Q: In shell-and-tube heat exchangers, the thickness of typical segmental baffles (with about 25–35% cut) is generally selected to be __________ the tube wall thickness to ensure rigidity and support.

Introduction / Context: Baffles in shell-and-tube exchangers serve multiple roles: they direct shell-side flow, enhance turbulence, and support tubes against vibration. Selecting an appropriate baffle thickness avoids warping and fretting. Given Data / Assumptions: Segmental baffles with about 25–35% cut are considered. We compare baffle plate thickness to typical tube wall thickness. Mechanical integrity and manufacturability guide thickness choice. Concept / Approach: Baffle plates must resist flow-induced vibration and maintain geometry under differential pressure. In practice, baffles are commonly thicker than individual tube walls to provide sufficient rigidity and to securely locate tubes with drilled holes. A practical rule is choosing baffle thickness on the order of about two times the tube wall thickness for many standard designs, subject to shell diameter and span. Step-by-Step Solution: Step 1: Recognize that tube walls are relatively thin (for thermal performance), while baffles must be stiffer structural elements.Step 2: Adopt a conservative sizing rule: baffle thickness ≈ 2 * tube wall thickness.Step 3: Select the corresponding option. Verification / Alternative check: Vendor standards often specify minimum baffle thickness increasing with shell diameter; in many cases these thicknesses exceed tube wall thickness by a factor near 2 or more, consistent with the heuristic. Why Other Options Are Wrong: Equal to/Half/One-third: Risk inadequate stiffness and increased vibration wear.Four times: Can be overly conservative, increasing weight and cost without proportional benefit in many standard sizes. Common Pitfalls: Ignoring shell diameter and baffle spacing effects; larger shells or wider spans may require thicker baffles than the simple rule. Final Answer: twice

Q: In multipass shell-and-tube heat exchangers, why are baffles provided on the shell side?

Introduction / Context: Shell-side baffles are key internal elements that shape the flow path across tube bundles. Their design affects heat-transfer coefficients, pressure drop, fouling propensity, and vibration behavior. Given Data / Assumptions: Typical segmental baffles creating alternating cross-flows. Reasonable baffle spacing (e.g., 0.2–1.0 times shell diameter). Design objective: enhance shell-side performance. Concept / Approach: By forcing the shell-side fluid to repeatedly cross the tube bundle, baffles increase velocity components normal to the tubes, which promotes turbulence and raises the film heat-transfer coefficient. Increased turbulence also tends to mitigate fouling. While pressure drop rises as a consequence, the primary purpose is heat-transfer enhancement, not merely pressure-loss generation or tube fixing. Step-by-Step Solution: Step 1: Recognize that cross-flow provides higher h than pure longitudinal flow.Step 2: Understand that baffles redirect flow to achieve alternating cross-flow zones.Step 3: Conclude the main purpose is to create turbulence and improve heat transfer. Verification / Alternative check: Performance correlations for shell-and-tube exchangers explicitly include baffle effects on shell-side coefficients and pressure drop, supporting this function. Why Other Options Are Wrong: Reduce scaling by lowering velocities: baffles actually increase local velocity.Increase pressure drop only: pressure drop is a side effect, not the goal.Fix tubes: Tube support is a secondary benefit; primary purpose is flow control.Allow thermal expansion: Managed by expansion joints or U-tubes, not baffles. Common Pitfalls: Over-baffling can cause excessive pressure drop or vibration; optimal spacing balances performance and hydraulics. Final Answer: To create cross-flow and turbulence for better heat transfer

Q: In a distillation column, for a saturated vapor feed, what is the value of q (the fraction of the feed that is liquid)?

Introduction / Context: The q-line in McCabe–Thiele analysis represents the thermal condition of the feed. The parameter q is defined as the fraction of the feed that is liquid at the feed tray conditions. Given Data / Assumptions: q = liquid fraction of the feed (on a molar basis). Feed is saturated vapor (no subcooling, no heat removal needed to vaporize). Concept / Approach: By definition, if the feed arrives as saturated vapor, the liquid fraction is zero. Hence q = 0. For saturated liquid, q = 1. For subcooled liquid, q > 1; for superheated vapor, q < 0. Step-by-Step Solution: 1) Identify feed state: saturated vapor.2) Apply definition: q = liquid fraction of feed.3) Saturated vapor has zero liquid present → q = 0. Verification / Alternative check: On the x–y diagram, a saturated-vapor q-line is horizontal (slope = q/(q−1) = 0/(−1) = 0), matching the textbook construction. Why Other Options Are Wrong: 1 is for saturated liquid, not saturated vapor. 1 corresponds to subcooled liquid. Common Pitfalls: Confusing thermal condition (q) with phase quality used in other contexts; forgetting that saturated states lie exactly at the dew/bubble curves. Final Answer: 0

Question 1

Riveted joints: identify the correct definition of “pitch” and related terms used in plate joining for process equipment.

Accepted Answer

Introduction / Context: Although welding is prevalent, riveted terminology still appears in legacy equipment, structural connections, and exam questions. Knowing basic terms ensures accurate interpretation of drawings and strength calculations. Given Data / Assumptions: Standard definitions from joints and fasteners. Linear rows of rivets in joined plates. Distinctions between lap and butt joints. Concept / Approach: Pitch is defined as the center-to-center spacing of adjacent rivets in the same row. A lap joint uses overlapping plates. A butt joint aligns plate edges with cover (strap) plates on one or both sides; the main plates do not overlap. Hole diameter is typically slightly larger than the rivet shank to allow insertion and expansion during riveting. Step-by-Step Solution: Step 1: Compare definitions with standards.Step 2: Confirm pitch = center-to-center spacing in a row.Step 3: Recognize lap vs butt joint distinctions.Step 4: Note that drilled hole usually exceeds nominal shank diameter. Verification / Alternative check: Handbooks consistently define pitch this way; rivet hole allowances ensure proper driving and residual interference fit. Why Other Options Are Wrong: A butt joint made in overlapping plates: incorrect; that is a lap joint.Hole smaller than rivet: generally false; holes are slightly larger.Pitch as edge distance: that is edge margin, not pitch.Butt joint without cover plates: not a standard butt joint definition. Common Pitfalls: Confusing pitch with gauge (distance between adjacent rows) and edge distance (from center to edge). Final Answer: Pitch is the center-to-center distance between two consecutive rivets in the same row.

Question 2

Condensation heat transfer: for identical operating conditions, the average film coefficient on a horizontal condenser tube is approximately __________ that for a vertical condenser.

Accepted Answer

Introduction / Context: During filmwise condensation on tubes, condensate drainage behavior influences film thickness and thus the heat-transfer coefficient. Tube orientation (horizontal vs vertical) changes how the condensate film develops and renews. Given Data / Assumptions: Laminar filmwise condensation on clean, smooth tubes. Same fluid, temperature difference, and average vapor conditions. No significant non-condensable gases. Concept / Approach: On a horizontal tube, condensate tends to drain around the circumference, keeping a relatively thinner average film on the upper regions and shedding at the bottom. On a vertical tube, the film thickens along the length as it flows downward, increasing thermal resistance. Consequently, the average heat-transfer coefficient on a horizontal tube is higher, commonly approximated as about twice that on a vertical tube under comparable conditions. Step-by-Step Solution: Step 1: Recognize that thinner condensate films yield higher coefficients (h ∝ 1/film thickness).Step 2: Compare drainage paths: circumferential (horizontal) vs axial thickening (vertical).Step 3: Use the heuristic that h_horizontal ≈ 2 * h_vertical.Step 4: Choose “twice.” Verification / Alternative check: Classical Nusselt film condensation analysis and design heuristics support the factor-of-two rule for comparable conditions, acknowledging real-world variations due to surface finish and condensate inundation. Why Other Options Are Wrong: Half: Opposite of observed trend.Thrice/Five times: Overstates typical orientation effect.The same: Ignores drainage differences. Common Pitfalls: For flooded bundles or with strong shear, the simple ratio can deviate; always validate with manufacturer data for critical services. Final Answer: twice

Question 3

Air-side film coefficient change: a single-pass air heater is replaced by a two-pass unit at the same air flow rate and conditions. If flow area per pass halves (velocity doubles), what is the approximate ratio of the new air-side film coefficient t…

Accepted Answer

Introduction / Context: In convective heat transfer for gases inside passages, the film coefficient increases with velocity. Reconfiguring a heater from single-pass to two-pass reduces the flow area per pass, increasing velocity and changing h. Given Data / Assumptions: Same total volumetric/mass flow rate of air. Two-pass arrangement halves area per pass → velocity doubles. Correlation scaling: h ∝ v^0.8 (typical turbulent gas heating estimate). Concept / Approach: With v_new / v_old = 2, the ratio of film coefficients is (2)^0.8. This reflects the sub-linear sensitivity of turbulent heat transfer to velocity, common in Dittus–Boelter-type behavior where Nu ∝ Re^m with m ≈ 0.8. Step-by-Step Solution: Step 1: Compute velocity ratio: v_ratio = 2.Step 2: Apply exponent: h_ratio = (v_ratio)^0.8.Step 3: Evaluate: 2^0.8 ≈ exp(0.8 * ln 2) ≈ exp(0.8 * 0.693) ≈ exp(0.554) ≈ 1.74.Step 4: Select 1.74 as the closest option. Verification / Alternative check: Direct calculation with a calculator confirms 2^0.8 ≈ 1.741. Design charts show similar scaling in relevant Reynolds regimes for air. Why Other Options Are Wrong: 1.26: Would imply v_ratio ≈ 1.3, not 2.2.00: Overstates increase; exponent less than 1.0.80: Would require a velocity decrease or exponent < 0.≈1.00: Ignores velocity impact of pass arrangement. Common Pitfalls: Assuming h scales linearly with velocity; turbulent convection typically follows a power < 1. Also ensure pressure-drop implications are acceptable when increasing passes. Final Answer: 1.74

Question 4

In shell-and-tube heat exchangers, the thickness of typical segmental baffles (with about 25–35% cut) is generally selected to be __________ the tube wall thickness to ensure rigidity and support.

Accepted Answer

Introduction / Context: Baffles in shell-and-tube exchangers serve multiple roles: they direct shell-side flow, enhance turbulence, and support tubes against vibration. Selecting an appropriate baffle thickness avoids warping and fretting. Given Data / Assumptions: Segmental baffles with about 25–35% cut are considered. We compare baffle plate thickness to typical tube wall thickness. Mechanical integrity and manufacturability guide thickness choice. Concept / Approach: Baffle plates must resist flow-induced vibration and maintain geometry under differential pressure. In practice, baffles are commonly thicker than individual tube walls to provide sufficient rigidity and to securely locate tubes with drilled holes. A practical rule is choosing baffle thickness on the order of about two times the tube wall thickness for many standard designs, subject to shell diameter and span. Step-by-Step Solution: Step 1: Recognize that tube walls are relatively thin (for thermal performance), while baffles must be stiffer structural elements.Step 2: Adopt a conservative sizing rule: baffle thickness ≈ 2 * tube wall thickness.Step 3: Select the corresponding option. Verification / Alternative check: Vendor standards often specify minimum baffle thickness increasing with shell diameter; in many cases these thicknesses exceed tube wall thickness by a factor near 2 or more, consistent with the heuristic. Why Other Options Are Wrong: Equal to/Half/One-third: Risk inadequate stiffness and increased vibration wear.Four times: Can be overly conservative, increasing weight and cost without proportional benefit in many standard sizes. Common Pitfalls: Ignoring shell diameter and baffle spacing effects; larger shells or wider spans may require thicker baffles than the simple rule. Final Answer: twice

Question 5

In multipass shell-and-tube heat exchangers, why are baffles provided on the shell side?

Accepted Answer

Introduction / Context: Shell-side baffles are key internal elements that shape the flow path across tube bundles. Their design affects heat-transfer coefficients, pressure drop, fouling propensity, and vibration behavior. Given Data / Assumptions: Typical segmental baffles creating alternating cross-flows. Reasonable baffle spacing (e.g., 0.2–1.0 times shell diameter). Design objective: enhance shell-side performance. Concept / Approach: By forcing the shell-side fluid to repeatedly cross the tube bundle, baffles increase velocity components normal to the tubes, which promotes turbulence and raises the film heat-transfer coefficient. Increased turbulence also tends to mitigate fouling. While pressure drop rises as a consequence, the primary purpose is heat-transfer enhancement, not merely pressure-loss generation or tube fixing. Step-by-Step Solution: Step 1: Recognize that cross-flow provides higher h than pure longitudinal flow.Step 2: Understand that baffles redirect flow to achieve alternating cross-flow zones.Step 3: Conclude the main purpose is to create turbulence and improve heat transfer. Verification / Alternative check: Performance correlations for shell-and-tube exchangers explicitly include baffle effects on shell-side coefficients and pressure drop, supporting this function. Why Other Options Are Wrong: Reduce scaling by lowering velocities: baffles actually increase local velocity.Increase pressure drop only: pressure drop is a side effect, not the goal.Fix tubes: Tube support is a secondary benefit; primary purpose is flow control.Allow thermal expansion: Managed by expansion joints or U-tubes, not baffles. Common Pitfalls: Over-baffling can cause excessive pressure drop or vibration; optimal spacing balances performance and hydraulics. Final Answer: To create cross-flow and turbulence for better heat transfer

Question 6

In distillation design using the McCabe–Thiele graphical method, what simplifying assumptions are made so that operating lines are straight and liquid/vapor flow rates remain almost constant across trays?

Accepted Answer

Introduction / Context: McCabe–Thiele is a foundational graphical method for estimating theoretical stages in binary distillation. To keep the construction simple and linear, it relies on a small set of thermodynamic and energy-balance simplifications that together are commonly referred to as the constant-molar-overflow assumption. Given Data / Assumptions: Binary or pseudo-binary separation with equilibrium represented by an x–y diagram. Column operates near constant pressure. Feed, distillate, and bottoms have similar latent heats. Heat effects between trays are modest. Concept / Approach: When sensible-heat changes are small and latent heats are similar, the enthalpy lines in an H–x diagram are nearly parallel. Material and energy balances then imply constant liquid (L) and vapor (V) rates between trays, which yields straight operating lines on the McCabe–Thiele plot. Step-by-Step Solution: 1) Assume small temperature change per tray → small sensible-heat differences.2) Assume heats of vaporization for components are similar → latent-heat terms cancel in energy balances.3) With 1) and 2), total molar flow in each phase changes little from tray to tray → L and V roughly constant.4) Constant L and V give linear operating lines that intersect the q-line and the equilibrium curve in the usual construction. Verification / Alternative check: Rigorous enthalpy–composition (Ponchon–Savarit) calculations relax these assumptions. When they produce operating lines close to straight, constant-molar-overflow is justified. Why Other Options Are Wrong: (a) Correct: small sensible-heat differences are assumed.(b) Correct: similar latent heats (Trouton-type behavior) support constant-molar-overflow.(c) Correct: the practical outcome is nearly constant L and V.(d) Therefore all are simultaneously true. Common Pitfalls: Confusing constant molar overflow with constant molar holdup; assuming it holds at very high reflux ratios with large heat leaks; applying it to strongly nonideal, high-pressure, or highly endothermic/exothermic systems without checking. Final Answer: All of the above.

Question 7

In a distillation column, for a saturated vapor feed, what is the value of q (the fraction of the feed that is liquid)?

Accepted Answer

Introduction / Context: The q-line in McCabe–Thiele analysis represents the thermal condition of the feed. The parameter q is defined as the fraction of the feed that is liquid at the feed tray conditions. Given Data / Assumptions: q = liquid fraction of the feed (on a molar basis). Feed is saturated vapor (no subcooling, no heat removal needed to vaporize). Concept / Approach: By definition, if the feed arrives as saturated vapor, the liquid fraction is zero. Hence q = 0. For saturated liquid, q = 1. For subcooled liquid, q > 1; for superheated vapor, q < 0. Step-by-Step Solution: 1) Identify feed state: saturated vapor.2) Apply definition: q = liquid fraction of feed.3) Saturated vapor has zero liquid present → q = 0. Verification / Alternative check: On the x–y diagram, a saturated-vapor q-line is horizontal (slope = q/(q−1) = 0/(−1) = 0), matching the textbook construction. Why Other Options Are Wrong: 1 is for saturated liquid, not saturated vapor.< 0 corresponds to superheated vapor.> 1 corresponds to subcooled liquid. Common Pitfalls: Confusing thermal condition (q) with phase quality used in other contexts; forgetting that saturated states lie exactly at the dew/bubble curves. Final Answer: 0

Question 8

For a 90° medium-radius elbow and a 90° square (sharp) elbow in single-phase pipe flow, the typical equivalent lengths (in diameters) are approximately what values, respectively?

Accepted Answer

Introduction / Context: Minor (local) losses in piping due to fittings are often represented using equivalent length, Le, expressed in pipe diameters, D. A 90° elbow causes additional head loss that can be approximated by Le/D values depending on geometry and radius. Given Data / Assumptions: Single-phase, turbulent flow in a standard industrial pipe. Comparison between a medium-radius 90° elbow and a square (sharp) 90° elbow. Concept / Approach: Sharp elbows disturb the flow more strongly than swept elbows, producing larger separation and eddy formation. Therefore, their equivalent length is higher. Typical handbooks list Le/D for 90° elbows from roughly 20–30 for medium radius and about 50–60 for square elbows. Step-by-Step Solution: 1) Identify the two fittings: medium-radius elbow vs sharp (square) elbow.2) Recall typical Le/D ranges: ~25 for medium radius, ~60 for square.3) Match the nearest option pair: 25 and 60. Verification / Alternative check: Using K-values, where K = f * (Le/D). If the pipe friction factor f is known, Le/D values that reproduce standard K data (e.g., K ≈ 0.9–1.5 for medium-radius elbows and ≈ 1.5–2.5 for sharp elbows) align with Le/D ≈ 25 and 60 in many practical cases. Why Other Options Are Wrong: 3 and 5 are too small to represent typical 90° elbows.100 and 250, and 250 and 600 are unrealistically large for standard elbows and would massively overpredict minor loss. Common Pitfalls: Confusing K with Le/D; using laminar-flow correlations; ignoring fitting radius—sharper turns raise loss substantially. Final Answer: 25 and 60

Question 9

In riveted joints, if a 25 mm hole is drilled in the plates, what shank diameter of rivet should be selected for proper fit and allowance?

Accepted Answer

Introduction / Context: Riveted joints require a practical clearance between the drilled hole and the rivet shank to permit insertion and account for slight hole enlargement or burrs. This is addressed by choosing a rivet shank slightly smaller than the drilled hole. Given Data / Assumptions: Drilled hole diameter in plates = 25 mm. Standard workshop allowances are assumed. Concept / Approach: Common practice keeps the rivet diameter marginally less than the hole (often by about 1–2 mm for sizes in this range) to allow driving or upsetting without excessive force and to ensure proper seating of the rivet head. Step-by-Step Solution: 1) Start with hole diameter: 25 mm.2) Apply typical clearance: choose rivet shank ≈ 1–2 mm smaller.3) From the options, 23 mm is the practical standard size below 25 mm. Verification / Alternative check: Shop practice and handbooks list nominal hole diameters larger than shank diameter to ease insertion and account for slight out-of-roundness. After driving, the shank plastically expands to tightly fill the hole. Why Other Options Are Wrong: 25 mm gives no insertion clearance.26 mm and 27 mm exceed the hole; insertion without reaming would be impractical. Common Pitfalls: Assuming interference is desirable before upsetting; forgetting that plastic expansion during riveting achieves tightness. Final Answer: 23 mm

Question 10

In shell-and-tube heat exchangers, why are smaller tube diameters often preferred for a given heat-transfer area?

Accepted Answer

Introduction / Context: Tube size selection in shell-and-tube exchangers affects heat-transfer coefficients, pressure drop, mechanical cleaning, layout, and cost. Designers often favor smaller tubes when cost and compactness are priorities. Given Data / Assumptions: Required total heat-transfer area is fixed. Material and tube length options are comparable. Concept / Approach: For a fixed area A, using smaller diameters permits more tubes within a given shell, enabling a smaller shell diameter. Shell diameter strongly influences material and fabrication costs; hence, smaller tubes can reduce total exchanger cost. However, mechanical cleaning becomes harder and fouling behavior depends on service—not inherently better for small tubes. Step-by-Step Solution: 1) Fix A = sum(pi * d_o * L) over all tubes.2) Decrease tube outside diameter d_o → more tubes fit in the same shell pitch.3) For the same A, shell diameter can be reduced → lower shell material and bundle cost. Verification / Alternative check: Preliminary sizing correlations and cost charts show strong sensitivity of cost to shell diameter. Small tubes give higher velocities at the same flow, often improving inside coefficients but raising pressure drop—trade-offs considered during detailed design. Why Other Options Are Wrong: Easier cleaning is usually associated with larger tubes, not smaller.Fouling tendency depends on fluid chemistry, temperature, and velocity, not simply tube diameter. Common Pitfalls: Ignoring pressure-drop limits; overemphasizing compactness when the service demands frequent mechanical cleaning. Final Answer: They allow a smaller shell diameter for the same area, reducing exchanger cost.

Question 11

For cylindrical pressure vessels, which head/closure type is structurally the weakest and therefore requires the greatest thickness for a given pressure?

Accepted Answer

Introduction / Context: Cylindrical vessels are closed by various head shapes. Head geometry controls membrane stresses under pressure; more curved shapes distribute stress more efficiently and need less thickness. Given Data / Assumptions: Thin-shell assumptions are applicable. Internal pressure loading dominates. Concept / Approach: For the same diameter and pressure, required thickness follows the stress efficiency of the head shape. Hemispherical is strongest (lowest thickness), then elliptical, then torispherical; flat plates (or shallow conical/flat closures) are weakest and need the greatest thickness and/or stiffeners. Step-by-Step Solution: 1) Compare curvature: hemisphere has continuous curvature → best stress distribution.2) Elliptical and torispherical are intermediate; knuckle and crown radii create stress concentrations but still curved.3) Flat/conical plates lack beneficial curvature → higher bending stresses → greater required thickness. Verification / Alternative check: Design codes (e.g., common vessel design formulas) assign lower required thickness to hemispherical and higher to flat heads for identical service. Why Other Options Are Wrong: Hemispherical, elliptical, and torispherical heads are stronger per unit thickness than flat plates. Common Pitfalls: Ignoring external pressure or vacuum, where buckling may reverse some practical choices; forgetting stiffener rings for flat covers. Final Answer: Conical or flat plate head

Question 12

During hydrostatic testing of welded pipe for strength and tightness, the minimum test pressure is typically what multiple of the internal design pressure?

Accepted Answer

Introduction / Context: Hydrostatic tests are proof tests applied to piping to demonstrate structural integrity and leak tightness. Industry practice sets the test pressure as a multiple of the design pressure to provide safety margin. Given Data / Assumptions: Standard welded pipe designed for internal pressure. Ambient temperature, incompressible test fluid (water) used. Concept / Approach: Common standards specify a hydrostatic test pressure of about 1.5 times the internal design pressure for proof testing. This level reveals gross defects without approaching yield for properly designed systems. Step-by-Step Solution: 1) Identify the reference: internal design pressure.2) Apply rule of thumb: P_test ≈ 1.5 * P_design.3) Select the matching option. Verification / Alternative check: Many piping codes and manufacturer hydrotest recommendations fall in the 1.3–1.5 range; 1.5 is a widely taught value. Why Other Options Are Wrong: 0.5 is below operating proof; 2.5 and 3.5 are excessive and risk plastic deformation. Common Pitfalls: Confusing design pressure with maximum allowable working pressure; ignoring temperature-dependent allowable stresses. Final Answer: 1.5 times the design pressure

Question 13

In a pressure vessel wall containing a circular opening, beyond what approximate radial distance from the hole edge does stress concentration become negligible?

Accepted Answer

Introduction / Context: Openings in pressure components disturb the membrane stress field and create local stress concentration. Designers need a rule-of-thumb distance beyond which the disturbance is small enough to neglect in primary stress evaluation. Given Data / Assumptions: Thin-wall approximation and linear elasticity. Single isolated circular opening. Concept / Approach: The perturbation caused by a hole decays with distance from the discontinuity. In many handbooks, the disturbed region is considered local, and primary membrane stress can be assumed recovered beyond a few diameters from the edge. Step-by-Step Solution: 1) Define characteristic length: hole diameter, D.2) Recognize decay of stress concentration with radial distance r from the edge.3) Apply common guideline: effects are negligible beyond about 5D. Verification / Alternative check: Finite-element studies and classic elasticity solutions show steep decay; practical design rules often choose 3–5D. A conservative rounded value used in many texts is roughly five diameters. Why Other Options Are Wrong: 2D can still show noticeable concentration.10D and 20D are overly conservative for typical thin-wall cases. Common Pitfalls: Applying this rule near multiple closely spaced openings where interactions increase disturbed zones; ignoring reinforcement pads and local thickness changes. Final Answer: Five times the hole diameter

Question 14

Piping components: identify the incorrect statement about common parts and supports.

Accepted Answer

Introduction / Context: Piping and heat-exchanger terminology can be confusing. This question tests recognition of standard component functions and common shop terms. Given Data / Assumptions: Industrial piping practice and exchanger fabrication methods. Concept / Approach: Reducers change diameter (from larger NPS to smaller or vice versa), not direction. Direction changes are made with elbows or bends. Tube rolling is the expansion process used to fix tubes into a tubesheet. The pipe vs tube distinction in industry is primarily about specification system (NPS schedule and nominal ID for pipe; actual OD and wall for tube), not simply “thicker wall.” Step-by-Step Solution: 1) Examine each statement for its core function.2) Identify that reducers alter cross-sectional size, not flow direction.3) Therefore the statement claiming reducers change direction is incorrect. Verification / Alternative check: Catalogs and P&ID symbols show reducers inline on the same centerline, whereas elbows have a change in angle (e.g., 45°, 90°). Why Other Options Are Wrong: (a) Lug supports are simple attachments and often cheaper than custom shoes/frames.(b) True nuance: pipe vs tube is a specification/nominal system difference, not “pipe has thicker wall.”(d) Tube rolling is the correct name for mechanical expansion into a tubesheet. Common Pitfalls: Assuming reducers behave like elbows; oversimplifying pipe vs tube differences. Final Answer: A reducer is used to change direction in a straight run of pipe.

Question 15

Tray efficiencies in distillation: which of the following can exceed 100% under certain conditions?

Accepted Answer

Introduction / Context: Tray or plate efficiencies quantify how closely actual mass transfer on a tray approaches ideal equilibrium behavior. Several efficiency definitions are used, including point, Murphree, and overall (column) efficiencies. Given Data / Assumptions: Binary distillation framework. Definitions based on vapor composition change across a tray and the corresponding equilibrium compositions. Concept / Approach: The Murphree vapor-phase efficiency compares the actual change in vapor composition across a tray to the change that would occur if the exiting vapor reached equilibrium with the exiting liquid. Because composition profiles can “overshoot” the local equilibrium point due to hydraulics and contact patterns, the Murphree efficiency can mathematically exceed 100%. Step-by-Step Solution: 1) Define E_M = (y_out − y_in) / (y_eq,liquid_out − y_in).2) If y_out lies beyond y_eq,liquid_out due to non-ideal contacting, the numerator exceeds the denominator.3) Hence E_M > 1 (i.e., > 100%) is possible even though it does not imply super-equilibrium mass transfer. Verification / Alternative check: Point efficiencies reflect local transfer and remain ≤ 100% by definition. Overall efficiencies average performance across many trays and hydraulic regimes and are also ≤ 100% in standard interpretations. Why Other Options Are Wrong: Overall plate efficiency is an averaged measure and does not exceed 100%.Point efficiency is defined locally relative to equilibrium and stays ≤ 100%.Therefore “None of these” is incorrect. Common Pitfalls: Interpreting E_M > 100% as physical violation; confusing Murphree with overall efficiency used for shortcut design. Final Answer: Murphree plate efficiency

Question 16

Elastic failure criteria: under the maximum tensile strain theory, failure is predicted to occur when the maximum tensile strain equals what value? (Here, σ = yield strength in simple tension, and E = Young's modulus of elasticity.)

Accepted Answer

Introduction / Context: This question tests the classical elastic failure criteria used in strength of materials and machine design. Specifically, it focuses on the maximum tensile strain theory, sometimes associated with Saint-Venant’s hypothesis, and asks which threshold indicates the onset of elastic failure when multiaxial stresses act on a ductile member. Given Data / Assumptions: σ denotes the yield strength obtained from a uniaxial tensile test. E denotes Young's modulus (modulus of elasticity) of the material. Material behavior is linearly elastic up to yield, and small-strain theory applies. Concept / Approach: In the maximum tensile strain theory, failure is predicted when the largest principal strain under a complex state equals the uniaxial yield strain. The uniaxial yield strain from a tensile test is simply σ/E, because Hooke’s law gives strain = stress / E in the elastic region. Step-by-Step Solution: 1) In a uniaxial test at yield: strain_y = σ/E.2) Under general loading: compute principal strains; the maximum of these governs.3) The theory states failure initiates when max tensile strain = strain_y.4) Therefore, the threshold strain equals σ/E. Verification / Alternative check: Under simple tension, the multiaxial case reduces to the test condition. The criterion must reproduce the same yield point as the test: σ/E, confirming consistency. Why Other Options Are Wrong: E: This has units of stress; not a strain threshold. σ: Stress value, not a strain; theory uses strain equality. E/σ: Inverse of the correct ratio; wrong dimensionally. None of these: Incorrect because σ/E is present. Common Pitfalls: Confusing stress-based (e.g., maximum principal stress) and strain-based criteria; mixing units; forgetting that Hooke’s law links stress and strain by E in the elastic regime. Final Answer: σ/E

Question 17

Pick out the wrong statement.

Accepted Answer

none of these.

Question 18

Pipe fittings: for a tee used as an elbow (flow entering the run), the approximate equivalent length is how many pipe diameters (L/D)?

Accepted Answer

Introduction / Context: In fluid mechanics and piping system design, minor (local) losses from fittings are commonly estimated using either a loss coefficient K or an equivalent-length method. This question asks for the typical equivalent length, expressed in diameters (L/D), for a tee used as an elbow when the flow enters the run (a relatively high-loss configuration). Given Data / Assumptions: Fitting: 90-degree turn accomplished by a tee, with flow entering the run. Fully turbulent, steady, incompressible flow; standard commercial fittings. Equivalent length method: L_eq = (K * D) / f, typically tabulated as L/D for fittings. Concept / Approach: Handbooks correlate fittings to L/D values to convert local losses into an equivalent straight-pipe length. Among tee configurations, “flow entering run” exhibits relatively larger loss than a smooth elbow. Typical tabulated values place this case around a few dozen diameters, commonly near 60 diameters, for design estimates. Step-by-Step Solution: 1) Identify the configuration: tee used as elbow, flow entering run.2) Recall standard tables: this case has high local loss.3) Typical L/D values for design are ≈ 60 (order-of-magnitude dozens, not single digits or hundreds).4) Therefore, select 60 as the practical estimate. Verification / Alternative check: Cross-checking common references shows L/D for this configuration in the range of several tens of diameters, with 60 a frequently cited nominal value for preliminary design. Why Other Options Are Wrong: 5 and 15: Much too small for a high-loss tee used as an elbow. 200 and 350: Excessively large for standard tees; would overpredict loss. Common Pitfalls: Confusing different tee orientations (branch vs run), or substituting values for long-radius elbows which have significantly lower losses. Final Answer: 60

Question 19

A minimum clearance of about __________ mm is maintained between the distillation column wall and the bubble cap.

Accepted Answer

38

Question 20

Riveted joints: the distance from the centre of a rivet hole to the nearest free edge of the plate is called what?

Accepted Answer

Introduction / Context: Basic terminology of riveted joints is foundational in machine design and fabrication. Correctly identifying geometric parameters helps ensure adequate strength and prevents failures like tearing at the edges or shear-out. Given Data / Assumptions: We consider a flat plate with rivet holes near a free edge. Standard nomenclature is used for riveted joint dimensions. Concept / Approach: Key distances include: pitch (centre-to-centre spacing of adjacent rivets along a row), back pitch or gauge (distance between parallel rows), and margin (also called edge distance), which is from the hole centre to the nearest free edge. Adequate margin prevents tearing along the edge under load. Step-by-Step Solution: 1) Identify which distance is being asked: centre of hole to nearest plate edge.2) By definition, this is the margin (edge distance).3) Therefore, the correct term is “Margin (edge distance)”. Verification / Alternative check: Texts in machine design consistently define margin as the edge distance, commonly specified as a multiple of plate thickness to avoid tearing failures. Why Other Options Are Wrong: Pitch: centre-to-centre spacing of adjacent rivets, not to the edge. Back pitch (gauge): spacing between rows, not to the plate edge. Gauge: often synonymous with back pitch; not edge distance. None of these: incorrect because the correct term is listed. Common Pitfalls: Mixing up pitch and margin; using too small an edge distance that leads to tearing or poor hole quality after drilling. Final Answer: Margin (edge distance)

Process Equipment and Plant Design Questions