PROBLEM 9.11KNOWN: Vertical plate experiencing free convection with quiescent air at atmospheric pressureand film temperature 400 K.FIND: Form of correlation for average heat transfer coefficient in terms of T and characteristiclength.SCHEMATIC:

ASSUMPTIONS: (1) Air is extensive, quiescent medium, (2) Perfect gas behavior.PROPERTIES: Table A-6, Air (Tf = 400K, 1 atm): = 26.41 10-6 m2/s, k = 0.0338 W/mK, = 38.3 10-6 m2/s.ANALYSIS: Consider the correlation having the form of Eq. 9.24 with RaL defined by Eq. 9.25.

nNu h L/k CRaLL L (1)where

3 2 3g T T L 9.8 m/s 1/400 K T Ls 7 3Ra 2.422 10 T L .L 6 2 6 226.41 10 m /s 38.3 10 m / s (2)

Combining Eqs. (1) and (2), n0.0338 W/m Kn 7 3h k/L CRa C 2.422 10 TL .L L L

(3)From Fig. 9.6, note that for laminar boundary layer conditions, 104 < RaL < 109, C = 0.59 and n =1/4. Using Eq. (3), 1 / 41 / 4 T1 3h 1.40 L T L 1.40 L

<For turbulent conditions in the range 109 < RaL < 1013, C = 0.10 and n = 1/3. Using Eq. (3), 1 / 31 3 1/3h 0.98 L T L 0.98 T .L

<COMMENTS: Note the dependence of the average heat transfer coefficient on T and L forlaminar and turbulent conditions. The characteristic length L does not influence hL for turbulentconditions.

PROBLEM 9.12KNOWN: Temperature dependence of free convection coefficient, 1/ 4h C T , for a solid suddenlysubmerged in a quiescent fluid.FIND: (a) Expression for cooling time, tf, (b) Considering a plate of prescribed geometry and thermalconditions, the time required to reach 80 C using the appropriate correlation from Problem 9.10 and (c)Plot the temperature-time history obtained from part (b) and compare with results using a constant ohfrom an appropriate correlation based upon an average surface temperature i fT T T 2 .SCHEMATIC:

ASSUMPTIONS: (1) Lumped capacitance approximation is valid, (2) Negligible radiation, (3)Constant properties.PROPERTIES: Table A.1, Aluminum alloy 2024 i fT T T 2 400 K : = 2770 kg/m3, cp =925 J/kg K, k = 186 W/m K; Table A.4, Air ( filmT = 362 K): = 2.221 10-5 m2/s, k = 0.03069 W/m K,

= 3.187 10-5 m2/s, Pr = 0.6976, = 1/ filmT .ANALYSIS: (a) Apply an energy balance to a control surface about the object, out stE E , andsubstitute the convection rate equation, with 1/ 4h C T , to find

5/ 4sCA T T d/dt VcT . (1)Separating variables and integrating, find

5/ 4sdT/dt CA Vc T Tf fi

T ts5 / 4T 0CAdT dtVcT T

fi

T1/ 4 s fTCA4 T T tVc

1/ 4 1/ 4f f is4 Vct T T T TCA

1/ 4i1/ 4 fs i4 Vc T T 1T TCA T T . (2) <

(b) Considering the aluminum plate, initially at T(0) = 225 C, and suddenly exposed to ambient airat T 25 C , from Problem 9.10 the convection coefficient has the form

1/ 4i th 1.40 L

1/ 4ih C T

where C = 1.40/L1/4 = 1.40/(0.150)1/4 = 2. 2496 3/ 42W m K . Using Eq. (2), findContinued...

PROBLEM 9.19KNOWN: Room and ambient air conditions for window glass.FIND: Temperature of the glass and rate of heat loss.SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible temperature gradients in the glass, (3)Inner and outer surfaces exposed to large surroundings.PROPERTIES: Table A.4, air (Tf,i and Tf,o): Obtained from the IHT Properties Tool Pad.ANALYSIS: Performing an energy balance on the window pane, it follows that in outE E , or

4 4 4 4i ,i sur o ,osur,iT T h T T T T h T T

where ih and oh may be evaluated from Eq. 9.26.2

1/ 6LL 8/ 279 /160.387RaNu 0.825

1 0.492 Pr

Using the First Law Model for an Isothermal Plane Wall and the Correlations and Properties Tool Padsof IHT, the energy balance equation was formulated and solved to obtain

T = 273.8 K <The heat rate is then qi = qo, or

2 4 4i isur,iq L T T h T T 174.8 W <COMMENTS: The radiative and convective contributions to heat transfer at the inner and outersurfaces are qrad,i = 99.04 W, qconv,i = 75.73 W, qrad,o = 86.54 W, and qconv,o = 88.23 W, with correspondingconvection coefficients of ih = 3.95 W/m2 K and oh = 4.23 W/m2 K. The heat loss could be reducedsignificantly by installing a double pane window.

PROBLEM 9.29KNOWN: Dimensions, interior surface temperature, and exterior surface emissivity of a refrigeratordoor. Temperature of ambient air and surroundings.FIND: (a) Heat gain with no insulation, (b) Heat gain as a function of thickness for polystyreneinsulation.SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible thermal resistance of steel andpolypropylene sheets, (3) Negligible contact resistance between sheets and insulation, (4) One-dimensional conduction in insulation, (5) Quiescent air.PROPERTIES: Table A.4, air (Tf = 288 K): = 14.82 10-6 m2/s, = 20.92 10-6 m2/s, k = 0.0253W/m K, Pr = 0.71, = 0.00347 K-1.ANALYSIS: (a) Without insulation, Ts,o = Ts,i = 278 K and the heat gain is

4 4wo s s,i s sur s,iq hA T T A T Twhere As = HW = 0.65 m2. With a Rayleigh number of RaH = 3s,ig T T H = 9.8 m/s2(0.00347K-1)(20 K)(1)3/(20.92 10-6 m2/s)(14.82 10-6 m2/s) = 2.19 109, Eq. 9.26 yields

21/ 69H 8/ 279 /16

0.387 2.19 10Nu 0.825 156.61 0.492 0.71

2Hh Nu k H 156.6 0.0253W m K 1m 4.0 W m K2 2 8 2 4 2 4 4 4woq 4.0 W m K 0.65 m 20 K 0.6 5.67 10 W m K 0.65 m 298 278 K

woq 52.00 42.3 W 94.3W <(b) With the insulation, Ts,o may be determined by performing an energy balance at the outer surface,where conv rad condq q q , or

4 4 is,o sur s,o s,o s,ikh T T T T T TLUsing the IHT First Law Model for a Nonisothermal Plane Wall with the appropriate Correlations andProperties Tool Pads and evaluating the heat gain from

Continued...

PROBLEM 9.29 (Cont.)i sw s,o s,ik Aq T TL

the following results are obtained for the effect of L on Ts,o and qw.

The outer surface temperature increases with increasing L, causing a reduction in the rate of heat transferto the refrigerator compartment. For L = 0.025 m, h = 2.29 W/m2 K, hrad = 3.54 W/m2 K, qconv = 5.16 W,qrad = 7.99 W, qw = 13.15 W, and Ts,o = 21.5 C.COMMENTS: The insulation is extremely effective in reducing the heat load, and there would be littlevalue to increasing L beyond 25 mm.

PROBLEM 9.34KNOWN: Surface temperature of a long duct and ambient air temperature.FIND: Heat gain to the duct per unit length of the duct.SCHEMATIC:

ASSUMPTIONS: (1) Surface radiation effects are negligible, (2) Ambient air is quiescent.PROPERTIES: Table A-4, Air (Tf = (T + Ts)/2 300K, 1 atm): = 15.89 10-6 m2/s, k =0.0263 W/mK, = 22.5 10-6 m2/s, Pr = 0.707, = 1/Tf.ANALYSIS: The heat gain to the duct can be expressed as s t b s t b sq 2q q q 2 h H h W h W T T . (1)Consider now correlations to estimate s t bh , h , and h . From Eq. 9.25, for the sides with L H,

33 2s 7L 6 2 6 2g T T L 9.8m/s 1/300K 35 10 K 0.2mRa 1.827 10 .15.89 10 m / s 22.5 10 m / s

(2)

Eq. 9.27 is appropriate to estimate sh ,

L

1/471/4L 4 /9 4/99/16 9/160.670 1.827 100.670RaNu 0.68 0.68 34.29

1 0.492/Pr 1 0.492/0.707

L 2sh Nu k / L 34.29 0.0263W/m K/0.2m 4.51W/m K. (3)

For the top and bottom portions of the duct, L As/P W/2, (see Eq. 9.29), find the Rayleigh numberfrom Eq. (2) with L = 0.1 m, RaL = 2.284 106. From the correlations, Eqs. 9.31 and 9.32 for the topand bottom surfaces, respectively, find

1/31/3 6 2t Lk 0.0263W/m Kh 0.15Ra 0.15 2.284 10 5.17W/m K.W / 2 0.1m (4)

1/46 2b k 0.0263W/m Kh 0.27 2.284 10 2.76W/m K.W / 2 0.1m (5)

The heat rate, Eq. (1), can now be evaluated using the heat transfer coefficients estimated from Eqs.(3), (4), and (5). 2 2 2q 2 4.51W/m K 0.2m 5.17W/m K 0.2m 2.76W/m K 0.2m 35 10 K

q 84.8W/m. <COMMENTS: Radiation surface effects will be significant in this situation. With knowledge of theduct emissivity and surroundings temperature, the radiation heat exchange could be estimated.

PROBLEM 9.38KNOWN: Width and thickness of sample material. Rate of heat dissipation at bottom surface ofsample and temperatures of top and bottom surfaces. Temperature of quiescent air and surroundings.FIND: Thermal conductivity and emissivity of the sample.SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in sample, (3) Quiescent air, (4)Sample is small relative to surroundings, (5) All of the heater power dissipation is transferred throughthe sample, (6) Constant properties.PROPERTIES: Table A-4, air (Tf = 335.5K): = 19.5 10-6 m2/s, k = 0.0289 W/m K, = 27.8 10-6 m2/s, Pr = 0.703, = 0.00298 K-1.ANALYSIS: The thermal conductivity is readily obtained by applying Fourier’s law to the sample.Hence, with q = Pelec, 22elec

1 270 W / 0.250mP / Wk 0.560 W / m KT T / L 50 C / 0.025m <

The surface emissivity may be obtained by applying an energy balance to a control surface about thesample, in which case

4 4 2elec conv rad 2 sur2P q q h T T T T W2elec 24 4sur2

P / W h T TT T

With L = As/P = W2/4W = W/4 = 0.0625m, RaL = g (T2 - T ) L3/ = 9.86 105 and Eq. 9.30yields

L 1/ 41/ 4 5 2LNu k k 0.0289 W / m Kh 0.54 Ra 0.54 9.86 10 7.87 W / m KL L 0.0625m <

Hence,2 2

8 2 4 4 4 470 W / 0.250m 7.87 W / m K 75 C 0.8155.67 10 W / m K 373 298 K <

COMMENTS: The uncertainty in the determination of is strongly influenced by uncertaintiesassociated with using Eq. 9.30. If, for example, h is overestimated by 10%, the actual value of would be 0.905.

PROBLEM 9.40KNOWN: Horizontal, circular grill of 0.2m diameter with emissivity 0.9 is maintained at a uniformsurface temperature of 130C when ambient air and surroundings are at 24C.FIND: Electrical power required to maintain grill at prescribed surface temperature.SCHEMATIC:

ASSUMPTIONS: (1) Room air is quiescent, (2) Surroundings are large compared to grill surface.PROPERTIES: Table A-4, Air (Tf = (T + Ts)/2 = (24 + 130)C/2 = 350K, 1 atm): = 20.92 10-6 m2/s, k = 0.030 W/mK, = 29.9 10-6 m2/s, = 1/Tf.ANALYSIS: The heat loss from the grill is due to free convection with the ambient air and toradiation exchange with the surroundings.

4 4q A h T T T T .s s s sur (1)Calculate RaL from Eq. 9.25,

3Ra g T T L /L s c where for a horizontal disc from Eq. 9.29, Lc = As/P = (D2/4)/D = D/4. Substituting numericalvalues, find

329.8m/s 1/350K 130 24 K 0.25m/4 6Ra 1.158 10 .L 6 2 6 220.92 10 m / s 29.9 10 m / s

Since the grill is an upper surface heated, Eq. 9.30 is the appropriate correlation, 1 / 41/4 6Nu h L / k 0.54Ra 0.54 1.158 10 17.72L cL L 2h Nu k / L 17.72 0.030W/m K / 0.25m/4 8.50W/m K.L cL (2)

Substituting from Eq. (2) for h into Eq. (1), the heat loss or required electrical power, qelec, is

W W8 42 4 4q 0.25m 8.50 130 24 K 0.9 5.67 10 130 273 24 273 K2 2 44 m K m K

q 44.2W 46.0W 90.2W. <COMMENTS: Note that for this situation, free convection and radiation modes are of equalimportance. If the grill were highly polished such that 0.1, the required power would be reduced bynearly 50%.

PROBLEM 9.45KNOWN: Diameter, thickness, emissivity and initial temperature of silicon wafer. Temperature ofair and surrounding.FIND: (a) Initial cooling rate, (b) Time required to achieve prescribed final temperature.SCHEMATIC:

ASSUMPTIONS: (1) Negligible heat transfer from side of wafer, (2) Large surroundings, (3) Wafermay be treated as a lumped capacitance, (4) Constant properties, (5) Quiescent air.PROPERTIES: Table A-1, Silicon ( T = 187 C = 460K): = 2330 kg/m3, cp = 813 J/kg K, k = 87.8W/m K. Table A-4, Air (Tf,i = 175 C = 448K): = 32.15 10-6 m2/s, k = 0.0372 W/m K, = 46.8 10-6 m2/s, Pr = 0.686, = 0.00223 K-1.SOLUTION: (a) Heat transfer is by natural convection and net radiation exchange from top andbottom surfaces. Hence, with As = D2/4 = 0.0177 m2,

4 4s t b i suriq A h h T T 2 T Twhere the radiation flux is obtained from Eq. 1.7, and with L = As/P = 0.0375m and RaL = g (Ti -T ) L3/ = 2.30 105, the convection coefficients are obtained from Eqs. 9.30 and 9.32. Hence,

1/ 4 2t Lk 0.0372 W / m K 11.8h 0.54Ra 11.7 W / m KL 0.0375m

1/ 4 2b Lk 0.0372 W / m K 5.9h 0.27 Ra 5.9 W / m KL 0.0375m

2 2 8 2 4 4 4 4q 0.0177 m 11.7 5.9 W / m K 300K 2 0.65 5.67 10 W / m K 598 298 K2 2q 0.0177 m 5280 8845 W / m 250 W <

(b) From the generalized lumped capacitance model, Eq. 5.15,4 4s t b sur sdTcA h h T T 2 T T Adt

4 4t b surT tTi 0

h h T T 2 T TdT dtc

Continued …..

PROBLEM 9.45 (Cont.)Using the DER function of IHT to perform the integration, thereby accounting for variations in thand bh with T, the time tf to reach a wafer temperature of 50 C is found to be

ft T 320K 181s <

As shown above, the rate at which the wafer temperature decays with increasing time decreases due toreductions in the convection and radiation heat fluxes. Initially, the surface radiative flux (top orbottom) exceeds the heat flux due to natural convection from the top surface, which is twice the fluxdue to natural convection from the bottom surface. However, because radq and cnvq decayapproximately as T4 and T5/4, respectively, the reduction in radq with decreasing T is morepronounced, and at t = 181s, radq is well below cnv,tq and only slightly larger than cnv,bq .COMMENTS: With 2 2 2r,i i sur i surh T T T T 14.7 W / m K, the largest cumulativecoefficient of 2tot r,i t,ih h h 26.4 W / m K corresponds to the top surface. If this coefficient isused to estimate a Biot number, it follows that 4totBi h / 2 / k 1.5 10 1 and the lumpedcapacitance approximation is excellent.

PROBLEM 9.53KNOWN: Horizontal rod immersed in water maintained at a prescribed temperature.FIND: Free convection heat transfer rate per unit length of the rod, qconvSCHEMATIC:

ASSUMPTIONS: (1) Water is extensive, quiescent medium.PROPERTIES: Table A-6, Water (Tf = (Ts + T)/2 = 310K): = 1/vf = 993.0 kg/m3, = / =695 10-6 Ns/m2/993.0 kg/m3 = 6.999 10-7 m2/s, = k/c = 0.628 W/mK/993.0 kg/m3 4178J/kgK = 1.514 10-7 m2/s, Pr = 4.62, = 361.9 10-6 K-1.ANALYSIS: The heat rate per unit length by free convection is given as q h D T T .conv D s (1)To estimate hD , first find the Rayleigh number, Eq. 9.25,

32 6 13 9.8m/s 361.9 10 K 56 18 K 0.005mg T T Ds 5Ra 1.587 10D 7 2 7 26.999 10 m /s 1.514 10 m / s

and use Eq. 9.34 for a horizontal cylinder,

21 / 60.387RaDNu 0.60D 8/279/161 0.599/Pr

21 / 650.387 1.587 10Nu 0.60 10.40D 8/279/161 0.599/4.62

2h Nu k / D 10.40 0.628W/m K/0.005m 1306W/m K.D D (2)

Substituting for hD from Eq. (2) into Eq. (1), 2q 1306W/m K 0.005m 56 18 K 780W/m.conv <

COMMENTS: (1) Note the relatively large value of hD ; if the rod were immersed in air, the heat transfer coefficientwould be close to 5 W/m2K.(2) Eq. 9.33 with appropriate values of C and n from Table 9.1 could also be used to estimate hD . Find 0.25n 5Nu CRa 0.48 1.587 10 9.58D D

2h Nu k / D 9.58 0.628W/m K/0.005m 1203W/m K.D D By comparison with the result of Eq. (2), the disparity of the estimates is ~8%.

PROBLEM 9.54KNOWN: Horizontal, uninsulated steam pipe passing through a room.FIND: Heat loss per unit length from the pipe.SCHEMATIC:

ASSUMPTIONS: (1) Pipe surface is at uniform temperature, (2) Air is quiescent medium, (3)Surroundings are large compared to pipe.PROPERTIES: Table A-4, Air (Tf = (Ts + T)/2 = 350K, 1 atm): = 20.92 10-6 m2/s, k = 0.030W/mK, = 29.9 10-6 m2/s, Pr = 0.700, = 1/Tf = 2.857 10-3 K-1.ANALYSIS: Recognizing that the heat loss from the pipe will be by free convection to the air and byradiation exchange with the surroundings, we can write

4 4q q q D h T T T T .conv rad D s s sur (1)To estimate hD , first find RaL, Eq. 9.25, and then use the correlation for a horizontal cylinder, Eq. 9.34,

33 2g T T D 9.8m/s 1/350K 400 300 K 0.150m 7sRa 1.511 10L 6 2 6 220.92 10 m /s 29.9 10 m / s

21 / 60.387RaLNu 0.60D 8/279/161 0.559/Pr

21 / 670.387 1.511 10Nu 0.60 31.88D 8/279/161 0.559/0.700

2h Nu k / D 31.88 0.030W/m K/0.15m 6.38W/m K.D D (2)

Substituting for hD from Eq. (2) into Eq. (1), find 2 8 2 4 4 4 4q 0.150m 6.38W/m K 400 300 K 0.85 5.67 10 W / m K 400 300 K

q 301W/m 397W/m 698W/m. <COMMENTS: (1) Note that for this situation, heat transfer by radiation and free convection are ofequal importance.(2) Using Eq. 9.33 with constants C,n from Table 9.1, the estimate for hD is 0.333n 7Nu CRa 0.125 1.511 10 30.73LD

2h Nu k / D 30.73 0.030W/m K/0.150m 6.15W/m K.D D The agreement is within 4% of the Eq. 9.34 result.

PROBLEM 9.62KNOWN: Dissipation rate of an immersion heater in a large tank of water.FIND: Surface temperature in water and, if accidentally operated, in air.SCHEMATIC:

ASSUMPTIONS: (1) Quiescent ambient fluid, (2) Negligible radiative exchange.PROPERTIES: Table A-6, Water and Table A-4, Air:

T(K) k103(W/mK) 107(/,m2/s) 107(k/cp,m2/s) Pr 106(K-1)Water 315 634 6.25 1.531 4.16 400.4Air 1500 100 2400 3500 0685 666.7ANALYSIS: From the rate equation, the surface temperature, Ts, is sT T q / DL h (1)where h is estimated by an appropriate correlation. Since such a calculation requires knowledge ofTs, an iteration procedure is required. Begin by assuming for water that Ts = 64C such that Tf =315K. Calculate the Rayleigh number,

32 6 13 6D 7 2 7 29.8m/s 400.4 10 K 64 20 K 0.010mg TDRa 1.804 10 .6.25 10 m / s 1.531 10 m /s

(2)Using the Churchill-Chu relation, find

D

21/6D 8/279/16

0.387RahDNu 0.60k 1 0.559/Pr

(3)

21/6628/279/16

0.387 1.804 100.634W/m Kh 0.60 1301W/m K.0.01m 1 0.559/4.16

Substituting numerical values into Eq. (1), the calculated value for Ts in water is

2sT 20 C 550W/ 0.010m 0.30m 1301W/m K 64.8 C. <Continued …..

PROBLEM 9.78KNOWN: Sphere with embedded electrical heater is maintained at a uniform surface temperaturewhen suspended in various media.FIND: Required electrical power for these media: (a) atmospheric air, (b) water, (c) ethylene glycol.SCHEMATIC:

ASSUMPTIONS: (1) Negligible surface radiation effects, (2) Extensive and quiescent media.PROPERTIES: Evaluated at Tf = (Ts + T)/2 = 330K:

106, m2/s k103, W/mK 106, m2/s Pr 103, K-1

Table A-4, Air (1 atm) 18.91 28.5 26.9 0.711 3.03Table A-6, Water 0.497 650 0.158 3.15 0.504Table A-5, Ethylene glycol 5.15 260 0.0936 55.0 0.65ANALYSIS: The electrical power (Pe) required to offset convection heat transfer is

2conv s s sq h A T T h D T T . (1)The free convection heat transfer coefficient for the sphere can be estimated from Eq. 9.35 using Eq.9.25 to evaluate RaD.

D1/4 3D D4/99/16 11D

Pr 0.70.589Ra g T DNu 2 Ra .

1 0.469/Pr Ra 10

(2,3)

(a) For air 32 3 14D 6 2 6 2

9.8m/s 3.03 10 K 94 20 K 0.025mRa 6.750 1018.91 10 m / s 26.9 10 m / s

D

1/44 2D 4/99/160.589 6.750 10k 0.02285W/m Kh Nu 2 10.6W/m KD 0.025m 1 0.469/0.711

22convq 10.6W/m K 0.025m 94 20 K 1.55W.

Continued …..

PROBLEM 9.78 (Cont.)(b,c) Summary of the calculations above and for water and ethylene glycol:

Fluid RaD 2Dh W / m K q(W)

Air 6.750 104 10.6 1.55 <Water 7.273 107 1299 187 <Ethylene glycol 15.82 106 393 57.0 <COMMENTS: Note large differences in the coefficients and heat rates for the fluids.

PROBLEM 9.79KNOWN: Surface temperature and emissivity of a 20W light bulb (spherical) operating in room airFIND: Heat loss from bulb surface.SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Quiescent room air, (3) Surroundings much largerthan bulb.PROPERTIES: Table A-4, Air (Tf = (Ts + T)/2 = 348K, 1 atm): = 20.72 10-6 m2/s, k = 0.0298W/mK, = 29.6 10-6 m2/s, Pr = 0.700, = 1/Tf.ANALYSIS: Heat loss from the surface of the bulb is by free convection and radiation. The rateequations are

4 4conv rad s s s s surq q q h A T T A T T where As = D2. To estimate h for free convection, first evaluate the Rayleigh number.

323 5D 6 2 6 29.8m/s 1/348K 125 25 K 0.040mg T DRa 2.93 10 .20.72 10 m / s 29.6 10 m / s

Since Pr 0.7 and RaD < 1011, the Churchill relation, Eq. 9.35, is appropriate.

D

1/451/4D 4/9 4/99/16 9/160.589 2.93 100.589RaNu 2 2 12.55

1 0.469/Pr 1 0.469/0.700

D 2h Nu k / D 12.55 0.0298W/m K /0.040m 9.36W/m K. Substituting numerical values, the heat loss from the bulb is,

8 42 4 42 2 4W Wq 0.040m 9.36 125 25 K 0.80 5.67 10 125 273 25 273 K

m K m K

q 4.70 3.92 W 8.62W. <

COMMENTS: (1) The contributions of convection and radiation to the surface heat loss arecomparable.(2) The remaining heat loss (20 – 8.62 = 11.4 W) is due to the transmission of radiant energy (light)through the bulb and heat conduction through the base.

PROBLEM 9.80KNOWN: A copper sphere with a diameter of 25 mm is removed from an oven at a uniformtemperature of 85 C and allowed to cool in a quiescent fluid maintained at 25 C.FIND: (a) Convection coefficients for the initial condition for the cases when the fluid is air andwater, and (b) Time for the sphere to reach 30 C when the cooling fluid is air and water using twodifferent approaches, average coefficient and numerically integrated energy balance.SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions for part (a); (2) Low emissivity coating makesradiation exchange negligible for the in-air condition; (3) Fluids are quiescent, and (4) Constantproperties.PROPERTIES: Table A-4, Air (Tf = (25 + 85) C/2 = 328 K, 1 atm): = 1.871 10-5 m2/s, k =0.0284 W/m K, = 2.66 10-5 m2/s, Pr = 0.703, = 1/Tf; Table A-6, Water (Tf = 328 K): = 5.121

10-7 m2/s, k = 0.648 W/m K, = 1.57 10-7 m2/s, Pr = 3.26, = 4.909 10-4 K-1; Table A-1,Copper, pure ( T = (25 + 85) C/2 = 328 K): = 8933 kg/m3, c = 382 J/kg K, k = 399 W/m K.ANALYSIS: (a) For the initial condition, the average convection coefficient can be estimated fromthe Churchill-Chu correlation, Eq. 9.35,

D1/ 4D D

4 / 99 /16h D 0.589 RaNu 2k 1 0.469 / Pr

(1)

3sD g T T DRa (2)with properties evaluated at Tf = (Ts + T )/2 = 328 K. Substituting numerical values find theseresults:

Fluid Ts( C) Tf(K) RaD NuD Dh (W/m2 K)Air 85 328 5.62 104 8.99 10.2 <

Water 85 328 5.61 107 46.8 1213 <(b) To establish the validity of the lumped capacitance (LC) method, calculate the Biot number for theworst condition (air).

2 4Dh D / 3Bi 10.2 W / m K 0.025 m / 3 / 399 W / m K 2.1 10kSince Bi << 0.1, the sphere can be represented by this energy balance for the cooling process

sin out st cv dTE E E q Mc dtsD s s dTh A T T Vc dt (3)

where As = D2 and V = D3/6. Two approaches are considered for evaluating appropriate values for Dh .Average coefficient. Evaluate the convection coefficient corresponding to the average temperature ofthe sphere, sT = (30 + 85) C/2 = 57.5 C, for which the film temperature is Tf = ( sT + T )/2. Usingthe foregoing analyses of part (a), find these results.

Continued …..

PROBLEM 9.80 (Cont.)Fluid sT C Tf(K) RaD NuD Dh (W/m2 K)Air 57.5 314 3.72 104 8.31 9.09

Water 57.5 314 1.99x107 37.1 940Numerical integration of the energy balance equation. The more accurate approach is to numericallyintegrate the energy balance equation, Eq. (3), with Dh evaluated as a function of Ts using Eqs. (1)and (2). The properties in the correlation parameters would likewise be evaluated at Tf, which varieswith Ts. The integration is performed in the IHT workspace; see Comment 3.Results of the lumped-capacitance analysis. The results of the LC analyses using the two approachesare tabulated below, where to is the time to cool from 85 C to 30 C:

to (s)Approach Air Water

Average coefficient 3940 39Numerical coefficient 4600 49

COMMENTS: (1) For these condition, the convection coefficient for the water is nearly two ordersof magnitude higher than for air.(2) Using the average-coefficient approach, the time-to-cool, to, values for both fluids is 15-20%faster than the more accurate numerical integration approach. Evaluating the average coefficient at

sT results in systematically over estimating the coefficient.(3) The IHT code used for numerical integration of the energy balance equation and the correlations isshown below for the fluid water.