MET330_Unit 1 REVIEW ASSIGNMENT_

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Feb 20, 2024

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MET330 Applied Fluids Week 1 Review Assignment I pledge to support the Honor System of ECPI. I will refrain from any form of academic dishonesty or deception, such as cheating or plagiarism. I am aware that as a member of the academic community, it is my responsibility to turn in all suspected violators of the honor code. I understand that any failure on my part to support the Honor System will be turned over to a Judicial Review Board for determination. I will report to the Judicial Review Board hearing if summoned. Type your name here: Date: Review Assignment Page 1
2.4. Which would have the greater dynamic viscosity, a cold lubricating oil or fresh water? Explain your answer. Cold lubricating oil has greater dynamic viscosity than fresh water. Considering cold lubricating oil and fresh water, if we stir the fluids with a rod, a velocity gradient is developed in both the fluids. A greater force is required to stir the cold oil, which indicates the higher shear stress in cold oil. It is a known fact that the dynamic viscosity of a fluid is directly proportional to its shear stress. Thus, we can conclude that the cold lubricating oil has a greater dynamic viscosity. 2.14. Define a  Newtonian fluid . As a fluid moves between a moving surface and a stationary surface, a shear stress is developed in the fluid. The shear stress developed is directly proportional to the measure of velocity change. 2.17. Give four examples of the types of fluids that are non-Newtonian. 1. Molasses 2. Starch in water 3. Paint 4. Mustard 2.36. Define  viscosity index  ( VI ). A measure of how greatly the viscosity of a fluid changes with temperature. 2.51. If you were asked to check the viscosity of an oil that is described as SAE 40, at what temperatures would you make the measurements? You would measure at either 40°C (100°F) or 100°C(212°F) Page 2
3.8 The pressure in a certain tank is −55.8 Pa(abs). (True? Or False? Explain your answer. False, the reason for this is because this is a absolute pressure (abs) which is always positive. 3.10. The pressure in a certain tank is −150 kPa(gage). (True? Or False? Explain your answer. True, because looking at the unit this is a gauge pressure and assuming the standard atmospheric pressure is 101.3kPA if the gauge pressure is below this it will be negative or sometimes called a vacuum. Review Assignment Problems a. Identify and list all variables from the problem. b. W rite the equation(s) you will use that are found in the textbook . Do not use any other equations. c. S ubstitute your variables into the equation. YOU MUST INCLUDE UNITS IN ALL PLACES OF YOUR EQUATIONS. ( m 2 , N m 3 , m s 2 ,kg, psi ,etc ) d. You must show all of your work, even “simple” calculations. During your calculations, please use 3 decimal places or 3 significant decimal places. This does not apply to financial cost calculations. e. Clearly identify your answer with correct units . CORRECT METHOD: p = γ liquid g = 2000 N m 2 9.81 m s 2 = 203.874 kg m 3 Page 3
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p = γ liquid g = 2000 N / m 3 9.81 m / s 2 = 203.874 kg / m 3 INCORRECT METHOD: p2 = (W * a)/g + p1 = a^2 - Ap – μ = 0 1.52. The maximum pressure that can be developed for a certain fluid power cylinder is 15.0 MPa. Compute the required diameter for the piston if the cylinder must exert a force of 30 kN. Known values: P = 15 x 10 6 N / m 2 F = 30 x 10 3 N Governing equations: 𝐹 = 𝑃 A Calculation The Diameter of the piston is 0.0505m 1.86. Glycerin has a specific gravity of 1.258. How much would 0.50 m 3  of glycerin weigh? What would be its mass? Known values: Sg of glycerin = 1.258 Page 4
Volume of glycerin V= 0.5 m 3 Governing equations: 𝑠𝑔 ¿ Yglycerin Yw@ 4 ° C Y = w v W = mg Calculations: Mass of glycerin = 629kg 1.93. A storage vessel for gasoline (sg = 0.68) is a vertical cylinder 10 m in diameter. If it is filled to a depth of 6.75 m, calculate the weight and mass of the gasoline. Known values: Page 5
Gasoline sg = 0.68 Diameter of cylinder D = 10m Depth of gasoline h = 6.75m Governing equations: 𝑉 = 𝐴ℎ 𝑤 = 𝛾𝑉 𝑘𝑔 𝑠𝑔 = 1000 𝑚 3 Calculations: Page 6
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Weight of gasoline is 3.48. For the tank in  Fig. 3.22 , compute the depth of the oil if the depth of the water is 2.80 m and the gage at the bottom of the tank reads 52.3 kPa(gage). Known values: Depth of water h2 = 2.8m Oil sg = 0.86 Pressure at bottom of tank Pbottom = 52.3kPa (gage) Governing equations: ∆𝑝 = 𝛾ℎ 𝑇 = 𝛾 𝑤 𝑤 + 𝛾 𝑜𝑖𝑙 𝑜𝑖𝑙 Calculations: Page 7 Fig for P3.48 The mass of gasoline is Also, Weight,
The depth of the oil in the tank is 2.94m 3.50. Figure 3.22  represents an oil storage drum that is open to the atmosphere at the top. Some water was accidentally pumped into the tank and settled to the bottom as shown in the figure. Calculate the depth of the water  h 2  if the pressure gage at the bottom reads 158 kPa(gage). The total depth  h T  = 18.0 m. Known values: h 2 willbe depthof water Governing equations: ∆𝑝=𝛾ℎ 𝑇 =𝛾 𝑤 𝑤 +𝛾 𝑜𝑖𝑙 𝑜𝑖𝑙 Calculations: The depth of water in the tank is 4.44m Page 8 Fig for P3.50
3.54. Figure 3.23  shows a closed tank that contains gasoline floating on water. Calculate the air pressure above the gasoline. Known values: Governing equations: 𝑝 𝑎𝑡𝑚 = 𝛾 𝑚𝑒𝑟𝑐𝑢𝑟𝑦 𝑚𝑒𝑟𝑐𝑢𝑟𝑦 + 𝛾 𝑤𝑎𝑡𝑒𝑟 𝑤𝑎𝑡𝑒𝑟 + 𝛾 𝑔𝑎𝑠 𝑔𝑎𝑠 = 𝑝 𝑎𝑖𝑟 Calculations: Air pressure above the gasoline is 43.81 kPa Page 9 Fig for P3.54
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3.67. For the compound manometer shown in  Fig. 3.31 , calculate the pressure at point A. Known values: Governing equations: 𝑝 𝑎𝑡𝑚 + 𝛾 𝑚𝑒𝑟𝑐𝑢𝑟𝑦 𝑚𝑒𝑟𝑐𝑢𝑟𝑦 − 𝛾 𝑤𝑎𝑡𝑒𝑟 𝑤𝑎𝑡𝑒𝑟 + 𝛾 𝑚𝑒𝑟𝑐𝑢𝑟𝑦 𝑚𝑒𝑟𝑐𝑢𝑟𝑦 − 𝛾 𝑜𝑖𝑙 𝑜𝑖𝑙 = 𝑝 𝐴 Calculations: 4.14. A rectangular gate is installed in a vertical wall of a reservoir, as shown in  Fig. 4.26 . Compute the magnitude of the resultant force on the gate and the location of the center of pressure. Also compute the force on each of the two latches shown. Known values: Governing equations: Page 10 Fig for P3.67 Fig for P4.14 F R
𝐹 𝑅 = 𝛾 𝑤 𝐴 𝑐𝑒𝑛𝑡𝑒𝑟= h 3 Calculations: 4.17. If the wall in  Fig. 4.29  is 4 m long, calculate the total force on the wall due to the oil pressure. Also determine the location of the center of pressure and show the resultant force on the wall. Known values: Governing equations: Calculations: Page 11 L Fig for P4.17 2
4.35. For the water tank shown in  Fig. 4.43 , compute the magnitude and location of the total force on the inclined wall. Known values: Governing equations: Calculations: 4.52. Use  Fig. 4.52 . The surface is 1.50 m long. Compute the magnitude of the horizontal component of the force and compute the vertical component of the force exerted by the fluid on that surface. Then compute the magnitude of the resultant force and its direction. Show the resultant force acting on the curved surface. Known values: Page 12 Fig for P4.35
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Governing equations: Calculations: 5.7. A concrete block with a specific weight of 23.6 kN/m 3  is suspended by a rope in a solution with a specific gravity of 1.15. What is the volume of the concrete block if the tension in the rope is 2.67 kN? Known values: Governing equations: Calculations: Page 13 T Fig for P4.35
5.9. A steel cube 100 mm on a side weighs 80 N. We want to hold the cube in equilibrium under water by attaching a light foam buoy to it. If the foam weighs 470 N/m 3 , what is the minimum required volume of the buoy? Known values: Governing equations: Calculations: 5.12. Figure 5.20  shows a cube floating in a fluid. Derive an expression relating the submerged depth  X , the specific weight of the cube, and the specific weight of the fluid. Known values: Governing equations: Page 14 Fig for P5.9 w c F b Fig for P5.12
Calculations: 5.40. A container for an emergency beacon is a rectangular shape 30.0 in wide, 40.0 in long, and 22.0 in high. Its center of gravity is 10.50 in above its base. The container weighs 250 lb. Will the box be stable with the 30.0-in by 40.0-in side parallel to the surface in plain water? Known values: Governing equations: Calculations: 5.58. If the barge in  Problem 5.57  is loaded with 240 000 lb of loose coal having an average density of 45 lb/ft 3 , how much of the barge would be below the water? Is it stable? Known values: Page 15 Fig for P5.40 w c F b 20ft m barge = 210,000 lb m coal = 240,000 lb ? ? ? Fig for P5.58
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Governing equations: MB= I Vd Calculations: Page 16