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ENTC 303: Announcements

ENTC 303: Announcements. Homework assignment No. 1 From Mott : 1.45, 1.49, 1.55, 1.83, 1.86, 1.89, 2.19, 2.29, 2.48, 2.58, 2.68. From Esposito: 2-2, 2-3, 2-27, 2-39. Due next Tuesday, September 16 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. 1 • FromMott: 1.45, 1.49, 1.55, 1.83, 1.86, 1.89, 2.19, 2.29, 2.48, 2.58, 2.68. • From Esposito: 2-2, 2-3, 2-27, 2-39. • Due next Tuesday, September 16th before 3:35 pm • For more information, go to: • http://etidweb.tamu.edu/classes/entc303/

  2. Pascal’s Law(Esposito 3.3) • “Pressure applied to a confined fluid is transmitted undiminished in all directions throughout the fluid and acts perpendicular to the surfaces in contact with the fluid” • In a simple hydraulic jack application: http://science.howstuffworks.com/transport/engines-equipment/hydraulic1.htm

  3. Pressure and Elevation • Change in pressure in homogeneous liquid at rest due to a change in elevation:

  4. Pressure and Elevation • Change in pressure in homogeneous liquid at rest due to a change in elevation: DP = gh DP = Pbottom – Pabove = gh Where, DP = change in pressure, psi or kPa • = specific weight, N/m3, lb/ft3 h = change in elevation, m or ft Note: Pressure increases (+) with depth Pressure decreases (-) with elevation

  5. Pressure-Elevation Relationship • Valid for homogeneous fluids at rest (static) P2 = Patm + rgh Free Surface Free Surface P2 P1 P1 > P2

  6. Static Fluids: Same elevation and same fluid → same pressure P2 = Patm + rgh 1-15

  7. Pascal’s Paradox h

  8. Examples • Compute the change in water pressure from a free surface to a depth of 5m • Tank of oil has one side open and the other one sealed with air. Oil specific gravity is 0.9. Calculate gage pressure at points A, B, C, D, E and F.

  9. D B Example Air Air Air Already in static equilibrium F Open A E Oil C - Vertical distance between A-B and A-D is 3.0 m - Vertical distance between A-C is 6.0 m - Vertical distance difference between A-F is 1.5 m - The oil’s specific gravity is 0.9

  10. Already in static equilibrium too, but different conditions: Both sides are open to the atmosphere Oil B D

  11. Manometers • Used to measure pressure • Takes advantage of DP = gh relationship http://www.efunda.com/formulae/fluids/manometer.cfm

  12. How to calculate pressure using a manometer • Start from convenient point (open end, if possible) • Using DP = gh, write expressions for the changes in pressure from one end to the other (watch the sign and the type of fluid) • Pressure increases (positive +) with depth • Pressure decreases (negative -) with elevation • Equate expression from previous step • Substitute known values and solve

  13. Example: Manometer • Calculate pressure (psig) or kPa (gage) at Point A. Open end is at atmospheric pressure. A 0.15 m Water 0.4 m Hg: SG = 13.54

  14. Pressure Measurement Devices Manometers Highly sensitive inclined manometers for systems demanding precise measurement of low pressures

  15. Pressure Measurement Devices Gages Transducer: More information: www.coleparmer.com www.omega.com Others by web search

  16. Barometer and Atmospheric Pressure Patm = rgh Patm = 14.psi, 1 atm

  17. Example • Determine atmospheric pressure when the barometric reading is 740 mm Hg, g = 9.81 m/sec2, rHg is 13,570 kg/m3. • The standard atmospheric pressure is 101.3 kPa. Determine the height of mercury column for this pressure. • Hint: gHg = 132.8 kN/m3

  18. Forces due to Static Fluids • Pressure =Force/Area (definition) • Force = Pressure*Area • Example: • If a cylinder has an internal diameter of 2 in and operates at a pressure of 300 psig, calculate the force on the ends of the cylinder.

  19. Example: Cylinder 3 m Calculate the force on the bottom of the container. 2.4 m Oil, S.G. = 0.90 1.5 m Water, S.G. =1.0

  20. Example: Cone Shaped Container 1.2 m Calculate the force on the bottom of the container. 2.4 m Oil, S.G. = 0.90 1.5 m Water, S.G. =1.0 3 m

  21. Force-Pressure: Rectangular Walls Patm Vertical wall DP = g*h d

  22. Resultant Force on Vertical Wall

  23. Forces due to Pressure dy dz dx

  24. Derivation of Resultant Force (FR) on a rectangular (vertical) wall Patm • FR = g*(d/2)*A = Pavg*A • Pavg = g*(d/2) • LP = d/3 DP = g*h d FR Center of Pressure (CP) d/3

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