can write: dP = dz = or Now to find pressure at any point a depth of of the fluid as shown in the figure. Pressure variation with depth = Since h o = constant for incompressible fluid = h o = P = = pressure head g The pressure exerted by a fluid is dependent on the vertical head of the fluid and its specific weight. or The density of incompressible fluid does not remain constant i.e. constant. Now as per perfect gas equation: = m m but = density = = V V Now as per hydrostatic law, the pressure gradient is: Area = a Area = a Area = a Area = a dP dz P RT g dP dz R P T P RT ­ m dP = g dz ­ Mercury Pressure High High pressure line Gauge pressure Atm Low 0 Atmospheric pressure line Atmospheric pressure Absolute pressure Vacuum gauge pressure Low pressure line Absolute pressure Absolute zero vacuum (zero pressure) line Different Pressures 12. Besides manometers, what are the other types of pressure measuring instruments? ­ 13. What are the characteristics of mechanical gauges? Describe Bourdon tube gauge. The manometers work on the relationship between the pressure and the column of the fluid balanced by it. The simplest form of manometers is the pressure tube or piezometer as shown in the figure. The pressure tube consists of a single vertical tube open at the top connected to the vessel or pipe containing the fluid under pressure. Due to pressure of the fluid above the atmospheric pressure, the fluid rises in the tube to a height depending upon its pressure. If fluid rises meter, then we have gauge pressure of the fluid as: Vessel with high pressure fluid (Pabs) ­ Pabs f gh f Pabs Patm Pgauge If l l ghg l ghg Pgauge l ghg f l ghg f Patm gh f f gh f Pabs h f fg hv Patm Pabs h f Pvacuum h f If l lg fg fg Patm hv hv lg lg f Pvacuum hv lg a dh small 1 0 h alarge 100 ­ g h dh h g(h 0) gh h1 h2 1 2 Flow Flow Two Piezometer Tubes Method P h h h 1 2 g or P gh h1 g 1 ml ­ (rml, Sml) dP h Hg g h g h g Hg 1 h g SHg 1 g hS 1 k P h g Hg 1 k g Hg ­ ­ a h 0 A 2 P1 f P2 g h1 dh 1 gh 2 P1 P2 1 2 g z g h2 h dh 2 g h2 h dh 2 f g h1 dh A O2 H2 B C ­ h1 Hg g 760 13.6 10 3 9.81 1000 h2 Hg g 660 13.6 10 3 9.81 1000 1014. 88.06 10 12 13.34 10 12 1112 m 3 3 ­ ­­ ­ ­ Pr = 230 kpa Air h Water Hg 2 1 120 cm P1 = 140 kpa ­ P1 120 . Hg gh 3 g Pair . 13.6 10 9.81 h 103 9.81 230 400 120 103 103 160 9.81 h 230 h 10 102 . m 9.8 Let = absolute pressure of the water in the pipeline during equilibrium Pressure in the left limb from datum = Pressure in the right limb from datum P1 100 1000 w g 100 100 1000 Hg g Patm 3 . 103 g 9.81 01 P1 Patm 0.2 13.6 10 9.81 1000 1000 Pgauge 26.68 0.981 Pgauge 25.7 kpa 27. The figure shows a conical vessel having a U tube manometer attached to its outlet at . When the vessel is empty the reading of the manometer is given in the figure. Find the reading of manometer when the vessel has been completely filled with water. A When the vessel is empty, the pressure in both limbs is same from datum line Pressure left limb = Pressure in right limb 0.24 12 . 103 g h 0.8 103 g 0.288 + 0.8 0.2 = = = = = (0.24 h)1 103 0.24 + 0.078 0.39 m 390 mm h1 Equating pressure on left and right limb from datum line Pa (2 h h1 ) 1 g = Pb h 2g h1 1g 0.5 105 (2 h) 103 9.8 = 0.2 105 h 13.6 103 9.8 0.3 102 (2 h )9.81 = 13.6 9.81 h 49.62 0.401 m 126.6 PA 1 10 3 9.81 sin 30 h1 h 13.6 10 3 h 9.81 1 10 3 h1 h 9.81 PB 2 2 3 PA PB 10 9.81(13.6 .14 1 0.14) 9.83 10 3 (0.764) PA 0.24 0.8 10 3 g P1 0.10 .8 10 3 g 0.12 13.6 10 3 g P1 0.15 13.6 10 3 g or P1 PA 1.88 10 3 16 10 3 0.10 0.8 10 3 g PB PA 14.12 10 3 or P1 PB 20.01 10 3 32. A multi tube manometer is employed to determine the pressure in a pipeline. The levels inside the tubes are as shown in the figure. What would be the length of single mercury U-tube to record this pressure? ­ 3 PA 0.5 13.6 10 g P1 3 PA 66.7 10 P1 P1 .5 1 10 3 g .5 13.6 10 3 g P2 P1 61.8 10 3 P2 66.7 10 3 Patm P2 0.5 1 10 3 g Patm .5 13.6 10 3 g P2 Patm 61.8 10 3 61.8 10 3 Patm 123.6 10 3 61.8 10 3 200.3 10 3 Patm P1 Patm 123.6 10 3 PA Patm Pgauge 200.3 kPa In case we have ample U tube manometer, then, = 200.3 10 = 13.6 10 9.81 = 200.3 = 1.5 m 13.6 9.81 PA h1 2 g P1 from eqn (1) PB 2 hg 1hg P1 P1 PB hg ( 2 1 ) eqn (1) PA h1 2 g PB hg( 2 1 ) PA PB ( 2 1 )gh 2 h1 g PA h1 2 g h2 1 PA PC (h1 2g g PC h2 1g) ­­ PB h1 1 10 3 g h .875 10 3 g P1 PA (h1 h) 1 10 3 g P1 PA (h1 h) 1 103 g PB h .875 103 g h1 1 103 g 0.12 10 3 9.81 0.125 PA h 115 . 10 3 g P1 .20 0.92 10 3 g P1 (.50 h) 115 . 10 3 g PB PA h 115 . 10 3 g 0.2 0.92 10 3 g . 10 3 g PB (.5 h) 115 PA PB 0.2 0.92 10 3 g 0.5 115 . 10 3 g 1.8 10 3 5.64 10 3 384 . 103 kPa 3 3.843 10 1 10 9.81 391 mm of water ­ PM P1 Y 0.8 10 3 g P1 0.75 0.8 10 3 9.81 P1 Patm Z 13.6 10 3 g P1 Patm 0.25 13.6 10 3 9.81 PM 0.75 0.8 103 9.81 Patm 0.25 13.6 103 9.81 0.25 13.6 103 9.81 0.75 0.85 103 9.81 (33.35 6.25) 10 3 39.6 10 3 Pa PA PB 1 a r2 r1 gh A (0.06) 2 0.0028 4 15 1000 5.36 106 N/m2 0.0028 4 .000314 5.36 10 6 1684 N (0.02) 2 F 30 1684 30 400 400 C 70 cm ­ ­ ­ ­ 7.85 kN2 m ­ ­ 3 0.9 10 39.81 2.0 6.867 10 17.658 6.87 24.528 kN2 m l1 l2 Pair 1 g P1 0.4 0.8 103 9.81 P1 103 314 . P1 Patm 3 Hg g P1 3 P1 Patm 0.9 13.6 310 9.81 10 Patm 120.07 ­ ­ h a l 1 0.8 0.04 m A 20 h l sin 0.8 1 2 ­ dy dA g y y dy dp 1000 g 1 100 1000 y y y 200 3 10 9810 80 6400 512 10 200 3 10 1 y y 500 1000 2 2 2 1 2 80 3 6 0 3 6 ρ s = 1030 1500 mm ρ =? d dP K dP d d dP K (1030) 9.81 1500 2.4 109 . 106 9.81 15 . 103 106 2.4 109 6.5 kg/m3 ­ ­ ­­ ­ Water Oil Horizontal plane Pipes Horizontal plane Mercury ­ ­­ ­ ­­ ­ ­­ ­ ­­ ­ ­ ­ ­ H h S+ h1 h Dh ­­