MM303 FLUID MECHANICS I PROBLEM SET 1 (CHAPTER 2) FALL v=by 2 =-6 (1/2) 2 = -3/2 m/s

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MM303 FLUID MECHANICS I PROBLEM SET 1 (CHAPTER ) FALL 018 1) For the velocity fields given below, determine: i) Whether the flow field is one-, two-, or three-dimensional, and why.

1 MM303 FLUID MECHANICS I PROBLEM SET 1 (CHAPTER ) FALL 018 1) For the velocity fields given below, determine: i) Whether the flow field is one-, two-, or three-dimensional, and why. ii) Whether the flow is steady or unsteady, and why bt a) V ax e i V ae i bx j bx b) c) V ax t i by j d) V axy i byztj NOTE: The quantities a and b are constants. ) The velocity field V=axyi+by j, where a= 1/m s, b=-6 1/m s, and the coordinates are measured in meters. Calculate the velocity at the point (, ½). Find an expressionfor for the streamline passing through this point. Plot several streamlines in the first quadrant, including the one passes through the point (x,y)=(,1/) Solution: At point (, ½), the velocity components are u=axy= 1/= m/s v=by =-6 (1/) = -3/ m/s Steamline is tangent to the velocity vector. Hence, for streamline, dy v by by dx u axy ax Separating the variables, dy y bdx ax Integrating, we get, b ln( y) ln( x) c y Cx y Cx a ba / 3 The streamline passing through point (,1/) is obtained as C C C 4 y 3 x

2 3) Velocity field of a flow is given as V=Axi+Ayj, where A= s -1. Verify that the parametric equations for particle motion are x p=c 1e At and y p=c e At. Obtain the equation for the pathline of the particle located at the point (x,y)=(,) at time t=0. Compare the pathline with the streamline through the same point. Solution: dx dx dx 1 1 For pathlines up Ax Adt ln( x) At c1 x e x e e C1e dt dt x dy dy dy At c c At At vp Ay Adt ln( y) At c y e y e e Ce dt dt y At c c At At At Hence the parametric equations of the motion are x C1e and y C e At Equation of the pathlines can be obtained by combining the parametric equations as follows: x C1e e subtituting this into the following equation, we get, C At At x x y C e y C y Cx 1 At C1 y 1 At point (,), C. x The path line of the particle at point (,) at time t=0 is y 1 x For streamline, dy v Ay y dx u Ax x Separating the variable and integrating, we get, dy dx y int egrating ln( y) ln( x) c ln c y Cx y x x y 1 At point (,), C. x The steamline passing through point(,) is y 1 x The streamline passing through point (,) and the pathine that started at point (,) coincide because the flow is steady.

3 4) Velocity field of a flow is given by V=ai+bxj, where a= 3 m/s and b= s -1. Coordinates are measured in meters. a) Obtain the equation for the streamline passing through point (, 4). b) At t=5 s, what are the coordinates of the particle that passed through point (0, 4) at time t=0 s? c) What conclusion can you draw about the pathline, streamline and streakline for this flow?

4 5) Crude oil, with specific gravity SG=0.85 and viscosity =0.1 Ns/m, flows steadily down a surface inclined =45 0 in a film of thickness h=.5 mm. The velocity profile is given by the expression below. (Coordinates x is along the surface and y is normal to the surface.) Plot the velocity profile. Determine the magnitude and direction of the shear stress that acts on the surface. u = ρg y (hy ) sinθ μ

5 6) A block weighing 45 N and having dimensions 50 mm on each edge is pulled up on an inclined surface on which there is a film of SAE 10W oil at 37 0 C. If the speed of the block is 0.6 m/s and the oil film is 0.05 mm thick, find the force required to pull the block. Assume that velocity profile in the oil film is linear. The surface is inclined at an angle of 5 0 from the horizontal.

6 7) A concentric cylinder viscometer is driven by a falling mass M=0.0 kg connected by a cord and pulley to the inner cylinder, as shown. The liquid to be tested fills the annular gap of width a=0.4mm and height H=160 mm. After a brief starting transient, the mass falls at constant speed V m=60 mm/s. Develop an algebraic expression for the viscosity of the liquid in the device in terms of M, g, V m, r, R, a and H. Evaluate the viscosity of the liquid. Note: r=50 mm, R=100 mm.

7 8. At a depth of 7.5 km in the ocean, the pressure is 75 Mpa. Assume a specific weight of 10 KN/m 3 at the surface and an average bulk modulus of elasticity of.5 GPa for that pressure range. Find (a) the change in specific volume between the surface and 7.5 km depth, (b) the specific volume at 7.5 km (c) the specific weight at 7.5 km. Solution: Pressure at 7.5km (P ) = 75 Mpa = 75x106 N/m Specific weight at the surface (γ)= 10 KN/m 3 = 10x1000 = N/m 3 Bulk modulus of elasticity at the surface (E v) =.5 GPa =.5x109 N/m a) Density at the surface (ρ 1 ) = 10000/9.81 = kg/m 3 Specific volume at the surface (v s1)= 1/ρ 1 =1/1019.4= m 3 /kg Bulk modulus in terms of specific volume is Ev = P V s V s1

.5 10 9 = 7.5 106 0 V s 0.000981 V s = 0.000094 m 3 /kg b) v s = v s1 + V s = 0.000981-0.000094 = 0.000951 m 3 /kg c) Density at 7.5 km (ρ ) = 1/v s = 1051.5 kg/m 3 γ = ρ g = 1051.5 9.

8 = V s V s = m 3 /kg b) v s = v s1 + V s = = m 3 /kg c) Density at 7.5 km (ρ ) = 1/v s = kg/m 3 γ = ρ g = = N/m 3 9. In a fluid the velocity measured at a distance of 75 mm from the boundary is 1.15 m/s. The fluid has a dynamic viscosity of Ns/m and a specific gravity (SG) of 0.9. What is the velocity gradient and the shear stress at the boundary assuming a linear velocity distribution. Also calculate the kinematic viscosity. Solution: Assumptions: The fluid is Newtonian The velocity profile is linear Change in velocity ( u) = = 1.15 m/s Change in distance ( y) = 75 0 = 75 mm = 75/1000 m = m Dynamic viscosity (µ) = Ns/m specific gravity (SG)= 0.9 velocity gradient (du/dy) =? Shear stress (τ ) =? Kinematic viscosity (v) =? Velocity gradient on the boundary can be calculated as du/dy= u/ y=1.15/0.075=15 1/s τ = μ du = = 0.7 N/m dy Densitiy of fluid= SGxρ water =0.9x1000=900 kg/m 3

v = μ ρ = 0.048 900 = 5.3x10 5 m /s 10) A piston of weight 90 N slides in a lubricated pipe. The clearance between piston and pipe is 0.05 mm. If the piston decelerates at a rate of 0.

9 v = μ ρ = = 5.3x10 5 m /s 10) A piston of weight 90 N slides in a lubricated pipe. The clearance between piston and pipe is 0.05 mm. If the piston decelerates at a rate of 0.6 m/s when its speed is 0.5 m/s, what is the viscosity of the oil? Solution: Assumptions: - The oil is Newtonian - Velocity of the oil in the gap is linear. Weight of piston (W) = 90 N Clearance ( y)= 0.05 mm = m Deceleration (a) = 0.6 m/s Change in velocity ( u) = 0.5 m/s Diameter of piston (D) = 15 mm = 0.15m Length of piston (L) = 130mm = 0.13 m Viscosity of oil (µ) =? Frictional force acting on piston (F) = Shear stress(τ) at the piston surface x surface area of piston(a) F = μ du dy (πdl) = μ 0.5 (πx0.15x0x13) = 101x0μ Note: Since velocity profile in the gap is linear, on the piston surface du/dy= u/ y Applying Newtons secon la to the piston, we can write W F = W g a μ = ( 0.6) μ = Ns/m 11) A flat plate 0.3 m in area moves horizantally through oil between two large fixed parallel walls. The distance between the walls is 10 cm. If the velocity of the moving plate is 0.6 m/s and the oil has a kinematic viscosity of m /s and specific gravity 0.8, calculate the drag force when

05 m y = 10-.5 = 7.5 cm = 0.075 m Total force (F) = Force on side1 (F1) + Force on side (F) = τ1 A + τ A b) If the plate is equidistant, y1 = y = y = 10/ = 5 cm = 0.

10 (a) the plate is.5 cm from one of the walls and ( y1=.5). (b) the plate is equidistant from both the walls( y1= y). Solution Assumptions - The oil is Newtonian. - The velocity profile between the moving plate and the stationary walls is linear. a) y1 =.5 cm = 0.05 m y = = 7.5 cm = m Total force (F) = Force on side1 (F1) + Force on side (F) = τ1 A + τ A b) If the plate is equidistant, y1 = y = y = 10/ = 5 cm = 0.05 m Total force (F) = Force on side1 (F1) + Force on side (F) = τ1 A + τ A = (τ1 + τ) A 1) The tip of glass tube with an internal diameter of mm is immersed to a depth of 1.5 cm into a liquid of specific gravity Air is forced into the tube to form a spherical bubble just at the lower end of the tube. Estimate surface tension of liquid if the pressure in the bubble is 00 Pa.

11 Solution: Radius of bubble (r) = 1 mm = 0.001m Pressure inside bubble (Pi) = 00 Pa Depth of liquid (h) = 1.5 cm = m Specific weight of liquid= SGxγ water = 0.85x9810 = N Surface Tension (σ)=? Pressure outside the bubble (P o) = x h= x0.015 = Pa The difference between the pressure in the bubbleoutside of the bubble ( P) = = Pa The force actin on the surface of the bubble due this pressure difference is balanced by the surface tension force. Hence we can write dp = σ r σ = dpr = N/m

Homework of chapter (1) (Solution)

Homework of chapter (1) (Solution) بسم اهلل الرمحن الرحيم The Islamic University of Gaza, Civil Engineering Department, Fluid mechanics-discussion, Instructor: Dr. Khalil M. Al Astal T.A: Eng. Mohammed H El Nazli Eng. Sarah R Rostom First