Heat and Mass Transfer MCQs 2026

80 questions with detailed answers · 28 from past papers · 8 quiz batches available

📚 Mechanical Engineering 📄 28 Past-Paper Qs ✓ Free · No Login Needed
🎯 Mock Test

Read each question, think about the answer, then click Show Answer to reveal the correct option and explanation. Load 10 at a time so it stays manageable — perfect for one-topic study sessions on the bus or during a break.

Page 1 of 1 Questions 110 of 80
  1. Q1 easy

    Shell-and-tube heat exchanger baffles serve to

    1. A block all flow completely
    2. B reduce heat transfer area intentionally as primary purpose
    3. C increase shell-side turbulence and heat transfer
    4. D measure viscosity only
    💡 Explanation:

    Baffles redirect shell flow across tubes.

  2. Q2 Past Paper · PPSC/FPSC/NTS medium

    Parallel flow heat exchanger LMTD is

    1. A always larger than counterflow LMTD always
    2. B zero always
    3. C generally smaller than counterflow for same inlet outlet states
    4. D independent of inlet outlet temperatures
    💡 Explanation:

    Counterflow usually gives higher ΔT_lm.

  3. Q3 medium

    Heat exchanger effectiveness ε is defined as

    1. A LMTD divided by UA as effectiveness definition
    2. B UA divided by C_min alone always as effectiveness
    3. C actual heat transfer divided by maximum possible heat transfer
    4. D pressure drop ratio as effectiveness
    💡 Explanation:

    ε = Q/Q_max.

  4. Q4 medium

    Fourier number Fo is

    1. A α t / L²
    2. B h L / k which is Bi labeled as Fo
    3. C k / h L as Fo
    4. D ρ v L / μ which is Re as Fo
    💡 Explanation:

    Fo characterizes transient conduction time scale.

  5. Q5 Past Paper · PPSC/FPSC/NTS hard

    Heisler charts are used for

    1. A transient heat conduction in solids with convection boundary
    2. B fatigue S-N curves
    3. C steam tables only as Heisler charts purpose
    4. D Mohr circle
    💡 Explanation:

    Graphical solution for Bi and Fo.

  6. Q6 hard

    Transient conduction in semi-infinite solid with sudden surface temperature uses

    1. A error function solution
    2. B Euler buckling formula
    3. C Mohr circle
    4. D Rankine formula
    💡 Explanation:

    Similarity variable η = x/(2√(αt)).

  7. Q7 hard

    Chilton-Colburn j-factor relates

    1. A stress and strain
    2. B pressure and velocity as j-factor primary
    3. C heat and mass transfer coefficients dimensionlessly
    4. D entropy and enthalpy as j-factor
    💡 Explanation:

    j_H = j_D for analogous transfer.

  8. Q8 Past Paper · PPSC/FPSC/NTS medium

    Evaporation rate from liquid surface increases with

    1. A higher air velocity and lower ambient vapor concentration
    2. B lower temperature difference always reducing driving force as increase always
    3. C zero concentration difference always beneficial
    4. D infinite humidity always beneficial
    💡 Explanation:

    Mass transfer driving force depends on concentration difference.

  9. Q9 hard

    Lewis number Le is ratio of

    1. A mass to thermal diffusivity as Le definition if reversed
    2. B Re to Pr as Le
    3. C thermal diffusivity to mass diffusivity
    4. D Nu to Sh always as Le
    💡 Explanation:

    Le = α/D.

  10. Q10 Past Paper · PPSC/FPSC/NTS hard

    Schmidt number Sc equals

    1. A ν / D
    2. B D / ν as Sc
    3. C α / k as Sc
    4. D h / k as Sc
    💡 Explanation:

    Sc = ν/D = μ/(ρ D).

  11. Q11 hard

    Sherwood number Sh is analogous to

    1. A Reynolds only as Sherwood analog alone
    2. B Euler number as Sherwood analog
    3. C Weber number as Sherwood analog
    4. D Nusselt number for mass transfer
    💡 Explanation:

    Sh = k_m L / D_AB.

  12. Q12 medium

    Mass diffusivity D has units similar to

    1. A dynamic viscosity μ only as D units similarity context
    2. B thermal diffusivity α
    3. C thermal conductivity k only
    4. D heat transfer coefficient h only
    💡 Explanation:

    Both m²/s.

  13. Q13 Past Paper · PPSC/FPSC/NTS medium

    Fick's first law of diffusion states mass flux is proportional to

    1. A negative concentration gradient
    2. B positive concentration gradient only always
    3. C temperature gradient only as Fick first law
    4. D pressure gradient only as Fick first law
    💡 Explanation:

    J″ = −D dC/dx.

  14. Q14 hard

    Wien's displacement law states λ_max T equals

    1. A σ T⁴ as Wien law
    2. B k ΔT as Wien law
    3. C h L/k as Wien law
    4. D constant approximately 2898 μm·K
    💡 Explanation:

    Peak wavelength shifts with temperature.

  15. Q15 hard

    Kirchhoff's law for radiation at thermal equilibrium relates

    1. A reflectivity and transmissivity only always as Kirchhoff primary alone
    2. B h and k always
    3. C emissivity and absorptivity for a surface
    4. D Nu and Re always
    💡 Explanation:

    ε = α for gray diffuse surfaces at equilibrium.

  16. Q16 Past Paper · PPSC/FPSC/NTS easy

    Stefan-Boltzmann constant σ appears in

    1. A Fourier conduction law only
    2. B radiation heat transfer calculations
    3. C Newton viscosity law only
    4. D Hooke law only
    💡 Explanation:

    σ ≈ 5.67×10⁻⁸ W/m²K⁴.

  17. Q17 medium

    Natural convection heat transfer depends strongly on

    1. A only forced velocity always required for natural convection primary
    2. B Mach number primarily for natural convection
    3. C Froude number primarily for natural convection
    4. D Grashof and Prandtl numbers
    💡 Explanation:

    Buoyancy drives flow; Gr = g β ΔT L³/ν².

  18. Q18 hard

    Grashof number in natural convection is analogous to

    1. A Prandtl number as Gr analog
    2. B Nusselt number as Gr analog
    3. C Reynolds number with buoyancy driving force replacing inertia
    4. D Peclet number as Gr analog
    💡 Explanation:

    Gr characterizes buoyant flow regime.

  19. Q19 easy

    Greenhouse effect analogy in heat transfer refers to

    1. A Fick law only as greenhouse analogy primary
    2. B transparent cover transmitting shortwave and blocking longwave radiation
    3. C Darcy law only
    4. D Hooke law only
    💡 Explanation:

    Spectral radiation properties matter.

  20. Q20 Past Paper · PPSC/FPSC/NTS medium

    Solar collector efficiency involves

    1. A absorbed radiation minus losses by convection and radiation
    2. B only Fourier conduction in insulation as sole efficiency factor alone
    3. C only bolt preload
    4. D only torsion in shaft
    💡 Explanation:

    Optical absorption and thermal losses.

  21. Q21 medium

    Cooling tower performance depends on

    1. A only conduction through concrete shell as performance primary alone
    2. B only nuclear fission
    3. C evaporative heat and mass transfer between water and air
    4. D only beam bending
    💡 Explanation:

    Wet bulb approach measures performance.

  22. Q22 medium

    Distillation column separation relies on

    1. A difference in volatility causing mass transfer between phases
    2. B only conduction through metal tray as separation mechanism primary alone
    3. C only radiation between trays
    4. D only centrifugal force as distillation mechanism
    💡 Explanation:

    Vapor-liquid equilibrium and staged contact.

  23. Q23 Past Paper · PPSC/FPSC/NTS medium

    Adsorption mass transfer at surface differs from absorption by

    1. A both identical always
    2. B accumulation on surface vs penetration into bulk phase
    3. C adsorption always in bulk only
    4. D absorption always surface only reversed always
    💡 Explanation:

    Adsorption is surface phenomenon.

  24. Q24 hard

    Knudsen number important in mass transfer indicates

    1. A transition between continuum and molecular flow regimes
    2. B turbulent vs laminar heat transfer only as Knudsen meaning alone always
    3. C elastic vs plastic stress only
    4. D subsonic vs supersonic only as Knudsen
    💡 Explanation:

    Kn = λ/L.

  25. Q25 medium

    Dryer design uses

    1. A only Rankine cycle analysis as dryer design primary tool alone
    2. B heat and mass transfer to remove moisture from solids
    3. C only Mohr circle
    4. D only belt friction
    💡 Explanation:

    Convective drying coupled with diffusion in material.

  26. Q26 Past Paper · PPSC/FPSC/NTS medium

    Humidification involves simultaneous

    1. A heat and mass transfer at air-water interface
    2. B only conduction in steel wall as humidification primary
    3. C only torsion in shaft
    4. D only buckling in column
    💡 Explanation:

    Evaporation cools water; affects air enthalpy.

  27. Q27 medium

    Mass convection over flat plate analog to heat transfer uses

    1. A only Darcy law as mass convection model primary
    2. B only Euler column formula
    3. C only Mohr circle
    4. D similar boundary layer equations with concentration instead of temperature
    💡 Explanation:

    Similarity between heat and mass transfer.

  28. Q28 Past Paper · PPSC/FPSC/NTS hard

    Recovery factor relates

    1. A fin efficiency labeled as recovery factor
    2. B heat exchanger effectiveness labeled as recovery factor
    3. C bolt tightening factor labeled as recovery factor
    4. D actual recovery of kinetic energy to total enthalpy in boundary layer
    💡 Explanation:

    r = (T_aw − T)/(T0 − T).

  29. Q29 hard

    Stagnation temperature in high-speed flow includes

    1. A only radiation temperature as stagnation T components alone
    2. B only wet bulb
    3. C static temperature plus kinetic energy contribution converted to enthalpy
    4. D only dew point as stagnation T
    💡 Explanation:

    T0 = T + V²/(2cp) approx for calorically perfect gas.

  30. Q30 medium

    Gray body reflects radiation such that for opaque surface

    1. A ε + k = 1 mixing emissivity and conductivity
    2. B h + k = 1
    3. C Nu + Pr = 1
    4. D α + ρ = 1
    💡 Explanation:

    Energy balance on incident radiation.

  31. Q31 Past Paper · PPSC/FPSC/NTS medium

    Radiation shield between two surfaces reduces

    1. A conduction through vacuum always increases as shield effect primary
    2. B mass diffusivity always increases as shield effect
    3. C Reynolds number always increases as shield effect
    4. D net radiant heat exchange
    💡 Explanation:

    Additional surface resistance lowers q.

  32. Q32 medium

    Nucleate boiling heat flux increases rapidly with

    1. A decreasing wall temperature below saturation as increase flux always
    2. B superheat ΔT_excess
    3. C zero pressure always beneficial alone
    4. D infinite thermal conductivity of vapor alone
    💡 Explanation:

    Bubble formation enhances convection.

  33. Q33 hard

    Pool boiling critical heat flux (CHF) marks

    1. A onset of conduction only as CHF
    2. B start of freezing
    3. C transition from efficient nucleate boiling to film boiling regime
    4. D laminar flow inception as CHF
    💡 Explanation:

    Burnout point on boiling curve.

  34. Q34 Past Paper · PPSC/FPSC/NTS medium

    Dropwise condensation compared to filmwise generally gives

    1. A lower coefficient always
    2. B same coefficient always
    3. C higher heat transfer coefficient
    4. D zero heat transfer
    💡 Explanation:

    Drops shed quickly exposing surface.

  35. Q35 hard

    Condensation heat transfer on vertical plate often uses

    1. A Fourier law alone without convection as condensation model primary
    2. B Nusselt film condensation theory
    3. C Darcy law
    4. D Bernoulli only as condensation
    💡 Explanation:

    Film theory predicts h ∝ (k³ρ²g/μLΔT)^¼.

  36. Q36 easy

    Fourier's law of heat conduction states that heat flux is proportional to

    1. A positive temperature gradient only always
    2. B pressure gradient
    3. C negative temperature gradient
    4. D velocity gradient
    💡 Explanation:

    q″ = −k dT/dx for one-dimensional conduction.

  37. Q37 Past Paper · PPSC/FPSC/NTS easy

    Thermal conductivity k has SI units

    1. A J/kg
    2. B W/(m·K)
    3. C Pa·s
    4. D W/m²
    💡 Explanation:

    k measures ability to conduct heat.

  38. Q38 easy

    Steady one-dimensional conduction through plane wall is

    1. A Q = h A ΔT only without conduction path
    2. B Q = σ A ΔT/L
    3. C Q = m cp ΔT only as conduction through wall
    4. D Q = k A ΔT / L
    💡 Explanation:

    Linear temperature profile in homogeneous wall.

  39. Q39 medium

    Thermal resistance of conduction layer is

    1. A k A / L
    2. B L / (k A)
    3. C h A only
    4. D 1/(h A) which is convection resistance
    💡 Explanation:

    R_cond = L/(kA) analogous to electrical resistance.

  40. Q40 Past Paper · PPSC/FPSC/NTS medium

    Composite wall in series has total thermal resistance

    1. A product of resistances
    2. B always zero
    3. C equal to smallest layer only
    4. D sum of individual layer resistances
    💡 Explanation:

    Heat flux same; resistances add.

  41. Q41 easy

    Newton's law of cooling for convection is

    1. A q″ = k dT/dx only always as convection law
    2. B q″ = σ T⁴ only as convection
    3. C q″ = h (T_s − T_∞)
    4. D q″ = D dC/dx only as convection
    💡 Explanation:

    h is convective heat transfer coefficient.

  42. Q42 medium

    Nusselt number Nu is defined as

    1. A h L / k
    2. B k / h L
    3. C ρ v L / μ which is Reynolds
    4. D μ cp / k which is Prandtl
    💡 Explanation:

    Nu compares convection to conduction across length L.

  43. Q43 Past Paper · PPSC/FPSC/NTS medium

    Prandtl number Pr equals

    1. A α / ν
    2. B ν / α or μ cp / k
    3. C h L / k which is Nusselt
    4. D g β ΔT L³/ν² which is Grashof
    💡 Explanation:

    Pr = momentum diffusivity / thermal diffusivity.

  44. Q44 medium

    Reynolds number in forced convection characterizes

    1. A radiation to conduction ratio
    2. B ratio of inertial to viscous forces
    3. C mass diffusivity to thermal diffusivity as Re
    4. D surface tension effects only always as Re
    💡 Explanation:

    Re = ρ v L / μ.

  45. Q45 easy

    Stefan-Boltzmann law for gray surface gives emitted power per area as

    1. A ε σ T⁴
    2. B σ T only linear
    3. C k ΔT only
    4. D h ΔT only as radiation law
    💡 Explanation:

    E = ε σ T⁴.

  46. Q46 Past Paper · PPSC/FPSC/NTS medium

    Emissivity ε of real surface is

    1. A always equal to 1 for all materials
    2. B ratio of surface emissive power to blackbody emissive power at same T
    3. C ratio of absorptivity to reflectivity always as definition of ε alone
    4. D always zero for metals
    💡 Explanation:

    0 ≤ ε ≤ 1.

  47. Q47 hard

    View factor F_12 between two surfaces represents

    1. A conductive heat flux fraction
    2. B convective coefficient ratio
    3. C mass transfer coefficient
    4. D fraction of radiation leaving surface 1 intercepted by surface 2
    💡 Explanation:

    Geometric radiation exchange factor.

  48. Q48 Past Paper · PPSC/FPSC/NTS hard

    Radiation heat exchange between large parallel gray plates includes factor

    1. A 1/(1/ε1 + 1/ε2 − 1) multiplying σ A (T1⁴ − T2⁴)
    2. B h ΔT only
    3. C k/L only
    4. D D/L only as radiation exchange
    💡 Explanation:

    Electrical network analogy for surface resistances.

  49. Q49 medium

    Fin effectiveness increases with

    1. A higher convection coefficient and higher fin thermal conductivity
    2. B lower k always beneficial
    3. C zero surface area always beneficial
    4. D infinite length always without diminishing returns
    💡 Explanation:

    Fins reduce convection resistance side.

  50. Q50 medium

    Fin efficiency compares

    1. A actual heat transferred by fin to heat if entire fin at base temperature
    2. B fin cost to weight only
    3. C Reynolds to Prandtl only without fin context
    4. D LMTD to NTU only as fin efficiency
    💡 Explanation:

    η_f ≤ 1 due to temperature drop along fin.

  51. Q51 Past Paper · PPSC/FPSC/NTS hard

    Infinite fin heat transfer rate for uniform h and k is

    1. A Q = h A only without fin geometry
    2. B Q = k A/L only as infinite fin
    3. C Q = √(h P k A_c) (T_b − T_∞)
    4. D Q = σ T⁴ only as fin law
    💡 Explanation:

    Exponential decay solution boundary.

  52. Q52 medium

    Biot number Bi is defined as

    1. A k / h L_c
    2. B h L_c / k
    3. C h L / ρ v cp
    4. D D / ν as Bi
    💡 Explanation:

    Bi compares internal conduction resistance to external convection.

  53. Q53 medium

    Lumped capacitance method valid when Bi is

    1. A less than about 0.1
    2. B greater than 10 always as lumped criterion
    3. C equal to Reynolds number
    4. D infinite always required
    💡 Explanation:

    Small Bi ⇒ uniform internal temperature.

  54. Q54 Past Paper · PPSC/FPSC/NTS medium

    LMTD for counter-flow heat exchanger uses

    1. A arithmetic mean always without log as LMTD
    2. B maximum difference only always as LMTD
    3. C minimum difference only always as LMTD
    4. D log mean of hot and cold end temperature differences
    💡 Explanation:

    ΔT_lm = (ΔT1 − ΔT2)/ln(ΔT1/ΔT2).

  55. Q55 hard

    NTU method in heat exchangers relates

    1. A only pressure drop in pipes alone as NTU method scope
    2. B only radiation view factors
    3. C size (UA) and heat capacity rates to effectiveness
    4. D only Fick diffusion only as NTU
    💡 Explanation:

    ε = f(NTU, C_r) for various flow arrangements.

  56. Q56 medium

    Overall heat transfer coefficient U in plane wall includes

    1. A only radiation always alone without convection conduction
    2. B conduction and convection resistances in series
    3. C only mass transfer coefficient
    4. D only Darcy friction factor
    💡 Explanation:

    1/U = 1/h1 + L/k + 1/h2.

  57. Q57 Past Paper · PPSC/FPSC/NTS hard

    Critical radius of insulation for cylinder occurs when

    1. A Bi = 0 always as critical radius condition
    2. B Bi = 1 for cylinder definition r_c = k/h
    3. C Re = 2300 always
    4. D Pr = 0.7 always as critical radius
    💡 Explanation:

    Adding insulation can increase heat loss below r_c.

  58. Q58 easy

    Electrical analogy for conduction uses

    1. A pressure as heat flow always in analogy
    2. B velocity as temperature
    3. C temperature difference as driving force and heat flow as current
    4. D mass flux as voltage
    💡 Explanation:

    Q = ΔT/R_th.

  59. Q59 Past Paper · PPSC/FPSC/NTS medium

    Contact resistance between two solids increases with

    1. A perfectly smooth surfaces always as increase contact R always
    2. B higher pressure always increases contact R
    3. C higher k always increases contact R
    4. D surface roughness and lower contact pressure
    💡 Explanation:

    Air gaps and roughness impede conduction.

  60. Q60 medium

    Heat pipe transfers heat effectively by

    1. A solid conduction only through vacuum as heat pipe mechanism
    2. B evaporation and condensation of working fluid in closed cycle
    3. C radiation only through vacuum as primary mechanism alone
    4. D forced air only without phase change as heat pipe
    💡 Explanation:

    Phase change carries latent heat.

  61. Q61 hard

    Thermal boundary layer thickness grows along flat plate because

    1. A radiation only at leading edge as sole cause
    2. B phase change only
    3. C mass diffusion only as thermal BL growth cause alone
    4. D momentum and thermal diffusion from wall
    💡 Explanation:

    Blasius/similarity solutions describe growth.

  62. Q62 Past Paper · PPSC/FPSC/NTS hard

    Dittus-Boelter equation estimates Nu for

    1. A turbulent flow in smooth tubes
    2. B laminar tube flow always as Dittus-Boelter primary regime
    3. C natural convection on vertical plate always as Dittus-Boelter primary
    4. D radiation between plates always as Dittus-Boelter
    💡 Explanation:

    Nu = 0.023 Re^0.8 Pr^n.

  63. Q63 hard

    For laminar fully developed flow in circular tube with constant heat flux, Nu equals

    1. A 0.023 Re^0.8 Pr^n which is turbulent correlation for laminar fully developed
    2. B 4.36
    3. C 3.66 for constant wall temperature case confused
    4. D 1.0 always
    💡 Explanation:

    Classic laminar tube Nu for constant heat flux.

  64. Q64 Past Paper · PPSC/FPSC/NTS hard

    Log-mean temperature difference correction factor F for shells is used when

    1. A Bi < 0.1 only as F factor purpose
    2. B radiation only as F factor purpose
    3. C mass transfer only as F factor purpose
    4. D flow arrangement deviates from true counterflow or multipass
    💡 Explanation:

    F ≤ 1 corrects LMTD for configuration.

  65. Q65 medium

    Countercurrent flow heat exchanger achieves

    1. A always lower effectiveness always compared to parallel
    2. B same effectiveness always regardless of arrangement
    3. C zero heat transfer
    4. D higher effectiveness than parallel flow for same UA generally
    💡 Explanation:

    Maintains higher ΔT along length.

  66. Q66 medium

    Mass transfer coefficient k_m relates

    1. A heat flux to temperature difference only as k_m definition
    2. B momentum flux to velocity gradient only as k_m
    3. C pressure drop to flow rate only as k_m
    4. D mass flux to concentration difference
    💡 Explanation:

    N″ = k_m (C_s − C_∞).

  67. Q67 Past Paper · PPSC/FPSC/NTS hard

    Equimolar counter-diffusion in gases has molar flux proportional to

    1. A temperature gradient only as equimolar diffusion driving force alone
    2. B pressure squared only alone always
    3. C concentration gradient and inversely to thickness
    4. D velocity cubed
    💡 Explanation:

    J = −D dC/dx for binary equimolar.

  68. Q68 easy

    Molecular diffusion in solids is generally

    1. A much slower than in gases
    2. B faster than gases always
    3. C identical rate always
    4. D independent of temperature always
    💡 Explanation:

    D much smaller in solids.

  69. Q69 medium

    Turbulent eddy diffusivity enhances

    1. A only laminar sublayer without effect always
    2. B both heat and mass transfer in turbulent flow
    3. C only radiation
    4. D only conduction in solids as turbulent eddy effect
    💡 Explanation:

    Mixing increases effective transport.

  70. Q70 Past Paper · PPSC/FPSC/NTS hard

    Peclet number Pe equals

    1. A Re × Pr or v L / α
    2. B Nu × Pr as Pe always
    3. C Gr × Pr which is Rayleigh labeled as Pe
    4. D Sh × Sc labeled as Pe
    💡 Explanation:

    Pe compares advection to diffusion.

  71. Q71 hard

    Rayleigh number Ra equals

    1. A Re × Pr which is Peclet labeled as Ra
    2. B Nu × Pr labeled as Ra
    3. C Sh × Sc labeled as Ra
    4. D Gr × Pr
    💡 Explanation:

    Ra governs natural convection transition.

  72. Q72 hard

    Stanton number St relates

    1. A only mass diffusivity as St primary alone
    2. B only radiation emissivity as St
    3. C heat transfer coefficient to flow properties in dimensionless form
    4. D only fin efficiency as St
    💡 Explanation:

    St = Nu/(Re Pr) = h/(ρ v cp).

  73. Q73 Past Paper · PPSC/FPSC/NTS hard

    Colburn analogy gives relation between

    1. A stress and strain in solids as Colburn analogy scope
    2. B entropy and enthalpy as Colburn analogy scope
    3. C bolt preload and torque only as Colburn analogy scope
    4. D Stanton and friction factor f/2 for similar transfer mechanisms
    💡 Explanation:

    St Pr^(2/3) = f/2 approximately.

  74. Q74 medium

    Thermal diffusivity α equals

    1. A ρ cp / k as α definition
    2. B k cp / ρ
    3. C h / k as α
    4. D k / (ρ cp)
    💡 Explanation:

    α measures temperature propagation rate.

  75. Q75 easy

    Insulation purpose on hot pipe is to

    1. A increase heat loss always as insulation purpose
    2. B increase pipe stress only as primary purpose
    3. C measure flow rate only
    4. D reduce heat loss to surroundings
    💡 Explanation:

    Lower q by increasing thermal resistance.

  76. Q76 Past Paper · PPSC/FPSC/NTS easy

    Extended surface (fin) material should have

    1. A high thermal conductivity
    2. B low conductivity always beneficial for fin material choice
    3. C zero conductivity ideally
    4. D low density only without conductivity consideration alone always as sole criterion
    💡 Explanation:

    High k minimizes temperature drop along fin.

  77. Q77 medium

    Adiabatic tip fin boundary condition means

    1. A fixed temperature at tip always as adiabatic condition
    2. B zero heat flux at tip
    3. C infinite convection at tip as adiabatic
    4. D constant heat generation at tip as adiabatic
    💡 Explanation:

    Insulated tip: dT/dx = 0 at end.

  78. Q78 hard

    Heat generation in solid modifies conduction equation by adding

    1. A only convection term as sole modification always
    2. B radiation only as sole modification
    3. C q_gen term to energy balance
    4. D mass flux only as sole modification in conduction eqn
    💡 Explanation:

    ∇·(k∇T) + q_gen = 0 steady.

  79. Q79 Past Paper · PPSC/FPSC/NTS hard

    Two-dimensional conduction shape factor S is used when

    1. A only turbulent pipe flow as shape factor use case
    2. B only radiation view factors conflated always
    3. C analytical series solutions cumbersome for complex geometries
    4. D only fatigue analysis
    💡 Explanation:

    Q = k S ΔT for conduction paths.

  80. Q80 hard

    Hydrodynamic and thermal entry lengths in tube flow differ because

    1. A Reynolds alone makes them identical always
    2. B Prandtl number affects thermal development rate
    3. C pressure alone determines both equally
    4. D roughness alone determines both equally
    💡 Explanation:

    Lt ~ L Re Pr for thermal, L ~ L Re for hydrodynamic scaling.