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GATE MCQ MECHANICAL ENGINEERING by NODIA

VOLUME 1.

ENGINEERING MECHANICS

1. Equilibrium of Forces 2. Structure 3. Friction 4. Virtual Work 5. Kinematics of Particle 6. Kinetics of Particles 7. Plane Kinematics of Rigid body 8. Plane Kinetics of Rigid body

STRENGTH OF MATERIALS

1. Stress and Strain 2. Axial Loading 3. Torsion 4. Shear Force and Bending Moment 5. Transformation of Stress and Strain 6. Design of Beams and Shafts 7. Deflection of Beams and Shafts 8. Column 9. Energy Methods

THEORY OF MACHINES

1. Analysis of Plane Mechanism 2. Velocity and Acceleration 3. Dynamic Analysis of Slider - Crank and Cam 4. Gear – Trains 5. Fly Wheel 6. Vibration

MACHINES DESIGN

1. Static and Dynamic Loading 2. Joints 3. Shaft and Shaft Components 4. Spur Gears 5. Bearings 6. Clutch and Brakes

VOLUME 2

FLUID MECHANICS

1. Basic Concepts and Properties of Fluids 2. Pressure and Fluid Statics 3. Fluid Kinematics & Bernouli Equation 4. Flow Analysis Using Control Volumes 5. Flow Analysis Using Differential Method 6. Internal Flow 7. External Flow 8. Open Channel Flow 9. Turbo Machinery

HEAT TRANSFER 1. Basic Concepts & Modes of Heat-Transfer 2. Fundamentals of Conduction 3. Steady Heat Conduction 4. Transient Heat Conduction 5. Fundamentals of Convection 6. Free and Force Convection 7. Radiation Heat Transfer 8. Heat Exchangers

THERMODYNAMICS 1. Basic Concepts and Energy Analysis

2. Properties of Pure Substances

3. Energy Analysis of Closed System

4. Mass and Energy Analysis of Control Volume

5. Second Law of Thermodynamics

6. Entropy

7. Gas Power Cycles

8. Vapor and Combined Power Cycles

9. Refrigeration and Air Conditioning

VOLUME 3

Manufacturing Engineering

1. Engineering materials and Heat treatment

2. Metal casting

3. Forming process

4. Sheet metal working

5. Joining

6. Machining and machine tool operations

7. Non‐traditional machiching process

8. Metrology and Inspection

9. Computer Integrated Manufacturing

Industrial Engineering 1. Production Planning and Control

2. Inventory Control

3. Network flow Models

Operations Research 1. Linear programming

2. Transportation

3. Assignment

4. Simple Quering Model

5. PERT and CPM

VOLUME 4

ENGINEERING MATHEMATICS

1. Linear Algebra

2. Differential Calculus

3. Integral Calculus

4. Directional Derivatives

5. Differential Equation

6. Complex Variable

7. Probability and Statistics

8. Numerical Methods

VERBAL ANALYSIS

1. Synonyms

2. Antonyms

3. Agreement

4. Sentence Structure

5. Spellings

6. Sentence Completion

7. Word Analogy

8. Reading Comprehension

9. Verbal Classification

10. Critical Reasoning

11. Verbal Deduction

QUANTITATIVE ANALYSIS

1. Number System

2. Surds, Indices and Logarithm

3. Sequences and Series

4. Average, Mixture and Alligation

5. Ratio, Proportion and Variation

6. Percentage

7. Interest

8. Time, Speed & Distance

9. Time, Work & Wages

10. Data Interpretation

11. Number Series

GATE TOPICWISE SOLVED PAPER(2013-2001)

1. Engineering Mathematics

2. Engineering Mechanics

3. Strength of Materials

4. Theory of Machines

5. Machine Design

6. Fluid Mechanics

7. Heat Transfer

8. Thermodynamics

9. Refrigeration and Air-Conditioning

10. Manufacturing Engineering

11. Industrial Engineering

12. General Aptitude

FM 6INTERNAL FLOW

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FM 6.1 Consider a fully developed laminar pipe flow. If the pipe diameter is reduced by half while the flow rate and pipe length are held constant, the head loss will be(A) Increase by a factor of 2 (B) Increase by a factor of 4

(C) Increase by a factor of 16 (D) Remains same.

FM 6.2 Consider a flow through a 15 m long horizontal pipe at the laminar transition point. The fluid is oil with 890 /kg m3r = and 0.07 /kg m sm -= . If the power delivered to the flow is 1 hp, the flow rate will be(A) 2420 /cm s3 (B) 4840 /cm s3

(C) 3630 /cm s3 (D) 484 /cm s3

FM 6.3 Glycerin at 40 Cc with 1252 /kg m3r = and 0.27 /kg m sm -= is flowing through a 5 cm diameter horizontal smooth pipe with an average velocity of 3.5 /m s. What will be the pressure drop per unit length of the pipe ?(A) 121 kPa (B) 1.21 kPa

(C) 12.1 kPa (D) 0.121 kPa

Common Data For Q. 4 and 5Water at 15 Cc ( 999.1 /kg m3r = ) is flowing steadily in a 45 m long and 4 cm diameter horizontal pipe made of stainless steel at a rate of 8 10 /m s3 3

#- . The

friction factor .f 0 01573= .

FM 6.4 What will be the head loss ?(A) 36.6 m (B) 3.66 m

(C) 366.0 m (D) 0.366 m

FM 6.5 The pumping power requirement to overcome pressure drop is(A) 1.5 kW (B) 4.5 kW

(C) 3 kW (D) 6.0 kW

Common Data For Q. 6 and 7A light liquid 950 /kg m3r =^ h flows through a horizontal smooth tube of diameter 5 cm at an average velocity of 10 /m s. The fluid pressure measured at 2 m intervals along the pipe is as given below:

( )mx 0 2 4 6

( )kPap 304 255 226 200

FM 6.6 The wall shear stress in the fully developed section of the pipe is (A) 163 Pa (B) 325 Pa

(C) 650 Pa (D) 81.5 Pa

GATE MCQ MECHANICAL ENGINEERING (4 volumes)

Sample Chapter

FM 212 Internal Flow FM 6


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FM 6.7 What will be the overall friction factor ?(A) 0.000183 (B) 0.183

(C) 0.00183 (D) 0.0183

FM 6.8 The piston shown in figure below is pushed steadily by a force F , which causes flow rate of 0.15 /cm sv 3=o through the needle. If fluid has 900 /kg m3r = and

0.002 /kg m sm -= , the force F will be

(A) 2.0 N (B) 3.6 N

(C) 1.35 N (D) 4.0 N

FM 6.9 A compressor that draws in air ( 1.149 / , 1.802 10 / )kg m kg m s3 5r m # -= = - from the outside, through an 12 m long, 20 cm diameter duct. The compressor takes in air at a rate of 0.27 /m s3 . If the friction factor is to be 0.0211, the useful power used by the compressor to overcome the frictional losses in the duct is(Disregarding any minor losses)

(A) 14.5 W (B) 15.4 W

(C) 51.4 W (D) 41.5 W

FM 6.10 In fully developed laminar flow in a circular pipe, the velocity at . R0 5 (midway between the wall surface and the center-line) is(A) u2 max (B) . u0 5 max

(C) . u0 75 max (D) Not changed

(where umax is the maximum velocity)

FM 6.11 The velocity profile in fully developed laminar flow in a circular pipe of inner radius 4 cmR = in /m s is given by

( )u r 1Rr4 2

2

= -c m

The maximum velocity in the pipe and the volume flow rate respectively, are(A) 4 / ,m s 0.01005 /m s3 (B) 0.01005 / ,m s 4 /m s3

(C) 0.01005 / ,m s3 4 /m s (D) 4 / ,m s3 0.01005 /m s



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FM 6.12 Consider a flow between two smooth parallel horizontal plates of 3 cm apart. If the fluid is 10SAE oil and 2 /m sV = 870 / 0.104 /kg m and kg ms3r m= =^ h, the head loss per meter is(A) 0.430 /m m (B) 0.487 /m m

(C) 0.325 /m m (D) 0.163 /m m

FM 6.13 Consider laminar flow of a fluid through a rectangular concrete channel with the smooth surfaces of friction factor ( / )Ref 58= . If the average velocity of the fluid is doubled, the change in the head loss of fluid in percentage is (Assume the flow regime remains same)(A) Decrease by 50% (B) Increase by 50%

(C) Increase by 100% (D) Decrease by 100%

FM 6.14 Water at 20 Cc flows from a tank by the pressurized air at a rate of 60 /m h3 as shown in figure below. If coefficient of friction .f 0 0136= , what gage pressure p1 is needed to drive the pipe flow ?

(A) 2.38 MPa (B) 1.2 MPa

(C) 0.238 MPa (D) 0.12 MPa

FM 6.15 A single 6 cm diameter tube consists of seven 2 cm diameter smooth thin tubes packed tightly as shown in figure below. Air at about 20 Cc and 1 atm (

1.2 /kg m3r = , 1.8 10 /kg m s5m # -= - ), flows through this system at 150 /m h3 . What will be the pressure drop per meter length of the pipe ? (Take .f 0 0250= )

(A) 202.5 Pa (B) 90 Pa

(C) 270 Pa (D) 27.0 Pa

FM 6.16 Oil with a density of 850 /kg m3 and kinematic viscosity of 6 10 /m s4 2#

- flows in a 5 mm diameter and 40 m long horizontal pipe, from a storage tank open to the atmosphere. If the height of the liquid level above the center of the pipe is 3 m and the flow is fully developed laminar, the flow rate of oil through the pipe is(A) 1.8 10 /m s8 6 3

#- (B) 1.8 10 /m s8 4 3

#-

(C) 1.8 10 /m s8 7 3#

- (D) 1.8 10 /m s8 8 3#

-



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FM 6.17 A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length l2 . The flow is fully developed laminar and the pressure drop for the first pipe is 1.44 times greater than for the second pipe. If the diameter of the first pipe is D , the diameter D3 of the second pipe is(A) . D1 64 (B) 1.37D

(C) . D1 095 (D) . D1 92

FM 6.18 A capillary viscometer measures the time of t 6D = seconds required for a 8 cm3 of water at 20 Cc to flow through a D diameter glass tube as shown in figure below. If 12 cmL = , 2 cml = and flow is laminar with no entrance and exit losses, the capillary diameter D will be (Take 0.001 /kg m sm -= )

(A) . mm1 5 (B) 15 mm

(C) 0.15 mm (D) 0.015 mm

FM 6.19 Oil with 894 /kg m3r = and 2.33 /kg m sm -= , flows at 0.5 /m s through 300 m long and 40 cm diameter cast iron pipe. Neglect minor losses. The pumping power required to overcome the pressure losses, is(A) 0.45 kW (B) 5. kW0

(C) 4 kW5 (D) 4.5 kW

FM 6.20 SAE 30 oil at 20 Cc . /kg m s0 29m -=^ , /kg m891 3r = h flows upward in a 3 cm diameter pipe through a pump from A to B at a rate of 3 /kg s as shown in figure below. At %100 efficiency, what pump power is required ?

(A) 4.8 kW (B) 4 kW

(C) 0.63 kW (D) 3.5 kW

FM 6.21 Oil with 910 /kg m3r = and 0.01 /kg m sm -= flows through a 1.2 m- diameter pipe at a rate of 3 /m s3 . The pressure drop along the pipe and friction factor are



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7.6 MPa and .0 0157 respectively. If the pump is %88 efficient, the power required and the length of the pipe respectively, are(A) 26 ,136.5MW km (B) 19.5 , 182MW km

(C) 19.5 , 136.5MW km (D) 26 ,182MW km

FM 6.22 The pump adds 25 kW to the water as shown in figure and causes a flow rate of 0.04 /m s3 . For either case .f 0 016= and neglect minor losses. What will be the flow rate expected when the pump is removed from the system ?

(A) 0.0289 /m s3 (B) 2.89 /m s3

(C) 0.289 /m s3 (D) 0.00289 /m s3

FM 6.23 Consider the pitot-static pressure arrangement as shown in figure below. Air at 20 Cc is flowing through the pitot tube 1.2 /kg m3r =^ , 1.8 10 /kg m s5m # -= -

h and the manometer fluid is colored water at 20 Cc 998 /kg m3r =^ , 0.001 /kg m sm -= h. If the friction factor of the flow is .f 0 0175= and .V V0 85avg CL= , the pipe volume flow rate and the wall shear stress respectively, are

(A) 0.109 /m s3 , 1.7 Pa (B) 0.109 /m s3 , 1.233 Pa

(C) 0.128 /m s3 , 1.233 Pa (D) 0.128 /m s3 , 1.7 Pa

FM 6.24 Glycerin at 20 Cc ( 1260 /kg m3r = , 1.50 /N s m2m -= ) flows upward in a vertical 75 mm diameter pipe with a centerline velocity of 1.0 /m s. The head loss and pressure drop in a 10 m length of the pipe respectively, are(A) 8.2 m, 225 kPa (B) 0.11 m, 125 kPa

(C) 6.75 m, 207 kPa (D) 3.43 m, 166 kPa

Common Data For Q. 25 and 26Oil ( 876 /kg m3r = and 0.24 /kg m sm -= ) is flowing through a 1.5 cm diameter pipe that discharges into the atmosphere at 98 kPa. The absolute pressure 15 m before the exit is measured to be 145 kPa.



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FM 6.25 If the pipe is horizontal, the flow rate of oil through pipe is(A) 1.62 10 /m s5 3

#- (B) 162 10 /m s5 3

#-

(C) 16.2 10 /m s5 3#

- (D) 162 10 /m s4 3#

-

FM 6.26 The flow rate of oil through the pipe, if the pipe is inclined at 8c upward from the horizontal, is(A) 100 10 /m s5 3

#- (B) 1.00 10 /m s5 3

#-

(C) 0. 10 /m s10 5 3#

- (D) 10.0 10 /m s5 3#

-

FM 6.27 Consider two types of drinking straws, one with a square cross-sectional shape and the other type the typical round shape. The amount of material in each straw and the length of the perimeter of the cross section of each shape are same . Assume the drink is viscous enough to ensure laminar flow and neglect gravity. What is the ratio of the flow rates v

vsquare

round

oo

_ i through the straws for a given pressure drop ? (For square cross section .Ref 56 9h = and for round shape Ref 64h = ).(A) 0.183 (B) 0.55

(C) 5.5 (D) 1.83

FM 6.28 Water flows from tank A to tank B with the valve closed as shown in figure. If the friction factor is .0 02 for all pipes and all minor losses are neglected, what will be the flow rate into tank B when the valve is opened to allow water to flow into tank C also ?

(A) 0.180 /m s3 (B) 0.00180 /m s3

(C) 0.0180 /m s3 (D) 1.80 /m s3

FM 6.29 Water at 20 Cc flows through a multiple parallel-plate passages heat exchanger as shown in figure below. The available pressure drop is 2 kPa and plate walls are hydraulically smooth. If the desired total flow rate is 0.25 /m s3 , the appropriate number of passages are .f 0 028=^ h

(A) 50 passagesN = (B) 30 passagesN =(C) 72 passagesN = (D) 42 passagesN =



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FM 6.30 Oil at 20 Cc ( 888.1 / , 0.8374 / )kg m kg m s3r m -= = is flowing through a vertical glass funnel as shown in figure. The funnel consists of 20 cm high cylindrical reservoir and a 1 cm diameter, 20 cm high pipe. The funnel is always maintained full by the addition of oil from the tank. Neglect entrance losses. What will be the ratio of the actual flow rate through the funnel to the maximum flow rate for the “Frictionless” case ?

(A) 43.91 (B) 0.0232

(C) 2.32 (D) 0.232

FM 6.31 Water at 20 Cc flows upward through an inclined 6 cm diameter pipe at 4 /m s is shown in figure. A mercury manometer has a reading of 135 mmh = . The pipe length between points (1) and (2) is 5 m and point (2) is 3 m higher than point (1). What will be the friction factor of the flow ?

(A) 0.114 (B) 0.07

(C) 0.025 (D) 0.044

FM 6.32 Viscous oil ( . . 0.85S G = , 0.10 Pa sm -= ) flows from tank A to tank B through the six rectangular slots as shown in figure below. If minor losses are negligible and the total flow rate is 30 /mm s3 , the pressure in tank A will be (Take f 3250= )



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(A) 1.54 kPa (B) 1.15 kPa

(C) 2.31 kPa (D) 1.92 kPa

FM 6.33 A 2 mm diameter and 20 cm long straw delivers the water at 10 Cc with a rate of 3 /cm s3 . If the flow is vertically up, what will be the axial pressure gradient

/p x2 2 ?(Take 1.307 10 /kg m s3m # -= - , 1000 /kg m3r = )(A) 2 /kPa m (B) 10 /kPa m

(C) 4 /kPa m (D) 20 /kPa m

FM 6.34 A tank of water has a 1.5 cm diameter hole at the bottom, where water discharges to the atmosphere. The water level is 3 m above the outlet. Disregarding the effect of the kinetic energy correction factor. If the entrance of the hole is sharp edged, the flow rate of water through the hole is (loss coefficient KL for sharp-edged .0 5= )(A) 1.11 10 /m s3 3

#- (B) 111 10 /m s3 3

#-

(C) 11.1 10 /m s3 3#

- (D) .111 10 /m s0 3 3#

-

FM 6.35 Water at a rate of 0.04 /m s3 , flows in a 0.12 m diameter pipe that contains a sudden contraction to a 0.06 m diameter pipe. If the loss coefficient .K 0 40L = , the pressure drop across the contraction section is(A) 99.75 kPa (B) 33 kPa

(C) 166.25 kPa (D) 133 kPa

FM 6.36 The water pipe system shown in figure below consists of 1200 m long cast-iron .f 0 0315=^ h pipe of 5 cm diameter, two 45c and four 90c flanged long-radius

elbows, a fully open flanged globe valve and a sharp exit into a reservoir. The minor losses coefficient for the pipe system is as follows

45c long-radius elbow : .K 0 2, 90c long-radius elbow : 0.K 3, Open flanged globe valve : .K 8 5, Sharp exit valve : .K 1 0,If the elevation at point 1 is 400 m, what gage pressure is required at point 1 to deliver 0.005 /m s3 of water at 20 Cc ( 0.001 / skg mm -= ) into the reservoir ?(A) .54 MPa4 (B) 3. MPa46

(C) . MPa1 43 (D) .4 MPa6

FM 6.37 Kerosine is to be withdrawn from a 15 cm high kerosine tank by drilling a well rounded 3 cm diameter hole with negligible loss at the bottom surface and attaching a horizontal 90c bend of negligible length. The kinetic energy correction factor is 1.05. What will be the flow rate of water through the bend, respectively



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if (a) the bend is a flanged smooth bend and (b) the bend is miter bend without vanes ?

(A) 8.08 / ,L s 4.78 /L s (B) 4.78 / ,L s 6.03 /L s

(C) 6.03 / ,L s 4.78 /L s (D) 8.08 / ,L s 6.03 /L s

FM 6.38 A horizontal pipe has an sudden expansion from 6 cmD1 = to 12 cmD2 = . The water is flowing at 10 /m s and 300 kPap1 = in the small section and the flow is turbulent. If the kinetic energy correction factor to be 1.06 at both inlet and outlet, the downstream pressure is

(A) 300 kPa (B) 278 kPa

(C) 377 kPa (D) 322 kPa

FM 6.39 A 4.5 m diameter tank is initially filled with water 2 m above the centre of a sharp edged 15 cm diameter orifice. The tank water surface is open to the atmosphere and the orifice drains to the atmosphere. Neglecting the effect of the kinetic energy correction factor. The time required to empty the tank is (loss coefficient for sharp edge .K 0 5L = )

(A) 9.6 min1 (B) . min26 4

(C) . min0 264 (D) . min2 64

Common Data For Linked Answer Q. 40 and 41Water at 20 Cc flows through a 10 cm diameter smooth pipe which contains an orifice plate with 5 cm-diameter. The measured orifice pressure drop is 75 kPa. Discharge coefficient 0.605Cd = and non- recoverable head loss coefficient 1.8K = .



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FM 6.40 What will be the flow rate in /m hr3 ?(A) 54 (B) 50

(C) 60 (D) 209

FM 6.41 What will be the non recoverable head loss ?(A) 33.4 kPa (B) 6.9 kPa

(C) 52. kPa4 (D) 26.4 kPa

Common Data For Linked Answer Q. 42 and 43Water at 20 Cc ( 9 8 / , 1.002 10 / )kg m kg m s9 3 3r m # -= = - flows through a 50 cm diameter pipe. The flow rate of water is measured with an orifice meter to be 0.25 /m s3 . The diameter ratio b and discharge coefficient Cd are .0 60 and .0 61 respectively.

FM 6.42 The pressure difference indicated by orifice meter is(A) 1.9 kPa (B) 19.0 kPa

(C) 14 kPa6 (D) 14.6 kPa

FM 6.43 What will be the head loss ?(A) 2.207 m (B) 2.0421 m

(C) 0.7734 m (D) 0.940 m

FM 6.44 A 5 cm diameter smooth pipe contains an orifice plate of .0 6b = and it is monitored by a mercury manometer /kg m13550 3r =^ h as shown in figure below. What will be the h when the flow rate is 0.334 /minm3 ?(Take .C 0 613d = )

(A) 75.75 cm (B) 7.52 cm

(C) 57 cm (D) 1.72 cm

FM 6.45 Air at 20 Cc ( 1.204 / )kg m3r = flows at high speed through a venturi-meter monitored by a water manometer as shown in figure below. If 40 cmh = , what will be the maximum mass flow rate of air that venturi can measure ? (Take discharge coefficient .C 0 98d = )



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(A) 2.73 /kg s (B) 0.273 /kg s

(C) 27.3 /kg s (D) 0.0273 /kg s

FM 6.46 Consider the flow of air at high speed through a venturi monitored by a mercury manometer 13550 /kg mHg

3r =^ h as shown in figure below. Discharge coefficient Cd and Expansion factor Y for this flow are .0 985 and .0 76 respectively. The upstream conditions are 150 kPa and 353 K. If 37 cmh = , the mass flow rate for flow to be compressible is

(A) 0.40 /kg s (B) 3.23 /kg s

(C) 7.27 /kg s (D) 0.90 /kg s

FM 6.47 Ethanol at 20 Cc /kg m789 3r =^ , 0.0012 /kg m sm -= h flows through a 5 cm diameter smooth pipe at a rate of 7 /m hr3 . Three piezometer tubes are installed as shown in figure below. If the pipe contains a thin plate orifice of diameter

3 cmd = , the piezometer levels h2 and h3 will be .Take K 1 5=^ and .f 0 023= h

(A) 1.7 mh2 = , 1.58 mh3 = (B) 0.42 mh2 = , 0.54 mh3 =(C) 1.58 mh2 = , 1.7 mh3 = (D) 1.58 mh2 = , 1.47 mh3 =

FM 6.48 Consider the parallel-pipe system as shown in figure below. The SAE 10 oil at 20 Cc 870 /kg m3r =^ and 0.104 /kg m sm -= h is flowing laminarly through the pipe system with pressure drop 21 kPap p1 2- = . What will be the total flow rate between 1 and 2 ?

(A) 0.0005 /m s3 (B) 0.0022 /m s3

(C) 0.0027 /m s3 (D) 0.0032 /m s3

FM 6.49 Consider the parallel-pipe system of two identical length and material pipe as shown in figure below. The diameter of pipe A is half of the diameter of pipe B



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. If the friction factor to be same in both case and disregarding minor losses, the flow rates in pipes A and B would be

(A) Remains same

(B) Flow rate of A increased by a factor of 0.177.

(C) Flow rate of B increased by a factor of 0.177.

(D) Flow rate of A decreased by a factor of 0.177.

Common Data For Linked Answer Q. 50 and 51Three pipes of same material .f 0 0275=^ h are laid in parallel with these dimensions:

Pipe 1 : 900 mL1 = 10 cmd1 =

Pipe 2 : 800 mL2 = 12 cmd2 =

Pipe 3 : 600 mL3 = 8 cmd3 =The total flow rate is 0.056 /m s3 of water at 20 Cc .

FM 6.50 The flow rate in each pipe is (A) 0.0166 /m sv1

3=o , 0.0277 /m sv23=o , 0.0116 /m sv3

3=o

(B) 0.0166 /m sv13=o , 0.0116 /m sv2

3=o , 0.0277 /m sv33=o

(C) 0.0277 /m sv13=o , 0.0166 /m sv2

3=o , 0.0116 /m sv33=o

(D) 0.0116 /m sv13=o , 0.0166 /m sv2

3=o , 0.0277 /m sv33=o

FM 6.51 The pressure drop across the system will be(A) 56 kPa (B) 55 kPa

(C) 550 kPa (D) 137.5 kPa

FM 6.52 For the Series -Parallel system of pipes shown in figure below, each pipe is 8 cmdiameter cast iron ( .f 0 0022, ) and the pressure drop 750 kPap p1 2- = . If the minor losses are neglected, what will be the resulting flow rate for water at 20 Cc ?

(A) 101 /m hr3 (B) 23 /m hr3

(C) 62 /m hr3 (D) 39 /m hr3



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FM 6.53 Water at 80 Cc ( 3.65 10 /m s7 2n #= - ) flows with an average velocity of 2 /m s through a 120 mm diameter pipe. If the pipe wall roughness is small enough so that it does not protrude through the laminar sublayer and the pipe is to be considered as smooth ( .f 0 0125= ), what will be the largest roughness allowed to classify this pipe as smooth ?(A) 23.1 mm

(B) 0.0231 mm

(C) 0.00231 mm

(D) 0.231 mm

FM 6.54 The three water-filled tanks are connected by pipes as shown in figure. If minor losses are neglected, the flow rate in /m s3 in each pipe is

(A) .v 0 01441 =o , .v 0 02842 =o , .v 0 01413 =o

(B) .v 0 01411 =o , .v 0 01442 =o , .v 0 02843 =o

(C) .v 0 02841 =o , .v 0 01412 =o , .v 0 01443 =o

(D) .v 0 02841 =o , .v 0 01442 =o , .v 0 01413 =o

FM 6.55 A highly viscous liquid flows under the action of gravity from a large container through a small diameter pipe in laminar flow as shown in figure below. Disregarding entrance effects and velocity heads, the variation of fluid depth in the tank with time, is

(A) 32 lnk hH

b l (B) lnk hH64 b l

(C) lnk hH128 b l (D) lnk h

Hb l

where kgdLD

4

2

n=

FM 6.56 A triangular passages ( 52.9/ )Ref = of heat exchanger with 60 cmL = and an isosceles triangle cross section of side length 2 cma = and included angle 80cb = is shown in figure below. If the oil 870 / , 0.104 /kg m kg ms3r m= =^ h at 20 Cc flows at 2 /m s, the pressure drop will be



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(A) 11.5 kPa (B) 23 kPa

(C) 2.3 kPa (D) 1.15 kPa

FM 6.57 An oil ( . . 0.87S G = and 2.2 10 /m s4 2n #= - ) flows at a rate of 4 10 /m s4 3#

- through a vertical pipe as shown in figure. The manometer reading h will be

(A) 18.5 m- (B) 13.87 m

(C) 13.87 m- (D) 18.5 m

FM 6.58 The water velocity at several locations along a cross section of 5 cm radius pipe is given in table below.

,cmr , /m sV

0 6.4

1 6.1

2 5.2

3 4.4

4 2.0

5 0.0

What will be the flow rate of water ?(A) 0.297 /m s3 (B) 0.0297 /m s3

(C) 2.97 /m s3 (D) 29.7 /m s3

FM 6.59 Oil ( 8900 /N m3g = , 0.10 /N s m2m -= ) flows through a 23 mm diameter horizontal tube as shown in figure. A differential U-tube manometer is used to measure the



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pressure drop along the tube. What will be the range of h for laminar flow ?

(A) 0.51 mh # (B) 0.51 0.51m mh# #-(C) 0 0.51 mh# # (D) 0.51 mh $

FM 6.60 The water at 20 Cc flows from the tank as shown figure below, through the 3 cm long horizontal plastic pipe attached to the bottom of the tank. What time it will take to empty the tank completely, assuming the entrance to the pipe is well-rounded with negligible loss ? (Take the friction factor of the pipe to be 0.022.)

(A) 8.66 hours (B) 82 hours

(C) 86.6 hours (D) 8.2 hours

***********



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SOLUTIONS

FM 6.1 Option (C) is correct.In a fully developed laminar pipe flow, head loss is given by

hL /RefD

Lg

VDL

gV

VD DL

gV

264

264

22 2 2

n= = =

D DL

gV642

n= D g

LDv64

24

2 2n

p= o

> H V

Dv

42p

= o

g D

Lv1284p

n= o

where n = Kinematic viscosity

L = Pipe Length

D = Diameter of pipe

vo = Volume flow rate

If diameter of pipe is reduced by half, then D2 D2=

So that hL2 16 128

g DLv

g DLv

2

1284 4

pn

pn

#= =o o

b l

h16 L=

Hence Reducing the pipe diameter by half increase the head loss by a factor of

16.

FM 6.2 Option (B) is correct.For laminar flow at transition point

Re Vdm

r= 2300=

or Vd890#m 2300= .Vd 890

2300 0 07& #= 0.181 /m s2= ...(i)

Power 1 hp 745.7 W v p minla arD#= = o

.745 7 AV p d VdLV

4322

2# #p mD# #= = a ck m

.745 7 8 LV 2p m#= 8 0.07 15 .d

0 181 2

#p# #= b l

or .745 7 .d

0 8652= .

.d 745 70 8652& =

d 0.034 m=

From eq. (i) V ... 5.32 /m sd

0 1810 0340 181= = =

Hence vo d V42p

# #= . .4 0 034 5 322p# #= ] ]g g

0.00484 / 4840 /m s cm s3 3= =

FM 6.3 Option (C) is correct.Reynolds number for this flow



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Re VDm

r= .. . 811.50 27

1252 3 5 0 05# #= =

Re .811 5 2300<=Hence the flow is laminar and friction factor for this Re is

f . .Re64

811 564 0 07887= = =

Then the pressure drop per unit length ( 1 )mL =

pLD fDL V

2

2

#r= 0.0788 .

( . )0 051

21252 3 5

100012

## # #=

12.1 kPa=

FM 6.4 Option (A) is correct.

Head loss hL fDL

gV2

2

#=

fDL

g Dv

21 4

2

2

p# #= o; E fD

Lg D

v2

162 4

2

# #p#= o V

Dv

42p= o

fgDLv

216

2 5

2

p#= o

hL . ( . )

. ( )2 9 81 0 04

0 01573 16 45 8 102 5

3 2

# # #

# # # #

p=

-

36.5 m9= 36.6 m,

FM 6.5 Option (C) is correct.Pressure drop is given by

pD f DL V

2

2

# #r= 4f D

LDv

2 2

2rp# # #= o; E

f DL

Dv f

DL v

216

216

2 4

2

2 5

2

# # # #r

p pr= =o o

pD .( . )

. ( )0 01573

2 0 04 100016 45 999 1 8 10

2 5

3 2

## # #

# # # #

p=

-

358.3 kPa= 359 kPa-

Hence the power requirements to overcome this pressure drop is

Ppump 8 10 359v p 3D # #= = -o

2.8 kW7= 3 kW,

FM 6.6 Option (A) is correct.The Wall shear stress in the fully-developed region is defined as

Lp

fully developedDD p p

d2 226000 4 w4 6 t= - = =

or wt . 162.5 Pad4 2

260008

26000 0 05## #= = = 163 Pa,

FM 6.7 Option (D) is correct.

The overall head loss for 0zD =

hf . 11.2 mgp

950 9 81304000 200000

#rD= = - =

The overall friction factor is defined as

foverall h Ld

Vg2

,f overall 2# #= 11.2 . .6

0 0510

2 9 812

##= b bl l

0.0183=



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FM 6.8 Option (D) is correctThe velocity of exit from the needle is

V Av

1 =o

.

. 306 /cm s

4 0 0250 15

2#

p= =] g

The energy equation gives

gp

gV z2

2 22

2r + + gp

gV z h h2 f f

1 12

1 1 2r= + + + + ...(i)

With z z1 2= , V 02 - , 0hf 2 , and hgDLV32

f 1 2rm= (laminar flow), equation (i)

becomes

gp p2 1

r- h g

V2f 1

12

= + gD

L Vg

V3222

12

# # #

rm= +

or gp

rD

. .32 0.002 0.015 3.06

..

900 9 81 0 00025 2 9 813 06

2

2

# # ## # #= +

]]

gg

5.32 0.48 5.8 m= + =

Then F p ApistonD #= 5.8 .g 4 0 01 2r p# # #= ] g

9.81 900 5.8 . 4.0 N4 0 01 2 ,p# # #= ] g

FM 6.9 Option (A) is correct.The average velocity for this flow

V ( . )

. 8.594 /m sAv

Dv 4

0 20 27 4

c2 2

#

#

#p p

= = = =o o

The pressure drop in the duct

pD p fDL V

2L

2rD #= =

Substitute f 0.0211, 1.149 / , 12 , 8.594 /kg m m m sL V3r= = = = and D 0.2 m=

So pLD . .. ( . )

0 0211 0 2012

21 149 8 594 2

# ##= 53.72 Pa=

Then the required pumping power becomes

Ppump 0.27 53.72v pD #= =o 14.5 W=

FM 6.10 Option (C) is correct.The velocity profile in fully developed laminar flow in a circular pipe is given by

( )u r uRr1max 2

2

= -; E

At /r R 2= ( / )u R 2 ( / )

uR

Ru1

21 4

1max max2

2

= - = -; :E D

.u u43 0 75max max= =

FM 6.11 Option (A) is correct.The general velocity profile in fully developed laminar flow is

( )u r uRr1max 2

2

= -; E

We have ( )u r uRr1 2

2

= -; E

By comparing these two, we get



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umax 4 /m s=Now the volume flow rate

vo V A V Ravg c avg2p#= =

Since Vavg 2.00 /m su2 2

4max= = =

Hence vo ( . )2 0 04 2# #p= 0.010048 /m s3= 0.01005 /m s3

b

FM 6.12 Option (C) is correct.The half-distance between plates is called “h”, then Hydraulic diameter

Dh 4 4 1.5 6perimeter cmA h4#= = = =

ReDh .. . 1004VD

0 104870 2 0 0 06h # #

mr= = = (Laminar flow)

For laminar flow between two parallel horizontal plates, the head loss and Pressure drop per meter flow is given by

pD .

. .hVL3

0 0153 0 104 2 0 1

2 2# # #m= =

]]

gg

2773.33 / 2770 /Pa m Pa m-=

Hence hf . 0.325 /m mgp

9 81 8702770#r

D= = =

FM 6.13 Option (C) is correct.The friction factor of given channel

f Re58= where Re VDH

mr= Hydraulic diameterDH =

Then Head loss hL1 RefDL

gV

DL

gV

258

2H H

2 2

# # #= =

VD DL

gV582H H

2#

# #rm=

H29

gDLV

2rm=

If the average velocity is doubled, then

V2 2V=

Hence hL2 gDL V

gDLV29 2 2 29

H H2 2

#

rm

rm= = = G 2hL1=

Therefore percentage change in Head loss is

(%)hD %hh h2 100 100

L

L L

1

1 1#= - = Increase

FM 6.14 Option (A) is correct.For water at 20 Cc , take 998 /kg m3r = and 0.001 /kg m sm -=

V ./

8.49 /m sAv

dv

dv

4

40 05

4 60 36002 2 2

# # #

#

p p p= = = = =o o o

]

^

g

h

The energy equation between points (1) (the tank) and (2) (the open jet) :

10gp

g201

2

r + + 80g gV

h20

2pipe

f

2

= + + +

or gp1

r 80 10gV

h2pipe

f

2

= + + - ...(i)

where hf 0.0136 . ..fD

Lg

V2 0 05

60 80 302 9 81

8 49pipe2 2

## # #= =

+ +^ h



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170 m,

then from equation (i),

gp1

r .. 80 170 102 9 81

8 49 2

#= + + -] g

or p1 ..g 2 9 81

8 49 80 170 102

#r= + + -] g

; E

998 9.81 .8.49

2 9 81 80 170 102

## #= + + -] g

; E

2385659 2.38 MPa,=

FM 6.15 Option (C) is correct.Assume all the flow goes through the seven 2 cm tubes. Thus each tube takes one - seventh of the flow rate :

V .

/18.95 /m sA

v7 0 01

150 3600,tubes7

2# #

,p

= =o] g

Re .

. . . 25300Vd1 8 10

1 2 18 95 0 025

#

# #m

r= = =- (Turbulent flow)

Hence pD 0.0250 .. .f d

L V2 0 021

21 2 18 952 2r

# # # #= = ] g

269.32 270 Pa,=

FM 6.16 Option (D) is correct.The pressure at the bottom of the tank is

p ,gage1 ghr= .1000

850 9 81 3# #= 25.02 /kN m2=

The pressure loss across the pipe, disregarding minor losses is

pD p p p patm1 2 1= - = - 25.02 /kN mp ,gage12= =

Pressure loss for fully developed laminar flow

pD DLV

DLv32 128

2 4m

pm= =

o v V D4

2p#=o

D

L v1284

# # # #

pr n=

o

Then vo Lp D

128

4

# # #r npD=

128 850 6 10 4025.02 10 ( . )0 005

4

3 4# #p

# # # #

#= -^

^

h

h 1.8 10 /m s8 8 3

#= -

FM 6.17 Correct option is (C)

For laminar flow vo lD p128

4

mp D=

where v v1 3=o o and .p p1 44 2 31 2D D=- -

Thus v1o lD p v l

D p128 128

1 2 2 34

334

mp

mpD D= = =- -o D D1 =

or D3 ( . ) .D pp D D1 44 1 095

2 3

1 2/

/1 4

1 4

DD= = =

-

-

c m



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FM 6.18 Option (A) is correct.Assume no pressure drop and neglect velocity heads. The energy equation reduces to

gp

gV z2

1 12

1r + + gp

gV z h2 f

2 22

2r= + + +

or 0 0 L l+ + +^ h 0 0 0 hf= + + + hf L l= +

For laminar flow hf gDLv128

4prm=

o and for uniform draining vo t

vD=

or tvvD = o

Lh gD

vgD h

Lv

128

128f f

4 4#

#

mpr pr

m= =

tD ( )gD L lLv128

4prm=

+ ( )h L lf = +

or 6 . ( . . )

. .D998 9 81 0 12 0 02

128 0 001 0 12 8 104

6

# # # #

# # # #p

=+

- .

D2 85 10

4

11#=

-

D4 4.75 10 12#= -

or D 0.00148 0.0015 1.5m mm- , =

FM 6.19 Option (D) is correct.Reynolds number for this flow

Re .. . .VD

2 33894 0 5 0 4 76 7# #

mr= = =

Since .Re 76 7 2300<= . Hence the flow is laminar.

The pressure loss for laminar flow is given by

pD fDL V

2

2r=

Since f . .Re64

76 764 0 834= = =

Then pD . .( . )

0 834 0 4300

2894 0 5

100012

# ##

#= 69.9 kPa=

Therefore pumping power required to overcome this pressure drop

Ppump v p V D p42

# pD D#= =o

. ( . )

.40 5 0 4

69 92

# ##

p= 4.39 4.5 kW,=

FM 6.20 Option (D) is correct.The pipe velocity

V .

4.76 /m sAm

891 0 0153

2# #r p

= = =o] g

Check Red .. . 439VD

0 29891 4 76 0 03# #

mr= = = (Laminar flow)

Apply the steady flow energy equation between A and B

gp

gV z2

A AA

2

r + + gp

gV z h h2

B BB f p

2 2

r= + + + -

or .891 9 81500000#

. 15 h h891 9 81180000

f p#

= + + - V V VA B= = 0and zA =



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h hf p- . . 15891 9 81500000

891 9 81180000

# #= - -

or h hf p- 21.61 m=

Where hf . .. 15 20 .

gdLV32

9 81 891 0 0332 0 29 4 76

2 2

2 2

# #

# # #

rm= = +

]] ]

gg g 140.4 m=

Now hp 21.6hf= - 140.4 21.16= - 119. m2=The pump power is then given by

P gvhpr= o mghp= o 3 9.81 119.2# #= 3508 3500 Watts-= or 3.5 kW

FM 6.21 Option (D) is correct.The flow rate gives the velocity and Reynolds number

V .

2.65 sm

Av

4 1 23

2#

p= = =o

] g

Red .. . 289380Vd

0 01910 2 65 1 2# #

mr= = = (turbulent flow)

Since pD f dL V2

2r=

or 7600000 0.0157 . .L1 2 2

910 2 65 2# # #= b ]l g

L 181800 182m km-=

Hence P . 2.6 10 Wattsv p0 88

3 7600000 7#hD

#= = =o

26 MW=

FM 6.22 Correct option is (A).

Applying the Bernoulli’s equation at section (1) and (2)

pg

V z h212

p1

1g + + + pg

V z fDl

gV

2 22 2

2

2

2

#g= + + +

Where p p 01 2= = , z h1 = , z 02 = and V 01 =

V2 ( . ). 31.8 /m sA

v

4 0 040 04

2 2p= = =o

c m

V ( . ). 14.15 /m sA

v

4 0 060 04

2#

p= = =o

h hp+ .( . )

0.016 . .( . )

133.2 m2 9 8131 8

0 0630

2 9 8114 152 2

## #

#= + =b l

Also hp ( . ) .63. m

vP

9 80 10 0 0425 10 73

3

# #

#g= = =o

Hence h 133.2 63.7 69.5 m= - =



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Without the pump h 0p = and z gV fD

lg

V2 21

22 2

#= + ...(i)

Where 69.5 mh z1= =

and V2 AAV

DD V

2 2

2= = b l or .V V V40

60 2 252

2= =b l

Thus from equation (i)

.69 5 .

( . ) . .V V

2 9 81

2 25 0 016 0 06302 2

#

#=

+ b l or 10.22 /m sV =

So that vo (0.06) 10.22 0.0289 /m sAV 42 3#

p= = =

FM 6.23 Option (B) is correct.The manometer reads

p p0 - ghwater airr r= -^ h . 9.81 . 391 Pa998 1 2 0 040#= - =] ]g g

Therefore, velocity at centre line

VCL . 25.5 /m sp21 2

2 391#rD= = =

Now average velocity

Vavg 0.85VCL= 0.85 25.5 21.675 /m s#= =Thus the flow rate

vo 21.675 . 0.109 /m sV A 4 0 08 2 3p# # #= = =] g

and wall shear stress

wt f V8 avg2r#=

. 1.2 .80 0175 21 675 2

# #= ] g 1.233 Pa=


For laminar flow in a pipe

Vaverage 1 0.5 /m sV21

21

max #= = =



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Thus Re .. . 31.5 2100VD

1 501260 0 5 0 075 <# #

mr= = =

The flow is laminar.

So that V ( )sin

lp l D

32

2

mg qD= -

Where 90cq =

pD D

lV l322

m g= + sin90 1=c

( . ). . 9.81 1260 100 075

32 1 50 10 0 52

# # ## #= + gg r=

1.66 10 166Pa kPa5- # =Also, applying Bernoulli’s equation at section (1) and (2),

pg

V z21 1

2

1g + + pg

V z h2 L2 2

2

2g= + + +

With p p p1 2 D= + , V V1 2= , z z l2 1- =

hL .. 10p l 9 81 1260

1 66 105

##

gD= - = - 3.43 m=

FM 6.25 Option (A) is correct.Flow rate can be determined from

vo Lp D

128

4

mpD=

Since p p p1 2D = - 145 98 47 kPa= - =

Hence vo .( . )

1.62 10 /m s128 0 24 1547000 0 015 4

5 3

# #

# #p#= = -

FM 6.26 Option (B) is correct.For uphill flow with an inclination of 8c,

vuphillo ( )sin

Lp gL D

128

4

mr q pD= -

.[ ( . )] ( . )sin

128 0 24 1547000 876 9 81 15 8 0 015 4

# #

# # # #c p= -

1.00 10 /m s5 3#= -


p1D p2D=So hL1 hL2= p hLgD =

f Dl

gV2h

11 1

2

1

# f Dl

gV2h

22 2

2

2

#=

where l l1 2= , D Dh1 = and perimeterD Aaa a4

44

h2

22

2 = = =

Thus Df V1 1

2

af V2 2

2

= ...(i)

Since the perimeters are equal

Dp a4= Therefore a D4p= ...(ii)



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and Reh1 V D V Dh1 11

n n= =

Reh2 V D V a V D

4h2 2 22

n n np= = = from equation (ii)

Thus, from equation (i)

D

V DV64

112

nb l

.

D

V DV

4

4

56 92

22

pn

p=

b l or .V V1 4411 2=

Also v1o A V D V41 12

1#p= =

and v2o A V a V D V162 22

2

22

2p= = =

So that vv

2

1oo

1.441 1.83D VD V

VV4 4

162

2

42

1

2

12 p p #= = = =p

p

vroundo . v1 83 square= o

FM 6.28 Option (C) is correct.

From continuity equation

v1o v v2 3= +o o

Since D1 D D2 3= = , it follows that

V1 V V2 3= + ...(i)

Also for fluid flowing from A to B ,

pg

V z2A A

A

2

g + + pg

V z f Dl

gV f D

lg

V2 2 2

B BB

2

11

1 12

22

2 22

# #g= + + + +

Where p p 0A B= = , 0V VA B= = , 15 mzA = and z 0B =

Thus zA f Dl

gV f D

lg

V2 21

1

1 12

22

2 22

= + ...(ii)

or 15 . .. [80 40 ]V V0 1 2 9 81

0 0212

22

# #= +

or .18 4 .V V0 512

22= + , ...(iii)

Similarly, for fluid flowing from A to C ,

pg

V z2A A

A

2

g + + pg

V z f Dl

gV f D

lg

V2 2 2

C CC

2

11

1 12

33

3 32

g= + + + +

where 0p pA C= = , 0V VA C= = , 15 mzA = and 0zC =

Thus zA f Dl

gV f D

lg

V2 21

1

1 12

33

3 32

= + ...(iv)



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Comparing equation (ii) and (iv), we get

f Dl

gV22

2

2 22

f Dl

gV23

3

3 32

=

Since f f2 3= and D D2 3= l V2 2

2 l V3 32=

40V22 75V3

2= V2 1.369V3= ...(v)

Solve equation (i), (iii) and (v) for V1, V2 and V3. From equation (i) and (v)

V1 1.369 2.369V V V3 3 3= + =from eq (iii) 18.4 ( . ) . ( . )V V2 369 0 5 1 3693

23

2= + V3 1.676 /m s=from eq (v) V2 1.369 1.676 2.29 /m s#= =

Hence v2o (0.1) 2.29A V 42 22

# #p= = 0.0180 /m s3-

FM 6.29 Option (C) is correct.If there are N passages, then cmb 50= for all N and the passage thickness is

0.5/H N= . The hydraulic diameter is 2D Hh = . The velocity in each passage is related to the Pressure drop.

pD f DL V2h

2r# # #= where .f f 0 028smooth= =

or 2000 0.028 ..

N

V2 0 5

2 02

998 2

## # #=

Where V . .

. 1 /m sAv

0 5 0 50 25

c #= = =o

^ h

Thus 2000 0.028 .2.0 N

2 0 5 2998 1 2

## # #= ] g

N . .. 71.57 72 passages0 028 2 0 998

2000 2 0 5 2# #

# # # ,= =

FM 6.30 Option (B) is correct.

The flow velocity for the frictionless case.

V ,max2 . . 2.801 /m sgz2 2 9 81 0 401 # #= = =Then maximum volume flow rate



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vo .( . )

V A V D4 2 801 4

0 01, ,max max2 2 2

22 2

# # ##p p= = =

2.20 10 /m s4 3#= -

Now the actual flow rate can be determined by

vactualo ( )sin

Lp gL D

128

4

mr q pD= -

Since the flow is vertically downwards, so 90cq =- and

p p pinlet outletD = - ( )p gh p ghatm cylinder atm cylinderr r= + - =(because at inlet total pressure becomes patm and pressure due to oil in cylinder

ghcylinderr and at exit atmospheric pressure patm is there)

Therefore vactualo ( )sin

Lgh gL D

128cylinder

4

mr r q p= -

( )

Lg h L D

128cylinder

4

mr p= +

( 90 ) 1sin c- =-

. .. . ( . . ) ( . )

128 0 8374 0 20888 1 9 81 0 20 0 20 0 01 4

# #

# # # #p= +

5.1 10 /m s6 3#= -

So, the ratio of actual flow rate through the funnel to the maximum flow rate is

vv

max

actualoo

..

2 20 105 1 10

4

6

#

#= -

- 0.0232=

FM 6.31 Option (C) is correct.By moving through the manometer, we obtain the pressure change between points (1) and (2).

p gh gh g zw m w1 r r r D+ - - p2=or p p1 2- gh gh g zm w wr r r D= - + gh g zm w wr r r D= - +^ h

9.81 0.135 998 9.81 313568 998# # #= - +^ h

16647 29371= + 46018 46000 Pa-=

Since hf . 3gp z 998 9 81

46000w #rD D= - = -

.4 7 3= - 1.7m=

The friction factor f h Ld

Vg2

f 2# #=

1.7 . .5

0 064

2 9 812

## #= 0.0250=


For each slot, flow rate becomes

vo 30 10 5 10 /m s61 9 9 3# # #= =- -

So that V ( )

1.67 10 /m sAv

3 1 105 10

6

93

# #

##= = =-

--o

Also r . . 0.85 1000 850 /kg mS G H O3

2# #= = =r

and Dh ( )( )

1.5perimeter mmA46 2

4 3 1# #= = + =

The energy equation gives



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pg

V z2A A

A

2

g + + pg

V z f Dl

gV

2 2B B

Bh

2 2

g #= + + +

with z zA B= , p 0B = , V V 0A B= =

pA f Dl V2

1h

2r= b bl l

3250 .. 850 (1.67 10 )0 0015

0 621 3 2

# # ## #= -

1540 / 1.54N m kPa2- =

FM 6.33 Option (D) is correct.For water at 10 Cc : 1000 /kg m3r = 1.307 10 /kg ms3m #= -

Re 4VD DAv

Dv D

2#m

rmr

m pr

#= = =o o V A

vDv4

2p= =o o

Re . .D

v41 307 10 0 002

4 1000 3 103

6

# # #

# # #mpr

p= = -

-o 1462= (Laminar flow)

Thus hf . .. .

gDLv128

1000 9 81 0 002128 1 307 10 0 2 3 10

4 4

3 6

# # #

# # # # #pr

mp

= =- -o

] g

0.204 m,

The axial pressure gradient if the flow is vertically up

LpD L

g h zfr D=

+^ h .

. . .0 2

1000 9 81 0 204 0 2# #= +^ h z LD =

19816 / 20 /Pa m kPa m,=


The energy equation for a control volume between two points

gp

gV z h2 pump

11

12

1r a+ + + gp

gV z h h2 turbine L

22

22

2r a= + + + +

Since p p1 2= ,patm= 0,z2 = V 01 = and h h 0pump turbine= =

z1 gV h2 L2

22

a= + gV K g

V2 2L2

22

22

a= + h K gV2L L

2

#=

V2 Kgz2

L2

1

a= +Since ,12a = then volume flow rate becomes

v V Ac2=o DK

gz4 1

2hole

L

21p

#= +Substituting the numerical values, we get

vo ( . )

..

40 015

1 0 52 9 81 32

##

# #p= + 1.11 10 /m s3 3#= -



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Applying the bernoulli’s equation at section (1) and (2),

pg

V z21 1

2

1g + + pg

V z K gV

2 2L2 2

2

222

g= + + + ...(i)

Where V1 ( . ). 3.54 /m sA

v

4 0 120 04

1 2#

p= = =o

V2 ( . ). 14.1 /m sA

v

4 0 060 04

2 2#

p= = =o and z z1 2=

Hence equation (i) becomes,

p p1 2- [ ]K V V V21

L 22

22

12r= + -

999 [0.40(14.1) (14.1) (3.54) ]21 2 2 2# #= + -

or p p1 2- . . 13 kPa499 5 265 80 3# -=

FM 6.36 Option (B) is correct.For water, take 998 /kg m3r = and 0.001 /kg m sm -= . With the flow rate known, we can compute

V .

. 2.55 /m sAv

4 0 05

0 0052p

= = =o

d ]n g

The energy equation is written from point 1 to the reservoir surface

gp

gV z2

1 12

1r + + gp

gV z h K K K K g

V2, ,f valve elbow elbow exit

2 22

2 45 90

2

r r= + + + + + + +c c6 @

gp

gV z2

1 12

1r + + 0 0 z h K K K K gV2, ,f valve elbow elbow exit2 45 90

2

= + + + + + + +c c6 @

or gp1

r z z gV f D

Lg

V2 22 1

12 2

# #= - - + c m

K K K K gV2, ,valve elbow elbow exit45 90

2

+ + + +c c6 @

500 400 .2.55 . . .

2.552 9 81 0 0315 0 05

12002 9 81

2 2

## #

#= - - +] ]

cg g

m

. . . ..8 5 2 0 2 4 0 3 1 2 9 81

2 55 2

# ##

#+ + + +] ]]

g gg

6 @

gp1

r 500 400 0.331 250.56 3.68= - - + + 353.9 354,=

p1 354 998 9.81 354gr # # #= = 346 3.46Pa MPa5794 ,=




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The energy equation for a control volume between two points

gp

gV z2

11

12

1r a+ + gp

gV z h2 L

22

22

2r a= + + +

Here 0, 0 andV z p p patm1 2 1 2= = = =

So z1 gV h2 L2

22

a= + gV K g

V2 2L2

22

22

a= + h K gV2L L

2

=

V2 Kgz2

L2

1

a= +

Then the flow rate becomes.

vo A Vpipe 2#= DK

gz4

2L

2

2

1pa= +

(a) For Flanged smooth bend ( . )K 0 3L =

vo ( . )

. ..

40 03

1 05 0 32 9 81 52

##

# #p= +

0.006 3 / 6.03 /m s L s0 3= =(b) For Miter bend without vanes ( . )K 1 1L =

vo ( . )

. ..

40 03

1 05 1 12 9 81 52

##

# #p= +

0.00478 /m s3= 4.78 /L s=


From continuity equation for incompressible flow.

A V1 1 A V2 2=

V2 //

AA V

DD

VDD V

44

2

11

2212

12212

1pp

# # #= = == ;G E

( . )( . )

10 2.5 /m s0 120 06

2

2

#= =

The loss coefficient for sudden expansion

KL ( . )( . )

.AA

DD1 1 1

0 120 06

0 5625argl e

small2

2212 2

2

2 2

= - = - = - =; ; =E E G



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and Head loss hL 0.5625 .( )

2.87 mK gV2 2 9 81

10L

12 2

## #= = =

The energy equation for the expansion section

gp

gV z2

11

12

1r a+ + gp

gV z h2 L

22

22

2r a= + + +

Since z z1 2= , then

gp

gV2

11

12

r a+ gP

gV h2 L

22

22

r a= + +

p2 p V V gh2 L11 1

22 2

2

r a a= + - -: D

300 100. ( ) . ( . )

0 21 06 10 1 06 2 52 2

# ##= + -

;

. .9 81 2 87 10001

# #- @

321.53 kPa= 322 kPa,


The average discharge velocity through the orifice at any given time at any

time, in general V2 Kgz

12

L= +

We denote the diameter of the orifice by Do and the diameter of tank by D .The amount of water flows through the orifice during a time interval dt is

dv vdt=o o DK

gz dt4 12o

L

2p= + ...(i)

and the decrease in the volume of water in the tank

dv ( )Area dz= - D dz42p= - ...(ii)

From mass conservation, equation (i) must be equal to equation (ii)

DK

gz dt4 12o

L

2p+ D dz4

2p=- ( ve- sign shows decrease in volume)

dt ( )DD

gK z dz2

1 /

o

L2

21 2=- + -

Then the draining time is

dt0t

tf

=# ( )

DD

gK z dz2

1 /

o

L

z2

21 2

2

0=- + -

=#

tf DD

gK z2

1 2 /

o

L2

21 2

2

0#=- +

6 @ ( 2 )DD

gK

21 2

o

L2

2

# #=- + -

2DD

gK

21 2

o

L2

2

# #= + DD

gK2 1

o

L2

2

#= +



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2( . )( . )

..

0 14 5

9 811 0 5

2

2

# #= + 1584 26.4sec min= =

FM 6.40 Option (A) is correctSince flow rate

vo C A p12

d t 4r bD

#=-_ i

where 0.5Dd

105b = = =

0.605 .998 1 0.54 0 05 2 75000

42 #p

# # #=-]]

gg6 @

0.0150 / 54 /m s m h3 3-=


The velocity Vt .

. 7.64 /m sAv

4 0 050 0150

t 2#

p= = =o

] g

Since head loss due to pressure difference is known as non recoverable head loss. So

plossD K V2 t2r

# #= 1.8 .2998 7 64 2

# #= ] g

52427 52400 52.4Pa kPa-= =

FM 6.42 Option (D) is correct.For a pressure drop of pD across the orifice plate, the flow rate is expressed as

vo ( )

A C p12

o d 4r bD=-

...(i)

Since b 0.6Dd d

50= = =

d 30 cm=

Then Ao ( . )

0.07069 md4 4

0 32 22# #p p= = =

Therefore from equation (i)

.0 25 (0.07069) (0.61)( ) 1 0.6

p998

24

#

D# #=

- ] g6 @

pD 14600 /kg m s2-= 14.6 kPa=

FM 6.43 Option (D) is correct.The head loss between two measurement section is estimated by energy equation.

hL gp p

gV V

2w

1 2 22

12

r= - - - gp

gV V

2w

22

12

rD= - - z z1 2= ...(i)

Since for constant volume flow rate,

V D41

2

#p V d

42

2p=

V2 dD V

2

1#= b l

V1 ( . ). 1.27 /m sA

v0 5

0 25 4c

2#

#p

= = =o

Substituting in equation (i)

hL gp

gdD V V

gp

gdD V

2 2

1

w w

2

1

2

12

4

12

#

r rD D= -

-= -

-b bl l; ;E E



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hL . .

.

998 9 8114600

2 9 813050 1 1 27

42

# #

#

= --b l; E

1.49 0.5521 0.9379= - = 0.940 m,


Since b 0.6 Dd d

5= = = or 0.6 5 3 cmd #= =

Now vo . C A p60

0 33412

d 4r bD

# #= =-_ i

C dgh

4 1

2d

water

Hg water24

pr br r

# #=-

-

_

^

i

h

where p ghHg wr rD #= -^ h

or .60

0 334 0.613 .998 1 0.6

. h4 0 03

2 13550 998 9 814

2 # #p# #=

--

]]

^g

g

h

6 @

.60

0 334 0.00044 16.8 h4#=

0.00557 0.0074 h=

or h .. 0.57 m0 0074

0 00557 2

= =b l 57 cm=

FM 6.45 Option (B) is correct.The pressure drop across the venturi-meter is

pD p p gh ghw air1 2 r r= - = -

( )gh gh1w air airair

wr r r rr= - = -b l

Then flow rate vo ( )

( )( )

A Cp p

A Cgh

12

1

2 1o d

airo d

air

air

wair

41 2

4

#

r b r brr r

=-- =

-

-b l

( )

A Cgh

1

2 1o d

air

w

4brr

=-

-b l

where Ao ( . )

0.002827 md4 4

0 062 22#p p= = =

and b 0.40Dd

156= = =

Thus vo 0.002827 0.981 0.40

. . .2 9 81 0 4 1 2041000 1

4

# #

# #=-

-

]

b

g

l

6 @

0.2265 /m s3=Then the maximum mass flow rate that venturi-meter can measure is

m vr=o o 1.204 0.2265#= 0.273 /kg s=

FM 6.46 Option (A) is correct.The upstream density is

1r 1.48 /kg mRTp

287 353150000

1

1 3

#= = =

and b Dd

64

32= = =

The pressure difference measured by the mercury manometer



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p p1 2- . 9.81 0.37gh 13550 1 48Hg airr r #= - = -^ ]h g

pD 49200 Pa=Now the compressive venturi formula thus predicts :

mo C YAp p

1

2d t 4

1 1 2

br

#=--^ h

0.985 0.76 . .4 0 04

1 32

2 1 48 4920024#

#p# # #=

- b

]

l

g9 C

0.40 /kg s=

FM 6.47 Option (C) is correct.The flow velocity

V .

. 0.99 /m sAv

D4

36007

0 050 00195 4

2 2#

#p p

= = = =o] g

Also Vt .. 2.75 /m sA

v0 03

0 00195 4t

2#

# ,p

= =o] g

For 0.6Dd

53b = = = and 1.5K = , the head loss across the office is

hD 1.5 2 9.812.75 0.58 mh h K g

V2

2t

2 1

2

# ##

= - = = =] g; E

Hence h2 0.58 0.58 1 1.58 mh1= + = + =Then the piezometer change between (2) and (3) is due to friction loss

h h hf3 2- = f DL

gV2

2

# #= 0.023 . .. 0.115 m0 05

52 9 81

0 99 2

## #= =] g

or h3 0.115 1.58 0.115h2= + = + 1.695 1.7 m,=


From the pressure drop relation, the flow rates are

paD 21 1000dL v128a

a a4p

m#= =

o

or vao .. 0.0027 /m s128 0 104 76

21 10 0 083 43

# #

# # #p= =] g

and pbD 21000dL v128b

b b4p

m= =o

or vbo .. 0.0005 /m s128 0 104 61

21000 0 05 43

# #

# #p= =] g

For parallel pipe system

vo v va b= +o o

0.0027 0.0005= + 0.0032 /m s3=

FM 6.49 Option (D) is correct.For parallel system of pipe, head loss for each pipe must be same.When the minor losses are disregarded, head loss is



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hL fDL

gV2

2

= c m f DL

g Dv

21

42

2

# # # p= o

> H

f DL

D gv

216

2 4

2

#

#p# #= o

fgD

L v8 2 5

2

##

p= o

Thus vo fLh g D8

.L2

2 5#p#=

When the pipe length, friction factor and head loss is constant, the flow rate

becomes proportional to the D .2 5 in parallel connection.

vo kD .2 5= k= constant of proportionality

Let diameter of pipe B DB= and diameter of pipe A DA= .

Therefore vBo ( )k D .B

2 5=

vAo ( ) ( . )k D k D0 5. .A B

2 5 2 5= = ( )givenD D2AB=

vAo 0.177 ( ) 0.177k D v.B B

2 5# #= = o

Hence the flow rate of pipe A is decreased by a factor of 0.177.


For parallel pipe system the head losses are the same for each pipe

hf gdfL v

gdfL v

gdfL v8 8 8

215

1 12

225

2 22

215

3 32

p p p= = =

o o o

v v v1 2 3+ +o o o 0.056 /m s3= ...(i)

or hf . ..

. ..v v

9 81 0 18 0 0275 900

9 81 0 128 0 0275 800

2 512

2 522

# #

# # #

# #

# # #p p

= =o o

] ]g g

. .

. v9 81 0 08

8 0 0275 6002 5

32

# #

# # #p

=o

] g

hf 204501 73053 416059v v v12

22

32= = =o o o

or v22o 2.8v v73053

20450112

12= =o o or 1.673v v2 1=o o

v22o 0.492v v416059

20450112

12= =o o or 0.7014v v3 1=o o

Now from equation (i),

1.673 0.7014v v v1 1 1+ +o o o 0.056=

or 3.374v1o .0 056= ] g

v1o ..

3 3740 056= or 0.0166 /m sv1

3=o

and v2o =1.673 1.673 0.0186 0.0277 /m sv13

#= =o

v3o 0.7014 0.7014 0.0166 0.0116 /m sv13

#= = =o


Since hf . .. .

g df L v8

9 81 0 18 0 0275 900 0 0166

215

1 1 12

2 5

2

# #

# # #

# #

# # #

p p= =

o

]]g

g

56.35 m=Then pressure drop

p ghfrD = 998 9.81 56.35# #=

551 . 550 kPa687 9 ,=



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FM 6.52 Option (A) is correct.The head loss is the same through pipes AC and BC (parallel system)

gp

rD h h h hfA fC fB fC= + = +

f dL

gV f d

Lg

V fdL

gV fd

Lg

V2 2 2 2

A

A

B

C

B

B

C

C

2 2 2 2

# # # #= + = +c c c cm m m m

998750000 . . . .

V V0 022 0 08250

2 0 022 0 08150

2A C2 2

# # # #= +b bl l ...(i)

and 998750000 . . . .

V V0 022 0 08100

2 0 022 0 08150

2B C2 2

# # # #= +b bl l ...(ii)

From equation (i) and (ii),

751.50 34.375 20.625V VA C2 2= + ...(iii)

751.50 13.75 20.625V VB C2 2= + ...(iv)

From equation (iii) and (iv),

VA2 .

. ..

. . .V34 375

751 50 20 62534 375

751 50 20 625 5 35C2 2

# #= - = - ] g

VA 2.16 /m s=

and VB2 .

. . .13 75

751 50 20 625 5 35 2#= - ] g

VB 3.42 /m s= Now for parallel pipe A and B,

vABo v vA B= +o o

vABo V A V AA A B B= + V V AA B= +^ h D DA B=

. . .2 16 3 42 4 0 08 2p# #= +^ ]h g

0.0280 /m s3=

And for Pipe in series

vABo 0.0280 /m svC3= =o

Hence the total volume flow rate

v v vAB C= =o o o 0.0280 /m s3=

101 /m hr3-

FM 6.53 Correct option is (B)Let roughnessh = . Thus h sd=

Where sd uV5*= ...(i)

And u* lD p

4

/ /w

1 2 1 2

#rt

rD= =c dm n where l

D p4wtD=

u* fV8

/2 1 2

= c m since p fDl V2

1 2# rD =

Hence u* 80.0125 (2)

0.0791 /m s2 /1 2

#= =; E

and from eq (i), sd .( . )

2.31 10 0.0231m mm0 07915 3 65 10 7

5##= = =

--

If the roughness element is smaller than 0.0231 mm it lies within the laminar sublayer.



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FM 6.54 Correct option is (D)

Assume that fluid flows from A to B and A to C .

v1o v v2 3= +o o

or (0.1) V42

1#p (0.08) (0.08)V V4 4

22

23# #

p p= +

V1 0.64 0.64 . ( )V V V V0 642 3 2 3= + = + ...(i)

For fluid flowing from A to B with p p 0A B= = and V V 0A B= =

zA z f Dl

gV f D

lg

V2 2B 1

1

1 12

22

2 22

# #= + +

60 20- 0.015 . . 0.020 . .V V

0 1200

2 9 81 0 08200

2 9 8112

22

# ## # # #= +

or 40 . .V V1 529 2 5512

22= + ...(ii)

Similarly, for fluid flowing from A to C with 0p pA C= = and V V 0A C= =

zA z f Dl

gV f D

lg

V2 2C 1

1

1 12

33

3 32

= + +

60 0- 0.015 . ( . )0.020 . ( . )

V V0 1200

2 9 81 0 08400

2 9 8112

32

## # #= +

or 60 . .V V1 529 5 1012

32= + ...(iii)

Solve equation (i), (ii) and (iii) for ,V V1 2 and V3. From equation (i) and (iii)

60 1.529 (0.64) ( ) 5.10V V V22 3

232

#= + + 60 0. ( ) .V V V626 5 102 3

232+ +

.95 8 ( ) .V V V8 142 32

32= + + ...(iv)

Subtract equation (ii) from equation (iii)

60 40- 5.10 2.55V V32

22= - or .V V2 7 842 3

2= - ...(v)

Thus, from equation (iv) and (v)

0 8.14 ( . ) 95.8V V V2 7 8432

32

32= + - + -

This can be simplified to

.V V2 2 7 843 32 - . . V103 6 11 14 3

2= - ...(vi)

Squaring both the sides and rearrange to get

1 . 92.5V V9 6334

32- + 0= which can be solved by the quadratic formula

to give

V32

. ( . ) .2

19 63 19 63 4 92 52! #= - -

.11 77= or .7 86

Thus V3 3.43 /m s= or 2.80 /m sV3 =



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Note : The value 3.43 /m sV3 = is not a solution of the original equations,

equation (i), (ii) and (iii). With this value the right hand side of equation (vi) is

negative (i.e . . . . ( . ) .V103 6 11 14 103 6 11 14 3 43 24 532 2- = - =- ). As seen from the

left hand side of equation (vi), this cannot be. This extra root was introduced

by squaring equation (vi).

Thus v3o (0.08) 2.80 0.0141 /m sA V 43 32 3

# #p= = =

Also, from equation (iii)

60 1.529 5.10 (2.80)V12 2

#= + or 3.62 /m sV1 =

or v1o (0.10) 3.62 0.0284 /m sA V 41 12 3

# #p= = =

and from equation (i), we get

3.62 0.64 0.64 2.80V2 #= + or 2.86 /m sV2 =

or v2o (0.08) 2.86 0.014 /m sA V 4 42 22 3p

# #= = =

FM 6.55 Option (A) is correct.We take point (1) at the free surface of tank and point (2) at the exit of the pipe. Then, the energy equation between these two points.

gp

gV z2

11

12

1r a+ + gp

gV z h2 L

22

22

2r a= + + +

Since ,p p patm1 2= = ,V 01 , 0,z2 = and V 02 = (Velocity head negligible)

Above equation becomes

z1 hL= and h hL=where h is the liquid height in the tank at any time t .

Now hL f dL

gV2

2

#=

For fully developed laminar flow

f /Re Vd

64 64n= =

Thus hL Vd d

Lg

V642

2

n# #=

dL

gV6422

n#=

The average velocity

V Av

dv

dv

4

4c

2 2

p p= = =o o o

hL d

Lg d

v6421 4

2 2n

p# #= o

dL

d gv

g dLv64

24 128

2 2 4#

np p

n#= =o o

h hL = g d

Lv1284p

n= o ...(i)

From mass conservation, above equation must be equal to change of liquid height in the tank.

vo Ddtdh

42p=-

Now equation (i) becomes

h g d

L Ddtdh128

44

2

pn p

# #=- gdLD

dtdh32

4

2n=-



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dt gdLD

hdh32

4

2n#=-

Integrating from t 0= when h H= to t t= when h h=

t lngdLD

hH32

4

2n= b l


The bottom side of the triangle is

2 2 40 2.57sin cmc,#=

and A 2.57 cos21 2 40c# #= ] g 1.97 cm2=

Perimeter 2 2 2.57 6.57 cm= + + =Then Hydraulic diameter

Dh .. 1.20perimeter cmA4

6 574 1 97#= = =

Re VDD

hh m

r= .. 2010 104

870 2 0 0120# # ,= (Laminar flow)

Then f . . 0.263Re52 9

20152 9= = =

Hence pD f DL V2h

2r# # #=

. ..0 263 0 012

0 62

870 2 2# # #= ] b b ]g l l g

23000 23Pa kPa- =

FM 6.57 Correct option is (A)

V ( . )

1.27 /m sAv

4 0 024 10

2

4

#

#p= = =

-o



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So that Re .. . 115 2100VD VD2 2 101 27 0 02 <4

#

#m

rn= = = =- r

m n=

The flow is laminar and flow rate is given by ( 90cq = )

vo ( )

lp l D128

4

mp gD= -

or p p pD

lv l1281 2 4p

m gD = - = +o

...(i)

Thus g . . 0.87 9.81 8.53 /kN mS G H O 32# #= = =g

and m ( . . ) 2.2 10 0.87 999 0.191 /N s mS G H O4 2

2nr n r # # # -= = = =-

Equation (i) gives

pD ( . )

. 8.53 10 40 020

128 0 191 4 4 104

43

#

# # # ##p #= +

-

. 10 / . /N m kN m1 119 111 95 2 2#= = ...(ii)

From manometer equation,

p2 p h h hm1 1 2g g g= - + -

Where mg g. . 1.3 9.81 12.74 /kN mS G H Om3

2 #= = =and h2 h l h1= + - or h h h l2 1+ = +Thus p p1 2- g( ) ( )p h h h h lm m2 1 g g g gD= = + - =- - +

...(iii)Combine equation (ii) and (iii), we get

.111 9 (12.74 8.53) 8.53 4h #=- - +or h 18.5 m=-Note: Since h 0< , the manometer is displaced in the direction opposite that shown in the original figure.

FM 6.58 Option (B) is correct.The divided cross section of the pipe into 1 cm thick annual regions is shown in table.The flow rate is to be determined by using midpoint velocity values for each section. Therefore

vo [ ]V dA V r ravg cA

avg out in2 2

c

pS= = -#

. . (0.01) 0 . . 0.02 0.0126 4 6 1

26 1 5 22 2 2p p# # #= + - + + -b b ] ]l l g g6 6@ @

. . . . . . 0.04 0.0325 2 4 4 0 03 0 02 2

4 4 2 0 2 22 2p p# # #+ + - + + -b ] ] b ] ]l g g l g g6 6@ @

. 0.05 0.04022 0 0 2 2p# #+ + -b ] ]l g g6 @

vo 0.0297 /m s3=

FM 6.59 Correct option is (C)

For laminar flow Re 2100# or VD 2100#mr



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Thus the minimum h is h 0= (no flow) and the maximum h is for Re 2100= .

2100 .. .V

0 19 818900 0 023# #

=b l

gr g=

or V 10.06 /m s=For the flowing fluid, Bernoulli’s equation gives

pg

V z21 1

2

1g + + pg

V z fDl

gV

2 22 2

2

2

2

g= + + + z z1 2=

and V V V1 2= =

Thus p p1 2- f Dl

gV2

2

# # g=

(i)

And for laminar flow

f Re64= or .f 2100

64 0 0305= =

From equation (i),

pD 0.0305 ..

.( . )

8900p p 0 0230 5

2 9 8110 06

1 2

2

## # #= - =

304 /N m39 2=From manometer equation, we get

( ) . .p H h S G h HH Ooil oil1 2g g+ + - -g p2= p p p1 2D = - ( . . )S G hH O2 g= -g

or h ( )

0.5 m7 9800 8900

30439 10#

= - =

Hence 0.5 mh0 10# #


We take point (1) at free surface of the tank and point (2) at the reference level at exit. By applying energy equation for a control volume between these two points

gp

gV z h2 pump

11

12

1r a+ + + gp

gV z h h2 turbine L

22

22

2r a= + + + +

Since p p patm1 2= = , 0,z2 = h 0turbine = , and 0V1 ,

z1 gV h2 L2

22

a= + hL fDL

gV2

22

#=

So z1 gV fD

Lg

V2 22

22

22

a #= +

V2 /fL Dgz2

2

1

a=

+ ^ h

/fL Dgz

12=

+ ^ h 12a =



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where z is the water height relative to the center of the orifice at that time.

Now, the flow rate of water through the pipe during a time interval dt is

dvo 1 /

vdt A V dt DfL Dgz dt4

2c

o

o2

2p# #= = =

+o

^ h

where Do is the diameter of orifice.

From conservation of mass

Flow rate of water through the pipe tandecrease in the volume of water k=

1 /

DfL Dgz dt4

2o

o

2p# + ^ h

( )A dz,c tank= - D dz42

#p=-

dt 1 /

DD

gzfL D

dz2o

o2

2

=-+ ^ h

1 /

DD

gfL D

z dz2/

o

o2

21 2

#=-+ -^ h

By integrating above equation from t 0= when z z1= to t tf= when z 0=

(completely drained tank)

dtt

t

0

f

=# /

DD

gfL D z dz2

1 /

o

oz z

2

21 2

0

1

#=- + -

=^ h #

tf /DD

gfL D z2

1

21/

o

o

z

2

2 1 2 0

1

#=- + ^ h

/DD

gz fL D2 1

o

o2

21#= + ^ h$ .

Substituting the numerical values, we get

tf ( . )10

.

. .0 039 81

2 2 1 0 022 0 0325

2

2# ##= +]

bg

l

311965 seconds= 86.6 hours,

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