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Tutorial 1

Fluid properties
1. A reservoir of glycerin has a mass of 1100kg and a volume of 0.9 m 3. Calculate its weight, mass
density, specific weight and specific gravity.
Solution:
Mass of glycerin (m) = 1100kg
Volume (V) = 0.9 m3
Weight (W) = ?
Mass density (ρ) = ?
Specific weight (γ) = ?
Specific gravity (S) = ?
W = mg = 1100x9.81 = 10791N
ρ = m/V = 1100/0.9 = 1222.22 Kg/m3
γ = ρg = 1222.22x9.81 = 11990 N/m3
S = γglycerine/ γwater = 11990/9810 = 1.22
2. A liquid compressed in a cylinder has a volume of 2000 cm3 at 2MN/m2 and a volume of 1990 cm3 at
4MN/m2. What is its bulk modulus of elasticity?
Solution:
Initial volume (V) = 2000 cm3
Final volume (V1) = 1990 cm3
Change in volume ( ) = 1990-2000 = -10 cm3
Change in pressure( ) = 4-2 = 2MN/m2
Bulk modulus of elasticity (K) = ?
= 400 MN/m2

3. If the bulk modulus of elasticity of water is 2.2 Gpa (GN/m2), what pressure is required to reduce a
volume by 0.8%?
Solution:
Bulk modulus of elasticity (K) = 2.2 Gpa = 2.2x109 Pa
Reduction in volume (
) = -0.8% = -0.008
Pressure (P) = ?

Assume a specific weight at the surface of 10 KN/m2 and an average bulk modulus of elasticity of 2.000981 m3/kg Bulk modulus in terms of specific volume is = -0.5km (vs2) = ? Specific weight at 7.5km.5km (b) the specific volume at 7. The surface tension of mercury and water at 600 c are 0. What capillary height change will occur in these two fluids when they are in contact with air in a glass tube of radius 0.0000294 m3/kg (b) vs2 = vs1 + = 0. the pressure is 75Mpa.5 kg/m3 = 1051.4 kg/m3 Specific volume at the surface (vs1) = = 1/1019.0003m Surface tension for mercury ( ) = 0.5x109 N/m2 Change in specific volume ( ) = ? Specific volume at 7.0000294 = 0.5km ( ) = = 1/0.5x9.000981-0.47N/m and 0.0662N/m respectively.3/1000 = 0.5 Gpa for that pressure range.30mm? Use θ = 1300 for mercury and 00 for water. At a depth of 7. Solution: Pressure at 7.5km and (c) the specific weight at 7.30 mm = 0.0662N/m θ = 1300 for mercury and 00 for water Capillary height change for mercury (hm) = ? .5km in the ocean.5km (P2) = 75 Mpa = 75x106 N/m2 Specific weight at the surface ( ) = 10 KN/m2 = 10x1000 = 10000 N/m2 Bulk modulus of elasticity at the surface (K) = 2.000951 = 1051.5 Gpa = 2.47N/m Surface tension for water ( ) = 0.P = 17600 KPa 4. Solution: Radius of tube (r) = 0.5km ( ) = ? (a) Density at the surface ( ) = = 10000/9.4 = 0.81 = 10315 N/m3 5.000951 m3/kg (c) Density at 7. Find (a) the change in specific volume between the surface and 7.81 = 1019.

3 m? Solution: u = 0.9mm = 0.125-0 = 1.075m Absolute viscosity (µ) = 0.125/0.048 NS/m2 and relative density 0.9(0.85y .125m/s Change in distance (dy) = 75 – 0 = 75mm = 75/1000 m = 0.048/900 = 5.9 NS/m2 Shear stress (τ) at plate (for y = 0 ) = ? Shear stress (τ) for y = 0.048 NS/m2 relative density (S)= 0.9.125m/s.85-2y For y = 0 Shear stress (τ) = 0. Also calculate kinematic viscosity.9 velocity gradient (du/dy) = ? Shear stress ( ) = ? Kinematic viscosity (υ) = ? du/dy = 1.Capillary height change for water (h) = ? = -0.85y .075 =15 s-1 = 0.5mm 6.765 N/m2 .y2 Dynamic viscosity (µ) = 9 Poise =9/10 = 0.3m = ? du/dy = d(0.9x1000 = 900kg/m3 υ = µ/ρ = 0. Solution: Change in velocity (du) =1.0445m = 44. What is the velocity gradient and shear stress at the boundary assuming a linear velocity distribution.y2 (u is velocity in m/s and y is the distance from the plate in m).0149m = -14. The velocity distribution of a viscous liquid (dynamic viscosity = 9 Poise) flowing over a fixed plate is given by u = 0. The fluid has absolute viscosity 0.72N/m2 Density of fluid (ρ) = Sxρwater = 0. What are the shear stresses at the plate surface and at y=0.85-2x0) = 0. In a fluid the velocity measured at a distance of 75mm from the boundary is 1.3x10-5 m2/s 7.85y .y2)/dy = 0.

What is the force required to maintain this motion? 75mm 5.85-2x0.For y = 0. what is the viscosity of the oil? 125mm 130mm .7mm 125mm Solution: Speed of piston = 5.075m dv = 5.7)/2 mm = 0.1247m Length of piston (L) = 75mm = 0.7mm = 0.95 NS/m2 Diameter of piston (D) =124.15mm = 0.9(0. The clearance between piston and pipe is 0.7 m/s 124.7N 9.3) = 0.225 N/m2 8.5m/s. If the piston decelerates at 0.025mm. The film of oil separating the piston from the cylinder has a viscosity of 0.7m/s as shown in fig. A piston of weight 90N slides in a lubricated pipe.6m/s2 when the speed is 0.3m Shear stress (τ) = 0.00015m Frictional Force (F) = ? F = Shear stress(τ) at the piston surface x surface area of piston (A) = 1060.7m/s dy = dr = (125-124.7m/s Viscosity of oil (µ) = 0.95 Ns/m2. A piston is moving through a cylinder at a speed of 5.

13m Viscosity of oil (µ) = ? Frictional force (F) = Shear stress(τ) at the piston surface x surface area of piston (A) Summing the forces ∑ µ = 0.15KN and 250mm on an edge slides down an incline on a film of oil 6µm thick.25m Thickness (dy) = 6µm = 6x10-6m Viscosity of oil (µ) = 0.5m/s Diameter of piston (D) = 125mm = 0.006mm W 200 Solution: Weight of block (W) = 1.6m/s2 Change in velocity (du) = 0.125m Length of piston (L) = 130mm = 0. calculate the terminal speed of the block.007 NS/m2 Terminal velocity (u) = ? Frictional force (F) = Shear stress(τ) at block surface x surface area of the block (A) Component of W in the direction of F is WSin20 .15KN F 0.000025m Deceleration (a) = 0.Solution: Weight of piston (W) = 90N Clearance (dy)= 0.007 NS/m2. A square block weighing 1.025mm = 0.15KN =1150N Side of block (L) = 250mm = 0. 1. The viscosity of the oil is 0.094 NS/m2 10. Assuming a linear velocity profile in the oil.

Take viscosity of oil = 5 NS/m2. equilibrium occurs.9x1000 = 900kg/m3 Dynamic viscosity (µ) = υρ = 0.0375m Length of shaft (L) = 200mm = 0.9 Density of oil (ρ) = 0.1mm = 0.5 NS/m2 Clearance (dr) = dy = (70.2mm at 200rpm. assumed uniform. The clearance. The length of the shaft is 200mm.0001m Force exerted by the oil on the shaft (F) = ? F = Shear stress(τ) at the shaft x surface area of shaft (A) = 990N 12.2-70)/2 mm = 0.4 m/s 11. Shaft Sleeve 0.1mm = 0.At the terminal condition. Determine the resisting torque exerted by the oil and the power required to rotate the shaft. Find the force exerted by the oil on the shaft. F = Wsin20 72. is filled with oil of kinematic viscosity 0.4m/s through a bearing sleeve 70.25m Change in velocity (du) = 0.2-75)/2 mm = 0.2mm in diameter and 250mm long.075/2 = 0.4m/s Kinematic viscosity of oil (υ) = 0.005 m2/s Sp gr of oil (S) = 0.9u = 1150xSin20 u = 5.2m Speed of shaft (N) = 200rpm Dynamic viscosity of oil (µ) = 5 NS/m2 Clearance (dr) = dy = (75.075m Radius of shaft (r) = 0. A shaft 75mm in diameter is fixed axially and rotated inside a sleeve of diameter 75.0001m Torque exerted by the oil on the shaft (T) = ? Power required to rotate shaft (P) = ? . Solution: Diameter of shaft (D) = 75mm = 0.9.005x900 = 4.4 m/s Solution: Diameter of shaft (D) = 70mm = 0.005 m2/s and sp gr 0.07m Length of shaft (L) = 250mm = 0. A shaft 70mm in diameter is being pushed at a speed of 0.

6cm = 0.94rad/s = 0.9Nm N= 60rpm Viscosity of the liquid (µ) = ? Angular velocity ( ) = = 6. Determine the viscosity of the liquid that fills the space between the cylinders if a torque of 0.3m Torque (T) = 0.6x7.75m/s du = 0.28 = 0.Angular velocity ( ) = Tangential velocity (u) = du = 7.5cm from one of the planes and (ii) the plate is equidistant from both the planes.6N = 1849.6x0. Both cylinders are 0.265 NS/m2 14.28rad/s Tangential velocity of inner cylinder (u) = = 0.3m2 in area moves edgewise through oil between large fixed parallels 10cm apart.3m long. A flat plate 0.85m/s = 20.0375x20.0375= 69.94 = 0.785m/s Frictional force (F) = Shear stress(τ) at the shaft x surface area of shaft = 1849.12x6. .126m Radius of inner cylinder (r2) = 12cm = 0.12m dr =dy = 0.75m/s Frictional force (F) = Shear stress(τ) x surface area (A) µ = 0.6m/s and the oil has a kinematic viscosity of 0. A cylinder of 12cm radius rotates concentrically inside a fixed cylinder of 12. Solution: Radius of outer cylinder (r1) = 12.12 = 0.006m Length of cylinder (L) = 0.126-0. calculate the drag force when (i) the plate is 2. If the velocity of plate is 0.8.45 stokes and specific gravity 0.6cm radius.9 Nm is required to maintain an angular velocity of 60rpm.9 KW (or use P = Tω) 13.4Nm = 1849.85 = 145194W = 1451.

Solution: Area of plate (A) = 0. gr.075m Total force (F) = Force on side1 (F1) + Force on side2 (F2) = τ1 A + τ2 A = (τ1 + τ2) A ( ) ( ) = 0.5cm = 0.05m Total force (F) = Force on side1 (F1) + Force on side2 (F2) = τ1 A + τ2 A = (τ1 + τ2) A ( ) = 0. dy1 = dy2 = dy = 10/2 = 5cm = 0.5cm Density of fluid (ρ) = 0. If a 1400N force is applied. of fluid (S) = 0.26N 15.6m/s dv = 0. Solution: Force (F1) = 780N Velocity (u1) = 2m/s Force (F2) = 1400N Velocity (u2) = ? Also.8 2. When a force of 780N is applied to the sleeve parallel to the shaft.5 = 7.345N (I) If the plate is equidistant.6m/s Kinematic viscosity (υ) = 0. u2 = 3.5cm 7.8x1000 = 800kg/m3 Dynamic viscosity (µ) = υρ = 0.036 NS/m2 (I) dy1 = 2. Therefore.6m/s . = const. A Newtonian fluid fills the gap between a shaft and a concentric sleeve.45 stokes = 0.5cm = 0. the sleeve attains a speed of 2m/s.45x10-4 m2/s Sp.3m2 Velocity of plate (u) = 0.45x10-4x800 = 0.025m dy2 = 10-2. what speed will the sleeve attain? The temperature of the sleeve remains constant.

0725N/m Pressure outside droplet (Po) = 1.015m Specific weight of liquid ( ) = Surface tension ( ) = ? Pressure outside the bubble (Po) = = 0.032x105 = 117700 N/m2 17.0375N/m 2mm dia.5cm = 0.5Pa = 8338.032x105 N/m2).85x9810 = 8338.5cm into a liquid of sp.015 = 125.0725N/m.01x10-3 m Surface tension ( ) = 0. Solution: Radius of bubble (r) = 1mm = 0.001m Pressure inside bubble (Pi) = 200Pa Depth of liquid (h) = 1.02mm is atmospheric (1.032x105 N/m2 Pressure within droplet (Pd) = ? Pressure inside droplet in excess of outside pressure (P) = = 14500 N/m2 Pd = P+Po = 14500+1.4x0.16.93Pa = 0.07Pa Pressure within droplet in excess of outside pressure (P) = 200-125. Estimate surface tension of liquid if the pressure in the bubble is 200Pa. 0.5cm . The tip of glass tube with an internal diameter of 2mm is immersed to a depth of 1. The pressure outside the droplet of water of diameter 0. Solution: Diameter of droplet = 0.07 = 74. Calculate the pressure within the droplet of water. 1. Air is forced into the tube to form a spherical bubble just at the lower end of the tube.02x10-3 m Radius of droplet (r) = 0. gr. The surface tension of water in contact with air is 0.85.

045 Pa-s sp gr = 0.13 m/s . A fluid of absolute viscosity of 0.15m.153m dy1 = r2-r1 = 0.5236= 0.5m length are placed co-axially and the central tube is rotated at 5 rpm applying a torque of 6 Nm.15m Radius of cylinder2 (r2) = 0. F2 = Viscous force on tube 2 For small space between cylinders. dy2 = r3-r2 = 0. and (b) parabolic velocity distribution with vertex at point 70mm away from the surface.5m Applied torque to cylinder2 (T) = 6 Nm rpm of cylinder2 (N) = 6 Viscosity of oil ( ) =? 12 3 r1 r2 r3 Torque is transmitted through oil layer from tube 2 to tube 1 and tube 3.18. ra = (r1+r2)/2 = 0. Solution: Absolute viscosity ( ) =0.038 NS/m2 19. 0. Angular velocity of tube 2 ( ) = = 0. Three cylindrical tubes of 0. Calculate the shear stress at the solid boundary and at points 20mm above the plate considering (a) linear velocity distribution.152x0.13m/s.002m = 1.91 Velocity at 70mm from the plate = 1.5236rad/s Tangential velocity tube2 (u2) = = 0. r2 and r3 as 0.045 Pa-s and sp gr of 0.07958m/s du = 0. Solution: Radius of cylinder1 (r1) = 0. Take r 1. Determine the viscosity of oil which fills the space between tubes.154m.07958m/s F1 = Viscous force on tube 1.152m and 0. The velocity of fluid at 70mm height over the plate is 1. the velocity gradient may be assumed to be a straight line and average radius r can be taken.91 flows over a flat plate.002m.151m.152m Radius of cylinder3 (r3) = 0. rb = (r2+r3)/2 = 0.154m Length of cylinders (L) = 0.

Solution: . On one side of the plate is oil of viscosity and on the other side oil of viscosity . a thin plate of large extent is pulled at a velocity V.726 Pa (throughout) (b) Parabolic velocity distribution Equation: At y = 0.14a At y = 0.045x23.61 = 32.0656 = 1.02m. Shear stress ( = 0.13/0.7 = 16.13m/s At y = 0. u = 0 c=0 Y u = 1. and (b) the pull required to drag the plate is minimum.453 Pa = 0.0656 = 0.1y Y u = 1.13m/s y = 0. = 23. shear stress is also constant throughout.(a) Linear velocity distribution Equation: u = ay a = tanθ = 1. At y = 0.29 Hence. Calculate the position of the plate so that (a) the shear force in the two sides of the plate is equal.07m.1 = 0.02m.61 b = -0.07m X As the velocity gradient is constant.29 = 1. Through a narrow gap of height h.07m.07m X Substituting the values of b and c a =-230. At y = 0.13m/s y = 0. u = 1.045x32.1 u = 16.07+b = 0 b = -0.045x16. Shear stress at y = 0.038 Pa 20.14x-230. Shear stress at y = 0. 2ax0.

Assume the velocity gradient in the oil film to be linear. Find the viscosity of the oil if the torque required to rotate the disc at 60 rpm is 4x10-4 Nm. . A 120mm circular disc rotates on a table separated by an oil of film of 2mm thickness. dF/dy = 0 ( ) Solving for h/y and neglecting –ve root √ √ 21.h V y (a) shear stress on upper face = shear stress on lower face = (b)F = pull required to drag the plate per unit area For F to be minimum.

A disk of radius R rotates at an angular velocity inside an oil bath of viscosity as shown in figure.06m Thickness (h) = 2mm = 2x10-3m Torque (T) = 4x10-4 Nm N = 60 rpm Angular velocity ( ) = = 6. Assuming a linear velocity profile and neglecting shear on the outer disk edges. derive an expression for the viscous torque on the disk.Oil Fixed h R r dr Solution: Radius of disc (R) = 120/2 = 60mm = 0. Torque on the element (dT) = Shear force x r Total torque (T) = ∫ = 0. h Oil h R Radius of disc = R r dr .0062 Ns/m2 22.28rad/s Viscosity of oil ( ) = ? Consider an elementary ring or disc at radius r and having a width dr.

074N/m θ = 00 for water Capillary height change for water (h) is = 0. An oil of viscosity and thickness t fills the gap between the cone and the housing. R r Oil thickness = t ds 2θ Fixed Solution: dr θ . Distilled water stands in a glass tube of 10mm diameter at a height of 25mm.074 N/m and angle of contact = 00. Derive an expression for the torque required and the rate of heat dissipation in the bearing.003m = 3mm True static height = 25-3 = 22mm 24. Torque on the element (dT) = Shear force x r Total torque (T) = ∫ 23. Solution: Diameter of tube (d) = 10 mm = 0.Clearance = h on both sides Torque = T N = rpm Angular velocity = ( ) = Viscosity of oil = Consider an elementary ring or disc at radius r and having a width dr. A solid cone of maximum radius R and vertex angle is to rotate at an angular velocity . What is the true static height? Take surface tension of water = 0.01m Surface tension of water ( ) = 0.

Torque on the element (dT) = Shear force x r ( ) Total torque (T) = ∫ Rate of heat dissipation = power utilized in overcoming resistance = .Maximum radius of cone = R thickness of oil = t Angular velocity = ( ) = Viscosity of oil = Consider an elementary area dA at radius r of the cone.