HW_2 - 82 mama 3 I FLUID srnncs 61‘ 3.14 An inverted...

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Unformatted text preview: 82 mama 3 I FLUID srnncs 61‘ 3.14 An inverted cylindrical container is lowered slowly beneath the surface of a p00| of water. Air trapped in the container is compressed isothermally as the hydrostatic pressure increases. Develop an expression for the water height, y, inside the container in terms of the container height. H. and depth of submersion, h. Plot y/H versus h/H. 3.15 You close the top of your straw using your thumb and lift it out of your glass containing Coke. Holding it vertically. the total length of the straw is I? in., but the Coke held in the straw is in the bottom 6 in. What is the pressure in the straw just below your thumb? ignore any surface tension effects. 3.16 A water tank filled with water to a depth of 5 111 has an in- spection cover (2.5 cm X 2.5 cm square) at its base, held in place by a plastic bracket. The bracket can hold a load of 40 N. [s the bracket strong enough? If it is, what would the water depth have to be to cause the bracket to break? 3.17 A container with two circular vertical tubes of diameters d. = 39.5 mm and d3 = 12.7 mm is partially filled with mercury. The equilibrium level of the liquid is shown in the left diagram. A cylindrical object made from solid brass is placed in the larger tube so that it floats, as shown in the right diagram. The object is D = 37.5 mm in diameter and H = 76.2 mm high. Calculate the pressure at the lower surface needed to float the object. Determine the new equilibrium level. h. of the mercury with the brass cylinder in place. l‘fl “i Mercury P3.l7 3.18 A partitioned tank as shown contains water and mercury. What is the gage pressure in the air trapped in the left chamber? What pressure would the air on the left need to be pumped to in order to bring the water and mercury free surfaces level? P3. [8. 3.19 P320 3.19 in the tank of Problem 3.18. if the opening to atmosphere on the right chamber is first sealed, what pressure would the air on the left now need to be pumped to in order to bring the water and mercury free surfaces level? (Assume the air trapped in the right chamber behaves isothermally.) 3.20 A manometer is formed from glass tubing with uniform in- side diameter. D = 6.35 mm. as shown. The U-tube is partially filled with water. Then ti = 3.25 cm3 of Meriam red oil is added to the left side. Calculate the equilibrium height. H. when both legs of the U-tube are open to the atmosphere. 3.21 Consider the two-fluid manometer shown. Calculate the ap- plied pressure difference. P1 P2 Pt P2 air-"'- 15-: Jfivzflew‘}. -/ Carbon tetrachloride "-' Lt. ‘. a P3.2| P322 3.22 The manometer shown contains two liquids. Liquid A has 50 = 0.88 and liquid B has SG = 2.95. Calculate the deflection. h. when the applied pressure difference is pl — p2 = 18 lbffftz. ,4 3.23 The manometer shown contains water and kerosene. With both tubes open to the atmosphere. the free-surface elevations dif- fer by Hu = 20.0 mm. Determine the elevation difference when a pressure of 98.0 Pa (gage) is applied to the right tube. Liquid .4 ',..e- ’ _ e:me ea Kerosene Liquid 8 133.23 P324 3.24 Determine the gage pressure in psig at point a, if liquid A has SG = 0.75 and liquid 3 has S0 = 1.20. The liquid surrounding point a is water and the tank on the left is open to the atmosphere. 3.25 The NIH Corporation‘s engineering department is evaluat- ing a sophisticated $80,000 laser system to measure the difference in water level between two large water storage tanks. It is import- ant that small differences be measured accurately. You suggest that the job can be done with a $200 manometer arrangement. An oil less dense than water can be used to give a 10 : l amplification of meniscus movement; a small difference in level between the tanks will cause 10 times as much deflection in the oil levels in the manometer. Determine the specific gravity of the oil required for 10: l amplification. - ‘ \ "32x \ ‘ 3.26 Consider a tank containing mercury. water. benzene. and air as shown. Find the air pressure (gage). If an opening is made in the top of the tank. find the equilibrium level of the mercury in the manometer. It: 0.25 p__l__m .H 3.27 Water flows downward along a pipe that is inclined at 30‘ below the horizontal. as shown. Pressure difference pA e p3 is due partly to gravity and partly to friction. Derive an algebraic expression for the. pressure difference. Evaluate the pressure difference it'L = 5 ft and It = 6 in. 3.28 A rectangular tank, open to the atmosphere, is filled with water to a depth of 2.5 m as shown. A U-tube manometer is connected to the tank at a location 0.7 m above the tank bot- tom. II‘ the zero level of the Meriam blue manometer fluid is P3 .29 PROBLEMS 83 0.2 m below the connection. determine the deflection I after the manometer is connected and all air has been removed from the connecting leg. 3.29 A reserVoir manometer has vertical tubes of diameter D = l8 mm and d = 6 mm. The manometer liquid is Meriam red oil. Develop an algebraic expression for liquid deflection L in the small tube when gage pressure A]; is applied to the reservoir. Evaluate the liquid deflection when the applied pressure is equiv- alent to 25 mm of water (gage). 3.30 The manometer fluid of Problem 3.28 is replaced with mer- cury (same zero level). The tank is sealed and the air pressure is in- creased to a gage pressure o‘l't).5 atm. Determine the deflection I. 3.3] A reservoir manometer is calibrated for use with a liquid of specific gravity 0.827. The reservoir diameter is 5/3 in. and the (verti- cal) tube diameter is 3/16 in. Calculate the required distance between marks on the vertical scale for l in. of water pressure difference. 3.32 The inclined-tube manometer shown has D = 76 mm and d = 8 mm. and is filled with Meriam red oil. Compute the angle. H. that will give a l5«cm oil deflection along the inclined tube for an applied pressure of 25 mm of water (gage). Determine the sen— sitivity of this manometer. P332. 3.33 3.33 The inclined-tube manometer shown has D = 96 mm and d = 8 mm. Determine the angle. 6. required to provide a 5 : | in— crease in liquid deflection. L. compared with the total deflection in a regular U-lube manometer. Evaluate the sensitivity of this inclined-tube manometer. @6134 A student wishes to design a manometer with better sensi- tivity than a water-filled U-tube of constant diameter. The stu- dent‘s concept involves using tubes with different diameters and two liquids. as shown. Evaluate the deflection h of this manometer. if the applied pressure difference is Ap = 250 ij2. Determine the sensitivity of this manometer. Plot the manometer sensitivity as a function of the diameter ratio dgfdl. f’atm Farm Patm + AP Patrn llflil It Oil \tsc = 0.85) P334 3.35 A barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of 84 CHAPTER 3 I FLUID STATIGS a vacuum at the top of the barometer). On a day when the tem~ perature is 70”F. the mercury column height is 28.35 inches (cor- rected for thermal expansion). Determine the barometric pressure in psia. [fthe ambient temperature increased to 85F and the baro- metric pressure did not change, would the mercury column be longer, be shorter, or remain the same length? Justify your answer. 3.36 If the tank of Problem 3.28 is sealed tightly and water drains slowly from the bottom of the tank. determine the deflec- tion, 1, after the system has attained equilibrium. 3.37 A water column stands 50 mm high in a 2.5-mmvdiameter glass tube. What would be the column height if the surface tension were zero? What would be the column height in a l.0—min-diameter tube? 3.38 Consider a small diameter open-ended tube inserted at the interface between two immiscible fluids of different densities. Derive an expression for the height difference Ah between the in- terface level inside and outside the tube in terms of tube diameter D, the two fluid densities, p1 and p2, and the surface tension 0 and angle 0 for the two fiuids‘ interface. [f the two fluids are water and mercury, find the tube diameter such that Ah < 10 mm. 3.39 You have a manometer consisting of a tube that is 1.1-cm ID. On one side the manometer leg contains mercury. 10 cc of an oil ($0 = 1.67). and 3 cc of air as a bubble in the oil. The other leg just contains mercury. Both legs are open to the atmosphere and are in a static condition. An accident occurs in which 3 cc of the oil and the air bubble are removed from the one leg. How much do the mercury height levels change? \9‘340 Based on the atmospheric temperature data of the US. Standard Atmosphere of Fig. 3.3, compute and plot the pressure variation with altitude, and compare with the pressure data of Table A.3. 6:33.41 Two vertical glass plates 300 mm X 300 mm are placed in an open tank containing water. At one end the gap between the plates is 0.1 mm. and at the other it is 2 mm. Plot the curve ofwatcr height between the plates from one end of the pair to the other. 3.42 Compare the height due to capillary action of water exposed to air in a circular tube of diameter D = 0.5 mm, and between two infinite vertical parallel plates of gap a 2 0.5 mm. $3.43 On a certain calm day, a mild inversion causes the atmo— spheric temperature to remain constant at 85“]3 between sea level and 16.000 ft altitude. Under these conditions, (a) calculate the elevation change for which a 2 percent reduction in air pressure occurs. (b) determine the change of elevation necessary to effect a [0 percent reduction in density, and (c) plot py’p. and pg/p. as a function of A2. 6363.44 The Martian atmosphere behaves as an ideal gas with mean molecular mass of 32.0 and constant tentperature of 200 K. The atmospheric density at the planet surface is p = 0.015 kg/m'l and Martian gravity is 3.92 m/sz. Calculate the density of the Martian atmosphere at height a : 20 km above the surface. Plot the ratio of density to surface density as a function of elevation. Compare with that for data on the earth’s atmosphere. 6163.45 At ground level in Denver. Colorado. the atmospheric pressure and temperature are 83.2 kPa and 25C. Calculate the pressure on Pike’s Peak at an elevation of 2690 in above the city assuming (a) an incompressible and (b) an adiabatic atmosphere. Plot the ratio of pressure to ground level pressure in Denver as a function of elevation for both cases. 3.46 A door 1 m wide and 1.5 m high is located in a plane verti~ cal wall of a water tank. The door is hinged along its upper edge, which is 1 m below the water surface. Atmospheric pressure acts on the outer surface of the door. (a) If the pressure at the water surface is atmospheric. what force must be applied at the lower edge of the door in order to keep the door from opening? lb) if the water surface gage pressure is raised to 0.5 atm. what force must be applied at the lower edge of the door to keep the door from opening? (c) Find the ratio FIFO as a function of the surface pres- sure ratio p./p,,[,,,. {Fm is the force required when p,, = palm.) .347 A hydropneumatic elevator consists of a pistonrcylinder assembly to lift the elevator cab. Hydraulic oil. stored in an accu- mulator tank pressurized by air, is valved to the piston as needed to lift the elevator. When the elevator descends, oil is returned to the accumulator. Design the least expensive accumulator that can satisfy the system requirements. Assume the lift is 3 floors, the maximum load is [0 passengers, and the maximum system pres- sure is 800 kPa (gage). For column bending strength. the piston diameter must be at least 150 mm. The elevator cab and piston have a combined mass of 3000 kg, and are to be purchased. Per- fomt the analysis needed to define. as a function of system operat- ing pressure. the piston diameter, the accumulator volume and diameter. and the wall thickness. Discuss safety features that your company should specify for the complete elevator system. Would it be preferable to use a completely pneumatic design or a com- pletely hydraulic design? Why? ' 6363.48 A door I m wide and 1.5 m high is located in a plane verti- cal wall of a water tank. The door is hinged along its upper edge, which is l m below the water surface. Atmospheric pressure acts on the outer surface of the door and at the water surface. (a) Deter- mine the magnitude and line of action of the total resultant force from all fluids acting on the door. (b) If the water surface gage pressure is raised to 0.3 atm. what is the resultant force and where is its line of action? (c) Plot the ratios HF” and _v'/_v,. for different values of the surface pressure ratio pv/pmm. (F0 is the resultant force when p, = pwm.) 3.49 Find the pressures at points A. B. and C. as shown, and in the two air cavities. P3 .49 3.50 A triangular access port must he provided in the side ofa form containing liquid concrete. Using the coordinates and dimen- sions shown, determine the resultant force that acts on the port and its point of application. u—ias-ru in it 1.1. trial a H' n hit-Ft ...
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HW_2 - 82 mama 3 I FLUID srnncs 61‘ 3.14 An inverted...

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