Food Engineering II: Mechanical Operations, Thermal Operations and Psychrometrics
2. Homogenisation, settling and centrifugation
A high-pressure homogeniser is a piston pump that forces a liquid through a narrow adjustable gap, the homogenising valve. The sudden pressure release creates intense turbulence, shear and cavitation, which break fat globules and droplets. Milk fat globules (average about 3 to 4 µm) are reduced to about 1 µm or less at roughly 10 to 25 MPa with the milk warm and its fat liquid; a second stage at much lower pressure breaks up clusters of the new globules. Homogenised milk does not cream, looks whiter and tastes fuller.
The reason is Stokes' law for the terminal velocity of a small sphere in creeping flow (particle Reynolds number below about 1): v = d²(ρ_p − ρ_f)g/(18μ). A negative value means the particle rises, as fat does in milk. Velocity scales with d², so cutting globules from 4 µm to 1 µm slows creaming sixteen-fold. Gravity settling tanks clarify liquids; warming lowers μ and speeds settling.
A centrifuge replaces g by the centripetal acceleration ω²r: v = d²(ρ_p − ρ_f)ω²r/(18μ), and the relative centrifugal force is RCF = ω²r/g. At 5000 rpm (ω = 2π × 5000/60 = 523.6 rad/s) and r = 0.15 m, ω²r = 41,100 m/s², about 4190 g. At a fixed radius the velocity scales with the square of the speed. In a disc-stack cream separator the conical discs shorten the settling path: the denser skim milk moves outward and leaves at the periphery, while the lighter cream moves inward and leaves near the axis.
3. Filtration, mixing and agitation
In cake filtration a slurry is pushed through a cloth or screen and the retained solids form a cake that becomes the main filtering layer. At constant pressure the filtrate volume V collected in time t obeys t/V = [μαc/(2A²ΔP)]V + μR_m/(AΔP), where α is the specific cake resistance, c the dry cake mass per volume of filtrate and R_m the medium resistance; a plot of t/V against V is a straight line whose slope gives α and intercept R_m. With R_m negligible, t ∝ V², so twice the filtrate takes four times as long. Compressible cakes such as fruit pulp grow more resistant as pressure rises, so a filter aid such as diatomaceous earth is pre-coated or added to keep the cake porous.
Mixing and agitation of liquids uses an impeller in a tank. Propellers (axial flow, high speed) and flat-blade turbines (radial flow, high shear) suit thin liquids; anchors and helical ribbons turn slowly and sweep the wall in viscous products. Without baffles (usually four vertical strips) a central impeller in a thin liquid only swirls the contents and draws a vortex. Power is correlated through the impeller Reynolds number Re = ρND²/μ (N in rev/s, D the impeller diameter) and the power number Np = P/(ρN³D⁵). In the laminar range (Re below about 10) Np × Re is constant, so P ∝ μN²D³; in a baffled tank in the turbulent range Np is constant, so P = Np ρN³D⁵. Worked example: a turbine with Np = 5, D = 0.3 m and N = 2 rev/s in juice of ρ = 1000 kg/m³ draws 5 × 1000 × 2³ × 0.3⁵ = 97 W.
4. Blanching, pasteurisation and sterilisation
Blanching briefly heats vegetables in hot water or steam (typically 70 to 100 °C for a few minutes) before freezing, drying or canning. Its main aim is to inactivate enzymes (peroxidase, catalase, lipoxygenase, polyphenol oxidase) that would cause off-flavours, browning and loss of colour and vitamins in storage; it also expels tissue air and lowers the microbial load. Peroxidase is among the most heat-stable of these enzymes, so a negative peroxidase test is the usual check. Water blanching leaches more water-soluble vitamins and solids than steam blanching.
Pasteurisation is a mild heat treatment that kills vegetative pathogens and reduces spoilage organisms; the product still needs refrigeration. For milk the classic combinations are 63 °C for 30 min (batch, low temperature long time) and 72 °C for 15 s (high temperature short time, HTST); both destroy Coxiella burnetii, the most heat-resistant non-spore-forming pathogen commonly found in milk. A negative alkaline phosphatase test confirms adequate pasteurisation, because this native milk enzyme is slightly more heat-resistant than that pathogen. An HTST plant is a plate exchanger with a regeneration section (hot pasteurised milk preheats cold raw milk), heating and cooling sections, a holding tube sized so that the fastest element stays for the full holding time (in laminar flow v_max = 2v̄), and a flow-diversion valve that returns under-heated milk.
Commercial sterilisation of low-acid foods (pH above 4.6) must destroy spores of Clostridium botulinum; the 12D concept asks for twelve decimal reductions of its spores. Lethality is expressed as the F₀ value, equivalent minutes at 121.1 °C with z = 10 °C: the lethal rate at temperature T is L = 10^((T − 121.1)/z), and F₀ = Σ L Δt over the temperature history of the cold point, the slowest-heating point (the geometric centre in conduction-heating packs, lower on the vertical axis in convection-heating ones). Worked example: 20 min at 115 °C gives F₀ = 20 × 10^(−0.61) = 4.9 min. UHT processing (about 135 to 150 °C for a few seconds, then aseptic packaging) loses little quality because quality factors have much larger z values than spores, so higher temperatures speed spore death far more than quality damage.
5. Evaporation of liquid foods
Evaporation concentrates milk, juices, tomato pulp and sugar solutions by boiling off water. It runs under vacuum so the liquid boils at about 40 to 70 °C, protecting colour and flavour. A solution boils above pure water at the same pressure; this boiling-point elevation (BPE), estimated with Dühring's rule (solution boiling point is linear in water boiling point), reduces the useful temperature difference. Common designs are falling-film (a thin film down long tubes, short residence; standard for milk and juice), forced-circulation for viscous or fouling liquids, and agitated thin-film for very viscous, heat-sensitive concentrates.
For a single effect at steady state: F = P + V and F x_F = P x_P (solids are the tie component); neglecting BPE and losses, the energy balance is S λ_s = V λ_v + F c_p (T₁ − T_F), with S the steam and T₁ the boiling temperature. Steam economy = V/S. Worked example: 1000 kg/h of juice at 10% solids is concentrated to 40%: P = 1000 × 0.10/0.40 = 250 kg/h and V = 750 kg/h. With the feed at its boiling point, λ_v = 2350 kJ/kg and λ_s = 2200 kJ/kg, S = 750 × 2350/2200 = 801 kg/h, an economy of 0.94.
In a multiple-effect evaporator the vapour from each effect heats the next, which runs at lower pressure and temperature, so one kilogram of steam evaporates roughly n kilograms of water in n effects (somewhat less in practice). The overall temperature drop between steam and condenser is shared among the effects, so with the same area per effect the capacity is about that of a single effect of one effect's area: extra effects buy economy, not throughput. Forward feed flows from high to low pressure without pumps and leaves the concentrate in the coolest effect, which suits heat-sensitive foods; backward feed needs pumps but handles cold feeds and viscous concentrates better. Vapour recompression also raises economy: thermal (TVR, a steam-jet ejector) or mechanical (MVR, an electric compressor).
6. Hot-air, spray and freeze drying
In hot-air drying the air supplies heat and removes vapour. In the constant-rate period the surface stays wet like free water, the product sits at the air's wet-bulb temperature, and the rate is set by external conditions (air temperature, humidity and velocity), R_c = h(T_a − T_w)/λ_w. At the critical moisture content X_c the surface begins to dry and the falling-rate period starts: internal moisture movement by diffusion and capillary flow controls, the rate falls and the product warms toward the air temperature, when heat damage and case hardening occur. Drying stops at the equilibrium moisture content X_e fixed by the air's relative humidity.
With dry-solid mass M_s, drying area A and dry-basis moisture X, the constant-rate time is t_c = M_s(X₀ − X_c)/(A R_c). If the falling rate is proportional to (X − X_e), the falling-rate time is t_f = [M_s(X_c − X_e)/(A R_c)] ln[(X_c − X_e)/(X − X_e)]. Worked example: with M_s/A = 10 kg/m² and R_c = 2 kg m⁻² h⁻¹, drying from X₀ = 3.0 to X_c = 1.0 takes t_c = 10 × 2.0/2 = 10 h.
Spray drying atomises a liquid (rotary disc or pressure nozzle) into hot air and turns it into powder within seconds (milk, coffee, egg, flavours). Evaporative cooling holds each droplet near the wet-bulb temperature, far below the inlet air temperature, so heat-sensitive solids survive; cyclones collect the powder. Freeze drying first freezes the product, then in primary drying removes the ice by sublimation under vacuum, below the triple point of water (about 611 Pa and 0.01 °C), with shelf heat supplying the latent heat; secondary drying removes bound water by desorption at a higher shelf temperature. The porous product rehydrates quickly and keeps aroma and colour, but freeze drying is slow and the costliest method.
7. Freezing and crystallisation
A freezing curve shows cooling, slight supercooling before nucleation, then a near-plateau as latent heat is removed at the initial freezing point, which lies below 0 °C because solutes depress it. As ice forms the remaining solution concentrates and its freezing point keeps falling, so some water stays unfrozen even at low temperatures. Fast freezing passes quickly through the zone of maximum ice-crystal formation (about −1 to −5 °C) and forms many small crystals, inside cells as well as between them; slow freezing forms large extracellular crystals that damage tissue and cause drip loss on thawing.
Plank's equation estimates the freezing time: t_F = [ρλ/(T_F − T_a)](Pa/h + Ra²/k), with ρ, k and λ the density, thermal conductivity and latent heat of fusion (per kg) of the frozen food, T_F the initial freezing point, T_a the medium temperature, h the surface coefficient and a the full thickness of a slab (P = 1/2, R = 1/8) or the diameter of an infinite cylinder (P = 1/4, R = 1/16) or sphere (P = 1/6, R = 1/24). Texts that use the half-thickness change P and R, so a question states its form. Plank ignores sensible heat removed above and below T_F, so it underestimates the time. Worked example: a 5 cm slab (ρ = 1000 kg/m³, λ = 250 kJ/kg, k = 1.5 W m⁻¹ K⁻¹, T_F = −1.5 °C) in air at −30 °C with h = 25 W m⁻² K⁻¹ needs (1000 × 250,000/28.5)(0.5 × 0.05/25 + 0.125 × 0.05²/1.5) = 10,600 s, about 2.9 h.
Crystallisation separates a solute (sucrose, salt, lactose) or a fat from solution or melt. Its driving force is supersaturation, a concentration above the solubility curve reached by cooling or by evaporating solvent. In the metastable zone just above solubility, existing crystals grow but few new nuclei form, so seeding with fine crystals controls their number and size; beyond it, spontaneous nucleation gives showers of fine crystals. Examples: seeding sweetened condensed milk with fine lactose to prevent sandiness, and tempering chocolate so that cocoa butter sets as glossy form V crystals.
8. Psychrometrics, humidification and dehumidification
Psychrometrics treats moist air as dry air plus water vapour at total pressure P (101.325 kPa at sea level), and works per kg of dry air because that flow stays constant through a process. Humidity ratio W = 0.622 p_v/(P − p_v) kg water per kg dry air; relative humidity φ = p_v/p_s, the vapour pressure as a fraction of the saturation pressure at the dry-bulb temperature. The dew point is where air becomes saturated when cooled at constant W. The wet-bulb temperature (a thermometer with a wet wick in moving air) nearly equals the adiabatic saturation temperature for air–water. Enthalpy per kg dry air is often taken as h = 1.006T + W(2501 + 1.86T) kJ/kg with T in °C; texts round the constants slightly differently.
| Process | Path on the chart | What changes |
|---|---|---|
| Sensible heating | Horizontal, to the right | T rises; W and dew point constant; φ falls |
| Cooling and dehumidifying | Left, then down once below the dew point | T and W fall; water condenses on the coil |
| Adiabatic (evaporative) humidification | Along a constant wet-bulb line | T falls; W and φ rise; h nearly constant |
| Desiccant dehumidification | Toward lower W, roughly at constant enthalpy | W falls; T rises as heat of adsorption is released |
Worked example: air at 30 °C and 60% RH (p_s = 4.246 kPa) has p_v = 2.548 kPa and W = 0.622 × 2.548/(101.325 − 2.548) = 0.0160 kg/kg dry air. Heated to 70 °C for a dryer (p_s = 31.20 kPa) at the same W, its RH falls to 2.548/31.20 = 8.2%, which is why heated air dries well; over wet product it then cools along a constant wet-bulb line while gaining moisture. Humidification (water sprays, steam injection) keeps ripening rooms and cold stores humid to limit weight loss; dehumidification (cooling coils below the dew point, or desiccants such as silica gel) supplies dry air for drying and dry stores.
Key takeaways
- Rittinger (fine grinding) E ∝ 1/x₂ − 1/x₁; Kick (coarse) E ∝ ln(x₁/x₂); Bond E = 10Wᵢ(1/√x₂ − 1/√x₁) with x in µm.
- Stokes: v = d²Δρg/(18μ), with ω²r replacing g in a centrifuge. Filtration with negligible medium resistance: t ∝ V². Turbulent agitator power P = Np ρN³D⁵.
- Blanching inactivates enzymes (peroxidase test); milk is pasteurised at 63 °C/30 min or 72 °C/15 s (phosphatase test); sterility is judged by F₀ = Σ10^((T − 121.1)/10)Δt at the cold point.
- Evaporation: F = P + V, F x_F = P x_P, S λ_s = V λ_v + F c_p(T₁ − T_F). n effects multiply economy about n-fold but not capacity.
- Drying: constant rate at the wet-bulb temperature, externally controlled; falling rate below X_c, internally controlled. Freeze drying sublimes ice below 611 Pa.
- Plank: t_F = [ρλ/(T_F − T_a)](Pa/h + Ra²/k) with P, R = 1/2, 1/8 (slab) or 1/6, 1/24 (sphere). W = 0.622p_v/(P − p_v); heating lowers RH at constant W.
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.
Which size-reduction law assumes that the energy required is proportional to the new surface area created?
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Answer: C — Rittinger's law, E = K_R(1/x₂ − 1/x₁)
New surface per unit mass is inversely proportional to particle size, so energy ∝ (1/x₂ − 1/x₁): Rittinger's law (n = 2), suited to fine grinding. Kick's law, the tempting distractor, makes energy depend on the size-reduction ratio and suits coarse crushing; Bond's law is the intermediate n = 1.5 case; Stokes' law gives a settling velocity, not grinding energy.A spice mill grinds 10 t/h of material whose Bond work index is 12 kWh/t. The feed and product sizes (80% passing) are 5000 µm and 500 µm. Using Bond's law, E = 10Wᵢ(1/√x₂ − 1/√x₁) with x in µm and E in kWh/t, the power required, in kW (rounded to one decimal place), is ____.
Numerical answer — type the value.
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Answer: 36.7
E = 10 × 12 × (1/√500 − 1/√5000) = 120 × (0.044721 − 0.014142) = 120 × 0.030579 = 3.6695 kWh/t. Power = 3.6695 kWh/t × 10 t/h = 36.7 kW.A food material obeys Rittinger's law. Grinding it from 1.0 mm to 0.5 mm needs E₁ kWh per tonne, and grinding that product further from 0.5 mm to 0.25 mm needs E₂ kWh per tonne. The ratio E₂/E₁ is
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Answer: D — 2
Rittinger: E ∝ (1/x₂ − 1/x₁). E₁ ∝ 1/0.5 − 1/1.0 = 1 mm⁻¹ and E₂ ∝ 1/0.25 − 1/0.5 = 2 mm⁻¹, so E₂/E₁ = 2. The tempting answer 1 is what Kick's law gives, because both steps have the same reduction ratio of 2; Rittinger charges more for the finer step because it creates more new surface.A milk fat globule of diameter 4 µm and density 920 kg/m³ rises through skim milk of density 1035 kg/m³ and viscosity 1.5 mPa·s. Using Stokes' law with g = 9.81 m/s², the creaming velocity, in mm per hour (rounded to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 2.41
v = d²(ρ_f − ρ_p)g/(18μ) = (4 × 10⁻⁶)² × (1035 − 920) × 9.81/(18 × 1.5 × 10⁻³) = 1.6 × 10⁻¹¹ × 115 × 9.81/0.027 = 6.685 × 10⁻⁷ m/s. Multiplying by 3600 s/h and 1000 mm/m gives 2.41 mm/h, upward because fat is lighter than skim milk.A slurry is filtered at constant pressure through a medium whose resistance is negligible compared with that of the cake. The first 2 m³ of filtrate are collected in 10 min. With the same slurry and pressure, the total time from the start to collect 4 m³ of filtrate is
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Answer: C — 40 min
With R_m negligible, t = [μαc/(2A²ΔP)]V², so t ∝ V². Doubling V multiplies the time by 4: 10 × (4/2)² = 40 min from the start, of which the second 2 m³ take 30 min. 20 min assumes t ∝ V and ignores the growing cake; 28.3 min corresponds to t ∝ V^1.5 and 80 min to t ∝ V³.A liquid food is agitated by a turbine in a baffled tank in the fully turbulent range, where the power number is constant. A geometrically similar larger tank uses an impeller of twice the diameter run at half the speed, with the same liquid. The impeller power becomes
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Answer: A — 4 times the original
P = Np ρN³D⁵ with Np constant, so the ratio is (1/2)³ × 2⁵ = 32/8 = 4. The answer 2 is what the laminar relation P ∝ μN²D³ would give, (1/2)² × 2³ = 2, but the stated regime is turbulent; 16 treats power as proportional to N instead of N³, and no change at all assumes the two changes cancel.Which of the following statements about mechanical separations and agitation are correct?
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Answer: B — Homogenisation slows creaming because the Stokes velocity is proportional to the square of the globule diameter; D — A filter aid such as diatomaceous earth is used to keep a compressible cake porous
Creaming velocity ∝ d², so reducing globules from about 4 µm to 1 µm slows it about sixteen-fold. Filter aids such as diatomaceous earth build a rigid, porous layer that stops a compressible cake from sealing the medium. In a cream separator the lighter cream moves inward and leaves near the axis while the denser skim milk leaves at the periphery. Baffles are fitted to prevent swirl and vortexing, not to promote them.A negative test for which native milk enzyme is used to confirm that milk has been adequately pasteurised?
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Answer: B — Alkaline phosphatase
Alkaline phosphatase is a native milk enzyme slightly more heat-resistant than the target pathogens, so its absence after heating shows that the pasteurisation conditions were met. Pectin methylesterase is the usual index enzyme for citrus-juice pasteurisation, a tempting confusion; polyphenol oxidase browns cut fruit, and lipoxygenase is a target of vegetable blanching. None of these three is the milk test.The cold point of a can is held at 111.1 °C for 10 min; heating and cooling lags are ignored. With the reference temperature 121.1 °C and z = 10 °C, the F₀ value delivered is
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Answer: A — 1.0 min
L = 10^((111.1 − 121.1)/10) = 10⁻¹ = 0.1, so F₀ = 0.1 × 10 = 1.0 min: 10 min at 10 °C below the reference is worth 1 min at 121.1 °C. 0.1 min mistakes the lethal rate for F₀, 10 min treats the hold as if it were at the reference temperature, and 100 min multiplies by 10 instead of dividing, as if a lower temperature were more lethal.The main reason vegetables such as peas are blanched before freezing is to
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Answer: B — inactivate enzymes that would cause off-flavours and colour loss in frozen storage
Blanching is chiefly an enzyme-inactivation step: without it peroxidase, lipoxygenase and similar enzymes keep acting slowly in the freezer, causing off-flavours and loss of colour and vitamins. Its mild heat reduces microbes but is far from commercial sterility, the tempting misconception; it does not raise the freezing point, and it leaches ascorbic acid rather than adding it.A single-effect evaporator concentrates 5000 kg/h of juice from 12% to 60% total solids. The feed enters at 30 °C and the juice boils at 55 °C. Take the specific heat of the feed as 3.9 kJ kg⁻¹ K⁻¹, the latent heat of the vapour formed as 2370 kJ/kg and the latent heat released by the condensing heating steam as 2200 kJ/kg. Neglect boiling-point elevation and heat losses. The steam consumption, in kg/h (rounded to the nearest whole number), is ____.
Numerical answer — type the value.
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Answer: 4531
Solids balance: P = 5000 × 0.12/0.60 = 1000 kg/h, so V = 4000 kg/h. Heat needed = Vλ_v + Fc_p(T₁ − T_F) = 4000 × 2370 + 5000 × 3.9 × (55 − 30) = 9,480,000 + 487,500 = 9,967,500 kJ/h. S = 9,967,500/2200 = 4530.7, so 4531 kg/h (economy 0.88). Forgetting to heat the feed to its boiling point gives 4309 kg/h.Which of the following statements about the evaporation of liquid foods are correct?
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Answer: A — Boiling-point elevation reduces the temperature difference available between the heating steam and the boiling liquid; B — In forward feed, the most concentrated liquid is in the effect that operates at the lowest temperature
The solution boils above the saturation temperature of pure water at the same pressure, so BPE eats into the steam-to-liquid temperature difference. In forward feed the liquid flows with falling pressure, so it is most concentrated in the last, coolest effect. Sharing the same overall temperature drop among three effects leaves the capacity roughly that of one effect; what triples is the steam economy. A steam-jet ejector is thermal vapour recompression; mechanical recompression uses an electrically driven compressor.A food with 8 kg of dry solid per m² of drying surface is dried in a tray dryer. The constant drying rate is 2.0 kg water m⁻² h⁻¹, the critical moisture content is 1.20 and the equilibrium moisture content is 0.05 kg water per kg dry solid. In the falling-rate period the drying rate is proportional to (X − X_e). The time to dry from the critical moisture content to 0.15 kg water per kg dry solid, in hours (rounded to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 11.23
t_f = [M_s(X_c − X_e)/(A R_c)] ln[(X_c − X_e)/(X − X_e)] = (8 × 1.15/2.0) × ln(1.15/0.10) = 4.6 × ln 11.5 = 4.6 × 2.4423 = 11.23 h. Applying the constant-rate formula to this stretch, 8 × (1.20 − 0.15)/2.0 = 4.2 h, badly underestimates the time because the rate keeps falling.During the constant-rate period of hot-air drying of a wet food slab, with heat supplied only by convection from the air, the surface of the slab stays close to the
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Answer: D — wet-bulb temperature of the air
In the constant-rate period the surface behaves like free water: all the heat arriving by convection is used to evaporate water, and this balance holds the surface at the wet-bulb temperature. The surface approaches the dry-bulb temperature only in the falling-rate period, once it dries, which is the tempting confusion. At the dew point, which is lower, vapour would condense rather than evaporate, and convective air drying never raises the surface to the boiling point.Which of the following statements about drying and freezing of foods are correct?
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Answer: A — In primary freeze drying, ice is removed by sublimation at a water-vapour pressure below that of the triple point of water; C — Slow freezing forms large extracellular ice crystals and increases drip loss on thawing
Below the triple-point pressure of about 611 Pa ice cannot melt, so it sublimes directly to vapour. Slow freezing lets water migrate out of cells to grow a few large extracellular crystals that damage tissue, so more drip appears on thawing. Spray-dried droplets stay near the wet-bulb temperature while they evaporate, well below the inlet air temperature. Plank's equation considers only latent heat and ignores sensible heat, so it underestimates the freezing time.Green peas, treated as spheres of 1.0 cm diameter, are frozen in a fluidised-bed freezer with air at −35 °C and a surface heat-transfer coefficient of 100 W m⁻² K⁻¹. For the frozen peas take density 1050 kg/m³, thermal conductivity 1.2 W m⁻¹ K⁻¹, latent heat of fusion 260 kJ per kg of peas and initial freezing point −1.2 °C. Use Plank's equation t_F = [ρλ/(T_F − T_a)](Pa/h + Ra²/k) with a equal to the sphere diameter, P = 1/6 and R = 1/24. The freezing time, in minutes (rounded to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 2.71
ρλ/(T_F − T_a) = 1050 × 260,000/(−1.2 + 35) = 2.73 × 10⁸/33.8 = 8.077 × 10⁶. Pa/h = 0.01/(6 × 100) = 1.667 × 10⁻⁵ and Ra²/k = 0.01²/(24 × 1.2) = 3.47 × 10⁻⁶, sum 2.014 × 10⁻⁵. t_F = 8.077 × 10⁶ × 2.014 × 10⁻⁵ = 162.7 s = 2.71 min. Using the radius as a with these P and R gives 74 s, less than half the correct time.Ambient air at 101.325 kPa, 30 °C and 50% relative humidity is heated to 70 °C, without any change in moisture content, before it enters a dryer. Take the saturation vapour pressure of water at 30 °C as 4.246 kPa, the humidity ratio W = 0.622p_v/(P − p_v) and the moist-air enthalpy h = 1.006T + W(2501 + 1.86T) kJ per kg dry air, with T in °C. The heat required, in kJ per kg of dry air (rounded to two decimal places), is ____.
Numerical answer — type the value.
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Answer: 41.23
p_v = 0.50 × 4.246 = 2.123 kPa; W = 0.622 × 2.123/(101.325 − 2.123) = 0.01331 kg/kg dry air, unchanged by heating. Δh = (1.006 + 1.86W)(70 − 30) = (1.006 + 0.02476) × 40 = 41.23 kJ per kg dry air (h = 64.21 kJ/kg before and 105.44 kJ/kg after). Ignoring the vapour term gives 40.24 kJ/kg.Air passes through a spray chamber in which recirculated water evaporates into it adiabatically, with no external heat added or removed. Which set of changes does the air undergo?
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Answer: C — Dry-bulb temperature falls, humidity ratio rises and wet-bulb temperature stays nearly constant
With no external heat, the latent heat of evaporation comes from the air's own sensible heat: the dry-bulb temperature falls while vapour is added, so W rises, and the state moves along a line of nearly constant wet-bulb temperature and enthalpy. A constant dry-bulb temperature with rising W would need heat input, as in steam humidification; falling W and dew point describe a cooling coil below the dew point; rising temperature at constant W is sensible heating.