Applied Fluid Mechanics (7th Edition)
Applied Fluid Mechanics (7th Edition)
7th Edition
ISBN: 9780132558921
Author: Robert L. Mott, Joseph A. Untener
Publisher: PEARSON
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Chapter 9, Problem 9.1PP

Compute points on the velocity profile from the pipe wall to the centerline of a 2 -in Schedule 40 steel pipe if the volume flow rate of castor oil at 77 ° F is 0.25   f t 3 / s . Use increments of 0.20 sssssssin and include the velocity at the centerline.

Expert Solution & Answer
Check Mark
To determine

To compute: the velocity profile from the pipe wall to the centerline of a 2 in schedule 40 steel pipe.

Answer to Problem 9.1PP

The velocity profile from the pipe wall to the centerline of a 2 in schedule 40 steel pipe v=10.72ft/s, NR=253 Laminar

Explanation of Solution

Given:

To consider the value 2-in schedule 40 steel pipe chart D=2.067in=0.1723ft, Kinematic viscosity of fluid ν=7.31×103, volume flow rate of castor oil Q=0.25ft3/s, Full radius ro=D/2=1.034 in;

Concept Used:

Annuity problem requires the use of the local velocity for Laminar flow equation as follows:

   U=2v[1( r r o )2]

Here,

Velocity =

   U, Average velocity= v, Radius= r, ro= full radius

Calculation:

As per the given problem

   Q=0.25ft3/s, D=2.067in=0.1723ft

Annuity problem requires the use of this formula

   A=πD24

Substitute these values in the formula

   A=π( 0.1723ft)24=0.02333ft2

   A=0.02333ft2

Annuity problem requires the use of this formula

   Q=0.25ft3/s, A=0.02333ft2

   v=Q/A

Substitute these values in the formula

   v=0.25ft3/s0.02333ft2=10.72ft/s

   v=10.72ft/s

Annuity problem requires the use of the local velocity for Laminar flow equation as follows:

   U=2v[1( r r o )2]

Average velocity v=10.72ft/s, ro=D/2=1.034in.

To find out velocity at different radius

   r=0.0

Substitute these values in the formula

   U=2(10.72)[1( 0.0in 1.0338in)2]=21.44ft/s

   U=21.44ft/s

Annuity problem requires the use of this formula

   r=0.20in, Average velocity v=10.72ft/s, ro=D/2=1.034in.

   U=2v[1( r r o )2]

Substitute these values in the formula

   U=2(10.72)[1( 0.20in 1.0338in)2]=20.64ft/s

   U=20.64ft/s

Annuity problem requires the use of this formula

   r=0.40in, Average velocity v=10.72ft/s, ro=D/2=1.034in.

   U=2v[1( r r o )2]

Substitute these values in the formula

   U=2(10.72)[1( 0.40in 1.0338in)2]=18.23ft/s

   U=18.23ft/s

Annuity problem requires the use of this formula

   r=0.60in, Average velocity v=10.72ft/s, ro=D/2=1.034in.

Substitute these values in the formula

   U=2(10.72)[1( 0.60in 1.0338in)2]=14.22ft/s

   U=14.22ft/s

Annuity problem requires the use of this formula

   r=0.80in, Average velocity v=10.72ft/s, ro=D/2=1.034in.

Substitute these values in the formula

   U=2(10.72)[1( 0.80in 1.0338in)2]=8.60ft/s

   U=8.60ft/s

Annuity problem requires the use of this formula

   r=1.00in, Average velocity v=10.72ft/s, ro=D/2=1.034in.

   U=2v[1( r r o )2]

Substitute these values in the formula

   U=2(10.72)[1( 1.00in 1.0338in)2]=1.38ft/s

   U=1.38ft/s

Annuity problem requires the use of this formula

   r=1.0338in, Average velocity v=10.72ft/s, ro=D/2=1.034in.

   U=2v[1( r r o )2]

Substitute these values in the formula

   U=2(10.72)[1( 1.0338in 1.0338in)2]=0.00ft/s

   U=0.0ft/s

Annuity problem requires the use of this formula

   v=10.72ft/s, D=2.067in=0.1723ft, ν=7.31×103

   NR=vDρ/η

   NR=vD/ν

   NR Reynolds Number, ν Kinematic viscosity of fluid

Substitute these values in the formula

   NR=vD/ν

   NR=vD/ν=(10.72)(0.1723)/7.31×103=253

   NR=253 Laminar

Conclusion:

The velocity profile from the pipe wall to the centerline of a 2 in schedule 40 steel pipe,, NR=253 Laminar

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Chapter 9 Solutions

Applied Fluid Mechanics (7th Edition)

Ch. 9 - Prob. 9.11PPCh. 9 - Prob. 9.12PPCh. 9 - Prob. 9.13PPCh. 9 - Prob. 9.14PPCh. 9 - Using Eq. (9-4), compute the ratio of the average...Ch. 9 - Prob. 9.16PPCh. 9 - Repeat Problem 9.16 for the same conditions,...Ch. 9 - Prob. 9.18PPCh. 9 - A shell-and-tube heat exchanger is made of two...Ch. 9 - Figure 9.14 shows a heat exchanger in which each...Ch. 9 - Figure 9.15 shows the cross section of a...Ch. 9 - Air with a specific weight of 12.5N/m3 and a...Ch. 9 - Carbon dioxide with a specific weight of...Ch. 9 - Water at 90F flows in the space between 6 in...Ch. 9 - Refer to the shell-and-tube heat exchanger shown...Ch. 9 - Refer to Fig. 9.14, which shows two DN 150...Ch. 9 - Refer to Fig. 9.15, which shows three pipes inside...Ch. 9 - Water at 10C is flowing in the shell shown in Fig....Ch. 9 - Figure 9.19 shows the cross section of a heat...Ch. 9 - Figure 9.20 shows a liquid-to-air heat exchanger...Ch. 9 - Glycerin ( sg=1.26 ) at 40C flows in the portion...Ch. 9 - Each of the square tubes shown in Fig. 9.21...Ch. 9 - A heat sink for an electronic circuit is made by...Ch. 9 - Figure 9.23 shows the cross section of a cooling...Ch. 9 - Prob. 9.35PPCh. 9 - The blade of a gas turbine engine contains...Ch. 9 - For the system described in Problem 9.24. compute...Ch. 9 - For the shell-and-tube heat exchanger described in...Ch. 9 - For the system described in Problem 9.26 compute...Ch. 9 - For the system described in Problem 9.27 compute...Ch. 9 - For the shell-and-tube heat exchanger described in...Ch. 9 - For the heat exchanger described in Problem 9.29...Ch. 9 - For the glycerin described in Problem 9.31 compute...Ch. 9 - For the flow of water in the square tubes...Ch. 9 - If the heat sink described in Problem 9.33 is 105...Ch. 9 - Compute the energy loss for the flow of water in...Ch. 9 - In Fig. 9.26 ethylene glycol ( sg=1.10 ) at 77F...Ch. 9 - Figure 9.27 shows a duct in which methyl alcohol...Ch. 9 - Prob. 9.49PPCh. 9 - Figure 9.29 shows a system in which methyl alcohol...Ch. 9 - A simple heat exchanger is made by welding...Ch. 9 - Three surfaces of an instrument package are cooled...Ch. 9 - Figure 9.32 shows a heat exchanger with internal...

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