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PROCESSING TECHNOLOGY 1

PROCESSING TECHNOLOGY 1. Fluid Statics. PRESSURE. Pressure, P = Force/area Unit of pressure is the N/m 2 also called the Pascal (Pa), 1 Pa = 1 N/m 2 1000 N/m2  1 kN/m2 1000000  1 MN/m 2 etc 100000 N/m 2  10 5 N/m 2  1 bar  1 atm Atmospheric pressure at sea level is

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PROCESSING TECHNOLOGY 1

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  1. PROCESSING TECHNOLOGY 1 Fluid Statics

  2. PRESSURE Pressure, P = Force/area Unit of pressure is the N/m2 also called the Pascal (Pa), 1 Pa = 1 N/m2 1000 N/m2  1 kN/m2 1000000  1 MN/m2 etc 100000 N/m2 105 N/m2 1 bar  1 atm Atmospheric pressure at sea level is 101.3 kPa or 1.013 bar.

  3. Gauge and Absolute Pressures Many instruments show gauge pressure, i.e., the amount by which the pressure exceeds the atmospheric pressure, the sum of the two gives the absolute pressure. Absolute pressure, Pabsolute = Pgauge + Patmospheric

  4. A A P1 P2 Pressure Variation In A Static Fluid Horizontal change Consider a tank of stagnant fluid within which there exists a horizontal element of fluid of area A at either end with pressures P1 and P2 exerted on the ends of the element.

  5. Calculation of Forces • End 1: F1 = P1A End 2: F2 = P2A • Net force (F1 – F2) = 0 • = P1A - P2A = A(P1 - P2) • A(P1 - P2) = 0 • P1 = P2 • i.e. There is nochange of pressure at • constantheight in a static fluid.

  6. Vertical change P2 A Consider a vertical cylindrical element of fluid, area A on each end with pressure P1 and P2 at the ends, which are at heights z1 and z2. Z2 Z1 A P1

  7. Vertical change Mass m = density x volume = rV (1) Volume V = A(z1-z2) (2) Mass m = rA(z1-z2) Weight W = mass x g W = rA(z1-z2)g (3)

  8. Vertical change The weight of the element will act downwards, due to the pull of gravity. The downward force on the upper end of the element, F2 is given by, F2 = P2A The upward acting force F1 at the lower end of the element is given by, F1 = P1A

  9. Vertical change The vertical forces on the element must balance, Fdownward-acting = Fupward -acting F1 + W = F2 P1A + rA(z1-z2)g = P2A (P2 - P1) = rg(z1 - z2)

  10. Vertical change Assuming a constant fluid density r, the change in pressure DP due to a vertical difference in height is given by, DP = rg(z1 - z2)

  11. Pressure Measurement Pressure measurement is an essential part of any factory. It is used to ensure that a tank doesn’t exceed design pressure; or as an indicator of another factor, e.g. pressure difference across an orifice plate can be used to indicate fluid flowrate.

  12. P at z P Piezometer tube A piezometer is the simplest example of a pressure gauge or manometer. This is simply an open-ended tube connected to a pipe carrying liquid under pressure.

  13. P at z P Piezometer tube With the pipe under pressure, the fluid will rise up the tube until the column of liquid balances the pipe-line pressure.

  14. Piezometer tube The pressure difference due to the column of liquid within the piezometer is, DP = rgz

  15. Piezometer tube The pressure within the pipe-line pushing upwards on the column of liquid within the piezometer is P and the pressure pushing downwards on the column is atmospheric pressure Pat. Hence the pressure difference DP is, DP = P - Pat P at z P

  16. Piezometer tube Equating the two expressions for DP gives, P - Pat = rgz Hence the pipe-line pressure is given by, P = rgz + Pat

  17. Reservoir 2 @ P2 Reservoir 1 @ P1 z Manometer fluid Differential or U-tube Manometer

  18. Reservoir 1 @ P1 Reservoir 2 @ P2 z Manometer fluid Differential or U-tube Manometer Used to measure the pressure differential between two fluid reservoirs. The unknown pressure is indicated by the difference between the levels of fluid in the two arms of the U-tube.

  19. y x Reservoir 1 @ P1 Reservoir 2 @ P2 z B A Differential or U-tube Manometer

  20. y x Reservoir 1 @ P1 Reservoir 2 @ P2 z A B Differential or U-tube Manometer Draw a datum (A-B) across the manometer level with the lowest liquid interface. As we consider the manometer to be at steady state, the downward pressure at the left hand of the datum A is balanced by the downward pressure at the right hand datum B

  21. y x Reservoir 1 @ P1 Reservoir 2 @ P2 z A B Differential or U-tube Manometer The pressure at A consists of the summation of the reservoir pressure P1 and the pressure due to the height of reservoir fluid within the manometer leg x, = rRgx where rR is the density of the reservoir fluid

  22. y x Reservoir 1 @ P1 Reservoir 2 @ P2 z A B Differential or U-tube Manometer The pressure at B is a summation of the reservoir pressure P2 the pressure due to the height of reservoir fluid within the manometer leg y, = rRgy, and the pressure due to the height of manometer fluid within the manometer leg z, = rMgz, where rM is the density of the manometer fluid.

  23. Differential or U-tube Manometer PA = PB P1 + rRgx = P2 + rRgy + rMgz

  24. y x Reservoir 1 @ P1 Reservoir 2 @ P2 z B A Differential or U-tube Manometer x = y + z

  25. Differential or U-tube Manometer P1 + rRg(y + z) = P2 + rRgy + rMgz P1 + rRgy + rRgz = P2 + rRgy + rMgz P1 + rRgz = P2 + rMgz P1 - P2 = rMgz - rRgz P1 - P2 = (rM - rR)gz

  26. Bourdon Gauge Pivots Movement C-shaped oval cross Rigid mount section tube Pressure

  27. Pressure or Head? The terms pressure and head are often used inter-changeably. However, the units used are completely different i.e metres of fluid for head and N/m2 for pressure.

  28. Pressure or Head? It is common to describe a pressure as “being equivalent to a head of 10m of water.” This means that the actual pressure in N/m2 (or Pa) is given as, P = rgh = 1000 x 9.81 x 10 = 98.1 kPa Where h is the head of liquid, 10m in this case.

  29. Pressure or Head? Similarly a pressure P of 3 bar is equivalent to a head h of mercury (r=13600 kg/m3), P = rgh h = P/rg = 3x105/(13600 x 9.81) h = 2.25m

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