The Reynolds number for flow in a pipe, Re = ρv̄D/μ, represents the ratio of

inertial forces to viscous forces in the flowing fluid

viscous forces to inertial forces in the flowing fluid

buoyancy forces to viscous forces in the flowing fluid

momentum diffusivity to thermal diffusivity of the fluid

inertial forces to viscous forces in the flowing fluid

For fully developed laminar flow of a Newtonian liquid in a straight circular pipe, which of the following statements are correct?

The maximum velocity, on the pipe axis, is twice the mean velocity

The Darcy friction factor equals 64/Re

The maximum velocity, on the pipe axis, is twice the mean velocity

The Darcy friction factor equals 64/Re

The friction factor depends on the relative roughness of the pipe wall

At a fixed volumetric flow rate, doubling the viscosity halves the pressure drop

Milk is to be heated from 10 °C to 70 °C by hot water that cools from 90 °C to 60 °C in a tubular heat exchanger. Which statement is correct?

Only counter-current flow can achieve these temperatures, and its log-mean temperature difference is about 32.7 °C

Co-current flow can achieve these temperatures and is preferred, since its log-mean temperature difference, about 40 °C, is the larger

Either flow arrangement can achieve these temperatures, and both give the arithmetic-mean difference of 35 °C

Only counter-current flow can achieve these temperatures, and its log-mean temperature difference is about 32.7 °C

Only counter-current flow can achieve these temperatures, and its log-mean temperature difference is exactly 35 °C

Which of the following statements about radiation and overall heat transfer are correct?

By Kirchhoff's law, a surface's absorptivity equals its emissivity at the same wavelength and temperature

For a thin wall, the overall coefficient U is always smaller than the smaller of the two film coefficients

By Kirchhoff's law, a surface's absorptivity equals its emissivity at the same wavelength and temperature

By Wien's displacement law, a hotter surface emits its peak radiation at a longer wavelength

For a thin wall, the overall coefficient U is always smaller than the smaller of the two film coefficients

A fouling deposit on the heat-transfer surface raises the overall heat-transfer coefficient

In the solution–diffusion model of gas transport through a non-porous polymer packaging film, the permeability coefficient of the film for the gas equals the product of

the diffusivity of the gas in the polymer and its solubility in the polymer

the diffusivity of the gas in the polymer and the thickness of the film

the solubility of the gas in the polymer and the partial-pressure difference across the film

the convective mass-transfer coefficient at the film surface and the area of the film

Which of the following statements about molecular diffusion and convective mass transfer are correct?

The Sherwood number plays the role in mass transfer that the Nusselt number plays in heat transfer

Diffusivities of small molecules in gases are typically about four orders of magnitude larger than in liquids

At steady state, doubling the thickness of a film doubles the rate of diffusion through it

The Sherwood number plays the role in mass transfer that the Nusselt number plays in heat transfer

The Schmidt number is the ratio of thermal diffusivity to mass diffusivity

Diffusivities of small molecules in gases are typically about four orders of magnitude larger than in liquids

Food Engineering I: Mass and Energy Balances, Fluid Flow, Heat Transfer and Mass Transfer

This chapter answers the first four sub-items of Section XL5.4, Food Engineering, which belongs to XL5 Food Technology, one of the five optional sections of the Life Sciences (XL) paper (the Engineering Sciences (XE) paper sets the same four headings as its section XE6 Food Technology, one of nine, so this chapter serves a candidate of either paper): “Mass and energy balance. Momentum Transfer: Flow rate and pressure drop relationships for Newtonian fluids flowing through pipe, Reynolds number. Heat transfer: heat transfer by conduction, convection, radiation, heat exchangers. Mass Transfer: Molecular diffusion and Fick's law, conduction and convective mass transfer, permeability through single and multilayer films.” It builds the transport toolkit that the second chapter applies to unit operations: balances, the Reynolds number and pipe pressure drop (Hagen–Poiseuille, Darcy–Weisbach), conduction, convection, radiation and the log-mean temperature difference, and Fick's law with film permeability. The numericals are blending and drying balances, steam requirements, pressure drops, heat leaks through walls, exchanger areas and gas transmission through laminates; every property they need is stated in the question.

1. Mass and energy balances

Every food process obeys conservation of mass: input − output + generation − consumption = accumulation. In a continuous process at steady state nothing accumulates, and without reaction the balance is simply input = output, for the total mass and separately for each component. The skill lies in choosing the component: a tie component passes through unchanged (dry solids in drying or evaporation, fat in cream separation, nitrogen in air) and links streams whose total masses differ. Choose a basis (say 100 kg of feed) and write one independent equation per unknown.

Moisture is reported on two bases. Wet basis m = water ÷ total mass; dry basis X = water ÷ dry solids, so X = m/(1 − m) and m = X/(1 + X). A product at 80% moisture (wet basis) holds 4 kg of water per kg of dry solid. Worked example: 100 kg of carrot at 80% moisture (wet basis) is dried to 10%. The dry solids, 20 kg, are the tie component; the dried product weighs 20/0.90 = 22.2 kg, so 100 − 22.2 = 77.8 kg of water is evaporated, not the 70 kg that subtracting percentages suggests.

The energy balance reads: energy in with streams + heat supplied = energy out with streams + losses. Sensible heat changes temperature, Q = m cp ΔT; latent heat changes phase at constant temperature, Q = m λ. Worked example: 2000 kg/h of milk (cp = 3.9 kJ kg⁻¹ K⁻¹) is heated from 5 to 72 °C by steam that releases its latent heat of 2200 kJ/kg and leaves as condensate at its saturation temperature. Heat needed = 2000 × 3.9 × 67 = 522,600 kJ/h = 145.2 kW; steam = 522,600/2200 = 237.5 kg/h. If a stated fraction of the steam's heat is lost, divide the heat needed by (1 − loss fraction) before dividing by λ.

🧠 Blending in one line

Mixing problems (standardising milk fat, blending juices) are two balances, total and component. For x kg at concentration a blended with y kg at b to give concentration c, x/y = (c − b)/(a − c). Always check that c lies between a and b.

2. Momentum transfer: viscosity, Reynolds number and flow regimes

In momentum transfer a moving fluid layer drags the next through viscous shear. A Newtonian fluid obeys τ = μ (du/dy): shear stress is proportional to shear rate, so the viscosity μ (Pa·s) is constant at a given temperature (water, milk, clear juices, sugar syrups, edible oils). Non-Newtonian foods such as ketchup, purees and melted chocolate have a shear-dependent apparent viscosity and fall outside the formulas here. Liquid viscosity falls steeply as temperature rises, which is why viscous products are pumped warm. Kinematic viscosity is ν = μ/ρ (m²/s).

Volumetric flow rate Q = A v̄, with A = πD²/4 and v̄ the mean velocity; mass flow rate ṁ = ρQ. For an incompressible liquid the continuity equation A₁v̄₁ = A₂v̄₂ means that halving the diameter quadruples the mean velocity. The Reynolds number Re = ρv̄D/μ = v̄D/ν is the ratio of inertial to viscous forces and decides the regime. In the convention of this chapter, pipe flow is laminar for Re < 2100, turbulent above about 4000 and transitional in between; some texts put the lower limit at 2000 or 2300, so a question that depends on it states it. In laminar flow layers slide past one another without mixing; in turbulent flow eddies mix the core and flatten the velocity profile.

🎯 Why the regime matters in a plant

Turbulence improves heat transfer and cleaning, so cleaning-in-place runs at turbulent velocities. In laminar flow the axial fluid moves fastest and spends the least time in a holding tube, so holding tubes are sized on the fastest element.

3. Pressure drop in pipes: Hagen–Poiseuille and Darcy–Weisbach

For fully developed laminar flow of a Newtonian fluid the velocity profile is parabolic, u = u_max[1 − (r/R)²], with u_max = 2v̄, and the pressure drop over a length L is given by the Hagen–Poiseuille equation: ΔP = 32μLv̄/D² = 128μLQ/(πD⁴). At a fixed Q, ΔP is proportional to μ and to 1/D⁴, so halving the pipe diameter raises the pressure drop sixteen-fold.

For any regime the Darcy–Weisbach equation gives ΔP = f_D (L/D)(ρv̄²/2), where f_D is the Darcy friction factor. Many food-engineering texts use the Fanning factor f instead, with ΔP = 4f (L/D)(ρv̄²/2), so f_D = 4f. In laminar flow f_D = 64/Re (Fanning f = 16/Re), and substituting gives back Hagen–Poiseuille. In turbulent flow the factor depends on Re and the relative roughness ε/D (the Moody chart); for smooth pipes with 4000 < Re < 10⁵ the Blasius correlation gives f = 0.079 Re^(−0.25) in Fanning form, that is f_D = 0.316 Re^(−0.25). Adding elevation and fitting losses through the mechanical energy balance, pump power = QΔP_total/η.

Worked example: a sugar syrup (ρ = 1200 kg/m³, μ = 0.05 Pa·s) flows at v̄ = 0.5 m/s through 20 m of 25 mm pipe. Re = 1200 × 0.5 × 0.025/0.05 = 300, laminar, so ΔP = 32 × 0.05 × 20 × 0.5/0.025² = 25,600 Pa = 25.6 kPa. Q = (π/4)(0.025)²(0.5) = 2.45 × 10⁻⁴ m³/s, and the ideal power to overcome this friction is QΔP = 6.3 W. Milk (ρ = 1030 kg/m³, μ = 2 mPa·s) at 1.5 m/s in a 50 mm pipe has Re = 38,600, turbulent; Blasius gives f = 0.0056, about 0.52 kPa lost per metre.

⚠️ Three classic slips

Using a Fanning factor in the Darcy form (a four-fold error); holding v̄ constant when the question fixes Q and changes D; and using kinematic viscosity where dynamic viscosity is needed. Compute Re first: Hagen–Poiseuille holds only for laminar flow.

4. Conduction and thermal resistance

Conduction is heat flow through a material by molecular contact, governed by Fourier's law: q = −kA dT/dx, where k is the thermal conductivity (W m⁻¹ K⁻¹). Typical values are about 0.025 for still air, 0.02 to 0.03 for polyurethane foam, 0.6 for water, 2.2 for ice, 16 for stainless steel and 200 for aluminium. Ice conducts nearly four times better than water, one reason why thawing is slower than freezing: the thawed outer layer insulates the frozen core.

At steady state through a plane wall Q = kA(T₁ − T₂)/L, written Q = ΔT/R with the thermal resistance R = L/(kA), exactly like Ohm's law. Layers in series add: R_total = L₁/(k₁A) + L₂/(k₂A) + …, and a surface film adds 1/(hA). The same heat crosses every layer, so the largest temperature drop occurs across the largest resistance, the insulation. For a cylindrical shell such as pipe insulation, R = ln(r₂/r₁)/(2πkL), because the area grows with radius.

Worked example: a freezer wall is 0.10 m of polyurethane (k = 0.025 W m⁻¹ K⁻¹) between thin steel skins of negligible resistance. Inside air is at −25 °C with h = 10 W m⁻² K⁻¹; outside air is at 35 °C with h = 15 W m⁻² K⁻¹. Per m², R = 1/10 + 0.10/0.025 + 1/15 = 0.100 + 4.000 + 0.067 = 4.167 m² K/W, so U = 0.24 W m⁻² K⁻¹ and q = 0.24 × 60 = 14.4 W/m². The foam supplies 96% of the resistance.

5. Convection and radiation

Convection carries heat between a surface and a moving fluid, described by Newton's law of cooling: q = hA(T_s − T_∞), where h (W m⁻² K⁻¹) depends on the fluid, its velocity and the geometry. In natural (free) convection buoyancy drives the motion; in forced convection a pump or fan does. h comes from dimensionless correlations: the Nusselt number Nu = hD/k_fluid, the Prandtl number Pr = cpμ/k (momentum ÷ thermal diffusivity), Re for forced flow and the Grashof number for free convection. For turbulent flow inside a tube a common correlation is Dittus–Boelter, Nu = 0.023 Re^0.8 Pr^n, with n = 0.4 when the fluid is heated and 0.3 when it is cooled.

Orders of magnitude of the convective coefficient h
Situation	Typical h, W m⁻² K⁻¹
Free convection, gases	2 to 25
Free convection, liquids	50 to 1000
Forced convection, gases	25 to 250
Forced convection, liquids	100 to 20,000
Boiling and condensation	2500 to 100,000

Radiation needs no medium: every surface above 0 K emits electromagnetic energy. A black body emits E_b = σT⁴, with σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴ and T in kelvin; a real grey surface emits εσT⁴, where the emissivity ε lies between 0 and 1 and, by Kirchhoff's law, equals the absorptivity at the same wavelength and temperature. A small body of emissivity ε in large surroundings exchanges q = εσA(T_s⁴ − T_sur⁴). Wien's law, λ_max T = 2898 µm·K, shifts peak emission to shorter wavelengths for hotter sources. Worked example: a dough surface at 300 K (ε = 0.9) in an oven with walls at 500 K receives 0.9 × 5.67 × 10⁻⁸ × (500⁴ − 300⁴) = 2.78 kW/m². Doubling the absolute temperature raises emission sixteen-fold, so radiation dominates in ovens and infrared heaters.

⚠️ Exam trap

Convert to kelvin before raising anything to the fourth power; T⁴ in °C is meaningless.

6. Overall coefficient, LMTD and heat exchangers

In a heat exchanger heat passes from the hot fluid through a film, the wall and a second film into the cold fluid. The resistances add, so for a plane or thin-walled tube 1/U = 1/h_i + Δx/k_w + 1/h_o + R_f, where R_f is the fouling resistance of deposits such as milk-stone and burnt protein. U is always smaller than the smaller of h_i and h_o, so improving the poorer film pays most. Fouling lowers U, which is why exchangers handling milk are cleaned in place after every run.

The duty is Q = UA ΔT_lm, with the log-mean temperature difference ΔT_lm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂), where ΔT₁ and ΔT₂ are the differences at the two ends. In counter-current flow the fluids enter at opposite ends; in co-current (parallel) flow they enter at the same end. For the same terminal temperatures counter-current flow gives the larger ΔT_lm, and it is the only arrangement in which the cold fluid can leave hotter than the hot fluid leaves. Worked example: juice (1.0 kg/s, cp = 3.9 kJ kg⁻¹ K⁻¹) is heated from 20 to 70 °C counter-currently by water cooling from 95 to 75 °C, with U = 1000 W m⁻² K⁻¹. Q = 1.0 × 3.9 × 50 = 195 kW; ΔT₁ = 95 − 70 = 25 K and ΔT₂ = 75 − 20 = 55 K, so ΔT_lm = 30/ln 2.2 = 38.0 K and A = 195,000/(1000 × 38.0) = 5.1 m². The arithmetic mean (40 K) would overstate the driving force.

Heat exchangers in food plants
Type	Construction	Typical use
Plate	Gasketed corrugated plates; thin channels give a high U	Milk and juice pasteurisers with a regeneration section
Tubular (double-pipe, shell-and-tube)	Product in tubes, heating or cooling medium outside	Products with fibres or particles; high pressures
Scraped-surface	Rotating blades scrape the heated or cooled wall	Viscous foods: purees, ice-cream mix, margarine

⚠️ Exam trap

A co-current answer is impossible whenever the cold outlet is above the hot outlet. When ΔT₁ = ΔT₂ the formula becomes 0/0, and ΔT_lm simply equals that common difference.

7. Mass transfer: diffusion, convection and film permeability

Molecular diffusion moves a species from high to low concentration by random molecular motion. Fick's first law is the mass-transfer twin of Fourier's law: J_A = −D_AB dC_A/dx, with J the flux, D_AB the diffusivity (m²/s) and C the concentration. Orders of magnitude: about 10⁻⁵ m²/s in gases, 10⁻⁹ m²/s in liquids and 10⁻¹⁰ m²/s or less in solids. Steady diffusion through a layer of thickness L, the 'conduction' of mass, gives J = D(C₁ − C₂)/L, and layer resistances add as in heat conduction. The time for diffusion to penetrate a layer scales as L²/D, so halving the thickness of a slice cuts a diffusion-controlled drying or brining time about four-fold.

Convective mass transfer carries a species between a surface and a moving fluid: N_A = k_c A(C_s − C_∞), where k_c (m/s) is the mass-transfer coefficient. The correlations mirror heat transfer: the Sherwood number Sh = k_c L/D replaces Nu and the Schmidt number Sc = μ/(ρD) (momentum ÷ mass diffusivity) replaces Pr, so Nu = f(Re, Pr) becomes Sh = f(Re, Sc), the heat–mass transfer analogy. The Lewis number Le = α/D = Sc/Pr links the two; for air and water vapour the ratio h/(k_y c_s) is close to 1 (the Lewis relation), which is why the wet-bulb temperature nearly equals the adiabatic saturation temperature.

Gas and water vapour cross a non-porous polymer film by solution–diffusion: the permeant dissolves at the high-pressure face, diffuses across and desorbs at the other face. The permeability coefficient P̄ = D × S (diffusivity × solubility), and the steady rate is q = P̄AΔp/L, with Δp the partial-pressure difference and L the thickness; doubling L halves the rate. In a multilayer laminate the layers act as resistances in series, L_T/P̄_T = L₁/P̄₁ + L₂/P̄₂ + …, so the best barrier dominates. Worked example (illustrative oxygen permeabilities in cm³ µm m⁻² day⁻¹ atm⁻¹): 50 µm of LDPE with P̄ = 2.0 × 10⁵ transmits 2.0 × 10⁵/50 = 4000 cm³ m⁻² day⁻¹ atm⁻¹; adding 10 µm of EVOH with P̄ = 5 gives 1/(50/200,000 + 10/5) ≈ 0.50 cm³ m⁻² day⁻¹ atm⁻¹, an 8000-fold cut.

ℹ️ Barrier behaviour of common films

Polyolefins (LDPE, HDPE, PP) are good water-vapour barriers but poor oxygen barriers; EVOH is the reverse and loses its oxygen barrier when wet, so it is buried between polyolefin layers. WVTR and OTR mean something only at a stated temperature and humidity.

Key takeaways

At steady state input = output for the total and each component; balance on a tie component and keep wet basis m and dry basis X = m/(1 − m) apart. Energy balances add sensible heat m cp ΔT and latent heat m λ.
Re = ρv̄D/μ (inertial ÷ viscous); here laminar means Re < 2100, with u_max = 2v̄ and ΔP = 32μLv̄/D² = 128μLQ/(πD⁴). Darcy f_D = 4 × Fanning f = 64/Re in laminar flow.
Conduction resistance is L/(kA) for a slab and ln(r₂/r₁)/(2πkL) for a cylinder; films add 1/(hA); series resistances add.
Convection: q = hAΔT, h from Nu = f(Re, Pr). Radiation: q = εσA(T₁⁴ − T₂⁴) in kelvin; emissivity equals absorptivity (Kirchhoff).
1/U = 1/h_i + Δx/k + 1/h_o + R_f, and Q = UA ΔT_lm with ΔT_lm = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂). Only counter-current flow lets the cold outlet exceed the hot outlet.
Fick: J = −D dC/dx; convective N = k_c A ΔC, with Sh and Sc in place of Nu and Pr. Films: q = P̄AΔp/L with P̄ = D × S, and for laminates L_T/P̄_T = Σ Lᵢ/P̄ᵢ.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A batch of fresh peas contains 75% moisture on a wet basis. What is its moisture content on a dry basis?
1. 0.75 kg water per kg dry solid
2. 1.33 kg water per kg dry solid
3. 3.00 kg water per kg dry solid
4. 0.25 kg water per kg dry solid
Show answer
Answer: C — 3.00 kg water per kg dry solid
Each kg of peas holds 0.75 kg of water and 0.25 kg of dry solid, so X = 0.75/0.25 = 3.00 kg/kg. The value 0.75 kg/kg is the tempting slip of keeping the wet-basis fraction without changing the basis; 1.33 is 1/0.75 and 0.25 is the solids fraction.
A dairy blends 800 kg of whole milk containing 4.6% fat with skim milk containing 0.1% fat to obtain standardised milk containing 3.0% fat (all percentages by mass). The mass of skim milk required, in kg (rounded to one decimal place), is ____.
Numerical answer — type the value.
Show answer
Answer: 441.4
Let S be the mass of skim milk. Fat balance: 800(0.046) + S(0.001) = (800 + S)(0.030), so 36.8 + 0.001S = 24.0 + 0.030S, giving 0.029S = 12.8 and S = 441.4 kg. Check with the blending rule: whole/skim = (3.0 − 0.1)/(4.6 − 3.0) = 2.9/1.6, so S = 800 × 1.6/2.9 = 441.4 kg.
Tomato juice (1500 kg/h, specific heat 3.8 kJ kg⁻¹ K⁻¹) is heated from 20 °C to 90 °C in an indirect heater by saturated steam. The steam condenses completely and leaves as saturated liquid, releasing 2230 kJ per kg of steam. 5% of the heat released by the steam is lost to the surroundings. The steam consumption, in kg/h (rounded to one decimal place), is ____.
Numerical answer — type the value.
Show answer
Answer: 188.3
Heat taken up by the juice = 1500 × 3.8 × (90 − 20) = 399,000 kJ/h. This is 95% of the heat released by the steam, so the steam must release 399,000/0.95 = 420,000 kJ/h, and steam = 420,000/2230 = 188.3 kg/h. Multiplying by 1.05 instead of dividing by 0.95 gives 187.9 kg/h, which is wrong because the loss is defined as a fraction of the steam's heat, not of the juice's.
1000 kg of grapes containing 82% moisture (wet basis) are dried to raisins containing 15% moisture (wet basis). How much water is removed?
1. 670.0 kg
2. 788.2 kg
3. 211.8 kg
4. 820.0 kg
Show answer
Answer: B — 788.2 kg
Dry solids = 1000 × 0.18 = 180 kg, unchanged by drying. Raisins = 180/0.85 = 211.8 kg, so water removed = 1000 − 211.8 = 788.2 kg. The tempting 670.0 kg subtracts the percentages (82 − 15 = 67% of 1000 kg), which is wrong because the two percentages refer to different total masses; 211.8 kg is the mass of raisins and 820.0 kg is all the water originally present.
The Reynolds number for flow in a pipe, Re = ρv̄D/μ, represents the ratio of
1. viscous forces to inertial forces in the flowing fluid
2. buoyancy forces to viscous forces in the flowing fluid
3. momentum diffusivity to thermal diffusivity of the fluid
4. inertial forces to viscous forces in the flowing fluid
Show answer
Answer: D — inertial forces to viscous forces in the flowing fluid
Re = ρv̄D/μ compares inertia (scaling as ρv̄²) with viscous stress (scaling as μv̄/D); a large Re means inertia dominates and turbulence develops. The inverse ratio is the tempting distractor, but it would fall as velocity rises. Buoyancy to viscous forces is the Grashof number, and momentum to thermal diffusivity is the Prandtl number.
A Newtonian sugar syrup (density 1300 kg/m³, viscosity 0.08 Pa·s) is pumped at 0.5 L/s through a straight horizontal pipe of internal diameter 40 mm and length 15 m. Take the flow to be laminar if Re < 2100, and in that case use the Hagen–Poiseuille equation. The frictional pressure drop, in kPa (rounded to two decimal places), is ____.
Numerical answer — type the value.
Show answer
Answer: 9.55
Mean velocity v̄ = Q/A = 5 × 10⁻⁴/(π × 0.04²/4) = 0.398 m/s. Re = 1300 × 0.398 × 0.04/0.08 = 259, so the flow is laminar. ΔP = 32μLv̄/D² = 32 × 0.08 × 15 × 0.398/0.0016 = 9549 Pa = 9.55 kPa, the same as 128μLQ/(πD⁴).
A Newtonian liquid flows in laminar flow through a pipe. The pipe is replaced by one of half the internal diameter and the same length, and the volumetric flow rate is kept unchanged; the flow remains laminar. The frictional pressure drop becomes
1. 2 times the original value
2. 4 times the original value
3. 16 times the original value
4. 32 times the original value
Show answer
Answer: C — 16 times the original value
At fixed Q, Hagen–Poiseuille gives ΔP = 128μLQ/(πD⁴), proportional to 1/D⁴, so halving D multiplies ΔP by 2⁴ = 16. The tempting 4 comes from ΔP ∝ v̄/D² with v̄ held constant, but at fixed Q the mean velocity itself rises four-fold.
A friction-factor chart that uses the Fanning definition, ΔP = 4f (L/D)(ρv̄²/2), gives f = 0.005 at the Reynolds number of interest. If the pressure drop is computed with the Darcy–Weisbach form ΔP = f_D (L/D)(ρv̄²/2), the value of f_D to use is
1. 0.00125
2. 0.005
3. 0.010
4. 0.020
Show answer
Answer: D — 0.020
Comparing the two forms term by term, f_D = 4f = 4 × 0.005 = 0.020. Using 0.005 directly in the Darcy form, the commonest slip, underestimates the pressure drop four-fold; 0.00125 divides by 4 instead of multiplying, and 0.010 confuses the factor of 4 with 2.
For fully developed laminar flow of a Newtonian liquid in a straight circular pipe, which of the following statements are correct?
1. The maximum velocity, on the pipe axis, is twice the mean velocity
2. The Darcy friction factor equals 64/Re
3. The friction factor depends on the relative roughness of the pipe wall
4. At a fixed volumetric flow rate, doubling the viscosity halves the pressure drop
Show answer
Answer: A — The maximum velocity, on the pipe axis, is twice the mean velocity; B — The Darcy friction factor equals 64/Re
The laminar profile is parabolic, u = u_max[1 − (r/R)²], and its mean is u_max/2. Substituting Hagen–Poiseuille into Darcy–Weisbach gives f_D = 64/Re (Fanning 16/Re). Roughness affects the friction factor only in turbulent flow; in laminar flow it depends on Re alone. ΔP is directly proportional to μ, so doubling the viscosity doubles the pressure drop.
For steady one-dimensional conduction through a plane wall of thickness L, thermal conductivity k and face area A, the thermal resistance is
1. L/(kA)
2. kA/L
3. 1/(hA)
4. ln(r₂/r₁)/(2πkL)
Show answer
Answer: A — L/(kA)
Fourier's law gives Q = kA(T₁ − T₂)/L = ΔT/R, so R = L/(kA) in K/W. kA/L is the conductance, the reciprocal of the resistance; 1/(hA) is a convective film resistance; and ln(r₂/r₁)/(2πkL) is the resistance of a cylindrical shell.
A cold-store wall consists of 115 mm of brick (k = 0.80 W m⁻¹ K⁻¹) and 80 mm of polyurethane foam (k = 0.025 W m⁻¹ K⁻¹) in series. Outside air is at 32 °C with a surface coefficient of 20 W m⁻² K⁻¹; the store air is at −18 °C with a surface coefficient of 8 W m⁻² K⁻¹. Assume steady one-dimensional heat flow. The heat gain per square metre of wall, in W/m² (rounded to two decimal places), is ____.
Numerical answer — type the value.
Show answer
Answer: 14.21
Resistances per m² in series: outside film 1/20 = 0.0500; brick 0.115/0.80 = 0.14375; foam 0.080/0.025 = 3.2000; inside film 1/8 = 0.1250; total = 3.51875 m² K/W. q = (32 − (−18))/3.51875 = 50/3.51875 = 14.21 W/m². The foam supplies about 91% of the resistance.
Milk is to be heated from 10 °C to 70 °C by hot water that cools from 90 °C to 60 °C in a tubular heat exchanger. Which statement is correct?
1. Co-current flow can achieve these temperatures and is preferred, since its log-mean temperature difference, about 40 °C, is the larger
2. Either flow arrangement can achieve these temperatures, and both give the arithmetic-mean difference of 35 °C
3. Only counter-current flow can achieve these temperatures, and its log-mean temperature difference is about 32.7 °C
4. Only counter-current flow can achieve these temperatures, and its log-mean temperature difference is exactly 35 °C
Show answer
Answer: C — Only counter-current flow can achieve these temperatures, and its log-mean temperature difference is about 32.7 °C
The milk leaves at 70 °C, above the water outlet of 60 °C. In co-current flow the cold outlet can never exceed the hot outlet, so only counter-current flow works. Its end differences are 90 − 70 = 20 °C and 60 − 10 = 50 °C, so ΔT_lm = (50 − 20)/ln(50/20) = 30/0.916 = 32.7 °C. The 35 °C arithmetic mean is the tempting value but overstates the driving force; the log mean is always smaller unless the two end differences are equal.
In a counter-current tubular heater, juice is heated from 15 °C to 65 °C by water that cools from 85 °C to 55 °C, at a heat duty of 100 kW. The juice-side film coefficient is 1500 W m⁻² K⁻¹, the water-side film coefficient is 6000 W m⁻² K⁻¹, the stainless-steel wall is 1.5 mm thick (k = 15 W m⁻¹ K⁻¹) and a fouling resistance of 0.0002 m² K W⁻¹ acts on the juice side. Treat the wall as thin (plane-wall form for U). The heat-transfer area required, in m² (rounded to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 3.93
1/U = 1/1500 + 0.0015/15 + 1/6000 + 0.0002 = 0.000667 + 0.000100 + 0.000167 + 0.000200 = 0.001133 m² K/W, so U = 882.4 W m⁻² K⁻¹. End differences: 85 − 65 = 20 K and 55 − 15 = 40 K, so ΔT_lm = 20/ln 2 = 28.85 K. A = Q/(UΔT_lm) = 100,000/(882.4 × 28.85) = 3.93 m².
The absolute temperature of a grey radiant heater, whose emissivity does not change with temperature, is raised from 300 K to 600 K. By what factor does the radiant energy it emits per unit area increase?
1. 8
2. 16
3. 2
4. 4
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Answer: B — 16
Emissive power εσT⁴ varies with the fourth power of absolute temperature: (600/300)⁴ = 2⁴ = 16. The answer 2 assumes a linear law, as in convection; 4 and 8 use the wrong exponent.
Which of the following statements about radiation and overall heat transfer are correct?
1. By Kirchhoff's law, a surface's absorptivity equals its emissivity at the same wavelength and temperature
2. By Wien's displacement law, a hotter surface emits its peak radiation at a longer wavelength
3. For a thin wall, the overall coefficient U is always smaller than the smaller of the two film coefficients
4. A fouling deposit on the heat-transfer surface raises the overall heat-transfer coefficient
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Answer: A — By Kirchhoff's law, a surface's absorptivity equals its emissivity at the same wavelength and temperature; C — For a thin wall, the overall coefficient U is always smaller than the smaller of the two film coefficients
Kirchhoff's law equates spectral emissivity and absorptivity at the same temperature. Because 1/U is the sum of 1/h_i, 1/h_o and positive wall and fouling terms, 1/U exceeds each 1/h, so U lies below the smaller h. Wien's law, λ_max T = 2898 µm·K, moves the peak to shorter wavelengths as T rises. Fouling adds a resistance and lowers U.
In the solution–diffusion model of gas transport through a non-porous polymer packaging film, the permeability coefficient of the film for the gas equals the product of
1. the diffusivity of the gas in the polymer and its solubility in the polymer
2. the diffusivity of the gas in the polymer and the thickness of the film
3. the solubility of the gas in the polymer and the partial-pressure difference across the film
4. the convective mass-transfer coefficient at the film surface and the area of the film
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Answer: A — the diffusivity of the gas in the polymer and its solubility in the polymer
The gas dissolves at the high-pressure face (solubility S), diffuses across (diffusivity D) and desorbs at the other face, so P̄ = D × S and the rate is P̄AΔp/L. Thickness and the partial-pressure difference appear in the rate equation, not in the coefficient, and D multiplied by L is not a meaningful property. The convective coefficient describes transfer in the gas film outside the polymer, not through it.
A pouch with 0.05 m² of permeable area is made of a laminate of 12 µm PET (oxygen permeability 1500 cm³ µm m⁻² day⁻¹ atm⁻¹) and 50 µm LDPE (oxygen permeability 2.0 × 10⁵ cm³ µm m⁻² day⁻¹ atm⁻¹). The pouch is flushed with nitrogen so that the oxygen partial pressure inside is effectively zero, and it is stored in air with an oxygen partial pressure of 0.21 atm. Treat the layers as resistances in series. The oxygen entering the pouch, in cm³ per day (rounded to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 1.27
For the laminate, L/P̄ = 12/1500 + 50/200,000 = 0.00800 + 0.00025 = 0.00825, so its permeance is 1/0.00825 = 121.2 cm³ m⁻² day⁻¹ atm⁻¹. Rate = 121.2 × 0.05 × 0.21 = 1.27 cm³/day. The PET layer supplies 97% of the resistance; adding the permeabilities instead of the resistances is the classic error.
Which of the following statements about molecular diffusion and convective mass transfer are correct?
1. At steady state, doubling the thickness of a film doubles the rate of diffusion through it
2. The Sherwood number plays the role in mass transfer that the Nusselt number plays in heat transfer
3. The Schmidt number is the ratio of thermal diffusivity to mass diffusivity
4. Diffusivities of small molecules in gases are typically about four orders of magnitude larger than in liquids
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Answer: B — The Sherwood number plays the role in mass transfer that the Nusselt number plays in heat transfer; D — Diffusivities of small molecules in gases are typically about four orders of magnitude larger than in liquids
Sh = k_cL/D is the mass-transfer counterpart of Nu = hL/k, so heat-transfer correlations carry over with Sc in place of Pr. Gas diffusivities are about 10⁻⁵ m²/s and liquid ones about 10⁻⁹ m²/s, a factor of about 10⁴. The flux is DΔC/L, so doubling L halves it. The Schmidt number is momentum diffusivity over mass diffusivity, μ/(ρD); thermal over mass diffusivity is the Lewis number.