HW8 Solutions.pdfsm8-027.pdfHW8 SolutionsPROBLEM 8.12 KNOWN: Internal flow with prescribed wall heat flux as a function of distance. FIND: (a) Beginning with a properly defined differential control volume, the temperature distribution, Tm(x), (b) Outlet temperature, Tm,o, (c) Sketch Tm(x), and Ts(x) for fully developed and developing flow conditions, and (d) Value of uniform wall flux ′′qs (instead of ′qs = ax) providing same outlet temperature as found in part (a); sketch Tm(x) and Ts(x) for this heating condition. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Incompressible liquid with negligible viscous dissipation. PROPERTIES: Table A.6, Water (300 K): cp = 4.179 kJ/kg⋅K. ANALYSIS: (a) Applying energy conservation to the control volume above, conv p mdq mc dT=& (1) where Tm(x) is the mean temperature at any cross-section and dqconv = qdx′⋅. Hence, mpdTax mcdx=&. (2) Separating and integrating with proper limits gives ()mm,ixTxpmx0 Ta xdx mc dT==∫∫& ()2mm,ipaxTxT2mc=+& (3,4)< (b) To find the outlet temperature, let x = L, then ()2mm,om,i pTLT T aL2mc==+&. (5) Solving for Tm,o , we find ()()()22m,o20W m 30mT27C2 450kg h 3600s h 4179J kg K=+×⋅o 27 C 17.2 C 44.2 C=+ =oo o. < (c) For linear wall heating, ′=qaxs, the fluid temperature distribution along the length of the tube is quadratic as prescribed by Eq. (4). From the convection rate equation, () () ()()ssmqhxDTxTxπ′=⋅ − (6) For fully developed flow conditions, h(x) = h is a constant; hence, Ts(x) - Tm(x) increases linearly with x. For developing conditions, h(x) will decrease with increasing distance along the tube eventually achieving the fully developed value. Continued...PROBLEM 8.12 (Cont.) (d) For uniform wall heat flux heating, the overall energy balance on the tube yields ()spm,om,iqqDLmcT Tπ′′== −& Requiring that Tm,o = 44.2°C from part (a), find ()()2s450 3600 kg s 4179J kg K 44.2 27 Kq 95.3/ D W mD30mπ×⋅−′′==× < where D is the diameter (m) of the tube which, when specified, would permit determining the required heat flux, sq′′. For uniform heating, Section 8.3.2, we know that Tm(x) will be linear with distance. Ts(x) will also be linear for fully developed conditions and appear as shown below when the flow is developing. COMMENTS: (1) Note that cp should be evaluated at Tm = (27 + 44)°C/2 = 309 K. (2) Why did we show Ts(0) = Tm(0) for both types of history when the flow was developing? (3) Why must Tm(x) be linear with distance in the case of uniform wall flux heating?PROBLEM 8.25 KNOWN: Inlet temperature and flowrate of oil flowing through a tube of prescribed surface temperature and geometry. FIND: (a) Oil outlet temperature and total heat transfer rate, and (b) Effect of flowrate. SCHEMATIC: ASSUMPTIONS: (1) Negligible temperature drop across tube wall, (2) Incompressible liquid with negligible viscous dissipation. PROPERTIES: Table A.5, Engine oil (assume Tm,o = 140°C, hence mT = 80°C = 353 K): ρ = 852 kg/m3, ν = 37.5 × 10-6 m2/s, k = 138 × 10-3 W/m⋅K, Pr = 490, μ = ρ⋅ν = 0.032 kg/m⋅s, cp = 2131 J/kg⋅K. ANALYSIS: (a) For constant surface temperature the oil outlet temperature may be obtained from Eq. 8.41b. Hence ()m,o s s m,ipDLTTTTexp hmcπ⎛⎞=− − −⎜⎟⎜⎟⎝⎠& To determine h , first calculate ReD from Eq. 8.6, ()()()D40.5kgs4mRe 398D 0.05m 0.032kg m sπμ π== =⋅& Hence the flow is laminar. Moreover, from Eq. 8.23 the thermal entry length is ()()()fd,t Dx 0.05DRe Pr 0.05 0.05m 398 490 486m≈= =. Since L = 25 m the flow is far from being thermally fully developed. Since Pr > 5, h may be determined from Eq. 8.57 DD2/3D0.0668GzNu 3.6610.04Gz=++. With GzD = (D/L)ReDPr = (0.05/25)398 × 490 = 390, it follows that D26Nu 3.66 11.95.12.14=+ =+ Hence, 2Dk 0.138W m Kh Nu 11.95 33W m KD0.05m⋅== = ⋅ and it follows that Continued...PROBLEM 8.25 (Cont.) ()()()2m,o0.05m 25mT 150 C 150 C 20 C exp 33W m K0.5kg s 2131J kg Kπ⎡⎤=− − − × ⋅⎢⎥×⋅⎣⎦ooo Tm,o = 35°C. < From the overall energy balance, Eq. 8.34, it follows that ()()pm,o m,iq mc T T 0.5kg s 2131J kg K 35 20 C=−=×⋅×−o& q = 15,980 W. < The value of Tm,o has been grossly overestimated in evaluating the properties. The properties should be re-evaluated at T = (20 + 35)/2 = 27°C and the calculations repeated. Iteration should continue until satisfactory convergence is achieved between the calculated and assumed values of Tm,o. Following such a procedure, one would obtain Tm,o = 36.4°C, ReD = 27.8, h = 32.8 W/m2⋅K, and q = 15,660 W. The small effect of reevaluating the properties is attributed to the compensating effects on ReD (a large decrease) and Pr (a large increase). (b) The effect of flowrate on Tm,o and q was determined by using the appropriate IHT Correlations and Properties Toolpads. 0.5 1 1.5 2Mass flowrate, mdot(kg/s)202428323640Outlet temperature, Tmo(C) 0.5 1 1.5 2Mass flowrate, mdot(kg/s)15000200002500030000Heat rate, q(W) The heat rate increases with increasing m& due to the corresponding increase in ReD and hence h. However, the increase is not proportional to &m, causing ()m,o m,i pTT qmc−=&, and hence Tm,o, to decrease with increasing &m. The maximum heat rate corresponds to the maximum flowrate (m& = 0.20 kg/s). COMMENTS: Note that significant error would be introduced by assuming fully developed thermal conditions and DNu = 3.66. The flow remains well within the laminar region over the entire range of m&.PROBLEM 8.27 KNOWN: Inlet and outlet temperatures and velocity of fluid flow in tube. Tube diameter and length. FIND: Surface heat flux and temperatures at x = 0.5 and 10 m. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Negligible heat loss to surroundings, (4) Incompressible liquid with negligible viscous dissipation, (5) Negligible axial conduction. PROPERTIES: Pharmaceutical (given): ρ = 1000 kg/m3, cp = 4000 J/kg⋅K, μ = 2 × 10-3 kg/s⋅m, k = 0.80 W/m⋅K, Pr = 10. ANALYSIS: With ()( )23m VA 1000 kg/m 0.2 m/s 0.0127 m / 4 0.0253 kg/sρπ== =& Eq. 8.34 yields ()()pm,o m,iq m c T T 0.0253 kg/s 4000 J/kg K 50 K 5060 W.=−= ⋅=& The required heat flux is then ()2ssq q/A 5060 W/ 0.0127 m 10 m 12,682 W/m .π′′== = < With ()3-3DRe VD/ 1000 kg/m 0.2 m/s 0.0127 m/2 10 kg/s m 1270ρμ== × ⋅= the flow is laminar and