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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Question
Chapter 8, Problem 133P
To determine

(a)

The flow rate of oil through pipe when the pipe is placed horizontally.

Expert Solution
Check Mark

Answer to Problem 133P

The flow rate of oil through pipe when the pipe is placed horizontally is 7.46×103m3/s.

Explanation of Solution

Given Information:

The diameter of the pipe is 6cm, the length of the pipe is

  33m, the inlet pressure of the oil is 745kPa and the pressure at outlet is 97kPa.

Write the expression to calculate the area of the pipe.

  A=πD24...... (I)

Here, the diameter of the pipe is D and the area of the pipe is A.

Write the expression to calculate change in pressure.

  ΔP=P1P2...... (II)

Here, the pressure at inlet is P1 and the pressure at outlet is P2.

Write the expression to calculate the volume flow rate in an inclined pipe.

  ˙=[ΔPρgLsinθ]πD4128μL...... (III)

Here, the volume flow rate is ˙, the viscosity of the oil is μ, the density of the oil is ρ, the angle of inclination or declination is θ and the length of the pipe is L.

Calculation:

Substitute 6cm for D in Equation (I).

  A=π ( 6cm )24=π [ ( 6cm )( 10 2 m cm )]24=113.09× 10 4m24=2.827×103m2

Substitute 745kPa for P1 and 97kPa for P2 in Equation (II).

  ΔP=745kPa97kPa=648kPa=(648kPa)( 10 3 Pa kPa)=648×103Pa

Refer to the Table A-7, "Properties of liquids" to obtain the value of ρ for oil at 20°C as 888.1kg/m3 and μ for oil at 20°C as 0.8374kg/ms.

Substitute 648×103Pa for ΔP, 888.1kg/m3 for ρ, 0.8374kg/ms for μ, 0° for θ, 6cm for D, 9.81m/s2 for g and 33m for L in Equation (III).

  ˙=[ [ 648× 10 3 Pa( 888.1 kg/ m 3 )( 9.81m/ s 2 )( 33m )sin0°] ×π ( 6× 10 2 cm ) 4 ]128( 0.8374 kg/ ms )( 33m)=[648× 10 3Pa( kg/ m s 2 Pa )0]π [ ( 6cm )( 10 2 m cm )]4128( 27.6342 kg/s )=0.00746m3/s=7.46×103m3/s

Conclusion:

The flow rate of oil through pipe when the pipe is placed horizontally is 7.46×103m3/s.

To determine

(b)

The flow rate of oil through pipe when the pipe is inclined 15° upward.

Expert Solution
Check Mark

Answer to Problem 133P

The flow rate of oil through pipe when the pipe is inclined 15° upward is 6.59×103m3/s.

Explanation of Solution

Calculation:

Substitute 648×103Pa for ΔP, 888.1kg/m3 for ρ, 0.8374kg/ms for μ, 15° for θ, 6cm for D, 9.81m/s2 for g and 33m for L in Equation (III).

  ˙=[ [ 648× 10 3 Pa( 888.1 kg/ m 3 )( 9.81m/ s 2 )( 33m )sin15°] ×π ( 6× 10 2 cm ) 4 ]128( 0.8374 kg/ ms )( 33m)=[648× 10 3Pa( kg/ m s 2 Pa )74411.6 kg/ m s 2 ]π [ ( 6cm )( 10 2 m cm )]4128( 27.6342 kg/s )=0.00659m3/s=6.59×103m3/s

Conclusion:

The flow rate of oil through pipe when the pipe is inclined 15° upward is 6.59×103m3/s.

To determine

(c)

The flow rate of oil through pipe when the pipe is inclined 15° downward and the verification of flow to be laminar.

The value of Reynolds number is 159.71 and the flow of the oil is laminar.

Expert Solution
Check Mark

Answer to Problem 133P

The flow rate of oil through pipe when the pipe is inclined 15° downward is 7.22×103m3/s.

Explanation of Solution

Write the expression to calculate the velocity of flow.

  V=˙A...... (IV)

Here, the velocity of the flow is V.

Write the expression to calculate Reynolds's number.

  Re=ρVDμ...... (V)

Here, the Reynolds's number is Re.

Calculation:

Substitute 648×103Pa for ΔP, 888.1kg/m3 for ρ, 0.8374kg/ms for μ, 15° for θ, 6cm for D, 9.81m/s2 for g and 33m for L in Equation (III).

  ˙=[ [ 648× 10 3 Pa( 888.1 kg/ m 3 )( 9.81m/ s 2 )( 33m )sin( 15° )] ×π ( 6cm ) 4 ]128( 0.8374 kg/ ms )( 33m)=[648× 10 3Pa( kg/ m s 2 Pa )( 74411.6 kg/ m s 2 )]π [ ( 6cm )( 10 2 m cm )]4128( 27.6342 kg/s )=0.00722m3/s=7.22×103m3/s

Substitute 7.22×103m3/s for ˙ and 2.827×103cm2 for A in Equation (IV).

  V=7.22× 10 3 m 3/s2.827× 10 3m2=2.51m/s

Substitute 888.1kg/m3 for ρ, 0.8374kg/ms for μ, 2.51m/s for V, 6cm for D and 33m for L in Equation (V).

  Re=888.1kg/ m 3×2.51m/s×6cm0.8374kg/ms=2.229× 103kg/ m 2s×( 6cm)( 10 2 m cm )0.8374kg/ms=159.71

Conclusion:

The flow rate of oil through pipe when the pipe is inclined 15° downward is

  7.22×103m3/s.

The value of Reynolds number is 159.71, therefore the flow of the oil is laminar.

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Chapter 8 Solutions

Fluid Mechanics: Fundamentals and Applications

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