MIDTERM EXAM 31 MARCH :00 ANFI G
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1 MIDTERM EXAM 31 MARCH :00 ANFI G
2 BRANCHING PIPES Junction Problems
3 Junction Problems Q 1 +Q +Q 3 0 (flows towards the junction are considered to be positive!) h j z j +p j /γ h l,i h f,i + h m,i z i -h j h l,i h f,i + h m,i z i -h j j j
4 Identify the problem Given Reservoir elevations, z 1,z, z 3 Pipe characteristics: L 1,L,L 3,D 1,D,D 3,ε 1, ε,ε 3 Determine The flow rate Q 1,Q, Q 3 and direction on each pipe Junction head: h j (piozometer head)
5 Solution Procedure Assume an appropriate direction for the flow in each pipe Assume a junction head If friction factor is unknown Make best assumption for the initial value of the friction factor (fully turbulent rough flow) Write energy equation from each reservoir to the junction Calculate the flow rate for each pipe by using energy equation Put into continuity equation the determined discharge values Check that the continuity equation is satisfied or not If not change the junction head
6 Example
7 Assume Q 1 and Q + and Q 3 - Neglect all minor losses Energy conservation b/w A&J: Energy conservation b/w B&J: Energy conservation b/w J&C: h A h J +h l,aj h J +h f,aj h B h J +h l,bj h J +h f,bj h J h C +h l,jc h C +h f,jc Assume h J and check if Q 1 + Q + Q 3 0. If not, iterate by assuming new h J until Q i 0 checks.
8 Assume h J 100m (elevation+50 m of pressure head) h f,aj 8fLQ π gd f 1 Q 1 h f,bj 0. f Q h f,cj 9.9 f 3 Q 3 hence b/w A and J b/w B and J f 1 Q f Q b/w J and C f 3 Q 3 * Assume hydraulically rough flow, ff(ε/d)
9 h J 100m Concrete pipe ε Q 1 +Q Q 3 ε/d f i Q i >>.57 Therefore increase h J h J 130m f i Q i h J 139m f i Q i > 5.9 Therefore increase h J
10 NETWORK OF PIPES
11
12
13 NETWORK OF PIPES B C C A H 1 I 4 3 G E F Conservation of Mass: Flow into each junction must be equal to the flow out of the junction LOOP + clockwise Energy Conservation: Algebraic sum of head losses around each and every loop must be zero. A,...,I:JUNCTIONS/ NODES EF:LINK/BRANCH
14 Head Loss Equations link link link l link m f 4 5 m f Q Q K h Q K Q K Q K g 8Q D g 8Q D L f h h h h link link + π + π + l l K m
15 Conservation of Energy around any loop N i o N i o o K Q Q KQ N 1 i 1 n o N 1 i n o loop N 1 i 1 n o N 1 i n o n o i N 1 İ n i i N 1 i lloop numberoflinkelements N 1 i Q n If o QQ nk KQ 0 h QQ nk KQ Q) K (Q K Q h KK n for Darcy-Weisbach
16 Solution Procedure 1) Assume flow distribution Condition: Flow distribution should satisfy conservation of mass at each node/junction Notation: Q o ) Calculate Q for each loop 3) Check Q small magnitude YES: stop NO: Step 4 4) Correct Q values Q o,new Q o,old + Q loop
17 Example Given is the network shown in figure below. Find the discharges in each and every link.
18 Initial Guess Continuity Flow into each junction must be equal flow out of the junction LOOP 1 K Q Q K Q AD 6*70* *6*70840 AC 3*35* *3*3510 DC *-5*30* *5*30300 Σ8575 Σ1350 LOOP K Q Q K Q AB 1*15*15 5 *1*15300 BC *-35* **35140 AC 3*-13.83* *3* Σ 799 Σ Q AC
19 Initial Guess LOOP 1 Q AC Flow Distribution After First Iteration LOOP 799 Q
20 Second iteration LOOP 1 K Q Q K Q AD 6*48.83* *6* AC 3*.77*.77 3 *3* CD 5*-51.17* *5* Σ141 Σ1114 LOOP K Q Q K Q AB 1*6.06* *1*6.065 BC *-3.94* ** AC 3*-1.656* *3* Q Σ 475 Σ AC Q
21 Final Distribution of Second Iteration LOOP Q LOOP 475 Q AC Attention: One more iteration is needed
22 HYDRAULICS AND OPERATION OF PUMPED DISCHARGE LINES Pump: adds energy to a fluid, resulting in an increase in pressure across the pump. Turbine: extracts energy from the fluid, resulting in a decrease in pressure across the turbine.
23 Turbine
24 Types of Pumps (a) Radial flow pump (b) Axial-flow pump
25 Centrifugal Pumps
26 The power gained P f γ Q Hp H p Q P f Head supply by pump (m) Discharge (m 3 /s) in Watts (Nm/s) (in practice hp horsepower- is used) 1hp Watts P f γ QH p 745.7
27 A PUMP IS A MECHANICAL DEVICE WHICH ADDS ENERGY TO THE FLUID ENERGY (Supply) The power required to derive the motor is bhpωt MOTOR bhp: brakehorsepower ω:shaft angular velocity T: shaft torque P f γ Q H p ENERGY into fluid S H A F T BLADES ROTOR Axial Radial Mixed TRANSFER There is a casing which provides passageway to the fluid.
28 Static Suction Lift Suction lift- Static head when the pump is located above the suction tank
29 Static Suction Head Static head when the pump is located below the suction tank
30 Pump Head Net Head Water horsepower Brake horsepower Pump efficiency
31 Efficiency If there were no losses : bhp P f Yet, there are losses: mechanical losses in bearings and seals losses due to leakage of fluid other frictional and minor losses η P f bhp P P f p γ Q H P p p
32 Pump Power and efficiency P p γ Q H η p where η is the efficiency. So η must be maximized over a wide range of Q.
33 Pump Performance P p η H p shut off head η rated capacity (normal/design lowrate) P p H p Q r Q Q (lt/s) H (m) η (%)
34 SELECTION OF A PUMP Valve Pump
35 SELECTION OF A PUMP Pump H p vs Q must be known! System hydraulics must be known!
36 Consider a system with two reservoirs; water is supplied to the reservoir with higher elevation by means of a pumped line! P V P V 1 1 H 1 H H p + h f z + + z H + KQ 1 p γ g γ g H p z + KQ z H p H s + KQ 1 H s Total Static Head ± 0 valve P Static Suction Head
37 Pumps in Parallel Head across each pipe (pump) is identical The total discharge through the pumping system is equal to the sum of individual discharges through each pump.
38 Pumps in Parallel η p γh D PP Q Head across each pipe (pump) is identical The total discharge through the pumping system is equal to the sum of individual discharges through each pump.
39 Pump in Series H A B System Demand Pump B Pump A Q
40 Construction of Pump Curve H H i,b A B Pumps A and B combined System Demand For each discharge the head of the pump A and B are added each other. Then A new curve is constructed H i(a+b) H i,a +H i,b H i,a Pump B Pump A Discharge through each pump is identical!!! Qi η P γ ( H ) P PP Q D
41 Pump in Series H A B Pumps A and B combined System Demand Pump B Pump A Q D Q A Q B
42 EXAMPLE : etermine the discharge and compute the power consumption. Pipeline Characteristics L000 m; D0. m; ε0.000m; ν1* 10-6 m /s ) when only one pump is operating 60 m Fig m ) when both pumps are operating in series P 100 m 60 m Fig. ) when both pumps are operating in parallel P P 100 m 60 m Fig. 3 Pump Charecteristics P P Q (lt/s) H (m) η (%)
43 Single Pump R e VD 4Q ν πdν h f L (m) 000 D (m) 0. ε (m) υ (m /s) f L D V g Q Re f hl system H(m) Case 1: Single Pump Q(m 3 /s) system Single Pump eff. Single Pump H Q
44 Case 1: Single Pump H(m) Q(m 3 /s) system Single Pump eff. SOLUTION Q35 lt/s η80% (for Q35 l/s) H54m P3.kW η p γh D PP Q
45 H(m) Case : Pump in Series Q(m 3 /s) system Single Pump eff. p. in s. Single Pump H Q H(m) SOLUTION Q56 l/s η75% (for Q56 l/s) H7m P5.8kW Q(m 3 /s)
46 Single Pump Pump in Parallel H Q H Q Case 3: Pump in Parallel H(m) Q(m 3 /s) system Single Pump eff. p.i.p SOLUTION Q46 l/s η68% (for Q3l/s) H61m P40.5kW
47 summary single series paralel Q (lt/s) (*3) η (%) H (m) P (kw) (*0.4) P/Q (kw.s/lt)
48 GRAVITY PIPELINES 0.5 m/s < V < m/s (p/γ) > 5m h valve control valve The pipelines through which flow is maintained by the action of gravity are known as gravity pipelines. A control valve might be used to control the discharge and pressure
49 Because of limits set forth on velocity and pressure head there are lower and upper bounds for the discharge through a gravity pipeline. Consider the following schematic representation of a gravity pipeline system shown below Q in spillway h max h res A za ±0 B;valve Energy Equation between points (A) & (B) is: H A H B +h L, or in open form: P V P V A A B B z + h + + z + h A res B B γ g γ g h L
50 P V P V A A B B z + h + + z + h h A res B B γ g γ g L P A P B 0, V A 0, and z B 0. h ma x h res Q in A z A spillwa y ±0 B;valve the maximum value of V m/s. V g Ü x9.81 B h f 8f π where K L 0. gd 8f 5 π m Q L gd ˆcan be 5 KQ, Ë h f neglected KQ
51 Therefore, Energy Equation becomes: h res + z A h f KQ out ËQ out [( h + z )/ K] res A 1/ (note that since V < m/s velocity heads are neglected!) A relationship between Q and the reservoir level can be obtained as: Q h res + K z A h ma x h res Q in A z A spillwa y ±0 B;valve
52 For a full reservoir h res h max QQ max h Q A max h max h max Q Q K + z A max ( ) since z and K are constants maximum discharge that may occur in the pipeline system is called The system capacity For empty reservoir h res 0 Q Q min Q 0 Q 0 is the minimum flow rate, which will pressurize the pipe, in other words, it is the minimum flow rate for a full pipe flow. Therefore Q 0 z A K
53 If Q max < Q in hh max ; Q spill Q in -Q max If Q min < Q in <Q max 0< h res <h max ; Q spill 0 If Q in < Q min to prevent free flow the valve is partially closed Q in A spill h max QQ max h res z A A Q in < Q< Q max ±0 QQ 0 B Valve
54 Valve and free surface The free surface flow may be prevented by use of a valve at the pipe exit. Valves control the flow rate by providing a mean to adjust the over all system loss coefficient to the desired value.
55 Consider the same system with a valve at the end of the pipe: A h A Datum z A B Energy Equation between points (A) & (B): H A H B +h f +h V, or in open form:
56 P V P V z + h + A + A z + h + B + B + h + A res γ g B B γ g f h V P A P B 0, V A V B 0, and z B 0. Also h f KQ & the local loss can also be written as h V K V Q Therefore: z A +h A (K+K V )Q, solving for Q: Q z A K + h + K res V Datum A h A z A B
57 Valve Conditions Q z A K + h + K res V when valve is closed K V Therefore Q0. Openining of the valve reduces the value of K V, producing a desired flow rate.
58 The head loss in valves is mainly a result of dissipation of kinetic energy of a high speed portion of flow. when Q in <Q min, the control valve at the pipe exit will pressurize the flow in the pipe by dissipating the excess kinetic energy, h V, and in turn increasing the water level in the reservoir. The magnitude of the valve loss depends on how much one needs to presurize the flow, that is the water depth in the reservoir. If the water level is to be kept constant to obtain a certain discharge, a control valve can be chosen accordingly.
59 Example Consider the reservoir-pipe system given below, with following values. Reservoir depth h max 5m, z A 5m, L000m, D0.8m, f0.0. Determine: ) The system capacity, Q max. ) Minimum flowrate, Q 0. ) Spill flowrate, if Q in 1.m 3 /s. ) Valve loss, h v, if Q in 0.5m 3 /s. Q A QQ max h max h A Q 0 <Q<Q max Z A QQ 0 B Datum valve
60 Solution If minor losses except hv (valve loss) are neglected, Q out 1/ hres + z 8fL A K K gπ D Q max / 1.0 m 3 / s Q min / 0.70 m 3 / s If inflow is 1. m 3 / s Q spill 0.0 m 3 / s h res + z A h V + KQ in h V z A KQ in h V (0.5) h V.48 m
61 Selection of Pipe Diameter: Cost In design of gravity pipelines, an optimum diameter is selected that minimizes the capital cost (The estimated total cost) and operation cost. CostCapital+Operation cost
62 Selection of Pipe Diameter: Velocity Depending on the type of pipe material a lower and an upper limits are set for the velocity as: 0.5 m/s < V < m/s;
63 Selection of Pipe Diameter: Pressure To prevent air entrainment minimum pressure head, (p/γ)min permitted along the pipeline is 5m. The Range of Pressure Transmission line City Network 0 30 P / γ P / γ 80 (m) (m)
64 Computation of Pipe Diameter For A Gravity Pipelines S 1 C S E S 3 pipeline B D F
65 A S 1 C S E S 3 pipeline B D F Determine the control points (C,E,F) and their topographic elevations z c, z E, z F Add the minimum required pressure head to these elevations (z c and z e ). Determine the minimum energy grade line slope, between the reservoir and control points
66 Minimum Energy Slope A S 1 C S E S 3 pipeline B D F S < < S S 1 3 S 1 H A (z c L + AC p min / γ)
67 Pipe Diameter for A-C Compute the pipe diameter for line A-C using Darcy-Weisbach equation and select the nearest commercially available dameter 8fQ D computed ( gπ S 1 )
68 Velocity Check Compute the Velocity Q V πd / 4 If V min < V < V max then the selected diameter is used in the project If V < V min a booster pump must be installed at the reservoir site to increse the velocity to V min
69 Booster Pump Head If booster pump is installed, the head supplied by booster pump will be computed as p min H + H z + + A P c γ f L D V g min
70 Maximum Velocity Check If V > V max then the pipe diameter is increased to reduce the velocity πd 4 V max Q
71 Pressure Check If p/γ < p max used If p/γ > p max installled then the selected pipe diameter is A pressure reduction pump must be
72 Determination of diameter(c-d-e) Compute the piozemeter head at C Calculate the diameter for C-D-E-F segment Note that E is the control point Compute the diameter of C-D-E Compute the diameter of the segment E-F
73 Control valve Install a control valve at point F and determine the necassary headloss at the valve to maintin the pressurized flow at segment E-F.
74
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