Introduction and Fundamental Concepts (Lectures 1-7)

Transcription

1 Introduction and Fundamental Concepts (Lectures -7) Q. Choose the crect answer (i) A fluid is a substance that (a) has the same shear stress at a point regardless of its motion (b) is practicall incompressible (c) cannot remain at rest under action of an shear fce (d) obes Newton s law of viscosit [Ans.(c)] (ii) F a Newtonian fluid (a) shear stress is proptional to shear strain (b) rate of shear stress is proptional to shear strain (c) shear stress is proptional to rate of shear strain (d) rate of shear stress is proptional to rate of shear strain [Ans.(c)] n du (iii) Shear stress f a general fluid motion is represented b τ + A, where n d and A are constants. A Newtonian fluid is given b (a) n > and A = (b) n = and A = (c) n > and A (d) n < and A = [Ans.(b)] (iv) If the relationship between the shear stress τ and the rate of shear strain du d is n du expressed asτ = m d. The fluid with the exponent n < is known as (a) Pseudoplastic fluid (b) Bingham fluid (c) Dilatant fluid (d) Newtonian fluid [Ans.(a)] (v) The increase in temperature (a) increases the viscosit of a liquid and decreases the viscosit of a gas (b) decreases the viscosit of a liquid and increases the viscosit of a gas (c) increases the viscosit of both a liquid and a gas (d) decreases the viscosit of both a liquid and a gas [Ans.(b)] (vi) The bulk modulus of elasticit f an ideal gas (equation of state p = ρrt, where p is the pressure, ρ is the densit, R is the characteristic gas constant and T is the absolute temperature) at constant temperature T is given b p (a) ρ

2 (b) RT (c) p (d) ρ RT [Ans.(c)] Q. A plate having an area of.4 m is sliding down the inclined plane at to the hizontal with a velocit of. m/s. There is a cushion of fluid mm thick between the plane and the plate. The weight of the plate is N. Assuming linear velocit profile in the film, find the viscosit of the fluid. The arrangement is shown in the figure below. Fluid τ. m/s W sin θ W W = N cos θ Component of weight along the slope isw sin. Velocit gradient is found to be du V V = = d h h where h is the thickness of the oil film and V is the velocit of the plate.. Viscous resistance F is given b F = τ A du V F A A d h At equilibrium, the viscous resistance to the motion should be equal to the component of the weight of the solid block along the slope. Thus, V A = Wsin h..4 = sin =. N-s/m

3 Q. A thin plate is placed between two flat surfaces h apart such that the viscosit of liquids on the top and bottom of the plate are and respectivel. Determine the position of the plate such that the viscous resistance to unifm motion of the plate is minimum. Let us assume that the velocit of the plate be V. Let F and F be the shear fces per unit area on the lower surface and upper surface of the thin plate respectivel. Let us also consider that the distance of the thin plate from the bottom wall is as shown in the figure below. h V From Newton s law of viscosit, shear stress on the bottom surface of the plate τ is given b du τ d V = Shear fce per unit area on the bottom surface of the plate is du V F d From Newton s law of viscosit, shear stress on the upper surface of the plate τ is given b du τ d where d = h ( Neglecting thickness of the plate) V τ h Shear fce per unit area on the upper surface of the plate is du V F d h Total viscous resistance to drag the plate is F = F+ F V V + h

4 F minimum F, we have df d = V V + = h ( ) ( h ) ( h ) ( h ) = = h = + Q4. A unifm film of oil. mm thick separates two discs, each of mm diameter, mounted coaxiall. Igning the edge effects, calculate the tque necessar to rotate one disc relative to other at a speed of 7 rev/s, if the oil has a viscosit of.4 Pa-s. At a radial distance r (measured from the axis of the discs) in the oil film ( figure below), the velocit v= π 7 r dv π 7 r = =.8 r d. ( is measured along the direction of the axis i.e., perpendicular to the discs) 7 rev/s R = mm Oil. mm r dr The fce acting on an elemental ption of the disc of thickness dr at a radial location r is given b df =.4.8 r π rdr ( ) 4

5 Cresponding tque Hence, ( ) dt =.4.8 r π rdr r ( ) R T = dt =.4.8 r πr dr ( where R, the radius of the disc = mm =. m) ( ) π. = = 7.4 N-m 4 Q. (a) Find the change in volume of. m of water at 6.7 C when subjected to a pressure increase of MN/m (The bulk modulus of elasticit of water at 6.7 C is 9.4 N/m ). (b) From the following test data, determine the bulk modulus of elasticit of water: at. MN/m, the volume was. m and at 4 MN/m, the volume was.99 m. (a) 6 Change in pressure p = N/m Initial volume of water V =. m Bulk modulus of elasticit is given b E p = = p E 6. = =.89 m 9.4 p = 4. =.MN/ m =. N/m (b) Change in pressure ( ) 6 Change in volume = (.99) =. m The bulk modulus of elasticit of water is 6 p. E = =. =. N/m. 9 Q6. A spherical soap bubble of diameter d coalesces with another bubble of diameter d to fm a single bubble of diameter d containing the same amount of air. Assuming isothermal process, derive an analtical expression f d as a function of d, d, the ambient pressure p and the surface tension of soap solution σ. If d = mm, d = 4 mm, p = kn/m and σ =.9 N/m, determine d.

6 From conservation of mass m+ m = m where m, m and m are the masses of air inside the bubbles of diameter d, d and d respectivel. F an isothermal process (considering air to behave as an ideal gas), the above leads to pd + pd = pd where p, p and p are the pressures inside the bubbles of diameter d, d and d respectivel. Now, 8σ p = p + d 8σ p = p + d 8σ p = p + d Hence, 8σ 8σ 8σ p + d + p + d = p + d d d d F the given values of d = mm, d = 4 mm, p = kn/m, σ =.9 N/m (.) + + (.4) = + d..4 d d 7. + d = 7. d which gives d = 4.6 m = 4.6 mm 6