Oil flows into the pipe at A with an average velocity of 1.5 m/s and through B with an average velocity of 0.75 m/s. Determine the maximum velocity vmax of the oil as it emerges from C if the velocity distribution is parabolic, defined by vc = Vmax (1 – 200r²), where r is the radius in meters measured from the centerline of the pipe. V, max 20mm 30mm 24mm A В 1.5 m/s 0.75 m/s

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
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Oil flows into the pipe at A with an average velocity of 1 m/s and through B with an average
velocity of 0.5 m/s. Determine the maximum velocity vmax of the oil as it emerges from C if the
velocity distribution is parabolic, defined by vc = vmax(1 – 200r2), where r is the radius in
meters measured from the centerline of the pipe.
V.
max
20mm
32mm
20mm
B
0.5 m/s
1 m/s
SOLUTION
The control volume considered is fixed as it contains the oil in the pipe. Also, the flow
is steady and so no local changes occur. Here, the density of the oil is constant. Then
+
dA = 0
at
CV
0 - VAAA + VBAB +
r0.01m
-(1m/s)[n(0.016m)²] + (0.5 m/s)[n(0.01m)²] +
Vmax(1 – 200r2)(2nrdr) = 0
3
Homework #4 Solutions
-2.06(10-4)am³ + vmax[7(0.01m)? – 1007(0.01m)*] = 0
-2.06(10-4)am³ + 9.9(10-5)nv,max = 0
Vmax = 2.8m/s
Note: The integral in the above equation is equal to the volume under the velocity profile, while
in this case is a paraboloid.
1
V,dA =Tr²h =n0.01²v,max = 5(10-5)avmax
Ac
A, V= 0.5 m/s
1 m/s
(b)
Rof. HI RRELED A 49
Transcribed Image Text:Oil flows into the pipe at A with an average velocity of 1 m/s and through B with an average velocity of 0.5 m/s. Determine the maximum velocity vmax of the oil as it emerges from C if the velocity distribution is parabolic, defined by vc = vmax(1 – 200r2), where r is the radius in meters measured from the centerline of the pipe. V. max 20mm 32mm 20mm B 0.5 m/s 1 m/s SOLUTION The control volume considered is fixed as it contains the oil in the pipe. Also, the flow is steady and so no local changes occur. Here, the density of the oil is constant. Then + dA = 0 at CV 0 - VAAA + VBAB + r0.01m -(1m/s)[n(0.016m)²] + (0.5 m/s)[n(0.01m)²] + Vmax(1 – 200r2)(2nrdr) = 0 3 Homework #4 Solutions -2.06(10-4)am³ + vmax[7(0.01m)? – 1007(0.01m)*] = 0 -2.06(10-4)am³ + 9.9(10-5)nv,max = 0 Vmax = 2.8m/s Note: The integral in the above equation is equal to the volume under the velocity profile, while in this case is a paraboloid. 1 V,dA =Tr²h =n0.01²v,max = 5(10-5)avmax Ac A, V= 0.5 m/s 1 m/s (b) Rof. HI RRELED A 49
Oil flows into the pipe at A with an average velocity of 1.5 m/s and through B with an average
velocity of 0.75 m/s. Determine the maximum velocity vmax of the oil as it emerges from C if
the velocity distribution is parabolic, defined by vc = Vmax(1 – 200r²), where r is the radius in
-
meters measured from the centerline of the pipe.
max
20mm
30mm
24mm
A
B
1.5 m/s
0.75 m/s
Transcribed Image Text:Oil flows into the pipe at A with an average velocity of 1.5 m/s and through B with an average velocity of 0.75 m/s. Determine the maximum velocity vmax of the oil as it emerges from C if the velocity distribution is parabolic, defined by vc = Vmax(1 – 200r²), where r is the radius in - meters measured from the centerline of the pipe. max 20mm 30mm 24mm A B 1.5 m/s 0.75 m/s
Expert Solution
Step 1

Given that,

The average velocity of oil through pipe A VA = 1.5 ms

The diameter of pipe A DA = 30 mm = 0.03 m

The average velocity of oil through pipe B VB = 0.75 ms

The diameter of pipe B DB = 24 mm = 0.024 m

The diameter of pipe C DC = 20 mm = 0.02 m

Velocity distribution in pipe C Vc = Vmax1 - 200r2

It is required to calculate the maximum velocity Vmax of oil as it emerges from pipe C.

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