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ENTC 303: Announcements

ENTC 303: Announcements. Homework assignment No. 4 From Mott : 6.36, 6.37, 6.61, 6.72, 6.78, 6.94, 7.3, 7.9, 7.17, 7.22, 7.23, 7.37. From Esposito : 3.59E and 3.62E. Due Thursday, October 9 th before 3:35 pm For more information, go to: http://etidweb.tamu.edu/classes/entc303/.

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ENTC 303: Announcements

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  1. ENTC 303: Announcements • Homework assignment No. 4 • FromMott: 6.36, 6.37, 6.61, 6.72, 6.78, 6.94, 7.3, 7.9, 7.17, 7.22, 7.23, 7.37. • FromEsposito: 3.59E and 3.62E. • Due Thursday, October 9th before 3:35 pm • For more information, go to: • http://etidweb.tamu.edu/classes/entc303/

  2. ENTC 303: Announcements • Exam I • Tuesday, Oct 14th • Mott’s 1-7 • Esposito’s 1-3 • Homeworks1-4 • Review (optional): Sun, Oct 12th at 7 PM (Thom 122) • Closed book/closed notes • Yes, an equation sheet will be allowed • Double-sided is ok!

  3. Conservation of Energy

  4. Pump Energy Energy PUMP = DPE + DKE + FE + LOSSES + REMOVED DPE = Potential Energy Differential DKE = Kinetic Energy Differential FE = Flow Energy

  5. Power Required by Pumps • PA = hA*W • W = g*Q • Power added to a fluid by a pump: PA = hA*g*Q • Previous example • Calculate the power the pump delivered to the fluid

  6. Example The flow rate through a pipe-pump system is 0.014 m3/sec. The fluid has a S.G. of 0.86. Calculate the energy per unit weight delivered to the system. 2 in Schedule 40 Steel pipe PB = 296 kPa hL = 1.86 N-m/N 1.0 m 3 in Schedule 40 Steel pipe PA Pump PA = -28 kPa

  7. Power • Power = Energy/Unit Time • Units: 1 hp = 550 lb-ft/sec • 1 hp = 745.7 W • Is 1 Hp > 1 kW or the other way around?

  8. Mechanical Efficiency of Pumps Pumps cannot be 100% efficient because of energy losses due mechanical friction within the pump Pump efficiency is usually within 50 to 90% Manufacturers provide eMvalues for each pump and should be part of the performance data We will come back to this in Chapter 13 (Mott)

  9. Example • Determine the mechanical efficiency of a pump whose power input is 3.85 hp and pumps 500 gpm (gallons per minute). g = 56.0 lb/ft3. 4-in SCH 40 6-in SCH 40 1 2 Y 20.4 in

  10. Power Delivered to Fluid Motors • Power added to a fluid by a pump: • PR = hR*W • W = g*Q • PR = hR*g*Q • Mechanical Efficiency of Fluid Motors

  11. Example • Calculate the power delivered by the oil to the fluid motor if the volume flow rate is 0.25 m3/sec. There is an energy loss of 1.4 N-m/N in the piping system. If the motor has an efficiency of 75%, calculate the power output. S.G.OIL = 0.86 10 m Motor 300 mm, inside diameter

  12. Fluid Losses (hL) • Frictional Losses (due to fluid friction in pipes) • Minor Losses (due to valves, fittings, etc.) • How to calculate fluid losses? • Need to identify type of flow • Laminar or Turbulent? • Must know flow conditions and piping system specifications (size, length, etc.)

  13. Fluid Losses Laminar Turbulent Fluid Characteristic Need to know: Velocity Pipe Diameter Viscosity Density Roughness Need to know: Velocity Pipe Diameter Viscosity Density

  14. Laminar Flow • Streamline flow, smooth velocity profile

  15. Laminar Flow Profile in a Pipe

  16. Turbulent Flow • Fluid particles randomly fluctuate along the streamwise direction Average Velocity Profile http://www.youtube.com/watch?v=NplrDarMDF8

  17. Turbulent Flow Profile in a Pipe

  18. Laminar Vs. Turbulent Flow

  19. Laminar vs. Turbulent Laminar Turbulent http://www.engineering.uiowa.edu/fluidslab/gallery/images/turb6im.gif

  20. Reynolds Number Re < 2000 Laminar Flow Re > 4000 Turbulent 2000 < Re < 4000 Critical Region or Transitional

  21. Example • Water flowing at 285 L/min, DPIPE = 0.02524 m, Area = 5.017 x 10-4 m2, n = 4.11 x 10-7 m2/sec (kinematic viscosity). Is the flow laminar or turbulent?

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