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Jan 2007 3 cm 0.25 m/s y V Plate Fixed surface 1.FIGURE Q1 shows a plate with an area of 9 cm 2 that moves at a constant velocity of 0.25 m/s. A Newtonian.

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Presentation on theme: "Jan 2007 3 cm 0.25 m/s y V Plate Fixed surface 1.FIGURE Q1 shows a plate with an area of 9 cm 2 that moves at a constant velocity of 0.25 m/s. A Newtonian."— Presentation transcript:

1 Jan 2007 3 cm 0.25 m/s y V Plate Fixed surface 1.FIGURE Q1 shows a plate with an area of 9 cm 2 that moves at a constant velocity of 0.25 m/s. A Newtonian fluid with a specific gravity of 0.92 and a kinematic viscosity of 4  10 -4 m 2 /s, is filled between the plate and a fixed surface at a constant thickness of 3 cm. Due to the ‘no- slip’ condition, the velocity of the fluid, V, at any thickness of the fluid, y, can be determined through the velocity profile, which is given as: where: V = Velocity at any thickness of the fluid y = Thickness of the fluid measured from the fixed surface FIGURE Q1: Velocity profile of a Newtonian fluid

2 Determine: a.The dynamic viscosity of the fluid. [3 marks] b. The magnitude and direction of shear stress developed on the surface of the plate. [4 marks] c. The shear stress at the fixed surface. [4 marks] d. The magnitude of force required to move the plate. [4 marks] e. If all values remain constant and the velocity of the fluid follows a linear profile, determine: i.The new velocity gradient. [2 marks] ii.The magnitude of force required to move the plate. [3 marks]

3 a.The dynamic viscosity of the Newtonian fluid. Dynamic viscosity,  =  = (4  10 -4 )(1000)(0.92) = 0.368 kg/m.s b. The magnitude and direction of shear stress developed on the surface. [4 marks] For Newtonian fluid, shear stress is proportional to velocity gradient. Direction : to the left

4 c. The shear stress at the fixed surface. [4 marks] At the fixed surface, y = 0 d. Magnitude of force required to move the plate. [4 marks] At the moving plate,  = 0.0123 Pa F =  A = (0.0123)(9  10 -4 ) = 1.107  10 -5 N

5 e.If all values remain constant and the velocity of the fluid follows a linear profile, determine: i. The new velocity gradient. [2 marks] Since it follows a linear profile, ii. The magnitude of force required to move the plate. [3 marks] F =  A = (3.067)(9  10 -4 ) = 2.76  10 -3 N

6 2.a.FIGURE Q2a shows a pipe carries water and tapers uniformly from a diameter of 0.5 m at section A to 0.8 m at section B. The pipe with a length of 2 m slopes upwards from section A to section B at an angle of 30 o. Pressure gages are installed at sections A, B and also at C, the midpoint of AB. The gage pressures recorded at A and B are 2.0 bar and 2.3 bar, respectively. A C  = 30 o B FIGURE Q2a: Tapering pipe carries water Assuming the pipe is frictionless, determine: i. The velocity at section A. [4 marks] ii. The volumetric flow rate through the pipe. [3 marks] iii. The pressure recorded at C. [4 marks]

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9 ii. The volumetric flow rate through the pipe. [3 marks] Solution: Q = Q A = A A V A = = 1.903 m/s iii. The pressure recorded at C. [4 marks] Solution: Taking point A and point C as reference. D C =

10 b(i). If the system is under equilibrium, calculate the required force, F, acting on the piston. [3 marks]

11 b(ii). Determine the pressure head, expressed in meter of water, acting on the bottom of the tank. [3 marks]

12 2.b(iii). Greater force is exerted to the piston resulting in a decrease of the oil level in the cylinder by an amount of h1 and an increase of oil level in the tank to a height h2. Find the relationship between the air pressure in the tank and the force exerted if h1 = h2. [3 marks]

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