T3b - MASTER - Pump flow system - operating point 2023.pptx
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T3b - MASTER - Pump flow system - operating point 2023.pptx
- 1. KV WORKED EXAMPLES {for energy conversion} Keith Vaugh BEng (AERO) MEng
- 2. The flow system used to test a centrifugal pump at a nominal speed of 1750 rpm is shown in the figure. The liquid water enters the pump through a 90Β° bend at 20 Β°C after being drawn from the sump through a hinged disk foot valve along 7 m long pipe. This fluid is then transferred along a 250 m long pipe and passes through a second standard 90Β° elbows as shown. The suction and discharge pipes diameters are 120 mm. Develop an expression to represent the Pressure at the Pump and an expression to represent the required Head at the Pump PUMP SYSTEM EXAMPLE 2 Pd Ps Zd Zs Elev. Surface of Water in Sump 6 m Elev. Eye of Impeller 8.5 m Height Foot Valve with Hinge Disk Sudden Enlargeme nt Standard 90Β° Elbows Elev. Surface of Water in Tank 60 m
- 3. GIVEN: Pump and piping system FIND Conduct an engineering assessment must develop expressions for the Pressure at the Pump and an expression to represent the Head at the Pump, calculate pressure at Pump (eye of the impeller) the required head at the pump for the system, the NPSHA, select pumps, examine the impact ageing of pipe have on the system flow and plot the System Curves vs. Pump Curve ASSUMPTIONS β’ Steady and incompressible flow β’Uniform flow at each section β’ U1 = U2 = 0, Upipe found from Volumetric flow β’ P1 = P2 = Patm Pd Ps Zd Zs Elev. Surface of Water in Tank Elev. Surface of Water in Sump Elev. Eye of Impeller Height Foot Valve with Hinge Disk Sudden Enlargement Standard 90Β° Elbows
- 4. GOVERNING EQUATIONS The energy equation for steady incompressible pipe flow can be written as; The governing equations given that in represents the inlet and out the outlet of the system π ππ + π2 2π + π§ π π’ππ‘πππ = π ππ + π2 2π + π§ πππ πβππππ + βπΏ β π» Total head loss is the summation of the major and minor losses in the system βπΏ = π πΏ π· πππππ 2 2π + βπ πΏπ π· πππππ 2 2π + βπΎ πππππ 2 2π π» = βπ π and π ππ + π2 2π + π§ π π’ππ‘πππ = π ππ + π2 2π + π§ πππ πβππππ + π πΏ π· πππππ 2 2π + βπ πΏπ π· πππππ 2 2π + βπΎ πππππ 2 2π β π» π = β1.8πππ10 π π· 3.7 1.11 + 6.9 π π β2 Friction factor
- 5. Develop an expression to represented the total Pressure at the Pump πππ ππ + πππ 2 2π + π§ππ = πππ’ππ ππ + πππ’ππ 2 2π + π§ππ’ππ + π πΏ π· + πΏπ π· πππππ 2 2π + πΎ πππππ 2 2π Pin = Patm, Uin = 0, Upump = 0 i.e. the fluid is entering the eye of the impeller π§ππ β πππ’ππ ππ + π§ππ’ππ = π πΏ π· + πΏπ π· πππππ 2 2π + πΎ πππππ 2 2π π§ππ β πππ’ππ ππ β π§ππ’ππ = π πΏ π· + πΏπ π· πππππ 2 2π + πΎ πππππ 2 2π β πππ’ππ ππ = π§ππ’ππ β π§ππ + π πΏ π· + πΏπ π· πππππ 2 2π + πΎ πππππ 2 2π πππ’ππ = βππ π§ππ’ππ β π§ππ + π πΏ π· + πΏπ π· πππππ 2 2π + πΎ πππππ 2 2π . . . . . . . steps omitted . . steps omitted
- 6. Develop an expression to represented the required head at the pump π ππ + π2 2π + π§ π π’ππ‘πππ = π ππ + π2 2π + π§ πππ πβππππ + π πΏ π· πππππ 2 2π + βπ πΏπ π· πππππ 2 2π + βπΎ πππππ 2 2π β π» Pin = Pout = Patm, Uin = Uout = 0 π§ππ β π§ππ’π‘ = π πΏ π· πππππ 2 2π + βπ πΏπ π· πππππ 2 2π + βπΎ πππππ 2 2π β π» π§ππ β π§ππ’π‘ = π πΏ π· πππππ 2 2π + π πΏπ π· πππππ 2 2π + πΎ πππππ 2 2π β π» π» = π§ππ’π‘ β π§ππ + π πΏ π· πππππ 2 2π + π πΏπ π· πππππ 2 2π + πΎ πππππ 2 2π π» = π§ππ’π‘ β π§ππ + π πΏ π· + πΏπ π· + πΎ πππππ 2 2π . . . . . . . steps omitted . . steps omitted
- 7. Determine the total pressure at the eye of the impeller and the required head at the pump Given Data: π· = 12ππ πππ‘π = 101.3πππ π Β· = 600πΏ/πππ and π = 0.046ππ taken from table From relevant tables water at 20 Β°C π = 998ππ/π3 ππ£ππππ’π = 2.34πππ π = 1.01 Γ 10β6 π2 /π At the specified flow rate, the speed of the fluid is: π Β· = π΄πππππ βΆ πππππ = π Β· π΄ = 4π Β· ππ·2 = 0.884π/π π π = ππππππ· π = 0.844π/π Γ 0.12π 1.01 Γ 10β6π2/π = 1.05 Γ 105 π π· = 0.046ππ 0.12π = 3.8 Γ 10β4 π = β1.8πππ10 3.8 Γ 10β4 3.7 1.11 + 6.9 1.05 Γ 105 β2 = 0.0194 1 π = β1.8πππ10 π π· 3.7 1.11 + 6.9 π π Therefore using Note for Clarification - kinematic Viscosity, π = π π
- 8. KV β’ Saturation temperature Tsat: The temperature at which a pure substance changes phase at a given pressure. β’ Saturation pressure Psat: The pressure at which a pure substance changes phase at a given temperature. β’ Vapor pressure (Pv): The pressure exerted by its vapour in phase equilibrium with its liquid at a given temperature. It is identical to the saturation pressure Psat of the liquid (Pv = Psat). β’ Partial pressure: The pressure of a gas or vapor in a mixture with other gases. For example, atmospheric air is a mixture of dry air and water vapour, and atmospheric pressure is the sum of the partial pressure of dry air and the partial pressure of water vapour. VAPOUR PRESSURE AND CAVITATION
- 9. KV β’ There is a possibility of the liquid pressure in liquid-flow systems dropping below the vapour pressure at some locations, and the resulting unplanned vaporisation. β’ The vapour bubbles (called cavitation bubbles since they form βcavitiesβ in the liquid) collapse as they are swept away from the low-pressure regions, generating highly destructive, extremely high-pressure waves. β’ This phenomenon, which is a common cause for drop in performance and even the erosion of impeller blades, is called cavitation, and it is an important consideration in the design of hydraulic turbines and pumps. Cavitation damage on a 16-mm by 23-mm aluminium sample tested at 60 m/s for 2.5 h. The sample was located at the cavity collapse region downstream of a cavity generator specifically designed to produce high damage potential.
- 10. CONSIDER THE LOSSES (Major & Minor) Pd Ps Zd Zs Elev. Surface of Water in Tank Elev. Surface of Water in Sump Elev. Eye of Impeller Height Foot Valve with Hinge Disk Sudden Enlargement Standard 90Β° Elbows Table 1: Representative Dimensionless Equivalent Lengths for Values and Fittings Fitting Type Equivalent Length, Valves (Fully Open) - Gate Valve 8 - Globe Valve 340 - Angle Valve 150 - Ball Valve 3 - Lift Check Valve: Globe Lift 600 - Lift Check Valve: Angle Lift 55 - Foot Valve with Strainer: Poppet Disk 420 - Foot Valve with Strainer: Hinged Disk 75 Standard Elbow: 90ΒΊ 30 Standard Elbow: 45ΒΊ 16 Return bend, close pattern 50 Standard Tee: Flow Through Run 20 Standard Tee: Flow Through Branch 60
- 11. CONSIDER THE LOSSES (Major & Minor) Pd Ps Zd Zs Elev. Surface of Water in Tank Elev. Surface of Water in Sump Elev. Eye of Impeller Height Foot Valve with Hinge Disk Sudden Enlargement Standard 90Β° Elbows
- 12. The total pressure at the eye of the impeller πππ’ππ = βππ π§ππ’ππ β π§ππ + π πΏ π· + πΏπ π· πππππ 2 2π + πΎ πππππ 2 2π From part (c) above πππ’ππ = βππ 8.5π β 6π + 0.0194 7π 0.12π + 105 Γ 0.12π 0.12π 0.884 2 π/π 2 Γ 9.81π/π 2 + 0.78 Γ 0.884 2 π/π 2 Γ 9.81π/π 2 πππ’ππ = β26015ππ(πππ’ππ) = β26πππ(πππ’ππ) The Required head at the Pump from equation developed in (d) π» = π§ππ’π‘ β π§ππ + π πΏ π· + πΏπ π· + πΎ πππππ 2 2π π» = 60π β 6π + 0.0194 257π 0.12π + 135π Γ 0.12π 0.12π + 1.78 0.884π 2 2 Γ 9.81π/π 2 π» = 55.83π Hinged Foot Valve plus One 90ΒΊ Elbow Just Re-entrant Re-entrant plus Sudden Enlargement
- 13. The Net Positive Suction Head Available (NPSHA) ππππ»π΄ = πππ’ππ + πππ‘π β ππ£ππππ’π ππ = β26πππ + 101.3πππ β 2.34πππ 998ππ/π3 Γ 9.81π/π 2 A pump would be selected by finding one for which the NPSHR is less than the NPSHA. Based on the data and the information in the pump selection chart, a 5AE8N or a 3AE14 pump would be capable of supplying the required head at the given flow rate. The pump should be operated at a speed of between 1750 and 3500 rpm, but the efficiency may not be acceptable. One should consult a complete catalog to make an informed decision. ππππ»π΄ = 7.4π Select a pump suitable for this application and provide a justification
- 14. 1 foot - 0.3048 meters 1 gpm - 0.0038 m3/min 1 gpm - 3.785 l/min Select these pumps for this case
- 15. Develop and format appropriately an excel worksheet which calculates the; β’Pump head for a range of volumetric flow rates β’calculates the impact that the ageing of the pipes have on the Pump Head at twenty years and forty years of service Assume a head at 600 L/min for 40 year old pipes is 80% of the maximum head of the pump, and that the pump curve has the form π» = π»π β π΄π Β· 2 . Using the calculations from the Excel worksheet, plot the Pump Curve and the system curves for new pipes, pipes at 20 years service and pipes at 40 years service. PUMP SYSTEM EXAMPLE Pt 2
- 16. Table 2 - Data given in question or sourced from fluids tables Given Data Value Units Source Water at 20 Degrees Pipe Diameter 12 cm Ξ΅ 4.6E-05 mm Patm 101.3 kPa Kinematic Viscosity 1.01E-06 m2/s Tables Pvapour 2.34 kPa Tables Density 998 kg/m3 z1 6 m z2 60 m Lsuction 7 m Side of pump Ldelivery 250 m Side of pump LT 257 m Equivalent Lengths Hinged Disk Foot Valve 75 Tables Angle Lift Valve 0 Tables Gate Valve 0 Tables Standard 90 deg Elbow 30 2 of these Tables Le 135 Note 2 Elbows K Reentrant 0.78 Tables Sudden Expansion 1 Tables KT 1.78 20 Years 5 Tables 40 Years 8.75 Tables Summarise the Data provide in a table. Begin by summarising all the relevant data from the question into a table. Data that is not provided in the question should be sourced from the relevant tables and resources available in the essential reading text book (Fundamentals of Thermal Fluid Sciences from semester I) or from online sources. Pay particular caution to units and ensure these are corrected to the SI unit system. The data in the summary table should be linked to the calculations in the data sheets developed from the governing equations created previously in the question
- 17. Table 3: Calculate results Volumetric Flow Rate (L/Min) Vel (m/s) Reynolds Number, Re Friction Factor, f New Pipes (m) 20 Year Pipe (m) 40 Year Pipe (m) Pump Curve (m) 0 0.000 0.000 0.0000 54.00 54.00 54.00 86.88 200 0.295 35017.589 0.0234 54.24 55.19 56.07 84.95 400 0.589 70035.178 0.0207 54.86 58.20 61.32 79.16 600 0.884 105052.768 0.0194 55.83 62.89 69.50 69.52 800 1.179 140070.357 0.0187 57.15 69.22 80.54 56.03 1000 1.474 175087.946 0.0182 58.80 77.19 94.43 38.68 1200 1.768 210105.535 0.0179 60.78 86.78 111.15 17.47 1400 2.063 245123.125 0.0177 63.11 97.99 130.69 -7.60 1600 2.358 280140.714 0.0175 65.77 110.81 153.04 -36.52 1800 2.653 315158.303 0.0173 68.76 125.24 178.20 -69.29 2000 2.947 350175.892 0.0172 72.09 141.29 206.17 -105.93 π»600 = 69.5π π»π = 69.5π 0.8 = 86.875π π» = π»π β π΄π2 Β· βΆ 69.5π = 86.875π β π΄ 6002 π΄ = 86.875π β 69.5π 6002 = 4.82 Γ 10β5 π/(πΏ/πππ) Assume that the heat at 600 L/min for 40 year old pipe is 80% of the maximum head for the pump, and that the pump curve has the form π» = π»π β π΄π2 Β· π» = π»π β π΄π2 Β·
- 18. 50.00 54.00 58.00 62.00 66.00 70.00 74.00 78.00 82.00 86.00 90.00 200 300 400 500 600 700 800 900 1000 Pump Head (m) Volumetric Flow Rate in L/min New Pipes (m) 20 Year Pipe (m) 40 Year Pipe (m)
Editor's Notes
- Let us consider a simple case of laminar flow between two parallel plates separated by a small distance d. The upper plate moves at a constant velocity U while the lower plate remains at rest. At the plate fluid interface in both cases there is no velocity due to the strong forces of attraction. Therefore the velocity profile in the fluid is given by
- Let us consider a simple case of laminar flow between two parallel plates separated by a small distance d. The upper plate moves at a constant velocity U while the lower plate remains at rest. At the plate fluid interface in both cases there is no velocity due to the strong forces of attraction. Therefore the velocity profile in the fluid is given by