CLOSED CONDUITS II CHAPTER 5 (2) -- ROBERSON ET AL., WITH ADDITIONS EXPLICIT EQUATIONS FOR PIPE FLOW DARCY-WEISBACH FRICTION COEFFICIENT BASED ON RELATIVE ROUGHNESS AND REYNOLDS NUMBER: f = 0.25 { log [ (ks /(3.7D)) + (5.74/Re0.9) ] } -2                              (Eq. 5-17) RESTRICTED TO 0.00001 < ks /D < 0.02 4000 < Re < 100,000,000 FOR EXAMPLE, IN PROBLEM 1 EXAMPLE (CHAPTER 5-1): ks /D = 0.00065 Re = 318,000 f = 0.25 { log [(0.00065/ (3.7)) + (5.74/3180000.9) ] } -2 f = 0.01907 ≅ 0.018       OK! DISCHARGE BASED ON RELATIVE ROUGHNESS AND HEAD LOSS: A* = D1.5(g hf /L)0.5       [UNITS OF KINEMATIC VISCOSITY]                              (Eq. 5-18a) Q = -2.22 A* D log { (ks /(3.7D)) + (1.78 ν /A*)}                              (Eq. 5-18b) FOR EXAMPLE, IN PROBLEM 2 EXAMPLE (CHAPTER 5-1): D = 0.2 M hf/L = 0.0116 ks/D = 0.00065 ν = 1.0 × 10-6 M2/S A* = 0.21.5 [(9.81) (0.0116)]0.5 = 0.0302 Q = -2.22 (0.0302) (0.2) × log [(0.00065/3.7) + ((1.78 × 1.0 × 10-6) / (0.0302))] Q = 0.0487 M3/S ≅ 0.05 M3/S       OK! PIPE DIAMETER BASED ON DISCHARGE, HEAD LOSS, AND ROUGHNESS: B* = g hf /L                              (Eq. 5-19a) D = 0.66 [(ks1.25 Q 9.5 / B* 4.75) + (ν Q 9.4 / B* 5.2) ] 0.04                              (Eq. 5-19b) THIS EQUATION IS DIMENSIONLESS! FOR EXAMPLE, IN PROBLEM 3 EXAMPLE (CHAPTER 5-1): Q = 0.05 M3/S hf/L = 0.0116 ks/D = 0.00065 ν = 1.0 × 10-6 M2/S FROM FIG. 5-5:   ks = 0.0004 FT = 0.00012 M THEN: B* = 9.81 (0.0116) = 0.1138 D = 0.66 [(0.000121.25 0.059.5/0.11384.75) + (0.000001) 0.059.4 /0.11385.2) ] 0.04 D = 0.205 M ≅ 0.2 M       OK! SUMMARY: f = 0.25 { log [ (ks /(3.7D)) + (5.74/Re0.9) ] }-2       [dimensionless]                              (Eq. 5-17) A* = D1.5(g hf /L)0.5       [viscosity: L2/T]                                                                          (Eq. 5-18a) Q = -2.22 A* D log { (ks /(3.7D)) + (1.78 ν / A*) }       [discharge: L3/T]                              (Eq. 5-18b) B* = g hf /L       [acceleration: L/T2]                                                                                          (Eq. 5-19a) D = 0.66 [(ks1.25 Q 9.5 / B* 4.75) + (ν Q 9.4 / B* 5.2) ] 0.04       [diameter: L]                              (Eq. 5-19b) THESE FORMULAS ARE DIMENSIONALLY CONSISTENT! EXAMPLE 5-5: ITERATIVE SOLUTION USING HEAD LOSS EQUATION DETERMINE THE DIAMETER OF STEEL PIPE NEEDED TO DELIVER WATER (20oC) AT Q = 2 M3/S FROM A RESERVOIR AT ELEV. 60 M, TO A RESERVOIR 200 M AWAY WITH AN ELEV. 30 M. ASSUME A SQUARE-EDGED INLET AND OUTLET AND NO BENDS IN THE PIPE. ASSUME THERE ARE TWO OPEN GATE VALVES IN THE PIPE, WITH Kv = 0.2. Writing the energy equation from the upper to the lower reservoir: p1/γ + V12/(2g) + z1 = p2/γ + V22/(2g) + z2 + ΣhL 0 + 0 + z1 = 0 + 0 + z2 + [Ke + 2Kv + KE + f L/D ] V2/(2g) z1 = z2 + [Ke + 2Kv + KE + f L/D ] V2/(2g) z1 = z2 + [Ke + 2Kv + KE + f L/D ] [(Q2/A2)/(2g)] z1 = z2 + [Ke + 2Kv + KE + f L/D ] Q2 /(A2 × 2g) Select: -- ks = 0.046 MM = 0.000046 M (STEEL) (FIG. 5-5) -- Ke = 0.5 (square-edged inlet) -- Kv = 0.2 (open gate) -- KE = 1.0 (square-edged) -- ASSUME f = 0.02 60 = 30 + [0.5 + 0.4 + 1.0 + 0.02 (200)/D] × (2)2 /[(0.25πD2)2 (2 × 9.81)] 30 = [1.9 + (4/D)] × 4/[(0.25πD2)2(2)(9.81)] 30 = [1.9 + (4/D)] × 4/[(0.252π2D4) (2) (9.81)] 30 = [1.9 + (4/D)] × [0.33/D4] SOLVE FOR DIAMETER D: D = 0.56 M A = (π/4) D2 = 0.246 M2 V = Q/A = 2/ (0.246) = 8.12 M/S Re = VD/ν = 8.12 (0.56) /(0.000001) = 4,547,200 f = 0.25 / {log [(ks /3.7D) + (5.74/Re0.9)]}2 ks /(3.7D) = 0.000046 / [(3.7) (0.56)] = 0.0000222 f = 0.25 / {log [0.0000222 + (5.74/4,547,2000.9)]}2 f = 0.25 / {log [0.0000222 + 0.000005847]}2 f = 0.012 SUBSTITUTE THIS VALUE BACK TO ENERGY EQUATION: 60 = 30 + [0.5 + 0.4 + 1.0 + 0.012 (200)/D] × (2)2 /[(0.25πD2)2 (2 × 9.81)] 30 = [1.9 + (2.4/D)] × 4/[(0.25πD2)2(2)(9.81)] 30 = [1.9 + (2.4/D)] × 4/[(0.252π2D4) (2) (9.81)] 30 = [1.9 + (2.4/D)] × [0.33/D4] SOLVE FOR DIAMETER D: D = 0.52 M A = (π/4) D2 = 0.2124 M2 V = Q/A = 2/ (0.2124) = 9.42 M/S Re = VD/ν = 9.42 (0.52) /(0.000001) = 4,898,400. f = 0.25 / {log [(ks /3.7D) + (5.74/Re0.9)]}2 ks /(3.7D) = 0.000046 / [(3.7) (0.52)] = 0.0000239 f = 0.25 / {log [0.0000239 + (5.74/4,898,4000.9)]}2 f = 0.25 / {log [0.0000239 + 0.000005468]}2 f = 0.012       OK! EXAMPLE 5-6: ITERATIVE SOLUTION USING DIAMETER EQUATION ASSUME f = 0.02 EQUIVALENT LENGTH OF PIPE Le FOR MINOR LOSSES: hL = f [(L + Le)/D] V2/(2g) hL = f (L/D) V2/(2g) + f (Le/D) V2/(2g) f (Le/D) V2/(2g) = [0.5 + (2×0.2) + 1.0] V2/(2g) f (Le/D) = 1.9 FROM WHICH EQUIVALENT LENGTH OF PIPE Le Le = 1.9 D/f L = L + Le = 200 + 1.9 (D / 0.02) = 200 + 95 D WITH hf = 30 M AND L = 200 + 95 D PLUG IN DIAMETER EQUATION 5-19b (AND GET D = 0.51 M BY ITERATION). B* = g hf /L                              (Eq. 5-19a) D = 0.66 [(ks1.25 Q 9.5 / B* 4.75) + (ν Q 9.4 / B* 5.2) ] 0.04                              (Eq. 5-19b) B* = (9.81 × 30) / (200 + 95 D)                              (Eq. 5-19a) D = 0.66 [0.0004061.25 × 2 9.5 / B* 4.75) + (1. × 10-6 × 2 9.4 / B* 5.2) ] 0.04                              (Eq. 5-19b) GET D = 0.51 M BY ITERATION. GET NEW Re = 5 × 106 GET NEW f (Eq. 5-17): f = 0.0122 GET NEW Le = 1.9D/0.0122 = 155.7 D PLUG IN DIAMETER EQUATION 5-19b AND GET D = 0.506 M BY ITERATION. GET NEW Re = 5 × 106 GET NEW f (Eq. 5-17): f = 0.0122 GET NEW Le = 1.9D/ (0.0122) = 156 D NO SIGNIFICANT CHANGE. D = 0.51 M       OK! HYDRAULIC AND ENERGY GRADE LINES THE LIQUID WOULD RISE TO A HEIGHT p/γ THUS, HYDRAULIC GRADE LINE (HGL) ENERGY GRADE LINE (EGL) IS ABOVE HGL A DISTANCE OF αV2/2g HINTS IF VELOCITY HEAD IS ZERO, EGL AND HGL COINCIDE. EGL WILL SLOPE DOWNWARD IN THE DIRECTION OF FLOW. PUMP SUPPLIES ENERGY TO THE FLOW. THUS, AN ABRUPT RISE IN EGL OCCURS FROM U/S TO D/S OF PUMP. TURBINE ABSORBS ENERGY FROM THE FLOW. THUS, AN ABRUPT DROP IN EGL OCCURS FROM U/S TO D/S OF TURBINE. GRADUAL EXPANSION AT THE OUTLET WILL CONVERT MUCH OF VELOCITY HEAD TO PRESSURE HEAD. WHEN THE PRESSURE IS ZERO, THE HGL WILL COINCIDE WITH THE WATER LEVEL. FOR STEADY FLOW IN A PIPE WITH UNIFORM CHARACTERISTICS, THE HEAD LOSS PER UNIT OF LENGTH (SLOPE OF EGL AND HGL) WILL BE CONSTANT. IF FLOW PASSAGE CHANGES DIAMETER, THE DISTANCE BETWEEN EGL AND HGL WILL CHANGE. WHEN VELOCITY INCREASES, EGL WILL INCREASE BECAUSE HEAD LOSS PER UNIT OF LENGTH INCREASES. IF HGL FALL BELOW PIPE, THE PRESSURE p/γ IS NEGATIVE. THIS INDICATES A PRESSURE LOWER THAN ATMOSPHERIC. IF THE NEGATIVE PRESSURE REACHES (OR COMES CLOSE TO) 1 ATMOSPHERE, CAVITATION WILL OCCUR. AT MEAN SEA LEVEL, THE ATMOSPHERIC PRESSURE IS: 1 ATM = 101.25 kPa = 101,325 N/M2 = 10.333 M 1 ATM = 14.69 PSI = 2,115 PSF = 33.900 FT. CAVITATION CAN CAUSE STRUCTURAL DAMAGE TO THE PIPE. CAVITATION OCCURRED IN GLEN CANYON DAM DURING THE FLOOD OF JUNE 1983, WHEN THE CLOSE-CONDUIT EMERGENCY SPILLWAY WAS USED FOR THE FIRST TIME, FOLLOWING WIDESPREAD FLOODING IN THE COLORADO RIVER AS A RESULT OF THE 1983 EL NIÑO EVENT. Cavitation is the formation of vapour cavities in a liquid  i.e. small liquid-free zones ("bubbles" or "voids")  that are the consequence of forces acting upon the liquid. It usually occurs when a liquid is subjected to rapid changes of pressure that cause the formation of cavities where the pressure is relatively low. When subjected to higher pressure, the voids implode and can generate an intense shock wave. What is cavitation? Glen Canyon Dam, Arizona. CAVITATION OCCURRED IN TARBELA DAM, PAKISTAN, IN 1974, WHEN THE OUTLET STRUCTURES WERE USED FOR THE FIRST TIME, FOLLOWING AN EARLY UNEXPECTED FILLING OF THE RESERVOIR. Tarbela Dam, Pakistan, the largest earth dam in the world. Spillway of Tarbela Dam, Pakistan. IF PRESSURE DECREASES TO VAPOR PRESSURE AND STAYS THAT LOW, A LARGE VAPOR CAVITY CAN FORM. ENGINEER SHOULD BE EXTREMELY CAUTIOUS ABOUT NEGATIVE PRESSURE HEADS IN PIPES. HEAD-DISCHARGE RELATIONS FOR PUMP OR TURBINE HEADS hp AND ht ARE A FUNCTION OF THE DISCHARGE FOR A MACHINE (PUMP OR TURBINE) OPERATING AT A GIVEN SPEED. TYPICAL PERFORMANCE IS GIVEN IN FIG. 5-10. PERFORMANCE OR CHARACTERISTIC OF MACHINE: HEAD IS INVERSELY RELATED TO DISCHARGE FOR A GIVEN RPM OF MACHINE. OTHER PERFORMANCE CURVES, SUCH AS EFFICIENCY AND POWER VS DISCHARGE ARE INCLUDED WITH MACHINE SPECIFICATIONS. IF Q IS GIVEN, THE HEAD IS TAKEN DIRECTLY FROM THE PERFORMANCE CURVE. IF H IS GIVEN, A SIMULTANEOUS SOLUTION OF THE ENERGY EQUATION AND THE PERFORMANCE CURVE WILL YIELD Q. MORE ON CHAPTER 8. BRANCHING PIPES CONSIDER FIG 5-11: THREE RESERVOIRS CONNECTED BY A BRANCHED PIPE SYSTEM. THE PROBLEM IS TO DETERMINE THE DISCHARGE IN EACH PIPE AND THE HEAD AT D. THERE ARE FOUR UNKNOWNS: VAD VBD VDC pD /γ A SOLUTION IS OBTAINED BY SOLVING THE ENERGY EQUATION FOR THE PIPES (NEGLECTING VELOCITY HEADS AND INCLUDING ONLY PIPE LOSSES) AND THE CONTINUITY EQUATION. zA = zD + pD/γ + fAD (LAD /DAD ) VAD2 / (2g) zB = zD + pD/γ + fBD (LBD /DBD ) VBD2 / (2g) zD + pD/γ = fDC (LDC /DDC ) VDC2 / (2g) VAD AAD + VBD ABD = VDC ADC X = VAD AAD + VBD ABD - VDC ADC ITERATIVE METHOD: FIRST ASSUME zA = zD + pD/γ      (NO FLOW IN AD, VAD = 0) pD/γ = zA - zD THEN DETERMINE VBD, VDC, QBD AND QDC (PIPE ROUGHNESS, DIAMETER, LENGTH, AND AREAS ARE KNOWN) CALCULATE X IF X IS POSITIVE , pD/γ WILL HAVE TO BE INCREASED; FLOW MUST BE DIRECTED TO RESERVOIR A. IF X IS NEGATIVE , pD/γ WILL HAVE TO BE DECREASED; FLOW MUST BE DIRECTED OUT OF RESERVOIR A. NEXT TRY ANOTHER pD/γ UNTIL X = 0. EXAMPLE Assume three reservoirs A, B, and C connected as shown below. The elevations (relative to C) are:   zA = 100 ft, zB = 150 ft, zD = 10 ft. Line AD is 4,000-ft long and 6-in diameter; BD is 10,000-ft long and 8-in diameter; DC is 5,000-ft long and 10-in diameter. Calculate the discharge in each pipe and the head at D. Assume f = 0.02 in all pipes. Neglect minor losses. SOLUTION Three energy balance equations and continuity equation: zA = zD + pD/γ + fAD (LAD /DAD ) VAD2 / (2g) zB = zD + pD/γ + fBD (LBD /DBD ) VBD2 / (2g) zD + pD/γ = fDC (LDC /DDC ) VDC2 / (2g) VAD AAD + VBD ABD = VDC ADC AAD = 0.1963 ft2 ABD = 0.3491 ft2 ADC = 0.5454 ft2 State continuity equation as follows:   VAD AAD + VBD ABD - VDC ADC = X The solution is by trial and error. Assume pD/γ and calculate resulting velocities using energy equations; then plug velocities and areas in continuity equation until X = 0. First assume VAD = 0. Then pD/γ = 90.0 ft. Start by assuming a somewhat lesser value: pD/γ = 80 ft:   X = -2.38 Then: pD/γ = 60 ft:   X = -1.33 pD/γ = 34 ft:   X = -0.05 pD/γ = 33.5 ft:   X = 0.001 The final velocities are: VAD = 5.5036 fps. VBD = 5.3432 fps. VDC = 5.3994 fps. QAD = 0.1963 × 5.5036 = 1.080 cfs. QBD = 0.3491 × 5.3432 = 1.865 cfs. QDC = 0.5454 × 5.3994 = 2.945 cfs. pD = 33.5 ft × 62.4 lb/ft 3 = 2,090 lb/ft2 Verify with ONLINEPIPE02 PARALLEL PIPES CONSIDER A PIPE THAT BRANCHES INTO TWO PARALLEL PIPES AND THEN REJOINS. THE PROBLEM MAY BE TO DETERMINE THE DIVISION OF FLOW IN EACH PIPE, GIVEN THE TOTAL FLOW RATE. HEAD LOSS MUST BE THE SAME IN EACH PIPE: hL1 = hL2 f1 (L1/D1) V12/(2g) = f2 (L2 /D2) V22/(2g) (V1/V2) 2 = (f2 L2 D1) / (f1 L1 D2) V1/V2 = [(f2 L2 D1) / (f1 L1 D2)] 1/2 IF f1 AND f2 ARE KNOWN, THE DIVISION OF FLOW CAN BE READILY ASCERTAINED. Q = V1A1 + V2A2 Q/V2 = (V1/V2) A1 + A2 V2 = Q / [(V1/V2) A1 + A2] Q2 = V2A2 Q2 = Q A2/ [(V1/V2) A1 + A2] Q2 = Q / { [(V1/V2) (A1/A2)] + 1} Q2 = Q / { [(V1/V2) (D1/D2)2] + 1} a = (V1/V2) (D1/D2)2 Q2 = Q / (a + 1) Likewise: Q = V1A1 + V2A2 Q/V1 = (V2/V1) A2 + A1 V1 = Q / [(V2/V1) A2 + A1] Q1 = V1A1 Q1 = Q A1/ [(V2/V1) A2 + A1] Q1 = Q / { { 1 / [(V1/V2) (A1/A2)]}   + 1 } Q1 = Q / { { 1 / [(V1/V2) (D1/D2)2]}   + 1 } a = (V1/V2) (D1/D2)2 Q1 = Q / [(1/a) + 1] = aQ /(a + 1) a = (V1/V2) (D1/D2)2 V1/V2 = [(f2 L2 D1) / (f1 L1 D2)] 1/2 a = [(f2/f1) (L2/L1)]1/2 (D1/D2)5/2 Q1 = aQ /(a + 1) Q2 = Q / (a + 1) Q = Q1 + Q2        OK! IF FRICTION FACTORS ARE FUNCTIONS OF Re, SOME TRIAL AND ERROR MAY BE REQUIRED. PIPE NETWORKS THE MOST COMMON PIPE NETWORK SYSTEMS ARE THE WATER DISTRIBUTION SYSTEMS FOR MUNICIPALITIES. SOURCES: WATER ENTERING THE NETWORK LOADS: WATER OUTFLOWING THE NETWORK FIG. 5-16 SHOWS A SIMPLIFIED SYSTEM WITH 2 SOURCES AND 7 LOADS. ENGINEER IS OFTEN ENGAGED TO DESIGN THE ORIGINAL SYSTEM OR TO RECOMMEND AN ECONOMICAL EXPANSION TO THE NETWORK. THE ENGINEER IS REQUIRED TO PREDICT PRESSURES THROUGHOUT THE NETWORK FOR VARIOUS OPERATING CONDITIONS, THAT IS, FOR VARIOUS COMBINATIONS OF SOURCES AND LOADS. THE SOLUTION MUST SATISFY THREE BASIC REQUIREMENTS: CONTINUITY: THE FLOW INTO A JUNCTION MUST EQUAL THE FLOW OUT OF THE JUNCTION. SINCE PRESSURE MUST BE CONTINUOUS, THE HEAD LOSS BETWEEN ANY TWO JUNCTIONS MUST BE THE SAME, REGARDLESS OF THE PATH TAKEN FROM ONE JUNCTION TO THE OTHER. THE FLOW AND HEAD LOSS MUST BE CONSISTENT WITH THE APPROPRIATE VELOCITY HEAD EQUATION. TRIAL AND ERROR SOLUTION. HARDY CROSS METHOD STEPS: -- FIRST, FLOW IS DISTRIBUTED THROUGH THE NETWORK SO THAT CONTINUITY IS SATISFIED. -- FIRST GUESS WILL NOT SATISFY REQUIREMENT OF EQUAL HEAD LOSS IN LOOP. -- A DISCHARGE CORRECTION IS APPLIED TO YIELD A ZERO NET HEAD LOSS AROUND THE LOOP. HEAD LOSS IS: hL = f (L/D) (V2/2g) hL = f (L/D) [Q2/(2gA2)] hL = f (L/D) [8Q2/(gπ2 D4)] hL = [(8 f L) /(gπ2D5)] Q2 THE HEAD LOSS IS PROPORTIONAL TO THE SQUARE OF THE DISCHARGE. hL = k Q2 k = (8 f L) /(gπ2D5) ∑ hL clockwise = ∑ hL counterclockwise ∑ k Q 2 clockwise = ∑ k Q 2 counterclockwise ∑ k Qc2 = ∑ k Qcc2 A ΔQ WILL HAVE TO BE APPLIED TO SATISFY THE HEAD LOSS REQUIREMENT. CASE 1:   IF ∑ k Qc2 > ∑ k Qcc2 ΔQ IS SUBTRACTED FROM THE CLOCKWISE FLOWS AND ADDED TO THE COUNTERCLOCKWISE FLOWS. CASE 2:   IF ∑ k Qc2 < ∑ k Qcc2 ΔQ IS SUBTRACTED FROM THE COUNTERCLOCKWISE FLOWS AND ADDED TO THE CLOCKWISE FLOWS. IN GENERAL: ∑ k (Q c - ΔQ) n = ∑ k (Qcc + ΔQ) n EXPANDING THE SUMMATION ON EITHER SIDE, AND INCLUDING ONLY TWO TERMS OF THE SERIES EXPANSION: (Qc - ΔQ) n = Qcn - nQcn-1 ΔQ EXAMPLE WITH n = 2, Qc = 100, ΔQ = 1: (100 - 1)2 = 9801 ≅ [10000 - 2 × 100 × 1 ] = 9800 ∑ k (Qcn - nQcn-1 ΔQ) = ∑ k (Qccn + nQccn-1 ΔQ) ΔQ = (∑ k Qcn - ∑ k Qccn) / (∑ nk Qcn-1 + ∑ nk Qccn-1)              (EQ. 5-29) IN THIS FORMULA, WHATEVER THE SIGN OF ΔQ IS, IT SHOULD BE SUBTRACTED FROM THE CLOCKWISE FLOWS! THE CORRECTION IS APPLIED IN A COUNTERCLOCKWISE SENSE! Pipe Network Calculator by LMNO Engineering EXAMPLE FOR THE GIVEN SOURCE AND LOADS SHOWN IN FIG. A, WHAT IS THE DISTRIBUTION OF FLOWS IN THE NETWORK AND WHAT IS THE PRESSURE AT THE LOAD POINTS IF THE PRESSURE AT THE SOURCE IS 60 PSI. ASSUME HORIZONTAL PIPES AND f = 0.012. SOLUTION: CALCULATION OF k FOR ALL PIPES: k = (8f L) /(gπ2D5) kAB = [8(0.012)(1000)] /[(32.2)(3.1416)2 (2)5] = 0.00944 LIKEWISE: kBC = 0.3021; kBD = 2.2940; kAD = 1.0590; kDE = 0.3021; kCE = 0.7516. ASSUME: QAB = 10 CFS; QAD = - 5 CFS (cc); QBD = 0 CFS; QBC = 10 CFS; QDE = - 5 CFS (cc); QCE = 0 CFS. IF SIGNS ARE ASSUMED, ΔQ SHOULD BE SUBTRACTED FROM ALL FLOWS, C, CC, AND NO FLOW. FOR n = 2, FORMULA FOR ΔQ SIMPLIFIES TO: ΔQ = (∑ k Qc2 - ∑ k Qcc2) / (∑ 2k Qc + ∑ 2k Qcc) FIRST TRIAL LOOP ABD: PIPE kQ2 2kQ AB (c) 0.944 0.189 AD (cc) -26.475 10.59 BD (no flow) 0.000 0.000 ∑ -25.531 10.78 ΔQ = -25.531/10.78 = -2.4 ΔQ TO BE SUBTRACTED FROM ALL FLOWS. THEN: QAB = 10 - (-2.4) = 12.4 QAD = -5 - (-2.4) = - 2.6 QBD = 0 - (-2.4) = 2.4 LOOP BCDE: PIPE kQ2 2kQ BC (c) 30.21 6.042 BD (no flow) 0.000 0.000 CE (no flow) 0.000 0.000 DE (cc) -7.55 3.02 ∑ 22.66 9.062 ΔQ = 22.66/9.062 = 2.5 ΔQ TO BE SUBTRACTED FROM ALL FLOWS. THEN: QBC = 10 - 2.5 = 7.5 QCE = 0 - 2.5 = - 2.5 QBD = -2.4 (from loop ABD) - 2.5 = - 4.9 QDE = -5 - 2.5 = - 7.5 SUMMARY: QAB = 12.4 QAD = - 2.6 QBD = 4.9 (-4.9)     (shared in the loop) QBC = 7.5 QCE = - 2.5 QDE = - 7.5 SECOND TRIAL LOOP ABD: PIPE kQ2 2kQ AB (c) 1.451 0.234 AD (cc) -7.159 5.507 BD (c) 55.079 22.481 ∑ 49.371 28.222 ΔQ = 49.371/28.222 = 1.7 ΔQ TO BE SUBTRACTED FROM ALL FLOWS. QAB = 12.4 - 1.7 = 10.7 QAD = -2.6 - 1.7 = - 4.3 QBD = 4.9 - 1.7 = 3.2 LOOP BCDE: PIPE kQ2 2kQ BC (c) 16.993 4.5315 BD (cc) -55.079 22.4812 CE (cc) -4.6975 3.7580 DE (cc) -16.993 4.5315 ∑ -59.7765 35.3022 ΔQ = -59.7765/35.3022 = -1.7 ΔQ TO BE SUBTRACTED FROM ALL FLOWS. QBC = 7.5 - (-1.7) = 9.2 QBD = - 3.2 (from loop ABD) - (-1.7) = - 1.5 QCE = -2.5 - (-1.7) = - 0.8 QDE = - 7.5 - (-1.7) = - 5.8 SUMMARY: QAB = 10.7 QAD = - 4.3 QBD = 1.5 (-1.5)     (shared in the loop) QBC = 9.2 QCE = - 0.8 QDE = - 5.8 THIRD TRIAL LOOP ABD: PIPE kQ2 2kQ AB (c) 1.0808 0.2020 AD (cc) -19.581 9.1074 BD (c) 5.1615 6.882 ∑ -13.3387 16.1914 ΔQ = -13.3387/16.1914 = -0.8 ΔQ TO BE SUBTRACTED FROM ALL FLOWS. QAB = 10.7 - (-0.8) = 11.5 QAD = -4.3 - (-0.8) = - 3.5 QBD = 1.5 -(-0.8) = 2.3 LOOP BCDE: PIPE kQ2 2kQ BC (c) 25.570 5.5586 BD (cc) -5.1615 6.8820 CE (cc) -0.4810 1.2026 DE (cc) -10.163 3.5043 ∑ 9.7645 17.1475 ΔQ = 9.7645/17.1475 = 0.5 ΔQ TO BE SUBTRACTED FROM ALL FLOWS. QBC = 9.2 - 0.5 = 8.7 QBD = - 2.3 (from loop ABD) - 0.5 = - 2.8 QCE = - 0.8 - 0.5 = - 1.3 QDE = - 5.8 - 0.5 = - 6.3 SUMMARY: QAB = 11.5 QAD = - 3.5 QBD = 2.8 (-2.8)     (shared in the loop) QBC = 8.7 QCE = - 1.3 QDE = - 6.3 FOURTH TRIAL LOOP ABD: PIPE kQ2 2kQ AB (c) 1.248 0.2171 AD (cc) -12.972 7.413 BD (c) 17.985 12.846 ∑ 6.261 20.4761 ΔQ = 6.261/20.4761 = 0.3 ΔQ TO BE SUBTRACTED FROM ALL FLOWS. QAB = 11.5 - 0.3 = 11.2 QAD = - 3.5 - 0.3 = - 3.8 QBD = 2.8 - 0.3 = 2.5 LOOP BCDE: PIPE kQ2 2kQ BC (c) 22.866 5.2565 BD (cc) -17.984 12.846 CE (cc) -1.2702 1.9542 DE (cc) -11.990 3.8065 ∑ -8.3742 23.8632 ΔQ = - 8.3742/23.8632 = - 0.3 ΔQ TO BE SUBTRACTED FROM ALL FLOWS. QBC = 8.7 - (-0.3) = 9.0 QBD = - 2.5 (from loop ABD) - (-0.3) = - 2.2 QCE = - 1.3 - (-0.3) = - 1.0 QDE = - 6.3 - (-0.3) = - 6.0 SUMMARY: QAB = 11.2 QAD = - 3.8 QBD = 2.2 (-2.2)     (shared in the loop) QBC = 9.0 QCE = - 1.0 QDE = - 6.0 FINAL VALUES: QAB = 11.4 QAD = - 3.6 QBD = 2.4 (-2.4)     (shared in the loop) QBC = 9.0 QCE = - 1.0 QDE = - 6.0 PRESSURES AT THE LOAD POINTS C AND E pC = pA - γ (kAB QAB2 + kBC QBC2) pC (lb/ft2) = 60 lb/in2 × 122 (in2/ft2) - 62.4 lb/ft3 [(0.00944 × 11.42) + (0.3021 × 9.02)] ft pC (lb/ft2) = 7037 pC (lb/in2) = 48.9 pE = pA - γ (kAD QAD2 + kDE QDE2) pE (lb/ft2) = 60 lb/in2 × 122 (in2/ft2) - 62.4 lb/ft3 [(1.059 × 3.6 2) + (0.3021 × 6.02)] ft pE (lb/ft2) = 7105 pE (lb/in2) = 49.3 Pipe Network Calculator by LMNO Engineering 160311 16:40