Group 2Volumetric Flowrate and

Average Velocity

Sage Ross

Randy Goll

Lisa Gourley

Marci Wyatt

Jennifer Oberlag

April 21, 2001

Problem Statement

Fluid flows through a conduit with a quarter circle cross-section. The profile of the velocity in the z-direction can beapproximated by:

])/4(1[ ])/(1[ )/( 22 RrRrvv Az

With the coordinate system as shown and where vA is a constant equal to 20 cm/s and R=30 cm. Find the Volumetric flowrate Q and the average velocity.

The circular shape of the conduit suggests that we use cylindricalcoordinates to solve the problem.

vA=20 cm/s R=30 cm

3 dimensional view of conduit

Quarter circle cross section

])/4(1[ ])/(1[ )/( 22 RrRrvv Az

We need to use the following equations to solve the problem:

A

nvQ dA )(

A

Qvave

A

dAnvQ )(

Solution

zvnv drrddA

R

zvQ0

4

4

dr dr

R

Rr

Rr

A rvQ0

4

43

162 dr - ]1[ 2

3

Volumetric flowrate across element dA: dAnv )(

To find the volumetric flow we must integrate:

Such that: and

The limits for integration can be found from geometry.

Substitute the given value for vz:

R

AvQ0

4

4

242

Rr

Rr dr d]r -][1-[1

Integrate with respect to

1 0

2

3

16

3

16

44 2

64

3

2

64

3

R

Rr

Rr

A rdrvQ

R R

RrrA

RrA

R

vdrr

R

vQ

0 0532

2

53

2

4

312122

22

53 cm 30cm/s 2045

2

15

2

332

53

Rv

R

vQ AR

RRA

sQ /cm 2500 3

Evaluate the integral at the limits:

Simplify and integrate with respect to r:

Simplify and evaluate:

2

3

24

1 cm) 30(

)/cm 2500(4

s

R

Q

A

Qvave

cm/s 5.3avev

The average velocity is found by diving volumetric flowrate by area.

vz

Vave represents the average of all of the velocities over the velocity profile.