Q: Mouthpieces – discharge for an internal mouthpiece running free For an internal (Borda’s) mouthpiece discharging freely, the approximate discharge Q is best represented by which expression (a = area of mouthpiece, H = head over the mouthpiece)?

Introduction / Context: Mouthpieces are short tubes attached to an orifice. An internal (Borda’s) mouthpiece may run free or full depending on downstream conditions. Each condition has a characteristic coefficient of discharge, which directly affects the flow rate. Given Data / Assumptions: Internal mouthpiece (Borda’s type). Running free (vena-contracta formed and jet not filling the tube). H is the static head over the centre of the mouthpiece. Concept / Approach: For a free-running internal mouthpiece, the coefficient of discharge Cd is approximately 0.5. Thus, the discharge is Q ≈ Cd * a * √(2gH) ≈ 0.5 * a * √(2gH). When the mouthpiece runs full, Cd increases to about 0.707, and for well-designed external mouthpieces values can be higher. Step-by-Step Solution: Start with the ideal orifice formula: Q_ideal = a * √(2gH).Apply discharge coefficient for internal free-running condition: Cd ≈ 0.5.Hence Q = Cd * Q_ideal = 0.5 * a * √(2gH). Verification / Alternative check: Textbook tables list typical Cd: internal free ≈ 0.5; internal full ≈ 0.707; external free ≈ 0.62–0.65; values corroborate the selection. Why Other Options Are Wrong: 0.707 corresponds to internal mouthpiece running full. 0.855 and 1.0 are too high for this condition; 1.33 is nonphysical for a coefficient. Common Pitfalls: Confusing “free” with “full” running; the flow behavior inside the mouthpiece changes contraction and thus Cd. Final Answer: 0.5 · a · √(2gH)

Q: Pressure measurement – appropriate use of a piezometer tube A simple piezometer tube connected to a pipeline is suitable only for measuring which range/type of pressures in liquids?

Introduction / Context: A piezometer tube is the simplest device for measuring pressure head in a liquid line by observing the height to which the fluid rises in a vertical transparent tube. Knowing its limitations helps choose the right instrument for a given application. Given Data / Assumptions: Device: open, vertical tube connected to a liquid line. Fluid: essentially incompressible liquid; no entrained gas assumed. Pressure indicated as static head h = p/γ, read as liquid column height. Concept / Approach: The tube displays positive gauge pressure as a rise above the tap point. Very high pressures would require impractically tall tubes. Gases cannot be measured accurately because their density is low and compressible. Negative gauge pressure (vacuum) cannot be read directly, since the tube would not sustain a stable column below the tap point; manometers are used instead. Step-by-Step Reasoning: Relate pressure to head: h = p/γ.For large p, h becomes very large → impractical tube lengths.For vacuum (p < 0 gauge), the tube would not show a downward “negative” column → alternate devices required. Verification / Alternative check: Common practice uses U-tube or inclined manometers for accurate small pressure differences and vacuum, and Bourdon gauges for higher pressures, confirming the piezometer’s limited role. Why Other Options Are Wrong: High, gas, vacuum, or impact pressures are beyond the practical capability of a simple piezometer. Common Pitfalls: Attempting to use a piezometer on gas lines or expecting it to indicate negative gauge pressure directly. Final Answer: Low pressures (small positive gauge heads) only

Q: Flow measurement structures – definition of a Venturi flume A Venturi flume is a specially shaped, flumed structure constructed across an open channel by restricting its width to measure discharge. Is this statement correct?

Introduction / Context: Venturi flumes (e.g., Parshall flumes) are widely used for open-channel flow measurement. They rely on a controlled contraction of the channel to create a known relationship between upstream head and discharge. Given Data / Assumptions: Open-channel flow under gravity. Flume geometry includes converging, throat, and diverging sections. Adequate approach conditions and submergence limits are respected. Concept / Approach: The narrowing (fluming) raises velocity and changes specific energy, establishing a head–discharge rating. Proper geometry ensures critical or near-critical conditions in the throat, making the discharge largely a function of upstream head over a wide range of downstream levels. Step-by-Step Reasoning: Install flume with contraction → velocity increases in the throat.Measure head at specified gage points → use calibration to compute Q.For allowable submergence ratios, the rating remains accurate without needing a downstream head measurement. Verification / Alternative check: Standard rating tables and empirical equations for Venturi/Parshall flumes confirm the definition: a constricted channel section used for discharge measurement. Why Other Options Are Wrong: “False” contradicts field practice. Pipeline or weir requirements are irrelevant; flumes are independent structures for open channels. Common Pitfalls: Confusing Venturi meters (for pipes) with Venturi flumes (for open channels). They share a principle but are used in different flow regimes. Final Answer: True

Q: Reynolds number and critical velocity The velocity corresponding to a Reynolds number of about 2000 for internal pipe flow is commonly termed what?

Introduction / Context: Reynolds number Re characterizes the ratio of inertial to viscous forces in a flow. In internal flows through pipes, specific Re thresholds indicate transitions between laminar, transitional, and turbulent regimes. The velocity associated with Re ≈ 2000 has a special name. Given Data / Assumptions: Internal, fully developed, Newtonian flow in a circular pipe. Reynolds number Re = ρ V D / μ. Empirical transition: laminar for Re ≲ 2000, turbulent for Re ≳ 4000, transitional in between. Concept / Approach: The velocity at which laminar flow tends to become unstable is called the lower critical velocity (corresponding to Re ≈ 2000). A higher threshold, sometimes termed higher critical Re (≈ 4000), marks where turbulence is fully sustained. Step-by-Step Solution: Set Re = ρ V D / μ ≈ 2000.Solve for V → this V is termed the lower critical velocity.This is unrelated to sonic or supersonic speeds, which concern compressible flows and Mach number. Verification / Alternative check: Experimental pipe-flow charts and Moody diagrams demarcate laminar, transitional, and turbulent zones near these Re values. Why Other Options Are Wrong: Sub-sonic/super-sonic relate to Mach number, not Reynolds number. “Higher critical velocity” aligns with Re ≈ 4000, not 2000. Common Pitfalls: Confusing Re thresholds with compressibility criteria; they are independent nondimensional groups. Final Answer: Lower critical velocity

Q: Siphons – pressure requirement to avoid cavitation For a siphon to operate satisfactorily, the minimum pressure anywhere in the pipe should be __________ the vapour pressure of the liquid.

Introduction / Context: A siphon conveys liquid over a summit higher than the source free surface by maintaining continuity of liquid and pressure below atmospheric at the crest. Cavitation risk constrains the allowable minimum pressure within the siphon. Given Data / Assumptions: Incompressible liquid with known vapour pressure p_vap. Barometric pressure sufficiently high to sustain the siphon column. Negligible dissolved gas release for this conceptual check. Concept / Approach: To avoid cavitation (formation of vapour cavities), the local absolute pressure must not fall to the vapour pressure. Therefore, the minimum absolute pressure in the siphon must be greater than p_vap. At the summit, pressure is lowest due to elevation head and head losses. Step-by-Step Solution: Evaluate total head H_total = z + p/(γ) + v^2/(2g).At the summit: z is large → p/(γ) must decrease to maintain energy balance.Require p_abs(min) > p_vap to prevent vapour formation and flow breakage. Verification / Alternative check: Practical design limits the summit elevation so the computed p_abs remains above p_vap even after accounting for head losses and temperature-dependent vapour pressure. Why Other Options Are Wrong: Equal to/less than vapour pressure invites cavitation and siphon failure. “Independent of” is untrue; vapour pressure sets the limiting condition. “Twice” is arbitrary and not a governing criterion. Common Pitfalls: Ignoring temperature effects on vapour pressure or neglecting head losses, leading to underestimation of cavitation risk at the crest. Final Answer: more than

Question 1

Floating bodies – condition for stable equilibrium using metacentre A body floating in a liquid is said to be in stable equilibrium if its metacentre coincides with its centre of gravity. Is this statement correct?

Accepted Answer

Introduction / Context: The stability of floating bodies (ships, pontoons, buoys) is assessed using the relative positions of the centre of gravity (G), centre of buoyancy (B), and the metacentre (M). Knowing the correct condition for stability is vital in naval architecture and hydraulic design. Given Data / Assumptions: Small-angle stability analysis (metacentric theory). G is the centre of gravity; B is the centroid of displaced volume; M is the metacentre. Liquid is static; body floats freely without external constraints. Concept / Approach: For small heel (tilt), the restoring moment depends on the metacentric height GM = M − G (taking upward positive). The classical criterion states: stable equilibrium requires M above G (GM > 0). If M and G coincide (GM = 0), the body is neutrally stable; if M is below G (GM < 0), the body is unstable. Step-by-Step Solution: Define metacentric height: GM = BM − BG.Stable equilibrium → GM > 0 → M above G → restoring moment acts to return the body to upright.Neutral equilibrium → GM = 0 → no restoring moment; the body stays at the new angle.Unstable equilibrium → GM < 0 → overturning moment grows with heel. Verification / Alternative check: A simple float model shows that when G is lowered or B is raised to place M above G, the body rights itself after a disturbance. When M drops to G, it neither rights nor overturns—neutral. Why Other Options Are Wrong: “True” is incorrect; coincidence implies neutral equilibrium, not stable. Other options add conditions that are not the correct general criterion for stability. Common Pitfalls: Confusing the conditions for stability (M above G) with the condition for neutral equilibrium (M coincident with G). Final Answer: False

Question 2

Mouthpieces – discharge for an internal mouthpiece running free For an internal (Borda’s) mouthpiece discharging freely, the approximate discharge Q is best represented by which expression (a = area of mouthpiece, H = head over the mouthpiece)?

Accepted Answer

Introduction / Context: Mouthpieces are short tubes attached to an orifice. An internal (Borda’s) mouthpiece may run free or full depending on downstream conditions. Each condition has a characteristic coefficient of discharge, which directly affects the flow rate. Given Data / Assumptions: Internal mouthpiece (Borda’s type). Running free (vena-contracta formed and jet not filling the tube). H is the static head over the centre of the mouthpiece. Concept / Approach: For a free-running internal mouthpiece, the coefficient of discharge Cd is approximately 0.5. Thus, the discharge is Q ≈ Cd * a * √(2gH) ≈ 0.5 * a * √(2gH). When the mouthpiece runs full, Cd increases to about 0.707, and for well-designed external mouthpieces values can be higher. Step-by-Step Solution: Start with the ideal orifice formula: Q_ideal = a * √(2gH).Apply discharge coefficient for internal free-running condition: Cd ≈ 0.5.Hence Q = Cd * Q_ideal = 0.5 * a * √(2gH). Verification / Alternative check: Textbook tables list typical Cd: internal free ≈ 0.5; internal full ≈ 0.707; external free ≈ 0.62–0.65; values corroborate the selection. Why Other Options Are Wrong: 0.707 corresponds to internal mouthpiece running full. 0.855 and 1.0 are too high for this condition; 1.33 is nonphysical for a coefficient. Common Pitfalls: Confusing “free” with “full” running; the flow behavior inside the mouthpiece changes contraction and thus Cd. Final Answer: 0.5 · a · √(2gH)

Question 3

Pressure measurement – appropriate use of a piezometer tube A simple piezometer tube connected to a pipeline is suitable only for measuring which range/type of pressures in liquids?

Accepted Answer

Introduction / Context: A piezometer tube is the simplest device for measuring pressure head in a liquid line by observing the height to which the fluid rises in a vertical transparent tube. Knowing its limitations helps choose the right instrument for a given application. Given Data / Assumptions: Device: open, vertical tube connected to a liquid line. Fluid: essentially incompressible liquid; no entrained gas assumed. Pressure indicated as static head h = p/γ, read as liquid column height. Concept / Approach: The tube displays positive gauge pressure as a rise above the tap point. Very high pressures would require impractically tall tubes. Gases cannot be measured accurately because their density is low and compressible. Negative gauge pressure (vacuum) cannot be read directly, since the tube would not sustain a stable column below the tap point; manometers are used instead. Step-by-Step Reasoning: Relate pressure to head: h = p/γ.For large p, h becomes very large → impractical tube lengths.For vacuum (p < 0 gauge), the tube would not show a downward “negative” column → alternate devices required. Verification / Alternative check: Common practice uses U-tube or inclined manometers for accurate small pressure differences and vacuum, and Bourdon gauges for higher pressures, confirming the piezometer’s limited role. Why Other Options Are Wrong: High, gas, vacuum, or impact pressures are beyond the practical capability of a simple piezometer. Common Pitfalls: Attempting to use a piezometer on gas lines or expecting it to indicate negative gauge pressure directly. Final Answer: Low pressures (small positive gauge heads) only

Question 4

Flow measurement structures – definition of a Venturi flume A Venturi flume is a specially shaped, flumed structure constructed across an open channel by restricting its width to measure discharge. Is this statement correct?

Accepted Answer

Introduction / Context: Venturi flumes (e.g., Parshall flumes) are widely used for open-channel flow measurement. They rely on a controlled contraction of the channel to create a known relationship between upstream head and discharge. Given Data / Assumptions: Open-channel flow under gravity. Flume geometry includes converging, throat, and diverging sections. Adequate approach conditions and submergence limits are respected. Concept / Approach: The narrowing (fluming) raises velocity and changes specific energy, establishing a head–discharge rating. Proper geometry ensures critical or near-critical conditions in the throat, making the discharge largely a function of upstream head over a wide range of downstream levels. Step-by-Step Reasoning: Install flume with contraction → velocity increases in the throat.Measure head at specified gage points → use calibration to compute Q.For allowable submergence ratios, the rating remains accurate without needing a downstream head measurement. Verification / Alternative check: Standard rating tables and empirical equations for Venturi/Parshall flumes confirm the definition: a constricted channel section used for discharge measurement. Why Other Options Are Wrong: “False” contradicts field practice. Pipeline or weir requirements are irrelevant; flumes are independent structures for open channels. Common Pitfalls: Confusing Venturi meters (for pipes) with Venturi flumes (for open channels). They share a principle but are used in different flow regimes. Final Answer: True

Question 5

Reynolds number and critical velocity The velocity corresponding to a Reynolds number of about 2000 for internal pipe flow is commonly termed what?

Accepted Answer

Introduction / Context: Reynolds number Re characterizes the ratio of inertial to viscous forces in a flow. In internal flows through pipes, specific Re thresholds indicate transitions between laminar, transitional, and turbulent regimes. The velocity associated with Re ≈ 2000 has a special name. Given Data / Assumptions: Internal, fully developed, Newtonian flow in a circular pipe. Reynolds number Re = ρ V D / μ. Empirical transition: laminar for Re ≲ 2000, turbulent for Re ≳ 4000, transitional in between. Concept / Approach: The velocity at which laminar flow tends to become unstable is called the lower critical velocity (corresponding to Re ≈ 2000). A higher threshold, sometimes termed higher critical Re (≈ 4000), marks where turbulence is fully sustained. Step-by-Step Solution: Set Re = ρ V D / μ ≈ 2000.Solve for V → this V is termed the lower critical velocity.This is unrelated to sonic or supersonic speeds, which concern compressible flows and Mach number. Verification / Alternative check: Experimental pipe-flow charts and Moody diagrams demarcate laminar, transitional, and turbulent zones near these Re values. Why Other Options Are Wrong: Sub-sonic/super-sonic relate to Mach number, not Reynolds number. “Higher critical velocity” aligns with Re ≈ 4000, not 2000. Common Pitfalls: Confusing Re thresholds with compressibility criteria; they are independent nondimensional groups. Final Answer: Lower critical velocity

Question 6

Siphons – pressure requirement to avoid cavitation For a siphon to operate satisfactorily, the minimum pressure anywhere in the pipe should be __________ the vapour pressure of the liquid.

Accepted Answer

Introduction / Context: A siphon conveys liquid over a summit higher than the source free surface by maintaining continuity of liquid and pressure below atmospheric at the crest. Cavitation risk constrains the allowable minimum pressure within the siphon. Given Data / Assumptions: Incompressible liquid with known vapour pressure p_vap. Barometric pressure sufficiently high to sustain the siphon column. Negligible dissolved gas release for this conceptual check. Concept / Approach: To avoid cavitation (formation of vapour cavities), the local absolute pressure must not fall to the vapour pressure. Therefore, the minimum absolute pressure in the siphon must be greater than p_vap. At the summit, pressure is lowest due to elevation head and head losses. Step-by-Step Solution: Evaluate total head H_total = z + p/(γ) + v^2/(2g).At the summit: z is large → p/(γ) must decrease to maintain energy balance.Require p_abs(min) > p_vap to prevent vapour formation and flow breakage. Verification / Alternative check: Practical design limits the summit elevation so the computed p_abs remains above p_vap even after accounting for head losses and temperature-dependent vapour pressure. Why Other Options Are Wrong: Equal to/less than vapour pressure invites cavitation and siphon failure. “Independent of” is untrue; vapour pressure sets the limiting condition. “Twice” is arbitrary and not a governing criterion. Common Pitfalls: Ignoring temperature effects on vapour pressure or neglecting head losses, leading to underestimation of cavitation risk at the crest. Final Answer: more than

Question 7

Compressibility in fluids – blanket statement check “All gases have compressible flow and all liquids have incompressible flow.” Evaluate this statement.

Accepted Answer

Introduction / Context: Compressibility refers to density changes in response to pressure variations. Whether a flow should be modeled as compressible or incompressible depends on operating conditions, not just the substance being a gas or a liquid. Given Data / Assumptions: Mach number M = V / a as a compressibility indicator for gases. Liquids with small but non-zero bulk compressibility. Engineering approximations commonly used in pipelines and hydraulics. Concept / Approach: Gases can often be treated as incompressible when M ≲ 0.3 and density variations are negligible. Liquids, although far less compressible than gases, do show density changes under high pressure or rapid transients (water hammer), requiring compressible models. Thus, the blanket statement is not universally correct. Step-by-Step Reasoning: For gases: if M is small and temperature changes modest, Δρ/ρ is small → incompressible approximation acceptable.For liquids: finite bulk modulus means density increases under high pressure; transient events demand compressible analysis.Therefore, compressibility is a modeling choice based on conditions, not solely on phase. Verification / Alternative check: Classic practice treats air flows in HVAC ducts (low speed) as incompressible, while high-speed gas flows require compressible equations. Conversely, water hammer analysis for liquids uses wave speed and compressibility explicitly. Why Other Options Are Wrong: “Correct” is overly absolute. Conditional options provided are incomplete or irrelevant to the general statement. Common Pitfalls: Equating “gas” with “compressible” in every case or assuming “liquid” always incompressible, which can mislead design decisions. Final Answer: Incorrect

Question 8

Dimensional analysis – definition of Weber number (We) Select the correct definition of the Weber number used in free-surface and multiphase flows.

Accepted Answer

Introduction / Context: Nondimensional numbers organize complex fluid phenomena. The Weber number is particularly important when surface tension competes with inertia, such as in droplet formation, jets, splashing, and wave breaking. Given Data / Assumptions: Characteristic velocity V, length L, density ρ, and surface tension σ. Definition pertains to flows where interfaces and capillarity matter. Concept / Approach: Weber number compares inertia to surface tension effects: We = (ρ V^2 L) / σ. Large We implies inertia dominates and interfaces deform or break; small We indicates surface tension resists deformation and stabilizes the interface. Step-by-Step Solution: Identify inertia scaling: F_inertia ∝ ρ V^2 L^2 (up to constant factors).Identify capillary scaling: F_surface ∝ σ L.Form the ratio → We ∝ (ρ V^2 L) / σ. Verification / Alternative check: Related groups: Reynolds number Re = inertia/viscous, Froude number Fr = inertia/gravity. By analogy, Weber links inertia and surface tension, matching option D. Why Other Options Are Wrong: Elastic, gravity, or pressure-force ratios correspond to other dimensionless groups, not the Weber number. Common Pitfalls: Confusing We with Re or Fr; remembering the numerator ρ V^2 L (inertia) divided by σ (surface tension) helps avoid errors. Final Answer: Ratio of inertia force to surface tension force

Question 9

Weirs – effect of nappe condition on discharge Compared to a freely falling (aerated) nappe, the discharge over a sharp-crested weir with a depressed nappe is typically 6–7 percent __________. Choose the correct relation.

Accepted Answer

Introduction / Context: Nappe behavior over a sharp-crested weir influences the effective head and discharge. A “free” (aerated) nappe separates cleanly from the crest, while a “depressed” nappe clings due to sub-atmospheric pressure beneath it if aeration is inadequate. Given Data / Assumptions: Sharp-crested weir with insufficient nappe aeration leading to depression. Similar upstream head and geometry for comparison. Standard laboratory observations and rating corrections. Concept / Approach: When the nappe is depressed, the pressure under the nappe drops below atmospheric. This effectively increases the driving head over the crest, yielding a higher discharge than the free nappe at the same upstream head—typically by several percent (commonly quoted as about 6–7%). Step-by-Step Reasoning: Free nappe → aerated underside → pressure ≈ atmospheric → standard rating applies.Depressed nappe → sub-atmospheric underside → larger effective head → increased Q.Typical difference → Q_depressed ≈ 1.06 to 1.07 * Q_free (approximate). Verification / Alternative check: Many design manuals specify aeration vents to prevent nappe depression and maintain calibrated ratings, implying that lack of aeration artificially increases discharge readings. Why Other Options Are Wrong: “Less than” contradicts physical behavior. “Equal to” ignores the pressure change effect. “Indeterminate” is not accurate for the stated typical condition; the increase is well documented. “Zero” is physically impossible. Common Pitfalls: Using free-nappe coefficients when the nappe is not properly aerated, leading to systematic flow overestimation. Final Answer: more than

Question 10

In fluid mechanics, the Euler number (Eu) is defined as the ratio of pressure force to ______ force. Select the correct type of force from the options.

Accepted Answer

Introduction: Euler number (Eu) is a key dimensionless parameter in fluid mechanics used to compare pressure effects with inertia effects in a flow. It helps engineers judge when pressure differences dominate compared to the kinetic (inertial) behavior of a moving fluid. Given Data / Assumptions: Euler number is defined for a flowing fluid. Pressure force and inertial force are the compared effects. No special fluid properties are required beyond density and velocity for interpretation. Concept / Approach: By definition, Eu = pressure force / inertial force. In practice this is written as Eu = Δp / (rho * V^2), where Δp is a representative pressure difference, rho is fluid density, and V is a characteristic velocity. Therefore the denominator represents inertial effects. The missing word in the stem must be 'inertial'. Step-by-Step Solution: 1) Write the definition: Eu = pressure force / inertial force.2) Express in common engineering form: Eu = Δp / (rho * V^2).3) Identify the compared forces: numerator is pressure effect; denominator represents inertia (mass * velocity^2 scaling).4) Hence the blank corresponds to 'inertial' force. Verification / Alternative check: Other common dimensionless groups confirm the pattern: e.g., Reynolds number compares inertial to viscous, Froude compares inertial to gravity, and Weber compares inertial to surface tension. Euler uniquely involves pressure in the numerator over inertia in the denominator. Why Other Options Are Wrong: Gravitational: Gravity-based comparisons define the Froude number, not Euler. Viscous: Viscous comparisons define the Reynolds number, not Euler. Surface tension: Surface-tension comparisons define the Weber number. Common Pitfalls: Reversing the ratio (inertial over pressure) is incorrect for Euler. Confusing Eu with Re or Fr due to similar symbols and frequent use. Final Answer: inertial

Question 11

A pipe of diameter d1 suddenly contracts to diameter d2 at section B, and the downstream diameter d2 continues over BC. Assuming coefficient of contraction Cc = 0.62, the head loss due to sudden contraction is:

Accepted Answer

Introduction: Sudden contraction losses occur when flow passes abruptly from a larger pipe to a smaller one. The vena contracta forms downstream of the contraction, producing an energy loss modeled with the coefficient of contraction Cc. Given Data / Assumptions: Upstream diameter = d1, downstream diameter = d2. Coefficient of contraction Cc = 0.62. v2 is the mean velocity in the smaller pipe of diameter d2. Steady incompressible flow; elevation change across the fitting is negligible. Concept / Approach: The empirical head-loss relation for sudden contraction is h_c = ( (1/Cc) - 1 )^2 * (v2^2) / (2g). This comes from the dissipation associated with the jet contracting to the vena contracta and then expanding to fill the smaller pipe. Step-by-Step Solution: 1) Use h_c = ( (1/Cc) - 1 )^2 * (v2^2) / (2g).2) Substitute Cc = 0.62: (1/0.62) ≈ 1.612903.3) Compute (1.612903 - 1) = 0.612903.4) Square: 0.612903^2 ≈ 0.3756 ≈ 0.376.5) Therefore h_c ≈ 0.376 * (v2^2) / (2g). Verification / Alternative check: Textbook correlations consistently use the (1/Cc - 1)^2 factor for contraction loss. With Cc between 0.60 and 0.65, the coefficient typically falls near 0.35–0.45, matching our 0.376 estimate. Why Other Options Are Wrong: 0.62 * (v2^2)/(2g): Cc multiplies the velocity at the vena contracta, not the loss directly. (1 - 0.62)^2 * (v2^2)/(2g): Uses 1 - Cc instead of (1/Cc - 1), which is not the contraction-loss formula. (v1^2)/(2g): Ignores the contraction physics and the role of Cc and v2. Common Pitfalls: Using v1 instead of v2 in the loss expression. Confusing sudden contraction with sudden expansion; expansion loss uses a different relation with a different reference area. Final Answer: 0.376 * (v2^2) / (2g)

Question 12

Notches vs. weirs: Evaluate the statement — “The notch is of bigger size and the weir is of a smaller size.” Choose the correct option.

Accepted Answer

Introduction: Open-channel measurement uses notches (small openings in thin plates at tank edges) and weirs (barriers across full channels). Understanding their relative size and application is fundamental to flow measurement practice. Given Data / Assumptions: The statement claims a notch is bigger and a weir is smaller. Standard definitions from hydraulics are used. Concept / Approach: A notch is a calibrated opening, generally of smaller size, fitted to the side of a tank or reservoir to measure relatively small discharges. A weir is a dam-like obstruction across an open channel, usually larger in scale, used to measure or control open-channel flow. Step-by-Step Solution: 1) Recall definitions: notch = small opening in a plate; weir = barrier across channel width.2) Compare typical dimensions: notches are small; weirs span part or all of a channel.3) Therefore, the provided statement reversing their sizes is incorrect. Verification / Alternative check: Inspection of standard devices (rectangular/V-notch plates vs sharp-crested weirs in channels) confirms typical size relationships: notches are smaller apparatus; weirs are larger civil structures. Why Other Options Are Wrong: Agree: Contradicts standard definitions and typical applications. Partly correct in rare cases: Misleading; device classification is not based on atypical installations. Cannot be decided without discharge: Device type and size convention are independent of discharge once the device is selected. Common Pitfalls: Assuming “bigger discharge” automatically means “notch is bigger”. Device choice depends on site conditions and desired accuracy. Confusing V-notches with triangular weirs; the size convention still holds. Final Answer: Disagree

Question 13

A tank of liquid accelerates horizontally to the right with acceleration ax. How does the free surface of the liquid behave?

Accepted Answer

Introduction: When a container of liquid is accelerated horizontally, the free surface tilts due to the superposition of gravitational and inertial (accelerative) fields. Recognizing the direction of tilt is a classic statics-of-fluids concept. Given Data / Assumptions: Acceleration to the right with magnitude ax. Incompressible liquid, open to atmosphere. Quasi-static condition (no sloshing transients considered). Concept / Approach: In the accelerating frame, a pseudo-acceleration ax acts opposite to motion. The free surface becomes perpendicular to the resultant of gravity g (downward) and the pseudo-acceleration ax (to the left). This makes the surface a plane, not a curve, tilting so that it is lower at the front (direction of motion) and higher at the back. Step-by-Step Solution: 1) Resultant field has components: horizontal = ax (leftward in the accelerating frame), vertical = g (downward).2) The free surface is orthogonal to the resultant field vector.3) Geometry shows the surface drops toward the direction of acceleration and rises opposite to it.4) Therefore, with acceleration to the right, the front (right) end falls and the back (left) end rises. Verification / Alternative check: The slope of the free surface satisfies tan(theta) = ax / g, where theta is the inclination relative to the horizontal. Positive ax yields a downward slope in the forward direction, confirming the behavior described. Why Other Options Are Wrong: Remains horizontal: Only true for zero horizontal acceleration. Becomes curved: The surface is planar under uniform acceleration, not curved. Falls on the back end: Reverses the correct tilt direction. Common Pitfalls: Mixing inertial and accelerating frames and reversing the tilt. Assuming curvature due to viscosity; viscosity does not control static free-surface shape in this case. Final Answer: Falls on the front (right) end and rises at the back

Question 14

Re-entrant (Borda's) mouthpiece is classified as which type of mouthpiece?

Accepted Answer

Introduction: Mouthpieces are short tubes attached to orifices to control and measure discharge. Classification into internal and external types affects jet formation and discharge coefficients. The re-entrant or Borda's mouthpiece has a distinct geometry and flow pattern. Given Data / Assumptions: Name: Re-entrant (Borda's) mouthpiece. Short-tube element projecting into the reservoir. Concept / Approach: A re-entrant mouthpiece projects inward from the tank wall, creating a vena contracta inside the mouthpiece. Because it protrudes into the reservoir, it is called an internal mouthpiece. This geometry yields characteristic coefficients (coefficient of discharge often near 0.5 to 0.62 under running full orifice conditions, depending on regime). Step-by-Step Solution: 1) Identify geometry: tube projects inward (re-entrant).2) Classify based on projection direction: inward implies internal.3) Therefore, select 'internal' as the correct classification. Verification / Alternative check: Standard hydraulics texts depict Borda's mouthpiece as an internal projection creating a sharper contraction than external mouthpieces, confirming the classification. Why Other Options Are Wrong: External: Would project outward from the tank wall into the discharge region. Converging-diverging (Venturi): A long, smoothly varying nozzle/venturi, not a short mouthpiece. Bell-mouthed: Flared at entry, opposite of re-entrant behavior. Common Pitfalls: Confusing the flow regime 'running free' vs 'running full' with the geometric classification. Assuming all mouthpieces are external because the jet exits to atmosphere. Final Answer: internal

Question 15

Mach number (M) is defined as the ratio of the flow speed in the undisturbed stream to the speed of sound in the medium. Is this statement correct?

Accepted Answer

Introduction: Mach number is fundamental in compressible flow, indicating the importance of compressibility effects and shock phenomena. It compares the flow speed to the medium’s acoustic speed. Given Data / Assumptions: Undisturbed free-stream velocity = V. Local speed of sound in the medium = a. Idealized steady flow context. Concept / Approach: By definition, Mach number M = V / a. Classifications follow: subsonic (M < 1), transonic (around 1), supersonic (M > 1), and hypersonic (M ≫ 1). The definition does not require gas-only context, though in liquids a depends on bulk modulus and density and compressibility effects are usually weak at typical speeds. Step-by-Step Solution: 1) Write M = V / a.2) a depends on medium properties (e.g., a = sqrt(gamma * R * T) for ideal gases).3) The statement exactly matches the definition; therefore it is correct. Verification / Alternative check: Engineering practice in aerodynamics and gas dynamics universally uses M = V/a. Compressibility corrections, shock relations, and similarity laws are built around this definition. Why Other Options Are Wrong: No: Contradicts the standard definition. Only for gases: Liquids can also have a defined acoustic speed; the ratio is still meaningful even if less used. Only at STP: The definition applies at any state; a simply changes with state. Common Pitfalls: Confusing Mach number with Reynolds number or Froude number. Using a at sea-level conditions regardless of altitude or temperature, which skews M. Final Answer: Yes

Question 16

What is the specific weight (unit weight) of water in SI units, to the nearest standard value?

Accepted Answer

Introduction: Specific weight (unit weight) is weight per unit volume. For water at standard conditions, engineers commonly use a convenient rounded value for design and quick checks. Given Data / Assumptions: Standard reference conditions near 4°C to 20°C. Acceleration due to gravity g ≈ 9.81 m/s^2. Concept / Approach: Specific weight gamma = rho * g. For water, rho ≈ 1000 kg/m^3. Therefore gamma ≈ 1000 * 9.81 N/m^3 = 9810 N/m^3 = 9.81 kN/m^3. This is the widely accepted design value in SI. Step-by-Step Solution: 1) Take rho_w ≈ 1000 kg/m^3.2) Multiply by g: 1000 * 9.81 = 9810 N/m^3.3) Convert to kN: 9810 N/m^3 = 9.81 kN/m^3. Verification / Alternative check: Using more precise densities changes only the third decimal place of gamma, not the standard rounded 9.81 kN/m^3 used for hand calculations. Why Other Options Are Wrong: 981 kN/m^3: Off by a factor of 100; implies rho ≈ 100,000 kg/m^3. 9.81 N/m^3: Off by a factor of 1000; would imply extremely low density. 9810 kN/m^3: Off by a factor of 1000; numerically equals 9.81 MN/m^3, not water. Common Pitfalls: Mixing N/m^3 and kN/m^3 units. Using mass density (kg/m^3) directly as specific weight without multiplying by g. Final Answer: 9.81 kN/m^3

Question 17

“The density of air is the same at different heights (altitudes).” Evaluate the statement.

Accepted Answer

Introduction: Atmospheric density varies with altitude due to changes in pressure and temperature. Recognizing this variation is essential in aerodynamics, meteorology, and high-altitude engineering. Given Data / Assumptions: Standard atmosphere model (pressure decreases with height). Air treated as a compressible gas. Concept / Approach: Hydrostatic balance shows pressure decreases with height. Since density rho is related to pressure p and temperature T via the ideal gas relation rho = p / (R * T), changes in p and T with altitude cause rho to change as well. Therefore, density is not the same at different heights. Step-by-Step Solution: 1) Recognize dp/dz = -rho * g (hydrostatic relation).2) As z increases, p decreases, reducing rho unless temperature rises enough to compensate (which it usually does not).3) Hence, density varies with altitude; it decreases significantly in the troposphere and beyond. Verification / Alternative check: Standard atmosphere tables show sea-level density about 1.225 kg/m^3 dropping to roughly 0.736 kg/m^3 at 2500 m and far lower at high altitudes, confirming non-uniformity. Why Other Options Are Wrong: Correct: Contradicts well-established atmospheric physics. True only up to a few meters: Even tens of meters show measurable though small changes; the statement remains inaccurate. True in isothermal layers: Even if temperature is constant, pressure still falls with height so density changes. Common Pitfalls: Assuming air is incompressible in large-scale atmospheric problems. Using sea-level density for high-altitude performance estimates without correction. Final Answer: Incorrect

Question 18

Weber number (We) is defined as the ratio of inertia force to surface-tension force. Is this statement correct?

Accepted Answer

Introduction: Dimensionless numbers help compare competing physical effects in fluid flows. The Weber number measures the relative importance of inertia compared to surface tension, crucial in jet breakup, droplet formation, and atomization. Given Data / Assumptions: Flow with characteristic velocity V and length L. Fluid density rho and surface tension sigma. Concept / Approach: The Weber number is We = inertia force / surface-tension force. In formula form, We = (rho * V^2 * L) / sigma. A large We implies inertia dominates and interfaces deform or break; a small We implies surface tension resists deformation. Step-by-Step Solution: 1) Identify competing effects: inertia vs surface tension.2) Write We = (rho * V^2 * L) / sigma.3) The verbal statement matches the ratio definition; hence it is correct. Verification / Alternative check: In spray nozzles, increasing V or L increases We, promoting ligament breakup; increasing sigma lowers We, stabilizing the interface. Observations align with the definition. Why Other Options Are Wrong: Incorrect: Conflicts with standard definition in fluid mechanics. Only valid for water: The definition is general for any fluid, though sigma and rho differ. Valid only for capillary tubes: Weber number applies broadly to free-surface and multiphase flows, not just capillaries. Common Pitfalls: Confusing We with Reynolds number (inertia/viscous) or Capillary number (viscous/surface tension). Omitting the characteristic length L in calculations. Final Answer: Correct

Question 19

The rise, in water level, which occurs during the transformation of the unstable shooting flow to the stable streaming flow is called hydraulic jump.

Accepted Answer

Yes

Question 20

Fluid Mechanics – Stability of Floating Bodies When a floating body is given a very small angular displacement and it moves to a new position and then remains at rest in that new position, the body is said to be in which type of equilibrium?

Accepted Answer

Introduction: In hydrostatics, the stability of floating bodies is discussed using three standard categories of equilibrium: stable, unstable, and neutral (also called indifferent). The question tests recognition of the precise definition of neutral equilibrium for a floating body after a small angular displacement. Given Data / Assumptions: Body is floating in a liquid. A small angular displacement is applied. After displacement the body occupies a new position and then remains at rest there. Concept / Approach: Equilibrium of a floating body is judged by metacentric height GM, where G is the center of gravity and M is the metacenter. The nature is decided as: stable if GM > 0, unstable if GM < 0, and neutral if GM = 0. In neutral equilibrium the restoring moment is zero; hence the body has no tendency to return to the original position or move further away. It simply stays in the new position reached by the small disturbance. Step-by-Step Solution: Step 1: Apply a small angular displacement to the floating body.Step 2: Observe the subsequent behavior. The statement says the body now stays at rest in the new position, without coming back or overturning further.Step 3: Match the behavior with definitions. Remaining exactly where it is after the disturbance indicates zero restoring and zero overturning moment.Step 4: Conclude that GM = 0, which defines neutral (indifferent) equilibrium. Verification / Alternative check: For neutral equilibrium, small rotations do not change the potential energy to first order. The waterline and buoyant force adjust to pass through the center of gravity without producing a net moment. This matches the described behavior. Why Other Options Are Wrong: Stable: GM > 0 would create a restoring moment, returning the body to its original position, which is not described.Unstable: GM < 0 would lead to an overturning moment that increases angular displacement, also not described.Metastable: Not a standard classification for hydrostatic equilibrium of floating bodies in this context.Indifferent: Synonym of neutral; included as a distractor to test terminology, but the canonical term is neutral. Common Pitfalls: Learners sometimes confuse neutral with stable, assuming “remains at rest” means it returns. The keyword is “remains at rest in the new position,” not in the original one. Also, mixing up the sign of GM is common. Final Answer: neutral

Hydraulics and Fluid Mechanics Questions