Q: Energy Loss in Pipe Flow – Exit Loss to Atmosphere What is the loss of head at the exit of a pipe discharging to a large reservoir or atmosphere?

Introduction: Minor losses in pipe flow include losses at entrances, exits, bends, valves, and fittings. The exit loss is a standard case when a pipe discharges into a large tank or the atmosphere. Given Data / Assumptions: Pipe exits into a large reservoir or free atmosphere. Velocity in the reservoir is effectively zero. v denotes the average velocity inside the pipe at the exit section. Concept / Approach: The kinetic energy per unit weight carried by the jet inside the pipe is v^2 / (2g). On discharge to a large reservoir, this kinetic energy is dissipated as turbulence, so the corresponding head is lost. Hence the exit loss coefficient K_exit = 1. Step-by-Step Solution: Step 1: Write exit loss formula: h_exit = K_exit * v^2 / (2g).Step 2: For discharge to a large tank or atmosphere, K_exit = 1.Step 3: Therefore, h_exit = v^2 / (2g). Verification / Alternative check: Energy equation between a section just inside the pipe and a point in the reservoir at the same elevation shows the velocity head is not recovered in pressure but is dissipated, confirming the result. Why Other Options Are Wrong: 0.5 * v^2 / (2g): Entrance, not exit, typically uses approximately 0.5 for sharp edges.1.5 * v^2 / (2g) and 2 * v^2 / (2g): No standard basis for such high coefficients at a plain exit.Negligible: The velocity head is not recovered; the loss is significant. Common Pitfalls: Confusing entrance and exit coefficients or assuming kinetic energy is recuperated in pressure at the outlet to a large tank. Final Answer: v^2 / (2g)

Q: Types of Fluid Flow – Compressibility and Rotation Which of the following statements about compressible flow, incompressible flow, and rotational flow is correct?

Introduction: Classifying flows by compressibility and rotation is foundational in fluid mechanics. This question checks recognition of standard definitions and terminology. Given Data / Assumptions: Terms: compressible, incompressible, rotational. Context: general fluid flow behavior, small to moderate pressure changes unless otherwise noted. Concept / Approach: Compressible flow involves appreciable density (and thus volume) changes with pressure or temperature variations. Incompressible flow assumes density remains effectively constant. Rotational flow implies nonzero vorticity where fluid elements experience spin about their own axes. Step-by-Step Solution: Step 1: Evaluate statement A. If density changes significantly, the fluid element volume changes; this characterizes compressible flow. The statement is acceptable in general terms.Step 2: Evaluate statement B. If the volume, equivalently the density, remains constant, the flow is modeled as incompressible; this is the standard textbook definition.Step 3: Evaluate statement C. Rotation of particles indicates finite vorticity; by definition this is rotational flow.Step 4: Since A, B, and C are all correct, the choice is “all of the above”. Verification / Alternative check: Vorticity-based definitions and density-change criteria are consistent across common references for fluid kinematics and dynamics. Why Other Options Are Wrong: None of the above: Incorrect because each individual statement aligns with standard definitions. Common Pitfalls: Assuming liquids are always perfectly incompressible. Under very high pressures, even liquids can exhibit small compressibility; the definition here is about the model used, not the absolute impossibility of volume change. Final Answer: all of the above

Q: Open Channel Flow – Most Economical Rectangular Section For a rectangular channel, the most economical section is the one whose hydraulic radius equals which proportion of the flow depth?

Introduction: The most economical (most efficient) section minimizes wetted perimeter for a given area, maximizing discharge or minimizing frictional loss. In rectangular channels, there are simple geometric relations at optimum. Given Data / Assumptions: Prismatic, wide rectangular channel is not assumed; exact rectangle is considered. Uniform flow conditions with Chezy or Manning resistance concept. Hydraulic radius R = A / P, where A is area and P is wetted perimeter. Concept / Approach: For a rectangle of breadth b and depth y: A = b * y and P = b + 2y. At the most economical section, dP/dy under constant area or classical optimization yields b = 2y. Substituting b = 2y gives R = A / P = (2y^2) / (2y + 2y) = y/2. Thus R equals half the depth. Step-by-Step Solution: Step 1: Write A = b * y and P = b + 2y.Step 2: Optimize for maximum R = A / P, or equivalently minimum P for given A.Step 3: The condition for optimum gives b = 2y.Step 4: Compute R: R = (b * y) / (b + 2y) = (2y^2) / (2y + 2y) = y/2. Verification / Alternative check: Using Manning discharge Q proportional to A * R^(2/3) confirms maximizing R increases discharge for given slope and roughness, reinforcing the same geometry b = 2y. Why Other Options Are Wrong: Half the breadth: At optimum, R relates to depth, not breadth; breadth equals twice depth.Twice the depth / Twice the breadth / One-third of the depth: None match the derived R = y/2 relation. Common Pitfalls: Memorizing b = 2y but forgetting the associated R relationship. Also mixing up hydraulic mean depth with hydraulic radius for non-rectangular sections. Final Answer: half the depth

Q: Flow Classification – Unsteady and Non-Uniform A flow through an expanding tube at an increasing rate with time is best described as which type of flow?

Introduction: Flow can vary with time and space. The question tests whether the student can correctly classify a situation with both time variation and spatial variation. Given Data / Assumptions: Expanding tube means cross-section varies with position. Rate of flow is increasing with time. Concept / Approach: Uniform vs non-uniform refers to spatial variation at a fixed time. Steady vs unsteady refers to time variation at a fixed point. An expanding tube creates spatial variation in velocity distribution, and an increasing discharge with time creates temporal variation. Step-by-Step Solution: Step 1: Spatial check: Cross-section changes with x, so velocity differs along the tube, implying non-uniform.Step 2: Temporal check: Flow rate increases with time, so velocity at a point changes with time, implying unsteady.Step 3: Combine: The correct classification is unsteady and non-uniform. Verification / Alternative check: Continuity for incompressible flow, A(x) * V(x,t) = Q(t). If Q increases with time and A depends on x, then V depends on both x and t, confirming the classification. Why Other Options Are Wrong: Yes / No: These are not flow types and do not classify the physics.Steady uniform / Steady non-uniform: Both contradict the explicit time variation. Common Pitfalls: Assuming non-uniform automatically means unsteady. A diffuser with constant Q is non-uniform but steady; here Q is changing, so it is both. Final Answer: Unsteady and non-uniform flow

Q: Energy Loss in Pipe Flow – Entrance Loss for Sharp-Edged Inlet What is the head loss at the entrance of a pipe with a sharp-edged inlet (v is the average velocity in the pipe)?

Introduction: Entrance loss is one of the standard minor losses in internal flows. Its magnitude depends on how the fluid is introduced into the pipe, especially the inlet geometry. Given Data / Assumptions: Entrance is sharp-edged (square-edged). v is the uniform mean velocity inside the pipe. Discharge into the pipe from a large reservoir or plenum. Concept / Approach: The head loss due to entrance is expressed as h_e = K_e * v^2 / (2g). For a sharp-edged entrance, empirical data give K_e approximately 0.5. Streamline contraction and subsequent expansion to fill the pipe core create energy dissipation. Step-by-Step Solution: Step 1: Write the generic relation: h_e = K_e * v^2 / (2g).Step 2: For a sharp-edged entrance, take K_e ≈ 0.5.Step 3: Therefore, h_e ≈ 0.5 * v^2 / (2g). Verification / Alternative check: Smoother bell-mouth inlets have much smaller K_e (approaching 0.04 to 0.1), while square-edged inlets have K_e close to 0.5. This range corroborates the selected value. Why Other Options Are Wrong: v^2 / (2g) and 2 * v^2 / (2g): Too large for a plain sharp edge; typical only for combinations of fittings or exits.1.5 * v^2 / (2g): No standard entrance configuration yields such a high coefficient.Approximately zero: Only for well-rounded bell-mouth entrances, not sharp-edged. Common Pitfalls: Memorizing a single coefficient for all entrances, or mixing entrance and exit losses. Always link the coefficient to the geometry. Final Answer: 0.5 * v^2 / (2g)

Q: Orifices and Mouthpieces – Vena Contracta Location In a short cylindrical external mouthpiece running free, the vena contracta forms at a distance from the outlet equal to what fraction of the orifice diameter?

Introduction: The vena contracta is the section where a jet from an orifice or mouthpiece attains minimum cross-section and maximum velocity just downstream of the opening. Its location depends on the geometry. Given Data / Assumptions: Short cylindrical external mouthpiece. Running free (not drowned or submerged at the outlet). Diameter of the orifice is the reference scale. Concept / Approach: For a short external cylindrical mouthpiece discharging freely, experiments show that vena contracta typically occurs at about half a diameter downstream from the outlet plane of the orifice. Step-by-Step Solution: Step 1: Recognize configuration: short external mouthpiece with free discharge.Step 2: Use empirical rule-of-thumb for position: approximately 0.5 * d from the outlet.Step 3: Map to the option phrasing: one-half of the diameter. Verification / Alternative check: Classic hydraulics texts present diagrams placing the contracted jet minimum at about 0.5 d downstream for this geometry. Why Other Options Are Wrong: Equal to diameter / one-fourth / one-third / three-fourths: Do not match the standard empirical placement for short cylindrical mouthpieces running free. Common Pitfalls: Confusing mouthpieces with thin orifices or long pipes where contraction behavior differs. The fraction changes with geometry and whether the jet is submerged. Final Answer: one-half of the diameter

Q: Small Rectangular Orifice – Discharge Formula What is the discharge through a small rectangular orifice in a thin plate in terms of Cd, area a, and head h above the orifice centerline?

Introduction: Flow through small sharp-edged orifices is estimated using Torricelli's theorem corrected by a discharge coefficient Cd to account for contraction and viscous effects. Given Data / Assumptions: Small, sharp-edged rectangular orifice on a vertical tank wall. Head h is measured above the orifice center. a is the geometric area of the orifice opening. Cd is the coefficient of discharge accounting for contraction and friction. Concept / Approach: Ideal velocity from Torricelli: V_ideal = sqrt(2 * g * h). Ideal discharge would be a * V_ideal. Real discharge requires multiplying by Cd. Hence Q = Cd * a * sqrt(2 * g * h). Step-by-Step Solution: Step 1: Start with Torricelli: V = sqrt(2 * g * h).Step 2: Multiply by area to get ideal discharge: Q_ideal = a * sqrt(2 * g * h).Step 3: Apply coefficient of discharge: Q = Cd * Q_ideal = Cd * a * sqrt(2 * g * h). Verification / Alternative check: Dimensional check confirms units of Q are volume per unit time. Empirical values of Cd for sharp-edged orifices (about 0.60 to 0.65) support the corrected formula. Why Other Options Are Wrong: Cd * a * (2 * g * h): Missing the square root; dimensionally incorrect.Cd / a * sqrt(2 * g * h): Inverts area, giving wrong dimensions.a * sqrt(2 * g * h) / Cd: Dividing by Cd overestimates flow; Cd < 1.Cd * h * sqrt(2 * g * a): Mixes variables improperly; dimensionally wrong. Common Pitfalls: Forgetting the square root on 2 * g * h or using head to the wrong datum. Also confusing Cd with coefficients of contraction or velocity individually. Final Answer: Q = Cd * a * sqrt(2 * g * h)

Q: Accelerated Tanks – Free Surface Inclination An open tank of liquid is accelerated up an inclined plane. The inclination of the liquid free surface varies with the tank acceleration according to which relation?

Introduction: When a container of liquid undergoes linear acceleration, the free surface tilts until it becomes perpendicular to the resultant of gravitational acceleration and the imposed acceleration. This question asks how the inclination depends on the acceleration magnitude. Given Data / Assumptions: Open tank with a free surface. Tank accelerated along an incline with constant acceleration a. Gravity g acts vertically downward. Concept / Approach: The free surface aligns perpendicular to the resultant acceleration vector formed by g and the applied acceleration a. The tilt angle theta measured from the horizontal satisfies tan(theta) = a_parallel / g if a acts along the tank axis. Hence theta increases as a increases; for small angles, theta ≈ a / g, showing direct proportionality. Step-by-Step Solution: Step 1: Represent accelerations: gravity g downward and tank acceleration a along the incline.Step 2: The resultant acceleration magnitude is sqrt(g^2 + a^2), and the free surface is perpendicular to this vector.Step 3: The inclination angle theta relative to horizontal satisfies tan(theta) = a / g for the along-slope component.Step 4: Therefore, theta increases with a and is approximately proportional to a for small angles, matching “directly proportional”. Verification / Alternative check: Plotting theta vs a from tan(theta) = a / g shows monotonic increase; for a much less than g, theta ≈ a / g in radians, confirming proportionality. Why Other Options Are Wrong: Equal to the acceleration: Dimensionally meaningless; angle is not in acceleration units.Inversely proportional: Contradicts tan(theta) = a / g.Independent or proportional to acceleration squared: Both inconsistent with the tan relation. Common Pitfalls: Forgetting that the free surface is always perpendicular to the resultant acceleration and misinterpreting “inclination” as a length rather than an angle. Final Answer: directly proportional

Q: Pipe flow regime classification – threshold based on Reynolds number The flow in a pipe is considered turbulent when the Reynolds number is:

Introduction: Reynolds number (Re) classifies internal flows into laminar, transitional, and turbulent regimes. This question uses the conventional threshold values adopted in many practical handbooks for smooth pipes. Given Data / Assumptions: Reynolds number is defined as Re = (rho * V * D) / mu for pipe flow. Geometry: circular pipe; fluid properties treated as constant. Typical thresholds: laminar at low Re, turbulent at higher Re. Concept / Approach: In many textbooks, laminar is Re < 2000, transitional is 2000–~2800 (sometimes 2000–4000), and turbulent is above the upper threshold. The options in this question take 2800 as the practical cutoff. Step-by-Step Solution: 1) Identify regime ranges used in the options.2) Laminar: less than 2000 → not turbulent.3) Transitional: 2000 to 2800 → neither fully laminar nor fully turbulent.4) Above 2800 → turbulent by the convention implied in the options. Verification / Alternative check: Engineering references often cite laminar 4000, with a gray zone in between. Using 2800 as a cutoff is a simplified convention consistent with many MCQs. Why Other Options Are Wrong: Less than 2000: clearly laminar. Between 2000 and 2800: transitional, not definitively turbulent. None of these: incorrect because 'more than 2800' is acceptable for turbulent per the stated convention. Common Pitfalls: Forgetting that thresholds vary slightly by source; mixing criteria for pipes with external flows. Final Answer: more than 2800

Question 1

Compressible-Flow Basics – Definition of Subsonic Regime A flow is called subsonic if its Mach number falls in which range?

Accepted Answer

Introduction: Mach number M is the ratio of the local flow speed to the local speed of sound. Classifying regimes by M is fundamental in compressible flow and gas dynamics. This question checks the textbook definition of subsonic flow. Given Data / Assumptions: Definition uses Mach number M = V / a. Standard regime splits: subsonic, transonic, supersonic, hypersonic. Concept / Approach: By convention: subsonic M < 1, transonic roughly 0.8 to 1.2, supersonic M > 1 up to about 5, and hypersonic typically M > 5 or 6. The neat definition for subsonic is therefore M less than unity. Step-by-Step Solution: Step 1: Recall that unity means M = 1.Step 2: Subsonic implies the flow speed is lower than the speed of sound.Step 3: Therefore, select the option that states M is less than 1. Verification / Alternative check: Any aerodynamics text will define subsonic with M < 1, and reserve “unity” for sonic and “greater than 1” for supersonic/hypersonic, confirming the choice. Why Other Options Are Wrong: Unity: M = 1 corresponds to sonic flow, not subsonic.Between 1 and 6: This spans supersonic to hypersonic ranges.More than 6: Hypersonic; far from subsonic.Between 0.8 and 1 only: That is a subset of transonic, not the entire subsonic regime. Common Pitfalls: Confusing transonic margins with the exact definition, or assuming a single cutoff like 0.8. Always use the strict inequality for definitions. Final Answer: less than unity

Question 2

Energy Loss in Pipe Flow – Exit Loss to Atmosphere What is the loss of head at the exit of a pipe discharging to a large reservoir or atmosphere?

Accepted Answer

Introduction: Minor losses in pipe flow include losses at entrances, exits, bends, valves, and fittings. The exit loss is a standard case when a pipe discharges into a large tank or the atmosphere. Given Data / Assumptions: Pipe exits into a large reservoir or free atmosphere. Velocity in the reservoir is effectively zero. v denotes the average velocity inside the pipe at the exit section. Concept / Approach: The kinetic energy per unit weight carried by the jet inside the pipe is v^2 / (2g). On discharge to a large reservoir, this kinetic energy is dissipated as turbulence, so the corresponding head is lost. Hence the exit loss coefficient K_exit = 1. Step-by-Step Solution: Step 1: Write exit loss formula: h_exit = K_exit * v^2 / (2g).Step 2: For discharge to a large tank or atmosphere, K_exit = 1.Step 3: Therefore, h_exit = v^2 / (2g). Verification / Alternative check: Energy equation between a section just inside the pipe and a point in the reservoir at the same elevation shows the velocity head is not recovered in pressure but is dissipated, confirming the result. Why Other Options Are Wrong: 0.5 * v^2 / (2g): Entrance, not exit, typically uses approximately 0.5 for sharp edges.1.5 * v^2 / (2g) and 2 * v^2 / (2g): No standard basis for such high coefficients at a plain exit.Negligible: The velocity head is not recovered; the loss is significant. Common Pitfalls: Confusing entrance and exit coefficients or assuming kinetic energy is recuperated in pressure at the outlet to a large tank. Final Answer: v^2 / (2g)

Question 3

Types of Fluid Flow – Compressibility and Rotation Which of the following statements about compressible flow, incompressible flow, and rotational flow is correct?

Accepted Answer

Introduction: Classifying flows by compressibility and rotation is foundational in fluid mechanics. This question checks recognition of standard definitions and terminology. Given Data / Assumptions: Terms: compressible, incompressible, rotational. Context: general fluid flow behavior, small to moderate pressure changes unless otherwise noted. Concept / Approach: Compressible flow involves appreciable density (and thus volume) changes with pressure or temperature variations. Incompressible flow assumes density remains effectively constant. Rotational flow implies nonzero vorticity where fluid elements experience spin about their own axes. Step-by-Step Solution: Step 1: Evaluate statement A. If density changes significantly, the fluid element volume changes; this characterizes compressible flow. The statement is acceptable in general terms.Step 2: Evaluate statement B. If the volume, equivalently the density, remains constant, the flow is modeled as incompressible; this is the standard textbook definition.Step 3: Evaluate statement C. Rotation of particles indicates finite vorticity; by definition this is rotational flow.Step 4: Since A, B, and C are all correct, the choice is “all of the above”. Verification / Alternative check: Vorticity-based definitions and density-change criteria are consistent across common references for fluid kinematics and dynamics. Why Other Options Are Wrong: None of the above: Incorrect because each individual statement aligns with standard definitions. Common Pitfalls: Assuming liquids are always perfectly incompressible. Under very high pressures, even liquids can exhibit small compressibility; the definition here is about the model used, not the absolute impossibility of volume change. Final Answer: all of the above

Question 4

Open Channel Flow – Most Economical Rectangular Section For a rectangular channel, the most economical section is the one whose hydraulic radius equals which proportion of the flow depth?

Accepted Answer

Introduction: The most economical (most efficient) section minimizes wetted perimeter for a given area, maximizing discharge or minimizing frictional loss. In rectangular channels, there are simple geometric relations at optimum. Given Data / Assumptions: Prismatic, wide rectangular channel is not assumed; exact rectangle is considered. Uniform flow conditions with Chezy or Manning resistance concept. Hydraulic radius R = A / P, where A is area and P is wetted perimeter. Concept / Approach: For a rectangle of breadth b and depth y: A = b * y and P = b + 2y. At the most economical section, dP/dy under constant area or classical optimization yields b = 2y. Substituting b = 2y gives R = A / P = (2y^2) / (2y + 2y) = y/2. Thus R equals half the depth. Step-by-Step Solution: Step 1: Write A = b * y and P = b + 2y.Step 2: Optimize for maximum R = A / P, or equivalently minimum P for given A.Step 3: The condition for optimum gives b = 2y.Step 4: Compute R: R = (b * y) / (b + 2y) = (2y^2) / (2y + 2y) = y/2. Verification / Alternative check: Using Manning discharge Q proportional to A * R^(2/3) confirms maximizing R increases discharge for given slope and roughness, reinforcing the same geometry b = 2y. Why Other Options Are Wrong: Half the breadth: At optimum, R relates to depth, not breadth; breadth equals twice depth.Twice the depth / Twice the breadth / One-third of the depth: None match the derived R = y/2 relation. Common Pitfalls: Memorizing b = 2y but forgetting the associated R relationship. Also mixing up hydraulic mean depth with hydraulic radius for non-rectangular sections. Final Answer: half the depth

Question 5

Flow Classification – Unsteady and Non-Uniform A flow through an expanding tube at an increasing rate with time is best described as which type of flow?

Accepted Answer

Introduction: Flow can vary with time and space. The question tests whether the student can correctly classify a situation with both time variation and spatial variation. Given Data / Assumptions: Expanding tube means cross-section varies with position. Rate of flow is increasing with time. Concept / Approach: Uniform vs non-uniform refers to spatial variation at a fixed time. Steady vs unsteady refers to time variation at a fixed point. An expanding tube creates spatial variation in velocity distribution, and an increasing discharge with time creates temporal variation. Step-by-Step Solution: Step 1: Spatial check: Cross-section changes with x, so velocity differs along the tube, implying non-uniform.Step 2: Temporal check: Flow rate increases with time, so velocity at a point changes with time, implying unsteady.Step 3: Combine: The correct classification is unsteady and non-uniform. Verification / Alternative check: Continuity for incompressible flow, A(x) * V(x,t) = Q(t). If Q increases with time and A depends on x, then V depends on both x and t, confirming the classification. Why Other Options Are Wrong: Yes / No: These are not flow types and do not classify the physics.Steady uniform / Steady non-uniform: Both contradict the explicit time variation. Common Pitfalls: Assuming non-uniform automatically means unsteady. A diffuser with constant Q is non-uniform but steady; here Q is changing, so it is both. Final Answer: Unsteady and non-uniform flow

Question 6

Energy Loss in Pipe Flow – Entrance Loss for Sharp-Edged Inlet What is the head loss at the entrance of a pipe with a sharp-edged inlet (v is the average velocity in the pipe)?

Accepted Answer

Introduction: Entrance loss is one of the standard minor losses in internal flows. Its magnitude depends on how the fluid is introduced into the pipe, especially the inlet geometry. Given Data / Assumptions: Entrance is sharp-edged (square-edged). v is the uniform mean velocity inside the pipe. Discharge into the pipe from a large reservoir or plenum. Concept / Approach: The head loss due to entrance is expressed as h_e = K_e * v^2 / (2g). For a sharp-edged entrance, empirical data give K_e approximately 0.5. Streamline contraction and subsequent expansion to fill the pipe core create energy dissipation. Step-by-Step Solution: Step 1: Write the generic relation: h_e = K_e * v^2 / (2g).Step 2: For a sharp-edged entrance, take K_e ≈ 0.5.Step 3: Therefore, h_e ≈ 0.5 * v^2 / (2g). Verification / Alternative check: Smoother bell-mouth inlets have much smaller K_e (approaching 0.04 to 0.1), while square-edged inlets have K_e close to 0.5. This range corroborates the selected value. Why Other Options Are Wrong: v^2 / (2g) and 2 * v^2 / (2g): Too large for a plain sharp edge; typical only for combinations of fittings or exits.1.5 * v^2 / (2g): No standard entrance configuration yields such a high coefficient.Approximately zero: Only for well-rounded bell-mouth entrances, not sharp-edged. Common Pitfalls: Memorizing a single coefficient for all entrances, or mixing entrance and exit losses. Always link the coefficient to the geometry. Final Answer: 0.5 * v^2 / (2g)

Question 7

Orifices and Mouthpieces – Vena Contracta Location In a short cylindrical external mouthpiece running free, the vena contracta forms at a distance from the outlet equal to what fraction of the orifice diameter?

Accepted Answer

Introduction: The vena contracta is the section where a jet from an orifice or mouthpiece attains minimum cross-section and maximum velocity just downstream of the opening. Its location depends on the geometry. Given Data / Assumptions: Short cylindrical external mouthpiece. Running free (not drowned or submerged at the outlet). Diameter of the orifice is the reference scale. Concept / Approach: For a short external cylindrical mouthpiece discharging freely, experiments show that vena contracta typically occurs at about half a diameter downstream from the outlet plane of the orifice. Step-by-Step Solution: Step 1: Recognize configuration: short external mouthpiece with free discharge.Step 2: Use empirical rule-of-thumb for position: approximately 0.5 * d from the outlet.Step 3: Map to the option phrasing: one-half of the diameter. Verification / Alternative check: Classic hydraulics texts present diagrams placing the contracted jet minimum at about 0.5 d downstream for this geometry. Why Other Options Are Wrong: Equal to diameter / one-fourth / one-third / three-fourths: Do not match the standard empirical placement for short cylindrical mouthpieces running free. Common Pitfalls: Confusing mouthpieces with thin orifices or long pipes where contraction behavior differs. The fraction changes with geometry and whether the jet is submerged. Final Answer: one-half of the diameter

Question 8

Small Rectangular Orifice – Discharge Formula What is the discharge through a small rectangular orifice in a thin plate in terms of Cd, area a, and head h above the orifice centerline?

Accepted Answer

Introduction: Flow through small sharp-edged orifices is estimated using Torricelli's theorem corrected by a discharge coefficient Cd to account for contraction and viscous effects. Given Data / Assumptions: Small, sharp-edged rectangular orifice on a vertical tank wall. Head h is measured above the orifice center. a is the geometric area of the orifice opening. Cd is the coefficient of discharge accounting for contraction and friction. Concept / Approach: Ideal velocity from Torricelli: V_ideal = sqrt(2 * g * h). Ideal discharge would be a * V_ideal. Real discharge requires multiplying by Cd. Hence Q = Cd * a * sqrt(2 * g * h). Step-by-Step Solution: Step 1: Start with Torricelli: V = sqrt(2 * g * h).Step 2: Multiply by area to get ideal discharge: Q_ideal = a * sqrt(2 * g * h).Step 3: Apply coefficient of discharge: Q = Cd * Q_ideal = Cd * a * sqrt(2 * g * h). Verification / Alternative check: Dimensional check confirms units of Q are volume per unit time. Empirical values of Cd for sharp-edged orifices (about 0.60 to 0.65) support the corrected formula. Why Other Options Are Wrong: Cd * a * (2 * g * h): Missing the square root; dimensionally incorrect.Cd / a * sqrt(2 * g * h): Inverts area, giving wrong dimensions.a * sqrt(2 * g * h) / Cd: Dividing by Cd overestimates flow; Cd < 1.Cd * h * sqrt(2 * g * a): Mixes variables improperly; dimensionally wrong. Common Pitfalls: Forgetting the square root on 2 * g * h or using head to the wrong datum. Also confusing Cd with coefficients of contraction or velocity individually. Final Answer: Q = Cd * a * sqrt(2 * g * h)

Question 9

Accelerated Tanks – Free Surface Inclination An open tank of liquid is accelerated up an inclined plane. The inclination of the liquid free surface varies with the tank acceleration according to which relation?

Accepted Answer

Introduction: When a container of liquid undergoes linear acceleration, the free surface tilts until it becomes perpendicular to the resultant of gravitational acceleration and the imposed acceleration. This question asks how the inclination depends on the acceleration magnitude. Given Data / Assumptions: Open tank with a free surface. Tank accelerated along an incline with constant acceleration a. Gravity g acts vertically downward. Concept / Approach: The free surface aligns perpendicular to the resultant acceleration vector formed by g and the applied acceleration a. The tilt angle theta measured from the horizontal satisfies tan(theta) = a_parallel / g if a acts along the tank axis. Hence theta increases as a increases; for small angles, theta ≈ a / g, showing direct proportionality. Step-by-Step Solution: Step 1: Represent accelerations: gravity g downward and tank acceleration a along the incline.Step 2: The resultant acceleration magnitude is sqrt(g^2 + a^2), and the free surface is perpendicular to this vector.Step 3: The inclination angle theta relative to horizontal satisfies tan(theta) = a / g for the along-slope component.Step 4: Therefore, theta increases with a and is approximately proportional to a for small angles, matching “directly proportional”. Verification / Alternative check: Plotting theta vs a from tan(theta) = a / g shows monotonic increase; for a much less than g, theta ≈ a / g in radians, confirming proportionality. Why Other Options Are Wrong: Equal to the acceleration: Dimensionally meaningless; angle is not in acceleration units.Inversely proportional: Contradicts tan(theta) = a / g.Independent or proportional to acceleration squared: Both inconsistent with the tan relation. Common Pitfalls: Forgetting that the free surface is always perpendicular to the resultant acceleration and misinterpreting “inclination” as a length rather than an angle. Final Answer: directly proportional

Question 10

Fluid mechanics – force on immersed body and momentum change The force exerted by a moving fluid on an immersed body is directly proportional to the rate of change of momentum caused by the body's presence. This statement is called:

Accepted Answer

Introduction: The question tests your understanding of how fluid forces on bodies are interpreted using momentum principles. In many hydrodynamics applications (impact of jets on plates, drag on bluff bodies, and flow deflection), the fluid's change in momentum due to a body gives the resisting force experienced by that body. This concept is often summarized in practice as Newton's law of resistance. Given Data / Assumptions: Uniform stream of fluid interacts with a solid body. Measured force is the net effect required to change the fluid's momentum. Density and velocity may be considered steady for the interval of interest. Concept / Approach: Apply the momentum equation: the resultant force on a control volume equals the rate of change of linear momentum of the fluid across it. When a body is inserted, the fluid accelerates/decelerates and/or turns, creating momentum change. The equal and opposite reaction is the hydrodynamic force (resistance) on the body. Step-by-Step Solution: 1) Choose a control volume enclosing the body.2) Evaluate inflow and outflow momentum flux: m_dot * V for each face.3) Compute net change of momentum vector between outlets and inlets.4) The required external force on the control volume equals this momentum change.5) The reaction on the body is equal and opposite, i.e., the resistance/drag. Verification / Alternative check: Classic jet-impact problems show force F proportional to m_dot * (V_out - V_in). If a jet of density rho and area A strikes and is brought to rest, F = rho * A * V * (V - 0) = rho * A * V^2 which is exactly the momentum change. Why Other Options Are Wrong: Newton's law of motion: General mechanics statement, not the specific resistance formulation. Newton's law of cooling: Concerns heat transfer, not momentum/force. Newton's law of viscosity: Relates shear stress to velocity gradient in a viscous fluid, not overall momentum change on a body. Common Pitfalls: Confusing drag proportionality (sometimes ~ V^2 at high Reynolds number) with the broader momentum principle; mixing up viscosity law (shear in laminar layers) with overall resistance on a body. Final Answer: Newton's law of resistance

Question 11

Pipe flow regime classification – threshold based on Reynolds number The flow in a pipe is considered turbulent when the Reynolds number is:

Accepted Answer

Introduction: Reynolds number (Re) classifies internal flows into laminar, transitional, and turbulent regimes. This question uses the conventional threshold values adopted in many practical handbooks for smooth pipes. Given Data / Assumptions: Reynolds number is defined as Re = (rho * V * D) / mu for pipe flow. Geometry: circular pipe; fluid properties treated as constant. Typical thresholds: laminar at low Re, turbulent at higher Re. Concept / Approach: In many textbooks, laminar is Re < 2000, transitional is 2000–~2800 (sometimes 2000–4000), and turbulent is above the upper threshold. The options in this question take 2800 as the practical cutoff. Step-by-Step Solution: 1) Identify regime ranges used in the options.2) Laminar: less than 2000 → not turbulent.3) Transitional: 2000 to 2800 → neither fully laminar nor fully turbulent.4) Above 2800 → turbulent by the convention implied in the options. Verification / Alternative check: Engineering references often cite laminar < 2000 and turbulent > 4000, with a gray zone in between. Using 2800 as a cutoff is a simplified convention consistent with many MCQs. Why Other Options Are Wrong: Less than 2000: clearly laminar. Between 2000 and 2800: transitional, not definitively turbulent. None of these: incorrect because 'more than 2800' is acceptable for turbulent per the stated convention. Common Pitfalls: Forgetting that thresholds vary slightly by source; mixing criteria for pipes with external flows. Final Answer: more than 2800

Question 12

Sea water vs. pure water – comparison of specific weight The specific weight of sea water is __________ that of pure water:

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Introduction: Specific weight (gamma) equals rho * g. Because sea water contains dissolved salts, its density is higher than that of pure water under the same conditions, so its specific weight is higher. Given Data / Assumptions: Comparison at similar temperature and pressure. Sea water salinity increases density relative to fresh water. g (acceleration due to gravity) is effectively identical for both samples. Concept / Approach: Since gamma = rho * g and g cancels in the comparison, the fluid with higher density has higher specific weight. Sea water has rho slightly greater than 1000 kg/m^3 (often ~1025 kg/m^3). Step-by-Step Solution: 1) Identify property: specific weight depends on density.2) Sea water's dissolved salts increase rho.3) Therefore gamma_sea > gamma_pure. Verification / Alternative check: Typical values: pure water ~ 9.81 kN/m^3; sea water ~ 10.06 kN/m^3 at standard conditions, confirming the qualitative conclusion. Why Other Options Are Wrong: Same as: ignores salinity effect. Less than: opposite of physical reality. Not comparable: they are directly comparable at equal conditions. Common Pitfalls: Confusing specific gravity with specific weight; forgetting that temperature also affects density (but both compared at the same condition). Final Answer: more than

Question 13

Internal mouthpiece (orifice) – pressure head at vena contracta In an internal mouthpiece, the absolute pressure head at the vena contracta is __________ the atmospheric pressure head by an amount equal to the height of liquid above the vena contrac…

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Introduction: Flow through mouthpieces involves contraction and acceleration, creating a vena contracta where velocity is maximum and pressure is minimum. Understanding pressure at this section is key for cavitation checks and discharge estimation. Given Data / Assumptions: Steady incompressible flow. Internal mouthpiece with well-formed vena contracta. Negligible elevation change between reservoir free surface and jet centerline compared to head terms. Concept / Approach: Applying Bernoulli between the free surface and the vena contracta gives a pressure drop equal to the elevation (head) difference that accelerates the fluid. At the vena contracta, pressure may fall below atmospheric by approximately the liquid head above that section. Step-by-Step Solution: 1) Take points: (1) free surface (p ≈ p_atm, V ≈ 0), (2) vena contracta (p_vc, V_vc).2) Bernoulli: p_atm/gamma + H = p_vc/gamma + V_vc^2/(2g) + losses.3) The acceleration head H primarily converts to velocity head; thus p_vc/gamma ≈ p_atm/gamma − (H − losses).4) Therefore the absolute pressure head at vena contracta is less than atmospheric by roughly the height of liquid above it (allowing for small losses). Verification / Alternative check: Observed cavitation or air entrainment near sharp-edged orifices is consistent with low absolute pressure at the vena contracta. Why Other Options Are Wrong: More than: contradicts acceleration-induced pressure drop. Equal to: ignores conversion of head to kinetic energy. Indeterminate: while exact value needs losses, the trend (less than atmospheric) is well established. Common Pitfalls: Confusing gauge and absolute pressure; neglecting contraction coefficient effects on velocity head. Final Answer: less than

Question 14

Pipe flow regime – transitional (neither laminar nor turbulent) The flow in a pipe is neither laminar nor fully turbulent when the Reynolds number is:

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Introduction: The transitional regime bridges laminar and turbulent behavior. Disturbances may intermittently grow or decay, making friction factor and velocity profile uncertain. Given Data / Assumptions: Internal flow in a circular pipe. Reynolds number based on diameter and bulk velocity. Thresholds consistent with the option ranges provided. Concept / Approach: Common classification: laminar for Re < 2000; transitional for an intermediate window; turbulent beyond the upper threshold. Here, the question uses 2000–2800 as the transition band. Step-by-Step Solution: 1) Identify laminar threshold: Re < 2000.2) Identify turbulent threshold: Re > 2800 (per options).3) Therefore, 2000 ≤ Re ≤ 2800 → transitional → neither purely laminar nor fully turbulent. Verification / Alternative check: Laboratory pipe flows show intermittency, puffs, and slugs in this band; friction factor correlations become unreliable. Why Other Options Are Wrong: Less than 2000: laminar. More than 2800: turbulent. None of these: incorrect, since the transitional band is explicitly given. Common Pitfalls: Using a single universal cutoff (e.g., 4000) without checking the convention used in a given problem. Final Answer: between 2000 and 2800

Question 15

Capillarity in tubes – effect of tube size on rise/depression With an increase in the size (radius) of a capillary tube, the rise or depression of the liquid column due to surface tension will:

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Introduction: Capillary action causes rise or depression of liquids in narrow tubes due to surface tension and contact angle. The magnitude depends strongly on tube radius. Given Data / Assumptions: Tube is circular and clean; contact angle is characteristic of the liquid–solid pair. Gravitational acceleration is constant. Isothermal conditions for short observation times. Concept / Approach: The capillary rise formula is h = (2 * sigma * cos(theta)) / (rho * g * r) where sigma is surface tension, theta is contact angle, rho is density, g is gravity, and r is tube radius. Thus h is inversely proportional to r. Step-by-Step Solution: 1) Recognize that h ∝ 1/r.2) If r increases, 1/r decreases.3) Therefore, the capillary rise (or depression magnitude) decreases. Verification / Alternative check: In very large tubes (r → large), the capillary effect becomes negligible, consistent with everyday observation that buckets or pipes of large diameter do not show visible capillary rise. Why Other Options Are Wrong: Increase: contradicts inverse relation with r. Remain unchanged: ignores dependence on radius. Depend only on liquid temperature: temperature affects sigma, but radius remains a direct factor. Common Pitfalls: Confusing rise with meniscus curvature; forgetting contact angle sign can cause depression (e.g., mercury). Final Answer: decrease

Question 16

Energy grade lines in pipe flow – position of the Total Energy Line (TEL) The total energy line lies above the pipe centre line by an amount equal to:

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Introduction: Two reference lines describe energy in pipe flow: the Hydraulic Grade Line (HGL) and the Total Energy Line (TEL). Understanding their vertical positions relative to the pipe center line helps visualize pressure and kinetic energy distribution. Given Data / Assumptions: Steady incompressible flow in a pipe of constant elevation reference. Centre line chosen as z = 0 for convenience. No pump/turbine between the two nearby sections considered. Concept / Approach: Bernoulli's equation per unit weight is H = z + p/gamma + V^2/(2g). The HGL represents z + p/gamma, while the TEL represents the full H. If the centre line is the elevation datum (z = 0), the TEL is above it by p/gamma + V^2/(2g). Step-by-Step Solution: 1) Write H at the section: H = 0 + p/gamma + V^2/(2g).2) The vertical distance from centre line to TEL is H.3) Therefore, the distance equals pressure head plus velocity head. Verification / Alternative check: HGL sits below TEL by exactly the velocity head. Observing piezometer level (HGL) and total head (Pitot, TEL) in experiments confirms this spacing. Why Other Options Are Wrong: Pressure head: misses the velocity head contribution. Velocity head: misses the pressure head contribution. Pressure head − velocity head: not consistent with Bernoulli representation. Common Pitfalls: Confusing HGL with TEL; forgetting that selecting the pipe centre line as datum sets z = 0, not removing other head terms. Final Answer: pressure head + velocity head

Question 17

Hydrostatics – total pressure on a horizontal plane area at depth x For a plane surface of area A immersed horizontally in a liquid of specific weight w with its centroid at depth x from the free surface, the total pressure on the surface equals:

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Introduction: Total hydrostatic force on a plane surface equals the pressure at the centroid times the area when the surface is horizontal, because pressure is uniform over that surface at a given depth. Given Data / Assumptions: Surface is horizontal; its entire area is at the same depth x. Fluid is incompressible with specific weight w. Gravity is constant; edge effects are negligible. Concept / Approach: Pressure at depth x is p = w * x. For a horizontal plane at constant depth, pressure is uniform; hence total force F is pressure times area. Step-by-Step Solution: 1) Compute pressure: p = w * x.2) Multiply by area: F = p * A.3) Therefore, F = (w * x) * A = w * A * x. Verification / Alternative check: Integrating pressure over the area yields the same result because p is constant over a horizontal plane at given depth: F = ∬ p dA = p ∬ dA = p * A. Why Other Options Are Wrong: w.A: missing the depth factor x. w x: missing the area A. w A / x: incorrect inverse dependence on depth. Common Pitfalls: Confusing horizontal with vertical surfaces (where pressure varies with depth and centroid/center-of-pressure concepts differ). Final Answer: w A x

Question 18

Fluid Mechanics – Laminar Flow in Circular Pipes Definition and Criterion: The internal flow in a circular pipe is called laminar (streamline) when the Reynolds number, Re = (rho * V * D) / mu, is less than 2000. This empirical threshold separates o…

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Introduction: The question tests the fundamental criterion for classifying internal flow in circular pipes into laminar, transitional, or turbulent regimes using the Reynolds number. Understanding this threshold is essential for selecting formulas for head loss, velocity profiles, and mass/heat transfer correlations. Given Data / Assumptions: Newtonian fluid flowing steadily in a circular pipe. Reynolds number Re = (rho * V * D) / mu, where rho is density, V is mean velocity, D is pipe diameter, and mu is dynamic viscosity. Conventional engineering threshold values are used. Concept / Approach: Reynolds number compares inertia forces to viscous forces. Low Re implies viscous forces dominate, producing orderly layers of fluid (laminar flow). Empirically, for flow in smooth circular pipes, laminar flow persists when Re is below about 2000; between roughly 2000 and 4000 is transitional; above about 4000 is turbulent for most practical conditions. Step-by-Step Solution: 1) Use definition: Re = (rho * V * D) / mu.2) Interpret regimes: laminar if viscous effects dominate.3) Apply empirical threshold widely adopted in design: laminar for Re < 2000.4) Therefore the statement 'The flow in a pipe is laminar, when Reynolds number is less than 2000' is correct. Verification / Alternative check: In laminar flow, the theoretical velocity profile is parabolic and head loss is given by the Hagen–Poiseuille relation. These relations agree with experiments only when Re is sufficiently low (well below transitional values), reinforcing the Re < 2000 guideline. Why Other Options Are Wrong: False: Contradicts the accepted threshold for circular pipes. Only for inviscid fluids: Inviscid assumption removes viscosity, making Re undefined and the classification meaningless. Only for open channels: Reynolds criteria given here applies to internal pipe flow, not open-channel regimes. Common Pitfalls: Confusing exact universal limits with practical thresholds; misusing hydraulic diameter for noncircular ducts; ignoring surface roughness and disturbances that can trigger early transition even below 2000 in extreme cases. Final Answer: True

Question 19

Open Channel Flow – Economical Trapezoidal Section Key Property: For a most economical trapezoidal channel section (minimum perimeter for a given discharge and area), the hydraulic mean depth equals one-half of the flow depth.

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Introduction: This question checks the characteristic proportion of a most economical (efficient) trapezoidal channel section. The most economical section minimizes wetted perimeter for a given discharge, thereby reducing frictional losses. Given Data / Assumptions: Prismatic trapezoidal channel with side slope z:1 (generic). Uniform, steady open-channel flow conditions. Conventional definition: hydraulic mean depth m = A / T, where A is area of flow and T is top width. Concept / Approach: For the most economical section, geometric conditions arise from minimizing wetted perimeter P for given A. These lead to proportional relationships between depth and widths. A key result typically cited is that the hydraulic mean depth equals half the depth of flow for the economical trapezoidal section. Step-by-Step Solution: 1) Let depth be y, bottom width b, side slope z:1.2) Area A = (b + z y) * y.3) Top width T = b + 2 z y.4) Hydraulic mean depth m = A / T.5) Applying the optimality conditions obtained from dP/d(b or y) = 0 leads to the relation m = y / 2 for the most economical trapezoidal section. Verification / Alternative check: Textbook results also show companion conditions such as the half of the top width equalling the length of one sloping side in the optimal case. Using these and substituting into m = A / T reproduces m = y / 2. Why Other Options Are Wrong: 1/2 breadth: Breadth does not directly govern m; m depends on A and T. 1/2 sloping side: Side length influences wetted perimeter but m is not half of a side length. 1/4(depth + breadth): No standard optimality condition gives this relation. Common Pitfalls: Confusing hydraulic radius R = A / P with hydraulic mean depth m = A / T; mixing properties of rectangular, triangular, and trapezoidal optimal sections. Final Answer: 1/2 depth

Question 20

Rotating Liquid in a Cylindrical Vessel When a cylindrical container filled with liquid is revolved at a constant angular speed, the free surface assumes a paraboloid of revolution due to the balance of pressure, gravity, and centrifugal effects.

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Introduction: The shape of a rotating liquid surface is a classic application of hydrostatics in a rotating reference frame. Recognizing the resulting geometry is important for metering, mixing, and rotating-tank experiments. Given Data / Assumptions: Rigid-body rotation of a liquid in a vertical cylindrical vessel. Steady angular speed; negligible wave motion. Newtonian fluid; free surface exposed to atmosphere. Concept / Approach: In a rotating frame, the liquid experiences an effective body force with radial component proportional to r * omega^2 and vertical component due to gravity. The condition for a free surface is that pressure remains atmospheric along it; integrating the equilibrium equations gives a quadratic dependence of elevation on radius. Step-by-Step Solution: 1) Set differential condition: dp = rho * (g dz - omega^2 r dr).2) On the free surface, p is constant (atmospheric).3) Integrate: dz/dr = (omega^2 r) / g.4) Integrate again: z = (omega^2 / (2 g)) * r^2 + constant, which is a parabola in r.5) Rotating this curve about the axis yields a paraboloid of revolution. Verification / Alternative check: Laboratory demonstrations show a deepened rim and depressed center forming a smooth parabolic surface; measuring z versus r^2 gives a straight line, confirming the parabolic relation. Why Other Options Are Wrong: A triangle: Free surfaces are smooth and cannot form polygonal facets in this context. An ellipse: Would require a quadratic constraint in both axes not produced by this balance. None of these: Incorrect because the surface is demonstrably parabolic. Common Pitfalls: Confusing cross-sectional parabolic shape with full 3D form; neglecting that the result is a paraboloid, not just a plane curve. Final Answer: a paraboloid

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