10.7-Solving Turbulent Flow Prob, fluid mech
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Solving Turbulent Flow Problems
10.7 Solving Turbulent Flow Problems
This section describes how to solve problems that involve turbulent flow
*
in a pipe, emphasizing how to classify
problems as case 1, 2, or 3. Classification is important because cases 2 and 3 usually require either an iterative
approach or they require computer programs that can solve coupled nonlinear equations. Some useful computer
programs include TK Solver, EES, MathCAD, and MatLab.
To recognize problems that require iterative approaches or computer solutions, engineers classify problems into
three cases based on the goal of the problem and based on what information is known.
Case 1
is when the goal is to find the
head loss
, given the pipe length, pipe diameter, and flow rate. This
problem is straightforward because it can be solved using algebra; see Example 10.3.
Case 2
is when the goal is to find the
flow rate
, given the head loss (or pressure drop), the pipe length, and
the pipe diameter. This problem usually requires an iterative approach or solver program; see Examples 10.4
and 10.5.
Case 3
is when the goal is to find the
pipe diameter
, given the flow rate, length of pipe, and head loss (or
pressure drop). This problem usually requires an iterative approach or a solver program; see Example 10.6.
There are several approaches that sometimes eliminate the need for an iterative approach. For case 2, an iterative
approach can sometimes be eliminated by using an explicit equation developed by Swamee and Jain 7:
(10.40)
Using Eq. (10.40) is equivalent to using the top of the Moody diagram, which presents a scale for Re
f
1/2
. For
case 3, one can sometimes use an explicit equation developed by Swamee and Jain 7 and modified by Streeter
and Wylie 8:
(10.41)
Example 10.3 shows an example of a case 1 problem.
EXAMPLE 10.3 HEAD LOSS I A PIPE (CASE 1)
Water (
T
= 20°C) flows at a rate of 0.05 m
3
/s in a 20 cm asphalted castiron pipe. What is the head
loss per kilometer of pipe?
PROBLEM DEFINITION
Situation:
Water is flowing in a pipe.
Find:
Head loss (in meters) for
L
= 1000 m.
Assumptions:
Fully developed flow.
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Properties:
Water (20°C), Table A.5: ν = 1 × 10
6
m
2
/s.
Sketch:
PLAN
Since this is a case 1 problem (head loss is the goal), the solution is straightforward.
1. Calculate the mean velocity using the flow rate equation (5.8).
2. Calculate the Reynolds number using Eq. (10.2).
3. Calculate the relative roughness and then look up
f
on the Moody diagram.
4. Find head loss by applying the DarcyWeisbach equation (10.12).
SOLUTION
1. Mean velocity
2. Reynolds number
3. Resistance coefficient
·
Equivalent sand roughness (Table 10.4):
k
s
= 0.12 mm
·
Relative roughness:
·
Look up
f
on the Moody diagram for Re = 3.18 × 10
5
and
k
s
/
D
= 0.0006:
4. DarcyWeisbach equation
Example 10.4 shows an example of a case 2 problem. Notice that the solution involved application of the scale
on the top of the Moody diagram; thereby avoiding an iterative solution.
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EXAMPLE 10.4 FLOW RATE I A PIPE (CASE 2)
The head loss per kilometer of 20 cm asphalted castiron pipe is 12.2 m. What is the flow rate of
water through the pipe?
PROBLEM DEFINITION
Situation:
This is the same situation as Example 10.3 except that the head loss is now specified and
the discharge is unknown.
Find:
Discharge (m
3
/s) in the pipe.
PLAN
This is a case 2 problem because flow rate is the goal. However, a direct (i.e., noniterative) solution is
possible because head loss is specified. The strategy will be to use the horizontal scale on the top of
the Moody diagram.
1. Calculate the parameter on the top of the Moody diagram.
2. Using the Moody diagram, find the friction factor
f
.
3. Calculate mean velocity using the DarcyWeisbach equation (10.12).
4. Find discharge using the flow rate equation (5.8).
SOLUTION
1.
Compute the parameter
.
2. Determine resistance coefficient.
·
Relative roughness:
·
Look up
f
on the Moody diagram for
3. Find
V
using the DarcyWeisbach equation.
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4. Use flow rate equation to find discharge.
REVIEW
Validation. The calculated flow rate matches the value from Example 10.3. This is expected because
the data are the same.
When case 2 problems require iteration, there are several methods that can be used to find a solution. One of the
easiest ways is a method called “successive substitution,” which is illustrated by Example 10.5.
EXAMPLE 10.5 FLOW RATE I A PIPE (CASE 2)
Water (
T
= 20°C) flows from a tank through a 50 cm diameter steel pipe. Determine the discharge of
water.
PROBLEM DEFINITION
Situation:
Water is draining from a tank through a steel pipe.
Find:
Discharge (m
3
/s) for the system.
Assumptions:
1. Flow is fully developed.
2. Include only the head loss in the pipe.
Sketch:
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Properties:
1.
Water (20°C), Table A.5: ν = 1 × 10
6
m
2
/s.
2. Steel pipe, Table 10.4, equivalent sand roughness:
k
s
= 0.046 mm. Relative roughness (
k
s
/
D
) is
9.2 × 10
5
.
PLAN
This is a case 2 problem because flow rate is the goal. An iterative solution is used because
V
is
unknown, so there is no direct way to use the Moody diagram.
1. Apply the energy equation from section 1 to section 2.
2. First trial. Guess a value of
f
and then solve for
V.
3. Second trial. Using
V
from the first trial, calculate a new value of
f
.
4. Convergence. If the value of
f
is constant within a few percent between trials, then stop.
Otherwise, continue with more iterations.
5. Calculate flow rate using the flow rate equation (5.8).
SOLUTION
1. Energy equation (reservoir surface to outlet)
or
(1)
2. First trial (iteration 1)
·
Guess a value of
f
= 0.020.
·
Use eq. (1) to calculate
V
= 8.86 m/s.
·
Use
V
= 8.86 m/s to calculate Re = 4.43 × 10
6
.
·
Use Re = 4.43 × 10
6
and
k
s
/
D
= 9.2 × 10
5
on the Moody diagram to find that
f
= 0.012.
·
Use eq. (1) with
f
= 0.012 to calculate
V
= 10.7 m/s.
3. Second trial (iteration 2)
·
Use
V
= 10.7 m/s to calculate Re = 5.35 × 10
6
.
·
Use Re = 5.35 × 10
6
and
k
s
/
D
= 9.2 × 10
5
on the Moody diagram to find that
f
= 0.012.
4. Convergence. The value of
f
= 0.012 is unchanged between the first and second trials.
Therefore, there is no need for more iterations.
5. Flow rate
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