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BASIC PRINCIPLES
OF PIPE FLOW
Dr/ Ahmed safwat
Amir Ashraf sayed
Pipe Flow Under Siphon Action
A pipeline that rises above its hydraulic gradient line is termed a siphon
head loss due to friction is large and the form losses can be neglected. Thus,
the hydraulic gradient line is a straight line that joins the water surfaces at
points A and B.
Pipe Flow Under Siphon
Action
If the hydraulic gradient line is above the centerline of pipe:
the water pressure is above atmospheric.
if it is below the centerline of the pipe:
the pressure is below atmospheric.
1- points b and c, the water pressure is atmospheric.
2- between b and c it is less than atmospheric.
3- At the highest point e, the water pressure is the lowest.
4- If the pressure head at point e is less than - 2.5 m,
the water starts vaporizing.
Example 2.2A.
A pumping system with different pipe fittings is shown in Fig. 2.15. Calculate
residual pressure head at the end of the pipe outlet if the pump is generating an
input head of 50 m at 0.1 m3
/s discharge. The CI pipe diameter 𝐷 is 0.3 m. The
contraction size at point 3 is 0.15 m; pipe size between points 6 and 7 is 0.15 m;
and confusor outlet size 𝑑 = 0.15 m. The rotary valve at point 5 is fully open.
Consider the following pipe lengths between points:
Points 1 and 2 = 100 m, points 2 and 3 = 0.5 m; and points 3 and 4 = 0.5 m
Points 4 and 6 = 400 m, points 6 and 7 = 20 m; and points 7 and 8 = 100 m
Head loss between points 1 and 2.
Pipe length 100 m, flow 0.1 m3
/s, and pipe diameter 0.3 m.
Using Eq. (2.4b), 𝑣 for 20∘C is 1.012 Γ— 10βˆ’6 m2/s, similarly using Eq. (2.4c),
Reynolds number 𝐑 = 419,459. Using Table 2.1 for CI pipes, πœ€ is 0.25 𝐦𝐦. The
friction factor 𝑓 is calculated using Eq. (2.6b) = 0.0197. Using Eq. 2.3b the head
loss β„Žπ‘“12 in pipe (1 βˆ’ 2) is
π’‰π’‡πŸπŸ =
πŸ–π’‡π‘³π‘ΈπŸ
π…πŸπ’ˆπ‘«πŸ“
=
πŸ– Γ— 𝟎. πŸŽπŸπŸ—πŸ• Γ— 𝟏𝟎𝟎 Γ— 𝟎. 𝟏𝟐
πŸ‘. πŸπŸ’πŸπŸ“πŸ—πŸ Γ— πŸ—. πŸ–πŸ Γ— 𝟎. πŸ‘πŸ“
= 𝟎. πŸ”πŸ•πŸŽ 𝐦
Solution
2. Head loss between points 2 and 3 (a contraction transition).
For 𝐷 = 0.3, 𝑑 = 0.15, and transition length = 0.5 m, the contraction angle 𝛼𝑐
can be calculated using Eq. (2.13b):
𝛼𝑐 = 2tanβˆ’1
𝐷1 βˆ’ 𝐷2
2𝐿
= 2tanβˆ’1
0.3 βˆ’ 0.15
2 Γ— 0.5
= 0.298 radians .
Using Eq. (2.13a), the form-loss coefficient is
π‘˜π‘“ = 0.315𝛼𝑐
1/3
= 0.315 Γ— 0.2981/3
= 0.210
Using Eq. (2.12b), the head loss β„Žπ‘š23 = 0.193 m.
3. Head loss between points 3 and 4 (an expansion transition).
For 𝑑 = 0.15, 𝐷 = 0.3, the expansion ratio π‘Ÿ = 2, and transition length = 0.5 m.
Using Eq. (2.13d), the expansion angle 𝛼𝑒 = 0.298 radians. Using Eq. (2.13c), the
form-loss coefficient = 0.716. Using Eq. (2.12b), the head loss β„Žπ‘š34 = 0.657 m.
4. Headloss between points 4 and 6.
Using Eq. (2.4c), with 𝜈 = 1.012 Γ— 10βˆ’6
m2
/s, diameter 0.3, and discharge 0.1 m3
/s,
the Reynolds number = 419,459. With πœ€ = 0.25 mm using Eq. (2.6b), 𝑓 = 0.0197.
Thus, for pipe length 400 m, using Eq. (2.3b), head loss β„Žπ‘“ = 2.681 m.
5. Head loss at point 5 due to rotary valve (fully open).
For fully open valve 𝛼 = 0. Using Eq. (2.11), form-loss coefficient π‘˜π‘“ = 0 and using
Eq. (2.7b), the form loss β„Žπ‘š = 0.0 m.
6. Head loss at point 6 due to abrupt contraction.
For 𝐷 = 0.3 m and 𝑑 = 0.15 m, using Eq. (2.14b), the form-loss coefficient
π‘˜π‘“ = 0.5 1 βˆ’
0.15
0.3
2.35
= 0.402.
Using Eq. (2.12b), the form loss β„Žπ‘š = 0.369 m.
Example 2.2A continued
7. Head loss in pipe between points 6 and 7.
Pipe length = 20 m, pipe diameter = 0.15 m, and roughness height = 0.25 mm.
Reynolds number = 838,914 and pipe friction factor 𝑓 = 0.0227, head loss β„Žπ‘“67 =
4.930 m.
8. Head loss at point 7 (an abrupt expansion).
An abrupt expansion from 0.15 m pipe size to 0.30 m.
Using Eq. (2.14a), π‘˜π‘“ = 1 and using Eq. (2.12b), β„Žπ‘š = 0.918 m.
Head loss in pipe between points 7 and 8.
Pipe length = 100 m, pipe diameter = 0.30 m, and roughness height = 0.25 mm.
Reynolds number = 423,144 and pipe friction factor 𝑓 = 0.0197.
Head loss β„Žπ‘“78 = 0.670 m.
Head loss at outlet point 8 (confusor outlet).
Using Eq. (2.17), the form-loss coefficient
π‘˜π‘“
= 4.5
𝐷
𝑑
βˆ’ 3.5 = 4.5 Γ—
0.30
0.15
βˆ’ 3.5 = 5.5.
Using Eq. (2.12 b), β„Žπ‘š = 0.560 m.
Total head loss hL = 0.670 + 0.193 + 0.657 + 2.681 + 0.369 + 0 + 4.930 + 0.918 +
0.670 + 0.560 = 11.648 m: Thus, the residual pressure at the end of the pipe outlet =
50 - 11.648 = 38.352 m
Example 2.2B.
Design an cxpansion for the pipc diametcrs 1.0 m
and 2.0 m over a distance of 2 m for Fig. 2.9.
Solution
Calculating of optimal transition profile.
The geometry profile is 𝐷1 = 1.0 m, 𝐷2 = 2.0 m,
and 𝐿 = 2.0 m.
Substituting various values of π‘₯, the corresponding
values of 𝐷 and are tabulated in Table 2.3.
π‘₯ 𝐷 (optimal) 𝐷 (linear)
0.0 1.000 1.000
0.2 1.019 1.100
0.4 1.078 1.200
0.6 1.180 1.300
0.8 1.326 1.400
1.0 1.500 1.500
1.2 1.674 1.600
1.4 1.820 1.700
1.6 1.922 1.800
1.8 1.981 1.900
2.0 2.000 2.000
TABLE 2.3. Pipe Transition
Computations 𝒙 versus 𝑫
2.3. PIPE FLOW PROBLEMS
Nodal Head Problem
Discharge Problem
Diameter Problem
Analysis problem
Design and ansysis problem
Analysis problem
2.3.1.Nodal Head Problem
In the nodal head problem, the known quantities
are 𝐿, 𝐷, β„ŽπΏ, 𝑄, πœ€, 𝑣, and π‘˜π‘“. Using Eqs. (2.2b) and
(2.7b), the nodal head β„Ž2 (as shown in Fig. 2.1) is
obtained as
π’‰πŸ = π’‰πŸ + π’›πŸ βˆ’ π’›πŸ βˆ’ π’Œπ’‡ +
𝒇𝑳
𝑫
πŸ–π‘ΈπŸ
π…πŸπ’ˆπ‘«πŸ’
.
(2:20)
2.3.2. Discharge Problem
For a long pipeline, form losses can be neglected. Thus, in this case the known
quantities are 𝐿, 𝐷, β„Žπ‘“, πœ€, and 𝜈. Swamee and Jain (1976) gave the following
solution for turbulent flow through such a pipeline:
Equation (2.21a) is exact. For
laminar flow,
𝑸 = βˆ’πŸŽ. πŸ—πŸ”πŸ“π‘«πŸ
π’ˆπ‘« 𝒉𝒇 𝑳 π₯ 𝒏
𝜺
πŸ‘. πŸ•π‘«
+
𝟏. πŸ•πŸ–π’—
𝑫 π’ˆπ‘« 𝒉𝒇 𝑳
𝑸 =
π…π’ˆπ‘«πŸ’π’‰π’‡
πŸπŸπŸ–π’—π‘³
.
(2:21a)
(2:21b)
For laminar
flow
Swamee and Swamee (2008) gave the
following equation for pipe discharge that
is valid under laminar, transition,
and turbulent flow conditions:
𝑸 = π‘«πŸ
π’ˆπ‘« 𝒉𝒇 𝑳
πŸπŸπŸ–π’—
𝝅𝑫 π’ˆπ‘« 𝒉𝒇 𝑳
πŸ’
𝟏 +𝟏. πŸπŸ“πŸ‘
πŸ’πŸπŸ“π’—
𝑫 π’ˆπ‘« 𝒉𝒇 𝑳
πŸ–
βˆ’ π₯ 𝒏
𝜺
πŸ‘. πŸ•π‘«
+
𝟏. πŸ•πŸ•πŸ“π’—
𝑫 π’ˆπ‘« 𝒉𝒇 𝑳
βˆ’πŸ’ βˆ’πŸŽ.πŸπŸ“
Equation (2.21c) is almost exact as the maximum error in the equation is 0.1%.
𝑸 =
(2:21c)
2.3.3. Diameter Problem
In this problem. the known quantities are L.h_f,Ξ΅,Q. and v. For a pumping main,
head loss is not known, and one has to select the optimal value of head loss
by minimizing the cost. This has been dealt with in Chapter 6. However, for
turbulent flow in a long gravity main, Swamee and Jain (1976) obtained the
following solution for the pipe diameter:
𝑫 = 𝟎. πŸ”πŸ” 𝜺𝟏.πŸπŸ“
π‘³π‘ΈπŸ
π’ˆπ’‰π’‡
πŸ’.πŸ•πŸ“
+ π’—π‘ΈπŸ—.πŸ’
𝑳
π’ˆπ’‰π’‡
πŸ“.𝟐 𝟎.πŸŽπŸ’
(2:22a)
In general, the errors involved in Eq. (2.22a) are less than 1.5%. However, the
maximum error occurring near transition range is about 3%. For laminar flow, the
Hagen Poiseuille equation gives the diameter as
𝑫 =
πŸπŸπŸ–π’—π‘Έπ‘³
π…π’ˆπ’‰π’‡
𝟎.πŸπŸ–
(2:22b)
2.3.3. Diameter Problem
Swamee and Swamee (2008) gave the following equation for pipe diameter that
is valid under laminar, transition, and turbulent flow conditions
Equation (2.22c) yields 𝐷 within 2.75%. However, close to transition range,
the error is around 4%.
𝑫 = 𝟎. πŸ”πŸ” πŸπŸπŸ’. πŸ•πŸ“
𝒗𝑳𝑸
π’ˆπ’‰π’‡
πŸ”.πŸπŸ“
+ 𝜺𝟏.πŸπŸ“
π‘³π‘ΈπŸ
π’ˆπ’‰π’‡
πŸ’.πŸ•πŸ“
+ π’—π‘ΈπŸ—.πŸ’
𝑳
π’ˆπ’‰π’‡
πŸ“.𝟐 𝟎.πŸŽπŸ’
(2:22c)
Example 2.3.
As shown in Fig. 2.16, a discharge of 0.1 m3 /s flows through a CI pipe main of
1000 m in length having a pipe diameter 0.3 m.
A sluice valve of 0.3 m size is placed close to point B.
The elevations of points A and B are 10 m and 5 m, respectively. Assume water
temperature as 20C. Calculate:
(A) Terminal pressure h2 at point B and head loss in the pipe if terminal
pressure h1 at point A is 25 m.
(B) The discharge in the pipe if the head loss is 10 m.
(C) The CI gravity main diameter if the head
loss in the pipe is 10 m and a
discharge of 0.1 m3 /s flows in the pipe.
solution
A) The terminal pressure h2 at point B can be calculated using Eq. (2.20). The
friction factor f can be calculated applying Eq. (2.6a) and the roughness height of
CI pipe = 0.25 mm is obtained from Table 2.1. The form-loss coefficient for sluice
valve from Table 2.2 is 0.15. The viscosity of water at 208C can be calculated
using Eq. (2.4b). The coefficient of surface resistance depends on the Reynolds
number R of the flow:
𝑹 =
πŸ’π‘Έ
π’Žπ’—π‘«
= πŸ’πŸπŸ—, πŸ’πŸ“πŸ—.
Thus, substituting values in Eq. (2.6a), the friction factor
𝒇 =
πŸ”πŸ’
𝐑
πŸ–
+ πŸ—. πŸ“ π₯ 𝒏
𝜺
πŸ‘. πŸ•π‘«
+
πŸ“. πŸ•πŸ’
π‘πŸŽ.πŸ—
βˆ’
πŸπŸ“πŸŽπŸŽ
𝐑
πŸ” βˆ’πŸπŸ” 𝟎.πŸπŸπŸ“
= 𝟎. πŸŽπŸπŸ—πŸ•
solution
Using Eq- (2.20), the terminal head β„Ž2 at point 𝐡 is
π’‰πŸ = π’‰πŸ + π’›πŸ βˆ’ π’›πŸ βˆ’ π’Œπ’‡ +
𝒇𝑳
𝑫
πŸ–π‘ΈπŸ
π…πŸπ’ˆπ‘«πŸ’
π’‰πŸ = πŸπŸ“ + 𝟏𝟎 βˆ’ πŸ“ βˆ’ 𝟎. πŸπŸ“ +
𝟎. πŸŽπŸπŸ—πŸ• Γ— 𝟏𝟎𝟎𝟎
𝟎. πŸ‘
πŸ– Γ— 𝟎. 𝟏𝟐
πŸ‘. πŸπŸ’πŸπŸ“πŸ—πŸ Γ— πŸ—. πŸ–πŸ Γ— 𝟎. πŸ‘πŸ“
π’‰πŸ =
= πŸ‘πŸŽ βˆ’ (𝟎. πŸŽπŸπŸ“ + πŸ”. πŸ•πŸŽπŸ’) = πŸπŸ‘. πŸπŸ–πŸ 𝐦.
= 𝑸 = βˆ’πŸŽ. πŸ—πŸ”πŸ“π‘«πŸ
π’ˆπ‘« 𝒉𝒇 𝑳 π₯ 𝒏
𝜺
πŸ‘. πŸ•π‘«
+
𝟏. πŸ•πŸ–π’—
𝑫 π’ˆπ‘«π’‰ 𝑳
= = βˆ’πŸŽ. πŸ—πŸ”πŸ“ Γ— 𝟎. πŸ‘πŸ
πŸ—. πŸ–πŸ Γ— (𝟏𝟎 𝟏𝟎𝟎𝟎 π₯ 𝒏
𝟎. πŸπŸ“ Γ— πŸπŸŽβˆ’πŸ‘
πŸ‘. πŸ• Γ— 𝟎. πŸ‘
. =
= = 𝟎. πŸπŸπŸ‘ π’ŽπŸ‘
𝒔
solution
(B) If the total head loss in the pipe is predecided equal to 10 m, the discharge in
Cl pipe of size 0.3 m can be calculated using Eq. ( 2.21a ):
+
𝟏. πŸ•πŸ– Γ— 𝟏. 𝟎𝟏𝟐 Γ— πŸπŸŽβˆ’πŸ”
𝟎. πŸ‘ )
πŸ—. πŸ–πŸ Γ— 𝟎. πŸ‘ Γ— (𝟏𝟎 𝟏𝟎𝟎𝟎
solution
(C) Using Eq. (2.22a), the gravity main diameter for preselected head loss of 10 m
and known pipe discharge 0.1 m3
/s is
𝑫 = 𝟎. πŸ”πŸ” 𝜺𝟏.πŸπŸ“
π‘³π‘ΈπŸ
π’ˆπ’‰π’‡
πŸ’.πŸ•πŸ“
+ π’—π‘ΈπŸ—.πŸ’
𝑳
π’ˆπ’‰π’‡
πŸ“.𝟐 𝟎.πŸŽπŸ’
= 𝟎. πŸ”πŸ” 𝟎. πŸŽπŸŽπŸŽπŸπŸ“πŸ.πŸπŸ“
𝟏𝟎𝟎𝟎 Γ— 𝟎. 𝟏𝟐
πŸ—. πŸ–πŸ Γ— 𝟏𝟎
πŸ’.πŸ•πŸ“
+ 𝟏. 𝟎𝟏𝟐 Γ— πŸπŸŽβˆ’πŸ”
= Γ— 𝟎. πŸπŸ—.πŸ’
𝟏𝟎𝟎𝟎
πŸ—. πŸ–πŸ Γ— 𝟏𝟎
πŸ“.𝟐 𝟎.πŸŽπŸ’
= =
Also, if head loss is considered = 6.72 m, the pipe diameter is 0.306 m and flow is
0.1 m3
/s.
= 𝟎. πŸπŸ–πŸ’ 𝐦.
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pipe lines lec 2.pptx

  • 1. BASIC PRINCIPLES OF PIPE FLOW Dr/ Ahmed safwat Amir Ashraf sayed
  • 2. Pipe Flow Under Siphon Action A pipeline that rises above its hydraulic gradient line is termed a siphon head loss due to friction is large and the form losses can be neglected. Thus, the hydraulic gradient line is a straight line that joins the water surfaces at points A and B. Pipe Flow Under Siphon Action
  • 3. If the hydraulic gradient line is above the centerline of pipe: the water pressure is above atmospheric. if it is below the centerline of the pipe: the pressure is below atmospheric.
  • 4. 1- points b and c, the water pressure is atmospheric. 2- between b and c it is less than atmospheric. 3- At the highest point e, the water pressure is the lowest. 4- If the pressure head at point e is less than - 2.5 m, the water starts vaporizing.
  • 5. Example 2.2A. A pumping system with different pipe fittings is shown in Fig. 2.15. Calculate residual pressure head at the end of the pipe outlet if the pump is generating an input head of 50 m at 0.1 m3 /s discharge. The CI pipe diameter 𝐷 is 0.3 m. The contraction size at point 3 is 0.15 m; pipe size between points 6 and 7 is 0.15 m; and confusor outlet size 𝑑 = 0.15 m. The rotary valve at point 5 is fully open. Consider the following pipe lengths between points: Points 1 and 2 = 100 m, points 2 and 3 = 0.5 m; and points 3 and 4 = 0.5 m Points 4 and 6 = 400 m, points 6 and 7 = 20 m; and points 7 and 8 = 100 m
  • 6. Head loss between points 1 and 2. Pipe length 100 m, flow 0.1 m3 /s, and pipe diameter 0.3 m. Using Eq. (2.4b), 𝑣 for 20∘C is 1.012 Γ— 10βˆ’6 m2/s, similarly using Eq. (2.4c), Reynolds number 𝐑 = 419,459. Using Table 2.1 for CI pipes, πœ€ is 0.25 𝐦𝐦. The friction factor 𝑓 is calculated using Eq. (2.6b) = 0.0197. Using Eq. 2.3b the head loss β„Žπ‘“12 in pipe (1 βˆ’ 2) is π’‰π’‡πŸπŸ = πŸ–π’‡π‘³π‘ΈπŸ π…πŸπ’ˆπ‘«πŸ“ = πŸ– Γ— 𝟎. πŸŽπŸπŸ—πŸ• Γ— 𝟏𝟎𝟎 Γ— 𝟎. 𝟏𝟐 πŸ‘. πŸπŸ’πŸπŸ“πŸ—πŸ Γ— πŸ—. πŸ–πŸ Γ— 𝟎. πŸ‘πŸ“ = 𝟎. πŸ”πŸ•πŸŽ 𝐦 Solution
  • 7. 2. Head loss between points 2 and 3 (a contraction transition). For 𝐷 = 0.3, 𝑑 = 0.15, and transition length = 0.5 m, the contraction angle 𝛼𝑐 can be calculated using Eq. (2.13b): 𝛼𝑐 = 2tanβˆ’1 𝐷1 βˆ’ 𝐷2 2𝐿 = 2tanβˆ’1 0.3 βˆ’ 0.15 2 Γ— 0.5 = 0.298 radians . Using Eq. (2.13a), the form-loss coefficient is π‘˜π‘“ = 0.315𝛼𝑐 1/3 = 0.315 Γ— 0.2981/3 = 0.210 Using Eq. (2.12b), the head loss β„Žπ‘š23 = 0.193 m.
  • 8. 3. Head loss between points 3 and 4 (an expansion transition). For 𝑑 = 0.15, 𝐷 = 0.3, the expansion ratio π‘Ÿ = 2, and transition length = 0.5 m. Using Eq. (2.13d), the expansion angle 𝛼𝑒 = 0.298 radians. Using Eq. (2.13c), the form-loss coefficient = 0.716. Using Eq. (2.12b), the head loss β„Žπ‘š34 = 0.657 m. 4. Headloss between points 4 and 6. Using Eq. (2.4c), with 𝜈 = 1.012 Γ— 10βˆ’6 m2 /s, diameter 0.3, and discharge 0.1 m3 /s, the Reynolds number = 419,459. With πœ€ = 0.25 mm using Eq. (2.6b), 𝑓 = 0.0197. Thus, for pipe length 400 m, using Eq. (2.3b), head loss β„Žπ‘“ = 2.681 m.
  • 9. 5. Head loss at point 5 due to rotary valve (fully open). For fully open valve 𝛼 = 0. Using Eq. (2.11), form-loss coefficient π‘˜π‘“ = 0 and using Eq. (2.7b), the form loss β„Žπ‘š = 0.0 m. 6. Head loss at point 6 due to abrupt contraction. For 𝐷 = 0.3 m and 𝑑 = 0.15 m, using Eq. (2.14b), the form-loss coefficient π‘˜π‘“ = 0.5 1 βˆ’ 0.15 0.3 2.35 = 0.402. Using Eq. (2.12b), the form loss β„Žπ‘š = 0.369 m.
  • 10. Example 2.2A continued 7. Head loss in pipe between points 6 and 7. Pipe length = 20 m, pipe diameter = 0.15 m, and roughness height = 0.25 mm. Reynolds number = 838,914 and pipe friction factor 𝑓 = 0.0227, head loss β„Žπ‘“67 = 4.930 m. 8. Head loss at point 7 (an abrupt expansion). An abrupt expansion from 0.15 m pipe size to 0.30 m. Using Eq. (2.14a), π‘˜π‘“ = 1 and using Eq. (2.12b), β„Žπ‘š = 0.918 m.
  • 11. Head loss in pipe between points 7 and 8. Pipe length = 100 m, pipe diameter = 0.30 m, and roughness height = 0.25 mm. Reynolds number = 423,144 and pipe friction factor 𝑓 = 0.0197. Head loss β„Žπ‘“78 = 0.670 m. Head loss at outlet point 8 (confusor outlet). Using Eq. (2.17), the form-loss coefficient π‘˜π‘“ = 4.5 𝐷 𝑑 βˆ’ 3.5 = 4.5 Γ— 0.30 0.15 βˆ’ 3.5 = 5.5. Using Eq. (2.12 b), β„Žπ‘š = 0.560 m. Total head loss hL = 0.670 + 0.193 + 0.657 + 2.681 + 0.369 + 0 + 4.930 + 0.918 + 0.670 + 0.560 = 11.648 m: Thus, the residual pressure at the end of the pipe outlet = 50 - 11.648 = 38.352 m
  • 12. Example 2.2B. Design an cxpansion for the pipc diametcrs 1.0 m and 2.0 m over a distance of 2 m for Fig. 2.9. Solution Calculating of optimal transition profile. The geometry profile is 𝐷1 = 1.0 m, 𝐷2 = 2.0 m, and 𝐿 = 2.0 m. Substituting various values of π‘₯, the corresponding values of 𝐷 and are tabulated in Table 2.3. π‘₯ 𝐷 (optimal) 𝐷 (linear) 0.0 1.000 1.000 0.2 1.019 1.100 0.4 1.078 1.200 0.6 1.180 1.300 0.8 1.326 1.400 1.0 1.500 1.500 1.2 1.674 1.600 1.4 1.820 1.700 1.6 1.922 1.800 1.8 1.981 1.900 2.0 2.000 2.000 TABLE 2.3. Pipe Transition Computations 𝒙 versus 𝑫
  • 13. 2.3. PIPE FLOW PROBLEMS Nodal Head Problem Discharge Problem Diameter Problem Analysis problem Design and ansysis problem Analysis problem
  • 14. 2.3.1.Nodal Head Problem In the nodal head problem, the known quantities are 𝐿, 𝐷, β„ŽπΏ, 𝑄, πœ€, 𝑣, and π‘˜π‘“. Using Eqs. (2.2b) and (2.7b), the nodal head β„Ž2 (as shown in Fig. 2.1) is obtained as π’‰πŸ = π’‰πŸ + π’›πŸ βˆ’ π’›πŸ βˆ’ π’Œπ’‡ + 𝒇𝑳 𝑫 πŸ–π‘ΈπŸ π…πŸπ’ˆπ‘«πŸ’ . (2:20)
  • 15. 2.3.2. Discharge Problem For a long pipeline, form losses can be neglected. Thus, in this case the known quantities are 𝐿, 𝐷, β„Žπ‘“, πœ€, and 𝜈. Swamee and Jain (1976) gave the following solution for turbulent flow through such a pipeline: Equation (2.21a) is exact. For laminar flow, 𝑸 = βˆ’πŸŽ. πŸ—πŸ”πŸ“π‘«πŸ π’ˆπ‘« 𝒉𝒇 𝑳 π₯ 𝒏 𝜺 πŸ‘. πŸ•π‘« + 𝟏. πŸ•πŸ–π’— 𝑫 π’ˆπ‘« 𝒉𝒇 𝑳 𝑸 = π…π’ˆπ‘«πŸ’π’‰π’‡ πŸπŸπŸ–π’—π‘³ . (2:21a) (2:21b) For laminar flow
  • 16. Swamee and Swamee (2008) gave the following equation for pipe discharge that is valid under laminar, transition, and turbulent flow conditions: 𝑸 = π‘«πŸ π’ˆπ‘« 𝒉𝒇 𝑳 πŸπŸπŸ–π’— 𝝅𝑫 π’ˆπ‘« 𝒉𝒇 𝑳 πŸ’ 𝟏 +𝟏. πŸπŸ“πŸ‘ πŸ’πŸπŸ“π’— 𝑫 π’ˆπ‘« 𝒉𝒇 𝑳 πŸ– βˆ’ π₯ 𝒏 𝜺 πŸ‘. πŸ•π‘« + 𝟏. πŸ•πŸ•πŸ“π’— 𝑫 π’ˆπ‘« 𝒉𝒇 𝑳 βˆ’πŸ’ βˆ’πŸŽ.πŸπŸ“ Equation (2.21c) is almost exact as the maximum error in the equation is 0.1%. 𝑸 = (2:21c)
  • 17. 2.3.3. Diameter Problem In this problem. the known quantities are L.h_f,Ξ΅,Q. and v. For a pumping main, head loss is not known, and one has to select the optimal value of head loss by minimizing the cost. This has been dealt with in Chapter 6. However, for turbulent flow in a long gravity main, Swamee and Jain (1976) obtained the following solution for the pipe diameter: 𝑫 = 𝟎. πŸ”πŸ” 𝜺𝟏.πŸπŸ“ π‘³π‘ΈπŸ π’ˆπ’‰π’‡ πŸ’.πŸ•πŸ“ + π’—π‘ΈπŸ—.πŸ’ 𝑳 π’ˆπ’‰π’‡ πŸ“.𝟐 𝟎.πŸŽπŸ’ (2:22a)
  • 18. In general, the errors involved in Eq. (2.22a) are less than 1.5%. However, the maximum error occurring near transition range is about 3%. For laminar flow, the Hagen Poiseuille equation gives the diameter as 𝑫 = πŸπŸπŸ–π’—π‘Έπ‘³ π…π’ˆπ’‰π’‡ 𝟎.πŸπŸ– (2:22b) 2.3.3. Diameter Problem
  • 19. Swamee and Swamee (2008) gave the following equation for pipe diameter that is valid under laminar, transition, and turbulent flow conditions Equation (2.22c) yields 𝐷 within 2.75%. However, close to transition range, the error is around 4%. 𝑫 = 𝟎. πŸ”πŸ” πŸπŸπŸ’. πŸ•πŸ“ 𝒗𝑳𝑸 π’ˆπ’‰π’‡ πŸ”.πŸπŸ“ + 𝜺𝟏.πŸπŸ“ π‘³π‘ΈπŸ π’ˆπ’‰π’‡ πŸ’.πŸ•πŸ“ + π’—π‘ΈπŸ—.πŸ’ 𝑳 π’ˆπ’‰π’‡ πŸ“.𝟐 𝟎.πŸŽπŸ’ (2:22c)
  • 20. Example 2.3. As shown in Fig. 2.16, a discharge of 0.1 m3 /s flows through a CI pipe main of 1000 m in length having a pipe diameter 0.3 m. A sluice valve of 0.3 m size is placed close to point B. The elevations of points A and B are 10 m and 5 m, respectively. Assume water temperature as 20C. Calculate: (A) Terminal pressure h2 at point B and head loss in the pipe if terminal pressure h1 at point A is 25 m. (B) The discharge in the pipe if the head loss is 10 m. (C) The CI gravity main diameter if the head loss in the pipe is 10 m and a discharge of 0.1 m3 /s flows in the pipe.
  • 21.
  • 22. solution A) The terminal pressure h2 at point B can be calculated using Eq. (2.20). The friction factor f can be calculated applying Eq. (2.6a) and the roughness height of CI pipe = 0.25 mm is obtained from Table 2.1. The form-loss coefficient for sluice valve from Table 2.2 is 0.15. The viscosity of water at 208C can be calculated using Eq. (2.4b). The coefficient of surface resistance depends on the Reynolds number R of the flow: 𝑹 = πŸ’π‘Έ π’Žπ’—π‘« = πŸ’πŸπŸ—, πŸ’πŸ“πŸ—. Thus, substituting values in Eq. (2.6a), the friction factor 𝒇 = πŸ”πŸ’ 𝐑 πŸ– + πŸ—. πŸ“ π₯ 𝒏 𝜺 πŸ‘. πŸ•π‘« + πŸ“. πŸ•πŸ’ π‘πŸŽ.πŸ— βˆ’ πŸπŸ“πŸŽπŸŽ 𝐑 πŸ” βˆ’πŸπŸ” 𝟎.πŸπŸπŸ“ = 𝟎. πŸŽπŸπŸ—πŸ•
  • 23. solution Using Eq- (2.20), the terminal head β„Ž2 at point 𝐡 is π’‰πŸ = π’‰πŸ + π’›πŸ βˆ’ π’›πŸ βˆ’ π’Œπ’‡ + 𝒇𝑳 𝑫 πŸ–π‘ΈπŸ π…πŸπ’ˆπ‘«πŸ’ π’‰πŸ = πŸπŸ“ + 𝟏𝟎 βˆ’ πŸ“ βˆ’ 𝟎. πŸπŸ“ + 𝟎. πŸŽπŸπŸ—πŸ• Γ— 𝟏𝟎𝟎𝟎 𝟎. πŸ‘ πŸ– Γ— 𝟎. 𝟏𝟐 πŸ‘. πŸπŸ’πŸπŸ“πŸ—πŸ Γ— πŸ—. πŸ–πŸ Γ— 𝟎. πŸ‘πŸ“ π’‰πŸ = = πŸ‘πŸŽ βˆ’ (𝟎. πŸŽπŸπŸ“ + πŸ”. πŸ•πŸŽπŸ’) = πŸπŸ‘. πŸπŸ–πŸ 𝐦.
  • 24. = 𝑸 = βˆ’πŸŽ. πŸ—πŸ”πŸ“π‘«πŸ π’ˆπ‘« 𝒉𝒇 𝑳 π₯ 𝒏 𝜺 πŸ‘. πŸ•π‘« + 𝟏. πŸ•πŸ–π’— 𝑫 π’ˆπ‘«π’‰ 𝑳 = = βˆ’πŸŽ. πŸ—πŸ”πŸ“ Γ— 𝟎. πŸ‘πŸ πŸ—. πŸ–πŸ Γ— (𝟏𝟎 𝟏𝟎𝟎𝟎 π₯ 𝒏 𝟎. πŸπŸ“ Γ— πŸπŸŽβˆ’πŸ‘ πŸ‘. πŸ• Γ— 𝟎. πŸ‘ . = = = 𝟎. πŸπŸπŸ‘ π’ŽπŸ‘ 𝒔 solution (B) If the total head loss in the pipe is predecided equal to 10 m, the discharge in Cl pipe of size 0.3 m can be calculated using Eq. ( 2.21a ): + 𝟏. πŸ•πŸ– Γ— 𝟏. 𝟎𝟏𝟐 Γ— πŸπŸŽβˆ’πŸ” 𝟎. πŸ‘ ) πŸ—. πŸ–πŸ Γ— 𝟎. πŸ‘ Γ— (𝟏𝟎 𝟏𝟎𝟎𝟎
  • 25. solution (C) Using Eq. (2.22a), the gravity main diameter for preselected head loss of 10 m and known pipe discharge 0.1 m3 /s is 𝑫 = 𝟎. πŸ”πŸ” 𝜺𝟏.πŸπŸ“ π‘³π‘ΈπŸ π’ˆπ’‰π’‡ πŸ’.πŸ•πŸ“ + π’—π‘ΈπŸ—.πŸ’ 𝑳 π’ˆπ’‰π’‡ πŸ“.𝟐 𝟎.πŸŽπŸ’ = 𝟎. πŸ”πŸ” 𝟎. πŸŽπŸŽπŸŽπŸπŸ“πŸ.πŸπŸ“ 𝟏𝟎𝟎𝟎 Γ— 𝟎. 𝟏𝟐 πŸ—. πŸ–πŸ Γ— 𝟏𝟎 πŸ’.πŸ•πŸ“ + 𝟏. 𝟎𝟏𝟐 Γ— πŸπŸŽβˆ’πŸ” = Γ— 𝟎. πŸπŸ—.πŸ’ 𝟏𝟎𝟎𝟎 πŸ—. πŸ–πŸ Γ— 𝟏𝟎 πŸ“.𝟐 𝟎.πŸŽπŸ’ = = Also, if head loss is considered = 6.72 m, the pipe diameter is 0.306 m and flow is 0.1 m3 /s. = 𝟎. πŸπŸ–πŸ’ 𝐦.