PROBEM 7. KNOWN: Teperature and velocity of engine oil. Teperature and length of flat plate. FIN: (a) Velocity and theral boundary layer thickness at trailing edge, (b) Heat flux and surface shear stress at trailing edge, (c) Total drag force and heat transfer per unit plate width, and (d) Plot the boundary layer thickness and local values of the shear stress, convection coefficient, and heat flux as a function of x for 0 x 1. ASSUMPTIONS: (1) Critical Reynolds nuber is 5 10 5, () Flow over top and botto surfaces. PROPERTIES: Table A.5, Engine Oil (T f = 333 K): ρ = 864 kg/ 3, ν = 86.1 10-6 /s, k = 0.140 W/ K, Pr = 1081. ANAYSIS: (a) Calculate the Reynolds nuber to deterine nature of the flow, u 0.1 s 1 Re = = = 1161 ν 86.1 10 6 s Hence the flow is lainar at x =, fro Eqs. 7.19 and 7.4, and 1/ δ = 5 Re 1/ = 5 1 1161 = 0.147 1/3 δt = δ Pr = 0.147 ( 1081) 1/3 = 0.0143 (b) The local convection coefficient, Eq. 7.3, and heat flux at x = are k 1/ 1/3 0.140 W K 1/ 1/3 h = 0.33Re Pr = 0.33( 1161) ( 1081) = 16.5W K 1 q x = h( Ts T ) = 16.5 W K( 0 100) C =1300 W $ Also, the local shear stress is, fro Eq. 7.0, ρ u 3 1/ 864 kg 1/ τs, = 0.664 Re = ( 0.1 s) 0.664( 1161) τ s, = 0.084 kg s = 0.084 N (c) With the drag force per unit width given by = τs, where the factor of is included to account for both sides of the plate, it follows that 1/ 3 1/ = ρ u 1.38 Re = 1 864 kg 0.1 s /1.38 1161 = 0.337 N For lainar flow, the average value h over the distance 0 to is twice the local value, h, h = h = 3.5 W K The total heat transfer rate per unit width of the plate is $ q = h( Ts T ) = ( 1) 3.5 W K( 0 100) C =500 W Continued...
PROBEM 7. (Cont.) (c) Using IHT with the foregoing equations, the boundary layer thickness, and local values of the convection coefficient and heat flux were calculated and plotted as a function of x. 5000 deltax*10, hx*100, -q''x 4000 3000 000 1000 0 0 0. 0.4 0.6 0.8 1 istance fro leading edge, x () B thickness, deltax * 10 () Convection coefficient, hx * 100 (N/^) Heat flux, - q''x (W/^) COMMENTS: (1) Note that since Pr >> 1, δ >> δ t. That is, for the high Prandtl liquids, the velocity boundary layer will be uch thicker than the theral boundary layer. () A copy of the IHT Workspace used to generate the above plot is shown below. // Boundary layer thickness, delta delta = 5 * x * Rex ^-0.5 delta_ = delta * 1000 delta_plot = delta_ * 10 // Scaling paraeter for convenience in plotting // Convection coefficient and heat flux, q''x q''x = hx * (Ts - Tinf) Nux = 0.33 * Rex^0.5 * Pr^(1/3) Nux = hx * x / k hx_plot = 100 * hx // Scaling paraeter for convenience in plotting q''x_plot = ( -1 ) * q''x // Scaling paraeter for convenience in plotting // Reynolds nuber Rex = uinf * x / nu // Properties Tool: Engine oil // Engine Oil property functions : Fro Table A.5 // Units: T(K) rho = rho_t("engine Oil",Tf) // ensity, kg/^3 cp = cp_t("engine Oil",Tf) // Specific heat, J/kg K nu = nu_t("engine Oil",Tf) // Kineatic viscosity, ^/s k = k_t("engine Oil",Tf) // Theral conductivity, W/ K Pr = Pr_T("Engine Oil",Tf) // Prandtl nuber // Assigned variables Tf = (Ts + Tinf) / Tinf = 100 + 73 Ts = 0 + 73 uinf = 0.1 x = 1 // Fil teperature, K // Freestrea teperature, K // Surface teperature, K // Freestrea velocity, /s // Plate length,
PROBEM 7.4 KNOWN: iquid etal in parallel flow over a flat plate. FIN: An expression for the local Nusselt nuber. ASSUMPTIONS: (1) Steady, incopressible flow, () δ δ t, hence u(y) u, (3) Boundary layer approxiations are valid, (4) Constant properties. ANAYSIS: The boundary layer energy equation is T T T u + v = α. x y y Assuing u(y) = u, it follows that v = 0 and the energy equation becoes T T T α T u = α or =. x x u y y Boundary Conditions: T(x,0) = T s, T(x, ) = T. Initial Condition: T(0,y) = T. The differential equation is analogous to that for transient one-diensional conduction in a plane wall, and the conditions are analogous to those of Fig. 5.7, Case (1). Hence the solution is given by Eqs. 5.57 and 5.58. Substituting y for x, x for t, T for T i, and α/u for α, the boundary layer teperature and the surface heat flux becoe T x,y T y s T T = erf s α x/u k Ts T ( παx/u ) 1/ q s =. 1/ h x qx s Hence, with Nux = k ( Ts T ) k find 1/ x ( xu ) 1 ρ u x cpµ Nux = = = 1/ 1/ 1/ ( παx/u ) 1/ π k/ ρ c π µ k ( p ) 1/ 1/ Nux = 0.564 ( RexPr) = 0.564 Pe where Pe = Re Pr is the Peclet nuber. COMMENTS: Because k is very large, axial conduction effects ay not be negligible. That is, the α T/ x ter of the energy equation ay be iportant. 1/
PROBEM 7.7 KNOWN: Parallel flow over a flat plate and two locations representing a short span x 1 to x where (x - x 1 ). FIN: Three different expressions for the average heat transfer coefficient over the short span x 1 to x, h 1. ASSUMPTIONS: (1) Parallel flow over a flat plate. ANAYSIS: The heat rate per unit width for the span can be written as q1 = h1 xx1 Ts T (1) where h1 is the average heat transfer coefficient over the span and can be evaluated in either of the following three ways: x = x1+ x /. If the span is very short, it is reasonable to assue that h1 hx () where h x is the local convection coefficient at the id-point of the span. (a) ocal coefficient at (b) ocal coefficients at x 1 and x. If the span is very short it is reasonable to assue that h 1 is the average of the local values at the ends of the span, h1 [ hx1 + h x] /. (3) (c) Average coefficients for x 1 and x. The heat rate for the span can also be written as q1 = q 0 q0 1 where the rate q 0-x denotes the heat rate for the plate over the distance fro 0 to x. In ters of heat transfer coefficients, find h1 ( x x1) = h x h1 x1 x x h 1 1 = h h1 x x (5) 1 x x1 where h 1 and h are the average coefficients fro 0 to x 1 and x, respectively. COMMENTS: Eqs. () and (3) are approxiate and work better when the span is sall and the flow is turbulent rather than lainar (h x ~ x -0. vs h x ~ x -0.5 ). Of course, we require that x c x 1, x or x c > x 1, x ; that is, the approxiations are inappropriate around the transition region. Eq. (5) is an exact relationship, which applies under any conditions. (4)
PROBEM 7.15 KNOWN: Teperature, pressure and Reynolds nuber for air flow over a flat plate of unifor surface teperature. FIN: (a) Rate of heat transfer fro the plate, (b) Rate of heat transfer if air velocity is doubled and pressure is increased to 10 at. ASSUMPTIONS: (1) Steady-state conditions, () Unifor surface teperature, (3) Negligible radiation, (4) Re xc = 5 10 5. PROPERTIES: Table A-4, Air (T f = 348K, 1 at): k = 0.099 W/ K, Pr = 0.70. ANAYSIS: (a) The heat rate is q = h( w ) ( TsT ). Since the flow is lainar over the entire plate for Re = 4 10 4, it follows that Hence h 1/ 1/3 Nu 1/ 1/3 = = 0.664 Re Pr = 0.664( 40,000) ( 0.70) = 117.9. k k W / K h = 117 9 = 117 9 0099... = 17. 6 W / K 0. W o q = 17.6 0.1 0. 100 50 C= 17.6 W. K and (b) With p = 10 p 1, it follows that ρ = 10 ρ 1 and ν = ν 1 /10. Hence u u Re 5, 10 = 0 Re,1 8 10 ν = = = ν 1 and ixed boundary layer conditions exist on the plate. Hence h 4/5 4/5 1/3 5 1/3 Nu = = ( 0.037 Re 871 ) Pr = 0.037 ( 8 10 ) 871 ( 0.70) k Nu = 961. W / K Hence, h = 961 0. 099 = 1436. W / K 0. W o q = 143.6 ( 0.1 0. ) ( 100 50) C= 143.6 W. K COMMENTS: Note that, in calculating Re,, ideal gas behavior has been assued. It has also been assued that k, µ and Pr are independent of pressure over the range considered.
PROBEM 7.5 KNOWN: Plate diensions and initial teperature. Velocity and teperature of air in parallel flow over plates. FIN: Initial rate of heat transfer fro plate. Rate of change of plate teperature. ASSUMPTIONS: (1) Negligible radiation, () Negligible effect of conveyor velocity on boundary layer developent, (3) Plates are isotheral, (4) Negligible heat transfer fro sides of plate, (5) 5 Rex,c 5 10, = (6) Constant properties. PROPERTIES: Table A-1, AISI 1010 steel (573K): k p = 49. W/ K, c = 549 J/kg K, ρ = 783 kg/ 3. Table A-4, Air (p = 1 at, T f = 433K): ν = 30.4 10-6 /s, k = 0.0361 W/ K, Pr = 0.688. ANAYSIS: The initial rate of heat transfer fro a plate is q = ha s( Ti T ) = h ( Ti T ) With and Hence, 6 5 Re u / ν 10/s 1/30.4 10 /s 3.9 10, = = = flow is lainar over the entire surface 1/ 1/3 Nu 1/ 1/3 5 = 0.664Re Pr = 0.664 3.9 10 0.688 = 336 h = ( k / ) Nu = ( 0.0361W / K /1) 336 = 1.1W / K q = 1.1W / K 1 300 0 C = 6780 W Perforing an energy balance at an instant of tie for a control surface about the plate, Eout = E st, we obtain (Eq. 5.), dt ρδc h Ti T dt i = ( ) dt 1.1W / K 300 0 C = = 0.6 C / s dt 3 i 783 kg / 0.006 549J / kg K COMMENTS: (1) With ( δ ) 4 Bi h / / kp 7.4 10, = = use of the luped capacitance ethod is appropriate. () espite the large plate teperature and the sall convection coefficient, if adjoining plates are in close proxiity, radiation exchange with the surroundings will be sall and the assuption of negligible radiation is justifiable.
PROBEM 7.8 KNOWN: ength, thickness, speed and teperature of steel strip. FIN: Rate of change of strip teperature 1 fro leading edge and at trailing edge. ocation of iniu cooling rate. ASSUMPTIONS: (1) Constant properties, () Negligible radiation, (3) Negligible longitudinal conduction in strip, (4) Critical Reynolds nuber is 5 10 5. PROPERTIES: Steel (given): ρ = 7900 kg/ 3, c p = 640 J/kg K. Table A-4, Air ( T = 750K, 1 at ): ν = 76.4 10-6 /s, k = 0.0549 W/ K, Pr = 0.70. ANAYSIS: Perforing an energy balance for a control ass of unit surface area A s riding with the strip, E& out = de st /dt hxas( T T ) = ρδac s p( dt/dt ) ( ) p 3 0 /s( 1) 5 hx T T 900K hx dt/dt) = = =0.119hx ( K/s ). ρδc 7900 kg/ 0.003 640 J/kg K Vx At x = 1, Rex = = =.6 10 Re -6 x,c. Hence, ν 76.4 10 /s 1/ 1/3 0.0549 W/ K 1/ 1/3 h 5 x = ( k/x) 0.33Rex Pr = ( 0.33)(.6 10 ) ( 0.70) = 8.9W/ K 1 and at x = 1, dt/dt) = -0.987 K/s. At the trailing edge, 7 Rex =.6 10 > Re x,c. Hence 4/5 1/3 0.0549 W/ K 7 4/5 1/3 hx = ( k/x) 0.096Rex Pr = ( 0.096) (.6 10 ) ( 0.70) = 1.4W/ K 100 and at x = 100, dt/dt) = -1.47 K/s. 5 The iniu cooling rate occurs just before transition; hence, for Rex,c = 5 10 5 6 5 5 10 76.4 10 /s xc = 5 10 ( ν /V) = = 1.91 0 /s COMMENTS: The cooling rates are very low and would reain low even if radiation were considered. For this reason, hot strip etals are quenched by water and not by air.
PROBEM 7.4 KNOWN: Conditions associated with air in cross flow over a pipe. FIN: (a) rag force per unit length of pipe, (b) Heat transfer per unit length of pipe. ASSUMPTIONS: (1) Steady-state conditions, () Unifor cylinder surface teperature, (3) Negligible radiation effects. PROPERTIES: Table A-4, Air (T f = 335 K, 1 at): ν = 19.31 10-6 /s, ρ = 1.048 kg/ 3, k = 0.088 W/ K, Pr = 0.70. ANAYSIS: (a) Fro the definition of the drag coefficient with A f =, find ρv F = CAf ' ρv F = C. With V 15 /s ( 0.05 ) Re 4 = = = 1.94 10 ν 19.31 10-6 /s fro Fig. 7.8, C 1.1. Hence F 3 = 1.1( 0.05 ) 1.048 kg/ ( 15 /s ) /= 3.4 N/. (b) Using Hilpert s relation, with C = 0.193 and = 0.618 fro Table 7., k 1/3 0.088 W/ K 4 0.618 1/3 h = C Re Pr = 0.193 1.94 10 0.70 0.05 h = 88 W/ K. Hence, the heat rate per unit length is o q = h( π ) ( Ts T ) = 88 W/ K( π 0.05 ) ( 100 5) C= 50 W/. COMMENTS: Using the Zhukauskas correlation and evaluating properties at T (ν = 15.71 10-6 /s, k = 0.061 W/ K, Pr = 0.707), but with Pr s = 0.695 at T s, 0.6 0.061 15 0.05 0.37 1/4 h = 0.6 ( 0.707) ( 0.707/0.695) = 10 W/ K. 0.05 15.71 10-6 This result agrees with that obtained fro Hilpert s relation to within the uncertainty norally associated with convection correlations.
PROBEM 7.43 KNOWN: Initial teperature, power dissipation, diaeter, and properties of heating eleent. Velocity and teperature of air in cross flow. FIN: (a) Steady-state teperature, (b) Tie to coe within 10 C of steady-state teperature. ASSUMPTIONS: (1) Unifor heater teperature, () Negligible radiation. PROPERTIES: Table A.4, air (assue T f 450 K): ν = 3.39 10-6 /s, k = 0.0373 W/ K, Pr = 0.686. ANAYSIS: (a) Perforing an energy balance for steady-state conditions, we obtain q conv = h ( π )( T T ) = Pelec = 1000 W With V ( 10 s) 0.01 Re = = = 3,087 ν 3.39 10 6 s the Churchill and Bernstein correlation, Eq. 7.57, yields 4/5 0.6 Re 1/ 1/3 5/8 Pr Re Nu = 0.3 + 1 + /3 1/4 8, 000 1+ ( 0.4 Pr) 1/ 1/3 5/8 4/5 0.6( 3087) ( 0.686) 3087 Nu = 0.3 + 1 + 8. /3 1/4 8, 000 = 1+ ( 0.4 0.686) k 0.0373W K h = Nu = 8. = 105. W K 0.010 Hence, the steady-state teperature is Pelec 1000 W T = T + = 300 K + = 603K π h π 0.01 105. W K (b) With Bi = hro k = 105. W/ K(0.005 )/40 W/ K = 0.00, a luped capacitance analysis ay be perfored. The tie response of the heater is given by Eq. 5.5, which, for T i = T, reduces to T= T + b a 1exp at Continued...
PROBEM 7.43 (Cont.) 7 7 = 0.0173 s -1 and b/a = where a = 4 hρ cp = 4 105. W K 0. 01 700 kg 3 900 J kg K π h = 1000 W ( 0.01 105. W K) Pelec π = 30.6 K. Hence, ( ) 593 300 K 1 exp( 0.0173t) = = 0.968 30.6 K t 00s COMMENTS: (1) For T = 603 K and a representative eissivity of ε = 0.8, net radiation exchange between the heater and surroundings at T sur = T = 300 K would be ( q 4 4 ) rad = εσ π T Tsur = 0.8 5.67 10-8 W/ K 4 (π 0.01 )(603 4-300 4 )K 4 = 177 W/. Hence, although sall, radiation exchange is not negligible. The effects of radiation are considered in Proble 7.46. () The assued value of T f is very close to the actual value, rendering the selected air properties accurate.
PROBEM 7.57 KNOWN: Person, approxiated as a cylinder, is subjected to prescribed convection conditions. FIN: Heat rate fro body for prescribed teperatures. ASSUMPTIONS: (1) Steady-state conditions, () Person can be approxiated by cylindrical for having unifor surface teperature, (3) Negligible heat loss fro cylinder top and botto surfaces, (4) Negligible radiation effects. PROPERTIES: Table A-4, Air (T = 68 K, 1 at): ν = 13.04 10-6 /s, k = 3.74 10-3 W/ K, Pr = 0.75; (T s = 97 K, 1 at): Pr = 0.707. ANAYSIS: The heat transfer rate fro the cylinder, approxiating the person, is given as q = has Ts T where As = π l and h ust be estiated fro a correlation appropriate to cross-flow over a cylinder. Use the Zhukauskas relation, h Nu n = = C Re Pr Pr/Prs k and calculate the Reynold s nuber, 1/4 V 15/s 0.3 Re = = = 345,09. ν 13.04 10-6 /s Fro Table 7-4, find C = 0.076 and = 0.7. Since Pr 10, n = 0.37, giving 1/4 0.7 0.37 0.75 Nu = 0.076 ( 345,09) 0.75 511 0.707 = k 511 3.74 10 3 W/ K h = Nu = = 40.4 W/ K. 0.3 The heat transfer rate is q 40.4W/ o = K 0.3 1.8 4 5 C= 1988 W. ( π ) COMMENTS: Note the teperatures at which properties are evaluated for the Zhukauskas correlation.
PROBEM 7.59 KNOWN: Mercury-in-glass theroeter ounted on duct wall used to easure air teperature. FIN: (a) Relationship for the iersion error, T i = T() - T as a function of air velocity, theroeter diaeter and length, (b) ength of insertion if T i is not to exceed 0.5 C when the air velocity is 10 /s, (c) For the length of part (b), calculate and plot T i as a function of air velocity for to 0 /s, and (d) For a given insertion length, will T i increase or decrease with theroeter diaeter increase; is T i ore sensitive to diaeter or velocity changes? ASSUMPTIONS: (1) Steady-state conditions, () Theroeter approxiates a one-diensional (glass) fin with an adiabatic tip, (3) Convection coefficient is unifor over length of theroeter. PROPERTIES: Table A.3, Glass (300 K): k g = 1.4 W/ K; Table A.4, Air (T f = (15 + 77) C/ 30 K, 1 at): k = 0.078 W/ K, ν = 17.90 10-6 /s, Pr = 0.704. ANAYSIS: (a) Fro the analysis of a one-diensional fin, see Table 3.4, T T 1 = Tb T cosh where P = π and A c = π /4. Hence, the iersion error is hp 4h = kga = c kg (1) Ti = T T = Tb T cosh. () Using the Hilpert correlation for the circular cylinder in cross flow, k 1/3 1/3 k V 1/3 kpr h = CRe 1 Pr = C Pr C V ν = (3) ν 1/3 1 kpr h = N V where N= C (4,5) ν Substituting into Eq. (), the iersion error is 1/ T i( V,,) ( Tb T ) cosh ( 4 kg) N V = (6) where k g is the theral conductivity of the glass theroeter. (b) When the air velocity is 10 /s, find V 10 s 0.004 Re = = = 35 ν 17.9 10 6 s Continued...
PROBEM 7.6 KNOWN: ong coated plastic, 0- diaeter rod, initially at a unifor teperature of T i = 5 C, is suddenly exposed to the cross-flow of air at T = 350 C and V = 50 /s. FIN: (a) Tie for the surface of the rod to reach 175 C, the teperature above which the special coating cures, and (b) Copute and plot the tie-to-reach 175 C as a function of air velocity for 5 V 50 /s. ASSUMPTIONS: (a) One-diensional, transient conduction in the rod, () Constant properties, and (3) Evaluate therophysical properties at T f = [(T s + T i )/ + T ] = [(175 + 5)/ + 350] C = 5 C = 500 K. PROPERTIES: Rod (Given): ρ = 00 kg/ 3, c = 800 J/kg K, k = 1 W/ K, α = k/ρc = 5.68 10-7 /s; Table A.4, Air (T f 500 K, 1 at): ν = 38.79 10-6 /s, k = 0.0407 W/ K, Pr = 0.684. ANAYSIS: (a) To deterine whether the luped capacitance ethod is valid, deterine the Biot nuber h( ro ) Bilc = k The convection coefficient can be estiated using the Churchill-Bernstein correlation, Eq. 7.57, 4/5 1/ 1/3 5/8 h 0.63 Re Pr Re Nu = = 0.3 + 1 + k /3 1/4 8, 000 1+ ( 0.4 Pr) Re V 6 = = 50 s 0.00 38.79 10 s = 5,780 ν (1) 4/5 1/ 1/3 5/8 0.63( 5, 780) ( 0.684) 1/4 /3 1+ ( 0.4 0.684) 0.0407 W K 5, 780 h = 0.3+ 1+ 0.00 8, 000 = 184 W/ K() Substituting for h fro Eq. () into Eq. (1), find Bi lc = 184 W K( 0.010 ) 1W K = 0.9 >> 0.1 Hence, the luped capacitance ethod is inappropriate. Using the one-ter series approxiation, Section 5.6., Eqs. 5.49 with Table 5.1, θ * = C * * 1exp ζ 1 Fo Jo ζ1r r = r ro = 1 ( o ) ( ) i ( $ ) * T r, t T 175 350 C θ = = = 0.54 T T 5 350 C $ Bi = hro k = 1.84 ζ1= 1.5308 rad C1= 1.3384 Continued...
PROBEM 7.6 (Cont.) 0.54 = 1.3384exp[-(1.5308rad) Fo]J o (1.5308 1) Using Table B.4 to evaluate J o (1.5308) = 0.4944, find Fo = 0.0863 where αt 7 o 5.68 10 s t Fo = = o = 5.68 10 3 t o ro ( 0.010 ) (6) to = 15.s (b) Using the IHT Model, Transient Conduction, Cylinder, and the Tool, Correlations, External Flow, Cylinder, results for the tie-to-reach a surface teperature of 175 C as a function of air velocity V are plotted below. 100 80 Tie, to (s) 60 40 0 0 0 10 0 30 40 50 Air velocity, V (/s) COMMENTS: (1) Using the IHT Tool, Correlations, External Flow, Cylinder, the effect of the fil teperature T f on the estiated convection coefficient with V = 50 /s can be readily evaluated. T f (K) 460 500 63 h (W/ K) 187 184 176 At early ties, h = 184 W/ K is a good estiate, while as the cylinder teperature approaches the airstea teperature, the effect starts to be noticeable (10% decrease). () The IHT analysis perfored for part (b) was developed in two parts. Using a known value for h, the Transient Conduction, Cylinder Model was tested. Separately, the Correlation Tools was assebled and tested. Then, the two files were erged to give the workspace for deterining the tie-to-reach 175 C as a function of velocity V.
PROBEM 7.68 KNOWN: Conditions associated with airflow over a spherical light bulb of prescribed diaeter and surface teperature. FIN: Heat loss by convection. ASSUMPTIONS: (1) Steady-state conditions, () Unifor surface teperature. PROPERTIES: Table A-4, Air (T f = 5 C, 1 at): ν = 15.71 10-6 /s, k = 0.061 W/ K, Pr = 0.71, µ = 183.6 10-7 N s/ ; Table A-4, Air (T s = 140 C, 1 at): µ = 35.5 10-7 N s/. ANAYSIS: The heat rate by convection is ( π ) ( s ) q = h T T where h ay be estiated fro the Whitaker relation k h = + 0.4 Re + 0.06 Re Pr / where V 0.5 /s 0.05 Re = = = 1591. ν 15.71 10-6 / s Hence, ( 1/ /3 ) 0.4 ( s ) 1/4 µ µ 1/4 0.061 W/ K 1/ /3 0.4 183.6 h = 0.4( 1591) 0.06( 1591 ) + + ( 0.71) 0.05 35.5 h = 11.4 W/ K and the heat rate is W q = 11.4 π ( 0.05 ) ( 140 5) C = 10.3 W. K $ COMMENTS: (1) The low value of h suggests that heat transfer by free convection ay be significant and hence that the total loss by convection exceeds 10.3 W. () The surface of the bulb also dissipates heat to the surrounding by radiation. Further, in an actual light bulb, there is also heat loss by conduction through the socket.
PROBEM 7.73 KNOWN: iaeter and initial and final teperatures of copper spheres quenched in a water bath. FIN: (a) Terinal velocity in the bath, (b) Tank height. ASSUMPTIONS: (1) Sphere descends at terinal velocity, () Unifor, but tie varying surface, teperature. PROPERTIES: Table A-1, Copper (350K): ρ = 8933 kg/ 3, k = 398 W/ K, c p = 387 J/kg K; Table A-6, Water (T = 80 K): ρ = 1000 kg/ 3, µ = 14 10-6 N s/, k = 0.58 W/ K, Pr = 10.6; (T s 340 K): µ s = 40 10-5 N s/. C 3 π /4 ρv / ρcu ρ g π /6, ANAYSIS: A force balance gives = ( ) 4 ρcu ρ 4 0.0 89331000 C V = g = 9.8 /s =.07 /s. 3 ρ 3 1000 An iterative solution is needed, where C is obtained fro Figure 7.8 with Re = V/ν = 0.0 V/1.4 10-6 /s = 14,085 V (/s). Convergence is achieved with V.1 /s for which Re = 9,580 and C 0.46. Using the Whitaker expression 0.4 1/4 Nu 1/ /3 = + 0.4 9,850 + 0.06 9,850 10.6 14/40 = 439 h = Nu k/= 439 0.58 W/ K/0.0 = 1,775 W/ K. Bi= h r o/3/kcu = 1,775 W/ K 0.01 /3 /398W/ K= 0.11. Applicability is arginal. Use Heisler charts, To deterine applicability of luped capacitance ethod, find To T 30 80-1 k αt θ f o = = = 0.5, Bi = = 3.1, Fo 0.88 =. T i T 36080 hro ro With α cu = k/ρc p = 398 W/ K/(8933 kg/ 3 ) (387 J/kg K) = 1.15 10-4 /s, find t 4 f = 0.88 0.01 /1.15 10 /s= 0.77 s. Required tank height is H = tf V = 0.77 s.1 /s= 1.6. θo = C 1 exp ζ1 Fo, we COMMENTS: If t f is evaluated fro the approxiate series solution, obtain t f = 0.76 s. Note that the terinal velocity is not reached iediately. Reduced V iplies reduced h and increased t f.