Fluid Mechanics

March 20, 2018 | Author: Sushil Kumar Singh | Category: Fluid Dynamics, Boundary Layer, Shear Stress, Pump, Navier–Stokes Equations
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GATE-MEPrevious Years Solved Paper Fluid Mechanics www.gatescore.in GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER - ME YEAR 2013 Fluid Mechanics ONE MARK Q. 1 For steady, fully developed flow inside a straight pipe of diameter D , neglecting gravity effects, the pressure drop Dp over a length L and the wall shear stress tw are related by ∆pD ∆pD2 (A) τw = (B) τw = 4L 4L2 ∆pD 4∆pL (C) τw = (D) τw = 2L D Q. 2 In order to have maximum power from a Pelton turbine, the bucket speed must be (A) equal to the jet speed (B) equal to half of the jet speed. (C) equal to twice the jet speed (D) independent of the jet speed. YEAR 2013 two MARKs Q. 3 Water is coming out from a tap and falls vertically downwards. At the tap opening, the stream diameter is 20 mm with uniform velocity of 2 m/s. Acceleration due to gravity is 9.81 m/s2 . Assuming steady, inviscid flow, constant atmospheric pressure everywhere and neglecting curvature and surface tension effects, the diameter in mm of the stream 0.5 m below the tap is approximately. (A) 10 (B) 15 (C) 20 (D) 25 Q. 4 A hinged gate of length 5 m, inclined at 30c with the horizontal and with water mass on its left, is shown in the figure below. Density of water is 1000 kg/m3 . The minimum mass of the gate in kg per unit width (perpendicular to the plane of paper), required to keep it closed is (A) 5000 (B) 6600 (C) 7546 (D) 9623 YEAR 2012 Q. 5 ONE MARK Oil flows through a 200 mm diameter horizontal cast iron pipe (friction factor, f = 0.0225 ) of length 500 m. The volumetric flow rate is 0.2 m3 /s . The head loss (in m) due to friction is (assume g = 9.81 m/s2 ) (A) 116.18 (B) 0.116 www.gatescore.in GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER - ME Fluid Mechanics (C) 18.22 (D) 232.36 Q. 6 The velocity triangles at the inlet and exit of the rotor of a turbomachine are shown. V denotes the absolute velocity of the fluid, W denotes the relative velocity of the fluid and U denotes the blade velocity. Subscripts 1 and 2 refer to inlet and outlet respectively. If V2 = W1 and V1 = W2 , then the degree of reaction is (A) 0 (C) 0.5 (B) 1 (D) 0.25 YEAR 2012 TWO MARKS Q. 7 An incompressible fluid flows over a flat plate with zero pressure gradient. The boundary layer thickness is 1 mm at a location where the Reynolds number is 1000. If the velocity of the fluid alone is increased by a factor of 4, then the boundary layer thickness at the same location, in mm will be (A) 4 (B) 2 (C) 0.5 (D) 0.25 Q. 8 A large tank with a nozzle attached contains three immiscible, inviscide fluids as shown. Assuming that the change in h1, h2 and h 3 are negligible, the instantaneous discharge velocity is (A) (C) 2gh 3 c1 + 2g c r1 h1 r2 h2 (B) + r3 h 3 r3 h 3 m r1 h1 + r2 h2 + r3 h 3 m (D) r1 + r2 + r3 YEAR 2011 Q. 9 2g (h1 + h2 + h 3) 2g r1 h2 h 3 + r2 h 3 h1 + r3 h1 h2 r1 h1 + r2 h2 + r3 h 3 ONE MARK A streamline and an equipotential line in a flow field (A) are parallel to each other (B) are perpendicular to each other (C) intersect at an acute angle (D) are identical www.gatescore.in GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER - ME Fluid Mechanics YEAR 2011 Q. 10 Figure shows the schematic for the measurement of velocity of air (density = 1.2 kg/m3 ) through a constant area duct using a pitot tube and a water tube manometer. The differential head of water (density = 1000 kg/m3 ) in the two columns of the manometer is 10 mm. Take acceleration due to gravity as 9.8 m/s2 . The velocity of air in m/s is (A) 6.4 (C) 12.8 Q. 11 TWO MARKS (B) 9.0 (D) 25.6 A pump handing a liquid raises its pressure from 1 bar to 30 bar. Take the density of the liquid as 990 kg/m3 . The isentropic specific work done by the pump in kJ/ kg is (A) 0.10 (B) 0.30 (C) 2.50 (D) 2.93 YEAR 2010 ONE MARK Q. 12 For the stability of a floating body, under the influence of gravity alone, which of the following is TRUE ? (A) Metacenter should be below centre of gravity. (B) Metacenter should be above centre of gravity. (C) Metacenter and centre of gravity must lie on the same horizontal line. (D) Metacenter and centre of gravity must lie on the same vertical line. Q. 13 The maximum velocity of a one-dimensional incompressible fully developed viscous flow, between two fixed parallel plates, is 6 ms-1 . The mean velocity (in ms-1 ) of the flow is (A) 2 (B) 3 (C) 4 (D) 5 Q. 14 A phenomenon is modeled using n dimensional variables with k primary dimensions. The number of non-dimensional variables is (A) k (B) n (C) n - k (D) n + k Q. 15 A hydraulic turbine develops 1000 kW power for a head of 40 m. If the head is www.gatescore.in 4k Q. Neglecting frictional effects. At section S2 (elevation : 12 m) the pressure is 20 kPa and velocity is 2 ms-1 . Q-W. Weber number S.53 m (C) flow is from S1 to S2 and head loss is 1. 20 You are asked to evaluate assorted fluid flows for their suitability in a given laboratory application.4j (D) i . Nusselt number R.8 ms-2 . S-U. Density of water is 1000 kgm-3 and acceleration due to gravity is 9. Heat convection Y. Free surface flow V. Pipe flow X.ME Fluid Mechanics reduced to 20 m. The fluid has a vapour pressure of 50 kPa and a specific weight of 5 kN/m3 . S-U. The pressure in the pipe at section S1 (elevation : 10 m) is 50 kPa. R-Z. T-X YEAR 2009 (B) P-W. 18 Match the following P.38 Q. Compressible flow U. v = 0 www. Boundary layer flow W. Reynolds number Q.k (C) i . R-Z. R-V. Froude number T. T-V TWO MARKS Q. The following three flow choices. T-V (D) P-Y.j (B) 4i . Q-X. Q-W. Q-X. 1) is (A) 4i .05 (B) 0.06 m Q. Skin friction coefficient (A) P-U. P : u = 2y. 17 A smooth pipe of diameter 200 mm carries water.06 m (D) flow is from S2 to S1 and head loss is 1.53 m (B) flow is from S2 to S1 and head loss is 0.16 (C) 0. 16 Velocity vector of a flow field is given as V = 2xyi − x2 zj . Which of the following is TRUE (A) flow is from S1 to S2 and head loss is 0. The vorticity vector at (1.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . The pressure in the 20 cm pipe just upstream of the reducer is 150 kPa. T-W (C) P-Y. 1. Mach number Z. are made available. v =− 3x Q : u = 3xy. 19 Consider steady. the power developed (in kW) is (A) 177 (B) 354 (C) 500 (D) 707 YEAR 2010 TWO MARKS Q. incompressible and irrotational flow through a reducer in a horizontal pipe where the diameter is reduced from 20 cm to 10 cm. R-Z. the maximum discharge (in m3 /s) that can pass through the reducer without causing cavitation is (A) 0. S-Z.gatescore.27 (D) 0. expressed in terms of the two dimensional velocity fields in the xy -plane. S-U.in . 07 m3 /s . which one of the following is a necessary condition ? (A) steady flow (B) irrotational flow (C) inviscid flow (D) incompressible flow YEAR 2008 Q.ME Fluid Mechanics R : u =− 2x. per unit mass flow rate of the incoming jet.02 and density of water is 1000 kg/m3 . If value of Darcy friction factor for this pipe is 0.8 (B) 17.R b l 8µ dx 4µ dx 2 2 dp dp (C) .in . as shown in the figure.0 km long. The average velocity of fluid in the pipe is dx 2 2 dp dp (A) . such that each stream is deflected by 120c as shown in the figure.4 (C) 20.I.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . having a density of 1000 kg/m3 . 22 The velocity profile of a fully developed laminar flow in a straight circular pipe. The wheel rotates at 10 rad/s. pipe of 200 mm diameter at the rate of 0. 21 Water at 25c C is flowing through a 1.gatescore.0 Q. the pumping power (in kW) required to maintain the flow is (A) 1. Magnitude of the torque exerted by the water on the wheel. is www. 23 For the continuity equation given by d : V = 0 to be valid.R b l (B) .R b l (D) . 24 ONE MARK TWO MARKS Water. issues from a nozzle with a velocity of 10 m/s and the jet strikes a bucket mounted on a Pelton wheel.5 (D) 41. is given by the expression 2 2 dp u (r) =− R b lc1 − r 2 m 4µ dx R dp Where is a constant. The jet is split into two equal streams by the bucket. where V is the velocity vector. The mean diameter of the wheel is 1 m. G.R b l 2µ dx µ dx YEAR 2008 Q. v = 2y Which flow(s) should be recommended when the application requires the flow to be incompressible and irrotational ? (A) P and R (B) Q (C) Q and R (D) R Q. Friction in the bucket may be neglected. Q. the fluid contained between the two plates flows out radially.gatescore. then (A) F < 1/2 (B) F = 1/2 (C) F = 1 (D) F > 1 www. 25 The radial velocity Vr at any radius r . as shown in the figure. 26 The radial component of the fluid acceleration at r = R is 2 2 (B) V R (A) 3V 2R 4h 4h2 2 2 (C) V R (D) V h2 2 2h 2R YEAR 2007 Q. In the process. If F is the ratio of the drag force on the front half of the plate to the drag force on the rear half. The fluid is assumed to be incompressible and inviscid. when the gap width is h .in .ME (A) 0 (N-m)/(kg/s) (C) 2. aligned with the direction of an incoming uniform free stream.5 (N-m)/(kg/s) Common Fluid Mechanics (B) 1. 27 ONE MARK Consider an incompressible laminar boundary layer flow over a flat plate of length L.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . is (A) Vr = Vr (B) Vr = Vr 2h h (C) Vr = 2Vh (D) Vr = Vh r r Q.25 (N-m)/(kg/s) (D) 3. as the circular plate comes down at a uniform speed V towards the stationary bottom surface.75 (N-m)/(kg/s) Data For Q 25 and 26 The gap between a moving circular plate and a stationary surface is being continuously reduced. The blades are so shaped that the tangential component of velocity at blade outlet is zero. The diameter of the model is half of that of the full scale turbine. The blade efficiency of the runner is (A) 25% (B) 50% (C) 80% (D) 89% Q. R: Velocity is directly proportional to the radius from the center of the vortex.5u 0 4u 0 Q. If N is the RPM of the full scale turbine. The time required for a fluid particle on the axis to travel from the inlet to the exit plane of the nozzle is (A) L (B) L ln 4 u0 3u 0 (C) L (D) L 2. The flow velocity remains constant throughout the blade passage and is equal to half of the blade velocity at runner inlet. Priming S. the RPM of the model will be (A) N/4 (B) N/2 (C) N (D) 2N Q. the flow velocity on the nozzle axis is given by v = u 0 (1 + 3x/L). Surging R.ME Fluid Mechanics Q. 33 Match List-I with List-II and select the correct answer using the codes given below the lists : List-I List-II P. where x is the distance along the axis of the nozzle from its inlet plane and L is the length of the nozzle. 29 Consider steady laminar incompressible anti-symmetric fully developed viscous flow through a straight circular pipe of constant cross-sectional area at a Reynolds number of 5.in . The ratio of inertia force to viscous force on a fluid particle is (A) 5 (B) 1/5 (C) 0 (D) 3 YEAR 2007 TWO MARKS Q. 28 In a steady flow through a nozzle. 32 Which combination of the following statements about steady incompressible forced vortex flow is correct ? P: Shear stress is zero at all points in the flow. (A) P and Q (B) R and S (C) P and R (D) P and S Q. Pure impulse www.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . S: Total mechanical energy per unit mass is constant in the entire flow field. 31 A model of a hydraulic turbine is tested at a head of 1/4 th of that under which the full scale turbine works. Kaplan turbine 4. Centrifugal pump 2. Axial flow Q. Centrifugal compressor 1. Q: Vorticity is zero at all points in the flow.gatescore. Pelton wheel 3. 30 The inlet angle of runner blades of a Francis turbine is 90c. 34 The velocity profile is uniform with a value of U 0 at the inlet section A. The velocity profile at section B downstream is Z y 0#y#d ]Vm d . Q.ME Fluid Mechanics Codes : P Q R S (A) 2 3 4 1 (B) 2 3 1 4 (C) 3 4 1 2 (D) 1 2 3 4 Common Data For Q 34 and 35 Consider a steady incompressible flow through a channel as shown below.in ONE MARK . ] d # y # H−d u = [Vm. 35 YEAR 2006 Q. 36 For a Newtonian fluid (A) Shear stress is proportional to shear strain (B) Rate of shear stress is proportional to shear strain (C) Shear stress is proportional to rate of shear strain (D) Rate of shear stress is proportional to rate of shear strain www.pB The ratio A (where pA and pB are the pressures at section A and B ) 1 rU 2 0 2 respectively.(d/H) 1 + (d/H) p .gatescore.1 1 + ( d/H ) 1 ( 2 d / H ) 6 @ (A) Q.1 [ 1 ( d /H )] 2 1 d / H 8 ^ hB 1 1 (C) (D) 2 . and r is the density of the fluid) is 1 1 (A) (B) 2 . H−d # y # H d \ The ratio Vm /U 0 is 1 (B) 1 1 .2 (d/H) 1 1 (C) (D) 1 .GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . ] H−y ]Vm . gatescore. the power developed is (A) 7. If the jet deflection angle is 120c and the flow is ideal.34 (B) 4.0 kW (C) 22.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . 41 Q.1 m3 /s . 40 The velocity profile in fully developed laminar flow in a pipe of diameter D is given by u = u 0 (1 − 4r2 /D2). the velocity at point P in the siphon tube is Q.68 (C) 9. Assuming ideal fluid and the reservoir is large. 38 Fluid Mechanics In a two-dimensional velocity field with velocities u and v along the x and y directions respectively. a 1 : 4 scale model of the turbine operates under a head of 10 m. the convective acceleration along the x -direction is given by (B) u2u + v2v (A) u2v + v2u 2x 2y 2x 2y (C) u2u + v2u (D) v2u + u2u 2x 2y 2x 2y In a Pelton wheel.5 kW YEAR 2006 TWO MARKS Q.h1) (D) 2g (h2 + h1) A large hydraulic turbine is to generate 300 kW at 1000 rpm under a head of 40 m.75 www.38 (D) 18. The equation of stream line is (A) x2 y = constant (B) xy2 = constant (C) xy = constant (D) not possible to determine Q. 39 A two-dimensional flow field has velocities along the x and y directions given by u = x2 t and v =− 2xyt respectively. For initial testing. where r is the radial distance from the center. where t is time. 42 (A) 2gh1 (B) 2gh2 (C) 2g (h2 .in . 37 Q. the bucket peripheral speed is 10 m/s.5 kW (B) 15. The power generated by the model (in kW) will be (A) 2. If the viscosity of the fluid is µ.5 kW (D) 37.ME Q. the water jet velocity is 25 m/s and volumetric flow rate of the jet is 0. the pressure drop across a length L of the pipe is µu 0 L 4µu 0 L (A) (B) 2 D D2 8µu 0 L 16µu 0 L (C) (D) 2 D D2 A siphon draws water from a reservoir and discharge it out at atmospheric pressure. 44 The mass flow rate (in kg/s) across the section q .15 Q. where y is the height from plate.10 (D) 0.0 kg/m3 .33 (C) 0. and follows a linear velocity distribution. is (A) 0. at the section r-s.r is (A) zero (B) 0.in .ME Q. 45 The integrated drag force (in N) on the plate.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . Steam tables show saturation pressure at 65cC is 25 kPa. between p-s. Assume that the boundary layer is thin. Then 2u/2x is equal to (B) -2v (A) 2v 2x 2x (C) 2v (D) -2v 2y 2y www.001020 m3 /kg . Q. respectively.s is 10 mm. 46 The velocity components in the x and y directions of a two dimensional potential flow are u and v . and specific volume of the saturated liquid is 0. 43 Fluid Mechanics A horizontal-shaft centrifugal pump lifts water at 65cC . two-dimensional.67 (B) 0. The pump Net Positive Suction Head (NPSH) in meters is (A) 24 (C) 28 Common (B) 26 (D) 30 Data For Q 44 and 45 A smooth flat plate with a sharp leading edge is placed along a gas stream flowing at U = 10 m/s .17 (D) zero year 2005 one mark Q.gatescore. The thickness of the boundary layer at section r . The pressure at this point equals 200 kPa gauge and velocity is 3 m/s. u = U (y/d). The suction nozzle is one meter below pump center line.05 (C) 0. the breadth of the plate is 1 m (into the paper) and the density of the gas r = 1. What will be the distance of the leaf from the centre when it has moved through half a revolution ? (A) 48 m (B) 64 m (C) 120 m (D) 142 m year 2004 one mark Q.4 # 10-7 m2 /s .5 mm between these plates. 47 A venturimeter of 20 mm throat diameter is used to measure the velocity of water in a horizontal pipe of 40 mm diameter. 7.0 m/s Q.4 m/s (D) 2. At a given instant. specific gravity. 49 A B A B to B to A to B to A and and and and pA − pB = 20 kPa pA − pB = 1. the flow velocity is (A) 0.4 kPa pB − pA = 20 kPa pB − pA = 1. If the top plate is moved with a velocity of 0. If the pressure difference between the pipe and throat sections is found to be 30 kPa then.in .4 kPa A leaf is caught in a whirlpool.5 m/s while the bottom one is held stationary. The density of mercury is 13600 kg/m3 and g = 9.ME Fluid Mechanics year 2005 two marks Q. 0. 48 A U-tube manometer with a small quantity of mercury is used to measure the static pressure difference between two locations A and B in a conical section through which an incompressible fluid flows.651 (C) 6.88) is held between two parallel plates. the mercury column appears as shown in the figure.651 # 103 www.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .2 m/s (B) 1. the shear stress in Pascals on the surfaces of top plate is (A) 0. neglecting frictional losses. 50 An incompressible fluid (kinematic viscosity.0 m/s (C) 1.gatescore.651 # 10-3 (B) 0. the leaf is at a distance of 120 m from the centre of the whirlpool. the fluid attains a linear velocity profile in the gap of 0.51 (D) 0. Which of the following is correct ? (A) Flow direction is (B) Flow direction is (C) Flow direction is (D) Flow direction is Q. The whirlpool can be described by the following velocity distribution: 3 3 Vr =−b 60 # 10 l m/s and Vθ = 300 # 10 m/s 2pr 2πr Where r (in metres) is the distance from the centre of the whirlpool. At a particular flow rate.81 m/s2 . 2 m having Darcy’s friction factor of 0.9 to 8. then absolute discharge pressure at the pump exit is (A) 0.5 to 7. d = 4.911 bar (D) 55. then thrust at the bottom of the cylinder is ρω2 R2 (A) πR2 ρgH (B) πR2 4 (C) πR2 (ρω2 R2 + ρgH) (D) πR2 c Q.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .1 to 8.5 # 10-5 m2 /s and density of 1.203 bar www.6 # 10-3 N/m2 (C) 4. 53 Q.23 kg/m3 . as U = 3y − 1 y 3 . is (A) 2.36 # 10-3 N/m2 (D) 2.64x U3 2d 2 a d k Re x If the free stream velocity is 2 m/s.5 8.26 litres/sec For a fluid flow through a divergent pipe of length L having inlet and outlet radii of R1 and R2 respectively and a constant flow rate of Q .7 Frequency 1 5 Mean flow rate of the liquid is (A) 8. velocity (U) and boundary layer thickness (d) can be expressed respectively.18 # 10-3 N/m2 Q. and air has kinematic viscosity of 1.503 bar (C) 44. the wall shear stress at x = 1 m.gatescore.R2) 2Q2 (R1 .16 litres/sec Q. 56 A centrifugal pump is required to pump water to an open water tank situated 4 km away from the location of the pump through a pipe of diameter 0. 55 12 ρω2 R2 + ρgH m 4 For air flow over a flat plate. the acceleration at the exit is 2Q (R1 . If it is to maintain a constant head of 5 m in the tank.06 litres/sec (D) 8. 2) is (A) x − 2y = 0 (B) 2x + y = 0 (C) 2x − y = 0 (D) x + 2y = 0 year 2004 two marks Q.R2) 2Q2 (R2 .5 to 8.R1) (C) (D) p2 LR 25 p2 LR 25 A closed cylinder having a radius R and height H is filled with oil of density r .9 7.36 # 102 N/m2 (B) 43. 52 The following data about the flow of liquid was observed in a continuous chemical process plant : Flow Rate (litres / sec) 7. If the cylinder is rotated about its axis at an angular velocity of w.00 litres/sec (C) 8.3 to 8. The average speed of water in the pipe is 2 m/s.1 8. neglecting other minor losses. assuming the velocity to be axial and uniform at any cross-section.3 8. 51 Fluid Mechanics A fluid flow is represented by the velocity field V = axi + ayj .01. where a is a constant.449 bar (B) 5. The equation of stream line passing through a point (1.ME Q.7 to 7. 54 35 17 10 (B) 8.in .R2) (A) (B) pLR 23 pLR 23 2Q2 (R1 .7 7. Microhydel plant 3. Draft tube 5. height H and density rs .1 m3 /s respectively. 59 Match List-I with List-II and select the correct answer using the codes given below the lists : List-I List-II P. and tied to the bottom with a string. available head and flow rate are 24.0 revolution per second (rps) with an overall efficiency of 90%.gatescore. low pressure ratio 6. High flow rate.01 bar (C) 5. Reciprocating pump 1. the suitable type of turbine for this site is (A) Francis (B) Kaplan (C) Pelton (D) Propeller Q. The value of unknown pressure p is (A) 1. 60 A cylindrical body of cross-sectional area A.ME Q.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . If the turbine to be installed is required to run at 4. 58 At a hydro electric power plant site.in .00 bar (B) 2. Backward curved vanes 4. Plant with power output between 100 kW to 1 MW R. Axial flow pump 2. 57 Fluid Mechanics The pressure gauges G1 and G2 installed on the system show pressure of pG1 = 5.5 m and 10. Positive displacement S. is immersed to a depth h in a liquid of density r.01 bar Q.00 bar and pG2 = 1.00 bar.01 bar (D) 7. Centrifugal pump impeller Codes : P Q R S P Q R S (A) 3 5 6 2 (B) 3 5 2 6 (C) 3 5 1 6 (D) 4 5 1 6 year 2003 one mark Q. Plant with power output below 100 kW Q. The tension in the string is www. Q = 120 (C) H = 60. Impulse water turbine 6. the force F in newtons required on the plunger to push out the water is (A) 0 (B) 0.rs)ghA (D) (rh . Pressure compounding 5.13 (D) 1. Velocity compounding S Francis 4.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .15 www.in . Water from a tap is falling vertically into the container with a volume flow rate of Q . 63 Match List-I with the List-II and select the correct answer using the codes given below the lists : List-I List-II P Curtis 1. Reaction steam turbine Q Rateau 2.r)ghA (C) (r . 64 P Q R S (A) 2 1 1 6 (B) 3 1 5 7 (C) 1 3 1 5 (D) 3 4 7 6 Assuming ideal flow. the velocity of the water when it hits the water surface is U . then the head H in metres and flow rate Q in litres per minute at maximum efficiency are estimated to be (A) H = 60. Q = 30 Q. At a particular instant of time the total mass of the container and water is m . Axial turbine 7. If the rpm is changed to 1000.04 (C) 0. The force registered by the weighing balance at this instant of time is (A) mg + rQU (B) mg + 2rQU (C) mg + rQU 2 /2 (D) rQU 2 /2 Q. 61 A water container is kept on a weighing balance. Centrifugal pump Codes : Q. Q = 120 (B) H = 120. Q = 480 (D) H = 120. Mixed flow turbine 8. Gas turbine R Kaplan 3.gatescore.ME Fluid Mechanics (A) rghA (B) (rs .rs H) gA year 2003 two marks Q. 62 A centrifugal pump running at 500 rpm and at its maximum efficiency is delivering a head of 30 m at a flow rate of 60 litres per minute. 1m πD s2 Dt year 2002 one mark Q. Assuming incompressible frictionless flow.n (D) m/n Q. The spring constant is k .3 (D) 4. 67 If there are m physical quantities and n fundamental dimensions in a particular process.gatescore. 69 Flow separation in flow past a solid object is caused by (A) a reduction of pressure to vapour pressure (B) a negative pressure gradient (C) a positive pressure gradient (D) the boundary layer thickness reducing to zero www. Air density is r .13 (B) 0.ME Fluid Mechanics Q. the laminar boundary layer thickness varies as (A) 1 (B) x 4/5 x (C) x2 (D) x1/2 Q.in . throat diameter is Dt . 65 Neglect losses in the cylinder and assume fully developed laminar viscous flow throughout the needle. the number of non-dimentional parameters is (A) m + n (B) m # n (C) m . The bottom surface of the piston is exposed to atmosphere. 68 If x is the distance measured from the leading edge of a flat plate.16 (C) 0. 66 Air flows through a venturi and into atmosphere. the force F in newtons required on the plunger is (A) 0. atmospheric pressure is pa . the Darcy friction factor is 64/Re. Due to the flow.1m πD s2 Dt 4 (D) (ρU 2 /8k) c D 4 . exit diameter is D and exit velocity is U . the piston moves by distance x .4 Q.0 # 10-3 kg/s-m. Where Re is the Reynolds number.1m πD s2 Dt 2 (C) (ρU 2 /2k) c D 2 .GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . x is (A) (ρU 2 /2k) πD s2 2 (B) (ρU 2 /8k) c D 2 . The throat is connected to a cylinder containing a frictionless piston attached to a spring. Given that the viscosity of water is 1. ME Q.gatescore. 70 Fluid Mechanics The value of Biot number is very small (less than 0.48 (C) 24.1 (B) Biot number > 0.0248 (B) 2. are (A) Fx = rghrw and Fy = 0 (B) Fx = 2rghrw and Fy = 0 2 (C) Fx = rghrw and Fy = rgwr /2 (D) Fx = 2rghrw and Fy = πρgwr2 /2 www. dynamic viscosity = 0. having a width w into the plane of figure.1523 # 10−2 N−s/m2 and thermal conductivity = 8. specific heat at constant pressure = 0.540 W/m −K .1 (C) Fourier number < 0.1393 kJ/kg −K . density = 13529 kg/m3 . 72 The SI unit of kinematic viscosity (u) is (A) m2 /s (B) kg/m-s (C) m/s2 (D) m3 /s2 Q. 75 The horizontal and vertical hydrostatic forces Fx and Fy on the semi-circular gate.8 (D) 248 year 2001 one mark Q. 73 A static fluid can have (A) non-zero normal and shear stress (B) negative normal stress and zero shear stress (C) positive normal stress and zero shear stress (D) zero normal stress and non-zero shear stress Q.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .1 (D) Fourier number > 0. 74 Lumped heat transfer analysis of a solid object suddenly exposed to a fluid medium at a different temperature is valid when (A) Biot number < 0. The Prandtl number of the mercury at 300 K is (A) 0.in .1 year 2001 two marks Q.01) when (A) the convective resistance of the fluid is negligible (B) the conductive resistance of the fluid is negligible (C) the conductive resistance of the solid is negligible (D) None of the above year 2002 two marks Q. 71 The properties of mercury at 300 K are. 76 The two-dimensional flow with velocity v = (x + 2y + 2) i + (4 − y) j is (A) compressible and irrotational (B) compressible and not irrotational (C) incompressible and irrotational (D) incompressible and not irrotational Q.gatescore. Which one of the following relationships between the hydrodynamic boundary layer thickness (d) and the thermal boundary layer thickness (dt) is true? (A) dt > d (B) dt < d (C) dt = d (D) cannot be predicted ********** www.ME Fluid Mechanics Q.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .in . 77 Water (Prandtl number = 6 ) flows over a flat plate which is heated over the entire length. 2 Sol.V 1 = Z − Z or 1 2 ^ p 1 = p 2h 2g or Z1 − Z2 = 0. 1 Sol. www. Applying the bernoulli’s equation at the tap opening and the 0. For steady.V 12 = ^Z1 − Z2h 2g or V 22 = 2 # 9.72 m/ sec Now applying the continuity equation A1 V1 = A2 V2 p d 2 V = p d 22 V2 4 1 1 4 or d 22 = V1 d 12 = 2 # ^20h2 3. where m is the mass of the gate. 3 Option (A) is correct. Option (B) is correct. fully developed flow inside a straight pipe.5 m V 22 .72 V2 d2 = 15 mm Sol. The force imposed by the jet on the runner is equal but opposite to the rate of momentum change of the fluid. the pressure drop and wall shear stress are related by Dp = 4Ltw D DpD or tw = 4L Option (B) is correct.gatescore. Power P = Fu = 2rQ ^Vi − u h u For maximum power dP = 2rQ V − 2u = 0 ^ i h du or u = Vi 2rQ ! 0 2 Hence bucket speed ^u h must be equal to half of the jet speed. 4 V1 = 2 m/ sec Option (D) is correct.ME Fluid Mechanics SOLUTION Sol.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .5 + ^2 h2 or V2 = 3.81 # 0. Here mg shows the weight of the gate.in .5 m below the tap 2 p1 V 12 p + + Z1 = 2 + V 2 + Z 2 pg 2g pg 2g 2 2 V 2 . F =− m ^Vf − Vi h =− rQ 6^− Vi + 2u h − Vi@ =− rQ ^− 2Vi + 2u h = 2rQ ^Vi − u h where u is the bucket speed and Vi is the jet speed. f = 0.0225 # 500 # 0.33 m .2 V = ν = = 6. W1 = V2 (V12 − V22) Hence R = 1 − 2 (V1 − V22) + (U 2 − U 2) + (V12 − V22) (V12 − V22) = 1− = 1 − 1 = 0. 7 R = 1 − Option (C) is correct.2 2 # 9. h = 0.(1) 2g D Given that h = 500 m..2 m .0225 1000 Since volumetric flow rate no = Area # velocity of flow (V) o 0. From Darcy Weischback equation head loss 2 h = f # L # V .116. 6 h = 116.91x = 4.37) 2 Hence.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .5 2 2 (V12 − V22) Sol. D = 200 = 0.2) 2 # 4 (6. 5 Option (A) is correct.in . Boundary layer thickness 1/2 d (x) = 4.gatescore. For flat plate with zero pressure gradient and Re = 1000 (laminar flow).37 m/s π Area (0. Degree of reaction (V12 − V22) (V12 − V22) + (U12 − U22) + (W22 − W12) where V1 and V2 are absolute velocities W1 and W2 are relative velocities U1 and U2 = U for given figure Given W2 = V1..ME Fluid Mechanics In equilibrium condition Torque due to pressure of water = Torque due to weight of the plate Now Torque due to pressure at distance s for infinitesimal length (pressure force acts normal to the surface) T0 = ^pgy ds h s = ρgs2 sin θds = ρg sin θs2 ds Torque due to weight of the gate is = mg # L cos q 2 L Thus ρg sin θs2 ds = mg # L cos q 2 0 2 2ρL tan θ or m = = 2 # 103 # ^5 h2 # tan 30c = 9623 kg 3 3 # Sol.91x = 4.18 m Option (C) is correct.81 Sol.91x Vx V Re x u u 1/2 x For a same location (x = 1) & d \ 1/2 V d \ (V ) -1/2 where V = velocity of fluid www. dy For Equipotential line. Option (C) is correct.1.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . 9 Sol.(i) v dx For stream function. =− u = Slope of equipotential line .1 + 1 1 + 2 2 E r3 r3 r3 h 3 r3 h 3 Option (B) is correct.gatescore.32 = 12.. g = 9.G \ Density of Liquid V = 2gh = 2 # 9. a = 10 # 10−3 :1000 − 1D = 8. x = 10 # 10−3 m .5 2 4 Option (A) is correct.. Given : p1 = 1 bar . 1 1 + 2 2 + h 3E = 2gh 3 # .2 kg/m3 . rw = 1000 kg/m3 . dy = v = Slope of stream line . 10 V2 = rh rh rh rh = 2g # .u # v =− 1 v u And.. + V 2 p1 + r1 gh1 + r2 gh2 + r3 gh 3 + 1 1 2g 2g At Reference level (2) z2 = 0 and V1 = 0 at point (1) Therefore 2 & p1 + r1 gh1 + r1 gh2 + r3 gh 3 = patm.(1) 2g Sol. 1 1 + 2 2 + h 3E r3 g r3 g Therefore Sol. when m1 m2 =− 1. r = 990 kg/m3 www. w − 1 = x rw − 1 h = x :SG : ra D D SG .ME Fluid Mechanics d1 = V1 −1/2 d2 bV2 l 1/2 1/2 d2 = bV1 l # d1 = b V1 l # 1 V2 = 4V1 (Given) V2 4V1 1/2 = b 1 l # 1 = 1 = 0.8 # 8.32 m 1. therefore the stream line and an equipotential line in a flow field are perpendicular to each other. Then .2 Weight density of liquid Where S. 2g # [r1 gh1 + r2 gh2 + r3 gh 3] r gh r gh = 2g # .(ii) u dx It is clear from equation (i) and (ii) that the product of slope of equipotential line and slope of the stream line at the point of intersection is equal to . p2 = 30 bar . 11 S.8 m/ sec2 If the difference of pressure head ‘h ’ is measured by knowing the difference of the level of the manometer liquid say x . + V 2 .. Takes point (1) at top and point (2) at bottom By Bernoulli equation between (1) and (2) 2 V 2 (p + p2 + p 3) = patm. .G = Weight density of water Velocity of air Sol. Given : pa = 1. Then lines are perpendicular.8 m/ sec Option (D) is correct...in . 8 Since Hence p1 = atmospheric pressure (because tank is open) p1 = p atm. from the unit power. when flow is fully developed. Stable equilibrium occurs when M is above G .gatescore. these two forces (Gravity force and Buoyant force) create a restoring moment and return the body to the original position.e. and thus GM is positive. Here n = dimensional variables k = Primary dimensions (M. A measure of stability for floating bodies is the metacentric height GM . T) So. and unstable if point M is below point G .93 kJ/kg Sol. As shown in figure above. T). H1 = 40 m . the ratio of Vmax and Vavg is a constant. and thus GM is negative. W = ndp = m dp r W = 1 dp = 1 (30 − 1) # 105 pascal r 990 # m ν=m ρ = 2929. non dimensional variables.in . 14 Sol. 15 Option (C) is correct. In case of two parallel plates.29 J/kg = 2. Vmax = 3 Vmax = 6 m/ sec 2 Vavg Vavg = 2 # Vmax = 2 # 6 = 4 m/ sec 3 3 Option (C) is correct. If point Bl is sufficiently far from B .ME Fluid Mechanics Isentropic work down by the pump is given by. Sol. From Buckingham’s p-theorem It states “If there are n variable (Independent and dependent variables) in a physical phenomenon and if these variables contain m fundamental dimensions (M. Given : P1 = 103 kW . which is the distance between the centre of gravity G and the metacenter M (the intersection point of the lines of action of the buoyant force through the body before and after rotation. the behavior of turbine can be easily known from the values of unit quantities i.m) dimensionless terms. & n . 12 Option (B) is correct.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . 13 Sol. H2 = 40 − 20 = 20 m If a turbine is working under different heads. www.k Option (B) is correct. then variables are arranged into (n . L.) A floating body is stable if point M is above the point G . L. 1). 1. 2 p1 V 12 p + + z 1 = 2 + V 2 + z 2 + h L r g 2g rg 2g p1 p From equation (i) + z 1 = 2 + z 2 + h L rg rg h L = b 3 ^50 − 20h # 10 p1 − p 2 + ( − ) = + (10 − 12) z z 1 2 rg l (1000 # 9.06 m Head at section (S1) is given by. A1 V1 = A2 V2 V1 = V2 D1 = D2 so A1 = A2 .. r = 1000 kg/m3 .6 .09 m rg 10 # 9.8 www. V2 = 2 m/ sec . = 2x 2y 2z 2xy − x2 z 0 = i :− 2 ^− x2 z hD − j :− 2 (2xy)D + k . 17 Option (C) is correct. Given : V = 2xyi − x2 zj P (1. 3 p H1 = 1 + Z1 = 503# 10 + 10 = 15. g = 9.gatescore. Given : p1 = 50 kPa . w = 0 i j k 2 2 2 So. Z2 = 12 m .058 − 2 = 1..in .GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . u = 2xy .8) = 3. i j k 2 2 2 Vorticity Vector = 2x 2y 2z u v w Substitute. 1. = i + k [− 2 − 2]= i − 4k Sol. 16 Fluid Mechanics Pu = P3/2 H P1 = P2 H 13/2 H 23/2 3/2 3/2 P2 = b H2 l # P1 = b 20 l # 1000 = 353. p2 = 20 kPa . v =− x2 z . Z1 = 10 m . 354 kW 40 H1 Option (D) is correct. 2 (− x2 z) − 2 (2xy)E 2z 2z 2x 2y = x2 i − 0 + k [− 2xz − 2x] Vorticity vector at P (1.ME So Sol. (i) Applying Bernoulli’s equation at section S1 and S2 with head loss hL .8 m/ sec2 Applying continuity equation at section S1 and S2 . 1) The vorticity vector is defined as. w = 5 kN/m3 = rg Consider steady. Nusselt number So. Here type of flow is related to the dimensionless numbers (Non-dimensional numbers). Mach number Q. incompressible and irrotational flow and neglecting frictional effect. So. For maximum discharge pV = p2 = 50 kPa Applying Bernoulli’s equation at point (1) and (2) 2 p1 V 12 p + + z 1 = 2 + V 2 + z 2 rg 2g rg 2g Here z1 = z2 for horizontal pipe and w = rg = 5 kN/m2 2 150 + V 12 = 50 + (4V1) From equation (i) V2 = 4V1 5 2g 5 2g www. A1 V1 = A2 V2 p (d ) 2 V = p (d ) 2 V 4 1 # 1 4 2 # 2 Substitute the values of d1 and d2 .GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . Heat convection V.in . Pipe flow U. First of all applying continuity equation at section (1) and (2). Boundary layer Z. T-V Sol.04 m rg 10 # 9. S-U..8 From H1 and H2 we get H1 > H2 . 3 p2 + Z2 = 203# 10 + 12 = 14.(i) 400V1 = 100V2 Cavitation is the phenomenon of formation of vapor bubbles of a flowing liquid in a region where the pressure of liquid falls below the vapor pressure [pL < pV ] So. 18 Option (D) is correct. Compressible flow Y. R-Z. we can say that maximum pressure in downstream of reducer should be equal or greater than the vapor pressure. So P. correct pairs are P-Y. Given : pV = 50 kPa . 19 Option (B) is correct. Skin friction coefficient S. flow is from S1 to S2 H 2 = Sol. Reynolds number T. we get p (20) 2 V = p (10) 2 V # 1 # 2 4 4 & V2 = 4V1 .gatescore. Q-W.ME Fluid Mechanics Head at section S2 . Free surface flow W. Weber number R.. .114 m/sec 15 Sol..3 .V = 0 R : u =− 2x.50 = 16V 12 − V 12 2g 2g 5 5 2 20 = 15V 1 2g V 12 = 40 # 9.5 ! 0 2u = 3y . 2u = 3x 2x 2y 2v = 0 .2u = 0 2 c2x 2y m 2v . 2y 2x 2u + 2v = 0 (Flow is incompressible) & 0 + 0 = 0 2x 2y 2v . Q max = A2 V2 = p (d2) 2 V2 = p (10 # 10−2) 2 # 20.V = 2y For incompressible fluid.ME 150 . Q and R 2u = 0 . 2u = 0 2x 2y 2v = 2 . z z = 1 c2v − 2u m 2 2x 2y 1 2v .46 = 0. 2u = 2 For P : u = 2y . 2x 2y 2v = 0 . 2u + 2v + 2w = 0 .16 m3 / sec 4 Option (D) is correct.46 m/ sec Maximum discharge..2 = 0 For Q : u = 3xy v = 0 2u + 2v = 0 2x 2y 2v .46 4 4 = p # 10−2 # 20.gatescore.(i) 2x 2y 2z For irrotational flow z z = 0 .2u = 0 2x 2y Or. Given : P : u = 2y.V =− 3x Q : u = 3xy. check P. 2v = 0 2y 2x (Rotational flow) (Compressible flow) (Rotational flow) .(ii) 2x 2y From equation (i) and (ii).2u = 0 .in & . 2v = 0 2y 2x & 3y = Y 0 & − 3x = Y 0 2u =− 2 .114 = 20.2u = 0 Or.. 0 . 2v =− 3 v =− 3x .3x = 0 For R : u =− 2x v = 2y www.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .81 = 5. 2x 2y . 20 Fluid Mechanics And V2 = 4V1 = 4 # 5. ME Fluid Mechanics 2u + 2v = 0 2x 2y − 2 + 2 = 0 2v . D = 200 mm = 0.07) = # 2 # (3. 22 Option (A) is correct.2) 5 # (9.61 m of water Pumping power required. 21 Option (A) is correct. 2 fLV 2 fL 16fLQ2 8fLQ2 4Q 2 Q = pD # V h f = = = = 2 c m 2 5 2 5 4 D # 2g D # 2g p D p D # 2g pDg 2 8 0. the velocity profile in fully developed laminar flow in a pipe is parabolic with a maximum at the center line and minimum at the pipe wall.R − R 2 E 2µ dx 2 2µ dx 2 4R 0 4R 2 2 dp dp =− 1 b l # R =− R b l 2µ dx 8µ dx 4 # Alternate Method : Now we consider a small element (ring) of pipe with thickness dr and radius r . We find the flow rate through this elementary ring.r − r 2 E =− 1 b l.81) = 0.07 M3 / sec f = 0. 0. 2 2 dp u (r) =− R b lc1 − r 2 m 4µ dx R Therefore.2 m .gatescore.2u = 0 2x 2y Or. 2 2 R R dp Vavg = u (r) rdr =− 22 # R b lc1 − r 2 m rdr R 0 4µ dx R 0 3 dp R =− 1 b l # cr − r 2 m dr 2µ dx 0 R R 2 4 2 4 dp dp =− 1 b l.07 # 2.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .30 P = rgQ # h f = 1000 # 9. The average velocity is determined from its definition. we can easily see that R is incompressible and irrotational flow.81 # 0. r = 1000 kg/m3 Head loss is given by. & 0 = 0 (Incompressible flow) (Irrotational flow) & 0 = 0 0 .752 kW . Sol. Given : L = 1 km = 1000 m .0 = 0 So.in . Put the value of u (r) dQ = (2pr) # dr # u (r) 2 2 dp dQ = (2πr) # dr # c− R mb lc1 − r 2 m 4µ dx R www.02 .61 = 1752.784 = 2.8 kW Sol. 1. Q = 0.14) # (0.02 1000 # (0.287 = 1. R = 0. q = 180 − 120 = 60c.R − R 2 E =− 2π R b l:R − R D 4µ dx 2 4 4µ dx 2 4R 2 2 4 dp dp =− 2π c R mb l:R D =− πR b l 8µ dx 4µ dx 4 So Now Sol. we have 2 2 4 2 2 2 dp dp Q =− 2π R b l. 4 2 Q dp dp Vavg. 2 (ru) + 2 (rv) + 2 (rw) = 0 2y 2z 2x For incompressible flow r =Constant r . Force exerted on the bucket Fx = ρAV 6V − (− V cos θ)@ = ρAV 61 + cos θ@ V Mass flow rate Q = rAV = Q (1 + cos q) V Torque. 2 − 4R2 E 0 Now put the limits. Given : r = 1000 kg/m3 . the above equation represents the incompressible flow.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . Sol. Q & 0 to Q and R & 0 to R dp # dQ =− 2π 4Rµ b dx l# 2 2 r c1 − r 2 m dr R 0 0 2 2 4 R R dp r r Q 6Q@0 =− 2π 4µ b dx l.5 # Q = 7.ME Fluid Mechanics Now for total discharge integrate both the rides within limit. 24 None of these is correct.5 m Initial velocity in the direction of jet = V Final velocity in the direction of the jet =− V cos q .2u + 2v + 2w E = 0 2x 2y 2z 2u + 2v + 2w = 0 2x 2y 2z d : V =0 So. 23 Q R Q = Area # Average velocity = A # Vavg. Tx = Fx # R = QV (1 + cos q) R Torque per unit mass flow rate Tx = V (1 + cos q) R = 10 (1 + cos 60c) 0.gatescore. = = − πR b l # 1 2 =− R b l 8µ dx 8µ dx A πR Option (D) is correct.5 N − m/kg/ sec www.in . The continuity equation in three dimension is given by. V = 10 m/sec . 27 Fy = ρAV (0 − V sin θ) =− QV sin θ T = T x2 + T y2 = Tx = 7..(i) Volume of fluid which flows out at radius r = Vr # 2pr # h . 26 Ty = Fy # R = 0 R=0 Option (A) is correct.(ii) Equating equation (i) and (ii). V # pr2 = Vr # 2prh Vr = 2Vr h & Vr = Vr 2h Alternate Method : Apply continuity equation at point (i) and (ii). Vr = Vr 2h Acceleration at radius r is given by ar = Vr # dVr = Vr # d :Vr D = Vr # V .5 N − m/kg/ sec Sol. A1 V1 = A2 V2 V # pr2 = Vr # 2prh Vr = Vr 2h Option (B) is correct. so we can say that Volume of fluid moving out radially = Volume of fluid displaced by moving plate within radius r Volume displaced by the moving plate = Velocity of moving plate # Area = V # pr2 . From previous part of question.. 25 Fluid Mechanics Option (D) is correct...in rAV 2 rAV 2 = 1. FD = CD # www.33 Re L ..GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .33 # 2 2 Re L CD = 1.ME And Torque in y -direction Total Torque will be Sol.gatescore. Here Gap between moving and stationary plates are continuously reduced.(i) 2h dr dr 2h 2 ar = VR # V = V R 2h 2h 4h2 At r = R Sol.. GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER ..gatescore.(i) = 1.F. 28 Option (B) is correct. Given figure shows the velocity triangle for the pelton wheel.33 ρV # 2 # µ µ 1 ρbV 2 L . F = FD/2 = FlD/2 _1 − FD 2 1 2 i FD = 1 >1 2 −1 v = u 0 b1 + 3x l L dx = u 1 + 3x = u 0 (L + 3x) 0b L Ll dt 1 dt = L # dx u 0 (L + 3x) On integrating both the sides within limits t & 0 to t and x & 0 to L .ME Fluid Mechanics ρVL 1 ρ bLV 2 Re L = = 1...33 ρV # 2 µ So from equation (i) FD \ L . 30 Option (A) is correct. FlD/2 = FD − FD/2 = c1 − 1 m FD 2 Now ratio of FD/2 and FlD/2 is Sol. we get L t #0 dt = uL0 #0 (L +1 3x) dx L L 6t @t0 = 3u 0 6ln (L + 3x)@ 0 t = L 6ln 4L − ln L@ = L ln 4 3u 0 3u 0 Given : Sol. www.in . 29 Sol.(ii) Drag force on front half of plate From Equation (ii) FD/2 = L = FD 2 2 Drag on rear half. Viscous force µ# #A L Option (C) is correct.. ρAV 2 ρVL Re = = Inertia force = = 5 = I.F. µ V V. For forced Vortex flow the relation is given by.ME Fluid Mechanics Given : Flow velocity at Inlet Vf 1 = flow velocity at outlet Vf 2 Vf1 = Vf 2 = u1 (blade velocity) 2 V2 = Vf2 u1 = Vw1 From Inlet triangle. V = rw . 31 q = 90c Option (C) is correct. S-1 www. 2gH H \ DN H = Constant DN So using this relation for the given model or prototype. . List-I List-II P.ρg (z . Surging Q. Sol. Dm = 1 D p .z ) = 0 2 1 2 1 2 & DK. Centrifugal compressor 2. H H c DN m = c DN m p m Hp Np Dm .. Centrifugal pump 3.r 2) .E. Kaplan Turbine 1.in .E. Axial Flow So. 33 Option (A) is correct. Pure Impulse S.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . = 0 Now total mechanical energy per unit mass is constant in the entire flow field.E = D P. D K.gatescore. N m = N Sol. correct pairs are P-2. 32 Option (B) is correct. N p = N 2 4 1D Hp N 2 p = 4#1 =1 = # 2 1H Nm Dp 4 p So. Priming R. 2 V 12 = (Vf1) 2 + (Vw 1) 2 = a u1 k + (u1) 2 = 5 u 12 2 4 5 2 2 2 u − 1 u2 Blade efficiency = V 1 −2V 2 # 100 = 4 15 24 1 # 100 = 80% V1 4 u1 Sol. Q-3.(i) From equation (i) it is shown easily that velocity is directly proportional to the radius from the centre of the vortex (Radius of fluid particle from the axis of rotation) And also for forced vortex flow.E.. u = pDN = 60 From this equation. Pelton wheel 4. R-4... 1 ρω2 (r 2 .D P.(i) = Hm # D p Nm Given : Hm = 1 H p . 34 Fluid Mechanics Option (C) is correct.in .pB V m2 Vm 2 2 − 1 = bU l − 1 2 = 1 0 U0 2 rU 0 Vm = 1 From previous part of question U0 1 − Hd pA . Let width of the channel = b From mass conservation Flow rate at section A = flow rate at B or Velocity A # Area of A = Velocity at B # Area of B U 0 # (H # b) = Velocity for (0 # y # d) # dy # b + velocity for (d # y # H − d) # dy # b + velocity for (H − d # y # H) # dy # b dy H−d H H−y or dy + Vm dy + Vm dy U 0 # H = Vm d 0 d d H−d or U 0 # H = Vm d + Vm (H − 2d) + Vm d 2 2 # # U 0 # H = Vm d + Vm (H − 2d) = Vm (d + H − 2d) Vm = H = H = 1 d U0 d + H − 2d H−d 1− H or Sol. Applying Bernoulli’s Equation at the section A and B . 36 From the Newton’s law of Viscosity. www.ME Sol. pA . the shear stress (t) is directly proportional to the rate of shear strain (du/dy).V m2 − 1E U0 = 2 2 Substitute.pB V B2 − V A2 = 2g rg pA . pA .pB V B2 − V A2 V m2 − U 02 = = 2 2 r VB = Vm and VA = U 0 U 02 .pB 1 2 −1 2 = 1 61 − d/H @ 2 rU 0 Option (C) is correct. Sol.gatescore. 35 # Option (A) is correct. 2 pA V A2 p + + z A = B + V B + z B rg 2g rg 2g Here.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . zA = zB = 0 So. 1 m3 / sec Jet deflection angle = 120c C f = 180c − 120c = 60c rQ [Vw1 + Vw2] # u P = kW . So. Convective acceleration along x -direction.(i) 1000 From velocity triangle..(ii) 2y 2 Sol.. Given : u = u1 = u2 = 10 m/ sec . 37 Sol..in 2 1 .ME Fluid Mechanics t \ du = µdu dy dy Where µ= Constant of proportionality and it is known as coefficient of Viscosity. Hence for given flow. called convective acceleration.5 kW 1000 Option (D) is correct. Vw1 = V1 = 25 m/ sec Vw = Vr cos f − u2 Vr 2 = Vr = V1 − u1 = 15 cos 60c − 10 = 25 − 10 = 15 m/ sec = 15 − 10 =− 2. ax = 2u + u2u + v2u + w2u 2t 2x 2y 2z In above equation term 2u/2t is known as local acceleration and terms other then this.. ax = u2u + v2u [w = 0] 2x 2y Option (C) is correct..GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . V1 = 25 m/ sec . v =− 2xyt The velocity component in terms of stream function are 2y = v =− 2xyt .1 [25 − 2. In Cartesian coordinates. Q = 0.. the components of the acceleration vector along the x -direction is given by. Sol. Convective Acceleration is defined as the rate of change of velocity due to the change of position of fluid particles in a fluid flow. 39 www.5] # 10 P = kW = 22. 38 Option (C) is correct.(i) 2x 2y =− u =− x2 t . The velocity triangle for the pelton wheel is given below.5 m/ sec 2 Now put there values in equation (i) 1000 # 0. Given : u = x2 t .gatescore. it is not possible to determine equation of stream line. there is no condition for t is constant.ME Fluid Mechanics Integrating equation (i).t y . Given : Dp = p1 − p2 = r 32µuL . K is a constant of integration which is independent of ‘x ’ but can be a function of ‘y ’ Differentiate equation (iii) w.r. # # www. w..(iii) Where..(i) 2 D (From the Hagen poiseuille formula) Where ur = average velocity 2 R R And ur = 22 u (r) rdr = 22 uo c1 − r 2 m rdr R 0 R 0 R 2 4 R 3 R = 2u2o r − r 2 m dr = 2u2o . − x2 yt + K1 = 0 K1 = x2 yt If ‘t ’ is constant then equation of stream line is. we get 2y =− x2 t + 2K 2y 2y 2y But from equation (ii). x 2 y = K1 = K 2 t But in the question.in # .r − r 2 E c R R 2 4R 0 R 0 R 2 2 4 = 2u2o .t ‘x ’.gatescore. So. Dp = 2 # 2 = D D2 Note : The average velocity in fully developed laminar pipe flow is one-half of the maximum velocity.R − R 2 E = 2u2o :R D 4R 0 R 2 R 4 u = uo 2 Substitute the value of u in equation(1) 32µL uo 16 µuo L So. =− x2 t 2y 2y Comparing the value of .. we get # y = (− 2xyt) dx =− x2 yt + K . Sol. Hence.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . we get 2y − x2 t + 2K =− x2 t 2y 2K = 0 2y K = Constant(K1) From equation (iii) y =− x2 yt + K1 The line for which stream function y is zero called as stream line. 2 2 u = uo c1 − 4r2 m = uo c1 − r 2 m D R Drop of pressure for a given length (L)of a pipe is given by. 40 Option (D) is correct.r. 95 m of water 1. NPSH = 22.34 4 40 d1 H1 Option (A) is correct. Given : P1 = 300 kW . H = 10 m 2 d1 4 Specific power for similar turbine is same.24 m of water www. the total energy at any point of the fluid is constant. (NPSH) = Pressure head + static head Pressure difference. V 22 = 2g (h2 − h1) V2 = 2g (h2 − h1) So.h1 = V 2 2g And So.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . H1 = 40 m d2 = 1 .in . 43 2g (h2 − h1) Option (A) is correct.33 m = 22.25 # 10. 42 Sol. Net positive suction head. P1 = P2 2 d 1 H 13/2 d 22 H 23/2 2 3/2 2 3/2 P2 = b d2 l b H2 l # P1 = b 1 l b 10 l # 300 = 2.95 .gatescore. N1 = 1000 rpm . we have P = Constant 2 d H 3/2 For both the cases. So from the relation.013 Static head = 1 m (Given) Now.ME Sol. So applying the Bernoulli’s Equation at section (1) and (2) 2 p1 V 12 p + + Z1 = 2 + V 2 + Z 2 rg 2g rg 2g V1 = 0 = Initial velocity at point (1) Z2 = 0 = At the bottom surface p1 = p2 = patm z1 = h 2 − h1 2 h 2 . 41 Fluid Mechanics Option (C) is correct.25 bar = 2. In a steady and ideal flow of incompressible fluid. Dp = 200 − (− 25)= 225 kPa (Negative sign shows that the pressure acts on liquid in opposite direction) Dp = 225 # 103 Pa = 2. velocity of fluid is same inside the tube Vp = V2 = Sol.95 + 1 = 23. ME Sol.1 − 0. integrating both sides within the limits.05 kg/ sec oqr = m m Option (D) is correct. 44 Fluid Mechanics Option (B) is correct. Von Karman momentum Integral equation for boundary layer flows is.y E δ 2 0 2 o = ρUB # δ = 1 ρUBδ m 2 2 δ o rs = 1 # 10−2 # 10 # 1 # 1 = 5 # 10−2 = 0.0 kg/m3 .1 kg/ sec m For mass flow through section r . 45 o pq − m o rs = 0. d = 10 mm = 10−2 meter .05 kg/ sec m 2 So. Then Mass flow rate through this element.s . a d 2x 0 d 2x # www. dm & 0 to m y & 0 to d δ m dy #0 dmo = #0 yb ρUB δ l 2 δ 6mo @m0 = ρUB . o = r # Volume = r # (A # u) dm y = ρ # u # B # (dy) = ρBU a k dy δ For total Mass leaving from rs .in # 0 d y y2 c d − d2 m dyG u =y U d . τo = 2 ρU 2 2x .gatescore. y B = 1 m and u = U a k d From the figure we easily find that mass entering from the side qp = Mass leaving from the side qr + Mass Leaving from the side rs m pq = (m pq − mrs) + mrs So. Mass leaving from qr Sol. r = 1. # 0 d u 1 − u dy Uk E Ua dy y 1 − k dyE = 2 = = 2. we have to take small element of dy thickness.05 = 0. τo = 2q ρU 2 2x d u 1 − u dy and q = momentum thickness = 9 UC U 0 # So.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . Given : U = 10 m/ sec . we get o pq = 1 # (1 # 10−2) # 10 = 0. firstly Mass flow rate entering from the side pq is o pq = r # Volume = r # (A # U) = 1 # (B # d) # U m Substitute the values. GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .p2 V 22 − V 12 = 2g rg ∆p V 22 − V 12 = 2 ρ 2 2 30 # 103 = (4V1) − V 1 2 1000 From equation (i) 2 2 2 30 = 16V 1 − V 1 = 15V 1 2 2 & V1 = 2 m/ sec V 12 = 30 # 2 = 4 15 www. The continuity equation for two dimensional flow for incompressible fluid is given by..in . A1 V1 = A2 V2 p 2 d V1 = c A2 m V2 = p4 22 # V2 A1 4 d1 2 2 = d 22 # V2 = b 20 l V2 = V2 40 4 d1 V2 = 4V1 . d1 = 40 mm = 0. 2 p1 V 12 p For horizontal pipe z1 = z2 + + z 1 = 2 + V 2 + z 2 rg 2g rg 2g p1 . Given : d2 = 20 mm = 0.040 m Dp = p1 − p2 = 30 kPa Applying continuity equation at section (1) and (2).020 m . 47 Option (D) is correct. we get 2 3 y2 y3 d = 2 >c − 2 m H = 2 =c d − d 2 mG = 2 : d D = 0 2x 2d 3d 0 2x 6 2x 2d 3d to = 0 And drag force on the plate of length L is. We know that potential flow (ideal flow) satisfy the continuity equation.ME Fluid Mechanics Integrating this equation. FD = L # t # b # dx = 0 0 o Sol.gatescore. 2 u + 2v = 0 2x 2y 2u =−2v 2x 2y Sol.(i) Now applying Bernoulli’s equation at section (1) and (2). 46 Option (D) is correct. 50 Option (B) is correct. 20 kPa Hence pA .01 # 103 Pa = 20.... 48 Option (A) is correct.gatescore.81 # 0. Given : Vr =−b 60 # 10 l m/ sec . r π #120 drr =− 15 #0 dθ r =− 1 6θ@ 0π 6ln r @120 5 ln r . 5 dt dt dr =− 1 dq r 5 On integrating both the sides and put limits.150 meter Static pressure difference for U -tube differential manometer is given by. In this manometer A and B at different level and the liquid contain in manometer has the same specific gravity (only mercury is fill in the manometer) Given : rmercury = 13600 kg/m3 . Sol.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .533 120 And Sol.150 = 20. Dh = 150 mm = 0..81 m/ sec2 . g = 9.(iii) 5 In this equation (iii) Vr = Radial Velocity = dr dt Vq = Angular Velocity = rω = r dθ dt dr =− 1 r dq So. we get Vr =− 60 # 103 2pr =− 1 2pr # 300 # 103 5 Vq Vr =−Vq . www. pA .01 kPa .(ii) 2pr Dividing equation (i) by equation (ii).ln 120 =− 1 [p − 0] =− p 5 5 ln r =− p 120 5 r = e− p/5 = 0.pB is positive and pA > pB .. between r & 120 to r and θ & 0 to π (for half revolution). Flow from A to B . 49 Option (B) is correct.. It is a U -tube differential Manometer.pB = rg (hA − hB) = ρg∆h = 13600 # 9.ME Fluid Mechanics Sol.533 # 120 = 64 meter .(i) 2pr 3 3 Vq = 300 # 10 m/ sec .in r = 0. 9 to 8.5 to 8..3 8. Sol. In this question we have to make the table for calculate mean flow rate : Flow rate litres/ sec.8 Mean flow rate.8 = 8. uy = ay and uz = 0 Substitute there values in equation (ii).3 to 8.9 7.88 .6 7. 2).16 litres/ sec 80 Σf ..4 8.5 8.651 Pa Sol.2 17 139.(iii) 1 = 2c & c = 1 2 y x = & 2x − y = 0 2 Option (C) is correct.(ii) uy uz ux From equation (i).4 12 100.4 # 10−7 # 880 = 6.88 # 1000 = 880 kg/m3 µ Dynamic viscosity Kinematic Viscosity u = = ρ Density of liquid µ = υ # ρ = 7. we get dx = dy ax ay dx = dy x y Integrating both sides.gatescore. 52 log x = log y + log c = log yc & x = yc ...6 10 86 Σf = 80 Σfx = 652. Given : V = axi + ayj .5 # 10−3 meter Density of liquid = S # density of water = 0.5 t = µ # u = 6.in x = Σfx = 652.. From equation (iii). S = 0.1 8. Mean flow rate x + xf x=b i 2 l Frequency f fx 7. 51 Option (C) is correct. www.7 8. we get # dxx = # dy y At point (1.8 8. ux = ax ..1 to 8.7 to 7.8 5 39 7.ME Fluid Mechanics Given : u = 7.5 # 10−3 = 0.4 # 10−7 m2 / sec .512 # 10−4 Pa− s From the Newton’s law of viscosity. y = 0.512 # 10−4 # = 0. 0.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .6512 N/m2 y 0.7 7.(i) The equation of stream line is. dx = dy = dz .5 mm = 0.0 35 280 8.5 to 7.6 1 7. R 12 R 22 E Q 2 (R + R2) (R1 − R2) = 2 2 = 1 G p R2 L R 12 R 22 Considering limiting case R1 " R2 2Q2 (R1 − R2) 2Q 2 Q 2 (R1 − R2) 2R2 Then. a = 2 2 = G = p2 R 5 L 6R1 − R2@ = p R2 L R 22 R 22 p2 R 25 L 2 Option (D) is correct. 53 Fluid Mechanics Option (C) is correct. 12 − 12 E p R 2 R1 Acceleration at the exit section.in .gatescore.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . Q = AV Q Q Q Inlet velocity. a = dV = V dV dt dx Flow rate. Q dV = V2 − V1 = .R 22 R 12 E pR 22 p 2 R 22 L . 54 A1 = p d 12 4 dV = V2 − V1 V = V2 and dx = L 2 Q Q 1 R 12 − R 22 1 = Q a = − # pL . resultant velocity will be. V2 = = A2 pR 22 Therefore. In this case So. = V1 = = p 2 A1 pR 12 (2R1) 4Q Q Outlet Velocity. Sol. Total thrust at the bottom of cylinder = Weight of water in cylinder + Pressure force on the cylinder www.ME Sol. 5 # 10−5 m2 /s .5 # 10−5 # 1. v = 1.(i) U3 2 d 2 a d k Re x U3 = U = 2 m/ sec . 55 m = ρν A = pR 2 ρω2 R2 + ρgH E 4 Option (C) is correct. µ u = ρ µ = υ # ρ = 1.. 2 ρω2 r2 p = ρω2 # :r − 0D = 2 2 Dividing whole area of cylinder in the infinite small rings with thickness dr .64 # 1 5 = 0. V = rw And 2p = ρω2 rdr Integrating both the sides within limits p between 0 to p and r between 0 to r .845 # 10−5 # 3U3 dy y = 0 2d www.33 # 105 µ 1. Sol.64x .23 kg/m3 .ME For rotating motion. t0 = µc dU m = 1. Fluid Mechanics rV 2 ρr2 ω2 2p = = ρω2 r = r r 2r p = Pressure.33 # 10 U = 3 y − 1 y 3 And 2 d 2a d k U3 2 dU = U d 3 y − 1 y 3 = U 3 1 − 3y # 3 3 . 4 4 So.0127 1.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . Given relation is. ρUx 1. Force on elementary ring ρω2 r2 Pressure intensity # Area of ring = 2πrdr 2 # Total force.23 # 2 # 1 = Re x = = 1. p #0 2p = #0 r ρω2 rdr 6p@0p = ρω : 2 D 0 2 r2 r For calculating the total pressure on the cylinder.in . F = #0 R ρω2 r2 2πrdr = πρω2 2 # #0 R 4 4 R r3 dr = πρω2 :r D = πρω2 R 4 0 4 Weight of water = mg = ρνg = ρπR2 # Hg = ρgHπR2 Net force = ρgHπR2 + ρω2 πR = πR2 . U = 3 y − 1 y 3 and d = 4.845 # 10−5 kg/m sec Reynolds Number is given as. L = x = 1 Kinematic viscosity.23 = 1. r = 1.gatescore..845 # 10−5 d = 4.2 d 2 d3 E dy dy :2 d 2 a d k D where U3 = Free stream velocity = U dU 3 3U3 c dy m = U3 :2d D = 2d y=0 We know that shear stress by the Newton’s law of viscosity. Sol.5) 51 < NS < 255 for francis turbine.01 # 4000 # (2) 2 fLV 2 h f = = = 40. V = 2 m/ sec .81 [40. p = H & p = Hrg rg = 1000 # 9.01 bar Absolute pressure of G1 = pG1 + pabs (G2) = 5. 56 Option (B) is correct.5 bar For water hatm = 10. pG = 1.2 2gd Now total pressure (absolute discharge pressure) to be supplied by the pump at exit = Pressure loss by pipe + Head pressure of tank + Atmospheric pressure head Total pressure.01 + 1. 59 . S-6 www.3] = 5. p = rgh f + rgH + rghatm p = rg [h f + H + hatm] Pressure head. 0. d = 0.00 + 2.5 m .01 = 7. Reciprocating pump 3.01. Given : pG = 5.81 # 0. 58 2 Option (A) is correct.90 # 1000 # 9.00 bar . correct pairs are P-3.77 + 5 + 10.01 bar Absolute pressure of G2 = Atmospheric pressure + Gauge pressure = 1.gatescore. h0 = 90% .in . 57 Option (D) is correct. Plant with power output between 100 kW to 1 MW S.74 = 205.77 m of water 2 # 9.00 = 2.1 # 24.1 m3 / sec .36 # 10−3 N/m2 Sol. we get = 1.74 kW For turbine Specific speed.845 # 10−5 # 3 # 2 2 # 0. Given : H = 24.81 # 10.5 1000 1000 rwater = 1000 kg/m3 = 2184.01 bar 1 Sol. Given : L = 4 km = 4 # 1000 = 4000 m .00 bar .5 # 105 N/m2 = 5. Microhydel plant 2.82 # 10−5 N/m2 = 4.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . Option (B) is correct. low pressure ratio R. R-2.ME Fluid Mechanics Substitute the values of U3 and d. H = 5 meter Head loss due to friction in the pipe. Positive Displacement Q. patm = 1. Backward curved vanes 6. Q-5.80 rpm NS = N 5/P4 = 240 2184 5/4 H (24. Centrifugal pump impeller So.3 m Sol.0127 = 435. Q = 10. Hence. List-I List-II P. High Flow rate. Axial flow pump 5.2 m f = 0. N = 4 rps = 4 # 60 = 240 rpm P h0 = Shaft Power in kW = Water Power in kW r#g#Q#H b l 1000 η ρ g Q H P = 0 # # # # = 0. GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . 61 m = ρν n = A#H Option (A) is correct. 60 Fluid Mechanics Option (D) is correct.gatescore.in . Given : Flow rate = Q Velocity of water when it strikes the water surface = U Total Mass (container + water) = m Force on weighing balance due to water strike = Change in momentum DP =Initial Momentum . total force on weighing Balance = rQU + mg www.ME Sol. Given : Cross section area of body = A Height of body = H Density of body = rs Density of liquid = r Tension in the string = T We have to make the FBD of the block. B = Buoyancy force From the principal of buoyancy. So. Weight of container and water = mg Therefore. Downward force = Buoyancy force T + mg = rhAg T + ρs νg = rhAg T + rs AHg = rhAg T = rhAg − rs AHg = Ag (rh − rs H) Sol.Final momentum Final velocity is zero = rQU − rQ (0) = rQU Weighing balance also experience the weight of the container and water. 63 List-I List-II P. correct pairs are P-3. 3/4 = H1 H 23/4 3/4 3/4 Q2 = N1 # b H2 l # Q1 = 500 # b 120 l # 60 1000 30 N2 H1 = 1 # (2) 3/2 # 60 2 Squaring both sides Q2 = 1 # 8 # 60 = 120 litre/ min 4 N s = Alternate : From unit quantities unit speed N u = N1 = N 2 H1 H2 H 2 = N 2 H1 N1 N1 = N 2 H1 H2 2 2 (1000) # 30 H2 = N 2 #2 H1 = = 120 m N1 (500) 2 Q Q2 Unit discharge Q u = 1 = H1 H2 Q1 Q2 = H1 H2 Q H2 or Q 2 = 1 = 60 # 120 = 120 litre/ min H1 30 None of these is correct.in . So D1 = D2 DN \ H & N\ N \ H H1 = N 12 H2 N 22 2 2 (1000) H2 = N 22 # H1 = # 30 = 120 m N1 (500) 2 The specific speed will be constant for centrifugal pump and relation is. N2 = 1000 rpm . www.ME Sol. Mixed flow turbine So. Curtis 3. Kaplan 6.gatescore. Axial flow turbine S. or Sol. Q1 = 60 litres per minute From the general relation. S-7. Pressure compounding R. Given : N1 = 500 rpm . H1 = 30 meter . Rateau 4.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . R-6. U = pDN = 2gH 60 H D Centrifugal pump is used for both the cases. Velocity compounding Q. Francis 7. N Q = Constant H 3/4 N1 Q 1 N2 Q2 For both cases So. Q-4. 62 Fluid Mechanics Option (B) is correct. 1 # (1) 2 fLV 22 h f = = = 0.81 # 0.3265 2 And = 499.965 = 3702.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .001 2gd2 Applying Bernoulli’s equation at section (1) and (2) with the head loss in account. 2 p1 V 12 p + + z 1 = 2 + V 2 + z 2 rg 2g rg 2g The syringe and the plunger is situated on the same plane so z1 = z2 .gatescore. V1 = 10 mm/ sec We have to take the two sections of the system (1) and (2). Given : L = 100 mm . 0. www.ME Sol. µ = 1 # 10−3 kg/s − m Re ρVd ρV2 d2 Re = = = 1000 # 1 #−30. D = 10 mm .95 + 3202. Take p2 = 0 = Atmospheric pressure (Outside the needle) 2 2 p1 = V 2 − V 1 2g rg r p1 = (V 22 − V 12) = 1000 [(1) 2 − (0. Q = flow rate 2 p/4 (0.95 # p (0.81 # 0.010 = 1 m/ sec A2 p/4 (0.3265 m of water 2 # 9.915 N/m2 Force required on plunger.01) 2] = 499.001 = 1000 For Needle µ µ 1 # 10 And f = 64 = 64 = 0. Apply continuity equation on section (1) and (2). 2 p1 V 12 p + + z 1 = 2 + V 2 + z 2 + h f rg 2g rg 2g At the same plane z1 = z 2 Atmospheric pressure p 2 = 0 2 2 p1 = cV 2 − V 1 m + h f 2g rg r p1 = (V 22 − V 12) + rgh f 2 = 1000 6(1) 2 − (0.04 N 4 Option (C) is correct.064 Re 1000 From the help of f we have to find Head loss in needle.001) 2 Again applying the Bernoulli’s equation at section (1) and (2). A1 V1 = A2 V2 Q = AV . 65 Fluid Mechanics Option (B) is correct.01) V2 = c A1 m # V1 = # 0.01) 2 = 0.95 N/m2 2 2 Force required on plunger.in . 64 Sol.064 # 0. F = p1 # A1 = 499.01) 2@ + 1000 # 9. Given : f = 64 . d = 1 mm . we get A1 V1 = A2 V2 p D 2 V = p D2 U V = U .p2 = (V 22 − V 12) .p2 = . 4 4 r r p1 . & m . Let at point (1) velocity = V 2 1 4 t 1 4 2 V1 = b D l # U .. “If there are m variables (Indepenent and dependent variables) in a physical phenomenon and if these variables contain n fundamental dimensions (M.b D l − 1E 8 Dt Sol. So.(i) 2 Apply continuity equation.3 N Sol.n .b Dt l Option (C) is correct. T) then variables are arranged into (m ..in .. we have Spring force = Pressure force due to air . L... www.gatescore. First of all we have to take two section (1) and (2) Applying Bernoulli’s equation at section (1) and (2).kx = As (p1 − p2) = p D s2 # (p1 − p2) 4 4 ρ From equation (iii) = π D s2 # U 2 ..GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .ME Fluid Mechanics F = p1 # A1 = 3702.1 − b D l E .U 2 − b D l U 2E = U 2 . 67 x = ρU 2 D 4 − 1E πD s2 8k .n ) dimensionless terms.915 # p # (0. non dimensional variables.01) 2 4 = 0.(iii) 2 2 Dt Dt From the figure.1 − b D l E 2 4 Dt 4 kx = π D s2 ρU 2 . From Buckingham’s p-theorem.(ii) Dt Substitute the value of V1 from equation (ii) into the equation (i). 2 p1 V 12 p + + z 1 = 2 + V 2 + z 2 rg 2g rg 2g 2 p1 V 12 p z1 = z 2 + = 2 + V 2 2 2 r r r p1 . 66 Option (D) is correct. ME Sol. The pressure is minimum at point C . Along the region CSD of the curved surface.gatescore. Conduction resistance << Convection resistance Sol.1523 # 10−2 N−s/m2 k = 8. Due to decrease of velocity. The laminar boundary layer generation along a flat plate for this flow. then d grows like the square root of x . Given : www. is d + 1 L Re L If we substitute x for L and for a laminar boundary layer on a flat plate. the pressure increases in the direction of flow and dp pressure gradient dp/dx is positive or >0 dx Sol. where V (x) = V = constant.in c p = 0. 70 Option (C) is correct. A small value of Bi implies that the fluid has a small conduction resistance i.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . Biot Number Bi = hl k where h = Convective heat transfer coefficient k = thermal conductivity l = linear dimension Biot Number gives an indication of the ratio of internal (conduction) resistance to the surface (convection) resistance. the area of flow increases and hence velocity of flow along the direction of Fluid decreases.e. 71 Option (A) is correct. d + 1 x Vx u Sol. 69 d + x 1 &δ\ x V υ Option (C) is correct. 68 Fluid Mechanics Option (D) is correct.540 W/m −K .1393 k −J/kg −K µ = 0. 1. there is no relative motion between adjacent fluid layers and thus there are no shear (tangential) stresses in the fluid trying to deform it. Biot number gives an indication of internal (conduction) resistance to the surface (convection) resistance. 75 Option (D) is correct. the topic of fluid statics has significance only in gravity field. 72 Sol. The SI unit of kinematic viscosity is m2 / sec .ME Sol.0248 Option (A) is correct.e. Therefore.gatescore. At the depth h . Biot Number < 0. The only stress in static fluid is the normal stress. pressure is given by. 74 Option (A) is correct.1523 # 10 # 0. In static fluid. 73 Fluid Mechanics µc p k Prandtl Number Pr = −2 3 Pr = 0. p = rgh then horizontal force. Sol. i.1. which is the pressure and the variation of pressure is due only to the weight of the fluid and it is always positive.540 Pr = 0. then lumped that transfer analysis is valid. Fx = A # p = (2r # w) # rgh www. Here F1 = weight of water column above the top surface. Fluid static deals with problems associated with fluids at rest.1393 # 10 8. Bi = hl k If the value of Biot number is less than 0.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER . dy µ τ du u = = τ = µ du ρ ρ dy dy F # du A u = r Substitute the units of all the parameters Newton m #m 2 / sec kgm m u = N= 3 kg/m sec3 kgm m sec kg 2 2 2 m = sec m3 = m = sec m 3 sec kg/m kg/m Option (C) is correct. Sol. F2 = weight of water column above the bottom surface.in . 82 dt d = 1. Fy = mg = a π r2 # w k ρg m = rv and v = A # w 2 Fy = Sol. when viewed in the direction of Fx Fx = 2rghrw Fy = F2 − F1 = weight of water contained in volume of semi circular gate. And for irrotational flow.ME Fluid Mechanics where A = Normal area. 76 πρgwr2 2 Option (D) is correct. Given : where v = (x + 2y + 2) i + (4 − y) j u = x + 2y + 2 . d = hydrodynamic boundary layer thickness dt = thermal boundary layer thickness d = (6) 1/3 = 1.in . 2v = 0 2y 2x 1 . v = 4 − y 2u = 1 . flow is incompressible.GATE Previous Year Papers Solved by Team GATESCORE GATE SOLVED PAPER .1 = 0 So. flow is not irrotational.2u = 0 & 2x 2y 0 . Sol.2 ! 0 So.82dt d = 1. 2u = 2 2x 2y We know. z z = 0 z z = 1 c2v − 2u m = 0 2 2x 2y 2v .82dt d > dt or dt < d *********** www.gatescore. 77 Option (B) is correct. for Incompressive flow 2u + 2v = 0 2x 2y 2v =− 1.2 = 0 . Pr = 6 So. d = (Pr) 1/3 dt where Given. The non-dimensional Prandtl Number for thermal boundary layer is given by.
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