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- 1. Chapter 3 Local Energy (Head) Losses3.1. IntroductionLocal head losses occur in the pipes when there is a change in the area of the cross-section ofthe pipe (enlargement, contraction), a change of the direction of the flow (bends), andapplication of some devices on the pipe (vanes). Local head losses are also named as minorlosses. Minor losses can be neglected for long pipe systems.Minor losses happen when the magnitude or the direction of the velocity of the flow changes.In some cases the magnitude and the direction of the velocity of the flow may changesimultaneously. Minor losses are proportional with the velocity head of the flow and isdefined by, V2 hL = ξ (3.1) 2gWhere ξ is the loss coefficient.3.2. Abrupt EnlargementThere is certain amount of head loss when the fluid moves through a sudden enlargement ina pipe system such as that shown in Fig. (3.1)Impuls-momentum equation is applied to the 1221 control volume along the horizontal flowdirection. The forces acting on to the control volume are, r a) Pressure force on the 1-1 cross-section, F1 = p1 A1 s b) Pressure force on the 2-2 cross-section, F2 = p2 A2 r c) Pressure force on the 3-3 cross-section, F3 = p3 ( A2 − A1 ) r Laboratory experiments have verified that p3 = p1 , F3 = p1 ( A2 − A1 ) r d) Momentum on the 1-1 cross-section, M 1 = ρQV1 r e) Momentum on the 2-2 cross-section, M 2 = ρQV2Neglecting the shearing force on the surface of the control volume, ρQV1 + p1 A1 + p1 ( A2 − A1 ) − p 2 A2 − ρQV2 = 0 ( p1 − p2 )A2 = ρQ(V2 − V1 ) γ ρ= g p1 − p 2 = Q (V2 − V1 ) γ gA2Applying the continuity equation, 1 Prof. Dr. Atıl BULU
- 2. Energy Line (V1 − V2 )2 hL = 2g V 22 V12 2g 2g H.G.L p2 p1 γ γ Flow 1 3 2 3 2 1 p2A2 ρQV2 ρQV1 p1A1 p1(A2 -A1) 3 2 Fig. 3.1. Q = V1 A1 = V2 A2 A1 V2 = V1 A2Substituting this, p1 − p 2 1 ⎛ A ⎞ = ⎜V1 A1V1 1 − V1 A1V1 ⎟ γ gA2 ⎜ ⎝ A2 ⎟ ⎠ p1 − p 2 V12 ⎡⎛ A ⎞ 2 A ⎤ = ⎢⎜ 1 ⎟ − 1 ⎥ ⎜ ⎟ (3.2) γ g ⎢⎝ A2 ⎠ A2 ⎥ ⎣ ⎦ 2 Prof. Dr. Atıl BULU
- 3. Applying Bernoulli equation between cross-sections 1-2 for the horizontal pipe flow, p1 V12 p 2 V22 + = + + hL γ 2g γ 2g p1 − p 2 V22 V12 = − + hL γ 2g 2gSubstituting Equ. (3.2), V12 ⎡⎛ A1 ⎞ A ⎤ V2 V2 2 ⎢⎜ ⎟ − 1 ⎥ = 2 − 1 + hL 2 g ⎢⎜ A2 ⎟ A2 ⎥ 2 g 2 g ⎣⎝ ⎠ ⎦ A V V1 A1 = V2 A2 → 1 = 2 A2 V1 V12 V22 V12 ⎡⎛ V2 ⎞ V2 ⎤ 2 hL = − + ⎢⎜ ⎟ − ⎥ 2 g 2 g g ⎢⎜ V1 ⎟ V1 ⎥ ⎣⎝ ⎠ ⎦ 2 2 2 V V V VV hL = 1 − 2 + 2 − 1 2 2g 2g g g V12 V1V2 V22 hL = − + 2g g 2g hL = (V1 − V2 )2 (3.3) 2gThis equation is known as Borda-Carnot equation. The loss coefficient can be derived as, V12 ⎛ V V2 ⎞ hL = ⎜1 − 2 2 + 22 ⎟ 2g ⎜⎝ V1 V1 ⎟ ⎠ V12 ⎛ A A2 ⎞ hL = ⎜1 − 2 1 + 12 ⎟ 2g ⎜⎝ A2 A2 ⎟ ⎠ 2 V12 ⎛ A ⎞ hL = ⎜1 − 1 ⎟ (3.4) 2 g ⎜ A2 ⎟ ⎝ ⎠The loss coefficient of Equ. (3.1) is then, 2 ⎛ A ⎞ ξ = ⎜1 − 1 ⎟ ⎜ A ⎟ (3.5) ⎝ 2 ⎠ 3 Prof. Dr. Atıl BULU
- 4. The loss coefficient ξ is a function of the geometry of the pipe and is not dependent on to theflow characteristics.Special case: Connecting to a reservoir, V12/2g Energy Line H.G.L. A2 A1 Fig. 3.2.Since the reservoir cross-sectional area is too big comparing to the area pipe cross- Asection, A2 >> A1 , 1 ≅ 0 , the loss coefficient will be equal to zero, ζ=0. The head loss is A2then, V12 hL = (3.6) 2g3.3. Abrupt ContractionFlow through an abrupt contraction is shown in Fig. 3.3. Dead Zone 1 a 2 A1 V1 pa A2 V2 2 1 a Fig. 3.3 4 Prof. Dr. Atıl BULU
- 5. Derivation of the ζ loss coefficient may be done by following these steps. a) Bernoulli equation between 1, C, and 2 cross-sections by neglecting the head loss between 1 and C cross-sections is for the horizontal pipe system, p1 V12 p C VC2 p 2 V 22 + = + = + + hL γ 2g γ 2g γ 2g b) Continuity equation of the system, Q = V1 A1 = VC AC = V 2 A2 c) Impuls-Momentum equation between C and 2 cross-sections, ( p C − p 2 )A2 = ρQ(V2 − VC )Solving these 3 equations gives us head loss equation as, (VC − V2 )2 hL = 2g 2 (3.7) ⎛ A2 ⎞ V 22 ⎜ hL = ⎜ − 1⎟ ⎟ 2g ⎝ AC ⎠Where, AC CC = = Coefficient of contraction (3.8) A2The loss coefficient for abrupt contraction is then, 2 ⎛ 1 ⎞ ξ =⎜ ⎜C − 1⎟ ⎟ (3.9) ⎝ C ⎠ V 22 hL = ξ 2gCC contraction coefficients, and ζ loss coefficients have determined by laboratory experimentsdepending on the A2/A1 ratios and given in Table 3.1. 5 Prof. Dr. Atıl BULU
- 6. Table 3.1. A2/A1 0 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 CC 0.62 0.62 0.63 0.64 0.66 0.68 0.71 0.76 0.81 0.89 1.00 ζ 0.50 0.46 0.41 0.36 0.30 0.24 0.18 0.12 0.06 0.02 0Example: At an abrupt contraction in a pressured pipe flow, Q = 1 m3/sec, D1 = 0.80 m, D2 =0.40 m. Calculate the local head when contraction coefficient is CC = 0.60.Solution: Since discharge has been given, 4Q 4 ×1 V2 = = = 7.96 m sec πD 2 π × 0.40 2 2 πD 2 2 π × 0.40 2 A2 = = = 0.1257m 2 4 4 AC = C c A2 = 0.60 × 0.1257 = 0.0754m 2 Q 1 VC = = = 13.26 m sec AC 0.0754 (VC − V2 )2 (13.26 − 7.96)2 hL = = = 1.43m 2g 19.62Or, 2 ⎛ 1 ⎞ V2 hL = ⎜ ⎜C − 1⎟ × 2 ⎟ 19.62 ⎝ C ⎠ 2 2 ⎛ 1 ⎞ 7.96 hL = ⎜ − 1⎟ × = 1.44m ⎝ 0.60 ⎠ 19.62Special case: Pipe entrance from a reservoir,Loss coefficient ζ depends on the geometry of the pipe connection to the reservoir. The losscoefficient ζ is 0.50 for square-edge entrance. Fig. (3.4) V2 h L = 0.5 (3.10) 2g 6 Prof. Dr. Atıl BULU
- 7. The head loss at the pipe fittings, bends and vanes is calculated with the general local lossequation (Equ. 3.1). The loss coefficients can be taken from the tables prepared for the bendsand vanes. 0.5V2/2g V2/2g R.E.L. H.G.L. V Fig. 3.4Example 3.1. Two reservoirs is connected with a pipe length L. What will the elevationdifference between the water surfaces of the reservoir? Draw the energy and hydraulic gradeline of the system.Solution: Bernoulli equation between reservoir (1) and (2), p1 V12 p V2 z1 + + = z 2 + 2 + 2 + hL γ 2g γ 2gSince the velocities in the reservoirs are taken as zero, V1 = V2 =0, and the pressure on thesurface of reservoirs is equal to the atmospheric pressure, z1 − z 2 = h LThe head losses along the pipe system are, V2 1) Local loss to the entrance of the pipe, h L1 = 0.5 2g f V2 2) Head loss along the pipe, h L2 = JL = L D 2g 7 Prof. Dr. Atıl BULU
- 8. V2 3) Local loss at the entrance to the reservoir, h L3 = 2g 4) The total loss is then will be equal to the elevation difference between the reservoir levels, 5) z1 − z 2 = h L V2 f V2 V2 6) z1 − z 2 = 0.5 + L+ 2g D 2g 2g V2 f V2 z1 − z 2 = 1.5 + L 2g D 2g 0.5V2/2g V2/2g E.L. ΔH=? H1 H.G.L. V H2 L= Pipe Length Fig. 3.5Numerical Application: If the physical characteristics of the system are given as below, whatwill be difference between the reservoir levels? Calculate the ratio of local losses to the totalhead loss and give a physical explanation of this ratio. Q = 0.2 m3/sec, f = 0.02, L = 2000 m, D = 0.30 mThe velocity in the pipe, 4Q 4 × 0 .2 V= = πD 2 π × 0 .3 2 V = 2.83 m secThe velocity head, 8 Prof. Dr. Atıl BULU
- 9. V 2 2.83 2 = = 0.41m 2 g 19.62The elevation difference is then, V2 f V2 z1 − z 2 = 1.5 + L 2g D 2g 0.02 z1 − z 2 = 1.5 × 0.41 + 0.41× × 2000 0.2 z1 − z 2 = 0.62 + 82.0 z1 − z 2 = 82.62m 0.5V2/2g=0.20m V2/2g=0.41m E.L. ΔH=82.62 z1 H.G.L. V z2 L= 2000m Figure 3.6The ratio of local losses to the total head loss is, 0.62 = 0.0075 82.62As can be seen from the result, the ratio is less than 1%. Therefore, local losses are generallyneglected for long pipes. 9 Prof. Dr. Atıl BULU
- 10. 3.4 Conduit SystemsThe other applications which are commonly seen in practical applications for conduit systemsare, a) The Three-Reservoir Problems, b) Parallel Pipes, c) Branching Pipes, d) Pump Systems.3.4.1 The Three-Reservoir ProblemsConsider the case where three reservoirs are connected by a branched-pipe system. Theproblem is to determine the discharge in each pipe and the head at the junction point D. Thereare four unknowns (VAD, VBC, VDC and pD/γ ), and the solution is obtained by solving theenergy equations for the pipes (neglecting velocity heads and including only pipe losses andnot the minor losses) and the continuity equation. The physical characteristics of the systemsuch as lengths, diameters and f friction factors of the pipes, the geometric elevations of thewater surfaces at the reservoirs and the piezometric head at the junction D are given. Thereservoirs are located as, zA> zB> zC. There are three possible solutions of this problem, ⎛ p⎞a) ⎜ z + ⎟ = z A ⎜ ⎝ γ ⎟D ⎠ zB E.L zA B A pD γ zC zD D C Figure 3.7 10 Prof. Dr. Atıl BULU
- 11. ⎛ p⎞Since ⎜ z + ⎟ = z A , there will be no flow from or to reservoir A, and therefore, ⎜ ⎝ γ ⎟D ⎠ Q AD = 0, V AD = 0, J AD = 0 Q BD = Q DCThe relations between the reservoir levels B, and D and the piezometric head at the junction Dare, pD z B − h LBD = z D + = z C + h LCD γEnergy line slope (JBD) of the BD pipe can be found as, pD zB − zD − γ z B − z A h LBD J BD = = = (3.11) L BD L BD L BDUsing the Darcy-Weisbach equation, fV 2 J= D2 gThe flow velocity in the BD pipe can be calculated by, 12 ⎛ 2 gD BD J BD ⎞ V BD =⎜ ⎜ ⎟ ⎟ (3.12) ⎝ f BD ⎠And the discharge is, Q BD = ABDV BDThe energy line slope, the velocity and the discharge of the CD pipe are then, 11 Prof. Dr. Atıl BULU
- 12. z A − z C h LCD J CD = = LCD LCD 12 ⎛ 2 gDCD J CD ⎞ VCD = ⎜ ⎜ ⎟ ⎟ ⎝ f CD ⎠ QCD = ACDVCd = Q BD ⎛ p⎞b) zC < z A < ⎜ z + ⎟ < z B ⎜ ⎝ γ ⎟D ⎠The continuity equation for this case is, Q Bd = Q AD + Q DCThe relations between the reservoir levels B, A, and C and the piezometric head at thejunction D are, ⎛ p⎞ z B − h LBD = ⎜ z + ⎟ = z A + h L AD = z C + h LCD ⎜ ⎝ γ ⎟D ⎠The energy line slopes of the pipes are, zB E.L. zA B pD A γ zC zD D C Figure 3.7 12 Prof. Dr. Atıl BULU
- 13. ⎛ p⎞ zB − ⎜ z + ⎟ ⎜ ⎝ γ ⎟ D h LBD ⎠ J BD = = L BD L BD ⎛ p⎞ ⎜z + ⎟ − zA ⎜ ⎝ γ ⎟D ⎠ hL J AD = = AD L AD L AD ⎛ p⎞ ⎜ z + ⎟ − zC ⎜ ⎝ γ ⎟D ⎠ hL J CD = = CD LCD LCD ⎛ p⎞c) zC < ⎜ z + ⎟ < z A < z B ⎜ ⎝ γ ⎟D ⎠ zB E.L. zA B A pD γ zC zD D C Figure 3.8The continuity equation is, Q BD + Q AD = QCDThe relations between the reservoir levels B, A, and C and the piezometric head at thejunction D are, 13 Prof. Dr. Atıl BULU
- 14. ⎛ p⎞ ⎜ z B − h LBD = ⎜ z + ⎟ = z A − h L AD = z C + h LCD ⎝ γ ⎟D ⎠The energy line slopes of the pipes are, ⎛ p⎞ zB − ⎜ z + ⎟ ⎜ ⎝ γ ⎟ D h LBD ⎠ J BD = = L BD L BD ⎛ p⎞ zA −⎜z + ⎟ ⎜ ⎝ γ ⎟ D h L AD ⎠ J AD = = L AD L AD ⎛ p⎞ ⎜ z + ⎟ − zC ⎜ ⎝ γ ⎟D ⎠ hL J CD = = CD LCD LCDExample: The water surface levels at the A and C reservoirs are respectively zA=100 m andzC= 70 for the given three-reservoir system. Reservoirs A and B are feeding reservoir C andQA = QB . Calculate the water surface level of the reservoir B. Draw the energy line of thesystem. The physical characteristics of the pipes are, zB=89.47m E.L. zA=100m B A pD γ zC=70m zD D C Figure 3.9 Pipe Length (m) Diameter (mm) f AD 500 150 0.03 BD 1000 200 0.02 DC 1500 250 0.03 14 Prof. Dr. Atıl BULU
- 15. Solution: The head loss along the pipes 1 and 3 is, h L1 + h L3 = z A − z C = 100 − 70 = 30m (1)Since, Q1 = Q 2 = Q Q3 = 2QUsing the Darcy-Weisbach equation for the head loss, 2 fV 2 L fL ⎛ 4Q ⎞ hL = = ×⎜ ⎟ 2 gD 2 g ⎝ πD 2 ⎠ 8f LQ 2 hL = × 5 gπ 2 DUsing Equ. (1), 8 ⎛ f 1 L1Q12 f 2 L 2 Q 2 ⎞ 2 hL = ⎜ + ⎟ gπ 2 ⎜ D15 ⎝ D2 ⎟ 5 ⎠ 8 × 0.03 ⎡ 500 × Q 2 1500 × (2Q ) ⎤ 2 30 = ⎢ + ⎥ 9.81× π 2 ⎣ 0.15 5 0.25 5 ⎦ 30 = 31566Q 2 Q = 0.0308 m 3 sec = 30.8 lt sec Q1 = 0.0308 m 3 sec 4Q1 4 × 0.0308 V1 = = = 1.74 m sec πD 1 2 π × 0.15 2 Q3 = 2Q1 = 2 × 0.0308 = 0.0616 m 3 sec 4Q3 4 × 0.0616 V3 = = = 1.26 m sec πD 2 3 π × 0.25 2Head losses along the pipes are, f 1V12 L1 0.03 ×1.74 2 × 500 h L1 = = = 15.43m 2 gD1 19.62 × 0.15 f 3V32 L3 0.03 × 1.26 2 ×1500 h L3 = = = 14.57 m 2 gD3 19.62 × 0.25 hl1 + h L3 = 15.43 + 14.57 = 30.00mFor the pipe 2, 15 Prof. Dr. Atıl BULU
- 16. Q1 = Q 2 = 0.0308 m 3 sec 4Q 2 4 × 0.0308 V2 = = = 0.98 m sec πD 2 2 π × 0.20 2 f 2V 22 L2 0.02 × 0.98 2 × 1000 h L2 = = = 4.90m 2 gD 2 19.62 × 0.20Water surface level of the reservoir B is, ⎛ p⎞ ⎜ z + ⎟ = z A − hL1 = 100.00 − 15.43 = 84.57m ⎜ ⎝ γ ⎟D ⎠ ⎛ p⎞ ⎜ z + ⎟ = z C + h L3 = 70.00 + 14.57 = 84.57 m ⎜ ⎝ γ ⎟D ⎠ ⎛ p⎞ z B = ⎜ z + ⎟ + h L2 = 84.57 + 4.90 = 89.47m ⎜ ⎟ ⎝ γ ⎠D3.4.2 Parallel PipesConsider a pipe that branches into two parallel and then rejoins as in Fig….. . In pipelinepractice, looping or laying a pipeline parallel to an existing pipeline is a standard method ofincreasing the capacity of the line. A problem involving this configuration might be todetermine the division of flow in each pipe given the total flow rate. In such problems,velocity heads, and minor losses are usually neglected, and calculations are made on the basisof coincident energy and hydraulic grade lines. hL1 = hL2 E.L PA/γ Q1, L1, D1, f1 PB/γ Q Q A B Q2, L2, D2, f2 Figure 3.10 16 Prof. Dr. Atıl BULU
- 17. Evidently, the distribution of flow in the branches must be such that the same head loss occursin each branch; if this were not so there would be more than one energy line for the pipeupstream and downstream from the junctions.Application of the continuity principle shows that the discharge in the mainline is equal to thesum of the discharges in the branches. Thus the following simultaneous equations may bewritten. h L1 = h L2 (3.13) Q = Q1 + Q 2Using the equality of head losses, f 1 L1V12 f L V2 = 2 2 2 D1 2 g D2 2 g 12 V1 ⎛ f 2 L2 D1 ⎞ =⎜ ⎟ (3.14) V 2 ⎜ f 1 L1 D 2 ⎟ ⎝ ⎠If f1 and f2 are known, the division of flow can be easily determined.Expressing the head losses in terms of discharge through the Darcy-Weisbach equation, fV 2 L fL 16Q 2 hL = × D 2 g 2 gD π 2 D 4 ⎛ 16 fL ⎞ hL = ⎜ 2 5 ⎜ 2π gD ⎟×Q2 ⎟ ⎝ ⎠ (3.14) h L = KQ 2Substituting this to the simultaneous Equs (3.12), K 1Q12 = K 2 Q 2 2 (3.15) Q = Q1 + Q 2Solution of these simultaneous equations allows prediction of the division of a discharge Qinto discharges Q1 and Q2 when the pipe characteristics are known. Application of these 17 Prof. Dr. Atıl BULU
- 18. principles allows prediction of the increased discharge obtainable by looping an existingpipeline.Example: A 300 mm pipeline 1500 m long is laid between two reservoirs having a differenceof surface elevation of 24. The maximum discharge obtainable through this line is 0.15m3/sec.When this pipe is looped with a 600 m pipe of the same size and material laid parallel andconnected to it, what increase of discharge may be expected?Solution: For the single pipe, using Equ. (3.14), h L = KQ 2 24 = K × 0.15 2 K = 1067 1 24m B, LB=600m A 1 Figure 3.11K for the looped section is, L2 600 KA = K= ×1067 L 1500 K A = 427For the unlooped section, L − L2 K′ = K L 1500 − 600 K′ = ×1067 1500 K ′ = 640 18 Prof. Dr. Atıl BULU
- 19. For the looped pipeline, the headloss is computed first in the common section plus branch Aas, h L = K ′Q 2 + K A Q A 2 (1) 24 = 640 × Q 2 + 427 × Q A 2And then in the common section plus branch B as, 24 = 640 × Q 2 + 427 × Q B 2 (2)In which QA and QB are the discharges in the parallel branches. Solving these by eliminatingQ shows that QA=QB (which is expected to be from symmetry in this problem). Since, fromcontinuity, Q = Q A + QB Q QA = 2Substituting this in the Equ (1) yields, 24 = 640 × (2Q A ) + 427 × Q A 2 2 24 = 2987Q A 2 Q A = 0.09 m 3 sec Q = 0.18 m 3 secThus the gain of discharge capacity by looping the pipe is 0.03 m3/sec or 20%.Example: Three pipes have been connected between the points A and B. Total dischargebetween A and B is 200 lt/sec. Physical characteristics of the pipes are, f 1 = f 2 = f 3 = 0.02 L1 = L2 = L3 = 1000m D1 = 0.10m, D 2 = 0.20m, D3 = 0.30mCalculate the discharges of each pipe and calculate the head loss between the points A and B.Draw the energy line of the system.Solution: Head losses through the three pipes should be the same. 19 Prof. Dr. Atıl BULU
- 20. h L1 = h L2 = h L3 J 1 L1 = J 2 L2 = J 3 L3Since the pipe lengths are the same for the three pipes, hL= 13.33m 1 2 A B 3 Figure J1 = J 2 = J 3 f V2 f 1 16Q 2 8 f Q2 J= × = × × = × D 2 g D 2 g π 2 D 4 gπ 2 D 5And also roughness coefficients f are the same for three pipe, the above equation takes theform of, Q12 2 Q2 Q32 = = D15 5 D2 5 D3Using this equation, 52 2.5 Q1 ⎛ D1 ⎞ ⎛ 0.10 ⎞ =⎜ ⎟ =⎜ ⎟ = 0.177 Q2 ⎜ D2 ⎟ ⎝ ⎠ ⎝ 0.20 ⎠ Q1 = 0.177Q 2 52 2.5 Q3 ⎛ D3 ⎞ ⎛ 0.30 ⎞ =⎜ ⎟ =⎜ ⎟ = 2.76 Q2 ⎜ D2 ⎟ ⎝ ⎠ ⎝ 0.20 ⎠ Q3 = 2.76Q 2 20 Prof. Dr. Atıl BULU
- 21. Substituting these values to the continuity equation, Q1 + Q 2 + Q3 = 0.200 0.177Q 2 + Q 2 + 2.76Q 2 = 0.200 3.937Q 2 = 0.200 Q 2 = 0.051 m 3 sec Q1 = 0.177 × 0.051 = 0.009 m 3 sec Q3 = 2.76 × 0.051 = 0.140 m 3 sec Q1 + Q 2 + Q3 = 0.009 + 0.051 + 0.140 = 0.200 m 3 secCalculated discharges for the parallel pipes satisfy the continuity equation.The head loss for the parallel pipes is, 8 f Q32 h L AB = × 5 × L3 gπ 2 D3 8 × 0.02 0.14 2 h L AB = × × 1000 = 13.33m 9.81× π 2 0.30 53.4.2 Branching Pipes zA pc p min A ≥ pB γ γ zC γ C pD p min B ≥ γ γ zD D Figure 3.12Consider the reservoir A supplying water with branching pipes BC and BD to a city. Pressureheads (p/γ) are required to be over 25 m to supply water to the top floors in multistorybuildings. 21 Prof. Dr. Atıl BULU
- 22. pC pD ≥ 25m, ≥ 25m γ γUsing the continuity equation, Q AB = Q BC + Q BDTwo possible problems may arise for these kinds of pipeline systems; a) The physical characteristics of the system such as the lengths, diameters and friction factors, and also the discharges of each pipe are given. Water surface level in the reservoir is searched to supply the required (p/γ)min pressure head at points C and D. The problem can be solved by following these steps, 1) Velocities in the pipes are calculated using the given discharges and diameters. Q 4Q V= = A πD 2 2) Head losses for the BC and BD pipes are calculated. f V2 hL = × ×L D 2gPiezometric head at junction B is calculated by using the given geometric elevations of thepoints C and D. ⎛ p⎞ ⎛ p⎞ ⎜ z + ⎟ = ⎜ ⎟ + z C − h LBC ⎜ (1) ⎝ γ ⎠ B ⎜ γ ⎟ min ⎟ ⎝ ⎠ ⎛ p⎞ ⎛ p⎞ ⎜ z + ⎟ = ⎜ ⎟ + z D − h LBD ⎜ ⎜ (2) ⎝ γ ⎠ B ⎝ γ ⎟ min ⎟ ⎠Since there should be only one piezometric head (z+p/γ) at any junction, the largest valueobtained from the equations (1) and (2) is taken as the piezometric head for the junction B.Therefore the minimum pressure head requirement for the points C and D has been achieved. 22 Prof. Dr. Atıl BULU
- 23. 3) The minimum water surface level at the reservoir to supply the required pressure head at points C and D is calculated by taking the chosen piezometric head for the junction, ⎛ p⎞ z A = ⎜ z + ⎟ + h L AB ⎜ ⎝ γ ⎟B ⎠ b) The physical characteristics and the discharges of the pipes are given. The geometric elevations of the water surface of the reservoir and at points C and D are also known. The minimum pressure head requirement will be checked, 1) VAB , VBC and VBD velocities in the pipes are calculated. 2) Head losses along the pipes AB, BC and BD are calculated. 3) Piezometric heads at points B, C and D are calculated. ⎛ p⎞ ⎜ z + ⎟ = z A − h L AB ⎜ ⎝ γ ⎟B ⎠ ⎛ p⎞ ⎛ p⎞ ⎛ p⎞ ⎜ z + ⎟ = ⎜ z + ⎟ − h LBC ≥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜γ ⎟ ⎝ γ ⎠C ⎝ γ ⎠ B ⎝ ⎠ min ⎛ p⎞ ⎛ p⎞ ⎛ p⎞ ⎜ z + ⎟ = ⎜ z + ⎟ − h LBD ≥ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜γ ⎟ ⎝ γ ⎠D ⎝ γ ⎠B ⎝ ⎠ min 4) If the minimum pressure head requirement is supplied at points C and D, the pipeline system has been designed according to the project requirements. If the minimum pressure head is not supplied either at one of the points or at both points, a) The reservoir water level is increased up to supply the minimum pressure head at points C and D by following the steps given above. b) Head losses are reduced by increasing the pipe diameters.Example: Geometric elevation of the point E is 50 m for the reservoir system given. Therequired minimum pressure at point E is 300 kPa. The discharge in the pipe DE is 200 lt/sec.The physical characteristics of the reservoir-pipe system are given as, Pipe Length(m) Diameter(mm) f ABD 1000 200 0.03 ACD 1000 300 0.03 DE 1500 400 0.03 23 Prof. Dr. Atıl BULU
- 24. 116.59m 94.50m A 80m B 50m C D E Figure 3.13Calculate the minimum water surface level to supply the required pressure at the outlet E.Draw the energy line of the system. γwater= 10 kN/m3.Solution: The velocity and the head loss along the pipe DE are, 4Q 4 × 0.2 V De = = = 1.59 m sec πD 2 π × 0.4 2 2 fV DE L DE 0.03 × 1.59 2 × 1500 h LDE = = = 14.50m 2 gD 19.62 × 0.4Since, h L ABD = h L ACD 2 2 fV ABD L ABD fV ACD L ACD = 2 gD ABD 2 gD ACD 0.03 × V ABD × 1000 0.03 × V ACD × 1000 2 2 = 2 g × 0.2 2 g × 0.3 0.3 × V ABD = 0.2 × V ACD 2 2 V ABD = 0.816V ACDApplying the continuity equation, Q = Q ABD + Q ACD π 4 ( ) × 0.2 2 × V ABD + 0.3 2 × V ACD = 0.200 2 0.04 × V ABD + 0.09 × V ACD = 0.255 0.04 × 0.816 × V ACD + 0.09 × V ACD = 0.255 V ACD = 2.08 m sec V ABD = 0.816 × 2.08 = 1.70 m sec 24 Prof. Dr. Atıl BULU
- 25. The discharges of the looped pipes are, πD ABD 2 Q ABD = × V ABD 4 π × 0.2 2 Q ABD = × 1.70 4 Q ABD = 0.053 m 3 sec π × D ACD 2 Q ACD = × V ACD 4 π × 0.3 2 Q ACD = × 2.08 4 Q ACD = 0.147 m 3 secThe head loss along the looped pipes is, 2 fV ABD L ABD h L ACD = h L ABD = 2 gD ABD 0.03 × 1.70 2 ×1000 hL = 19.62 × 0.2 h L = 22.09mPiezometric head at the outlet E is, ⎛ p ⎞ 300 ⎜z+ ⎜ ⎟ = 50 + ⎟ = 80m ⎝ γw ⎠E 10The required minimum water surface level at the reservoir is, ⎛ p⎞ z A = ⎜ z + ⎟ + h LDe + h L ACD ⎜ ⎝ γ ⎟E ⎠ z A = 80 + 14.50 + 22.09 z A = 116.59mExample: The discharges in the AB and AC pipes are respectively Q1=50 lt/sec and Q2=80lt/sec for the pipe system given. The required pressure at the B and C outlets is 200 kPa and 25 Prof. Dr. Atıl BULU
- 26. the geometric elevations for these points are zB= 50 m and zc= 45 m. The physicalcharacteristics of the pipe system are, Pipe Length (m) Diameter (mm) f RA 2000 300 0.02 AB 1000 350 0.02 AC 1500 400 0.02Calculate the minimum water surface level of the reservoir R to supply the required pressureat the outlets. Draw the energy line of the system. γwater= 10 kN/m3. 93.79m 70.79m pB R = 20m γ 45m B pc = 24.22m A γ 50m C Figure 3.14Solution: Since the discharges in the pipes are given, 4Q AB 4 × 0.050 V AB = = = 0.52 m sec πD 2 AB π × 0.35 2 2 fV AB L AB 0.02 × 0.52 2 × 1000 h L AB = = = 0.79m 2 gD AB 19.62 × 0.35 ⎛ p⎞ ⎛ p⎞ ⎜ z + ⎟ = ⎜ z + ⎟ + h L AB ⎜ ⎟ ⎝ γ ⎠A ⎝ γ ⎟B ⎜ ⎠ ⎛ p⎞ ⎛ 200 ⎞ ⎜ z + ⎟ = ⎜ 50 + ⎜ ⎟ ⎟ + 0.79 = 70.79m ⎝ γ ⎠A ⎝ 10 ⎠ 26 Prof. Dr. Atıl BULU
- 27. 4Q AC 4 × 0.080 V AC = = = 0.64 m sec πD 2 AC π × 0.40 2 2 fV AC L AC 0.02 × 0.64 2 ×1500 h L AC = = = 1.57m 2 gD AC 19.62 × 0.40 ⎛ p⎞ ⎛ p⎞ ⎜ z + ⎟ = ⎜ z + ⎟ + h L AC ⎜ ⎝ γ ⎠ A ⎝ γ ⎟C ⎟ ⎜ ⎠ ⎛ p⎞ 200 ⎜ z + γ ⎟ = 45 + 10 + 1.57 = 66.57m ⎜ ⎟ ⎝ ⎠AIf we take the largest piezometric head of the outlets 70.79 m, the pressure will be 200kPa atthe outlet B, and the pressure at outlet C is,⎛ p⎞ ⎛ p⎞⎜ γ ⎟ = ⎜ z + γ ⎟ − h L AC − z C⎜ ⎟ ⎜ ⎟⎝ ⎠C ⎝ ⎠A⎛ p⎞⎜ ⎟ = 70.79 − 1.57 − 45.00 = 24.22m⎜ ⎟⎝ γ ⎠Cp = γ w × 24.22 = 10 × 24.22 = 242.2kPa > 200kPaWater surface level at the reservoir R is, Q RA = Q AB + Q AC = 0.05 + 0.08 = 0.13 m 3 sec 4Q RA 4 × 0.13 V RA = = = 1.84 m sec πD 2 RA π × 0.30 2 2 fV RA L RA 0.02 × 1.84 2 × 2000 h LRA = = = 23.00m 2 gD RA 19.62 × 0.30 ⎛ p⎞ z R = ⎜ z + ⎟ + h LRA = 70.79 + 23.00 = 93.79m ⎜ ⎝ γ ⎟A ⎠3.4.4. Pump SystemsWhen the energy (head) in a pipe system is not sufficient enough to overcome the head lossesto convey the liquid to the desired location, energy has to be added to the system. This isaccomplished by a pump. The power of the pump is calculated by, N = γQH (3.16)Where γ = Specific weight of the liquid (N/m3), Q = Discharge (m3/sec), H = Head to besupplied by pump (m), N = Power of the pump (Watt). Calculated pump power should bedivided by η = Efficiency factor of the pump. 27 Prof. Dr. Atıl BULU
- 28. Example: Reservoir D is fed by a pump and pipe system. Pressure heads in the pipe flow atthe entrance and at the outlet from the pump are respectively 5 and 105 m. The geometric ofthe pipe before the pump is 0.50 m. The efficient power of the pump is 75 kW. The frictionfactor at all the pipe system is f = 0.03. (Minor losses will be neglected). a) Calculate the discharge in pipe BC, b) What will be the discharges in the parallel C1D and C2D pipes? c) Calculate the water surface level of the reservoir D. d) Draw the energy line of the system. Pipe Length (m) Diameter (mm) f BC 1000 300 0.03 C1D 3000 400 0.03 C2D 1500 200 0.03 D 1 0.50m 2 Pump C A B Figure 3.Solution: a) The head to be supplied by the pump is, ΔH pump = 105 − 5 = 100mThe efficient power of the pump is, γQH N= η 28 Prof. Dr. Atıl BULU
- 29. Nη 75000 Q= = = 0.076 m 3 sec = 76 lt sec γH 1000 × 9.81×100b) The velocity in the pipe BC, 4Q 4 × 0.076 V BC = = = 1.08 m sec πD BC 2 π × 0.30 2The head loss is, f V2 h LBC = × BC × L BC D BC 2 g 0.03 1.08 2 h LBC = × × 1000 = 5.94m 0.30 19.62Since, h L1 = h L2 f V12 f V 22 × × L1 = × × L2 D1 2 g D2 2 g 0.03 V12 0.03 V 22 × × 3000 = × ×1500 0.40 2 g 0.20 2 g V12 = V 22 V1 = V 2 = VBy equation of continuity, Q = Q1 + Q 2 πD12 Q1 = × V1 4 πD 22 Q2 = ×V2 4 π × 0.4 2 0.16π Q1 = × V1 = ×V 4 4 π × 0.2 2 0.40π Q2 = × V2 = ×V 4 4 Q1 = 4Q 2 29 Prof. Dr. Atıl BULU
- 30. Q1 + Q 2 = 76 m 3 sec 5Q 2 = 76 lt sec Q1 = 15.2 lt sec Q 2 = 60.8 lt secc) The velocity and the head loss in the parallel pipes are, 4 × 0.0152 V1 = V 2 = V = = 0.48 m sec π × 0.20 2 f V2 0.03 0.48 2 h L1 = h L2 = × × L1 = × × 3000 = 2.64m D1 2 g 0.40 19.62Total head loss is, ∑h L = h LBC + h L1 ∑h L = 5.94 + 2.64 = 8.58mThe water surface level of the reservoir D is, ⎛ p⎞ z D = ⎜ z + ⎟ − ∑ hL ⎜ ⎝ γ ⎟B ⎠ z D = 0.50 + 105 − 8.58 = 96.92m 105.50m 99.56m 96.92m D p/ γ=100m 5.50m C 0.50m A pump B 30 Prof. Dr. Atıl BULU
- 31. Example: Reservoir A is feeding reservoirs B and C by a pump and pipe system. Thedischarge to reservoir C is Q2 = 0.10 m3/sec. If the efficient coefficient of the pump is η =0.70 n, what will be required power of pump? Draw the energy line of the system . Thephysical characteristics of the pipe system are, Pipe Diameter (mm) Length (m) f 1 300 400 0.020 2 150 300 0.015 3 200 1000 0.025 110m B 90m 2 65m C 3 A 1 Figure…..Solution: Beginning with pipe 2, Q 2 4Q 2 V2 = = A2 πD 22 4 × 0.10 V2 = = 5.66 m sec π × 0.15 2 f 2 V 22 h L2 = × × L2 D2 2 g 0.015 5.66 2 h L2 = × × 300 ≅ 49m 0.15 19.62Energy level at the outlet of the pump is, 31 Prof. Dr. Atıl BULU
- 32. ⎛ p⎞ H Poutlet = ⎜ z + ⎟ = z C + h L2 = 90 + 49 = 139m ⎜ ⎝ γ ⎟P ⎠Pipe 3:Head loss along the pipe 3 is, h L3 = H Poutlet − z B = 139 − 110 = 29m f 3 V32 h L3 = × × L3 D3 2 g 2 0.025 V3 29 = × × 1000 0.200 19.62 V3 = 2.13 m sec πD32 π × 0.20 2 Q3 = × V3 = × 2.13 = 0.067 m 3 sec 4 4Pipe 1: Q1 = Q 2 + Q3 Q1 = 0.100 + 0.067 = 0.167 m 3 sec 4Q1 4 × 0.167 V1 = = = 2.36 m sec πD 1 2 π × 0.30 2 f 1 V12 0.020 2.36 2 h L1 = × × L1 = × × 400 = 7.57 m D1 2 g 0.30 19.62Energy level at the entrance to the pump is, H Pent = z A − h L1 = 65.00 − 7.57 = 57.43mThe energy (head) to be supplied by pump is, ΔH p = H Poutlet − H Pent = 139.00 − 57.43 = 81.57 mThe required power of the pump is, γQΔH P NP = η 9.81× 1000 × 0.167 × 81.59 NP = = 190952W ≅ 191kW 0.70 32 Prof. Dr. Atıl BULU
- 33. 139m 110m B 90m 2 65m C 57.43m 3 A 1 PumpExample: Reservoir B is fed by reservoir A with 600 lt/sec discharge. The water surfacelevels in the reservoirs are zA = 100 m and zB = 90 m. If the pipe diameter is D = 0.50m, thefriction factor is f = 0.02, and the efficiency factor of the pump is η = 0.70, calculate therequired pump power. The lengths of the pipes are respectively L1 = 500 m and L2 = 1000 m.Draw the energy line of the system. 100m A pump 1 90m 2 B C FigureSolution: Since the discharge of the system is known, the head loss from reservoir A to thepump can be calculated as, 4Q 4 × 0.60 V= = = 3.06 m sec πD 2 π × 0.50 2 33 Prof. Dr. Atıl BULU
- 34. f V2 h L AC = × × L AC D 2g 0.02 3.06 2 h L AC = × × 500 = 9.54m 0.50 19.62The energy level at the entrance to the pump is, H Cent = 100 − 9.54 = 90.46mSince the pipe diameter is same along the pipe, and L2 = 2L1, the head loss along the pipe 2is, h L2 = 2hL1 = 2 × 9.54 = 19.08mThe energy level at the outlet of the pump should be, H Cout = 90.00 + 19.08 = 109.80mThe pumping is then, ΔH pump = 109.80 − 90.46 = 18.62mThe required power of the pump is, γQH 9810 × 0.600 ×18.62 N= = = 156568W ≅ 157kW η 0.70 109.08m 100m 90.46m A 90m C B 34 Prof. Dr. Atıl BULU

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