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Tutorial 2 Fluid pressure 1. A cylinder contains a fluid at a gauge pressure of 360 KN/m2. Express this pressure in terms of a head of (a) water, and (b) mercury of sp gr = 13.6 What would be the absolute pressure in the cylinder if atmospheric pressure is 760mm Hg. Solution: Pressure (P) = 360 KN/m2 = 360x103 N/m2 Head (h) = ? where ρ = Density of fluid a) Head in terms of water (ρ = 1000 kg/m3) = 36.7m b) Head in terms of mercury ρ = sp gr x density of water = 13.6x1000 = 13600 kg/m3 = 2.7m Atmospheric pressure (h) = 760mmhg = 0.76m hg ) Atmospheric pressure ( Absolute pressure (Pabs) = ? Pabs = Pgauge + Patm = 360+101.3 = 461.3KN/m2 = 101396N/m2 = 101.3KN/m2 2. What would the pressure in kN/m2 be if the equivalent head is measured as 400mm of (a) mercury (sp gr 13.6) (b) water (c) oil specific weight 7.9 kN/m3 (d) a liquid of density 520 kg/m3? Solution: Head (h) = 400mm = 0.4m Pressure (P) =? where ρ = Density of fluid a) In terms of mercury, ρ = sp gr x density of water = 13.6x1000 = 13600 kg/m3 = 13600x9.81x0.4 = 53366 N/m2 = 53.366 KN/m2 b) In terms of water, ρ = 1000 kg/m3 = 1000x9.81x0.4 = 3924 N/m2 = 3.924 KN/m2 c) In terms of oil of sp. wt. ( ) = 7.9 kN/m3 = 7.9x.4 = 3.16 KN/m2 d) In terms of liquid with ρ = 520 kg/m3 = 520x9.81x0.4 = 2040 N/m2 = 2.04 KN/m2 3. A manometer connected to a pipe indicates a negative gauge pressure of 50mm of mercury. What is the absolute pressure in the pipe in N/m2 if the atmospheric pressure is 1 bar? Solution: Atmospheric pressure (Patm) = 1 bar = 1x105 N/m2 Head (h) = -50mmhg = -0.05m hg Absolute pressure (pabs) = ? ρ of mercury = sp gr x density of water = 13.6x1000 = 13600 kg/m3 Gauge pressure (Pgauge) = -13600x9.81x0.05 = -6671 N/m2 Pabs = Pgauge + Patm = -6671+1x105 = 93329 N/m2 = 93.3KN/m2 4. An open tank contains 5.7m of water covered with 2.6m of kerosene (sp wt = 8KN/m 3). Find the pressure at the interface and at the bottom of the tank. Solution: Height of kerosene (h) = 2.6m Height of water (h1) = 5.7m Sp wt of kerosene ( ) = 8KN/m3 Pressure at interface (Pint) = ? Pressure at bottom (Pbottom) = ? = 8x2.6 = 20.8KN/m2 = 20.8+9.81x5.7 = 76.7 KN/m2 5. The closed tank in the fig. is at 200 C. If the pressure at point A is 96 Kpa absolute, what is the absolute pressure at point B? What percent error results from neglecting the specific weight of air? (Take sp wt of air = 0.0118 KN/m3) A 5m B Air Air 3m D C 5m 3m Solution: Water Sp wt of air ( ) = 0.0118 KN/m3 Sp wt of water ( ) = 9.81 KN/m3 Starting from A, PA + PAC -PCD-PDB = PB PB = 96 + 0.0118x5 – 9.81x2 – 0.0118x3 = 76.404 Kpa Neglecting air, PB = 96 – 9.81x2 = 76.38 Kpa Error = (76.404-76.38)/76.404 = 0.00031 = 0.031% 6. In the fig., the pressure at point A is 2900 N/m2. Determine the pressures at points B, C and D. (Take density of air = 1.2 kg/m3) Air Air 1m C 0.8m B A Air 1.5m 2m X Water 0.7m D Solution: Density of water (ρ) = 1000kg/m3 PA = 2900 N/m2 Density of air ( ) = 1.2 kg/m3 PB = ?, PC =?, PD = ? Starting from A, = 2900 – 1000x9.81x0.2 = 938 N/m2 = 2900 + 1000x9.81x1.3 -1.2x9.81x2.3 = 15626 N/m2 = 2900 + 1000x9.81x2 = 22520 N/m2 7. In the fig., the absolute pressure at the bottom of the tank is 233.5 Kpa. Compute the sp gr of olive oil. Take atmospheric pressure = 101.3 Kpa. Oil (sp gr = 0.9) 1.5m Water 2.5m Olive oil 2.9m Mercury 0.4m Solution: Absolute pressure at bottom (Pabs) =233.5 Kpa Atmospheric pressure (Patm) = 101.3 Kpa Sp wt of water ( ) = 9.81 KN/m3 Sp wt of oil ( ) = 0.9x9.81 KN/m3 = 8.829 KN/m3 Sp wt of mercury ( ) = 13.6x9.81 KN/m3 = 133.416 KN/m3 Sp gr of olive oil (S) = ? Pabs = Patm + Pgauge Pabs = Patm + Poil + Pwater + Polive oil + Pmercury KN/m3 = 1.44 8. The tube shown in the fig. is filled with oil of sp gr 0.82. Determine the pressure heads at A and B in meters of water. oil A 2.1m Air B 0.5m X oil oil Solution: sp gr of oil = 0.82 Sp wt of oil ( ) = 0.82x9810 = 8044.2 N/m3 Head in terms of water at a and B (hA and hB) = ? Take atmospheric pressure to be 0 for gauge pressure. = - 8044.2x2.6 = -20914.9 Pa = -2.132 m = -20914.9 + 8044.2x2.1= -4022.1 Pa = -0.41 m 9. Calculate the pressures at A, B, C and D in the fig. Air Air A C 0.4m X 0.4m Oil Sp gr 0.9 B 0.5m 1m Water Solution: Sp wt of water ( ) = 9810 N/m3 sp gr of oil = 0.9 Sp wt of oil ( ) = 0.9x9810 = 8829 N/m3 Take atmospheric pressure to be 0 for gauge pressure. = - 9810x0.8 = 7848 Pa = 9810x0.5 = 4905 Pa Neglecting air, PC =PB = 4905 Pa = 4905 + 8829x1.9 = 21680 Pa D 10. The tank in the fig. contains oil of sp gr 0.75. Determine the reading of gauge A in N/m2. Air Oil 0.2m X 3m Hg A Solution: sp gr of oil = 0.75 Sp wt of oil ( ) = 0.75x9810 = 7357.5 N/m3 Sp wt of mercury ( ) = 13600x9.81 N/m3 = 133416 N/m3 Take atmospheric pressure to be 0 for gauge pressure. Starting from X and neglecting air, PA = -133416x0.2+7357.5x3 = -4610.7 N/m2 11. In the left hand of the fig., the air pressure is -225mm of Hg. Determine the elevation of the gauge liquid in the right hand column at A. 20.4KN/m 2 Air 110.2m Air Oil 107.8m Sp gr = 0.8 Water 106m h A’ A Liquid (sp gr. = 1.6) Solution: Air pressure at the left hand tank = -225mm Hg = -0.225m Hg Sp wt of water ( ) = 9810 N/m3 Sp wt of oil ( ) = 0.8x9810 = 7848 N/m3 Sp wt of mercury ( ) = 13600x9.81 KN/m3 = 133416 KN/m3 Sp wt of liquid ( ) = 1.6x9810 = 15696 N/m3 = -0.225x133416 = -30018.6 N/m2 PA’= PA -30018.6 + 7848x(110.2-106) + 15696 xh = 20400 + 9810 x(107.8-106+h) h = 5.96m Elevation at A = 106-5.96 = 100.04m 12. Compartments B and C in the fig. are closed and filled with air. The barometer reads 99.98 Kpa. When gages A and D read as indicated, what should be the value of x for gage E? (Hg in each tube) 206.8Kpa A C 254mm B x D E Solution: PA = 206.8 KPa = 206800 Pa Sp wt of mercury ( ) = 13.6x9810 N/m3 = 133416 N/m3 Starting from A and neglecting air, 206800 – 133416 X + 133416x0.254 = 0 X = 1.8m 13. In the fig., the areas of the plunger A and cylinder B are 38.7 cm2 and 387 cm2, respectively, and the weight of B is 4500 N. The vessel and the connecting passages are filled with oil of specific gravity 0.75. What force F is required for equilibrium, neglecting the weight of A? F A 4.8m XL Oil (sp gr = 0.75) XR B Solution: Area of A (AA) =38.7 cm2, Area of B (AB) =387 cm2 Weight of B (WB) = 4500N Sp wt of water ( ) = 9810 N/m3 Sp wt of oil ( ) = 0.75x9810 N/m3 = 7357.5 N/m3 Pressure at XL = Pressure at XR PA + 7357.5x4.8 = 4500/387x10-4 PA = 80963 N/m2 F = PA AA = 80963 x 38.7x10-4 = 313N 14. For the open tank, with piezometers attached on the side, containing two different immiscible liquids as shown in the fig., find (a) the elevation of liquid surface in piezometer A, (b) the elevation of liquid surface in piezometer B, and (c) the total pressure at the bottom of the tank. A Elv 2.1m B Liquid A (sp gr = 0.72) Elv 0.3m Liquid B (sp gr = 2.36) Elv 0m Solution: Sp wt of liquid A ( ) = 0.72x9810 N/m3 = 7063.2 N/m3 Sp wt of liquid B ( ) = 2.36x9810 N/m3 = 23151.6 N/m3 a) Elevation of liquid surface in piezometer A = elevation of liquid A in the tank = 2.1 m b) Pressure due to A at the interface (PA) = = 7063.2 x (2.1-0.3) = 12713.8 N/m2 Equivalent head for liquid B due to PA is = 12713.8/23151.6 = 0.55m Elevation of liquid surface in piezometer B = Elv of liquid at B + hA = 0.3 + 0.55 = 0.85m c) Total pressure at the bottom = = 7063.2 x (2.1-0.3) + 23151.6x0.3 = 19659 N/m2 15. In the fig., gage A reads 290Kpa abs. What is the height of water h? What does gage B read? Air : 175Kpa abs Water Hg h B A Solution: Sp wt of water ( ) = 9.81 KN/m3 Sp wt of mercury ( ) = 13.6x9.81 = 133.416 KN/m3 PA =290Kpa abs h=? PB =? 290 = 175 + 9.81h+133.416x0.7 h = 2.2m ( ) PB = 175 + 9.81 (2.2+0.7) = 203.4 KN/m2 70cm 16. A manometer is attached to a tank containing three different fluids as shown in fig. What will be the difference in elevation of the mercury column in the manometer (i.e. y)? Elv 6m Air pressure = 35Kpa Elv 5m Oil (sp gr = 0.82) Elv 2m Water Elv 0m 1m A B Hg Solution: Sp wt of water ( ) = 9.81 KN/m3 Sp wt of mercury ( ) = 13.6x9.81 = 133.416 KN/m3 Sp wt of oil ( ) = 0.82x9.81 = 8.0442 KN/m3 PA = PB 35+ 8.0442x3+9.81x3= 133.416y y = 0.66m y 17. Determine the pressure difference between two points A and B in the fig. Kerosene Sp. gr. = 0.82 Air 3 Sp. wt. = 11.77 N/m Benzene Sp. gr. = 0.88 B 10cm A 41cm 21cm 15cm 9cm Water Mercury Sp. gr. = 13.6 Solution: Sp wt of water ( ) = 9.81 KN/m3 Sp wt of mercury ( ) = 13.6x9.81 = 133.416 KN/m3 Sp wt of benzene ( ) = 0.88x9.81 = 8.6328 KN/m3 Sp wt of kerosene ( ) = 0.82x9.81 = 8.0442 KN/m3 Sp wt of air ( ) = 0.01177 KN/m3 Starting from A, ( PA – PB = 10.22 Kpa ) ( ) 18. Fig. below shows a conical vessel having its outlet at A to which a U-tube manometer is connected. The reading of the manometer given in the fig. shows when the vessel is empty. Find the reading of the manometer when the vessel is completely filled with water. 2m 3m A Initial h y Water Final 20 cm h Hg Solution: Sp wt of water ( ) = 9.81 KN/m3 Sp wt of mercury ( ) = 13.6x9.81 = 133.416 KN/m3 When the vessel is empty, equating pressure at initial level 9.81 y = 133.416x0.2 y = 2.72m When the vessel is filled with water, let us say the mercury moves by height h Equating pressure at final level ( ) ( ) ( ) ( ) h = 0.1145m Deflection of mercury = 2h+0.2 =2x0.1145+0.2 = 0.429m 19. Compute the absolute pressure at point A in the fig. Water Oil Sp. gr. = 0.88 15cm A 23cm 17.5cm Water Mercury Sp. gr. = 13.6 Solution: Sp wt of water ( ) = 9.81 KN/m3 Sp wt of mercury ( ) = 13.6x9.81 = 133.416 KN/m3 Sp wt of oil ( ) = 0.88x9.81 = 8.6328 KN/m3 Starting from A, PA = 27.7 Kpa Absolute pressure at A = Patm +27.7 = 101.3+27.7 = 129Kpa 20. The fig. shows a 1cm diameter U-tube containing mercury. If now 20cc of water is poured into the right leg, find the levels of the free liquid surfaces in the two tubes. Water h y y E 18cm Mercury F Solution: With the addition of 20ml of water, Drop in mercury level in right leg = y Rise in mercury level in left leg = y Depth of water column = h ( h= Vol of water added/Cross sectional area = ) = 0.254m PE = PF 13.6x9810x2y = 9810x0.254 y = 0.0093m = 0.93cm Height of free mercury level in the left leg = 18+0.93 = 18.93cm Height of free water level in the right leg = 18-0.93+25.4 = 42.47cm 21. The diameters of the limbs A and B of a U-tube shown in fig. are 5mm and 20mm respectively. The limb A contains a liquid of sp. gr. 0.9 while the limb B contains a liquid of sp.gr. 1.3. The fig. shows the position of the liquids in the two limbs. Find what pressure should be applied on the surface of the heavier liquid in limb B so that the rise in level in the limb A is 10mm. A B 10mm P liquid of sp.gr. = 0.9 ha h hb E1 E F liquid of sp.gr. = 1.3 No external pressure Solution: Sp gr of liquid in A = 0.9 Sp. gr. of liquid in B = 1.3 Diameter of limb A = 5mm Diameter of limb B = 20mm 10mm External pressure P F1 a. When no external pressure is applied PE = PF (a) b. When external force p is applied on the surface of liquid B Rise in limb A = 10mm Fall in limb B = h Level EF shifts to new position is E1F1. Volume of liquid transferred to left limb = Volume of liquid fell in right limb c/s area of left limbx10mm = c/s of right limbxh ( ( ) ) ( h = 0.000625m PE1 = PF1 ( ( ) ) ) ( ) ( ( ) ) (b) From a and b ( P = 135.5 N/m ) 2 22. Find the pressure difference between containers A and B shown in the figure. A 20cm Air 0 45 Water B 50cm 30cm E Mercury F 15cm Solution: Pressure at E = pressure at F = -82633 N/m2 23. An 8cm diameter piston compresses manometer oil into an inclined 7mm diameter tube, as shown in figure below. When a weight W is added to the top of the piston, the oil rises an additional distance of 10cm up the tube. How large is the weight, in N? W 10cm D = 8cm h Piston L1 A Initial level X X dh Y Y Final level Oil, sp gr = 0.827 D = 7mm 0 15 Solution: Diameter of piston (D) = 8cm = 0.08m Diameter of tube (d) = 7mm = 0.007m h = 0.10Sin15 = 0.0258m When the manometer is not connected to the container, the mercury in the reservoir is at original level and at level A in the tube. Equating pressure at XX (a) Due to compression, the fluid in the container moves down by dh and the fluid in the tube moves up by 10cm. Volume of fluid fallen = Volume of fluid risen dh = 0.000766m Equating the pressure at new level (YY) ( From a and b ) (b) ( ) Neglecting dh W = 1.05 N 24. Figure below shows a pipe containing a liquid of sp gr 0.8 connected to a single column micromanometer. The area of reservoir is 60 times that of the tube. The manometer liquid is mercury. Find the pressure in the pipe. A Liquid of sp gr 0.8 0.25m y A h1 X dy Z X Z Hg Solution: When the manometer is not connected to the container, the mercury in the reservoir is at original level and at level A in the tube. Equating pressure at XX (a) Due to pressure, manometric liquid in the reservoir drops by dy and it will travel a distance of 0.25m in the tube. Volume of fluid fallen = Volume of fluid risen Ady=ax0.25 60ady = ax0.25 dy = 0.0041m Equating pressure at new level (ZZ) ( ) ( ) (b) From a and b ( ) ( ) = 33868 N/m2 25. Find the gauge readings at A and B if the atmospheric pressure is 755mmHg. Closed A Air Oil, sp. gr. = 0.8 Hg Vapor 1.5m Water 0.6m C 2m B D 0.2m Hg Solution: Atm. pr. = = 100729 N/m2 Neglect pressure due to mercury vapor and air (small) Writing pressure equation for gauge A, = 13.6x9810x0.6 = 80049.6 N/m2 (abs. pr.) Gauge pressure at A = Abs. pr. - Atm. pr. = 80049.6-100729 = -20679.5 N/m2 Writing pressure equation for gauge B starting from gauge A = 8750.5 N/m2 Gauge pressure at B = 8750.5 N/m2