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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) = ?

5km (b) the specific volume at 7.81 = 10315 N/m3 5.5x9. At a depth of 7. Solution: Pressure at 7.5km ( ) = ? (a) Density at the surface ( ) = = 10000/9.5km ( ) = = 1/0.P = 17600 KPa 4.4 kg/m3 Specific volume at the surface (vs1) = = 1/1019.5 Gpa = 2.0662N/m respectively.30 mm = 0.4 = 0.47N/m and 0.000981-0. Assume a specific weight at the surface of 10 KN/m2 and an average bulk modulus of elasticity of 2. Find (a) the change in specific volume between the surface and 7.5x109 N/m2 Change in specific volume ( ) = ? Specific volume at 7.5km and (c) the specific weight at 7.30mm? Use θ = 1300 for mercury and 00 for water.5km in the ocean. What capillary height change will occur in these two fluids when they are in contact with air in a glass tube of radius 0.5 kg/m3 = 1051.0000294 m3/kg (b) vs2 = vs1 + = 0.0662N/m θ = 1300 for mercury and 00 for water Capillary height change for mercury (hm) = ? .5km (vs2) = ? Specific weight at 7.0000294 = 0.000951 = 1051.5km. Solution: Radius of tube (r) = 0.3/1000 = 0.47N/m Surface tension for water ( ) = 0.0003m Surface tension for mercury ( ) = 0.81 = 1019. the pressure is 75Mpa. The surface tension of mercury and water at 600 c are 0.5 Gpa for that pressure range.000951 m3/kg (c) Density at 7.000981 m3/kg Bulk modulus in terms of specific volume is = -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.

85-2y For y = 0 Shear stress (τ) = 0.y2 Dynamic viscosity (µ) = 9 Poise =9/10 = 0. Solution: Change in velocity (du) =1.048 NS/m2 relative density (S)= 0.075 =15 s-1 = 0.3m = ? du/dy = d(0.9x1000 = 900kg/m3 υ = µ/ρ = 0.3 m? Solution: u = 0.y2 (u is velocity in m/s and y is the distance from the plate in m). Also calculate kinematic viscosity.125m/s Change in distance (dy) = 75 – 0 = 75mm = 75/1000 m = 0.85y .Capillary height change for water (h) = ? = -0.9 NS/m2 Shear stress (τ) at plate (for y = 0 ) = ? Shear stress (τ) for y = 0. The velocity distribution of a viscous liquid (dynamic viscosity = 9 Poise) flowing over a fixed plate is given by u = 0.125-0 = 1.85y .85y .0149m = -14.125m/s.9 velocity gradient (du/dy) = ? Shear stress ( ) = ? Kinematic viscosity (υ) = ? du/dy = 1.0445m = 44.9(0.048 NS/m2 and relative density 0.y2)/dy = 0. The fluid has absolute viscosity 0. In a fluid the velocity measured at a distance of 75mm from the boundary is 1.075m Absolute viscosity (µ) = 0.5mm 6.048/900 = 5.9mm = 0. 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=0.85-2x0) = 0.3x10-5 m2/s 7.9.125/0.72N/m2 Density of fluid (ρ) = Sxρwater = 0.765 N/m2 .

A piston of weight 90N slides in a lubricated pipe.00015m Frictional Force (F) = ? F = Shear stress(τ) at the piston surface x surface area of piston (A) = 1060.85-2x0.225 N/m2 8.1247m Length of piston (L) = 75mm = 0. If the piston decelerates at 0.3) = 0.9(0.For y = 0.7m/s dy = dr = (125-124.95 Ns/m2. The clearance between piston and pipe is 0. A piston is moving through a cylinder at a speed of 5.025mm.7N 9.7m/s Viscosity of oil (µ) = 0.15mm = 0. The film of oil separating the piston from the cylinder has a viscosity of 0. What is the force required to maintain this motion? 75mm 5.7m/s as shown in fig.95 NS/m2 Diameter of piston (D) =124.6m/s2 when the speed is 0.7 m/s 124. what is the viscosity of the oil? 125mm 130mm .5m/s.7)/2 mm = 0.7mm 125mm Solution: Speed of piston = 5.075m dv = 5.3m Shear stress (τ) = 0.7mm = 0.

007 NS/m2. Assuming a linear velocity profile in the oil.25m Thickness (dy) = 6µm = 6x10-6m Viscosity of oil (µ) = 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. calculate the terminal speed of the block.15KN F 0.006mm W 200 Solution: Weight of block (W) = 1.094 NS/m2 10.000025m 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 = 0.025mm = 0.15KN =1150N Side of block (L) = 250mm = 0. The viscosity of the oil is 0.125m Length of piston (L) = 130mm = 0.15KN and 250mm on an edge slides down an incline on a film of oil 6µm thick. A square block weighing 1.5m/s Diameter of piston (D) = 125mm = 0. 1.Solution: Weight of piston (W) = 90N Clearance (dy)= 0.

0375m Length of shaft (L) = 200mm = 0.9u = 1150xSin20 u = 5.9x1000 = 900kg/m3 Dynamic viscosity (µ) = υρ = 0. Solution: Diameter of shaft (D) = 75mm = 0.4m/s Kinematic viscosity of oil (υ) = 0.0001m Torque exerted by the oil on the shaft (T) = ? Power required to rotate shaft (P) = ? .4 m/s 11. Determine the resisting torque exerted by the oil and the power required to rotate the shaft.4m/s through a bearing sleeve 70. A shaft 70mm in diameter is being pushed at a speed of 0. The length of the shaft is 200mm. Take viscosity of oil = 5 NS/m2.005 m2/s and sp gr 0. Shaft Sleeve 0. The clearance. is filled with oil of kinematic viscosity 0.2-70)/2 mm = 0. F = Wsin20 72.2mm in diameter and 250mm long.0001m Force exerted by the oil on the shaft (F) = ? F = Shear stress(τ) at the shaft x surface area of shaft (A) = 990N 12.2m Speed of shaft (N) = 200rpm Dynamic viscosity of oil (µ) = 5 NS/m2 Clearance (dr) = dy = (75.075m Radius of shaft (r) = 0.2-75)/2 mm = 0. equilibrium occurs. A shaft 75mm in diameter is fixed axially and rotated inside a sleeve of diameter 75.075/2 = 0.005x900 = 4.07m Length of shaft (L) = 250mm = 0.At the terminal condition. Find the force exerted by the oil on the shaft.25m Change in velocity (du) = 0.005 m2/s Sp gr of oil (S) = 0.9 Density of oil (ρ) = 0.1mm = 0. assumed uniform.2mm at 200rpm.5 NS/m2 Clearance (dr) = dy = (70.4 m/s Solution: Diameter of shaft (D) = 70mm = 0.1mm = 0.9.

If the velocity of plate is 0. A cylinder of 12cm radius rotates concentrically inside a fixed cylinder of 12.4Nm = 1849. Determine the viscosity of the liquid that fills the space between the cylinders if a torque of 0. .75m/s du = 0. Both cylinders are 0.45 stokes and specific gravity 0. Solution: Radius of outer cylinder (r1) = 12.12x6.126m Radius of inner cylinder (r2) = 12cm = 0.006m Length of cylinder (L) = 0.265 NS/m2 14.0375x20.6N = 1849.85m/s = 20.6cm = 0.94 = 0.6cm radius.9Nm N= 60rpm Viscosity of the liquid (µ) = ? Angular velocity ( ) = = 6.3m long.6m/s and the oil has a kinematic viscosity of 0.126-0.94rad/s = 0.3m Torque (T) = 0.12 = 0.5cm from one of the planes and (ii) the plate is equidistant from both the planes.Angular velocity ( ) = Tangential velocity (u) = du = 7.9 Nm is required to maintain an angular velocity of 60rpm.85 = 145194W = 1451.75m/s Frictional force (F) = Shear stress(τ) x surface area (A) µ = 0.8. calculate the drag force when (i) the plate is 2.28rad/s Tangential velocity of inner cylinder (u) = = 0.3m2 in area moves edgewise through oil between large fixed parallels 10cm apart.9 KW (or use P = Tω) 13.6x0.785m/s Frictional force (F) = Shear stress(τ) at the shaft x surface area of shaft = 1849. A flat plate 0.12m dr =dy = 0.28 = 0.0375= 69.6x7.

dy1 = dy2 = dy = 10/2 = 5cm = 0.45x10-4 m2/s Sp. If a 1400N force is applied.6m/s .5cm = 0.05m Total force (F) = Force on side1 (F1) + Force on side2 (F2) = τ1 A + τ2 A = (τ1 + τ2) A ( ) = 0.5 = 7.025m dy2 = 10-2.3m2 Velocity of plate (u) = 0.8 2.6m/s Kinematic viscosity (υ) = 0. = const. gr.036 NS/m2 (I) dy1 = 2.8x1000 = 800kg/m3 Dynamic viscosity (µ) = υρ = 0. the sleeve attains a speed of 2m/s. A Newtonian fluid fills the gap between a shaft and a concentric sleeve.45 stokes = 0. of fluid (S) = 0. When a force of 780N is applied to the sleeve parallel to the shaft.5cm = 0. what speed will the sleeve attain? The temperature of the sleeve remains constant.26N 15.5cm Density of fluid (ρ) = 0. u2 = 3.345N (I) If the plate is equidistant. Solution: Force (F1) = 780N Velocity (u1) = 2m/s Force (F2) = 1400N Velocity (u2) = ? Also.Solution: Area of plate (A) = 0.075m Total force (F) = Force on side1 (F1) + Force on side2 (F2) = τ1 A + τ2 A = (τ1 + τ2) A ( ) ( ) = 0.45x10-4x800 = 0.6m/s dv = 0.5cm 7. Therefore.

015m Specific weight of liquid ( ) = Surface tension ( ) = ? Pressure outside the bubble (Po) = = 0.07Pa Pressure within droplet in excess of outside pressure (P) = 200-125.5cm = 0.032x105 N/m2).5cm .16.0725N/m.0725N/m Pressure outside droplet (Po) = 1. The pressure outside the droplet of water of diameter 0.032x105 N/m2 Pressure within droplet (Pd) = ? Pressure inside droplet in excess of outside pressure (P) = = 14500 N/m2 Pd = P+Po = 14500+1.02x10-3 m Radius of droplet (r) = 0. gr. Air is forced into the tube to form a spherical bubble just at the lower end of the tube. 0.01x10-3 m Surface tension ( ) = 0.001m Pressure inside bubble (Pi) = 200Pa Depth of liquid (h) = 1.5cm into a liquid of sp. 1.85.5Pa = 8338. Solution: Diameter of droplet = 0.032x105 = 117700 N/m2 17.0375N/m 2mm dia.93Pa = 0. Estimate surface tension of liquid if the pressure in the bubble is 200Pa. The surface tension of water in contact with air is 0. Calculate the pressure within the droplet of water.015 = 125.02mm is atmospheric (1. Solution: Radius of bubble (r) = 1mm = 0. The tip of glass tube with an internal diameter of 2mm is immersed to a depth of 1.07 = 74.4x0.85x9810 = 8338.

18. r2 and r3 as 0.151m. Determine the viscosity of oil which fills the space between tubes.154m.07958m/s du = 0. F2 = Viscous force on tube 2 For small space between cylinders. The velocity of fluid at 70mm height over the plate is 1.13 m/s .5m length are placed co-axially and the central tube is rotated at 5 rpm applying a torque of 6 Nm. rb = (r2+r3)/2 = 0.91 flows over a flat plate.15m Radius of cylinder2 (r2) = 0.002m.045 Pa-s sp gr = 0.15m. 0. Angular velocity of tube 2 ( ) = = 0. A fluid of absolute viscosity of 0.152m Radius of cylinder3 (r3) = 0.5236rad/s Tangential velocity tube2 (u2) = = 0. Calculate the shear stress at the solid boundary and at points 20mm above the plate considering (a) linear velocity distribution. the velocity gradient may be assumed to be a straight line and average radius r can be taken. Solution: Absolute viscosity ( ) =0.152x0.154m Length of cylinders (L) = 0. ra = (r1+r2)/2 = 0.002m = 1. 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 3.5236= 0.153m dy1 = r2-r1 = 0.91 Velocity at 70mm from the plate = 1.038 NS/m2 19.045 Pa-s and sp gr of 0. Three cylindrical tubes of 0. dy2 = r3-r2 = 0. Take r 1.13m/s.152m and 0.07958m/s F1 = Viscous force on tube 1. Solution: Radius of cylinder1 (r1) = 0.

07m X As the velocity gradient is constant. a thin plate of large extent is pulled at a velocity V.0656 = 1.038 Pa 20.1 u = 16.1 = 0. u = 0 c=0 Y u = 1.045x23.61 b = -0. and (b) the pull required to drag the plate is minimum.29 = 1.13m/s y = 0.07m. shear stress is also constant throughout.07m X Substituting the values of b and c a =-230.07+b = 0 b = -0.13m/s y = 0. Calculate the position of the plate so that (a) the shear force in the two sides of the plate is equal.0656 = 0. Solution: . At y = 0.14a At y = 0. Shear stress at y = 0.7 = 16. At y = 0.726 Pa (throughout) (b) Parabolic velocity distribution Equation: At y = 0.13m/s At y = 0.02m. = 23. On one side of the plate is oil of viscosity and on the other side oil of viscosity . Through a narrow gap of height h.02m.045x16.453 Pa = 0.045x32. Shear stress ( = 0. Shear stress at y = 0. 2ax0.(a) Linear velocity distribution Equation: u = ay a = tanθ = 1. u = 1.13/0.14x-230.61 = 32.07m.29 Hence.1y Y u = 1.

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. dF/dy = 0 ( ) Solving for h/y and neglecting –ve root √ √ 21. . 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.

Assuming a linear velocity profile and neglecting shear on the outer disk edges. h Oil h R Radius of disc = R r dr . Torque on the element (dT) = Shear force x r Total torque (T) = ∫ = 0.06m Thickness (h) = 2mm = 2x10-3m Torque (T) = 4x10-4 Nm N = 60 rpm Angular velocity ( ) = = 6.0062 Ns/m2 22. A disk of radius R rotates at an angular velocity inside an oil bath of viscosity as shown in figure.28rad/s Viscosity of oil ( ) = ? Consider an elementary ring or disc at radius r and having a width dr. 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.

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. Solution: Diameter of tube (d) = 10 mm = 0. Distilled water stands in a glass tube of 10mm diameter at a height of 25mm.003m = 3mm True static height = 25-3 = 22mm 24. An oil of viscosity and thickness t fills the gap between the cone and the housing.01m Surface tension of water ( ) = 0. Derive an expression for the torque required and the rate of heat dissipation in the bearing.074N/m θ = 00 for water Capillary height change for water (h) is = 0. What is the true static height? Take surface tension of water = 0. R r Oil thickness = t ds 2θ Fixed Solution: dr θ . Torque on the element (dT) = Shear force x r Total torque (T) = ∫ 23.074 N/m and angle of contact = 00. A solid cone of maximum radius R and vertex angle is to rotate at an angular velocity .

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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