Homework of chapter (1) (Solution)

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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.

1 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 semester, Homework of chapter (1) () 1. A certain gasoline weights 46.5 Ib/ft 3. What are its mass density,specific volume, and specific gravity? 2. A tank with dimensions of 3m length, 2m width, and 2.5m height is filled with a liquid which have a specific gravity of 2.5.Calculate the mass and weight of this fluid. V tank = V liquid = = 15m 3, ρ liquid = = 2500Kg/m 3. M liquid = V liquid ρ liquid = = 37,500 Kg W liquid = 37, = N = KN. 3. Calculate density, specific weight, and weight of one litere of petrol of (s.g. 0.7)? ρ petrol = = 700 Kg/m 3, V petrol = 1 liter = 0.001m 3 γ petrol = ρ petrol g = = 6867 N/m 3. W petrol = = N. 1

2 4. A 40- Ib, 0.8 ft diameter cylindrical tank slides slowly down a ramp with a constant speed of 0.1ft/s as shown in figure below. The uniform thickness oil layer on the ramp has a viscosity of 0.2 Ib.s/ft 2. Determine the angle,θ, of the ramp. The free body diagram of the tank is shown below: F x = sin(θ) = τa τ = 40sin(θ) A A = π 4 (0.8) 2 = ft 2 τ = τ = μ v y 40 sin (θ) = 79.57sin (θ) sin(θ) = θ = The viscosity of a fluid is to be measured by a viscometer constructed of two 75-cm-long concentric cylinders. The outer diameter of the inner cylinder is 15 cm, and the gap between the two cylinders is 0.12 cm. The inner cylinder is rotated at 200 rpm, and the torque is measured to be (0.8 N.m). Determine the viscosity of the fluid? Torque = F shear R F shear = τ = μ du dy = F As 0.8 = 5.34 N As = π = m 2 2

3 ω = 200 2π rad = s v = ω R = = m/s μ du dy = F As ====> μ = = pa. s 6.A 25-mm-diameter shaft is pulled through a cylindrical bearing as shown in the figure. The lubricant that fills the 0.3mm gap between the shaft and bearing is oil having a kinematic viscosity of m 2 /s and a specific gravity of Determine the force P required to pull the shaft at a velocity of 3m/s. Assume the velocity distribution in the gap is linear. P = F shear = τ A s τ = μ du dy ν = μ ρ μ = ρ ν = = Pa. s. du = velocity = 3m/s, dy = gap = 0.3mm = m. τ = = 7280 Pa A s = π d L = π = m 2 P = Shear force (F) = = N. 7. Two square flat parallel plates with dimensions of is 60cm x 60cm. The oil film with thickness of 12.5mm exists between the two plates, the upper plate which moves at 2.5m/s requires a force of 9.81N to maintain the speed and the lower plate is stationary. Determine the dynamic viscosity of oil in poise and the Kinematic viscosity of oil in stokes if the specific gravity of oil is A) τ = F A = μ (v y ) μ = F A y v = s μ = 0.136N. m 2 10Poise s = 1.36Poise. 1N. m 2 Where: 10Poise = 1N. s/m 2 B) G s = 0.95, ρ w = 1000Kg/m 3 ρ = G s ρ w = 950Kg/m 3 3

4 v = μ ρ = = m 2 /s v = St. Where: 10 4 St = 1m 2 /s. 8. For the flow condition shown in figure below if the top & bottom plates are fixed, determine the velocity at which the central plate of area 3 m 2 will move if a force of 150N is applied to it. The dynamic viscosities of the two oils are in the ratio of 1:3 and viscosity of the top oil =0.1N.s/m 2. τ = F A = = 50N/m2 τ = μ 1 ( v y 1 ) + μ 2 ( v y 2 ) where: y 1 = y 2 = m τ = 50 = v (μ 1 + μ 2 ) μ 1 = 0.1N. s/m 2 v ( ) μ 2 = 3(0.1) = 0.3N. s/m 2 v = m/s. 9. Two layers of fluid be dragged along by the motion of an upper plate as shown in figure below. The bottom plate is stationary. The top fluid puts a shear stress on the upper plate, and the lower fluid puts shear stress on the bottom plate. Determine the ratio of these two shear stresses. Assume that the thickness of fluid 1 = fluid 2 = 2cm For Fluid 1: τ 1 = μ 1 ( dv dy ) top surface dv = 3 2 = 1m, dy = 0.02m, μ s 1 = 0.4Pa. s τ 1 = = 20 Pa For Fluid 2: τ 2 = μ 2 ( dv dy ) bottom surface dv = 2 0 = 1m, dy = 0.02m, μ s 1 = 0.2Pa. s τ 2 = = 20 Pa

5 τ 1 τ 2 = = A Newtonian fluid having a specific gravity of 0.92 and a kinematic viscosity of 4x10-4 m 2 /s flows past a fixed surface. Due to the non-slip condition, the velocity of fluid at the fixed surface is zero. The fluid velocity profile near the surface is shown in below figure. Determine the magnitude and the direction of the shearing stress develop on the plate. Express your answer in term of U and, with U and expressed in unit of (m/s) and (m) respectively. τ = μ du dy ν = μ ρ μ = ρ ν = = pa. s u = U ( 3y 2δ y3 2δ 3) du dy = U ( 3 2δ 3y2 2δ 3) τ = U ( 3 2δ 3y2 2δ 3) 11.An open 2-mm-diameter tube is inserted into a pan of ethyl alcohol (ρ = 785 Kg/m 3 ) and a similar 4-mm diameter tube is inserted into a pan of water. In which tube will the height of the rise of the fluid column due to capillary action be the greatest? Assume the angle of contact is the same for both tubes. ρgd For ethyl alcohol: = σ cos(θ) For water: = σ cos(θ) Similar tube σ 1 = σ 2 = σ σ cos(θ) is the same for each fluid > the height in thefirst dube of 2 mm diameter is the greatest. 5

6 12.At 60 C the surface tension of mercury and water is 0.47 and N/m, respectively. What capillary height changes will occur in these two fluids when they are in contact with air in a clean glass tube of diameter 0.4 mm? ρgd For Water σ = N/m, θ = 0.0, and ρ = 1000Kg/m 3 4(0.0662) cos(0) = m (Rise). (1000)(9.81)(0.0004) For mercury σ = 0.47N/m, θ = 130, ρ = 13546Kg/m 3 4(0.47) cos(130) = 0.023m (Fall). (13546)(9.81)(0.0004) 13. Determine the bulk modulus of elasticity of a fluid that has a density increase of percent for a pressure increase of KN/m 2. K = ρ p ρ, ρ ρ = ρ ρ = 50,000, p = N K = 50, = KN/m (a) The water strider bug shown in Fig. is supported on the surface of a pond by surface tension acting along the interface between the water and the bug s legs. Deternine the minimum length of this interface needed to support the bug. Assume the bug weights 10-4 N and the surface tension force acts vertically upwards.(b) Repeat part (a) if surface tention were to support a person weighing 750 N. (σ for water = N/m ) For equilibrium W = σ L L = W σ 6

7 a- L = W σ = = m b- L = W σ = 750 N = m 7

Fluids and their Properties

Chapter (1) Fluids and their Properties Dr. KHALIL MAHMOUD ALASTAL Eng.Mohammed AbuRahma Eng.Reem Sbaih 2017 Newton s Law of Viscosity: - / Non-Newtonian Fluids: - Mass Density: - / Specific weight: -