Thermal Considerations in a Pipe FlowThermal ConditionsEnergy TransferEnergy BalanceExample (cont.)Convection CorrelationsEmpirical CorrelationsSlide 8Slide 9Temperature DistributionTemperature variation for constant heat fluxThermal Considerations in a Pipe Flow• Thermal conditions Laminar or turbulent entrance flow and fully developed thermal conditionThermal entrance region, xfd,tFor laminar flows the thermal entrance length is a function of the Reynolds number and the Prandtle number: xfd,t/D  0.05ReDPr, where the Prandtl number is defined as Pr = / and  is the thermal diffusitivity.For turbulent flow, xfd,t  10D.Thermal Conditions• For a fully developed pipe flow, the convection coefficient is a constant and is not varied along the pipe length. (as long as all thermal and flow properties are constant also.)xh(x)xfd,tconstant• Newton’s law of cooling: q”S = hA(TS-Tm)Question: since the temperature inside a pipe flow is not constant, what temperature we should use. A mean temperature Tm is defined.Energy TransferConsider the total thermal energy carried by the fluid as(mass flux) (internal energy)vAVC TdANow image this same amount of energy is carried by a body of fluid with the same mass flow rate but at a uniform mean temperature Tm. Therefore Tm can be defined asvAmvVC TdATmC&Consider Tm as the reference temperature of the fluid so that the total heat transfer between the pipe and the fluid is governed by the Newton’s cooling law as: qs”=h(Ts-Tm), where h is the local convection coefficient, and Ts is the local surface temperature. Note: usually Tm is not a constant and it varies along the pipe depending on the condition of the heat transfer.Energy BalanceExample: We would like to design a solar water heater that can heat up the water temperature from 20° C to 50° C at a water flow rate of 0.15 kg/s. The water is flowing through a 5 cm diameter pipe and is receiving a net solar radiation flux of 200 W per unit length (meter). Determine the total pipe length required to achieve the goal.Example (cont.)Questions: (1) How do we determine the heat transfer coefficient, h?There are a total of six parameters involving in this problem:h, V, D, , kf, cp. The last two variables are thermal conductivity and the specific heat of the water. The temperature dependence is implicit and is only through the variation of thermal properties. Density  is included in the kinematic viscosity, . According to the Buckingham theorem, it is possible for us to reduce the number of parameters by three. Therefore, the convection coefficient relationship can be reduced to a function of only three variables:Nu=hD/kf, Nusselt number, Re=VD/, Reynolds number, and Pr=, Prandtle number. This conclusion is consistent with empirical observation, that is Nu=f(Re, Pr). If we can determine the Reynolds and the Prandtle numbers, we can find the Nusselt number, hence, the heat transfer coefficient, h.Convection CorrelationsD sD sLaminar, fully developed circular pipe flow: Nu 4.36, when q " constant, (page 543, ch. 10-6, ITHT) Nu 3.66, when T constant, (page 543, ch. 10-6, ITHT)Note: tfhDk   mhe therma conductivity should be calculated at T .Fully developed, turbulent pipe flow: Nu f(Re, Pr),Nu can be related to Re & Pr experimentally, as shown. ln(Nu)ln(Re)slope mFixed Prln(Nu)ln(Pr)slope nFixed ReEmpirical Correlations4 / 5Ds m s mDDittus-Boelter equation: Nu 0.023Re Pr , (eq 10-76, p 546, ITHT)where n 0.4 for heating (T T ), n 0.3 for cooling (T T ).The range of validity: 0.7 Pr 160, Re 10, 000, / 10.nL D      Note: This equation can be used only for moderate temperature difference with all the properties evaluated at Tm.Other more accurate correlation equations can be found in other references. Caution: The ranges of application for these correlations can be quite different.D1/ 2 2 / 3DFor example, the Gnielinski correlation is the most accurate among all these equations:( / 8)(Re 1000) PrNu (from other reference)1 12.7( / 8) (Pr 1)It is valid for 0.5 Pr 2000 and 3000 Re 5 10Dff     6m.All properties are calculated at T .Example (cont.)In our example, we need to first calculate the Reynolds number: water at 35°C, Cp=4.18(kJ/kg.K), =7x10-4 (N.s/m2), kf=0.626 (W/m.K), Pr=4.8.4D1/ 2 2 / 34 4(0.15)Re 5460(0.05)(7 10 )Re 4000, it is turbulent pipe flow.Use the Gnielinski correlation, from the Moody chart, f 0.036, Pr 4.8( / 8)(Re 1000) Pr(0.Nu1 12.7( / 8) (Pr 1)DmDVD mADff           &&1/ 2 2 / 32036 / 8)(5460 1000)(4.8)37.41 12.7(0.036 / 8) (4.8 1)0.626(37.4) 469( / . )0.05fDkh Nu W m KD   Energy BalanceQuestion (2): How can we determine the required pipe length?Use energy balance concept: (energy storage) = (energy in) minus (energy out). energy in = energy received during a steady state operation (assume no loss)'( ) ( ),( )(0.15)(4180)(50 20)94( )' 200P out inP in outq L mC T TmC T TL mq   &&q’=q/LTinToutTemperature Distributions ssFrom local Newton's cooling law:q hA(T ) ' ( )(T ( ) ( ))' 200( ) ( ) 20 0.319 22.7 0.319 ( )(0.05)(469)At the end of the pipe, T ( 94) 52.7( )m ms mT q x h D x x T xqT x T x x x CDhx C                Question (3): Can we determine the water temperature variation along the pipe?Recognize the fact that the energy balance equation is valid for any pipe length x:'( ) ( ( ) )' 200( ) 20 20 0.319(0.15)(4180)It is a linear distribution along the pipeP ininPq x mC T x TqT x T x x xmC      &&Question (4): How about the surface temperature distribution?Temperature variation for constant heat fluxTmx( )Tsx( )x0 20 40 60 80 1002030405060Note: These distributions are valid only in the fully developed region. In the entrance region, the convection condition should be different. In general, the entrance length x/D10 for a turbulent pipe flow and is usually negligible as compared to the total pipe length.Constant temperaturedifference due to the constant heat