Fluid Mechanics Answer Key of Objective & Conventional Questions

019 MPROVEMENT Mechanical Engineering Fluid Mechanics Answer Key of Objective & Conventional Questions

1 Fluid Properties 1. (c). (b) 3. (c) 4. (576) 5. (3.61)(3.50 to 3.75) 6. (0.058)(0.05 to 0.06) 7. (b) 8. (d) 9. (c) 10. (a) 11. (b) Copyright: Subject matter to MADE EASY Publications, New Delhi. No part of this book may be reproduced or utilised in any form without the written permission. 1. (d) 13. (c) 14. (c) 15. (c) 16. (c) 17. (c) 18. (c) 19. (b) 0. (b) 1. (d) www.madeeasypublications.org Copyright

Rank Improvement Workbook 3 Solution : Solution : 3 Frictional resistance = 1553.33 N Depth of the sea = 8950 m Solution : 4 τ y =0 = 0.9 1 = 10.8 N/m τ y =0.15 = 0.9 6 = 5.4 N/m τ y =0.3 = 0.9 0 = 0 Solution : 5 Radial distance in m 0.6 0.4 Velocity profile 0. 0 0. 0.4 1 3 4 5 6 7 0.6 Velocity in m/s Velocity profile Radial distance in m 0.5 0.4 0.3 0. 0.1 0 0.1 0. 0.3 0.4 0.5 0.01 0.0 0.03 0.04 0.05 0.06 Shear stress in N/m Shear stress distribution Solution : 6 Solution : 7 y = 0, τ = 0.0474 N/m y = 3 mm, τ = 0.0335 N/m y = 6 mm, τ = 0 μ = 0.397 Pa.s Copyright www.madeeasypublications.org

4 Mechanical Engineering Fluid Mechanics Solution : 8 R 0 = 54.6 mm Force required to separate the bubble into two identical halves is given by ΔP = 5.86 N/m Solution : 9 p i = 9.15 N/m = 9.15 Pa www.madeeasypublications.org Copyright

Fluid Statics 1. (b). (c) 3. (b) 4. (b) 5. (c) 6. (b) 7. (c) 8. (c) 9. (d) 10. (a) 11. (a) 1. (d) Copyright: Subject matter to MADE EASY Publications, New Delhi. No part of this book may be reproduced or utilised in any form without the written permission. 13. (109.9)(109 to 110.5) 14. (a) 15. (c) 16. (d) 17. (a) 18. (a) 19. (0.338)(0.35 to 0.340) 0. (33.5)(33 to 34) 1. (a) Copyright www.madeeasypublications.org

6 Mechanical Engineering Fluid Mechanics Solution : Solution : 3 F 1 = 138814.95 N V = 5590.9 m 3 i.e. total volume weight of iceberg = ρ i V g = 915 5590.9 9.81 = 50.184 10 6 N. Solution : 4 The metacentric height is negative. Hence cylinder will not float with vertical axis. Solution : 5 Solution : 6 h A = 4 m of water H = x + 0.3 =.87 m Solution : 7 pm p N γ γ = 13.86 m pn p R γ γ = 5.04 m p M γ pr = 18.9 m γ Solution : 8 Resultant water pressure in the dam F = Fx + F y = 790.907 kn θ = F 1 y = tan 51 40 F x www.madeeasypublications.org Copyright

3 Fluid Kinematics 1. (b). (c) 3. (b) 4. (c) 5. (d) 6. (d) 7. (c) 8. (b) 9. (c) 10. (d) 11. (b) 1. (b) 13. (c) Copyright: Subject matter to MADE EASY Publications, New Delhi. No part of this book may be reproduced or utilised in any form without the written permission. 14. (b) 15. (8) 16. (8) 17. (c) 18. (8)(7.9 to 8.1) 19. (d) 0. (b) 1. (c). (a) 3. (a) 4. (0) Copyright www.madeeasypublications.org

8 Mechanical Engineering Fluid Mechanics Solution : 5 (i) V 1 = Q π 4 ( d ) 1 4Q = πd Q 4Q Q Similarly, V = = π and V d πd 3 = π d 4 4 (ii) V 1 =.39 m/s V = 9.55 m/s and V 3 = 0.679 m/s (iii) V 1 =.39 m/s V = 19.10 m/s V 3 = 0.5 m/s 1 3 Solution : 6 (i) Local Acceleration is defined as the rate of increase of velocity with respect to time at a given point in a flow field. In the equations a x = du = u u + v u + w u + u dt x y z t a y = a z = dv = u v + v v + w v + v dt x y z t dw = u w + v w + w w + w dt x y z t u v w,, are known as local accelerations t t t Convective acceleration is defined as the rate of change of velocity due to the changed position of fluid u v w particles in fluid flow. The expressions other than,, t t t convective accelerations. (ii) ω = 1i ˆ + 0 ˆ k+ ( 4) ˆ k = iˆ 4kˆ in the above equations are known as Solution : 7 P a P = 4 4 x y ρ x y + + x y = 4 4 x y ρ + + x y Solution : 8 N = 567. rpm www.madeeasypublications.org Copyright

Rank Improvement Workbook 9 Solution : 9 ρ ρ = exp[(1 y )cos ω t] / ω 0 Solution : 30 If u, v and w satisfy the equation of continuity, then it will be possible liquid motion. u v w + + =0 x y z 3x y y 3x yz. + yz. + 0= 0 3 3 ( x + y ) ( x + y ) This shows that the liquid motion is possible. For irrotational motion, we have v w =0 z y w u x z =0 and u v y x =0 So, Also, and y x y = 0 v w x = z y ( x + y ) ( x + y ) w u = x z xy xy + = 0 ( x + y ) ( x + y ) u v y x = xz(3 y x ) xz(3 y x ) = 0 3 3 ( x + y ) ( x + y ) which satisfy the condition. Hence the motion is irrotational. Solution : 31 The continuity equation becomes, u v + =6x 6x = 0 x y Hence, the flow is physically possible. To find rotational or irrotational: u v Because, the flow is rotational. y x To find angular velocity; Copyright www.madeeasypublications.org

10 Mechanical Engineering Fluid Mechanics ω z = 6xy Note at the origin (0, 0), because ω z = 0, the flow is rotational except at the origin. To find vorticity : vorticity, ξ = 1xy To find shear strain: γ xy = 6xy Note: The negative sign for angular velocity and shear strain indicates clockwise motion. To find dilatancy (linear strain): u x = = 6x x v y = = 6x y To find the circulation: Circulation Γ = 1xy(πa ) m /s where a is the radius of circle x + y ay = 0 www.madeeasypublications.org Copyright

4 Fluid Dynamics & Flow Measurement 1. (c). (d) 3. (1.11)(1 to 1.15) 4. (d) 5. (c) 6. (d) 7. (c) 8. (b) 9. (34.90) (34.85 to 35.5) 10. (a) 11. (d) 1. (c) Copyright: Subject matter to MADE EASY Publications, New Delhi. No part of this book may be reproduced or utilised in any form without the written permission. 13. (.47)(.40 to.50) 14. (80.79)(80 to 81) 15. (137500) 16. (d) 17. (a) 18. (d) 19. (d) 0. (5.56)(5.0 to 6.0) 1. (55.45)(55.0 to 55.90). (0.9848)(0.980 to 0.987) 3. (135)(134.50 to 135.50) 4. (b) 5. (b) 6. (8.76) Copyright www.madeeasypublications.org

1 Mechanical Engineering Fluid Mechanics Solution : 7 (i) h = 500 1 m 9.81 13600 1 1.1807 h = 11 mm 1 = 17.4 = 0.011 m 11517.6 ( 40 (ii) h = ) 1 9.81 13600 1 1.1807 m = 176400 9.81 11517.6 = 0.78 m or 780 mm Solution : 8 h =.03 = 0.175 m = 17.5 cm 1.6 Deflection : As the value of h comes positive, hence our assumed direction is correct i.e. the manometer limb connected to section 1 will be having a smaller column of mercury than the other limb. Solution : 9 Discharge: Q = AV 1 = 0.009734 m 3 /s = 9.734 litre/s e/s Solution : 30 H = V p z + + = 3.058 m g γ Solution : 31 Q = A 3 V 3 = 3.8 10 3 cumecs p = 40.61 kn/m Solution : 3 Hence, h =.46 4.5 + = 4.7999 m 9.81 Solution : 33 Equation of trajectory Maximum elevation of the jet Maximum horizontal distance s = =1.73x 0.7848x U0 sin θ 5 = (sin60) = 0.955 m g 9.81 0 = 1.73x 0.7848x or x(1.73 0.7848x) =0 That is, x =0 or x =. m So maximum horizontal distance =. m. www.madeeasypublications.org Copyright

Rank Improvement Workbook 13 Solution : 34 Solution : 35 stagnation pressure = 161.109 kpa gauge Velocity = 14.37 kmph Solution : 36 F = 810 N [In this problem, p = 0 has been assumed as if p 1 = 00 kpa = above p ] Copyright www.madeeasypublications.org

5 Dimensional Analysis 1. (d). (a) 3. (c) 4. (c) 5. (b) 6. (c) 7. (a) 8. (c) 9. (c) 10. (c) 11. (c) 1. (a) Copyright: Subject matter to MADE EASY Publications, New Delhi. No part of this book may be reproduced or utilised in any form without the written permission. 13. (c) 14. (d) 15. (a) 16. (b) 17. (d) 18. (a) 19. (d) www.madeeasypublications.org Copyright

Rank Improvement Workbook 15 Solution : 0 Buckingham s π-theorem states that If there are n-variable (Independent and dependent) in a physical phenomenon and if these variable contain m fundamental dimensions then the variables are arranged into (n m) dimensionless terms which are called π-terms. Let X 1, X, X 3... X n are the variables involved in a physical problem. Let X 1 be the dependent variable and X, X 3, X 4... X n are the independent variables on which X 1 depends. Mathematically it can be written as X 1 = f (X, X 3, X 4,... X n ) and, f (X 1, X, X 3, X 4... X n )=0 Hence, f (π 1, π, π 3... π n m )=0 Drag force, F = f (V, L, K, ρ, g) F (Drag force) = K gl ρlv φ, L V Solution:1 Total discharge, Q p = 377.54 m 3 /sec Discharge through each turbine Q p = 377.54 = 94.385 m 3 /sec 4 For Model Power developed by each model P m = 17.06 kw Np H p 60 Lr N = m H = m 5 = 3.46 From equation (i) and (ii), N m = 38.5 rpm Specific speed, (N s ) m = 131.754 rpm (N s ) p = 133.90 rpm Type of turbine runner is Francis. Solution : Importance of Dimensional Analysis: Dimensional analysis is an important tool in designing fluid machines, structures and various aspects of fluid flow on them. (i) In dimensional analysis, model of actual structure is analysed under various flow conditions in order to see various effects of flow on actual structure. (ii) Dimensional analysis predicts the performance of actual structure with the help of model. (iii) The model of actual structure made with the help of dimensionless analysis is the replica of actual structure. So it eliminates the chances of error, redesign and reduces the cost. Rayleigh s Method: This method is used only for simple problems involving 3 to 4 variables only and there is no calculation for dimension less groups. The dimensionless groups are formed directly. The power to maintain the rotation of cylinder is directly proportional to density ρ, kinematic viscosity v, and diameter D of cylinder. Copyright www.madeeasypublications.org

16 Mechanical Engineering Fluid Mechanics Solution : 3 V m = Lp ρp Vp = 451.89 m/ s L ρ Drag force F ρv L Drag on prototype F p = 1.095 400 = 437.86 N m m Solution : 4 The specific speed of a turbine is defined as the speed of operation of a geometrically similar model of the turbine which produces 1 kw power when operating under 1 m head. The expression for specific speed is derived as follows: We have N = f(ρ, gh, P, D) Wherein gh called shaft work is considered to be one variable only. Thus in all there are 5 variables involved in the phenomenon which can be completely described by 3 fundamental dimensions. As such the above functional relationship may be replaced by another one involving only dimensionless π-terms. Choosing ρ, gh and P as repeating variables, Thus substituting P = 1, H = 1 and N = N s for the model and assuming g and ρ to be same for the model and the prototype, we get N P N s = 5/ 4 H Solution : 5 Given: η is a function of ρ, μ, ω, D, Q η = f(ρ, μ, ω, D, Q) μ Q or η = f, 3 D ωρ D ω Solution : 6 Distorted Models: A model is said to be distorted if it is not geometrically similar to its prototype. For a distorted model different scale ratios for the linear dimensions are adopted. The models of river and harbours are made as distorted models as two different scale ratios, one for the horizontal dimensions and other for vertical dimensions are taken. If for the river and harbours, the horizontal and vertical scale ratios are taken to be same so that the model is undistorted, then the depth of water in the model of the river will be very-very small which may not be measured accurately. Therefore these are made as distorted models. Advantages: 1. The vertical dimensions of the model can be measured accurately.. The cost of the model can be reduced. 3. Turbulent flow in the model can be maintained. Disadvantages: 1. Due to different scales in the different directions, the velocity and pressure distribution in the model is not same as that in the prototype.. Waves are not simulated in distorted models. 3. The results of the distorted model cannot be directly transferred to its prototype. www.madeeasypublications.org Copyright

6 Flow Through Pipes 1. (d). (b) 3. (b) 4. (c) 5. (6.035)(6 to 6.10) 6. (0.151)(0.14 to 0.16) 7. (c) 8. (a) 9. (d) 10. (c) 11. (a) 1. (c) 13. (c) Copyright: Subject matter to MADE EASY Publications, New Delhi. No part of this book may be reproduced or utilised in any form without the written permission. 14. (c) 15. (4) 16. (4.8)(4.7 to 4.9) 17. (1.788)(1 to 13) 18. (0.3768)(0.31 to 0.33) 19. (d) 0. (b) 1. (c). (d) 3. (1.8)(1.81 to 1.83) Copyright www.madeeasypublications.org

18 Mechanical Engineering Fluid Mechanics Solution : 4 h s =.073 m Solution: 5 Consider a fluid (air) flowing in a pipe which has a sudden contraction in area 1 C PA 1 1 PA C 1 Eddies Formation (Turbulent Flow) Let the dynamic loss coefficient be K and coefficient of contraction be C C. K = Solution : 6 1 1 = 0.3756 0.6 = 0.3756 For maximum differential pressure, the ratio of the diameters is V corresponding loss of head, h e = 0.5 1 g and corresponding differential pressure head: ΔP = ρg 0.5 V 1 g D1 D = 1 Solution : 7 Solution : 8 p = 110.68 kpa τ = ρ ghf r0 = 183.486 N/m L π Q = 1 1 4 D V π = (0.06).4 = 6.848 lit/s 4 Solution : 9 Q = π 3 0.15 m /s 4 dv = www.madeeasypublications.org Copyright

Rank Improvement Workbook 19 Solution : 30 Solution : 31 Solution : 3 (h L ) 1 = 0.051 m; (h L ) 1-3 = 7.95 m p 1 = 190.65 kpa Q = 6.788 cumecs Q 3 = 18.70 cumecs Copyright www.madeeasypublications.org

7 Laminar & Turbulent Flow 1. (c). (a) 3. (b) 4. (d) 5. (d) 6. (c) 7. (b) 8. (d) 9. (a) 10. (c) 11. (6.4) 1. (.66) Copyright: Subject matter to MADE EASY Publications, New Delhi. No part of this book may be reproduced or utilised in any form without the written permission. 13. (c) 14. (d) 15. (a) 16. (16) 17. (0.00784)(0.00780 to 0.00785) 18. (1) 19. (d) 0. (c) 1 (c). (a) 3. (c) www.madeeasypublications.org Copyright

Rank Improvement Workbook 1 \ Solution : 4 Since ρ and Q are the same for both the pipes, the ratio of Cost of pumping 150 mm diameter pipe Cost of pumping 10 mm diameter pipe = hf 1 h f = 5 5 1 fd fd = 0.034 ( 10) 5 1 0.01448 ( 150) 5 = 5.8677 Solution : 5 Since Re < 000, flow is laminar. Pressure loss, Δp = 7019 10 5 N/m Maximum flow rate for laminar flow will be obtained for Re = 000, Solution : 6 V = 3 ν Re 0.7 10 000 = = 140 m/s d 0.01 Q max = 0.01099 m 3 /s = 10.99 l/s = 659.4 litres/minute π (i) Flow rate: Q = Au = D u = 0.003679 m 3 /s 4 The maximum velocity occurs at centre line of the pipe and it equals twice the average flow velocity. U max = u = 1.875 = 3.75 m/s (ii) Power required to maintain the flow in 100 m length of pipe: P P = 66. W (iii) Velocity at radius r is =.01 m/s and shear stress; τ = 153 N/m Solution : 7 μ = 0.11094 Ns/m Solution : 8 f = 0.017 Shear stress at the pipe surface is given by τ 0 = 110.065 N/m Shear velocity V* = 0.33 m/s v max = 7.64 m/s. Solution : 9 Distance from the boundary y = 0.93R Solution : 30 α = β = 4 1.33 3 = Copyright www.madeeasypublications.org

Mechanical Engineering Fluid Mechanics Solution : 31 The shear stress at the upper plate, τ = du μ dy = 108.7 N/m = y b and it must be resisting the plate motion. The discharge through the channel is given by Q = b udy 0 for no discharge, Ub 1 d = ( p +ρ ghb ) 3 1μ dx U = b d ( p +ρ gh ) 6μ dx = 0.183 m/s www.madeeasypublications.org Copyright

8 Boundary Layer Theory, Drag & Lift 1. (c). (b) 3. (c) 4. (d) 5. (c) 6. (c) 7. (a) 8. (d) 9. (c) 10. (b) 11. (b) 1. (c) Copyright: Subject matter to MADE EASY Publications, New Delhi. No part of this book may be reproduced or utilised in any form without the written permission. 13. (a) 14. (b) 15. (d) 16. (c) 17. (b) 18. (a) 19. (c) 0. (a) 1. (c). (b) Copyright www.madeeasypublications.org

4 Mechanical Engineering Fluid Mechanics Solution : 3 x = 0.03576 m or 35.76 mm To find maximum thickness of the boundary layer δ max = 4.37 10 4 m Solution : 4 When 0.3 m side is parallel to flow C D = 5.51 10 3 Total drag due to both sides F D =0.04 N When 1.0 m side is parallel to flow C D = 3.0 10 3 F D = 0.0133N Solution : 5 1. Blasius solution δ = 0.011376 m C D = 1.511 10 3. Approximate solution with assumption of cubic velocity profile δ = 0.010557 m C D = 1.469 10 3 Approximate solution deviate from the exact solution by = 7.% Boundary layer thickness =.78% Solution : 6 FD Therefore, 1 F D =.61 Solution : 7 Solution : 8 δ = 0.6 cm δ* = 0.0975 cm θ = 0.1393δ = 0.036 cm F = 0.766 N Solution : 9 δ = μx ρ U π 4 π and The displacement thickness, δ x = 1/ μπ 4.795 = 1/ (4 πρ ) U x Rex www.madeeasypublications.org Copyright

Rank Improvement Workbook 5 Shear stress, δ* = δ 1 = 0.363δ π θ = δ δ = 0.1366δ π τ w = μu 0.376 Re x 1/ x Solution : 30 L 1 L = 4 = 3 9 Copyright www.madeeasypublications.org

9 Hydraulic Machines 1. (d). (d) 3. (d) 4. (1.41)(1.30 to 1.50) 5. (66.57)(66 to 663) 6. (a) 7. (a) 8. (b) 9. (b) 10. (b) 11. (b) 1. (a) 13. (5.80) 14. (b) Copyright: Subject matter to MADE EASY Publications, New Delhi. No part of this book may be reproduced or utilised in any form without the written permission. 15. (109.5) (109 to 110) 16. (d) 17. (c) 18. (b) 19. (a) 0. (d) 1. (c). (c) 3. (d) 4. (b) 5. (d) 6. (d) www.madeeasypublications.org Copyright

Rank Improvement Workbook 7 Solution : 7 The maximum length of the draft tube = H s + h = 4.48 + 1 = 5.48 m Solution : 8 (a) Flow rate: Q = 0.4157 m 3 /s (b) Shaft power: Power developed by the runner = 131919.64 W = 131.919 kw or Shaft power: P = 1189.71 kw Solution : 9 (i) Guide vane angle at inlet: α 1 α i = 8.796 (ii) Runner vane angle at inlet: β i β i = 180 31.5 = 148.5 Solution : 30 Solution : 31 N = 303.5 rpm φ = 8.19 p A = 9810 0.841 = 850.1 N/m. p B = 9810.335 = 906.35 N/m =.906 kn/m =.90 kpa Solution : 3 Refer to figure to find the vane angles at the inlet and the outlet. u φ V r V = Vf V 1 V V r f 1 1 α = 30 θ u 1 (i) Vane angle at the inlet: Vane angle at the inlet, θ = 48.06 Relative velocity is given by V r 1 V w 1 = 40.4 m/s Copyright www.madeeasypublications.org

8 Mechanical Engineering Fluid Mechanics (ii) Vane angle at the outlet: Relative velocity of the jet and the vane at the outlet is V r = 34.357 m/s Vane angle at the outlet, φ = 43 18 Solutin : 33 To find diameter of wheel (D) D = 1.56 m To find diameter of the jet (d) d = 0.156 m To find the width of the bucket width = 0.78 m Depth of buckets = 0.187 m For number of buckets = 0 Solution : 34 (i) (ii) (iii) Solution : 35 If Discharge pressure N m = 1469.69 r.p.m. P m = 35.7 kw Q p = 69.4 m 3 /s. η m = 0.55 = 55.% P 1 = P atm = 1 bar P = P 1 + ρgh m = 4.78 bar Solution : 36 Q t = 4.4175 10 3 cumecs or 65.05 l/m Coefficient of discharge, C d = 0.943 Slip of the pump = 15.05 l/m Slip of the pump in percentage = 5.68% Solution : 37 Power required to drive the pump =.401 kw Solution : 38 Q t = 0.0044175 cumecs The percentage slip = 4.9% Theoretical power, P t = 1.083 kw Therefore, h ad = 0.75 m At the middle of the delivery stroke: Therefore, h ad = 0 Solution : 39 Speed of the pump without separation during suction stroke: N = 87.66 rpm Speed of the pump without separation during delivery stroke: But, N = 55.58 rpm www.madeeasypublications.org Copyright