Unformatted text preview: VI. VISCOUS INTERNAL FLOW
To date, we have considered only problems where the viscous effects were either:
a. known:
i.e.  known FD or
hf
b. negligible: i.e.  inviscid flow
This chapter presents methodologies for predicting viscous effects and viscous
flow losses for internal flows in pipes, ducts, and conduits.
Typically, the first step in determining viscous effects is to determine the flow
regime at the specified condition.
The two possibilities are:
a. Laminar flow
b. Turbulent flow
The student should read Section 6.1 in the text, which presents an excellent
discussion of the characteristics of laminar and turbulent flow regions.
For steady flow at a known flow rate, these regions exhibit the following:
Laminar flow: A local velocity constant with time, but which varies
spatially due to viscous shear and geometry. Turbulent flow: A local velocity which has a constant mean value but
also has a statistically random fluctuating component due
to turbulence in the flow. Typical plots of velocity time histories for laminar flow, turbulent flow, and
the region of transition between the two are shown below. Fig. 6.1 (a) Laminar, (b)transition, and (c) turbulent flow velocity time histories.  v v VI1  eText Main Menu  Textbook Table of Contents  Study Guide Principal parameter used to specify the type of flow regime is the
Reynolds number  Re = ρV D V D
=
µ
ν V  characteristic flow velocity
D  characteristic flow dimension
µ  dynamic viscosity υ  kinematic viscosity = µ
ρ We can now define the
Recr ≡ critical or transition Reynolds number
Recr ≡ Reynolds number below which the flow is laminar,
above which the flow is turbulent
While transition can occur over a range of Re, we will use the following for
internal pipe or duct flow: Re cr ≅ 2300 = ρVD VD =
µ cr
υ cr Internal Viscous Flow
A second classification concerns whether the flow has significant entrance region
effects or is fully developed. The following figure indicates the characteristics of
the entrance region for internal flows. Note that the slope of the streamwise
pressure distribution is greater in the entrance region than in the fully developed
region.
Typical criteria for the length of the entrance region are given as follows: Le
≅ 0.06 Re
D Laminar:  v v VI2  eText Main Menu  Textbook Table of Contents  Study Guide Le
≅ 4.4Re1/ 6
D Turbulent where: Le = length of the entrance region Note: Take care in neglecting entrance region effects.
In the entrance region, frictional pressure drop/length > the pressure
drop/length for the fully developed region. Therefore, if the effects of the
entrance region are neglected, the overall predicted pressure drop will be
low. This can be significant in a system with short tube lengths, e.g., some
heat exchangers. Fully Developed Pipe Flow
The analysis for steady, incompressible, fully developed, laminar flow in a
circular horizontal pipe yields the following equations: R2 dP r2 U (r ) = −
1 − 4 µ dx R 2  v v VI3  eText Main Menu  Textbook Table of Contents  Study Guide r2 U = 1 − 2 , Umax = 2Vavg
R
Umax and Q = A Vavg = π R2 Vavg
Key Points: Thus for laminar, fully developed pipe flow (not turbulent):
a.
b.
c.
d. The velocity profile is parabolic.
The maximum local velocity is at the centerline (r = 0).
The average velocity is onehalf the centerline velocity.
The local velocity at any radius varies only with radius, not on the
streamwise (x) location ( due to the flow being fully developed). Note: All subsequent equations will use the symbol V (no subscript) to represent
the average flow velocity in the flow cross section.
Darcy Friction Factor:
We can now define the Darcy friction factor f as: D ∆P
f L
f≡
V2
ρ
2 where ∆Pf = the pressure drop due to friction
only.
The general energy equation must still be used
to determine total pressure drop. Therefore, we obtain L V2
∆P f = ρ g h f = f ρ
D2
and the friction head loss hf is given as L V2
hf = f
D 2g
Note: The definitions for f and hf are valid for either laminar or turbulent flow.
However, you must evaluate f for the correct flow regime, laminar or turbulent.  v v VI4  eText Main Menu  Textbook Table of Contents  Study Guide Key Point: It is common in industry to define and use a “fanning” friction
factor ff . The fanning friction factor differs from the Darcy friction factor
by a factor of 4. Thus, care should be taken when using unfamiliar
equations or data since use of ff in equations developed for the Darcy friction
factor will result in significant errors (a factor of 4). Your employer will not
be happy if you order a 10 hp motor for a 2.5 hp application. The equation
suitable for use with ff I s L V2
h f = 4 ff
D 2g
Laminar flow:
Application of the results for the laminar flow velocity profile to the definition of
the Darcy friction factor yields the following expression: f= 64
Re laminar flow only (Re < 2300) Thus with the value of the Reynolds number, the friction factor for laminar flow is
easily determined.
Turbulent flow:
A similar analysis is not readily available for turbulent flow. However, the
Colebrook equation, shown below, provides an excellent representation for the
variation of the Darcy friction factor in the turbulent flow regime. Note that the
equation depends on both the pipe Reynolds number and the roughness ratio, is
transcendental, and cannot be expressed explicitly for f . 2.51
ε/D f = − 2 log + Re f 1/ 2
3.7 turbulent flow only (Re > 2300)  v v VI5  eText Main Menu  Textbook Table of Contents  Study Guide where ε = nominal roughness of pipe or duct being used. (Note: Take care with units for ε; (Table 6.1, text) ε /D must be nondimensional). A good approximate equation for the turbulent region of the Moody chart is given
by Haaland’s equation: 6.9 ε / D 1 .11 f = −1.8log + Re 3.7 −2 Note again the roughness ratio ε/D must be nondimensional in both equations.
Graphically, the results for both laminar and turbulent flow pipe friction are
represented by the Moody chart as shown below. Typical roughness values are shown in the following table:  v v VI6  eText Main Menu  Textbook Table of Contents  Study Guide Table 6.1 Average roughness values of commercial pipe Haaland’s equation is valid for turbulent flow (Re > 2300) and is easily set up on a
computer, spreadsheet, etc.
Key fluid system design considerations for laminar and turbulent flow
a. Most internal flow problems of engineering significance are turbulent, not
laminar. Typically, a very low flow rate is required for internal pipe flow to be
laminar. If you open your kitchen faucet and the outlet flow stream is larger
than a kitchen match, the flow is probably turbulent. Thus, check your work
carefully if your analysis indicates laminar flow.
b. The following can be easily shown:
Laminar flow: { &
Wf
Turbulent flow: −4 ∆Pf ~ µ , L, Q, D { } ~ µ, L,Q 2, D −4 } ∆Pf ~ {ρ , µ , L, Q , D
3/ 4 1/ 4 1.75 −4.75 &
Wf ~ { ρ , µ , L, Q , D
3/4 1/ 4 2.75 } −4.75 } Thus both pressure drop and pump power are very dependent on flow rate
and pipe/conduit diameter. Small changes in diameter and/or flow rate can
significantly change circuit pressure drop and power requirements.  v v VI7  eText Main Menu  Textbook Table of Contents  Study Guide Example ( Laminar flow):
Water, 20oC flows through a 0.6 cm
tube, 30 m long, at a flow rate of 0.34
liters/min. If the pipe discharges to the
atmosphere, determine the supply
pressure if the tube is inclined 10o
above the horizontal in the flow
direction. L
10 3 L = 30 m 1 Water Properties: D = 0.6 cm Energy Equation
2
P1 − P2 V2 − V12
=
+ Z2 − Z1 + h f − hp
ρg
2g 3 ρg = 9790 N/m ρ = 998 kg/m 2
o 2 ν = 1.005 E6 m /s
which for steadyflow in a
constant diameter pipe with P2 = 0
gage becomes,
−3 P
1
= Z2 −Z1 + hf = L sin 10o + hf
ρg 3 Q 0.34 E m / min*1min/ 60 s
V= =
= 0.2 m / s
A
π (0.3 /100 )2 m2
Re =
f= VD υ = 0.2 * 0.006
= 1197 → laminar flow
1.005 E −6 64
64
=
= 0.0535
Re 1197
2 2 0.2
LV
30 m
hf = f
= 0.0535*
= 0.545 m
D 2g
0.006 m 2 * 9.807m / s 2
P1
= 30 * sin10Þ+ 0.545 = 5.21 + 0.545
ρg
gravity
head
3 = 5.75m friction
head total head
loss 3 P1 = 9790 N/m *5.75 m = 56.34 kN/m (kPa) ~ 8.2 psig ans.  v v VI8  eText Main Menu  Textbook Table of Contents  Study Guide 1 Example: (turbulent flow)
3 2 Oil, ρ = 900 kg/m , ν = 1 E5 m /s,
3
flows at 0.2 m /s through a 500 m length
of 200 mm diameter, cast iron pipe. If
the pipe slopes downward 10o in the
flow direction, compute hf, total head
loss, pressure drop, and power required
to overcome these losses.
The energy equation can
be written as follows
where ht = total head
loss. 3 υ = L = 500 m D = 200 mm ht = Z2 − Z1 + h f Q 0.2 m / s
V= =
= 6.4 m / s
A π (.1)2 m 2
VD 2 L P1 − P2
V22 − V12
= ht =
+ Z 2 − Z1 + h f − hp
ρg
2g which reduces to Re = o 10 Table 6.1, cast iron, ε = 0.26 mm 6.4 *.2
= 128, 000 → turbulent flow ,
1 E −5 ε
D = 0.26
= 0.0013
200 Since flow is turbulent, use Haaland’s equation to determine friction factor (check your
work using the Moody chart).
−2 1.11 6.9 6.9 ε / D 1.11 0.0013 f = −1.8log + , f = −1.8log 128, 000 + 3.7 Re 3.7 2 2 −2 6.4
LV
500 m
f = 0.02257 h f = f
= 0.02257*
= 116.6 m ans.
D 2g
0.2 m 2 * 9.807m / s 2
ht = Z2 – Z1 + hf =  500 sin 0 + 116.6 =  86.8 + 116.6 = 29.8 m ans.  v v VI9  eText Main Menu  Textbook Table of Contents  Study Guide Note that for this problem, there is a negative gravity head loss (i.e. a head increase) and
a positive frictional head loss resulting in the net head loss of 29.8 m. ∆P = ρ g ht = 900 kg / m * 9.807 m / s * 29.8 m = 263 kPa ans.
3 2 &
W = ρ Q g ht = Q ∆P = 0.2 m3 / s * 273,600 N / m 2 = 54.7 kw ans. Note that this is not necessarily the power required to drive a pump, as the pump
efficiency will typically be less than 100%.
These problems are easily set up for solution in a spreadsheet as shown below. Make
sure that the calculation for friction factor includes a test for laminar or turbulent flow
with the result proceeding to the correct equation.
Always verify any computer solution with problems having a known solution.
FRICTIONAL HEAD LOSS CALCULATION
All Data are entered in S.I. Units e.g. (m, sec., kg), except as noted, ε
Ex. 6.7
Input Data Calculated Results L= 500 (m ) V= 6.37 D=
ε= 0.2
0.26 (m)
(mm) Re =
e/D = 127324
0.0013 ρ= 900 (kg/m^3) f= 0.02258 ν= 1.00E05 (m^2/sec) hf = 116.62 Q= 0.2 (m^3/sec) sum Ki = 0.00 D1 = 0.08 (m) hm = 0.00 (m) D2 = 0.08 (m) d KE = 0.00 (m) ht = 29.62 (m) dZ= 87 (m) (m)
P1P2 = 261.43 v v VI10   eText Main Menu (m/sec)  Textbook Table of Contents  Study Guide (kPa) Solution Summary:
To solve basic pipe flow frictional head loss problem, use the following
procedure:
1. Use known flow rate to determine Reynolds number.
2. Identify whether flow is laminar or turbulent.
3. Use appropriate expression to determine friction factor (w ε/D if necessary).
4. Use definition of hf to determine friction head loss.
5. Use general energy equation to determine total pressure drop. Unknown Flow Rate and Diameter Problems
Problems involving unknown flow rate and diameter in general require iterative/
trial & error solutions due to the complex dependence of Re, friction factor, and
head loss on velocity and pipe size.
Unknown Flow Rate:
For the special case of known friction loss hf a closed form solution can be
obtained for the problem of unknown Q.
The solution proceeds as follows:
Given: Known values for D, L, hf, ρ, and µ calculate V or Q. Define solution parameter: ς= 1
2 g D3 h f
f Re =
Lυ 2
2
D Note that this solution does not contain velocity and the parameter ζ can be
calculated from known values for D, L, hf, ρ, and µ. The Reynolds number and
subsequently the velocity can be determined from ζ and the following equations: Re D = − (8ς ) 1/ 2 Turbulent: ε / D 1.775 log + 3.7
ζ  v v VI11  eText Main Menu  Textbook Table of Contents  Study Guide Re D = Laminar: ς
32 and laminar to turbulent transition can be assumed to occur approximately at
ζ = 73,600 (check Re at end of calculation to confirm).
Note that this procedure is not valid (except perhaps for initial estimates) for
problems involving significant minor losses where the head loss due only to pipe
friction is not known.
For these problems a trial and error solution using a computer is best.
Example 6.9
3 2 Oil, with ρ = 950 kg/m and ν = 2 E5 m /s, flows through 100 m of a 30 cm
diameter pipe. The pipe is known to have a head loss of 8 m and a roughness ratio
ε/D = 0.0002. Determine the flow rate and oil velocity possible for these
conditions.
Without any information to the contrary, we will neglect minor losses and KE
head changes. With these assumptions, we can write: g D3 hf 9.807 m / s 2 * 0.303 m3 *8.0 m
ς=
=
= 5.3 E 7 > 73,600; turbulent
2
2
2
Lυ
100 m * 2 E − 5 m / s ( Re D = − (8*5.3 E 7) 1/ 2 ) 0.0002
1.775 log + = 72, 600 checks, turbulent
3.7
5.3 E 7 2 72,600*2 E −5 m / s
m
Re D =
, V=
= 4.84
ans.
υ
0.3 m
s
VD Q = A V = π .15 2 m 2 4.84 m/s = 0.342 m 3/s ans.
This is the maximum flow rate and oil velocity that could be obtained through the
given pipe and given conditions (hf = 8 m).  v v VI12  eText Main Menu  Textbook Table of Contents  Study Guide Note that this problem could have also been solved using a computer based trial
and error procedure in which a value is assumed for the fluid flow rate until a flow
rate is found which results in the specified head loss. Note also that with this
procedure, the problem being solved can include the effects of minor losses, KE,
and PE changes with no additional difficulty.
Unknown Pipe Diameter:
A similar difficulty arises for problems involving unknown pipe difficulty, except
a closed form, analytical solution is not available. Again, a trial and error solution
is appropriate for use to obtain the solution and the problem can again include
losses due to KE, PE, and piping components with no additional difficulty.
NonCircular Ducts:
For flow in noncircular ducts or ducts for which the flow does not fill the entire
crosssection, we can define the hydraulic diameter Dh as Dh = 4A
P where
A = crosssectional area of actual flow,
P = wetted perimeter, i.e. the perimeter
on which viscous shear acts Cross sectional area  A Perimeter  P With this definition, all previous equations for the Reynolds number Re friction
factor f and head loss hf are valid as previously defined and can be used on
both circular and noncircular flow cross sections.
Minor Losses
In addition to frictional losses for a length L of pipe, we must also consider
losses due to various fittings (valves, unions, elbows, tees, etc.). These losses are
expressed as V2
hm = Ki
2g  v v VI13  eText Main Menu  Textbook Table of Contents  Study Guide where
hm = the equivalent head loss across the fitting or flow component.
V = average flow velocity for the pipe size of the fitting
Ki = the minor loss coefficient for given flow component; valve, union, etc.
See Sec. 6.7, Table 6.5, 6.6, Fig. 6.19, 6.20, 6.21, 6.22, etc.
Table 6.5 shows minor loss K values for several common types of valves, fully
open, and for elbows and tees.
Table 6.5 Minor loss coefficient for common valves and piping components Figure 6.18 shows minor loss K values for several types of common valves.
Note that the K valves shown here are for the indicated fractional opening. Also,
fully open values may not be consistent with values indicated in Table 6.5 for fully
open valves or for the valve of a particular manufacturer. In general, use specific
manufacturer’s data when available.  v v VI14  eText Main Menu  Textbook Table of Contents  Study Guide Fig. 6.18 Average loss coefficients for partially open valves Note that exit losses are K
≅ 1 for all submerged
exits, e.g., fluid discharged
into a tank at a level below
the fluid surface.
Also, for an open pipe
discharge to the
atmosphere, there is no loss
coefficient when the energy
equation is written only to
the end of the pipe.
In general, do not take
point 1 for an analysis to be
in the plane of an inlet
having an inlet loss. You
do not know what fraction
of the inlet loss to consider. Fig. 6.21 Entrance and exit loss coefficients
(a) reentrant inlets; (b) rounded and beveled inlets  v v VI15  eText Main Menu  Textbook Table of Contents  Study Guide Note that the losses shown in Fig.
6.22 do not represent losses
associated with pipe unions or
reducers. These must be found in
other sources in the literature.
Also note that the loss coefficient
is always based on the velocity in
the smaller diameter (d) of the
pipe, irrespective of the direction
of flow.
Fig. 6.22 Sudden contraction and expansion
losses. Assume that this is also true for
reducers and similar area change
fittings. These and other sources of data now provide the ability to determine frictional
losses for both the pipe and other piping/duct flow components.
The total frictional loss now becomes L V2
V2
hf = f
+ ∑ Ki
2g
D 2g
or
2 f L + ∑K V
hf = i
D 2g These equations would be appropriate for a single pipe size (with average velocity
V). For multiple pipe/duct sizes, this term must be repeated for each pipe size.
Key Point: The energy equation must still be used to determine the total head
loss and pressure drop from all possible contributions.  v v VI16  eText Main Menu  Textbook Table of Contents  Study Guide Example 6.16 Water, ρ = 1.94 slugs/ft3 and ν = 1.1 E5 ft2/s, is pumped between two reservoirs
at 0.2 ft3/s through 400 ft of 2–in diameter pipe with ε/D = 0.001 having the
indicated minor losses. Compute the pump horsepower (motor size) required.
Writing the energy equation between points 1 and 2 (the free surfaces of the two
reservoirs), we obtain P − P V22 − V12
1
2
=
+ Z2 − Z1 + h f − hp
ρg
2g
For this problem, the pressure (P1 = P2) and velocity (V1 = V2 = 0) head terms are
zero and the equation reduces to
2
f L +∑K V
hp = Z2 − Z1 + h f = Z2 − Z1 + i
D 2g For a flow rate Q = 0.2 ft3/s we obtain  v v VI17  eText Main Menu  Textbook Table of Contents  Study Guide Q
0.2 ft 3 / s
V= =
= 9.17 ft / s
A π (1/ 12)2 ft 2
Re = With ε/D = 0.001 and VD ν = 9.17 ft / s ( 2 / 12) ft
= 139,000
1.1E − 5 ft 2 / s the flow is turbulent and Haaland’s equation can be used to determine the friction
factor: 1.11 6.9 .001
f = −1.8log + 3.7 139, 000 −2
= 0.0214 the minor losses for the problem are summarized in the following table:
Note: The loss for a
pipe bend is not the
same as for an elbow
fitting.
If there were no tank at
the pipe discharge and
point 2 were at the pipe
exit, there would be no
exit loss coefficient.
However, there would
be an exit K.E. term. Loss element
Sharp entrance (Fig. 6.21)
Open globe valve (Table 6.5)
12 " bend, R/D = 12/6 = 2 (Fig. 6.19)
Threaded, 90Þ, reg. elbow, (Table 6.5)
Gate valve, 1/2 closed (Fig. 6.18)
Submerged exit (Fig. 6.20)
Ki = Ki
0.5
6.9
0.15
0.95
2.7
1
12.2 Substituting in the energy equation we obtain 400 9.17 2 9.17 2 h p = (120 − 20) + 0.0214 + 12.2 2/12 64.4 64.4 h p = 100 + 67.1 + 15.9 = 183ft  v v VI18  eText Main Menu  Textbook Table of Contents  Study Guide Note the distribution of the total loss between static, pipe friction, and minor
losses.
The power required to be delivered to the fluid is give by
3 slug
ft
ft
Pf = ρ Qg h p = 1.94 3 32.2 2 0.2 183 ft = 2286 ft lbf
s
ft
s P=
f 2286ftlbf
= 4.2hp
550 ftlbf / s / hp If the pump has an efficiency of 70 %, the power requirements would be specified
by &
w p = 4.2 hp = 6 hp
0.70 Solution Summary:
To solve basic pipe flow pressure drop problem, use the following procedure:
1. Use known flow rate to determine Reynolds number.
2. Identify whether flow is laminar or turbulent.
3. Use appropriate expression to find friction factor (with ε/D if necessary).
4. Use definition of hf to determine friction head loss.
5. Tabulate and sum minor loss coefficients for piping components.
6a. Use general energy equation to determine total pressure drop, or
6b.Determine pump head requirements as appropriate.
7. Determine pump power and motor size if required.  v v VI19  eText Main Menu  Textbook Table of Contents  Study Guide MultiplePipe Systems
Basic concepts of pipe system analysis apply also to multiple pipe systems.
However, the solution procedure is more involved and can be iterative.
Consider the following:
a. Multiple pipes in series
b. Multiple pipes in parallel
Series Pipe System:
The indicated pipe system has a
steady flow rate Q through three
pipes with diameters D1, D2, & D3. a 3 2 1 b Two important rules apply to this
problem.
1. The flow rate is the same through each pipe section. For incompressible
flow, this is expressed as
2 2 2 Q1 = Q2 = Q3 = Q or D1 V1 = D2 V2 = D3 V3
2. The total frictional head loss is the sum of the head losses through the
various sections. h f ,a− b = h f ,1 + h f ,2 + h f ,3 Note: Be careful how you evaluate the transitions from one section to the
next. In general, loss coefficients for transition sections are based on
the velocity of the smaller section. VI20  v v h f ,a− b 2
2
2 f L + ∑ K V1 + f L + ∑ K V2 + f L + ∑ K V3
=
i
i
i
D
D
D
D1 2 g
D2 2 g
D3 2 g  eText Main Menu  Textbook Table of Contents  Study Guide Example: Given a pipe system as shown in the previous figure. The total
pressure drop is Pa – Pb = 150 kPa and the elevation change is Za – Zb = – 5 m.
Given the following data, determine the flow rate of water through the section.
Pipe
1
2
3 L (m)
100
150
80 D (cm)
8
6
4 The energy equation is written as e (mm)
0.24
0.12
0.2 e/D
0.003
0.02
0.005 Pa − Pb Vb2 − Va2
=
+ Zb − Za + h f − h p
ρg
2g where hf is given by the sum of
the total frictional losses for three
pipe sections. With no pump; hp is 0, Zb  Za =
Pa − Pb 150,000 N / m 2
ht =
=
= 15.3 m
 5 m and ht = 15.3 m for ∆P =
ρg
9790 N / m2
150 kPa
Since the flow rate Q and thus velocity is the only remaining variable, the
solution is easily obtained from a spreadsheet by assuming Q until ∆P = 150 kPa.
Fluid 1 2 3 100
0.08
0.24 150
0.06
0.12 80
0.04
0.20 ε/D= ρ( kg/m^3) =
ν(m^2/s) = 0.003 0.002 0.005 V(m/s)=
Re =
f=
hf = 0.56
44082.8
0.02872
0.58 1.00
58777.1
0.02591
3.30 2.25
88165.6
0.03139
16.18 Ki
hm= 0
0 0
0 0
0 hf(calc) = 0.58 3.30 16.18 1000
1.02E06 L(m)
D(m) =
ε(mm) = inlet & exit 5
0.08
0.04 dZ (m) =
Da(m) =
Db(m) =
Assume
Q (m^3/s)= 0.00283 Va(m/s)= 0.56
2.25
0.24
20.08
Actual
150 Vb(m/s)=
dKE(m) =
hf (net) =
Pa  Pb (kPa) Calculated
150.00 Q(m^3/hr) = 10.17  v v VI21  eText Main Menu  Textbook Table of Contents  Study Guide 3 Thus it is seen that a flow rate of 10.17 m /hr produces the indicated head loss
through each section and a net total ∆P = 150 kPa.
A solution can also be obtained by writing all terms explicitly in terms of a single
velocity, however, the algebra is quite complex (unless the flow is laminar), and
an iterative solution is still required. All equations used to obtained the solution
are the same as those presented in previous sections.
Parallel Pipe Systems
A flow rate QT enters the indicated
parallel pipe system. The total flow
splits and flows through 3 pipe
sections, each with different
diameters and lengths. 1 Q 2 a
T b 3 Q T Two basic rules apply to parallel
pipe systems;
1. The total flow entering the parallel section is equal to the sum of the flow
rates through the individual sections,
2. The total pressure drop across the parallel section is equal to the pressure
drop across each individual parallel segment.
Note that if a common junction is used for the start and end of the parallel section,
the velocity and elevation change is also the same for each section. Thus, the flow
rate through each section must be such that the frictional loss is the same for each
and the sum of the flow rates equals the total flow.
For the special case of no kinetic or potential energy change across the sections,
we obtain:
ht = ( hf + hm)1 = ( hf + hm)2 =( hf + hm)3
and
QT = Q1 + Q2 + Q3  v v VI22  eText Main Menu  Textbook Table of Contents  Study Guide Again, the equation used for both the pipe friction and minor losses is the same as
previously presented. The flow and pipe dimensions used for the previous
example are now applied to the parallel circuit shown above. Example: A parallel pipe section consists of three parallel pipe segments with the
lengths and diameters shown below. The total pressure drop is 150 kPa and the
parallel section has an elevation drop of 5 m. Neglecting minor losses and kinetic
energy changes, determine the flow rate of water through each pipe section.
The solution is iterative and is again presented in a spreadsheet. The net friction
head loss of 20.3 m now occurs across each of the three parallel sections.
Fluid 1 2 3 100
0.08
0.24 150
0.06
0.12 80
0.04
0.20 V(m/s)=
Re =
f=
hf = 0.003
62.54
3.46
271083.5
0.02666
20.30 0.002
25.95
2.55
149977.8
0.02450
20.30 0.005
11.41
2.52
98919.7
0.03129
20.30 Ki =
hm= 0
0.00 0
0.00 0
0.00 hf,net(m) =
hf + ∆z = ρ( kg/m^3) =
ν(m^2/s) = 20.30 20.30 20.30 15.30 15.30 15.30 62.54 25.95 11.41 1000
1.02E06 L(m)
D(m) =
ε(mm) =
ε/D= inlet & exit Db(m) = 5
0.08
0.04 Q (m^3/hr)= 99.91 dZ (m) =
Da(m) = Q(m^3/hr) = Assume 62.54
25.95 Q1 (m^3/s)=
Q2(m^3/s)= Pa  Pb (kPa) 150.13 ht(m) = 15.31 Q(m^3/hr) = Total Flow, Qt(m^3/hr) = 99.91 The strong effect of diameter can be seen with the smallest diameter having the
lowest flow rate, even though it also has the shortest length of pipe.  v v VI23  eText Main Menu  Textbook Table of Contents  Study Guide ...
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This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.
 Spring '08
 Sakar

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