Chapter (3)
Water Flow in Pipes
Water Flow in Pipes
Hydraulics
Bernoulli Equation
Pρg + 2gv + z = ρgP + 2gv + z h + h + h ∑ h h =Major Losses+Minor Losses Recall fluid mechanics course, the Bernoulli equation is:
Here, we want to study how to calculate the total losses
:
Major Losses
Occurs mainly due to the pipe friction and viscous dissipation in the flowing water. The major head loss is termed by .
h
There are several formulas have been developed to calculate major losses: 1. Darcy-Weisbach Formula: Is the most popular formula used to calculate major losses and it has the following form:
L V Q Q 8 f L Q h = f D 2g , V = A = π4 D → h = πg D LD == ldiength ngametth ofoerefrpiofpeppie mmpepe mm V2g = velocity head head mm Qf ==frflioctitiwonornataftaectmoror /s : 64 f = R smooth pipe → 0.316 f = R.
Friction Factor
For laminar flow (R e<2000) the friction factor depends only on R e :
5 If the pipe is smooth (e= 0) and the flow is turbulent with (4000
Page (2)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Water Flow in Pipes
Hydraulics
For turbulent flow (R e > 4000) the friction factor can be founded by Moody diagram. To use Moody diagram you need the followings: The Reynolds number: R e The
relative roughness:
ρ V D μ V D R = μ b ut ν = ρ → R = ν μ=dynami c vi s cosi t y Pa. s o r kg/m. s νV == kimean nmeeanmaivelvielcovicistcostcoy stinythme pi/pepe nema D=pipe diameter Reynolds Number: R e
Note:
Kinematic viscosity depends on the fluid temperature and can be calculated from the following formula:
− 497×10 ν = T+42.5. T: is fluid temperature in Celsius eandmm mmit depends = Roughne Roughnessss height ght internal roughness of the pipee Relative Roughness:
mainly on the pipe material. Table (3.1) in slides slides of Dr.Khalil shows the the value of e for different pipe materials. The following figure exhibits Moody diagram:
Page (3)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Water Flow in Pipes
Hydraulics
Note: Moody diagram is a graphical representation of Colebrook-White formula:
1√ f = 2 log e/D3.7 + R2.5√ 1f
Empirical Formulas for Friction Head Losses
These formulas gives exact value for friction head losses and each formula extensively used in a specific field, for example, Hazen-Williams formula is used extensively in water supply systems and most software’s (like WaterCAD) used it in analyzing and design of water networks. However, Manning equation us extensively used in open channel, and in designing of waste water networks. See these formulas from the slides of Dr.Khalil.
Page (4)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Water Flow in Pipes
Hydraulics
Minor Losses Occurs due to the change of the velocity of the flowing fluid in the magnitude or in the direction. So, the minor losses at:
Valves.
Tees.
Bends.
Contraction and Expansion.
h V h = K 2g K =minor losses coefficient and it depends on the type of fitting The minor losses are termed by
and have a common form:
Minor Losses Formulas
Minor Losses
h = K h = K ( − h = K ) h = K
Entrance of a pipe:
Sudden Contraction :
h = K hV= KV or h = 2g (−) = h = K
Exit of a a pipe:
Sudden Expansion :
Gradual Enlargement :
Gradual Contraction:
Bends in pipes:
Pipe Fittings :
The value of (K) for each fitting can be estimated from tables exist in slides of Dr.Khalil. You must save the following values of K: For sharp edge entrance: K = 0.5 For all types of exists: K = 1.
Page (5)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Water Flow in Pipes
Hydraulics
Problems 1. In the shown figure below, the smaller tank is 50m in diameter. Find the flow rate, Q. Assume laminar flow and neglect minor losses. Take
μ=1. 2 ×10− kg/m. s ρ=788 kg/m f= − ρ V D 788×2×10 R = μ = 1.2×10− ×V →R64 =1313.6433 V subst0.0it487ute in f f= R = 1313.33 V = V ≫1 Solution
1
For laminar
2
Now by applying Bernoulli’s equation from the free surface of upper to lower reservoir (Points 1 and 2).
Pρg + 2gv +z = ρgP + 2gv +z +h P =P =V =V =0.0 → 0+0+ 0.4+0.6 =0+0+0+h →h =1m h =1m=h i n t h e pi p e t h at t r ansport f l u i d f r om reservoi r 1 t o 2 L V h =f D2g L=0.4+0.8=1.2m , D=0.002m 1=f 0.1.020219.V62≫2→Substi t ute f r o m1i n 2→ 0. 0 487 1. 2 V 1= V ×π 0.00219.62→V=0.671 m/s Q=A×V= 4 ×0.002 ×0.671=2.1×10− m/s . Assume the datum at point (2)
✓
Page (6)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Water Flow in Pipes
Hydraulics
2.
℃ ν=1.003×10− m/s 10 V/2
A uniform pipeline, 5000m long, 200mm in diameter and roughness size of 0.03mm, conveys water at 20 between two reservoirs as shown in the figure. The difference in water level between the reservoirs is 50m. Include all minor losses in your calculations, determine the discharge. Note: the valve produces a head loss of and the entrance to and exit from the pipe are sharp. 1
2
Solution
Pρg + 2gv +z = ρgP + 2gv +z +h P =P =V =V =0.0 point 2 → 0+0+50=0+0+0+h →h =50m h =h + h =50 L V 5000 V h =f D2g=h =f× 0.2 19.62=1274.21 f V V h =K 2g For sharp edge entrance, the value ofK =0.5→ Applying Bernoulli’s equation between the two reservoirs
Assume the datum at lower reservoir
Major Losses:
Minor losses: For Entrance:
Page (7)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Water Flow in Pipes
Hydraulics
V h =0. 5 × 19.62 =0.0255 V V h =K 2g For exit: K =1→ h =1× 19.V62 =0.051 V V h =10× 2g =0.51V h = 0.0255+0.051+0.51V =0.5865 V 50 h =1274. 2 1 f V +0. 5 865 V =50→V = >1 1274. 2 1 f + 0. 5 865 eD = 0.20003 =0.00015 f For Exit:
For Valve:
In all problems like this (flow rate or velocity is unknown), the best initial value for can be found as following:
Draw a horizontal line (from left to right) starts from the value of till
intercept with the vertical axis of the Moody chart and the initial f value is the intercept as shown in the following figure:
Page (8)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Water Flow in Pipes
Hydraulics
f = 0. 0 128 50 Substitute in Eq.1→V = 1274.21×0.0128+0.5865 =2.96 →V=1. V72Dm/s 1.72×0.2 →R = ν = 1.003×10− =3. 4 ×10 and De =0.00015→Moody f=0.016→ Substitute in Eq. 1 →V=1.54→R =3. 1 ×10 →Moody→f≅0.016 V=1. 5 4 m/s Q=A×V= ×0.2 ×1.54=0.0483 m/s . So as shown in the above figure, the initial value of
So, the velocity is
✓
3.
The pipe shown in the figure below contains water flowing at a flow rate of 3 0.0065 m /s. The difference in elevation between points A and B is 11m. For the pressure measurement shown, a) What is the direction of the flow? b) What is the total head loss between points A and B? c) What is the diameter of the pipe? Given data:
ν=10− m/s ,e=0.015mm , Pipe length=50m ,1atm=10 Pa.
Page (9)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Water Flow in Pipes
Hydraulics
Solution a) Direction of flow?? To know the direction of flow, we calculate the total head at each point, and then the fluid will moves from the higher head to lower head. Assume the datum is at point A: Total head at point A:
Pρg + v2g +z = 2.9810 5×10 + v2g +0=25. 4 84+ 2gv + v +11=31. 3 9+ v Pρg + 2gv +z = 2×10 9810 2g 2g V =V Since there is the same diameter and flow at A and B . Pρg + 2gv +z = ρgP + 2gv +z + hL 31.39+ =25.484+ + ∑ h , But V =V → ∑ h =5.9 m . h =hL +Vh =5.9m no minor losses→h =h =5.9m h =f D2g 5. 9 =f 50D×19.V62 − Q 0. 0 065 0. 0 0827 6. 8 5×10 V= A = π4 ×D = D →V = D → − 50 6. 8 5×10 5. 9 =f D × 19.62 D →5.9=f× 0.000174 D Total head at point BA:
But,
So, the total head at B is larger than total head at A>> the water is flowing from B (upper) to A (lower) ✓ b) Head loss between A and B?? Apply Bernoulli’s equation between B and A (from B to A)
✓
c) Diameter of the pipe??
Page (10)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Water Flow in Pipes
Hydraulics
→D =2.0.0082796×10× D− f →D= 2.96×10− ×f →D=0.124 f 0. 0 0827 D R = 10− = 10− D f 0. 0 0827 →D=0. 1 24× 0. 0 2 =0. 0 567 m→R = 10− ×0.0567 =1. 4 6×10 → De = 0.56.0157 =0.00026→Moody→f≅0. 0 19 0. 0 0827 →D=0. 1 24× 0. 0 19 =0. 0 56 m→R = 10− ×0.056 =1. 4 7×10 → De = 0.56015 =0.00027→Moody→f≅0.019 .
Now, assume the initial value for is 0.02 (random guess)
So, the diameter of the pipe is 0.056 m = 56 mm ✓
4. In the shown figure , the connecting pipe is commercial steel 6 cm in diameter having a roughness height of 0.045mm. Determine the direction of -3 flow, and then calculate the flow rate. The fluid is water (μ = 1 × 10 kg/m.s). Neglect Minor losses.
Solution
Page (11)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Water Flow in Pipes
Hydraulics Assume the datum is at point 1: Total head at point A:
+0+0=20.387 m Pρg + 2gv +z = 200×10 9810 Pρg + 2gv +z =0+0+15=15 m . Pρg + 2gv +z = ρgP + 2gv +z +h 20.387=15+h →h =5.38 m h =h + h =5.38 m h =0given→h =5.38 m L V 50 V 5. 3 8=f D2g=h =f×0.0619.62=42.47 f V →V =ρ V0.D1f26 1000×V×0. →Eq.1 06 R = μ = 1×10− =60,000V→Eq.2 eD = 0.60045 =0.00075 Assume ful y turbulent flow→f=0.018→Sub. in 1→ V=2.64 m/s→Sub. in 2→R =1.58×10 → ⃑Moody →f=0.02→V=2.51→R =1. 5 ×10 → ⃑Moody →f≅0.02→V=2. 5 1 m/s Q=A×V= ×0.06 ×2.51=0.0071 m/s . Total head at point 2:
So, the total head at 1 is larger than total head at 2>> the water is flowing from 1 to 2 ✓ Applying Bernoulli’s equation between 1 and 2:
✓
Page (12)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Water Flow in Pipes
Hydraulics
5. Two reservoirs having a constant difference in water level of 66 m are connected by a pipe having a diameter of 225 mm and a length of 4km. The pipe is tapped at point C which is located 1.6km from the upper reservoir, 3 and water drawn off at the rate of 0.0425 m /s. Determine the flow rate at which water enters the lower reservoir. Use a friction coefficient of f = 0.036 for all pipes. Use the following K values for the minor losses: K ent = 0.5, K exit = 1, K v = 5
Solution
Pρg + v2g +z = ρgP + 2gv +z +h P =P =V =V =0.0 → 0+0+66=0+0+0+h →h =66m h =h + h =66
Applying Bernoulli’s equation between points A and E:
Assume the datum at point E
Major Losses: Since the pipe is tapped, we will divide the pipe into two parts with the same diameter ; (1) 1.6 km length from B to C, and (2) 2.4 km length from C to D.
Page (13)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Water Flow in Pipes
Hydraulics
L V h =f D2g 1600 V h, =0.036×0.22519.62=13.05 V 2400 V h, =0.036× 0.22519. 62=19.6 V h =13.05 V +19.6 V For part (1)
For part (2)
Minor losses: For Entrance: (due to part 1)
V h =K 2g K =0.5→ h =0. 5 × 19.V62 =0.0255 V V h =K 2g K =1→ h =1× 19.V62 =0.051 V V h =5× 2g=0.255V h =0.0255 V +0.051 V 0.255V =0.0255 V +0.306V h =(13.05 V+19.6 V)+(0. 0255 V +0.306V)=66 →66=13.0755 V +19.906 V ≫Eq.1 π4 ×0.225 ×V =0.0425+ π4 ×0.225 ×V V =1.068+V substitute in Eq. 1→ For Exit: (due to part 2)
For Valve: (valve exists in part 2)
+
Continuity Equation:
Page (14)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Water Flow in Pipes
Hydraulics
+19.906 V →V =0.89 m/s 66=13. 0 7551. 0 68+V Q = π4 ×0.225 ×V = π4 ×0.225 ×0.89=0.0354 m/s . ✓
Page (15)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Chapter (4)
Pipelines and Pipe Networks
Pipelines and Pipe Networks
Hydraulics
Flow through series pipes Is the same in case of single pipe (Ch.3), but here the total losses occur by more than 1 pipe in series. (See examples 3.9 and 4.1 in text book).
Flow through Parallel pipes Here, the main pipe divides into two or more branches and again join together downstream to form single pipe. The discharge will be divided on the pipes:
Q=Q +Q +Q +⋯ h =h, =h, =h, =⋯
But, the head loss in each branch is the same, because the pressure at the beginning and the end of each branch is the same (all pipes branching from the same point and then collecting to another one point).
Pipelines with Negative Pressure (Siphon Phenomena) When the pipe line is raised above the hydraulic grade line, the pressure (gauge pressure) at the highest point of the siphon will be negative. The highest point of the siphon is called Summit (S). If the negative gauge pressure at the summit exceeds a specified value, the water will starts liberated and the flow of water will be obstructed. The allowed negative pressure on summit is -10.3m (theoretically), but in practice this value is -7.6 m. If the pressure at S is less than or equal (at max.) -7.6, we can say the water will flow through the pipe, otherwise (P s >-7.6) the water will not flow and the pump is needed to provide an additional head. Note
P =P +P . × P =101.3 kPa= =10.3m P =10. 3 +P P P =10.3m as max, theo. The value of
So always we want to keep
positive (
To maintain the flow without pump.
Page (17)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
Problems: 1. Three pipes A,B and C are interconnected as shown. The pipe are as follows: Find: 1) The rate at which water will flows in each pipe. 2) The pressure at point M. Hint: Neglect minor losses.
Pipe A B C
D (m) L (m) 0.15 600 0.1 480 0.2 1200
f 0.02 0.032 0.024
Solution
Pρg + V2g +z = ρgP + V2g +z +h P =V =P =0.0 V, V =V =?? , h =?? 0+0+200=0+ 19.62 +50+h 0.0509 LV +h =150m→Eq. 1 V h =f D2g h, =h, Parallel Pipes take pipe A h =h, +h, h, =0.020.60015V2g =4.07V
Applying Bernoulli’s equation between the points 1 and 2.
Page (18)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
1200 V h, =0.024 0.2 2g =7.34V i t u t e i n Eq.1→ h =4. 07V +7. 34V subst 0.→7.050939VV +4.+4.007V7V+7.=150m→Eq. 34V =150m2 V and V?? Qπ×0.+Q15=QV + π ×0.1V = π ×0.2V 4→0.025V+0.401V =0.04V4 →Eq.3 h , =h, → h, =4.07V 480 calculVated above h, =0.032 0.1 2g =7. 8 2V 4.→V07V =0.=7.728V2V subst→Vitute=0.in Eq.3 52 V →0.→7.0325V90.805+0.V010.+4.720V7V=0.=150m→ 04V →VV=0.=4.81051 m/sV Subs.in Eq.2 →V→V =0.=0.8705×4. 2 ×4.111=2. 1 3 m/s 1=3. 3 m/s Q = ππ4 ×0.15 ×4.11=0.0726 m/s . Q = π4 ×0.1 ×2.13=0.0167 m/s . Q = 4 ×0.2 ×3. → 3 =0.103 m/s . Pρg + V2g +z = ρgP + V2g +z +h→ V =V hP→ =h, =7.34×3.3 =79.93m → 9810 +120=0+50+79. 9 3 →P =97413.3 Pa. . How we can find other relation between Continuity Equation
But,
)
✓
✓
✓
Pressure at M:
Bernoulli’s equation between the points M and 2.
✓
Page (19)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
2. Three pipes A,B and C are interconnected as shown. The pipe are as follows: Find: Pipe 1) The rate at which water will flows in A B each pipe. C 2) The pressure at point M. Solve the problem in the following two cases: a) The valve (V) is closed. b) The valve (V) is opened with K= 5
D (m) L (m) 0.15 600 0.1 480 0.2 1200
f 0.02 0.032 0.024
Solution Case (a): Valve is closed When the valve is closed, no water will flowing in pipe B because the valve prevents water to pass through the pipe. Thus the water will flowing
throughout pipe A and pipe C (in series) and the system will be as follows:
Page (20)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
Now, you can solve the problem as any problem (Pipes in series). And if you are given the losses of the enlargement take it, if not, neglect it. Case (b): Valve is Open Here the solution procedures will be exactly the same as problem (1) >> water will flow through pipe A and B and C, and pipes A and B are parallel to each other . But the only difference with problem (1) is:
TotTotaall head head lloossss iinnpipippeeA=h A=Tot,al head loss in pipe B Total head loss inVpi pe B=hV , +h, h, =K 2g =5 2g
Now, we can complete the problem, as problem 1.
3. The flow rate between tank A and tank B shown in the figure below is to be increased by 30% (i.e., from Q to 1.30Q) by the addition of a second pipe in parallel (indicated by the dotted lines) running from node C to tank B. If the elevation of the free surface in tank A is 7.5m above that in tank B, determine the diameter, D, of this new pipe. Neglect minor losses and assume that the friction factor for each pipe is 0.02
1
2
3
Solution
Page (21)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics Before addition of pipe:
Pρg + V2g +z = ρgP + V2g +z +h→ 0+0+7.5=0+0+0+h180+150 → →h→ =7.5 m h→ =7.5 m=0.π 02× 0.15 × V2g →V=1.828 m/s →Q =A×V= 4 ×0.15 ×1.828=0.0323 m/s Q,0.=1.04123 Q =1.3×0.0323=0.0412 m/s V = π4 ×0.15 =2.33 m/s 0+0+7.5=0+0+0+h→ →h→ =7.5 m h→ =h +h 180 2. 3 3 150 V 7.5=0.02× 0.15 ×π 2g +0.02× 0.15 × 2g →V =0.918 m/s →Q→Q =A=Q ×VQ ==0.4 ×0.04120. 15 ×0.0162=0. 918=0.0025162mm/s/s h =h150 0.918 150 V 0.02× 0.15 × 2g =0.π02× D × 2g →D =0.π178V Q→V =A=0. ×V99m/s→D →0.025==0.4 ×D178×0. ×V →0. 0 25= 4 ×(0.178V) ×V 99 =0.174 m =174 mm . Apply Bernoulli’s equation between reservoirs A and B.
After addition of pipe:
Apply Bernoulli’s equation between reservoirs A and B.
✓
Page (22)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
4. A 500 mm diameter siphon pipeline discharges water from a large reservoir. Determine: a. The maximum possible elevation of its summit, B, for a discharge of 2.15 3 2 m /s without the pressure becoming less than 20 kN/m absolute. b. The corresponding elevation of its discharge end (ZC). Neglect all losses.
Solution
P =P20×10 +P PP,=10.= 39810m =2.038 m P, =2.03810. 3 =8. 2 6m Q=AV→V= QA = π4 2.×0.155 =10.95 m/s Pρg + V2g +z = ρgP + V2g +z +h Datum at A , h =010.gi9ven5 0+0+0=8. 2 6+ 19.62 +z +0→z =2.15 m . Pρg + V2g +z = ρgP + V2g +z +h Datum at A , h10.=095gi ven 0+0+0=0+ 19.62 +z→z =6.11 m .
Applying Bernoulli’s equation between the points A and B.
✓
Calculation of ZC: Applying Bernoulli’s equation between the points A and C.
✓
Page (23)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
5. In the shown figure, if the length of the pipe between reservoir A and point B is 180 m and the absolute pressure at point B is 7.3 mwc. Calculate the minor loss coefficient (K) of the valve at point C. Given data: The entrance is sharp edged. Total length of pipe (L= 730 m), Pipe diameter (d = 150mm), friction factor (f = 0.02).
Solution
P, =7.3m , P =10.3m → P, =7.310.3=3m Pρg + V2g +z = ρgP + V2g +z +h→ V 0+0+16=0+ +0+h → 2g V 16= 2g +h→ →Eq.1 h→ =h +h 730 V h =0.020.152g V V h =0.5 2g +K 2g h→ =4.86V +0.0509KV →Sub.in 1→ V 16= 2g +4.86V +0.0509KV But V=?? Apply Bernoulli’s equation between the points A and D. Datum at (D):
Apply Bernoulli’s equation between the points A and B. Datum at (B): Page (24)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
Pρg + V2g +z = ρgP + V2g +z +h→ V 180 V 0+0+0. 8 =3+ +0+0. 0 2× × →V=1. 7 27 m/s 2g 0. 1 5 2g 1. 7 27 →16= 2g +4.86×1.727 +0.0509K ×1.727 → K =8.9 . ✓
6.
The difference in surface levels in two reservoirs connected by a siphon is 7.5m. The diameter of the siphon is 300 mm and its length 750 m. The friction coefficient is 0.025. If air is liberated from solution when the absolute pressure is less than 1.2 m of water, what will be the maximum length of the inlet leg (the portion of the pipe from the upper reservoir to the highest point of the siphon) in order the siphon is still run if the highest point is 5.4 m above the surface level of the upper reservoir? What will be the discharge.
Solution The graph is not given so you should understand the problem, and then drawing the system as follows:
PP,=P=1.2m+P,P =10.3m → P, =1.210.3=9.1m Applying Bernoulli’s equation between the points 1 and S. Datum at (1): Page (25)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
Pρg + V2g +z = ρgP + V2g +z +h→ V 0+0+0=9.1+ 2g +5. 4 +h→ →Eq.1 X V h→ =h→ =0.0250.319.62 But V=?? Pρg + V2g +z = ρgP + V2g +z +h→ 0+0+7.5=0+0+0+h→ →h → =7.5m h→ =7. 5 m=h→=0.0257500.319.V62→V=1.534m X 1. 5 34 h→ =0.0250.3 19.62 =0.01X Substitute in Eq. 1 0+0+0=9.1+ V2g +5. 4+0.01X V =V=1.534 0+0+0=9.1+ 1.52g34 +5. 4 +0. 0 1X→X=358m . Applying Bernoulli’s equation between the points 1 and 2. Datum at (2):
✓
Page (26)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
7. For the three-reservoir system shown. If the flow in pipe 1(from reservoir A 3 to junction J)is 0.4m /s and the friction factor for all pipes is 0.02, calculate:
The flow in the other pipes (pipe 2, 3 and 4).
The elevation of reservoir B (Z B).
Neglect minor losses.
Solution If we put a piezometer at point J, the will rise at height of Z p, which can be calculated by applying Bernoulli’s equation between A and J:
Pρg + V2g +z = ρgPJ + V2gJ +z+h→J 0+0+100=Z8fLQ+h →J 8×0. 0 2×400×0. 4 h→J = πgD = πg×0. 4 =10.32m →10010.Z =89. 3 2=Z68>Z=89. =80, 68 m Z Z =hJ→ 8×0. 0 2×300×Q 89.6880= πg×0.3 →Q =0.2177 m/s . Q@J =0. 0 →0. 4 =0. 2 177+Q +Q→Q +Q =0.1823 m/s
Note that So the flow direction is from J to C, and to calculate this flow we apply Bernoulli’s equation between J and C:
✓
Now, by applying continuity equation at Junction J:
Page (27)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
Since pipe 2 and 3 are in parallel, the head loss in these two pipes is the same:
8f L Q 8f L Q h, =h, → πgD = πgD 8×0. 0 2×200×Q 8×0. 0 2×180×Q → πg×0. 2 = πg×0.15 →Q =0.513Q /s . Q +Q =0. 1 823→Q +0. 5 13Q =0. 1 823→Q =0. 1 2 m Q =0.513×0.12=0.0618 m/s . ✓
But,
✓
Now we want to calculate the elevation ZB:
Apply Bernoulli’s equation between J and B:
Z Z =h8×0.J→02×200×0. take pipe 212 89.68Z = πg×0. 2 →Z =74.8 m . ✓
8.
In the shown figure, determine the total length of pipe 3 (L 3) and the head delivered by the pump ( h p). Neglect minor losses and take f = 0.032 for all pipes.
Page (28)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
Solution
8f L Q 8×0. 0 32×950×Q h, = πgD →40= π ×9.81×0.45 →Q =0.542 m/s 8f L Q 8×0. 0 32×450×Q h, = πgD →40= π ×9.81×0.45 →Q =0.352 m/s Q@J =0.0→Q =0.5420.352=0.19 m/s →direction shown in the figure Z =30+hJ→ →Z =30+8=38m 8×0. 0 32×L ×0. 1 9 Z =25.5+hJ→ →3825. 5 = π ×9.81×0.3 →L =318.22 m . 18=Z +h, h →h =38+4018=60m .
To find the total head at point J (ZP): apply Bernoulli equation between J and A:
To find the length of pipe 3 (L 3): apply Bernoulli between J and B:
✓
To find the pump head (h P ): apply Bernoulli between C and J: ✓
Page (29)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
9. The network ABCD is supplied by water from reservoir E as shown in the figure. All pipes have a friction factor f = 0.02. Calculate: 1. The flow rate in each pipe of the system using the Hardy Cross method. Consider the following: Assume for the first iteration that: 3 3 QCB = 0.15 m /s from C to B and QBA = 0.05 m /s from B to A.
Do only one iteration (stop after you correct Q)
2. The pressure head at node A. Given Also:
Nodes elevation
Pipes dimension
Page (30)
Node/Point Elevation(m)
Pipe Length (m) Diameter (m)
Dr.Khalil Al-astal
AB 400 0.30
A 20
B 25 BC 400 0.30
C 20 CD 400 0.30
Eng. Ahmed Al-Agha
D 30 DA 400 0.30
Reservoir E 50 BD 600 0.30
CE 200 0.40
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
Solution Firstly, from the given flows, we calculate the initial flow in each pipe and the direction of flow through these pipes using continuity equation at each note.
Always we assume the direction of the loop in clock wise direction
We calculate the corrected flow in each pipe, from the following tables: Loop
I
Pipe
L
D
AB
400
0.3
BD
600
DA
400
h Q h Q ∆ Q -0.05
-0.68
13.6
0.3
+0.05
+1.02
20.4
0.3
+0.1
+2.72
27.2
-0.025
-0.075
-0.025
+0.075
8f L Q h calculated for each pipe from the fol owing relation: πgD 3.062 =0.025 m/s ∆= 2 ∑Q∑hh = 2×61. Q =Q +∆ +3.06
61.2
Note that, we don’t correct pipe BD because is associated with the two loops, so we calculate the flow in the associated pipe after calculating the correction in each loop. Page (31)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
Loop
II
Pipe
L
D
BC
400
0.3
BD
600
CD
400
h Q h Q ∆ Q -0.15
-6.12
40.8
0.3
-0.05
-1.02
20.4
0.3
+0.15
+6.12
40.8
+0.005
-0.145
+0.005 +0.155
∑ 0 2 ∆= 2 ∑ Q hh = 1. =+0. 0 05 m /s 2×102 Q QQ,, =Q=+0.05+0. ∆ , +∆ 0250.005=+0.02m/s QQ,, =Q=0.05+(0. , +∆ ∆ 005 0.025)=0.02 m /s -1.02
Now, to calculate correction:
102
for pipe BD, we calculate the associated value for
Or:
Now, we put the corrected flow rate on each pipe of the network:
Note: If the sign of the corrected flow rate is the same sign of initial flow rate, the direction of flow will remain unchanged, however if the sign is changed, the flow direction must reversed.
✓✓✓✓✓✓✓✓
Page (32)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
To calculate the pressure head at point A, we must starts from point which have a known head, so we start from the reservoir. Important Note: The total head at any node in the network is calculated as following:
h =h +h →h = Pγ +Z +Z P 8f L Q 50+0+0= γ +0+20+ πgD 3 → Pγ =5020 8×0.π ×9.02×200×0. 81×0.4 =27.1m P 8f L Q 20+27. 1 +0= γ +0+25+ πgD 1 45 → P γ =20+27. 1 25 8×0.π0×9.2×400×0. 81×0.3 =16.38m P 8f L Q 25+16. 3 8+0= γ +0+20+ πgD P→ γ =25+16. 3 820 8×0.π0×9.2×400×0. 0 75 81×0.3 =19.84m .
By considering a piezometer at each node, such that the water rise on it distance:
and the velocity is zero.
Starts from reservoir at E:
Apply Bernoulli between E and C:
Apply Bernoulli between C and B:
Apply Bernoulli between B and A:
✓
Page (33)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
10. 3
For all pipe segments, the initial estimate of the flow in each pipe (m /s) is shown in the figure below.
a) Show on the sketch all sources and sinks (in and out flows) of water from the pipe network. Use an arrow to indicate direction (in or out of system) and give the magnitude of the flow. b) Use Hardy-Cross method to correct the flow rate through each pipe. Do only one iteration (stop after you correct Q), assume = 0.02 for all pipes and neglect minor losses. c) If the pressure head at A = 90m, determine the pressure head at E.
Solution
Page (34)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
a) By applying continuity equation at each junction, the in and out flow from the system are shown in the following figure:
✓✓✓✓✓✓✓✓
b) We calculate the corrected flow in each pipe, from the following tables:
Loop
I
Pipe
L
D
AB
250
0.5
BD
180
DC CA
Q h Qh ∆ Q +0.5
+3.305
6.61
0.2
-0.2
-37.18
185.9
200
0.6
-0.8
-2.72
3.4
+0.1028
-0.6972
350
0.6
-0.8
-4.76
5.95
+0.1028
-0.6972
-41.35
201
+0.1028 +0.6028
3 5 ∆= 2 ∑Q∑hh = 41. =+0. 1 028 m /s 2×201
Page (35)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics Loop
Pipe
L
D
BE
400
0.2
BD
180
DE
380
Q h Qh ∆ Q +0.2
+82.62
413.1
0.2
+0.2
+37.18
186
0.4
-0.4
-9.81
24.52
+110
623.525
-0.088
+0.112
-0.088
-0.488
II
+110525 =0.088 m/s ∆= 2∑∑hQ h = 2×623. QQ,, =Q=0.2+(0. , +∆ ∆ 10280.088)=0.0092 m /s QQ,, =Q=+0.2+0. ∆ ,0+∆ 880.1028=+0.0092 m/s Or:
.
The corrected flows are shown in the following figure:
✓✓✓✓✓✓✓✓
Page (36)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pipelines and Pipe Networks
Hydraulics
P @ point A: γ =90m and Z=20m→h, =90+20=110m P 8f L Q 110= γ +25+ πgD =80.2 m P→ γ =11025 8×0.π0×9.2×250×0. 6 03 81×0.5 P 8f L Q 80.2+25= γ +30+ πgD =49.29m . P→ γ =105.230 8×0.π0×9.2×400×0. 1 12 81×0.2 c)
Apply Bernoulli between A and B:
Apply Bernoulli between B and E:
✓
Page (37)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Chapter (5)
Pumps
Pumps
Hydraulics
1. The figure below shows a portion of a pump and pipe system. The 30-m long pipeline connecting the reservoir to the pump is 1m in diameter with friction factor f = 0.02. For the pump, the required net positive suction head ( NPSH R ) is 2 m.
If the flow velocity V is 2.5 m/s check the system for cavitation . Take the atmospheric pressure = 100 kPa and the vapor pressure = 2.4 kPa.
NPSH =±h h h + P γ P γ hve =sig105 =5m nL because t h e suppl y t a nk i s bel o w t h e pump V 30 2. 5 h =f D2g=0.02× 1 2×9.81=0.191 m h =0. 5 +2×2+3× V2g =7. 5 × 2×9.2.5 81 =2.39 m → NPSH =50.1912.39+ 9.10081 9.2.841 =2.36 m NPSH =2. 3 6> NPSH =2 →System is adequate for cavitation . Solution
.
✓
Page (39)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pumps
Hydraulics
2. A pump is required to supply water to an elevated tank through a 0.2m diameter pipe which is 300m long and has a friction factor f equal to 0.025. 2 Minor losses causes an additional head loss of 4V /2g where V is the velocity in the pipe in m/s. The static head between the pump wet well and the elevated tank is 40 m. The relationship between head, H, and flow, Q, for the pump is given by the following equation: 2 H= 50 - 600Q 3 Where Q in m /s and H in meters. a) under these conditions determine the flow in the pipeline b) if the maximum efficiency of the pump is achieved when the flow is 70 ℓ/s then how could the system be redesigned so that the pump would operate at this efficiency.
Solution 2
3
Pump Curve Equation is: H= 50 - 600Q and since Q in m /s we calculate the head at the following values of Q as shown in the following table: 3
Q (m /s)
0
0.05
0.1
0.15
0.2
0.25
H (m)
50
48.5
44
36.5
26
12.5
Now, we want to find the system head equation: To find the relation between the required head and flow for the system, the best way is to apply Bernoulli equation on the given system and then find the relation between pump head and flow rate (general case), or by applying the following equation (special case):
H=H +h H=H +h H =40 m given ∑ h =h +h
When the pump is used to transfer water from one reservoir to the other.
Page (40)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pumps
Hydraulics
8f L Q h = πgD +h v Q h =4 2g =4× 2gA , but A= π4 ×D = π4 ×0.2 =0.0314 →h = .7Q8×0.025×300Q →h = π ×9.81×0.2 +206.7Q =2143.26Q H=40+2143.26Q 206
So, the following equation is for system curve:
Now, we calculate the head at the same values of Q (in pump curve) as shown in the following table: 3
Q (m /s)
0
0.05
0.1
0.15
0.2
0.25
H (m)
40.0
45.4
61.4
88.2
125.7
174.0
Now, we draw the pump curve and the system curve, and the point of intersection between the two curves is the operating point:
3
As shown in figure above, the pump flow rate is 0.06 m /s (60 ℓ /s) and the pump head is 49m.
Page (41)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pumps
Hydraulics The required B, means if we get maximum efficiency if the operating point 3 gives flow rate of (70 ℓ/s) = 0.07 m /s, what should you modify in the system to meet this efficiency?? From the above graph, the corresponding head at Q = 0.07 is 47.5m. Now, we know the system curve equation:
H=40+h and Q=0. 0 7 →→ 47. 5 =40+h →h =7.5m At maximum efficiency, H = 47.5 m
Now the only, factor that we can change is the diameter of the pipe, or changing material of the pipe (change f value), but assume the same material, so change the diameter of the pipe as following:
8×0. 0 25×300×0. 0 7 0. 0 7 h =7. 5 = π ×9.81×D +4× 2×9. 8 1×π4 ×D →→D=0.214mto get maximum efficiency . ✓
3.
For the shown figure below, (a) what is the operation point for the system that shown in the figure below, the pump characteristics curve is given below (Assume f = 0.014), (b) If the efficiency of the system is 80%, what is the power required?, (c) What is the operation point in case of two identical pumps in parallel?
Pipe is 20cm in diameter
Page (42)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pumps
Hydraulics
Solution a) The pump curve is given as shown above, now we want to find the system curve. So firstly we find the system curve equation: By applying Bernoulli equation between the two reservoir, the equation will be:
H=H +h H =3.5 +1510=8.5 m h =h + h 8f L Q h = πgD +h v Q h = 0.4 +0.9 +1 2g =2. 3 × 2gA , but A= ×D = ×0.2 =0.0314→h = Q 8×0. 0 14×200+200+15Q →h = π ×9.81×0.2 +118.9Q =1619Q H=8.5+1619Q 118.9
So, the following equation is for system curve:
Page (43)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pumps
Hydraulics 3
But, in this equation Q is in m /s and the horizontal axis in the given graph is 3 in m /min., so each value on the graph must be divided by 60 (to transform 3 to m /s) and then we draw the system curve on the same graph as following:
3
. 9810×0. 0 5×12. 5 5 η= γ ×Q×H →P = P 0.8 =7694.7 watt .
3
The value of pump flow (at the operation point) is 3 m /min = 0.05 m /s✓ The value of pump head (at the operation point) is 12.55 m ✓
.
b)
✓
c) When we use two pumps in parallel, the system curve will never change, but the pump curve will change (The values of head remains constant but the values of Q multiplied by 2 ), so we draw the new pump curve by multiplying each value of Q by 2 at each head as following:
Page (44)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Pumps
Hydraulics
3
.
3
So the new value of Q is 4.1 m /min = 0.0683 m /s and the new value of H is 16m ✓
Page (45)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Chapter (6)
Open Channels
Open Channels
Hydraulics
Formulas used to describe the flow in open channels
1 / . V=n=manni ×R ×S n ng coefficient S=The slope of channel bed.Wetted Area A R =Hydraulic Radius= Wetted Perimeter = P The most common formula is Manning equation:
Most Economical Section of Channels The most efficient section is satisfied when the flow in the channel is maximum and thereby the minimum wetted perimeter . I.e. for most economical section:
dPdy =0.0 dBdP =0.0 such that, y is the depth of water in channel also known by normal depth v Q E =y+ 2g →E =y+ 2Ag for any cross section and B is the width of the channel.
Energy Principals in Open Channels
Special Case for rectangular channel: Since the width of the channel B is constant, we can calculate the discharge
q=q m/s →→E =y+ 2yg For rectangular channel V FFround Number= gD D =Hydraulic depth of channel= WaterAreasurfaofcefldeptow h T
per unit width;
Sub-critical, critical and supercritical flow:
Page (47)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Open Channels
Hydraulics
FF <1→Subcri t i c al =1→Cri t i c al F >1→Sper critical y E F =1 IIff y>y →Subcr i t i c al f l o wF <1 y=y →Cri t i c al f l o wF =1 If y1 V F = g y / q q y = g →y = g VE= =Eg ×y =1. 5 y Qg = AT To calculate y , v2g = D2 A E =E = y +0.5 T
Critical Depth : Is the depth of flow of liquid at which at which the specific energy is minimum and the flow corresponds to this point is called critical flow ( ). So, we can classify the flow also as following:
For rectangular channel:
The critical velocity is:
Non-Rectangular Shapes (All Shapes):
Page (48)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Open Channels
Hydraulics
Hydraulic Jump
yy a=upst r em dept h and y =downst r eam dept h nd y a re cal l e d . < yy =0.5 1+ 1 +8yy yy =0.5 1+ 1 +8yy yy =0.5 1+ 1 +8 F →F = Vgy yy =0.5 1+ 1 +8 F →F = Vgy y y =∆E= 4yy (()=y y )=6hJ For rectangular channel:
Or in terms of Fround number:
Height of Hydraulic jump
Length of Hydraulic jump
Page (49)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Open Channels
Hydraulics
1. 3
A trapezoidal channel is to be design to carry a discharge of 75m /s at maximum hydraulic efficiency . The side slopes of the channel are 1V:2H (1 vertical and 2 horizontal) and the Manning’s roughness n is 0.03. • If the maximum allowable velocity in the channel is 1.75m/s m what should be the dimensions of the channel (bottom width and height) • What should be the longitudinal slope of the channel if the flow is uniform?
Solution a) We draw the following graph (from the given data in the problem):
y + 2y =√ 5 y
A= QV = 1.7575 =42.86 m → → dPdy =0.0 Or dBdP =0.0 P=B+2×√ 5 y→P=B+4. 4 7y→→Eq. 1 1 A=42.42.8 6=By+2× ×2y×y→42. 8 6=By+2y 2 →B= y86 2y Substitute in Eq. 1→→→ P= 42.y86 2y+4. 4 7y→P= 42.y86 +2.47y dPdy = 42.y86 +2.47=0.0 for most economical At maximum hydraulic efficiency (most economical section)
Page (50)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Open Channels
Hydraulics
42. 8 6 → y =2.47→y=4.16 m . →B= 42.4.1866 2×4.16=1.98 m . V=V=1.1n7×R5 m/s/ ×S, . n=0.A03 given42.86 R = P = 1.98+4.47×4.16 =2.083 m →1. 7 5= 0.103 ×2.083/ ×S. →S=0.00104 m/m . ✓
✓
b) According Manning equation:
✓
2. 3
In the figure shown, water flows uniformly at a steady rate of 0.4 m /s in a very long triangular flume (open channel) that has side slopes 1:1. The flume is on a slope of 0.006, and n = 0.012; (a) Find normal depth (yn). (b) Determine wether the flow is subcritical or supercritical.
(c)Find the specific energy (Es) and the critical specific energy (Ec).
y +y =√ 2 y Page (51)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Open Channels
Hydraulics
Solution a) To find the normal depth, we use Manning equation:
1 A / / . . V=Q=0.n ×R4 m/s×S, n=0.→Q=AV= ×R ×S n 0121 , S=0.006 Gi vens A=areaA of tryiangle = 2 × 2y ×y=y R = P = 2× √ 2 y =0.353y y 0.4= 0.012 × 0.353y/ ×0.006. →y=0.457 m . F = gVD V= QAA= 0.y40.574 =1.0.457915 m/s D = T = 2y = 2×0.457 =0.2285 m 1.9152285 =1.28>1→ . →F = √ 9.81×0. Qg = AT → 9.0.841 = y2y →y =0.504 y =0.504>y=0.457→ . v 1. 9 15 E =y+ 2g =0.457+ 2×9.81 =0.644 m . 1 A 1 0. 5 04 E =y + 2 T=0.504+ 2 2×0.504=0.63 m . E =Emust be less than E . now, substitute by all above data in Manning’s equation:
✓
b) To determine the type of flow, there are two methods:
✓
Or, we calculate the critical depth:
Since
✓
c)
✓
✓
✓
Page (52)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
Open Channels
Hydraulics
3. You are asked to design a rectangular channel that has the minimum wetted perimeter and that conveys flow in critical condition. Find the relationship between the critical depth and the channel width if the flow discharge is constant. Your answer should look something like y c = mB , where m is constant and B is the channel width.
Solution
P=B+2y →→Eq.1 q Q Q y = g , q= B→y = Bg →B = yQg →B= y/Q g →B= Qg y−/ Substitute in Eq. 1→ P=dP y−/ +2y gy dydP =0.0 foQr most−economical Q − dy =1.5 g y +2=0.0→2=1.5 g y Q=AV =By × gy =B g y B g y B y B y − →2=1.5 g y → y =1.5 2 →y × y =1.5 2 →y =0.75 B . P=B+2y → dydP = dydB +2→ dydB =2→→Eq. 1 We want other relation, between B and yc For rectangular channel:
Remember: Critical velocity in rectangular open channel is:
But,
✓
We can solve this problem alternatively as following:
Now, we want to find other relation for
Page (53)
Dr.Khalil Al-astal
Eng. Ahmed Al-Agha
Eng. Ruba Awad
