Test Midterm 1 F2013 MULTIPLE-CHOICE PROBLEMS :(Two marks per answer) (Circle the Letter Beside the Most Correct nswer in the Questions Below.) 1. The absolute viscosity µ of a fluid is primarily a function of: a. Density b. Temperature c. Pressure d. Velocity e. None of the above 2. n ideal fluid a. Has zero viscosity and satisfies p = ρrt. b. Has constant density and viscosity. c. Is incompressible and satisfies p = ρrt. d. Has zero viscosity and is incompressible. e. Is one which obeys Newton's Law of Viscosity. 3. streamline is a. The line connecting the midpoints of the flow cross-sections. b. Defined only in uniform flow. c. lways fixed in space for steady flows. d. Drawn perpendicular to the velocity vector at each point in the flow. e. lways along the path of a particle. 4. The velocity field given by 2 2 2 V x i xt j x t k is a. One-dimensional and steady. b. Two-dimensional and unsteady. c. Three-dimensional and steady. d. One-dimensional and unsteady. e. Three-dimensional and unsteady. 5. If a uniform solid body weighs 50 N in air and 30 N in water, its specific gravity is: a. 1.25 b. 1.50 c. 1.67 d. 2.50 e. 3.00
Test Midterm 1 F2013 6. Two parallel plates, one moving at 4 m/s and the other stationary, are separated by a 5- mm-thick layer of oil with specific gravity of 0.80 and kinematic viscosity 1.25E-4 m 2 /s. What is the average shear stress in the oil? a. 60 Pa b. 80 Pa c. 100 Pa d. 125 Pa e. 160 Pa 7. Newton s law of viscosity relates a. Intensity of pressure and rate of angular deformation b. Shear stress and rate of angular deformation c. Shear stress, viscosity and temperature d. Viscosity and rate of angular deformation e. The normal stress and rate angular deformation 8. If the weight of a body immersed in a fluid exceeds the buoyant force, then the body will a. Rise until its weight equals the buoyant force b. Tend to move downward and it may finally sink c. Flip over d. Float e. None of the above 9. The location of the centre of pressure of the force caused by the liquid acting on a plane surface submerged in a liquid is a. lways at the centroid of the surface. b. Independent of the inclination of the surface. c. t the centroid if the surface is horizontal. d. lways below the centroid regardless of the value of the angle of the surface. e. lways above the centroid. 10.The horizontal component of the force acting on a submerged curved surface is equal to the a. Weight of the liquid vertically above the curved surface. b. Weight of the liquid displaced. c. Product of the pressure at the centroid and the curved surface area. d. Force on a projection of the curved surface on a vertical plane. e. Weight of the curved surface.
Test Midterm 1 F2013 FILL IN THE BLNKS: (One mark per blank) (Fill in the blanks with the most appropriate word/number or mathematical expression.) There are ten blanks to fill in. Each correct answer is worth one mark. 1. The shear stress τyx is directed in the X coordinate direction and acts on the plane whose normal is in the Y direction. 2. STREKLINE is a line joining the present location of all particles that have passed a given point. 3. Reynolds number is the ratio of the inertia force to the VISCOUS force. 4. The buoyant force has a magnitude equal to the weight of the fluid DISPLCED _ by the body and is directed vertically UPWRD. 5. fluid is a substance that deforms CONTINUOUSLY when subjected to a shear stress no matter how SMLL that shear stress may be. 6. The buoyancy force that acts on a submerged body is the difference between the weight of fluid above the LOWER surface of the body and the weight of fluid above the UPPER surface of the body. TRUE ND FLSE QUESTIONS: (One and one-half marks per answer) (Circle the correct response, TRUE or FLSE) 1. In unsteady flows, streamlines show which fluid particles have passed through a given point (TRUE/FLSE). 2. Viscosity is not important in a study of a fluid body at rest (TRUE/FLSE). 3. Random molecular motion makes a more important contribution to the viscosity of a gas than intermolecular cohesion (TRUE/FLSE). 4. For the continuum assumption to be valid the Reynolds number must be very small (TRUE/FLSE). 5. Turbulent flow generally occurs for cases involving very slow motions of very viscous fluids in very small diameter tubes (TRUE/FLSE). 6. Shear stresses are important within the boundary layer (TRUE/FLSE). 7. The Lagrangian Method is concerned with a region or a point in space (TRUE/FLSE). 8. Pipe flows are always laminar if the Reynolds number is less than 2000 (TRUE/FLSE). 9. The eddy viscosity of water at 20ᴼC is 1.00x10-3 kg / (m sec) (TRUE/FLSE). 10. If a perfect gas undergoes an isothermal process, the relationship between pressure and density is P = constant *ρ (TRUE/FLSE).
Test Midterm 1 F2013 SHORT NUMERICL PROBLEMS: (nswer these problems in the space allotted.) 1. Determine the value of (P - PB) in Pa. Where: h1 = 15 cm h2 = 20 cm h3 = 15 cm h4 = 25 cm (20 Marks) The Line of common pressure in the manometer U is m Bm. Therefore, for constant density liquids: From table: = 680 kg/m 3 g = 9.81 m/s 2 gasoline = 998 kg/m 3 = 13,550 kg/m 3 Hg Pm = PBm P P h m 1 gasoline and P P h h h Bm B 4 gasoline 3 2 Hg 0 P P h h h h B 1 gasoline 4 gasoline 3 2 Hg P P h h h h B 1 gasoline 4 gasoline 3 2 Hg P P () h h h h B 4 1 gasoline 3 2 Hg (0.25 0.15) g 0. 15 g 0.2 gasoline 0.10 680 0.15 998 0.213,550 9.81 68 149.7 1355 9.81 28720.7 28.7kPa Hg g
Test Midterm 1 F2013 2. The tank shown in the figure below has a hemispherical dome of 1 m radius as part of its top surface. The tank is completely closed and contains pressurized water at 20ᴼC. pressure gage is located on the top surface as shown and has a reading of 150 kpa gage pressure. Determine the net horizontal and vertical components of the force that the water on the inside and the air on the outside exert on just the dome portion of the top. (20 Marks) s the pressure gauge registers gauge pressure, the air pressure contribution on the inside of the tank is cancelled by the air pressure on the outside of the tank. s it is a hemispherical dome, the net horizontal component of the force applied by the pressurized water is zero since the dome is symmetric. The net vertical force applied by the pressurized water can be determined by finding the net vertical force created by a column of water of height so as to create a hydrostatic force equal to 150 kpa at the top of the tank. Or, 150000 g h H O 998 9.81 h or h = 15.32 m. Using 3 H 998 kg/m 2O from table The total vertical force will be equal to the weight of water that would fill a column of that height with a diameter of 2 metres plus the weight of water that would fill the inside of the hemisphere. 2 F V gv T Thus, the total volume of water is Total Volume Volume of Cylinder + Volume of hemisphere V R h 2 R T 2 3 / 3 48.129 2.094 50.223 Then the vertical force applied by the water on the dome = VT * ρ g = 50.233 * 998 * 9.81 Fv = 491702 Newtons = 492 kn in the upward direction.