Energy Engineering - Heat and Mass Transfer

Review exercises Black Body

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PROBLEM 1 Some bars (diameter D=40 mm, length L=1000 mm, initial temperature T i=550°C, density =4000kg/m 3, thermal conductivity k=46W/mK, specific heat capacity c=765J/kgK) undergo a 3 steps cooling process. Step 1: the bars are impinged normally to the axis by a water flow (velocity U w=1.2m/s, free stream temperature T w=20°C, dynamic viscosity =0.001kg/ms, thermal conductivity k=0.6W/mK) for a time interval t=12s. Step 2: the bars are removed from water and are thermally insulated till the thermal equilibrium is reached. Step 3: the thermal insulation is removed and the bars are impinged normally to the axis by an air flow (free stream temperature T a=22°C, heat transfer coefficient h a=60W/m 2K, density  a=1.2kg/m 3, dynamic viscosity  a=1.9·10 -5kg/ms, thermal conductivity k a=0.27W/mK). Assuming that the radiative heat transfer be negligible, compute 1) the maximum and the minimum temperature in the bars at the end of step 1; 2) the equilibrium temperature at the end of step 2; 3) the time required to get the bars at temperature T f=30°C; 4) the air velocity; Correlation list for the problem Nu=2+0.589Ra 0.25 [1+(0.469/Pr) 9/16]-4/9 Nu=C Re mPr 1/3 Nu=0.664Re 0.5 Pr 1/3 C m Nu=0.3+0.62Re 0.5 Pr 1/3[1+(Re/282000) 5/8]4/5/[1+(0.4/Pr) 2/3]1/4 Re [4·10 1; 4·10 3) 0.683 0.466 Nu=(f/8)(Re-1000)Pr/[1+12.7(f/8) 0.5(Pr 2/3-1)] Re [4·10 3; 4·10 4) 0.193 0.618 f=(0.79lnRe-1.64) -2 Re [ 4·10 4;+ ) 0.027 0.805 1-term approximation for cylinder (radius R) Fo=t/R 2 =[T(r,t)-T ]/[T i-T ]=C·exp(- 2Fo)J 0(y) D[r nJn(r)]/dr= r nJn-1 (r) y=r/R  m=[T i-T m(t)]/[T i-T ]=1-2·C·exp(- 2Fo)·J 1()/ mean value  Coefficients used in the one-term approximation Bi 0.2 0.4 0.6 0.8 1 2 4 6 Plane Wall 0.433 0.5933 0.7052 0.7908 0.8598 1.0776 1.2648 1.3491 C 1.0311 1.058 1.0814 1.1015 1.119 1.1787 1.2288 1.2478 Cylinder 0.6171 0.8514 1.0187 1.1493 1.2561 1.6003 1.908 2.0496 C 1.0483 1.0931 1.1346 1.1725 1.2072 1.3387 1.4697 1.5255 Sphere 0.7589 1.0531 1.2648 1.432 1.5723 2.0295 2.4559 2.6523 C 1.0591 1.1164 1.1714 1.2236 1.2738 1.4797 1.7203 1.833 Bessel J 0(x) table x 0.00 0.15 0.30 0.45 0.60 0.75 0.90 1.05 1.20 J0(x) 1.0000 0.9944 0.9776 0.9500 0.9120 0.8642 0.8075 0.7428 0.6711 x 1.35 1.50 1.65 1.80 1.95 2.10 2.25 2.40 2.55 J0(x) 0.5937 0.5118 0.4268 0.3400 0.2528 0.1666 0.0827 0.0025 -0.0729 Bessel J 1(x) table x 0.00 0.15 0.30 0.45 0.60 0.75 0.90 1.05 1.20 J1(x) 0.0000 0.0748 0.1483 0.2194 0.2867 0.3492 0.4059 0.4559 0.4983 x 1.35 1.50 1.65 1.80 1.95 2.10 2.25 2.40 2.55 J1(x) 0.5325 0.5579 0.5743 0.5815 0.5794 0.5683 0.5484 0.5202 0.4843 PROBLEM 2 An air flow (heat flux q s = 3000 W/m 2) cools the outer surface of the squared sight glass of a kiln (thermal conductivity k = 1 W/mK, cross- section area S = 0.01 m 2, thickness s = 6 cm, temperature of the inner surface T 0 = 600 K). The heating effect of radiation absorption, which is small but not negligible, can be dealt as a uniform power generation in glass (power generation per unit volume g = 500 W/m 3). Assuming one dimensional steady state conduction, compute: 1. the thermal power exchanged at the inner surface (please specify the sign convention); 2. the maximum temperature in the glass; 3. the surface temperature if the thermal conductivity is a function of the temperature: k(T) = A·T+B, (A = 2·10 -3 W/mK 2, B = -0.2 W/mK); 4. the temperature difference of the outer surface between the case of constant thermal conductivity and variable thermal conductivity. PROBLEM 3 Avacuum oven (wall temperature T w = 500°C) is endowed with a control device (maximum allowable temperature T max = 100°C) inserted in cooper case (inner diameter D i = 30 mm, outer diameter D o = 60 mm, thermal conductivity k = 400 W/mK, base temperature T 0 = 40°C), its surface has a special coating to reduce radiation heat transfer (opaque and diffuse surface, emissivity  s = 0.2). Assuming:  negligible emissive power of the case,  the tip and the inner surface of the case be adiabatic,  the oven is much larger than the case, compute: 1. the maximum length of the case not to exceed the temperature limit of the control device; 2. the power exchanged by the case with the air outside the oven; 3. if it is reasonable to neglect the emissive power of the case; 4. the total absorptivity of the surface if the spectral emissivity is �  L P0.4 � ∈> 0; 0.1; �I 0.1 � ∈> 0.1; 100; �I 0.2 � ∈> 100; E∞; �I xL 0 Do Di Tw VACUUM OVEN 0sxT T 0 g , k q s outer surface inner surface S S D , c, k t=0 ; T=T i g, k g, Pr g vg ; T g=T gi+bt t=0 ; T g=T gi t=t f ; T g=T gf PROBLEM 4 The superheater of a steam power plant is made of steel tubes (length L=25m, outer diameter D o=80mm,thickness s=5mm, emissivity:  t=0.7).Each tube is impinged, normally to its axis, by a gas flow (velocity v g=30m/s, temperature T g=700°C), while vapor flows inside (mass flow ratem g=1200kg/h, inlet temperature T v0=300°C). Assuming steady state operating conditions, compute: 1. the vapor outlet temperature; 2. the thermal power exchanged; 3. the inner wall temperature at the vapor inlet; 4. if the vapor inlet temperature is the reading of a spherical thermocouple (diameter D tc=6mm, emissivity  tc=1, T tc=T v0) which “sees” only the inlet of the tube, compute the real vapor inlet temperature T v0n , the real inner wall temperature at the inlet T w0n . Correlation list Nu=C Re m Pr 0.33 C m Nu=0.023Re 0.8 Pr 0.4 Re (40 ; 4000]0.683 0.4663 Nu=2+(0.44 Re 0.5 +0.066Re 2/3)Pr 0.4 Re (4000 ; 40000]0.193 0.6183 Nu=(0.037Re 0.8 -871)Pr 1/3 Re > 40000 0.027 0.805 Thermal properties vapor gas density [kg/m 3] 17 0.218 Specificheatcapacity [J/kgK] 2820 1224 Thermalconductivity [W/mK] 5·10 -2 6.5·10 -2 dynamic viscosity [kg/ms] 2.0·10 -5 4.0·10 -5 Prandtl number [-] 1.12 0.75 PROBLEM 5 A technological process requires a gas stream which temperature rise in time (free stream velocity: v g=10ms -1;T g=T gi+b·t; initial temperature: T gi=400°C, final temperature: T gf=500°C temperature change rate: b=0.05 Ks -1 thermal conductivity: k g=3·10 -2Wm -1K-1; kinematic viscosity:  g=4·10 -5m2s-1; Prandtl number: Pr g=0.70). A spherical thermocouple, introduced in the flow as the temperature ramp begins, is used to monitor the process (diameter: D=20mm; thermal conductivity: k=20Wm -1K-1; density: =8000kgm -3; specific heat: c=400Jkg -1K-1; emissivity: =0.1; initial temperature: T i=350°C). Assuming negligible the heat transfer by radiation: 1. the heat transfer coefficient on the surface of the thermocouple; 2. the reading of the thermocouple when the gas stream reaches T g=T gf; 3. the thermocouple time lag; 4. assuming the thermocouple much smaller than the surrounding (T w=400°C), check if the assumption about the heat transfer by radiation is correct. PROBLEM 6 A domestic geothermal heating system is made of a close loop in which a borehole heat exchanger (length L=1000m, annulus: inner diameter D i=6cm, outer diameter D i=7cm, inner tube thickness s=1cm, thermally insulated; negligible thermal resistance) exploits the geothermal gradient (linear temperature increase with the depth, surface temperature T g0=20°C, temperature gradient G=0.05K/m, T g=T g0+Gz, ground resistance per unit area r=0.05Km 2/W) to heat up liquid water (mass flow rate m=0.5kg/s, inlet temperature T i=20°C, downward motion in the annulus, thermal conductivity: k=0.6W/mK, dynamic viscosity: = 8.10- 4kg/ms). Assuming steady state, constant cross sectional areas and negligible entry length, compute: 1. the water outlet temperature; 2. thefrictional pressure drop between inlet and outlet of the borehole heat exchanger. PROBLEM 7 To fix the paint on one side of a metal disk (diameter: D 2=0.20m, temperature: T 2=127°C, flat, opaque, diffuse, gray, emissivity: 2 = 0.70), the disk is placed like a cap on top of a cylindrical oven (diameter: D 4=0.20m, length: H=0.030m, temperature: T 4=177°C, opaque, diffuse, gray, emissivity: 4=0.50)with the paint facing inside. In the oven base there is an electric heater (diameter: D 1=0.1m, temperature: T 1=427°C, flat, black surface) surrounded by a protective case (diameter: D 3=0.2m, flat, adiabatic surface). Assuming negligible the heat transfer by convection in the oven, compute: 1. the power required by the heater; 2. the case temperature. x=r 1/h ; y=r 2/h z=1+(1+y 2)/x 2 F12=0.5{z-[z 2-4(y/x) 2]0.5} Equation to compute the view factor HeaterCaseDisk R 1 R2 rDisk Heater Case Case vacuum Perspective view Side view A1 r1 r2 A2 h Do Di L s PROBLEM 8 In a test rig, designed to perform experiments involving refrigerants during phase change inside microfin tubes, the measure of the refrigerant heat transfer coefficient takes place takes place in a tube in tube counter flow heat exchanger, which is equipped with 4 thermocouples to record the main temperatures of the fluids. The refrigerant (inlet temperature: T r0 = 5.0°C, outlet temperature: T rL = 4.0°C) flows in the inner tube (inner diameter: D ti = 9.00 mm, outer diameter: D to = 10.00 mm, heat transfer length: L = 2.00 m, tube thermal conductivity: k t = 400.0 W·m −1·K −1) while in the annulus (outer tube: inner diameter: D ai = 14.00 mm, outer surface thermally insulated) there is water (inlet temperature: T wL = 10.0°C, outlet temperature: T w0 = 8.0°C, mass flow rate: m w = 70.00·10 −3 kg·s −1, thermal conductivity: k w = 0.600 W·m −1·K −1, dynamic viscosity:  w = 1.08·10 −3kg·m −1·s−1, Prandtl number: P rw = 7.53). Assuming steady state operating conditions, fully developed velocity and temperature profiles of the water flow: 1. compute the thermal power transferred between water and refrigerant; 2. compute the pressure drop in the annulus; 3. compute the pumping power required to flow the water in the annulus; 4. compute the refrigerant side heat transfer coefficient (related to the internal area of the tube); 5. check if the entry length is negligible (percentage of the heat transfer length lower than: Ld%max = 15%); 6. check if the conductive resistance of the tube is negligible (percentage contribution to the total resistance lower than R c%max = 5%). Blackbody radiation function T [m·K] 0 1000 2000 3000 4000 5000 6000 7000 F0 [-] 0.0000 0.0003 0.0667 0.2731 0.4807 0.6335 0.7376 0.8079 T [m·K] 8000 9000 10000 12000 14000 16000 18000 20000 F0 [-] 0.8561 0.8898 0.9140 0.9449 0.9627 0.9736 0.9806 0.9854 T [m·K] 25000 30000 35000 40000 45000 50000 55000 60000 F0 [-] 0.9920 0.9951 0.9968 0.9978 0.9984 0.9987 0.9990 0.9992 T [m·K] 65000 70000 75000 80000 85000 90000 95000 100000 F0 [-] 0.9993 0.9994 0.9995 0.9996 0.9996 0.9996 0.9997 0.9997 Coefficients used in the one-term approximation Plane Wall Cylinder Sphere Bi=h·L/k Bi=h·R/k Bi=h·R/k Bi C C C 0.02 0.1410 1.0033 0.1994 1.0050 0.2445 1.0060 0.04 0.1988 1.0066 0.2815 1.0099 0.3450 1.0120 0.06 0.2426 1.0098 0.3439 1.0149 0.4215 1.0179 0.08 0.2790 1.0130 0.3958 1.0197 0.4859 1.0239 0.10 0.3112 1.0161 0.4416 1.0246 0.5426 1.0298 0.20 0.4330 1.0311 0.6171 1.0483 0.7589 1.0591 0.40 0.5933 1.0580 0.8514 1.0931 1.0531 1.1164 0.60 0.7052 1.0814 1.0187 1.1346 1.2648 1.1714 0.80 0.7908 1.1015 1.1493 1.1725 1.4320 1.2236 1.00 0.8598 1.1190 1.2561 1.2072 1.5723 1.2738 2.00 1.0776 1.1787 1.6003 1.3387 2.0295 1.4797 4.00 1.2648 1.2288 1.9080 1.4697 2.4559 1.7203 6.00 1.3491 1.2478 2.0496 1.5255 2.6523 1.8330 8.00 1.3982 1.2571 2.1277 1.5523 2.7658 1.8922 10.00 1.4289 1.2620 2.1790 1.5675 2.8363 1.9249 20.00 1.4964 1.2699 2.2888 1.5921 2.9867 1.9784 40.00 1.5332 1.2724 2.3450 1.5992 3.0634 1.9942 60.00 1.5447 1.2728 2.3645 1.6007 3.0894 1.9974 80.00 1.5509 1.2730 2.3743 1.6012 3.1017 1.9984 100.00 1.5539 1.2731 2.3816 1.6015 3.1078 1.9989 + 1.5707 1.2733 2.4050 1.6018 3.1415 2.0000