MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAPALLI 6113 DEPARTMENT: CIVIL SUB.CODE/ NAME: CE6303/ MECHANICS OF FLUIDS SEMESTER: III UNIT-1 FLUID PROPERTIES TWO MARK QUESTIONS AND ANSWERS 1. Define fluid mechanics.(auc May/June 010) It is the branch of science, which deals with the behavior of the fluids (liquids or gases) at rest as well as in motion.. Define Mass Density..(AUC Nov/Dec 010) Mass Density or Density is defined as ratio of mass of the fluid to its volume (V) Density of water = 1 gm/cm 3 or 1000 kg / m 3. 3. Define Secific Weight..(AUC May/June 010) It is the ratio between weight of a fluid to its volume. Weight of fluid Mass of fluid w g g Volume of fluid w g Unit: N / m 3 Volume of fluid 4. Define Viscosity. (AUC Nov/Dec 009) Viscosity is defined as the roerty of fluid, which offers resistance to the movement of one layer of fluid over another adjacent layer of fluid. PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 1
When two layers move one over the other at different velocities, say u and u+, the viscosity together with relative velocity causes a shear stress acting between the fluid layers. The to layer causes a shear stress on the adjacent lower layer while the lower layer causes a shear stress on the adjacent to layer. This shear stress is roortional to the rate of change of velocity. Coefficient of namic viscosity (or) only viscosity / = rate of shear strain 5. Define Secific Volume. (AUC May/June 011) Volume er unit mass of a fluid is called secific volume Unit: m 3 / kg. 6. Define Secific Gravity. (AUC May/June 013) Secific gravity is the ratio of the weight density or density of a fluid to the weight density or density of standard fluid. It is also called as relative density. Unit : Dimension less. Denoted as: S S (forliquid) Weight density of liquid Weight density of water S (for gases)= 7. Calculate the secific weight, density and secific gravity of 1 litre of liquid which weighs 7 N. (AUC May/June 013) Solution: Given V 1ltre 1 1000 m 3 PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page
W = 7 N i. S. Weight (w) weight 7 N volume 1 m 3 1000 7000 N / m 3 ii Density () w 7000 N kg / m 3 9.81 m 3 713..5 Kg / m 3 g iii. S. Gravity (S) Density of liquid 713.5 Density of water 1000 (Density of water = 1000 kg / m 3) S = 0.7135 8. State Newton s Law of Viscosity. (AUC Nov/Dec 013) It states that the shear stress ( ) on a fluid element layer is directly roortional to the rate of shear strain. The constant of roortionality is called the co-efficient of viscosity 9. Name the Tyes of fluids. (AUC Nov/Dec 01) 1. Ideal fluid. Real fluid 3. Newtonian fluid 4. Non-Newtonian fluid. 5. Ideal lastic fluid 10. Define Kinematic Viscosity. (AUC Nov/Dec 011) It is defined as the ratio between the namic viscosity and density of fluid. PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 3
1 Stoke Cm 1 m 10 4 m / s. S 100 S Centistoke means 1 stoke 100 11. Determine the secific gravity of a fluid having viscosity 0.05 oise and Kinematic viscosity 0.035 stokes. Given: Viscosity, μ = 0.05 oise = (0.05 / 10) Ns / m. Kinematic viscosity ν = 0.035 stokes = 0.035 cm / s = 0.035 x 10-4 m / s 0.035 10 4 0.05 1 10 148.5kg / m 3 Density of liquid 148.5 1. Define Comressibility. Comressibility is the recirocal of the bulk molus of elasticity, K which is defined as the ratio of comressive stress to volumetric strain. 13. Define Surface Tension. Surface tension is defined as the tensile force acting on the surface of a liquid in contact with a gas or on the surface between two immiscible liquids such that the contact surface behaves like a membrance under tension. Unit: N / m. PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 4
14. Define Caillarity: Caillary is defined as a henomenon of rise of a liquid surface is a small tube relative to adjacent general level of liquid when the tube is held vertically in the liquid. The resistance of liquid surface is known as caillary rise while the fall of the liquid surface is known as caillary deression. It is exressed in terms of cm or mm of liquid. 15. The Caillary rise in the glass tube is not to exceed 0. mm of water. Determine its minimum size, given that surface tension of water in contact with air = 0.075 N/m Solution: Caillary rise, h = 0. mm = 0. x 10 3 m Surface tension 0.075N/m Let, Diameter of tube = d Angel for water = 0 Density for water = 1000 kg / m h 4 g d 0. 10 4 0.075 3 1000 9.81 d d 4 0.075 1000 9.81 0. 10 3 Minimum of the tube = 14.8 cm. 0.148m 14.8cm 16. Find out the minimum size of glass tube that can be used to measure water levelif the caillary rise in the tube is to be restricted to mm. Consider surface tension of water in contact with air as 0.073575 N/m. Solution: Caillary rise h =.0 mm =.0 10 3 m PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 5
Let, diameter = d Density of water = 1000 kg / m 3 0. 073575 N / m Angle for water 0 \ PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 6
h 4.0 10 3 4 0.073575 g d 1000 9.81 d d = 0.015 m = 1.5 cm. Thus the minimum diameter of the tube should be 1.5 cm. 17. Define Real fluid and Ideal fluid. Real Fluid: A fluid, which ossesses viscosity, is known as real fluid. All fluids, in actual ractice, are real fluids. Ideal Fluid: A fluid, which is incomressible and is having no viscosity, is known as an ideal fluid. Ideal fluid is only an imaginary fluid as all the fluids, which exist, have some viscosity. 18. Write down the exression for caillary fall. PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 7
19. Two horizontal lates are laced 1.5 cm aart. The sace between them being filled with oil of viscosity 14 oises. Calculate the shear stress in oil if uer late is moved with a velocity of.5 m/s. Solution: Given: Distance between the lates, = 1.5 cm = 0.015m. Viscosity μ = 14 oise = 14 / 10 Ns / m Velocity of uer late, u =.5 m/sec. Shear stress is given by equation as τ = μ ( / ). Where = change of velocity between the lates = u 0 = u =.5 m/sec. = 0.015m. τ = (14 /10) X (.5 / 0.015) = 80 N/m. PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 8
PART B 1.Calculate the caillary effect in millimeters a glass tube of 4mm diameter, whenimmersed in (a) water (b) mercury. The temerature of the liquid is 0 0 C and the values of the surface tension of water and mercury at 0 0 C in contact with air are 0.073575 and 0.51 N/m resectively. The angle of contact for water is zero that for mercury 130 0. Take secific weight of water as 9790 N / m 3 (AUC May/June 009) (AUC May/June 011) Given: Diameter of tube d = 4 mm = 4 10 3 m Caillary effect (rise or deression) h 4 cos g d Surface tension in kg f/m Angle of contact and = density i. Caillary effect for water 0.073575 N / m, 0 0 998 kg / m 3 @ 0 0 c 4 0.73575 Cos0 0 h 998 9.81 4 10 3 = 7.51 mm. 7.51 10 3 m Caillary effect for mercury : 0.51 N / m, 130 0 s gr 1000 13.6 1000 13600kg /m 3 PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 9
.46 10 3 m = -.46 mm. -Ve indicates caillary deression.. A cylinder of 0.6 m 3 in volume contains air at 50 0 C and 0.3 N/ mm absoluteressure. The air is comressed to 0.3 m 3. Find (i) ressure inside the cylinder assuming isothermal rocess (ii) ressure and temerature assuming adiabatic rocess. Take K = 1.4 (AUC May/June 010) Given: Initial volume 1 0.36m 3 Pressure P 1 = 0.3 N/mm 0.3 10 6 N / m Temerature, t 1 = 50 0 C T 1 = 73 + 50 = 33 0 K Final volume, 0.3m 3 K = 1.4 i. Isothermal Process: P Cons tan t (or ) Cons tan t 1 A A A 1 1 30 10 4 0.6 0.3 0.6 10 6 N / m = 0.6 N / mm PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 10
ii) adiabatic rocess K Cons tan t or K cons tan t. K 1 1 K K 0.6 1.4 1 1 K 30 10 4 30 10 4 1.4 0.3 0.791 10 6 N / m 0.791N / mm For temerature, RT, k cons tan t RT and RT k cons tan t RT k 1 Cons tan t T k 1 Cons tan t R is also cons tan t T V k 1 T V k 1 1 1 V 1 k 1 T T 33 1 V 0.6 1.4 1.0 0.3 33 0.4 46. 0 K t 46. 73 153. 0 C PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 11
3. If the velocity rofile of a fluid over a late is a arabolic with the vertex 0 cmfrom the late, where the velocity is 10 cm/sec. Calculate the velocity gradients and shear stress at a distance of 0,10 and 0 cm from the late, if the viscosity of the fluid is 8.5 oise. (AUC May/June 010) Given, conditions. Distance of vertex from late = 0 cm. Velocity at vertex, u = 10 cm / sec. Viscosity, 8.5 oise 8.5 Ns 10 m 0.85 Parabolic velocity rofile equation, u ay by C -------------- (1) Where, a, b and c constants. Their values are determined from boundary i) At y = 0, u = 0 ii) At y = 0cm, u = 10 cm/se. iii) At y = 0 cm, 0 Substituting (i) in equation (1), C = 0 Substituting (ii) in equation (1), 10 a 0 b 400 a 0b -----------() Substituting (iii) in equation (1), ay b solving 1 and, we get, 400 a 0 b 0 ( ) 40 a + b = 0 800 a 0 b 0 b = - 40 a 10 400 a 0 b 40 a 400 a 800 a 400 a 0 a 0 b 40a b-----------(3) a 10 3 400 10 0.3 PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 1
b 40 0.3 1. Substituting a, b and c in equation (i) u 0.3y 1y 0.3 y 1 0.6 y 1 Velocity gradient at y = 0, Velocity gradient, at y =10 cm, Velocity gradient, y 0 y 10 0.6 0 1 1 / s. 0.6 10 1 6 1 6 / s. at y = 0 cm, Velocity gradient, 0.6 0 1 1 1 0 Shear Stresses: Shear stresses is given by, i. Shear stress at y = 0, ii. Shear stress at y = 10, iii.shear stress at y = 0, y 0 y 0 y 10 y 0 0.85 1.0 10.N / m 0.85 6.0 5.1N / m 0.85 0 0 PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 13
4. A 15 cm diameter vertical cylinder rotates concentrically inside another cylinderof diameter 15.10 cm. Both cylinders are 5 cm high. The sace between the cylinders is filled with a liquid whose viscosity is unknown. If a torque of 1.0 Nm is required to rotate the inner cylinder at 100 rm determine the viscosity of the fluid. (AUC Nov/Dec 011) Solution: Diameter of cylinder = 15 cm = 0.15 m Diameter of outer cylinder = 15.10 cm = 0.151 m Length of cylinder L = 5 cm = 0.5 m Torque T= 1 Nm ; N = 100 rm. Viscosity = Tangential velocity of cylinder u DN 0.15 100 60 60 0.7854 m / s Surface area of cylinder A D L 0.15 0.5 = 0.1178 m u 0 u 0.7854 m / s 0.151 0.150 0.0005 m 0.7854 0.0005 Shear force, F Shear Stress Area 0.7854 0.0005 0.1178 Torque T F D 1.0 0.7854 0.15 0.1178 0.0005 1.0 0.0005 0.7854 0.1178 0.15 0.864 Ns / m PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 14
0.864 10 8.64 oise. 5. The namic viscosity of oil, used for lubrication between a shaft and sleeve is 6 oise. The shaft is of diameter 0.4 m and rotates at 190 rm. Calculate the ower lost in the bearing for a sleeve length of 90 mm. The thickness of the oil film is 1.5mm. (AUC Nov/Dec 013) Given, 6 oise D = 0.4 m N = 190 rm. 6 Ns 0.6 Ns 10 m m L 90 mm 90 10 3 m t 1.5 mm 1.5 10 3 m Power NT W 60 T force D Nm. F Shear stress Area DL N / m u DN m / s. 60 Tangential Velocity of shaft, u DN 0.4 190 60 60 3.98 m / s. = change of velocity = u 0 = u = 3.98 m/s. t 1.5 10 3 m. 3.98 10 1.5 10 3 159N / m Shear force on the shaft F = Shear stress x Area PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 15
F 159 D L 159 0.4 90 10 3 180.05N Torque on the shaft, T Force D 0.4 180.05 36.01 Ns. Power lost NT 190 36.01 60 60 716.48W 6.If the velocity distribution over a late is given byu y y in which U is the 3 velocity in m/s at a distance y meter above the late, determine the shear stress at y = 0 and y = 0.15 m. Take namic viscosity of fluid as 8.63 oise. (AUC May/June 013) Given: u y y 3 3 y 0 y 0 y 0.15 3 3 3 0.17 0.667 0.30 8.63 oise 8.63 SI units = 0.863 Ns / m 10 PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 16
i. Shear stress at y = 0 is given by 0 y 0 0.863 0.667 0.5756 N / m ii. Shear stress at y = 0.15 m is given by y 0.15 y 0.15 0.863 0.367 0.3167 N / m 7.The diameters of a small iston and a large iston of a hydraulic jack at3cm and10 cm resectively. A force of 80 N is alied on the small iston Find the load lifted by the large iston when: a. The istons are at the same level b. Small iston in 40 cm above the large iston. The density of the liquid in the jack in given as 1000 kg/m 3 (AUC Nov/Dec 01) Given: Dia of small iston d = 3 cm. Area of small iston, a d 4 4 3 7.068cm Dia of large iston, D = 10 cm P Area of larger iston, A 4 10 78.54cm Force on small iston, F = 80 N Let the load lifted = W a. When the istons are at the same level Pressure intensity on small iston P F 80 N / cm a 7.068 PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 17
This is transmitted equally on the large iston. Pressure intensity on the large iston 80 7.068 Force on the large iston = Pressure x area = 80 7.068 x 78.54 N = 888.96 N. b. when the small iston is 40 cm above the large iston Pressure intensity on the small iston F 80 N / cm a 7.068 Pressure intensity of section A A F ressure intensity e of height of 40 cm of liquid. P = gh. a But ressure intensity e to 40cm. of liquid g h 1000 9.81 0.4N / m 1000 9.81 0.4 N/cm 0.394N/cm 10 4 Pressure intensity at section 80 A A 0.394 7.068 = 11.3 + 0.394 = 11.71 N/cm Pressure intensity transmitted to the large iston = 11.71 N/cm Force on the large iston = Pressure x Area of the large iston 11.71 A 11.71 78.54 = 919. 7 N PREPARED BY K.M.SHAMEER (CE6303-MECHANICS OF FLUID) Page 18