Hagen 17971839 and j poiseuille 17991869 on the

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Unformatted text preview: t for a specified flow rate, the pressure drop and thus the required pumping power is proportional to the length of the tube and the viscosity of the fluid, but it is inversely proportional to the fourth power of the radius (or diameter) of the tube. Therefore, the pumping power requirement for a piping system can be reduced by a factor of 16 by doubling the tube diameter (Fig. 8–19). Of course the benefits of the reduction in the energy costs must be weighed against the increased cost of construction due to using a larger diameter tube. The pressure drop is caused by viscosity, and it is directly related to the wall shear stress. For the ideal inviscid flow, the pressure drop is zero since there are no viscous effects. Again, Eq. 8–47 is valid for both laminar and turbulent flows in circular and noncircular tubes. ⋅ Wpump = 16 hp D ⋅ Wpump = 1 hp 2D FIGURE 8–19 The pumping power requirement for a laminar flow piping system can be reduced by a factor of 16 by doubling the pipe diameter. dx mCpTx R4 P 8L PR2 R2 8L ave Ac dx r dr Temperature Profile and the Nusselt Number In the analysis above, we have obtained the velocity profile for fully developed flow in a circular tube from a momentum balance applied on a volume element, determined the friction factor and the pressure drop. Below we obtain the energy equation by applying the energy balance to a differential volume element, and solve it to obtain the temperature profile for the constant surface temperature and the constant surface heat flux cases. Reconsider steady laminar flow of a fluid in a circular tube of radius R. The fluid properties , k, and Cp are constant, and the work done by viscous stresses is negligible. The fluid flows along the x-axis with velocity . The flow is fully developed so that is independent of x and thus (r). Noting that energy is transferred by mass in the x-direction, and by conduction in the r-direction (heat conduction in the x-direction is assumed to be negligible), the steady-flow energy balance for a cylindrical shell element of thickness dr and length dx can be expressed as (Fig. 8–20) Qr · m CpTx mCpTx Qr FIGURE 8–20 The differential volume element used in the derivation of energy balance relation. dr · where m Ac after rearranging, · m CpTx dx · Qr · Qr dr 0 (8-49) (2 rdr). Substituting and dividing by 2 rdrdx gives, Cp Tx dx dx Tx · · 1 Q r dr Q r 2 rdx dr (8-50) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 435 435 CHAPTER 8 or · Q 2 Cp rdx r T x 1 (8-51) where we used the definition of derivative. But · Q r k2 rdx r Substituting and using T r 2 kdx r r T r (8-52) k/ Cp gives α T r dr r r T x (8-53) which states that the rate of net energy transfer to the control volume by mass flow is equal to the net rate of heat conduction in the radial direction. Constant Surface Heat Flux For fully developed flow in a circular pipe subjected to constant surface heat flux, we have, from Eq. 8–24, dTs dx T x · 2qs mCpR dTm dx constant (8-54) If heat conduction in the x-direction were considered in the derivation of Eq. 8–53, it would give an additional term 2T/ x2, which would be equal to zero since T/ x constant and thus T T(r). Therefore, the assumption that there is no axial heat conduction is satisfied exactly in this case. Substituting Eq. 8–54 and the relation for velocity profile (Eq. 8–41) into Eq. 8–53 gives · 4qs 1 kR r2 R2 dT 1d r dr r dr (8-55) which is a second-order ordinary differential equation. Its general solution is obtained by separating the variables and integrating twice to be T · qs 2 r kR r2 4R2 C1 r C2 (8-56) The desired solution to the problem is obtained by applying the boundary conditions T/ x 0 at r 0 (because of symmetry) and T Ts at r R. We get T Ts · qs R 3 k4 r2 R2 r4 4R4 (8-57) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 436 436 HEAT TRANSFER The bulk mean temperature Tm is determined by substituting the velocity and temperature profile relations (Eqs. 8–41 and 8–57) into Eq. 8–4 and performing the integration. It gives Tm · Combining this relation with qs h Ts · 11 qs R 24 k h(Ts Tm) gives 48 k 11 D 24 k 11 R (8-58) 4.36 k D (8-59) or Circular tube, laminar (q·x constant): Nu hD k 4.36 (8-60) Therefore, for fully developed laminar flow in a circular tube subjected to constant surface heat flux, the Nusselt number is a constant. There is no dependence on the Reynolds or the Prandtl numbers. Constant Surface Temperature A similar analysis can be performed for fully developed laminar flow in a circular tube for the case of constant surface temperature Ts. The solution procedure in this case is more complex as it requires iterations, but the Nusselt number relation obtained is equally simple (Fig. 8–21): Ts = constant 64 f = ––– Re D Nu = 3.66 max m Fully developed laminar flow FIGURE 8–21 In laminar flow in a tube with constant surface temperature, both the friction factor and the heat transfer coefficient remain constant in the fully developed region. Circular tube, laminar (Ts constant): Nu hD k 3.66 (8-61) The thermal conductivity k for use in th...
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This note was uploaded on 01/28/2010 for the course HEAT ENG taught by Professor Ghaz during the Spring '10 term at University of Guelph.

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