# CH-205: Fluid Dynamics

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1 CH-05: Fluid Dynamics nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering) Solutions of Mid Semester Examination Data Given: Density of water, ρ = 1000 kg/m 3, gravitational acceleration, g = 9.81 m/s Question # 1: [4+4+] Consider the flow of water through a clear tube (cross-section area A 1 ). It is sometimes possible to observe cavitation in the throat created by pinching off the tube to a very small diameter of cross-section area A. Let the average velocities and pressures at points 1 (inlet) and (throat) are (V 1, P 1 ) and (V, P ), respectively. Both the maximum velocity and minimum pressure occur at throat. Consider that the throat diameter is 1/0th of the inlet diameter. (a) estimate the minimum average inlet velocity at which cavitation is likely to occur for water entering at pressure of kpa and temperature of (a) 0 o C and (b) 50 o C, respectively. Explain why the required velocity in part (b) is higher or lower than that part (a). Under the incompressible flow with negligible gravitational effects and negligible irreversibilities conditions, both (V A) and [P +ρ(v /)] may be taken as constant along the flow. The saturation pressure of water may be taken as.34 and 1.35 kpa at 0 o C and 50 o C, respectively. Solution # 1 Given that the flow is incompressible with negligible gravitational effects and negligible irreversibilities, V A and [P + ρ(v /)] may be taken as constant along the flow. Therefore, applying the prescribed balances in between inlet (point 1) and throat (point ), we get and V 1 A 1 = V A or V = V 1 (A 1 /A ) or V = V 1 (D 1 /D ) (1) On substituting Eq. (1) in Eq. (), we get (P 1 P ) = ρ V 1 ( D 4 1 D 4 P 1 + ρ V 1 = P + ρ V ) 1 or V 1 = () ( (P 1 P ) D ρ D 4 1) (3) Given P 1 = kpa, D 1 /D = 0, ρ = 1000 kg/m 3 We understand that the the pressure (P ) anywhere in flow should not be allowed to drop below the vapor pressure (P v ) at the given temperature to avoid the cavitation. For a pure substance, the vapor pressure (P v ) equals to the saturation pressure (P sat ). Therefore, the cavitation is likely to occur at the throat when P P sat. (a) At 0 o C, P = P sat =.34 kpa After substituting the numerical values in Eq. (3), we get ( ) 10 3 V 1 = ( ) 1 = m/s (4) (b) At 50 o C, P = P sat = 1.35 kpa After substituting the numerical values in Eq. (3), we get ( ) 10 3 V 1 = ( ) 1 = m/s (5) Instructors: RPB & SC Page 1 of 6 Sept 11, 013 (10 a.m. - 11:30 a.m.)

2 CH-05: Fluid Dynamics nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering) The results show that at 50 o C, cavitation may occur at lower value of the inlet average velocity compared to that at 0 o C. It is simply because for the given decrease in the duct cross-sectional area, the velocity increases and the pressure decreases according to given flow continuity and mechanical energy balances. We also know that the vapor pressure increases with increasing value of temperature. Therefore, the pressure difference (P 1 P ) to be maintained to avoid the cavitation decreases with increasing temperature, In first case, the pressure (P 1 P ) is very large and, thus, the corresponding inflow velocity is larger compared to the second case. Question # : [ ] A metal cylinder (diameter D = 3 m and height L = 3 m) of specific weight of 5886 N/m 3 is required to float in water with its axis vertical. Determine the (i) depth of immersion h of cylinder in water, (ii) distance of center of buoyancy B and center of gravity G from the bottom point of cylinder O (iii) distance of center of gravity G from the center of buoyancy B (iv) volume of the cylinder submerged in the water V sub, (v) metacentric height (GM). Based on the above results, state whether the cylinder is in stable equilibrium. Solution # Given, the length of cylinder, L = 3 m, diameter of the cylinder D = 3m, specific weight of cylinder w c = 5886 N/m 3 Consider that h height of the cylinder is immersed in the liquid. Taking the force balance, Weight of cylinder = weight of water displaced where w is the specific weight of water. (i) the depth of immersion (h) of cylinder in water π 4 D L w c = π 4 D h w h = L w c w = = 1.8 m 9810 (ii) Distance of center of buoyancy (B) from the bottom center (O) OB = h = 0.9 m Distance of center of gravity (G) from the bottom center (O) OG = L = 1.5 m (iii) Distance of center of gravity (G) from center of buoyancy (B) (iv) volume of cylinder submerged in water BG = OG OB = 0.6 m V sub = π 4 D h = π = 1.73 m Instructors: RPB & SC Page of 6 Sept 11, 013 (10 a.m. - 11:30 a.m.)

3 CH-05: Fluid Dynamics nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering) (v) metacentric height, GM is a measure of stability for the floating bodies. It is the distance between the center of gravity (G) and the metacenter (M, the intersection point of the lines of action of buoyancy force through the body before and after rotations). where BM is metacentric radius. GM = BM BG The second moment of inertia (I xx,c ) would be related to the buoyant force for the the small angle of rotations due to displacement as follow w I xx,c = BM F B w I xx,c = BM w V sub BM = I xx,c V sub For the circular cross-sections, the second moment of inertia, I xx,c = πd4 64 BM = I xx,c V sub = D 16h = m GM = = m The negative value of the GM indicates that the metacentre is below the center of gravity. Thus, the cylinder is in unstable equilibrium. Question # 3(A): [3] Consider two identical glasses of water, one stationary and other moving on a horizontal plane with constant acceleration. Assuming no splashing or spilling occurs, give your to-the-point explanation that which glass will have a higher pressure at the (i) front (ii) midpoint and (iii) back of the bottom surface? Solution # 3(A) We know that the pressure in all cases is the hydrostatic pressure, which is directly proportional to the fluid height. The pressure at the bottom surface is constant when the glass is stationary. For a glass moving on a horizontal plane with constant acceleration, water will collect at the back but the water depth will remain constant at the center. Therefore, the pressure at the midpoint will be the same for both glasses. But the bottom pressure will be low at the front relative to the stationary glass, and high at the back (again relative to the stationary glass). Question # 3(B): [3++] A water tank is being towed on an uphill road (inclined by 0 o with the horizontal) with constant acceleration of 5 m/s in the direction of motion. Determine the angle the free surface of water makes with horizontal. What would your answer be if the direction of motion were downward on the same road with the same acceleration? If the water is replaced by oil (specific gravity 0.8) then what would be the new values of angle in both the cases? Solution # 3(B) The effects of splashing, breaking, driving over bumps, and climbing hills are assumed to be negligible and the acceleration remains constant. Instructors: RPB & SC Page 3 of 6 Sept 11, 013 (10 a.m. - 11:30 a.m.)

4 CH-05: Fluid Dynamics nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering) From geometrical considerations, the horizontal and vertical components of acceleration are a x = a cos α a z = a sin α The tangent of the angle the free surface makes with the horizontal is tan θ = a x g + a z = a cos α g + a sin α = 5 cos 0 o = θ =.0o sin 0o When the direction of motion is reversed, both a x and a z are in negative x- and z-directions, respectively, and thus become negative quantities, a x = a cos α a z = a sin α Then the tangent of the angle the free surface makes with the horizontal becomes tan θ = a x g + a z = a cos α g a sin α = 5 cos 0o = θ = 30.1o sin 0o The analysis is valid for any fluid with constant density, not just water. Since we used no information that pertains to water in the solution and so there will not be any change in the results after changing the fluid from water to oil. Question # 4(A): [++] Consider steady, incompressible, two-dimensional flow through a converging duct. A simple approximate velocity and pressure field for this flow are given as V = (u, v) = (U 0 + bx) i by j (6) P = P 0 ρ [ U0 bx + b (x + y ) ] (7) where U 0 and P 0 are the horizontal velocity and pressure, respectively, at x = 0. Note that this equation ignores viscous effects along the walls but is a reasonable approximation throughout the majority of the flow field. Calculate the material acceleration for fluid particles passing through this duct. Further, give your answer in two ways: (i) as acceleration components a x and a y, and (ii) as acceleration vector a. Also, obtain an expression for the rate of change of pressure following a fluid particle. Instructors: RPB & SC Page 4 of 6 Sept 11, 013 (10 a.m. - 11:30 a.m.)

5 CH-05: Fluid Dynamics nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering) Solution # 4(A) The velocity and pressure fields for steady, incompressible, two-dimensional flow through a converging duct is given as, V = (u, v) = (U 0 + bx) i by j P = P 0 ρ [ U0 bx + b (x + y ) ] The acceleration field components are obtained from its definition (the material acceleration) in Cartesian coordinates, The material acceleration components D V Dt = V t + V. V a x = Du Dt = u t + u u x + v u y + w u z = (U 0 + bx)b + ( by)0 = (U 0 + bx)b a y = Dv Dt = v t + u v x + v v y + w v z = (U 0 + bx)0 + ( by)( b) = b y In terms of the vector notations, material acceleration vector a = ax i + ay j = b(u0 + bx) i + b y j By definition, the material derivative, when applied to pressure, produces the rate of change of pressure following a fluid particle. Therefore, the rate of change of pressure following a fluid particle DP Dt = P t +u P P +v x y + w P z = (U 0+bx)( ρu 0 b ρb x)+( by)( ρb y) = ρ [ U0 b U 0 b x + b 3 (y x ) ] Question # 4(B): [+] Consider the following steady, two-dimensional velocity field V = (u, v) = [a (b cx) ] i + (c xy cby) j (8) Is there a stagnation point in this flow field? If so, calculate the location of stagnation point. The given velocity field, Solution # 4(B) V = (u, v) = [a (b cx) ] i + (c xy cby) j At a stagnation point, both u and v must equal zero. At any point (x,y) in the flow field, the velocity components u and v are given as u = a (b cx) v = c xy cby Instructors: RPB & SC Page 5 of 6 Sept 11, 013 (10 a.m. - 11:30 a.m.)

6 CH-05: Fluid Dynamics nd Year, B.Tech. & Integrated Dual Degree (Chemical Engineering) Setting these to zero and solving simultaneously for x and y yields the stagnation point u = 0 = a (b cx) x = b a c v = 0 = c xy cby y = 0 So, yes there is a stagnation point; its location is x = (b - a)/c, y = 0. If the flow were three-dimensional, then set w = 0 as well to determine the location of the stagnation point. Some flow fields may have more than one stagnation point. Instructors: RPB & SC Page 6 of 6 Sept 11, 013 (10 a.m. - 11:30 a.m.)

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