PTT 04/ Applied Fluid Mechanics Sem, Session015/016 ASSIGNMENT 1 CHAPTER AND CHAPTER 1. The air in an automobile tire with a volume of 0.0740 m is at 0 C and 140 kpa. Determine the amount of air that must be added to raise the pressure to the recommended value of 10 kpa gage. Assume the atmospheric pressure to be 100 kpa and the temperature and volume to remain constant. Assumptions 1 At specified conditions, air behaves as an ideal gas. The volume of the tire remains constant. Properties The gas constant of air is kj kpa m kpa m R 0.87 0.87 kg K kj kg K. The initial and final absolute pressures in the tire are P 1 = P g1 + P atm = 140 + 100 = 40 kpa Tire 0.0740 m 0C 140 kpa P = P g + P atm = 10 + 100 = 10 kpa Treating air as an ideal gas, the initial mass in the tire is m P1 V (40 kpa)(0.0740 m ) 1 RT 1 (0.87 kpa m /kg K)(0 K) 0.04 kg Noting that the temperature and the volume of the tire remain constant, the final mass in the tire becomes m PV (10 kpa)(0.0740 m ) RT (0.87 kpa m /kg K)(0 K) Thus the amount of air that needs to be added is m m m 1 0.68 0.04 0.0596kg 0.68 kg Discussion ideal gas law. Notice that absolute rather than gage pressure must be used in calculations with the
PTT 04/ Applied Fluid Mechanics Sem, Session015/016. A cylindrical tank of methanol has a mass of 40 kg and a volume of 51 L. Determine the methanol s weight, density and specific gravity. Also estimate how much force is needed to accelerate this tank linearly at 0.5 m/s? (Given: g=9.81 m/s). Assumptions 1 The volume of the tank remains constant. Properties The density of water is 1000 kg/m. The methanol s weight, density, and specific gravity are W = mg = 40 9.81 = 9.40 N ρ = m V = SG = 40 kg 51 L 1 m 1000 L ρ ρ H O The force needed to accelerate the tank at the given rate is = = 784 kg/m 784 kg/m 1000 kg/m = 0.784 F = ma = 40KG 0.5 m s = 10 N. Define cavitation and its causes to happen. In a piping system, the water temperature remains under 0C. Determine the minimum allowable pressure in the system to avoid cavitation. Cavitation is a structural damage that is caused by these following sequences: If P drops below P v, liquid is locally vaporized, creating vapor bubbles (called cavitation bubbles since they form cavities in the liquid). Cavitation bubbles collapse when local P rises above P v. Collapse of cavitation bubbles is a violent process which can damage machinery. Cavitation is noisy (cause by cavitation bubbles), and can cause structural vibrations. The vapor pressure of water at 0C is 4.46 kpa. (5 MARKS) To avoid cavitation, the pressure anywhere in the flow should not be allowed to drop below the vapor (or saturation) pressure at the given temperature. That is, P min P sat@0 C 4.46kPa Therefore, the pressure should be maintained above 4.46 kpa everywhere in flow. Discussion Note that the vapor pressure increases with increasing temperature, and thus the risk of cavitation is greater at higher fluid temperatures.
PTT 04/ Applied Fluid Mechanics Sem, Session015/016 4. Illustrate with aid of suitable graph on how does the dynamic viscosity of liquids and gases vary with temperature. Differentiate units for dynamic and kinematic viscosity. i. Absolute/dynamic viscosity, µ; unit Poise (1 Poise= 0.1 Pa.s)-M ii. Kinematic viscosity, =µ/; unit Stoke (1 Stoke= 1 cm /s)-m -In liquid, the viscosity reduces as the temperature increases, for gases it behaves another way around [The figure -M] In a liquid, the molecules possess more energy at higher temperatures, and they can oppose the large cohesive intermolecular forces more strongly. As a result, the energized liquid molecules can move more freely.-m In a gas, the intermolecular forces are negligible, and the gas molecules at high temperatures move randomly at higher velocities. This results in more molecular collisions per unit volume per unit time and therefore in greater resistance to flow.-m (10MARKS) 5. Differentiate between Newtonian and non-newtonian fluids. You may aid suitable diagrams to support your explanation. The rate of deformation (velocity gradient) of a Newtonian fluid is proportional to shear stress, and the constant of proportionality is the viscosity.-m Variation of shear stress with the rate of deformation for Newtonian and non-newtonian fluids (the slope of a curve at a point is the apparent viscosity of the fluid at that point).-m Newtonian fluid: Slope of the graph constant implies that the viscosity is constant. Eg: water, oil and air -1M 4M
PTT 04/ Applied Fluid Mechanics Sem, Session015/016 Non-newtonian: The slope is not constant, it does vary dependent on the slope of the curve. Eg: Pseudoplastic (blood plasma, molten polythelene, latexes, molasses), dilatants fluid (corn starch n ethylene glycol, starch in water, ingredient in paint) and Bingham fluids (chocolate, catsup, mustard, mayonnaise, toothpaste)=1m 6. (a) The absolute pressure in water at a depth of 8 m is read to be 175 kpa. Determine: i) The local atmospheric pressure Assumptions The liquid and water are incompressible. Properties The specific gravity of the fluid is given to be SG = 0.78. We take the density of water to be 1000 kg/m. Then density of the liquid is obtained by multiplying its specific gravity by the density of water, SG H O (0.78)(1000 kg/m ) 780 kg/m (a) Knowing the absolute pressure, the atmospheric pressure can be determined from P atm P gh (175 kpa) - (1000 kg/m )(9.81 m/s 96.5 kpa 96.5kPa 1kPa )(8 m) 1000 N/m P atm h ii) The absolute pressure at a depth of 8 m in a liquid whose SG is 0.78 at the same location. The absolute pressure at a depth of 8 m in the other liquid is P Patm gh (96.5 kpa) (780 kg/m )(9.81 m/s 157.7 kpa 158 kpa 1kPa )(8 m) 1000 N/m Discussion Note that at a given depth, the pressure in the lighter fluid is lower, as expected. (6) (b). A vacuum gauge connected to a tank reads 45 kpa at a location where the barometric reading is 755 mmhg. Determine the absolute pressure in the tank. Take SG Hg as 1.59. Solution The vacuum pressure reading of a tank is given. The absolute pressure in the tank is to be determined. Properties The density of mercury is given to be = 1,590 kg/m. The atmospheric (or barometric) pressure can be expressed as P abs 45kPa P atm gh 1 N 1 kpa (1,590 kg/m )(9.807 m/s )(0.755 m) 1 kg m/s 1000 N/m 100.6 kpa P atm = 755mmHg Then the absolute pressure in the tank becomes
PTT 04/ Applied Fluid Mechanics Sem, Session015/016 P abs P P 100.6 45 55.6 kpa atm vac Discussion kpa. The gage pressure in the tank is the negative of the vacuum pressure, i.e., P gage = 45 (0 MARKS) 7. Consider a double fluid manometer attached to an air pipe shown in Fig. 1 (Refer Fig. P-49). If the SG of one fluid is 1.55, determine the SG of the other fluid for the indicated absolute pressure of air. Take the atmospheric pressure to be 100 kpa. Assumptions 1 Densities of liquids are constant. The air pressure in the tank is uniform (i.e., its variation with elevation is negligible due to its low density), and thus the pressure at the air-water interface is the same as the indicated gage pressure. Properties The specific gravity of one fluid is given to be 1.55. We take the standard density of water to be 1000 kg/m. Starting with the pressure of air in the tank, and moving along the tube by adding (as we go down) or subtracting (as we go up) the gh terms until we reach the free surface where the oil tube is exposed to the atmosphere, and setting the result equal to P atm give P air gh gh P Pair Patm SGw gh SG1 w gh1 1 1 atm Rearranging and solving for SG, 76 100 kpa h1 Pair Patm 0. m 1000 kg m/s SG SG1 1.55 1.6 1.4 h w gh 0.40 m (1000 kg/m )(9.81 m/s )(0.40 m) 1 kpa m Air Fluid 1 SG 1 cm 40 cm Fluid SG 8. For a gate width of m into the paper (Refer Fig. P-66 in the textbook), determine the force required to hold the gate ABC at its location. Assumptions Atmospheric pressure acts on both sides of the gate, and thus it can be ignored in calculations for convenience. Properties Specific gravities are given in the figure.
PTT 04/ Applied Fluid Mechanics Sem, Session015/016 Since there are two different fluid layers it would be useful to convert one of them to another one to make the problem easier. The pressure at the interface is p 0.8698100.5 418. Pa Now, the question is how much fluid from the second one can make the same pressure. p 418. h SG 1. 0.5 m 5 cm 1.9810 1066. Therefore the system can be simplified as shown: Original level 45 0 SG = 0.86 50 cm A Hinge 10 cm 80 cm SG=1. F 1 C F F 40 cm B 0.1 0.8 F1 gh ca 86 9.81 0.4 gh A 1.9810 0.80 0.5 0.4 11101 0.1 10 9.81 0.5 0.8 1817 N N F c ycp I xx,c 0.9 0.9 1 ycg 0.4 0.99 m ycg A 0.85 (0.9 ) Take moment about hinge will give F 0.4 1817 0.4 1. 0.9911101 0 F = 14.05 kn *In solving this problem students only need to calculate F1 and F. (5 MARKS)