

  P=7.89Nm×125.66rad/s=991.45Nm/s×( 1W 1 Nm/s )=991.45W

Conclusion:

The power required to overcome the friction when the shaft rotates at a speed of 1200rpm is 991.45W.

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Chapter 2 Solutions

Fluid Mechanics: Fundamentals and Applications

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