Lecture 01

March 28, 2018 | Author: Muhammad Najam | Category: Pump, Fluid Dynamics, Civil Engineering, Gas Technologies, Soft Matter


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Pumps & Piping HydraulicsLecture-1 Objectives       Calculate & Select suitable line sizes for fluid Work up examples of circuits Hydraulics of a Tank-Pump-Tank Model Calculating Hydraulic Power, NPSHa, Determining Service of Pump (Flow/Pressure) Estimating  Suction Speed, Efficiency, Power, Motor Rating Recap  Driving Forces of Nature      Pressure. Temperature. Magnetic force Material handling involved material transfer to the point of use and run-down to storage Material handling costs range from 30-50% of total cost in an industry and sometime as high as 95% Pumps transfer liquid by directly or by converting centrifugal force/velocity to pressure This pressure creates the potential to transfer from A to B . Concentration. Voltage. Gravity. 6 ed. 1.Frictional Groups  R Moody’s Friction Factor f '  8 2 v R fD  2 2  Darcy or Fanning Friction Factor v   Stanton/Pannel R v 2 f '  4 f D  8 Chemical Engineering . ..Coulson & Richardson Vol. Moody’s Diagram . 1. .Stanton’s Diagram  Chemical Engineering .. 6 ed.Coulson & Richardson Vol. org/wiki/Darcy_friction_factor_ .Formulae  64 Laminar f  Re  Transient 1  2  log  2. This is the ColebrookWhite equation.51  f  Re f e   3.7 DH  [There are other equations for fully turbulent flow.wikipedia.] http://en. We will use primarily this equation for calculating friction loss. with pressure drop density also PM changes so calculations require some iteration  for longer runs of pipe ZRT .Bernoulli’s Equation 2 1 2 2 P1 v P2 v  z1    z2   hL g 2 g g 2g  Liquids do not experience a significant change in density so: P   z  hL g  For gas. Pressure Drop/Head Loss  The “magical” equation filled with so many concepts! Pf  LEq. v  hL  f  g 2 g DH  [Note again. the pressure drop will result in gas expansion or increasing velocity as gas flows] . if we have to calculate for gas. Practice #1  Calculate the pressure drop per 100 ft length in a 6” Sch-40 carbon steel pipe in which 400USGPM of water is flowing at 35°C (density = 62.024   v= 4. Pipe ID = 6.0018in) vD v  Re  A  D2  4 Q v  0.8cP. Re = 260621 .4 lb/ft³ viscosity = 0.4085  2 D  vD Re  124.442 ft/s. Q Qe= 0.065in. 7 D f H   Re f Take starting value of f = 0.065     12  2 P  62.Practice #1 contd…   2.442   100  1.017184264 0.01719  g 2g D 2  32.015 0.017191724 0.017321962 0.4  1.452 psi 144 2 .174  6.017191318 0. Pressure Drop  4.043 ft P v L  f  0.51 1 e   2  log    3 .043  0.015f0 f1 f2 f3 f4 f5  0.01719134 For 1 ft length. Press enter for each iteration .7D) = 0.51/Re = 2.065) = 8.Using Calculator (Casio 570MS/991ES etc)   Be in Computation mode Comp mode (Mode -1) For variable memory press [Alpha] Button A= 2.0212 x10-5 F= 0.51/260621 = 9.0018/(3.631x10-6 B= e/(3.7 x 6.015 Type the following in the calculator  2 log  A   2 F B F To store press [Shift+Sto] button. Buttons Alpha Button Shift Button Store Mode Button X Raised to the Power Y . alloys and low-abrasion resistant metals Settling Velocity ρv2 >100 [ρv2<148] v<10/ρ1/2 [v<12. Steam 10ft/s in fire water Intermittent services .6kPa/100m] High viscosity circuits Depressuring Circuits Erosion Velocity ρv2 ρv2 <10000 [ρv2<14884] v<100/ρ1/2 [v<122/ρ1/2] Intermittent service.g <20m/s Sat.2/ρ1/2] Rule of thumb for basic entrainment (Different materials have different settling rates) Service Velocity Specified in design practice e.Selecting Nominal Diameter/NPS Constraining Parameter Value or Limit Excluding Pressure drop per unit length <1psi/100 ft [<22. Quick Review Topics         Degree of Freedom Analysis Tips on liquid line size estimation Frequency & RPM Specific Suction Speed Pump Efficiency Types of Drives Piping & Instrumentation Diagrams Sparing of Pumps . either pressures is calculated Case-2: Pressures are known flow-rate is to be calculated .Piping Hydraulics  Solving Pressure and Flow Problems in Pipes Flow Rate (Q) Density (ρ) Pressure Viscosity (μ) (P1) Length (L) Diameter (D) Pressure (P2) Case-1: Flow Rate is known. Discharge Length: 2000m Vapour Pressure: 0.4lb/ft³. 15m Height Fill Tank: 30m Diameter. Data:      Suction Tank: 20m Diameter. Density: 62. 20m Height Suction Length: 60m. .Practice #2 You are required to pump water at 35°C from one atmospheric tank to the other at the rate of 400USGPM.8cP Size the system suitable for service. Viscosity = 0.82psia. 20m 400gp m 15m Height 60m 2000 m 0 psigDia.Sketch 0 psigDia. 30m 20m Height 0 psig . 3 RPM   Flow Rate  Specific Suction Number Ns  Head 3 / 4 1/ 2 .Formulae Reminder Hydraulic Power (hp)  Flow .rate (USGPM)  Differential Pressure (psi) 1714. 56 10.08 6.066 2.94 40.08 154.3 60.068 102.92 3.066 6.34 336.84 11.48 52.612 52.28 304.612 1.1 273.3 168.982 7.3 88.250 .049 1.068 3.4 48.64 26.066 202.54 13.56 254.940 12.1 323.64 1.982 254.026 4.022 303.48 2.74 7.92 77.022 10.049 40.74 202.6 ID (mm) ID (in) Sch-40 Sch-80 Sch-40 Sch-80 26.002 333.3 219.026 154.9 355.26 102.26 4.9 114.124 13.94 1.NPS/ID Data  NPS (in) OD (mm) 1 1½ 2 3 4 6 8 10 12 14 33.066 77. . . Margins & Motor Rating . Poles/RPM RPM  120  Frequency No. of Poles Approximate Electrical Motor Speed (RPM) No. Poles Speed with Rated Load Synchronous Speed (no Load) 60 hz 50 hz 60 hz 50 hz 2 Pole 3450 2850 3600 3000 4 Pole 1725 1425 1800 1500 6 Pole 1140 950 1200 1000 8 Pole 850 700 900 750 . 2 hp Electrical Rating Margin: 25% Electrical Rating: 34 hp Motor Selection: 40 hp .3hp NPSHa: 27 ft RPM: 1500 Specific Suction Speed: 662 Efficiency/ Pump Power: 60% / 27.Summary             Flow: 400 USGPM Suction Pressure: -0.9 psig Discharge Pressure/Head: 70psig / 162 ft Design Conditions: 400 USGPM/ 70 psig Hydrualic Power: 16.
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