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C HAPTER 7 & Internal Turbulent Flows 7.1 I NTRODUCTION This chapter is concerned with the prediction of the heat transfer rate from the wall o f a duct to a:fluid flowing through the duct, the flow in the duct being turbulent. The majority of the attentio,:! will be given to axi-symmetric flow through pipes and two-dimensional flow through plane ducts, i.e., essentially to flow between parallel plates. These two types of flow are shown in Fig. 7.1. Such flows effectively occur in many practical situations such as flows in heat exchangers. Attentiop-Will initially be given to fully developed flow, see References [1] to [12]. This will be followed by a discussion of developing duct flows. Lastly, a brief discussion of the numerical analysis of more complex duct flows will be presented. 7.2 ANALOGY SOLUTIONS F OR FULLY DEVELOPED P IPE F LOW As discussed in the previous chapter, most early efforts at trying to theoretically predict heat transfer rates in turbulent flow concentrated on trying to relate the waIl heat transfer rate to the wall shearing stress. I n the present section an attempt will be made to outline some of the simpler such analogy solutions for duct flows [13], [14],[15],[16],[17]. The ideas used in the previous chapter to derive analogy solutions for boundary layer flows can easily be extended to obtain such analogy solutions for turbulent 304 / CHAPTER 7: Internal Turbulent Flows 305 Flow Through a Pipe Flow Through a Plane Duct F IGURE7.l Flow in a pipe and in a plane duct. duct flows. The main difference between the solutions for boundary layer flows and those for'duct flows arises because in the latter case the Nusselt number is based on the difference between the wall temperature and the mean fluid temperature and the Reynolds number is based on the mean velocity; whereas in boundary layer flows the Nusselt number is based on the difference between the wall temperature and the . fluid temperature outside the boundary layer and the Reynolds number is based on the velocity outside the boundary layer. Consider, first, the simple Reynolds analogy for pipe flows. The pipe wall temperature will be assumed to be uniform. I f y is the distance from the wall to the point in the flow being considered as shown in Fig. 7.2, the total shearing stress and heat transfer rate are hgain written as: 7' = 7'm + 'TT= f.L iJy iJu - pv' u' (7.1) and: q = qm + f/.r = - k iJy •. iJT - ,-, - pCpv T (7.2) The above two equations can be written as: 7' iJu = p (v + e ). iJy (7.3) and: (7.4) ----------------R r --------------------of Pipe it Center: / y T FIGURE 7.2 Coordinate system used. 306 Introduction to Convective Heat Transfer Analysis T hese equations c an b e r earranged to give: au ay and: p (v 'T + e) (7.5) aT q - = - --::..-ay (7.6) I ntegrating these two equations outward from the w an, w here u = 0 a nd T == Tw, to a point a t a distance o f y f rom the wall where the mean velocity is u and the m ean temperature is T then gives: ' [Y U 'T == Jo p (v + e) d y (7.7) and: (7.8) T he s ame assumptions as were used i n d eriving the Reynolds analogy for boUndary l ayer flow a re now introduced, i.e., i t i s assiuned that: eH = e (7.9) t his is, o f course, equivalent to assuming that the turbulent P randtl number, P iT, is e qual to 1, a nd t hat Pr = 1, i.e., a = v (7.10) N o real fluids exist for which this i s true, but, as mentioned earlier, most gases have Prandtl numbers that are close to unity. I t i s also assumed that! qw 'Tw (7.11) w here 'Tw a nd q w the values o f t he shear stress a nd t he h eat t ransfer rate at the wall. U sing t he assumptions given i n E qs. (7.9) to (7.11), Eq. (7.8) c an b e rewritten as: are (7.12) I f t he following is then defined: G (y) = Ic y o p (v . 'T + e) dy (7.13) / Eqs. (7.7) a nd (7.12) can b e w ritten as: u = G (y) (7.14) CHAPI'BR 7: Internal Turbulent Flows 307 and: Tw - T= qw G(y) CpTw (7.15) This shows that the mean velocity and temperature profiles are similar when the Reynolds analogy assumptions are adopted. I f the average velocity and average temperature across the pipe are defined as usual by: u m = wR2Jo U2wrdr 1 rR and - m= T U mW 1 R2 i 0 R uT2wrdr then, because: r = R -y it follows that: 20 -.um = R21R u (R - y)dy (7.16) and: T m = foRuT(R - y)dy foRu(R - y)dy Using Eqs. (7.14) and (7.15) these give: (7..17) uin = ~2 foR G(R R y)dy (7.18) and: T _T w m = qw foR G2(R - y)dy . CpTw fo G(R - y) d y (7.19) D.ividing Eq. (7.19) by Eq. (7.18) then g~ve&: T w-Tm um qw fo G 2(R-'-y)dy 2[ R ]2 R2 fo G(R - y)dy R' = cpTw (7.20) This equap.?n can be written as: Tw - Tm = qw F un, CpTw . (7.21) 3 08 Introduction to Convective Heat Transfer Analysis where: F= foR G2(R - y)dy 2 R2 [ foR G(R - y)dy . ]2 (7.22) This factor, F, can be evaluated approximately by noting that in pipe ~ow the mean velocity profile is approximately given by: ' ~ = Ue (y)ln R (7.23) where ue is the mean velocity on the center line of the pipe. Comparing Eq. (7.23) with Eq. (7.14) then shows that the function G is approximately given by: G =uc(~r Substituting this into Eq. (7.22) then gives: from which it follows that: F = 150 147 i.e., F is effectively equal to 1 and Eq. (7.21) then gives: ---=-- Tw - Tm qw (7.24) This is the basic Reynolds analogy equation for pipe flow. It is identical to that for boundary layer flow except that the mean velocity and temperature occur in place of the free-stream values. Now, for fully developed pipe flow i t is usual to work with the friction factor, / , rather than the wall shearing stress. The friction factor is defined in terms of the pressure gradient by: / =_ d pldz pu~/2D (7.25) Um being, as before, the mean velocity in the pipe and D· is its diameter. I n order to find the relation between the wall shearing stress, 'Tw, and the friction factor, j , for the fully developed flow, the conservation o f momentum principle is applied to the control volume shown in Fig. 7.3. / C HAPfER 7: Internal Turbulent Flows 309 dz __ 2z a ------ -- --- ------- -1 ;~- --~, -- --- --------_. P -....I'£'w~ p +..1!dz Iii a Q Control Volume F IGURE 7.3 Control volume used i n relating wall shear stress to the friction factor. Because, in the fully developed flow, there is no change in the velocity profile with distance, z, along the pipe, there can be no change in fluid momentum through the control volume. The forces acting on the control must therefore balance, Le.: pA - ~ + ~~ d + = C - Tw A TwCdz where C = 'IT'D is the circumference of the pipe and A ' = 'IT'D214 is its crosssectional area. This equation gives on rearrangement: dp dz = = - Tw D 4 Substituting this equation into Eq. (7 .25) ~n gives the following relation between the friction factor and the wall shearing stress: Tw t -2 = sPUm (7.26) Substituting Eq. (7.26) into Eq. (7.24) then gives the following as the Reynolds analogy for pipe flow: Tw - u m Tm _ qw - 8 cp tpu~ (7.27) This equation can be rearranged in terms of the Nusselt number, N U D, based on the mean temperature difference and on the pipe diameter, and the Reynolds number, ReD, based on the mean v~locity and the diameter, to give, on noting that P r has already been assumed equal to 1: N UD = S Re D t, (7.28) Charts and equations describing the variation of the friction factor, t , with the Reynolds number, ReD, and wall roughness ratio, elD, where e is a measure of the roughness of the walls, are available [18],[19], [20]. A Moody chart that gives the friction factor variation is shown i n Fig. 7.4. ~ Alternatively, Colebrook gives the following equation for commercially rough pipes: 1 8.7] j1 = 1.74 - 210g10 [e + R eD.Jl J 2D (7.29) 310 Introduction to Convective Heat Transfer Analysis 0.1 0.09 0.08 0.07 0.06 0.05 Critical ~ flow zone Transition zone .1 Fullyrou IIII~I I II zone 11\ 0.05 0.04 0.03 0.02 !o. ..... .~ if II il~ 0.04 ~~ ~\$ II \ r--.. O.ot5 O.ot 0.008 0.006 0.004 J 0.03 0.025 0.02 ~ .:-... -11:1 r-.. -..... '"S ~ 0.002 0.001 0.0008 0.0006 0.0004 I ":I ~ ~ 0.015 ReQc 0.01 0.009 e(/Jm) 1.5 . CoIOlllel'Cial steel 46 Cast iron 260 Concrete 300-3000 2 3 4 5 6 B Ut 2 3 4 5 6 8 uf 2 DraWD~~ I1 Smooth pipes - r- -.. ,..:.. 0.0002 0.0001 ~ 0.000,05 r....:::.:: ,.,. ...... 0.008 3 4 S 6 8 106 2 3 4 ~ 6 8 10' - 0.000,01 74l!~101 \.... Reynolds number. ReD = u.: D =0.000.001 ~ =0.000,005 F IGURE 7.4 Moody chart. Some typical values of the roughness, e, for pipes that are likely to be encountered in heat transfer practice are listed in Table 7.1. For smooth pipes, i.e., for e = 0, Eq. (7.29) gives: v! 1 [-:; = 1.74 - 2log10 [ 1&7U ] ReD v ! :n = 2 l0g [Re lO DJI]- 0.8 (7.30) Alternatively, the following simple equation can be used to obtain approximate values for the friction factor for flow i n smooth pipes for Reynolds numbers less than T ABLE 7 .1 R oughness values f or c ommercial pipes Pipe material Wall roughness, e, mm 1.5 X 10-3 7.5 X 10-3 4.6 X 10-2 0.15 Drawn tubing Brass, glass Steel / Galvanized iron CHAPTER 7: Internal Turbulent Flows 311 1600 , . . - - - - - - - , - - - - - - - - - , 1200 NUD 800 400 . ; ;;.- ~ Experimental o~--------~--------~ ? 1 e+4 1 e+5 ReD 1 e+6 F IGURE 7.5 Variation of NUD with ReD for smooth pipe~. f= 0.305 Re~25 (7.31) The variation o f N UD with ReD for smooth pipes given by using Eq. (7.30) i n conjunction with Eq. (7.28) is shown in Fig. 7.5 together with experimental results for air. Air flows through a smooth pipe with a diameter of 50 m m at a mean velocity o f 35 rrils. The walls of the pipe are kept at a temperature of lOoe. At a certain section of the pipe, the mean air temperature is 4 0o e. Find the value of the heat transfer coefficient and the rate of heat transfer per meter length of pipe from the a ir to the walls at this section. E XAMPLE 7 .1. Solution. At the mean temperatqre of 400 e assuming that the pressure is near ambient, air has the following properties: v = 16.01 X 1 0-6 m2/s, k = 0.02638 W/m-oe Hence: ReD =V umD 35 x 0.05 n5 = 0.00001601 = 1.093 X lu- The flow is therefore turbulent. Because the pipe is smooth,! is given by Eq. (7.30), i.e., by: J1 1 .. = 2Iog 10 [ReD.Jj] - 0.8= 2Iog10 [109300.Jj] ~ 0.8 which can be solved to give: f Eq. (7.28) then gives: NUD = 0.0177 ! = gReD = 0.0177 8 X 109300 = 241.8 312 Introduction to Convective Heat Transfer Analysis Hence: h = N ;;k = 241.8 ~~502638 = 127.6 W/m20C The heat transfer rate per m length of the pipe is then given by: Qw = h7TD(Tw - T m) = 127.6 x 7T x 0.05 x (10 - 40) Le.: Qw = - 601.3 W/m This is negative because the heat transfer rate is taken as positive in the wall to air direction. The heat transfer rate from the a ir to the walls of the pipe is therefore 601.3 W/m. Therefore, the heat transfer coefficient is 127.6 W/m20C and the heat transfer from the air to the pipe walls per meter length is 601.3 W. I t will be seen from Fig. 7.5 that the agreement between the predicted values and experiment is quite good for the case considered because air has a Prandtl nqmber near 1. When the Prandtl number of the fluid involved is very different from one, however, the results given by Eq. (7.28) are unsatisfactory. For this reason, more refined analogy solutions have also been developed for pipe flows just as i n the caSe of boundary layer flows. Consideration will now be given to a thr~-layer analogy solution for pipe flows. I n this discussion, attention will be restricted to flow through a pipe in. which the wall heat flux is uniform. The analysis is, of course, similar to that given in Chapter 6 for boundary layer flow. . Now, i f the longitudinal heat transfer i n the fluid is neglected, the energy equation for fully developed constant fluid property turbulent pipe flow can be Written as: aT .1 1aZ 1 = - - r{eH+a)- a[ r ar aT] ar (7.32) I n writing this equation it has been noted that because the flow is fully developed, v is zero. Because the flow is fully developed, the form of the mean temperature profile is not changing with distance along the pipe, i.e., defining: Tw - T = F(!.-) Tw - Tc R (7.33) in fully developed flow the function, F, is independent of the distance along the pipe, z. In this equation, T c is the mean temperature on the center line and Tw is the wall temperature. Now, because F does not depend on z, i t follows that: ~[TW-T] = 0 az T w-T c CHAPTER 7: Internal Turbulent Flows 313 i.e., th;at: aT az aTw _ [TW - T ][dTw _ d Tc] T w-Tc dz dz az (7.34) In the situation here being considered, the heat flux at the wall, qw, is uniform and specified. Now, using Fourier's law, the heat transfer rate at the wall is given by: a T, qw = + ka r r =R (7.35) The positive sign arising because r is measured from the center toward the wall whereas the heat flux, qw, is taken as positive in the inward direction, i.e., in the wall to fluid direction. Eq. (7.35) can be written using the definition of the function, F, given in Eq. (7.33) as: _ _ k(Tw - Tc) d F, ' qw R d (r/R) r =R (7.36) Because the case where qw is uniform is being considered, this equation shows that Tw - Tc = constant from which it follows that: (7.37) Substituting this result into Eq. (7.34) then shows that when the wall heat flux is uniform: -=-- aT az d Tw dz (7.38) Therefore, the gradient o f temperature at all radii is the same. Hence, i f T m is the mean temperature at any section o f the pipe i t follows that: - aT az dT dT dTm =- w=- c=dz dz dz (7.39) I t should also b e noted that an overall energy balance applied to the control volume shown i n Fig. 7.6 gives: qw 7TD = _ 1T 2 [ P CpU m 4 D T m + dTzm - -m d T] I I dT . Z I 1. - ---------- ________ J ___ L ________ ~ ____________ _ Tm~. Tm+ d mdz _ iI ~.~, ~ qw Control Volume F IGURE 7.6 Control volume used in establishing energy balance. 314 Introduction to Convective Heat Transfer Analysis which can be rearranged to give: dTm 4qw 1T = dz pcpumD (7.40) Be.cause qw is constant, it follows from this that the gradient of temperature at all radii is constant. . Having established this result from the boundary conditions, attention can now be returned to the governing equation (7.32). Using Eq. (7.39), this equation gives: d 1 u - Tw = - -a [ r(€H dz r ar a. + a ) -T] ar (7.41) I t is convenient to rewrite Eq. (7.41) in terms of the distance from the wall, y, to the point being considered. I f, as above, R is the radius of the pipe~ y is, as diSCUSsed before, related to r by: y = R-r (7.42) In terms ofy, Eq. (7.41)becomes: dTm .1 a .[ u d z = (R _ y )cJy (R - Y)(€H + a) ay a T] (7.43) The boundary c?nditions on the solution to this equation are: , When r = 0. (L~., y = R): When r = R (Le., y = 0): ~~ = 0 (7.44) T = Tw The first of these conditions follows, o f course, from the reqtllrementthat the profile be symmetrical about the c~nter line. Integrating Eq. (7.43) from the center line where y = R to some point distance y from the wall and using the first boundary condition listed i n Eq. (7.44) then gives: (R...,.. y)(€H . adT. + a )T = - m ay dz I Y R u(R - y)dy i.e.: (R - y)(eH + a ) ~~ = dJzm [J: u(R - y)dy- C u(R - Y)d Y] (7.45) I t is convenient to define: J (y) = fY u(R Jo y) d y - r Jo . R u(R - 'y)dy (7.46) and i n terms of this function, Eq. (7.45) becomes: aT . J d Tm = ------ay (R - y)(€H + a ) d z / CHAPI'BR 7: Internal Turbulent Flows 315 Integrating this equation out from the wall and using the second of the boundary conditions given in Eq. (7.44) then gives: T - Tw = d T m dz Jo r y I ( R - Y)(EH + a) d Y (7.48) Using Eq. (7.40) allows this equation to be written as: T - Tw . = 4qw p cpumD Jo fY I (R - Y)(EH + a) dy (7.49) I f the velocity profile is known together with the distribution of E then, for any assumed relationship between the distributions of EH and E , Eq. (7.49) can be used to deduce the temperature profile. Once this has been obtained the relation between . the Nusselt and Reynolds numbers can be derived. Before illustrating this procedure, there is a simplifying assumption that can be introduced without any significant loss of accuracy. Because the velocity profile in turbulent pipe flow is relativ~ly flat over a large portion of the pipe cross-section, it is usually sufficiently accurate to replace U in the integral in the expression for I by the constant value Urn, i.e., to write: J (y) = = I: u (R - y) d y - , t R u (R - y) d y um[J:<R- y)dy - f<R - Y)d Y] (7.50) Substituting this approximate result into Eq. (7.49) then gives: T. - Tw , =- qw pCp Jo fY (1 (E H + a) ylR) d y (7.51) I t should clearly be understood that the above approximation is not essential to the analysis. I t is used because it leads to a considerable simplification and yet gives results that are of adequate accuracy for most purposes. \ Because of the nature of the parameters that are used to describe the distribution of E , it is convenient to write Eq. (7.51) as:, T - T - - (qw) w- pcp ~ fY+ ( 1- ylR) d + V~ Jo (EIPrT + lIPr) y y+ (7.52) where: E = E, v = y vVP f !i (7.53) . 316 Introduction to Convective Heat Transfer Analysis Now, because the flow is fully developed, the momentum equation can be written as: ! r ay ~(rT) = dp dz (7.54) where T is the total shearing stress which is given as before by: T = p (v au + E )ay (7.55) Integrating Eq. (7.54) then gives since the shearing stress is zero on the center line: r T= - - d p ,.z dz 2 Le.: T dp r = -- dz 2 = - -(R-y) 1 dp 2 dz (7.56) This shows that there is a linear variation of shear stress across the flow in fully developed pipe flow. Applying this to conditions at the wall then gives: Tw = - - . R dp 2 dz (7.57) a result that was previously derived in a somewhat different way. Dividing Eq.(7.56) by Eq. (7.57) then gives: Y -T = 1 -Tw R (7.58) Now Eq. (7.55) can be written as: 1+E = T lpv a ulay (7.59) Substituting Eq. (7.58) into Eq. (7.59) then gives on rearrangement: E=~-l au ay ~ (1 -1'. \ (7.60) Defining as before: u+ = u !P, VTw: (7.61) CHAPTER 7: Internal Turbulent Flows 317 Eq. (7.60) c an be written as: E=~-1 au+ ay+ ( '1 - L y, (7.62) Therefore i f u+ is a known function o f y+, the distribution o f E c an b e found using Eq. (7.62) and i f the value o f PrT is assumed to be known, the temperature profile can b e determined using Eq. (7.52) and this can then b e used to find the Nusselt number. In carrying out this procedure i t will here b e assumed, based on experimental mean velocity measurements and intuitive reasoning, that the flow can be split into three regions. The three layers will b e assumed to extend from y+ = 0 to 5, from y+ = 5 to 30, and from y+ = 3 0 to the pipe center line. Each o f these regions will now b e separately discussed. (i) Inner Layer: 0 < y+ < 5: .. In this region which lies adjacent to the wall, the turbulent shearing stress and turbulent heat transfer rate will b e assumed to b e negligible, i.e., i t will b e assumed that in this region: . E = 0 andEH = 0 (7.63) Therefore, for this region Eq. (7.52) gives:. T- T = - (qw) VTPPrJfoY+ (1 - R dy+ w / y) ;, p Cp Also, because the extent o f this region is small, y /R remains very small in this region and 1 - y/R is, therefore, effectively equal to 1 i n this region. Eq. (7.63) therefore gives: T- Tw = _(qw) VT;, / PPry+ p Cp (7.64) I f Ts is the t emperatuIeat the outer edge o f this layer, i.e., at y+ = 5 , then this equation gives: -s - Tw = - 5 · - · - Pr (qw)ft T p Cp Tw (7.65) (ii) Buffer Layer: 5 < y+ < 30: On the basis o f available. measurements, i t will b e assumed that i n this layer the mean velocity profile is given by: u+ = 5 In y+ au+ - 3.05 (7.66) Differentiating this equation gives: --ay+ y+ 5 / 318 Introduction to Convective Heat Transfer Analysis Substituting this result into Eq. (7.62) and assuming that in this region y lR also remains very small and 1 - ylR, therefore, effectively equal to 1 in this region. gives: E = 5- y+ 1 (7.67) Substituting this equation into Eq. (7.52) and assuming that the turbulent Prandtl number is 1 gives in the buffer layer: T - Ts ( )Ifi =qw P pCp Y+ Tw 5 y+ _ 1 + d Y+ ~ Pr (7.68) 5 i t having already been assumed that in this region, region y lR remains very small and1hat 1 - ylR is, therefore, effectively equal to i . Carrying out the integration, this equation gives: i.e.: T - Ts = -5(:;' )Ifln(Y; Pr -Pr + 1) = _ 5(qW) rE ln(5Pr+ 1) p Cp (7.69) I f the mean temperature at the outer edge of the buffer layer, i.e., at y+ is Tb, Eq. (7.69) gives: = 30, (7.70) Tb - Ts (iii) Outer Layer: y+ VTw > 30: In this outer region i t is assumed that the velocity profile is given by: u+ = 2 .5lny+ + 5.5 a u+ = 2.5 a y+ y+ (7.71) Differentiating this equation gives for the outer layer: (7.72) In this region it can also be assumed that the molecular shearing stress and heat transfer rate are negligible compared to the turbulent components, i.e., that E » v and E H » a and that in this region therefore, because the turbulent Prandtl number is being assumed to be equal to 1, Eqs. (7.52) and (7.62) become: T - Tb = -;;;p If f~ () + (l- Y) E R d y+ (7.73) / CHAPl'ER 7: Internal Turbulent Flows 319 and E = ( 1- ylR) au+ ay+ respectively. Using Eq. (7.72) i n Eq. (7.74) then gives: E= (7.74) (1- y)y+ R 2.5 (7.75) Substituting Eq. (7.75) into Eq. (7.73) then gives: T - Tb = - (qw) y"7C Jy+ 2.: dy+ /P ( pCp y , 30 (7.76) Carrying out the integration then gives: T - Tb = - 2.5(QW) IP pcp y~ In(Y+) 30 (7.77) O n the center line o f t he pipe, y = R. Hence, i f y t is the value o f y+ on the center line, it follows that: yt = R f !i v yp (7.78) Therefore, i f Tc is the temperature on the center line, Eq. (7.77) gives: -c -'Tb = T . - 2.5 - (qW )HP (-30v fiw) -In R pCp p 'Tw (7.79) This ,completes the analysis o f t he temperature distributions i n t he three layers. The overall temperature 9han.ges are obtained b y adding Eqs. (7.65), (7.70), and (7.79) together to eliminate the intermediate temperature Ts a nd Tb. T his gives: Tw - T, = ~~ )H[25In(3~vfi)+ 5 In(5Pr+ 1) + 5prJ (7.80) In t he discussion o f the Reynolds analogy given earlier it'was shown that: (7.26) Hence:. R v yp f !i = RUm v YB !1 = ReD Y32 fT (7.81) / where ReD is as before the Reynolds number based on D and on um • 320 Introduction to Convective Heat Transfer Analysis Using Eqs. (7.26) and (7.81) allows Eq. (7.80) to b e written as: TW - T= c (p!';.m )ft [2.5 In e;; J~} 51n(5Pr + 1) + 5pr] (7.82) To proceed further, the relation between ( T w - T m) a nd ( T w - T e) m ust be obtained. This could b e done b y integrating the temperature distributions obtained for each o f the three layers. However, sufficient accuracy for most purposes is obtained b y using the approximate 117th p ower law, i.e., by assuming that the velocity and temperature distributions are effectively given by: ~ =(~r and: (7.83) Tw - T ' = Tw - Te R (y)ln . (7.84) where ue is, as before, the mean velocity o n the center line o f the pipe. Therefore using: _ fORu(Tw - T)2'1rrdr Tw - T m = ~-R=-----li2'1rrdr fo i t follows that: (7.85) from which it follows that: Tw - Te =- (7.86) 6 Substituting this into Eq. (7.82) then gives: T W T m = Hp!Wum )ft [.5 In (R;; 2 Hz) + 1n(5Pr + 1 ) + 5pr] (7.87) This can b e rearranged to give: (7.88) / CHAPTER 7: Internal Turbulent Flows 321 For any value of R eD, this equation allows N UD to be found by using the methods of finding! that were discussed before. Eq. (7.88) gives results that are in good agreement with experiment for the flow of fluids with Prandtl numbers between about 0.5 and 30. For fluids with Prandtl numbers outside this range the effects of the various assumptions made in deriving Eq. (7.88), e.g., the neglect of the molecular heat transfer rate in the outer region, render it inaccurate. More refined analyses which overcome these deficiencies have been developed. E XAMPLE 7 .2. A ir flows through a smooth pipe with a diameter o f 65 mm at a mean velocity o f 3 0 m1s. T he walls o f the pipe are kept at a temperature o f 80°C. A t a certain section o f the pipe, the mean air temperature is 40°C. Using the three-layer solution, find the value o f the heat transfer coefficient and the rate o f heat transfer per m length o f pipe from the a ir to t he walls at this section. Solution. A tthe mean temperature o f 40°C assuming that th~ pressure is near ambient, air has the following properties: v = 16.01 X 10-6 m 2/s, k = 0.02638 W/m-oC Hence: ReD = umD v = 3 0 x 0.065 0.00001601 = 1.218 X 105 The flow is therefore turbulent. Because the pipe is s mooth,f is given by Eq. (7.30), i.e., by: 1 J I = 2log lO [ReD JI] - 0.8 = 2Ioglo [121800.jf] which can b e solved t o give: 0.8 f Eq. (7.88) gives: = 0.0173 hence 121800 X 0.7 X J O.0173 . 8. 5 6 [ 2 .5ln ( 121800 J O.0173) + 5 ln(5 X 0 .7+ 1) + 5 x 0.7] 30 32 Hence: = 211.7 . .The heat transfer rate per m length of pipe is then given by: Qw = h'lT'D(Tw - Tm) = 85.9 x 'IT' x 0.065 X (80 - 40) = 701.6 W lm 322 Introduction to Convective Heat Transfer Analysis Therefore, the heat transfer coefficient is 85.9 W /m2 °C and the heat transfer from the pipe walls to the air per m length is 701.6 W. 7.3 THERMALLY DEVELOPING P IPE F LOW Here the concern is with flow in a long pipe with a long unheated initial section. The unheated section is long enough for the velocity field to become fully developed before the heating begins. When heating begins, the temperature field develops and it is this thermal development region that is considered in this section [21],[22],[23]. Fluid properties will be assumed to be constant and the velocity field will not change in the thermal development region. The flow situation being considered is thus as shown in Fig. 7.7. . Because the velocity profile is fully developed, the radial velocity component, i.e., 11, is zero. The energy equation for the,situation here being considered is therefore as given in the previous section, Le.: aT u -az 1 =-- a[ aT] r ar .r(EH + a ) -ar (7.89) the longitudinal flux of heat again having been ignored compared to the. radial flux, i.e., the parabolic form of the governing equations having been used; The mean velocity, u, is here a function of r alone, i.e., does not vary with z. Eq. (7.89) can be written as: u~~ = ~:rH~ +;r)~a r;u / l.= D j U = 11m' E E (7.90) It is convenient to write·this equation in. dimensionless form. For .this purpose the following dimensionless varia~les are introduced: = v' Z = z/D ReDPr (7.91) . . Unheated WalL 'I' . Heated Wall ." . Temperature Prome Velocity Prome (Fully· . Developed) Temperature Prome Velocity Prome (Unchanged) / F IGURE 7.7 Thermally developing pipe flow. C HAPrER 7: Internal Turbulent Flows 323 Here, D is the diameter of the pipe and ReD is as usual defined by: Um being the average velocity across the pipe. Attention will be focused on the case where the wall temperature is unifonn and equal to Tw. The following dimensionless temperature is then introduced: f) = Ti. Tw - Ti T- (7.92) where T i is the initial temperature Of the fluid before the heating begins. In tenns of the variables defined in Eqs. (7.91) and (7.92), Eq. (7.90) becomes: Uof) = .R oR [R (E PPrT + 1) oR o !. ~ r o f)] Z Z (7.93) The boundary and initial conditions on the solution to this equation for f) are: = 0: f) = 0, R = 0.5: f) = 1, R = 0: of) oR =0 (7.94) Because the velocity field is fully developed, the variations of U and E with R are known. The solution to Eq. (7.93) can therefore be obtained using a similar procedure to that used in Chapter 4 to solve for thennally developing laminar pipe flow, i.e., using separation of variables. Here, however, a numerical finite-difference solution procedure will be, u~ed because it is more easily adapted to the situation where the wall temperature IS varying withZ. The nodal pointS1iefined in Fig. 7.8 are used. The following finite-difference approximations are introduced, it bebtg noted that U does .not change in the i-direction: U of) , a z.. I,J I = U j [(Ji,j - (Ji-l,j] AZ . ' (7.95) ._._-j. I· j +l A Rj + 1 I" ~"'-'--j TI A Rj i-I L . j-l I FIGURE7.S Nodal points used in obtaining finite-difference solution. Rj _ 1 324 Introduction to Convective Heat Transfer Analysis (7.96) Substituting these finite-difference approximations into Eq. (7.93) gives an equation of the form: A /hj + B j8 i,j+l + C /);,j-l 2 '{' = Pj (7.97) where the coefficients are given by U· Aj = - } !l.Z + -1 R j !l.R j +1 + !l.R j 2 [ Rj+l(Ej+1PrIPrr + 1) + R j{EjPrlPrr + 1)] ( 1 ) ' ! l.Rj+l + [Rj(EjPrIPrr + 1) + ~j-l(Ej-lprlprr + l)](~j)} 2 { B j = -' 1 , R j ! l.Rj+l + i lR j ,' }T [ Rj+l(Ej+lPr/pr (7.98) + 1) + R j(EjPrIPrT + 1 )]( 1 )~ 2 , , !l.Rj+1 ~ ' (7.99) , · -- 1 C a' Rj ! l.Rj+l + ! l.Rj { [ Rj(EjprIPrr D . - U/J;-l,j ,}!l.Z + 1) + ~j-l(Ej-lprlprr + 1) ](~j )} (7.100) (7.101) The boundary conditions give 8i,2 = 8 i,1 and 8i,N = 1. The use of these boundary conditions together with the application ofEq. (7.97) to all of the "internal" points on the i-line, i.e., j = 2,3,4, . .. , N - 2, N - 1, gives a set of N equations in the N unknown values of 8. This set of equations has the following form: 8i,1 - 8i,2 A28i,2 A38i,3 A N-18i,N-l =0 = D2 = D3 + B28i,3 + C28i,1 + B38i,4 + C38i,2 (7.102) / + BN-18i,N + CN-18i,N-2 8i,N = D N-l =1 C HAPTER 7: Internal Thrbulent Flows 325 i.e., has the form: 1 -1 A2 0 B2 A3 C2 0 0 0 0 0 0 B3 A4 C3 0 0 0 0 0 0 B4 C4 0 0 0 0 0 0 C N-l 0 0 0 0 A N -l 0 0 0 0 B N-l (J., 1 / 0 0 ( J'2 /, (J., 3 / ( J'4 /, - D2 D3 D4 D N -l 1 0 0 0 0 0 0 1 ( Ji,N-l ( J'N /, i.e., Q(J I.,J. = R (7.103) . where Q is a tridiagonal matrix. This equation can be solved using the standard tridiagonal matrix solver algorithm. The local heat transfer rate at any value df Z is obtained by noting that: qw = k- aT ar r =R i.e.: qw D k(Tw - Ti) _ a(J I (7.104) a R R=O.5 i.e.: N UiD = a(J a R R=O.5 I where N UiD is the local Nusselt~number based on the difference between the wall and the inlet temperatures, Le.: (7.105) But: .' a(J I i,1 = (Ji,N - (Jl,N-l aR ARN (7.106) Substituting Eq. (7.106) into Eq. (7.104) then gives the following: N UiD = (Ji;N - (Jl,N-l A RN (7.107) Because pipefiow is being cOilsidered, it is more convenient to use a Nusselt number based on the difference between the wall and local average temperatufes, i.e., to use: (7.108) . 3 26 Introduction to Convective Heat Transfer Analysis Comparing Eqs. (7.108) and (7.105) shows that: . where: Tw N UD = N UiD Tw - Ti = - Tm N UiD 8m (7.109) 8 m-- Tw - Ti Tw-Tm Now: (7.110) from which i t follows that: 8m = Jo (0.5 U 8dR / Jo (0.5 U RdR (7.111) . 8 m c an therefore b e obtained using the numerically determined variation of 8 with R a t any Z value. The value o f N UD a t this value o f Z c an then b e determined. In this way the variation o f N UD with Z c an b e obtained. I n order to u se the above solution procedure, the variations o f U and E ( = e/v) across the flow must b e specified. Here, the distribution o f E i n t he flow will be assumed to b e described b y t he following set o f equations: y+ < 5: E =0 + 5 y+ < y+ < 30: E = Y5 - 1 1 .6R(y+)6n - 1 (7.112) > 30: E = where: . y+ = I { !; = (0.5 - v vp R)Rev V8 f7 where f is the friction factor. I t will here b e assumed, as discussed i n t he previous section, that: .f = 0.305 (7.31) Re~25 this being applicable for flow in smooth pipes for Reynolds numbers less than about I~. . T he above equations describe the variation o f E with R. I t will b e a ssumed that the mean velocity distribution is adequately described . by: ~ Uc = (D - 2r)ln D (7.113) where Uc is the mean center-line velocity. CHAPTER 7: Internal Thrbulent Flows 327 Now recalling that: 8 um = D2 Jo (DI2 u rdr it follows that Eq. (7.32) gives: ~m = 8 (0.5(1 ~ Jo _ 2R)ln R dR= 49 00 (7.114) Using this result,Eq. (7.113) gives: U = 60 (1 - 2R)ln 49 (7.115) This equation describes the variation o f U with R. A computer program, TURDUCD, written in FORTRAN, that implements the procedure is available i n the way discussed in the Preface. Some results obtained using this program are shown in Figs. 7.9 and 7.10. 500~--m---~----~--~--~ 400 100 F IGURE 7.9 O L...---....L...----'--....;.--..........- --'-_----' l e-7 l e-6 le-5 Z le-4 l e-3 l e-2 Variation of Nusselt number with Z in thennal entrance region for P r = 0.7 for various values of ReD. 1500 ~--.---~-"""'--~-......,....-~ ] 1000 ~ 10 Pr 0.7 0.1 F IGURE 7.10 le-2 O '--..........- --'---"----L..----' l e-7 l e-6 le-5 Z le-4 le-3 Variation of Nusselt number with Z in thermal entrance region for ReD = 105 for various values o f Pro / 328 Introduction to Convective Heat Transfer Analysis E XAMPLE 7 .3. A ir flows through a long 5-cm diameter pipe at such a velocity that the Reynolds number is 1 05 • T he air enters the pipe at a temperature o f 20°C. The first portion o f the pipe is unheated and the velocity profile becomes fully developed in this portion o f t he pipe. The second portion o f the pipe, which is heated to a uniform wall temperature o f 40°C, h as a length o f 1.5 m. Determine how the wall heat transfer rate varies with z i n the heated portion o f t he pipe. Solution. T he flow situation being considered is shown in Fig. E7.3a. z is measured from the beginning o f the heated section and the maximum value of Z is, assuming that P r = 0.7, therefore given by: Zmax = zmaxlD = 1.510.05 = 0.0004286 ReDPr 105 X 0.7 T he computer program, TURDUCD, discussed above has therefore been run with the following inputs: Zmax = 0.0004286, P r = 0.7, ReD = ,100,000 and because the wall temperature is uniform, the wall boundary condition parameter has been set equal to 1 (the program also gives results for the case where the wall heat flux is uniform). The program gives the variations o f two Nusselt numbers with Z, these being N uDi, which is NUiD, a nd N uDa w hich is NUD. Now: NUiD( = N uDi) = k(T~w~ T j) I' 1 .5m -I FIGURE E7.3a ~r-------~------~------~ 3000 1000 oL-------~------~------~ 1.5 1.0 0.5 0.0 z -m FIGURE E7.3b CHAPI'ER 7: Internal Turbulent Flows 329 I f k is evaluated at (20 + 40)/2 = 30°C, it is equal to 0.02638 W /m- °c. Therefore, since Tw = 40°C a nd Ti = 20°C, i t follows that: qw Also: = N U,D k(TwD. z Ti ) = N U,D 0.02638(40 . 0.05 20) = 1 055NU,D WIm2 . . = 3500Z m = ZDReDPr = Z x 0.05 x 100000 x 0.7 Therefore, using the calculated variation o f N U iD with Z, the variation o f qw with z can be detennined. T he variation so found is shown in Fig. E7.3b. 7.4 DEVELOPING F LOW I N A PLANE D UCT The previous section was concerned with a flow in which only the temperature field was developing, the velocity field having reached the fully developed state before the heating began. In general, however, both the velocity and temperature fields develop simultaneously [24],[25]. In order to illustrate the nature of such flows, developing two-dimensional flow in a plane duct will be considered here. The flow situation considered is shown in Fig. 7.11. It will be assumed here that the flow enters the duct through a "shaped" unheated inlet section and that the velocity and temperature are therefore unifonn across the inlet plane as illustrated in Fig. 7.12. Temp. Profile F IGURE 7.11 Flow in a plane duct. and Temperature ~. Inlet Unifonn Velocity ~.-PI_anef===:::::lL_ v=o F IGURE 7.12 Assumed inlet plane conditions. / 330 Introduction to Convective Heat Transfer Analysis F IGURE 7.13 Assumed form of velocity distribution. The flow in the development region initially consists essentially of a boundary layer on each wall with a constant velocity, uniform temperature core ,between thes~ two boundary layers as illustrated in Fig. 7.13. These boundary layers grow until they meet on the center line of the duct. Following this, there is a region where the flow near the center line adjusts from that in the outer part of a boundary layer to that in a fully developed duct flow. However, the changes i n this second region are relatively small and only the first boundary layer region wi11be conside,red h~re. 1;'he .. analysis presented here is based on the use of the integral equation method discu~sed in Chapter 6. As the bOl1l1:dary layers grow on the wall, the velocity near the wall is decreased and, as a consequence, the velocity in the core region, U t, increases. Becauseth,e velocity on the inlet plane is uniform, the velocity on the inlet plane must be eqwi1 to the mean velocity, Um , in the duct. Continuity therefore requires, assuming that the density is constant, that: PUmW = p fow u dy , (7.116) where W is the width of the duct. Assuming that the flow is symmetrical about the center line of the duct aitd,that the velocity is unifonn in the core region between the two'boundary laye.rs as shoWn in Fig. 7.13, Eq. (7.116) can be written as: i.e.: Le.: (7.117) CHAPTER 7: Internal Turbulent Flows 331 Now the quantity: 8, u, J: (1 - ~) dy = (7.118) was defined i n Chapter 6 as the boundary layer displacement thickness. Therefore, Eq. (7.117) can be written i n terms of the displacement thickness as: Ur(W - 28t> = umW (7.119) The right-hand side o f this equation is a constant so this equation gives: dur d8r (W - 28d dz - 2ur dz =0 i.e.: --- = Ul 1 d UI 2 d8 r dz (W .:.. 281) dz (7.q,Q) This equation relates the gradient of the velocity in the core region to the rate of growth of the boundary layer displacement thickness. The integral equation analysis given in Chapter 6 solved for the boundary layer momentum thickness, 82 , which is related to the displacement thickness by the fonn factor, H , which is defined by: (7.121) Eq. (7.120) can therefore be written as: i.e.: ~ ~UI ur dz = . . (W - 2H82) 2 [8 ddH + H ddz2] 8 z 2 (7.122) Now the equations used in the integral equation analysis given in Chapter 6 can be written as: \ d82 + 82 (2 + H)~ dur dz ur dz Tw _ = Tw (7.123) p ut (7.124) 0.123 X 10-0.678H p ut (UI 82/ V )0.268 I (U82)1/6 82 d H = e5(H-l.4) [ _ (U I82)1/6 82~ dur v dz v U l dz - 0.0135(H - 1.4)] (7.125) 332 Introduction to Convective Heat Transfer Analysis Eqs. (7.122) to (7.125) can b e simultaneously solved to give the variations o f 82 , H, and U l with z. T he analogy solutions can then be used to obtain the heat transfer rate, the Reynolds analogy, for example, giving: ---= -'- Tw - Ti qw (7.126) i t being noted that the temperature in the core region between boundary layer will be equal to the temperature on the inlet plane, i.e., Ti. I t will be recalled that Eq. (7.126) was derived using the assumption that P r i s equal to 1. The solution procedure outlined above applies until the boundary layer reaches the center line o f the pipe, i.e., when: W 8 =2 (7.127) I f a power law velocity distribution is assumed in the boundary layer, i.e., i f it is assumed that: 11 = U1 (~)1/n 8 then using: i t follows that: i.e., that: 82 = 8 (n n + 1)(n + 2) (7.128) Similarly, using the definition o f the displacement thickness i t follows that: i.e., that: 81 = 1 8 ( n + 1) Dividing Eq. (7.129) by Eq. (7.128) gives: (7.129) H = (n + 2) / n CHAPI'ER 7: Internal Turbulent Flows 333 Le.: 2 n = H -1 Substituting this into Eq. (7.128) then gives: -=--- 82 8 H -1 H (H + 1) (7.130) For any value o f H; Eq. (7.130) allows 8 to b e found provided the value o f 82 has been detennined. Eq. (7.127) c an then b e u sed to detennine i f the boundary layer has reached the center line. I t is convenient to write the above equations i n dimensionless form before obtaining the solution. For this purpose, the following dimensionless variables are introduced: U =~, Um Z= - z W' 8 d= W (7.131) Eqs. (7.122) to (7.125) c an b e written i n terms o f these variables as: 1 dU U dZ = (1 - 2Hd 2) d 2. d Z 2 [ dH + H dz Tw d82] (7.132) (7.133) (7.134) dA2 1 dU d Z + d2(2 + H )U d Z Tw _ = pur 0.123 X 10-O.678H pur - Re~268(Ud2)O.268 Rew (Ud 2 2 d H l/6 )1/6/l dZ where: 1l6 = eS(H-l.4) [ -Re w (Ud 2 )1/6 d2 dd Z 00135(H - 14)] UU- · . (7.135) umW Rew = - v Eq. (7.127) indicates that the solution must b e e nded when: (7.136) d = 0.5 and Eq. (7.130) gives: A = H (H + 1) d ( H - 1) 2 (7.137) (7.138) Eq. (7.126) gives the heat transfer rate as: T, . N Uw = .....!!.. URewPrOA pur (7.139) 334 Introduction to Convective Heat Transfer Analysis an approximate correction to account for the fact that Pr is not equal to 1 having been applied. Here: Nuw = k(Tw - Ti) andRew = qwW umW -p- (7.140) The mean temperature, T m, across any section of the duct is given by: u",(Tm - Ti)W = 2 Iown u (T - Ti)dy i.e., since T is equal to Ti outside the boundary layer: um(Tm - Ti)W =2 I: u (T - Ti)dy i.e.: = 2A Jo . Tw - f1u(T - Ti Ti )d(Y) 8 (1.141) Again assuming that: and also assuming: :1=* T m- Ti _ T w - Ti (n _ ()lIn (y)lIn A (7.142) T - Ti Tw - T =I-. ~ i U Eq. (7.141) gives: 2n + 1)(n + 2 ) I :: f Il z; ~ z; :s 500 Re I 5 x 104 o~----~------~------~ 0.1 1 10 100 Z F IGURE 7.14 Variation of N Um with Z in entrance region for P r = 0.7 for various value!? . of Rew. CHAP1ER 7: Internal Turbulent Flows 3 35 111erefore, sUnce: 2 n = H -1 Eq. (7.142) allows ( Tm - Ti)/(Tw - Ti) to be found. The Nusselt number based on (Tw - Tm) can then be found using: NU m = Nuw Tw - Tw - Ti Nuw Tm = 1 - (Tm· - Ti)/(Tw - T;) A computer program, TURINDEV, that implements this procedure is available as discussed Un the Preface. 111is program is a modified version of that discussed in Chapter 6 for the calculation of external boundary layer flows . . Some typical results obtained using this procedure are shown Un Fig. 7.14. The values of dUl/dz are relatively small Un the entrance region and results obtained, assumbng t hatH is constant and equal toits flat plate value, agree quite closely with those obtained allowing for variations Un H. EXAMPLE 7 .4. I f t he l ength o f t he entrance region is taken as the distance from the inlet at which the Nusselt number based on the difference between the w alland t he m ean fluid temperatures reaches within 1% .of i ts fully developed value, use the program discussed above to determine how t he l ength o f the entrance region varies with Reynolds number for the flow o f air i n·a plane duct. Solution. T he computer program discussed above has been run for Reynolds numbers between 1Q4 a nd 3 X lOS for a Prandtl number o f 0.7. T he c alculated variation o f N Urn with Z was then considered. T he final value o f NUrn was multiplied by 1.01 a nd t he results scanned to find the value o f Z a t which N Urn was 1.01 times the final value. This value o f Z was taken a s ZentfW, Zent b eing t he a ssumed entrance length, i.e., the distance from the inlet a t w hich N Urn is within 1% o f its final value. T he results obtained i n this way are shown i n Fig. E7.4. 35 30 25 ZentlW 20 15 10 l e+4 le+5 Rew l e+6 FIGUREE7.4 T hese results c an b e approximately represented by: ZentfW = 0.95Re~28 3 36 Introduction to Convective Heat Transfer Analysis 7.5 SOLUTIONS TO THE FULL GOVERNING EQUATIONS Attention has in this chapter been devoted to fully developed turbulent duct flows or to flows in which the parabolic form of the governing equations can be used to obtain the solution. While the solutions obtained using these assumptions are applicable in many situations of great practical importance, there are many other important problems that cannot be adequately dealt with using these assumptions, see References [26] to [36]. For example, consider two-dimensional flow in a plane duct with a rectangular heated block mounted on one wall of the duct as shown in Fig. 7.15, this being a model of some situations that arise in electronic cooling. Because large regions o f recirculating flow can occur in such a situation, this flow cannot be adequately treated using the equations given in this chapter. Instead, the heat transfer rates in such flows must be obtained by numerically solving the full Adiabatic Walls of Plane Duct - . . Turbulent - . . Flow - .. F IGURE 7.15 Block Heated to a Uniform Temperature Turbulent flow over a heated block on the wall of a plane duct. STREAMLINE CONTOUR PLOT LEGEND A - .5000E + 01 B - .1000E+02 C - .1500E +02 D - .2000E + 02 E - .2500E + 02 F - .3000E + 02 G - .3100E+02 F F F IGURE 7.16 Streamlines for turbulent flow in a n axi-symmetric tum-around duct. (Reproduced with permission o f F luid Dynamics International) C HAPTER 7: Internal Turbulent Flows 337 governing equations. A number o f commercial computer codes based either on the use o f the finite-volume method or the finite element method are available that can be used to find such solutions. An example o f a solution given by one such code is shown in Fig. 7.16. 7.6 CONCLUDING R EMARKS Some simple methods o f determining heat transfer rates to turbulent flows in a duct have b een considered in this chapter. Fully developed flow in a pipe was first considered. Analogy solutions for this situation were discussed. I n such solutions, the heat transfer rate is predicted from a knowledge o f the wall shear stress. I n fully developed pipe flow, the wall shear stress is conventionally expressed i n terms o f the friction factor and methods o f finding the friction factor were discussed. The Reynolds analogy was first discussed. This solution really only applies to fluids with a PrandtI number o f 1. A three-layer analogy solution which applies for all PrandtI . numbers was then discussed. Attention was then turned to developing duct flows. A numerical solution for thermally developing flow in a pipe was first considered. Attention was then turned to plane duct flow when both the velocity and temperature fields are simultaneously developing. An approximate solution based o n t he use o f t he boundary layer integral equations was discussed. PROBLEMS 7.1. The walls of a smooth 75-mm diameter pipe are kept at a uniform temperature of 50°C. Air flows through the pipe at a mean velocity of 35 mls. At a certain section of the pipe, the mean a ir temperature is lOoC. Using the Reynolds analogy and assuming that the flow i n the pipe is fully developed, find the value of the heat transfer coefficient and the rate of heat transfer p~ I h length of pipe from the walls of the pipe to the air at this section. 7.2. The walls of the 50-mm diameter pipe are kept at a uniform temperature of 70°C. Air flows through the pipe at a mean velocity of 40 mls. At a certain section of the pipe, the mean air temperature is 30°C. Assuming that the flow in the pipe is fully developed, find the heat transfer coefficient and the rate of heat transfer per m length of pipe from the pipe to the a ir at this section using both the Reynolds analogy and the three-layer solution. 7.3. Assuming that i f y is the distance from the wall of a pipe, the mean velocity and mean temperature profiles are given by: !! = (y)ln Uc / R 338 Introduction to Convective Heat Transfer Analysis and: T - Tc = 1 _ (y )In Tw - Tc R where R is the radius of the pipe and subscript c refers to conditions on the center line, derive expressions for umluc and ( T m - T c)/(Tw - T c). Also recalling that in fully developed pipe flow: ~ = 1- Y 'Tw R find the variations of E lv and EHlv across the flow. Assume that the furbulent Prandtl number is 0.9. 7.4. Use the Reynolds analogy to derive an expression for the Nusselt number for fully developed turbulent flow in an annulus in which the inner wall is heated to a unifonn temperature and the outer wall is adiabatic. Assume that the friction factor can be derived by introducing the hydraulic diameter concept. 7.5. The so-called Taylor-Prandtl analogy was applied to boundary layer flow in Chapter 6. Use this analogy solution to derive an expression for the Nusselt number in fully developed turbulent pipe flow. 7.6. Consider the flow of water through a 65-mm diameter smooth pipe at a mean velocity of 4 mls. The walls of the pipe are kept at a uniform temperature of 40°C and at a certain section of the pipe the mean water temperature is 30°C. Find the heat transfer coefficient for this situation using both the Reynolds analogy and the three-layer analogy solution. 7.7. Consider thermally developing flow in a smooth 60-mm diameter pipe. Air, at an initial temperature of lOoC, flows through this pipe, the mean air velocity being 3Om/s. The first portion of the pipe is unheated and the velocity profile becomes fully developed in this portion of the pipe. The second portion of the pipe, which has a length of 2 m, is heated to a uniform temperature of 50°C. Determine how the ~a1l heat transfer rate in W1m2 varies with distance along the heated portion of the pipe. 7.8. I f the length of the thermal entrance region in turbulent pipe flow is taken as the distance from the beginning of the heated section at which the Nusselt number based on the difference between the wall and .the mean fluid temperatures reaches to within 1% of its fully developed value, use the computer program for thermally developing flow to determine how the length of the thermal entrance region varies with Reynolds number for the flow of air. 7.9. Air at a temperature of lOoC enters a plane duct with a distance of 6 cm between the two surfaces of the duct. The mean velocity of the air i n the duct is 40 mls. 'l11e walls of the duct are maintained at a uniform temperature of 40°C. Assuming that the velocity and temperature distributions are uniform at the entrance to the duct, ·deteI'lltine how the mean heat transfer coefficient varies with distance along the duct. 7.10. Consider the fully developed flow at a mean velocity of 9 mls through a 5-cm diarn- / eter smooth pipe. There is a uniform heat flux at the pipe wall. Find the heat-transfer CHAPI'ER 7: Internal Thrbulent Flows 339 coefficients for the following cases and discuss the reasons for the differences between the values obtained: (a) air at a mean temperature of 80°C and standard atmospheric pressure, (b) air at a mean temperature of 80°C and a pressure that is ten times the standard atmospheric pressure, (c) water at a mean temperature of 80°C. 7.11. Water flows at a rate of 1 kgls through a 3-cm diameter smooth pipe. The water enters the pipe at a temperature of 15°C and must leave the the pipe at a mean temperature of 50°C. The pipe wall is heated in such a way that the wall temperature is 14°C higher than the local mean water temperature at all points along the pipe. What length of pipe is required? Ignore en~ance region effects. 7.12. Water flows at a mean velocity of 5 mls through a tube in a condenser that has a diameter of 3 cm and a length of 1.2 m. Because steam is condensing on the tube, its wall surface temperature is constant and. equal to 90°C. I f the water enters the tube at a temperature of 35°C, what will be its mean temperature at the tube exit? Assume that the tube is smooth and ignore entrance region effects. 7.13. Water flows at a mean velocity of 1 mls through a I-cm diameter pipe, the mean water temperature being 20°C. The flow can be assumed to be fully developed and turbulent. . 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