Full exam for Heat and Mass Transfer Scambio Termico e di Massa in the Energy Engineering degree programme at Politecnico di Milano. The document covers: Politecnico di Milano – A.Y. 2025–2026 Master Programme in Energy/Nuclear Engineering Heat and Mass Transfer Examination – 12 January 2026 Surname: Name: Grade 1 -. . . . . . . . . /10 Student ID number: 2 -. . . . . . . . . /10 3 -. . . . . . . . . /10 Total . . . . . . . . .
Full exam for Heat and Mass Transfer Scambio Termico e di Massa in the Energy Engineering degree programme at Politecnico di Milano. The document covers: Politecnico di Milano – A.Y. 2025–2026 Master Programme in Energy/Nuclear Engineering Heat and Mass Transfer Examination – 12 January 2026 Surname: Name: Grade 1 -. . . . . . . . . /10 Student ID number: 2 -. . . . . . . . . /10 3 -. . . . . . . . . /10 Total . . . . . . . . .
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Politecnico di Milano – A.Y. 2025–2026 Master Programme in Energy/Nuclear Engineering Heat and Mass Transfer Examination – 12 January 2026 Surname: Name: Grade 1 -. . . . . . . . . /10 Student ID number: 2 -. . . . . . . . . /10 3 -. . . . . . . . . /10 Total . . . . . . . . . /30 The time available is 2.5 hours. Only the material provided as Supplementary Material may be used. No textbooks, notes, or electronic devices are allowed. The maximum score is 30/30. At the end of the exam, this text must be submitted with the numerical results written next to each question, and only one protocol sheet (four written pages) with the detailed solution must be handed in. Exercise 1 (10 points) An isothermal (Tc = 85 ◦ C) silicon chip of width W = 20 mm on a side is soldered to an aluminum heat sink (λ = 180 W/m K) of equivalent width. The heat sink has a base thickness of Lb = 3 mm and an array of 11 rectangular fins, each of length Lf = 15 mm. An air flow at T∞ = 25 ◦ C and with a velocity of u∞ = 8.9 m/s is maintained through the channels formed by the fins and a cover plate. The fin spacing is δ = 1.8 mm. The solder joint has a thermal resistance Rt,c ′′ = 6 · 10−5 m2 K/W. An adiabatic fin tip condition may be assumed, and airflow along the outer surfaces of the heat sink may be assumed to provide a convection coefficient equivalent to that associated with airflow through the channels. The air properties should be evaluated at T∞ . Find: 1. the mean convective heat transfer coefficient, assuming an almost uniform fin temperature 2. the power dissipated to the air 3. the temperature at the base of a fin and the effectiveness of a single fin Accounting for the increase in air temperature as it flows through the heat sink, find: 4. the outlet temperature of the air exiting each…
this was the easiest question of this part but try to understand and put the 2 sides of the surface into calculation and use and understand the question very well step by step and know to use which Temp for the each source. Exercise 2 (10 points) A laser–materials–processing apparatus encloses a sample in the form of a circular disk of diameter D = 25 mm and thickness w = 1 mm. The sample surface is diffuse, and the spectral distribution of the emissivity, ελ (λ), is prescribed. To reduce oxidation, an inert gas stream flows over both the upper and lower surfaces of the sample. The gas is at a temperature T∞ = 500 K and provides a uniform convection coefficient h = 50 W/m2 K. The apparatus enclosure is large compared to the sample with a wall temperature Tenc = 300 K. To maintain the sample at a steady operating temperature, a collimated laser beam irradiates the upper surface of the sample. The laser operates at a wavelength λ = 0.5 µm. The total emissivity of the sample is ε = 0.643. Find: 1. the temperature Ts of the sample 2. the total absorptivity of the sample for irradiation originating from the enclosure walls 3. the laser irradiation required to maintain the sample at Ts Consider a cool–down process in which both the laser irradiation and the inert gas flow are deactivated, and the sample cools down by heat exchange with the surrounding enclosure until an equilibrium is reached. Find: 4. the emissivity of the sample in this final state Exercise 3 (10 points) Consider a cylindrical cup of diameter D = 9 cm and height H = 7 cm filled with coffee, whose thermophysical properties may be assumed equal to those of liquid water. The cup is placed on a heated base such that the free surface of the coffee is maintained at a uniform temperature of Ts = 60 ◦ C. The cup…
Solution Exercise 1 1. The thickness of the fins is evaluated as: W − (N − 1)δ t= = 0.18 mm N The Reynolds number is calculated using the hydraulic diameter of a single channel, while the air properties are evaluated at T∞ : 4Ac 4 Lf δ u∞ Dh Dh = = = 3.214 mm, ReD = = 1800, ν∞ = 15.89 · 10−6 m2 /s P 2(Lf + δ) ν∞ The flow is laminar and it simultaneously develops thermally and hydrodynamically. The convective heat transfer coefficient is thus derived from the combined entry length correlation by assuming a uniform surface temperature of the fin (as specified in the text): P r = 0.707, GzD = (D/W )ReD P r = 204.6 h i−1 −1/3 −2/3 −1 3.66 tanh(2.264 GzD + 1.7 GzD ) + 0.0499 GzD tanh(GzD ) N uD = −1/6 = 12.23 tanh(2.432 Pr1/6 GzD ) N uD λ∞ h= = 100 W/m2 K , λ∞ = 26.3 · 10−3 W/m K Dh 2. The contact resistance and the conduction resistance through the base are: ′′ Lb Rt,c = Rt,c /W 2 = 0.15 K/W, Rλ,b = = 4.17 · 10−2 K/W, λW 2 The overall thermal resistance is calculated as: Af = 2W Lf = 6 · 10−4 m2 , Ab = W 2 − N (t W ) = 3.6 · 10−4 m2 , At = N Af + Ab = 6.96 · 10−3 m2 p tanh mLf N Af m = 2h/λt = 78.2 m−1 , ηf = = 0.703, ηo = 1 − (1 − ηf ) = 0.719 mLf At 1 Rt,o = = 2.0 K/W ηo hAt Thus, the heat dissipated to the air is: Tc − T∞ Q̇ = = 27.4 W Rt,c + Rλ,b + Rt,o 3. The temperature at the base of the fin is: Tb = Tc − Q̇(Rt,c + Rλ,b ) = 79.74 ◦ C The effectiveness of a single fin is: Q̇f √ εf = , Q̇f = M tanh mLf , M = θb 2hW 2 λt, Ab,f = W t hAb,f θb r 2λ εf = tanh mLf = 116 ht
4. The heat sink can be treated as a heat exchanger with a fluid that varies its temperature and the other that remains at constant temperature Tc . Accounting for the air flow within the heat sink only, the value of the minimum heat capacity rate is: Cmin = ρ∞ u∞ cp,∞ Lf δ(N − 1) = 2.81 W/K, ρ∞ = 1.1614 kg/m3 , cp,∞ = 1007 J/kg K The NTU and the effectiveness are, respectively: UA 1 NTU = = = 0.162, ε = 1 − exp(−N T U ) = 0.150 Cmin (Rt,c + Rλ,b + Rt,o )Cmin The resulting heat rate, Q̇1 , and outlet air temperature, T∞,o are (the inlet air temperature is T∞,i = 25 ◦ C): Q̇1 Q̇1 = εQ̇max = ε Cmin (Tc − T∞,i ) = 25.3 W, T∞,o = T∞,i + = 34 ◦ C Cmin Exercise 2 1. From the emissivity value, the temperature of the sample is determined as follows: ε − ε2 ε = ε1 F0→λ1 Ts + ε2 (1 − F0→λ1 Ts ) ⇒ F0→λ1 Ts = = 0.738 ε1 − ε2 From the table: λ1 Ts = 6000 µm K ⇒ Ts = 2000 K 2. The sample surface is opaque and diffuse so that ελ = αλ : α = ε1 F0→λ1 Tenc + ε2 (1 − F0→λ1 Tenc ) , F0→λ1 Tenc ≈ 0, ⇒ α = 0.2 3. The steady–state energy balance on the sample per unit surface area is: αlas Glas + 2αGenc − 2E − 2qh′′ = 0 4 αlas Glas + 2ασTenc − 2εσTs4 − 2h(Ts − T∞ ) = 0 where the “2” accounts for the upper and lower surface areas. Recognizing that αlas (0.5 µm) = 0.8: Glas = 1646 kW/m2 4. In the limit t → ∞, the sample reaches equilibrium with the enclosure, Ts (∞) = Tenc : ε = α = 0.2
Exercise 3 1. Thermophysical properties of coffee (A) and air (B) at Ts , T∞ and Tf = (Ts + T∞ )/2 = 43.5 ◦ C are: MA = 18 kg/kmol, MB = 28.97 kg/kmol, R = 8314.4 J/kmol K pA,s = pA,sat (Ts ) = 19.878 kPa, pB,s = p − pA,s = 81.447 kPa pA,s MA pB,s MB ρA,s = = 0.129 kg/m3 , ρB,s = = 0.852 kg/m3 RTs RTs pA,∞ = ϕ · pA,sat (T∞ ) = 0.5 · 3.531 kPa = 1.7655 kPa, pB,∞ = p − pA,∞ = 99.670 kPa pA,∞ MA pB,∞ MB ρA,∞ = = 0.0127 kg/m3 , ρB,∞ = = 1.156 kg/m3 RT∞ RT∞ νB,f = 1.7565 · 10−5 m2 /s, λB,f = 0.0275 W/m K The binary diffusion coefficient of coffee (water) in air is evaluated at Tf : 3/2 Tf DAB = DAB (298 K) = 2.848 · 10−5 m2 /s 298 K The surface, free stream and mean boundary layer densities of the binary mixture AB are: ρs = ρA,s + ρB,s = 0.981 kg/m3 , ρ∞ = ρA,∞ + ρB,∞ = 1.168 kg/m3 , ρ = (ρs + ρ∞ )/2 = 1.075 kg/m3 The Rayleigh number is calculated using the characteristic length L = As /P = D/4 and the Prandtl number at Tf : g(ρ∞ − ρs )L3 GrL = 2 = 63177, P r = 0.705, ⇒ RaL = GrL · P r = 44519 ρνB,f The Nusselt number and the convective heat transfer coefficient are: 1/4 N uL λB,f N uL = 0.54 RaL = 7.843, h= = 9.6 W/m2 K L 2. The analogy between heat and mass transfer is used to evaluate hm : νB,f 1/4 ShL DAB Sc = = 0.617, Ram,L = GrL ·Sc = 38966, ShL = 0.54 Ram,L = 7.587, hm = = 9.6 · 10−3 m/s DAB L 3. The height of the coffee in the cup decreases due to evaporation. The coffee evaporation rate is: nA = hm (πD2 /4)(ρA,s − ρA,∞ ) = 0.0256 kg/h The volume of coffee evaporated in 3 hours is thus: VA,evap = 3 nA /1000 = 7.68 · 10−5 m3 The height of the coffee in the cup after 3 hours is thus: VA,evap HA = H − = 5.79 cm (πD2 /4) 4. The heat rate supplied from the base must be equal to the sum of the heat losses due to convection, radiation, and evaporation. Q̇h = (πD2…
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