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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) = ?

000981-0.5km (b) the specific volume at 7.4 = 0.0000294 = 0.P = 17600 KPa 4.81 = 1019.000951 = 1051.5km and (c) the specific weight at 7.0000294 m3/kg (b) vs2 = vs1 + = 0.3/1000 = 0.0662N/m θ = 1300 for mercury and 00 for water Capillary height change for mercury (hm) = ? .5x9.5 Gpa = 2.5 Gpa for that pressure range.30mm? Use θ = 1300 for mercury and 00 for water.0662N/m respectively.000981 m3/kg Bulk modulus in terms of specific volume is = -0.5km in the ocean.47N/m Surface tension for water ( ) = 0.5 kg/m3 = 1051. Assume a specific weight at the surface of 10 KN/m2 and an average bulk modulus of elasticity of 2. The surface tension of mercury and water at 600 c are 0.0003m Surface tension for mercury ( ) = 0.000951 m3/kg (c) Density at 7. the pressure is 75Mpa.81 = 10315 N/m3 5.5km. Find (a) the change in specific volume between the surface and 7.47N/m and 0.5km (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) = 2.4 kg/m3 Specific volume at the surface (vs1) = = 1/1019.5x109 N/m2 Change in specific volume ( ) = ? Specific volume at 7. At a depth of 7.30 mm = 0.5km ( ) = = 1/0. Solution: Pressure at 7.5km ( ) = ? (a) Density at the surface ( ) = = 10000/9.5km (vs2) = ? Specific weight at 7. Solution: Radius of tube (r) = 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.

0445m = 44. The velocity distribution of a viscous liquid (dynamic viscosity = 9 Poise) flowing over a fixed plate is given by u = 0.075m Absolute viscosity (µ) = 0.125m/s.9 velocity gradient (du/dy) = ? Shear stress ( ) = ? Kinematic viscosity (υ) = ? du/dy = 1.048 NS/m2 relative density (S)= 0.125m/s Change in distance (dy) = 75 – 0 = 75mm = 75/1000 m = 0.125/0.9.y2)/dy = 0.y2 Dynamic viscosity (µ) = 9 Poise =9/10 = 0.3x10-5 m2/s 7.9mm = 0. In a fluid the velocity measured at a distance of 75mm from the boundary is 1.125-0 = 1. What is the velocity gradient and shear stress at the boundary assuming a linear velocity distribution. What are the shear stresses at the plate surface and at y=0.5mm 6.72N/m2 Density of fluid (ρ) = Sxρwater = 0.048/900 = 5.075 =15 s-1 = 0.85-2y For y = 0 Shear stress (τ) = 0.9 NS/m2 Shear stress (τ) at plate (for y = 0 ) = ? Shear stress (τ) for y = 0.85y . Solution: Change in velocity (du) =1.9(0.y2 (u is velocity in m/s and y is the distance from the plate in m). The fluid has absolute viscosity 0.765 N/m2 .3 m? Solution: u = 0.85y .0149m = -14.3m = ? du/dy = d(0.85y .85-2x0) = 0. Also calculate kinematic viscosity.Capillary height change for water (h) = ? = -0.9x1000 = 900kg/m3 υ = µ/ρ = 0.048 NS/m2 and relative density 0.

85-2x0.15mm = 0. what is the viscosity of the oil? 125mm 130mm .7N 9.6m/s2 when the speed is 0. A piston is moving through a cylinder at a speed of 5.025mm.5m/s.00015m Frictional Force (F) = ? F = Shear stress(τ) at the piston surface x surface area of piston (A) = 1060.3m Shear stress (τ) = 0. The film of oil separating the piston from the cylinder has a viscosity of 0.7)/2 mm = 0.7mm 125mm Solution: Speed of piston = 5.225 N/m2 8.3) = 0. A piston of weight 90N slides in a lubricated pipe.For y = 0.9(0.7 m/s 124. The clearance between piston and pipe is 0. What is the force required to maintain this motion? 75mm 5.7mm = 0. If the piston decelerates at 0.95 NS/m2 Diameter of piston (D) =124.7m/s as shown in fig.7m/s dy = dr = (125-124.075m dv = 5.1247m Length of piston (L) = 75mm = 0.7m/s Viscosity of oil (µ) = 0.95 Ns/m2.

125m Length of piston (L) = 130mm = 0.007 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 .6m/s2 Change in velocity (du) = 0.094 NS/m2 10. The viscosity of the oil is 0.Solution: Weight of piston (W) = 90N Clearance (dy)= 0.000025m Deceleration (a) = 0.13m 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. calculate the terminal speed of the block.007 NS/m2. Assuming a linear velocity profile in the oil. A square block weighing 1.25m Thickness (dy) = 6µm = 6x10-6m Viscosity of oil (µ) = 0. 1.15KN =1150N Side of block (L) = 250mm = 0.025mm = 0.15KN F 0.5m/s Diameter of piston (D) = 125mm = 0.006mm W 200 Solution: Weight of block (W) = 1.15KN and 250mm on an edge slides down an incline on a film of oil 6µm thick.

assumed uniform.005 m2/s and sp gr 0.0001m Force exerted by the oil on the shaft (F) = ? F = Shear stress(τ) at the shaft x surface area of shaft (A) = 990N 12.9. The clearance. A shaft 75mm in diameter is fixed axially and rotated inside a sleeve of diameter 75. Shaft Sleeve 0.005x900 = 4.07m Length of shaft (L) = 250mm = 0.4 m/s 11.9x1000 = 900kg/m3 Dynamic viscosity (µ) = υρ = 0. A shaft 70mm in diameter is being pushed at a speed of 0. Take viscosity of oil = 5 NS/m2.075/2 = 0.2mm at 200rpm.25m Change in velocity (du) = 0.005 m2/s Sp gr of oil (S) = 0.2m Speed of shaft (N) = 200rpm Dynamic viscosity of oil (µ) = 5 NS/m2 Clearance (dr) = dy = (75.4 m/s Solution: Diameter of shaft (D) = 70mm = 0.At the terminal condition.9u = 1150xSin20 u = 5.2mm in diameter and 250mm long. Solution: Diameter of shaft (D) = 75mm = 0. is filled with oil of kinematic viscosity 0.9 Density of oil (ρ) = 0.1mm = 0. Determine the resisting torque exerted by the oil and the power required to rotate the shaft.5 NS/m2 Clearance (dr) = dy = (70.1mm = 0.075m Radius of shaft (r) = 0. F = Wsin20 72.4m/s through a bearing sleeve 70.2-70)/2 mm = 0.2-75)/2 mm = 0.0001m Torque exerted by the oil on the shaft (T) = ? Power required to rotate shaft (P) = ? . equilibrium occurs.4m/s Kinematic viscosity of oil (υ) = 0.0375m Length of shaft (L) = 200mm = 0. The length of the shaft is 200mm. Find the force exerted by the oil on the shaft.

006m Length of cylinder (L) = 0.265 NS/m2 14.3m Torque (T) = 0. If the velocity of plate is 0.3m long.94rad/s = 0. calculate the drag force when (i) the plate is 2. Solution: Radius of outer cylinder (r1) = 12.6cm = 0.6x7.28rad/s Tangential velocity of inner cylinder (u) = = 0.0375= 69.75m/s Frictional force (F) = Shear stress(τ) x surface area (A) µ = 0.28 = 0. A cylinder of 12cm radius rotates concentrically inside a fixed cylinder of 12.12 = 0.85m/s = 20.5cm from one of the planes and (ii) the plate is equidistant from both the planes.94 = 0.45 stokes and specific gravity 0.8.12x6.6cm radius.85 = 145194W = 1451.3m2 in area moves edgewise through oil between large fixed parallels 10cm apart. Determine the viscosity of the liquid that fills the space between the cylinders if a torque of 0.9Nm N= 60rpm Viscosity of the liquid (µ) = ? Angular velocity ( ) = = 6.6m/s and the oil has a kinematic viscosity of 0. Both cylinders are 0.785m/s Frictional force (F) = Shear stress(τ) at the shaft x surface area of shaft = 1849.9 Nm is required to maintain an angular velocity of 60rpm.126m Radius of inner cylinder (r2) = 12cm = 0.12m dr =dy = 0.6N = 1849.4Nm = 1849.0375x20.6x0.Angular velocity ( ) = Tangential velocity (u) = du = 7.126-0.75m/s du = 0. .9 KW (or use P = Tω) 13. A flat plate 0.

5cm 7.036 NS/m2 (I) dy1 = 2.8 2. = const.5cm = 0.025m dy2 = 10-2.45x10-4x800 = 0.45 stokes = 0.6m/s Kinematic viscosity (υ) = 0.5 = 7.3m2 Velocity of plate (u) = 0. u2 = 3. the sleeve attains a speed of 2m/s.6m/s . Therefore.Solution: Area of plate (A) = 0. of fluid (S) = 0.345N (I) If the plate is equidistant. If a 1400N force is applied. When a force of 780N is applied to the sleeve parallel to the shaft.26N 15.5cm = 0.45x10-4 m2/s Sp.6m/s dv = 0.05m Total force (F) = Force on side1 (F1) + Force on side2 (F2) = τ1 A + τ2 A = (τ1 + τ2) A ( ) = 0. gr. A Newtonian fluid fills the gap between a shaft and a concentric sleeve.5cm Density of fluid (ρ) = 0. what speed will the sleeve attain? The temperature of the sleeve remains constant. Solution: Force (F1) = 780N Velocity (u1) = 2m/s Force (F2) = 1400N Velocity (u2) = ? Also. dy1 = dy2 = dy = 10/2 = 5cm = 0.8x1000 = 800kg/m3 Dynamic viscosity (µ) = υρ = 0.075m Total force (F) = Force on side1 (F1) + Force on side2 (F2) = τ1 A + τ2 A = (τ1 + τ2) A ( ) ( ) = 0.

0375N/m 2mm dia.85x9810 = 8338.5cm into a liquid of sp. The surface tension of water in contact with air is 0.01x10-3 m Surface tension ( ) = 0. Calculate the pressure within the droplet of water.0725N/m Pressure outside droplet (Po) = 1.16.07 = 74.85.032x105 = 117700 N/m2 17. 1. 0.015m Specific weight of liquid ( ) = Surface tension ( ) = ? Pressure outside the bubble (Po) = = 0.015 = 125.0725N/m. Solution: Radius of bubble (r) = 1mm = 0.032x105 N/m2). Solution: Diameter of droplet = 0. The tip of glass tube with an internal diameter of 2mm is immersed to a depth of 1. Estimate surface tension of liquid if the pressure in the bubble is 200Pa.032x105 N/m2 Pressure within droplet (Pd) = ? Pressure inside droplet in excess of outside pressure (P) = = 14500 N/m2 Pd = P+Po = 14500+1. The pressure outside the droplet of water of diameter 0. gr.02mm is atmospheric (1.02x10-3 m Radius of droplet (r) = 0.93Pa = 0.5cm = 0.4x0.001m Pressure inside bubble (Pi) = 200Pa Depth of liquid (h) = 1.5Pa = 8338. Air is forced into the tube to form a spherical bubble just at the lower end of the tube.5cm .07Pa Pressure within droplet in excess of outside pressure (P) = 200-125.

the velocity gradient may be assumed to be a straight line and average radius r can be taken. Calculate the shear stress at the solid boundary and at points 20mm above the plate considering (a) linear velocity distribution. Three cylindrical tubes of 0. The velocity of fluid at 70mm height over the plate is 1.5236rad/s Tangential velocity tube2 (u2) = = 0.152x0.91 Velocity at 70mm from the plate = 1.91 flows over a flat plate.002m = 1.152m and 0.07958m/s du = 0.154m Length of cylinders (L) = 0.045 Pa-s and sp gr of 0. rb = (r2+r3)/2 = 0.15m. dy2 = r3-r2 = 0. Solution: Absolute viscosity ( ) =0.154m.045 Pa-s sp gr = 0. 0.152m Radius of cylinder3 (r3) = 0.5m 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 3.13 m/s . F2 = Viscous force on tube 2 For small space between cylinders.002m. Angular velocity of tube 2 ( ) = = 0.07958m/s F1 = Viscous force on tube 1.5236= 0. r2 and r3 as 0. Determine the viscosity of oil which fills the space between tubes. A fluid of absolute viscosity of 0. and (b) parabolic velocity distribution with vertex at point 70mm away from the surface.13m/s.038 NS/m2 19.15m Radius of cylinder2 (r2) = 0. ra = (r1+r2)/2 = 0.18. Take r 1.151m.153m dy1 = r2-r1 = 0.5m length are placed co-axially and the central tube is rotated at 5 rpm applying a torque of 6 Nm. Solution: Radius of cylinder1 (r1) = 0.

1y Y u = 1.045x23.29 = 1.045x16. and (b) the pull required to drag the plate is minimum.(a) Linear velocity distribution Equation: u = ay a = tanθ = 1.13/0.726 Pa (throughout) (b) Parabolic velocity distribution Equation: At y = 0.61 b = -0.07m X Substituting the values of b and c a =-230. = 23.0656 = 0.07+b = 0 b = -0. Calculate the position of the plate so that (a) the shear force in the two sides of the plate is equal. Through a narrow gap of height h. a thin plate of large extent is pulled at a velocity V. u = 0 c=0 Y u = 1.02m.29 Hence.13m/s y = 0. 2ax0.038 Pa 20.1 = 0.13m/s At y = 0.14x-230. On one side of the plate is oil of viscosity and on the other side oil of viscosity . Shear stress at y = 0.07m.1 u = 16.7 = 16. Shear stress at y = 0. At y = 0.0656 = 1. u = 1.07m.13m/s y = 0.07m X As the velocity gradient is constant.453 Pa = 0. At y = 0. shear stress is also constant throughout.14a At y = 0. Solution: . Shear stress ( = 0.61 = 32.045x32.02m.

. 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. 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.

Oil Fixed h R r dr Solution: Radius of disc (R) = 120/2 = 60mm = 0.28rad/s Viscosity of oil ( ) = ? Consider an elementary ring or disc at radius r and having a width dr. Torque on the element (dT) = Shear force x r Total torque (T) = ∫ = 0. derive an expression for the viscous torque on the disk. A disk of radius R rotates at an angular velocity inside an oil bath of viscosity as shown in figure.0062 Ns/m2 22. Assuming a linear velocity profile and neglecting shear on the outer disk edges.06m Thickness (h) = 2mm = 2x10-3m Torque (T) = 4x10-4 Nm N = 60 rpm Angular velocity ( ) = = 6. h Oil h R Radius of disc = R r dr .

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

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. Torque on the element (dT) = Shear force x r ( ) Total torque (T) = ∫ Rate of heat dissipation = power utilized in overcoming resistance = .

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