Homework of chapter (1) (Solution)

1 The Islamic University of Gaza, Instructor: Dr. Khalil M. Al Astal Civil Engineering Department, : Eng. Mohammed H El Nazli Fluid mechanics-Discussion, Eng. Sarah R Rostom First semester, 2016. Homework of chapter (1) (Solution). 1. A certain gasoline weights Ib/ft3 . What are its mass density ,Specific volume, and specific gravity ? Solution 2. A tank with dimensions of 3m length, 2m width, and height is filled with a liquid which have a specific gravity of the mass and weight of this fluid. Solution Vtank = Vliquid = 3 2 = 15m3 , liquid = 1000 = 2500Kg/m3 . Mliquid = Vliquid liquid = 15 2500 = 37,500 Kg Wliquid = 37,500 = 367875 N = KN . 3. Calculate density , specific weight , and weight of one litere of petrol of ( ) ? Solution petrol = 1000 = 700 Kg/m3 , Vpetrol = 1 liter = petrol = petrol g = 700 = 6867 N/m3.

2 Wpetrol = 6867 = N . 1. 4. A 40- Ib, ft diameter cylindrical tank slides slowly down a ramp with a constant speed of as shown in figure below. The uniform thickness oil layer on the ramp has a viscosity of Determine the angle , , of the ramp. Solution The free body diagram of the tank is shown below: 40sin( ). Fx = 40sin( ) = A =. A.. A = ( ) 2 = ft 2. 4. 40 sin( ). = = ( ). v = sin( ) = = . y 5. 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 cm. The inner cylinder is rotated at 200 rpm, and the torque is measured to be ( ) . Determine the viscosity of the fluid ? Solution Torque = Fshear R Fshear = = N.

3 15 10 2. du . = =. dy . = 15 10 2 75 10 2 = 2. 2. 2 . = 200 = 60 . v = R = 15 10 2 = / . du 10 2. = ====> = = 10 3 .. dy 25-mm-diameter shaft is pulled through a cylindrical bearing as shown in the figure. The lubricant that fills the gap between the shaft and bearing is oil having a kinematic viscosity of 8 10-4 m2/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. Solution du P = Fshear = As = . dy . = = = 1000 8 10 4 = Pa. s.. du = velocity = 3m/s , dy = gap = = 3. = = 7280 Pa. As = d L = = m2. P = Shear force (F) = 7280 = N . 7. Two square flat parallel plates with dimensions of is 60cm x 60cm. The oil film with thickness of exists between the two plates, the upper plate which moves at requires a force of to maintain the speed and the lower plate is stationary.

4 Determine the dynamic viscosity of oil in poise and the Kinematic viscosity of oil in stokes if the specific gravity of oil is Solution F v A) = A = (y). F y = = . A v s 10 Poise = 2 s = . m 1N. 2. m Where: 10 Poise = 1N. s/m2. B) Gs = , w = 1000Kg/m3 = Gs w = 950Kg/m3. 3. v= = = 10 4 m2 /s 950. v = St . Where: 104 St = 1m2 /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 m2 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 = Solution F 150. = = = 50N/m2. A 3. v v = 1 (y ) + 2 (y ) where: y1 = y2 = 5 10 3 m 1 2. v = ( + 2 ) 1 = s/m2. 5 10 3 1.

5 V 50 = ( + ) 2 = 3( ) = s/m2. 5 10 3. 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. Solution Assume that the thickness of fluid 1 = fluid 2 = 2cm For Fluid 1: dv 1 = 1 ( ). dy top surface 1m dv = 3 2 = , dy = , 1 = s s 1. 1 = = 20 Pa. For Fluid 2: dv 2 = 2 ( ). dy bottom surface 1m dv = 2 0 = , dy = , 1 = s s 2. 2 = = 20 Pa. 4. 1 20. = = 1 . 2 20. Newtonian fluid having a specific gravity of and a kinematic viscosity of 4x10-4. m2/s flows past a fixed surface . Due to the non-slip condition, the velocity of fluid at the fixed surface is zero.

6 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. Solution du = . dy . = = = 1000 4 10 4 = .. 3 3. u = ( 3). 2 2 . du 3 3 2. = U ( 3). dy 2 2 . 3 3 2. = U ( ). 2 2 3. open 2-mm-diameter tube is inserted into a pan of ethyl alcohol ( = 785 Kg/m3 ). 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. Solution 4 cos( ). h=. gd For ethyl alcohol: 4 cos( ). h= = cos( ).

7 785 For water: 4 cos( ). h= = cos( ). 1000 Similar tube 1 = 2 = cos( ) is the same for each fluid > the height in thefirst dube of 2 mm diameter is the greatest . 5. 60 C the surface tension of mercury and water is 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 mm? Solution 4 cos( ). h=. gd = , = , and = 1000Kg/m3. 4( ) cos(0). h= = (Rise) . (1000)( )( ). = , = 130, = 13546Kg/m3. 4( ) cos(130). h= = (Fall) . (13546)( )( ). 13. Determine the bulk modulus of elasticity of a fluid that has a density increase of percent for a pressure increase of KN/m2. Solution p . K= , = = 50,000 , p = 103 N. 100 . K = 50,000 = 106 KN/m2 . 14. (a) The water strider bug shown in Fig.

8 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. ( for water = 10-2 N/m ). Solution For equilibrium = .. =.. 6. 10 4. a- = = = 10 3 . 10 2. 750 . b- = = = 104 . 10 2. 7.