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University study material for Heat and Mass Transfer Scambio Termico e di Massa in the Energy Engineering degree programme at Politecnico di Milano. The document covers: PROBLEM 1 Some bars (diameter D=40 mm, length L=1000 mm, initial temperature T i=550°C, density =4000kg/m3, thermal conductivity k=46W/mK, speci fic heat capacity c=765J /kgK) undergo a 3 step s cooling process. Step 1 : the bars are impinged normally to the axis by a water

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University study material for Heat and Mass Transfer Scambio Termico e di Massa in the Energy Engineering degree programme at Politecnico di Milano. The document covers: PROBLEM 1 Some bars (diameter D=40 mm, length L=1000 mm, initial temperature T i=550°C, density =4000kg/m3, thermal conductivity k=46W/mK, speci fic heat capacity c=765J /kgK) undergo a 3 step s cooling process. Step 1 : the bars are impinged normally to the axis by a water

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PROBLEM 1 Some bars (diameter D=40 mm, length L=1000 mm, initial temperature T i=550°C, density =4000kg/m3, thermal conductivity k=46W/mK, speci fic heat capacity c=765J /kgK) undergo a 3 step s 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 conductiv ity 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/m2K, density a=1.2kg/m3, dynamic viscosity a=1.9·10-5kg/ms, thermal conductivity ka=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 Tf=30°C; 4) the air velocity; Correlation list for the problem Nu=2+0.589Ra0.25[1+(0.469/Pr)9/16]-4/9 Nu=C Re mPr1/3 Nu=0.664Re0.5Pr1/3 C m Nu=0.3+0.62Re0.5Pr1/3[1+(Re/282000)5/8]4/5/[1+(0.4/Pr)2/3]1/4 Re[4·101; 4·103) 0.683 0.466 Nu=(f/8)(Re-1000)Pr/[1+12.7(f/8)0.5(Pr2/3-1)] Re[4·103; 4·104) 0.193 0.618 f=(0.79lnRe-1.64)-2 Re[ 4·104;+) 0.027 0.805 1-term approximation for cylinder (radius R) Fo=t/R2 =[T(r,t)-T]/[Ti-T]=C·exp(-2Fo)J0(y) D[rnJn(r)]/dr= rnJn-1(r) y=r/R m=[Ti-Tm(t)]/[Ti-T]=1-2·C·exp(-2Fo)·J1()/ 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…

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