PROBLEM 9.55 KNOWN: Diameter and outer surface temperature of steam pipe. Diameter, thermal conductivity, and emissivity of insulation. Temperature of ambient air and surroundings. FIND: Effect of insulation thickness and emissivity on outer surface temperature of insulation and heat loss. SCHEMATIC: See Example 9.4, Comment 2. ASSUMPTIONS: (1) Pipe surface is small compared to surroundings, (2) Room air is quiescent. PROPERTIES: Table A.4, air (evaluated using Properties Tool Pad of IHT). ANALYSIS: The appropriate model is provided in Comment 2 of Example 9.4 and includes use of the following energy balance to evaluate T s,2 , condconvrad qqqq ′′′′ =+≡ ( ) ( ) ( ) ( ) is,1s,2 44 2s,22sur s,2 21 2kTT h2rTT2rTT lnrr π πεπσ ∞ − =−+− from which the total heat rate q′ can then be determined. Using the IHT Correlations and Properties Tool Pads, the following results are obtained for the effect of the insulation thickness, with ε = 0.85. 0 0.01 0.02 0.03 0.04 0.05 Insulation thickness, t(m) 20 50 80 110 140 170 Surface temperature, Ts,2(C) 0 0.01 0.02 0.03 0.04 0.05 Insulation thickness, t(m) 0 100 200 300 400 500 600 700 800 Heat loss, q'(W/m) The insulation significantly reduces T s,2 and q′ , and little additional benefits are derived by increasing t above 25 mm. For t = 25 mm, the effect of the emissivity is as follows. Continued PROBLEM 9.55 (Cont.) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Emissivity, eps 34 36 38 40 42 44 Surface temperature, Ts,2(C) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Emissivity, eps 48 49 50 51 52 53 Heat loss, q'(W/m) Although the surface temperature decreases with increasing emissivity, the heat loss increases due to an increase in net radiation to the surroundings. PROBLEM 9.56 KNOWN: Dimensions and temperature of beer can in refrigerator compartment. FIND: Orientation which maximizes cooling rate. SCHEMATIC: ASSUMPTIONS: (1) End effects are negligible, (2) Compartment air is quiescent, (3) Constant properties. PROPERTIES: Table A-4, Air (T f = 288.5K, 1 atm): ν = 14.87 × 10 -6 m 2 /s, k =0.0254 W/m⋅K, α = 21.0 × 10 -6 m 2 /s, Pr = 0.71, β = 1/T f = 3.47 × 10 -3 K -1 . ANALYSIS: The ratio of cooling rates may be expressed as ( ) ( ) s vvv hhsh TT qhh DL . qhDLTTh π π ∞ ∞ − == − For the vertical surface, find ( ) ( ) ( )( ) 231 s 3393 L 6262 gTT9.8m/s3.4710K23C RaLL2.510L 14.8710m/s2110m/s β να −− ∞ −− −××° ===× ×× ( ) 3 96 L Ra2.5100.158.4410, =×=× and using the correlation of Eq. 9.26, ( ) ( ) L 2 1/6 6 8/27 9/16 0.3878.4410 Nu0.82529.7. 10.492/0.71 × =+= +           Hence L 2 Lv k0.0254W/mK hhNu29.75.03W/mK. L0.15m ⋅ ====⋅ For the horizontal surface, find ( ) ( ) 3s 395 D gTT RaD2.5100.065.410 β να ∞ − ==×=× and using the correlation of Eq. 9.34, ( ) ( ) D 2 1/6 5 8/27 9/16 0.3875.410 Nu0.6012.24 10.559/0.71 × =+= +           D 2 Dh k0.0254W/mK hhNu12.245.18W/mK. D0.06m ⋅ ====⋅ Hence v h q 5.03 0.97. q5.18 == < COMMENTS: In view of the uncertainties associated with Eqs. 9.26 and 9.34 and the neglect of end effects, the above result is inconclusive. The cooling rates are approximately the same. PROBLEM 9.57 KNOWN: Length and diameter of tube submerged in paraffin of prescribed dimensions. Properties of paraffin. Inlet temperature, flow rate and properties of water in the tube. FIND: (a) Water outlet temperature, (b) Heat rate, (c) Time for complete melting. SCHEMATIC: ASSUMPTIONS: (1) Negligible k.e. and p.e. changes for water, (2) Constant properties for water and paraffin, (3) Negligible tube wall conduction resistance, (4) Free convection at outer surface associated with horizontal cylinder in an infinite quiescent medium, (5) Negligible heat loss to surroundings, (6) Fully developed flow in tube. PROPERTIES: Water (given): c p = 4185 J/kg⋅K, k = 0.653 W/m⋅K, µ = 467 × 10 -6 kg/s⋅m, Pr = 2.99; Paraffin (given): T mp = 27.4°C, h sf = 244 kJ/kg, k = 0.15 W/m⋅K, β = 8 × 10 -4 K -1 , ρ = 770 kg/m 3 , ν = 5 × 10 -6 m 2 /s, α = 8.85 × 10 -8 m 2 /s. ANALYSIS: (a) The overall heat transfer coefficient is io 111 . Uhh =+ To estimate i h, find D 6 4m40.1kg/s Re10,906 D 0.025m46710kg/sm πµ π − × === ×××⋅ & and noting the flow is turbulent, use the Dittus-Boelter correlation ( ) ( ) 4/50.3 4/50.3 D D Nu0.023RePr0.02310,9062.9954.3 === 2 D i Nuk54.30.653W/mK h1418W/mK. D0.025m ×⋅ ===⋅ To estimate o h, find ( ) ( ) ( ) ( ) 3 241 3 s D 6282 9.8m/s810K5527.4K0.025m gTTD Ra 510m/s8.8510m/s β να −− ∞ −− ×− − == ××× 6 D Ra7.6410 =× and using the correlation of Eq. 9.34, ( ) D 2 1/6 D 8/27 9/16 0.387Ra Nu0.6035.0 10.559/Pr =+= +           D 2 o k0.15W/mK hNu35.0210W/mK. D0.025m ⋅ ===⋅ Alternatively, using the correlation of Eq. 9.33, Continued … PROBLEM 9.57 (Cont.) n DD D NuCRawithC0.48,n0.25Nu25.2 ==== 2 o 0.15W/mK h25.2151W/mK. 0.025m ⋅ ==⋅ The significant difference in h o values for the two correlations may be due to difficulties associated with high Pr applications of one or both correlations. Continuing with the result from Eq. 9.34, 32 io 11111 5.46710mK/W Uhh1418210 − =+=+=×⋅ 2 U183W/mK. =⋅ Using Eq. 8.46, find m,o 2 m,ip TT DL0.025m3mW expUexp183 TTmc0.1kg/s4185J/kgK mK ππ ∞ ∞  −  ×× =−=−   −×⋅ ⋅   & ( ) ( ) m,om,i TTTT0.90227.427.4600.902C ∞∞  =−−=−−°  m,o T56.8C. =° < (b) From an energy balance, the heat rate is ( ) ( ) pm,im,o qmcTT0.1kg/s4185J/kgK6056.8K1335W=−=×⋅−= & < or using the rate equation, ( ) ( ) ( ) 2 m 6027.4K56.827.4K qUAT183W/mK0.025m3m 6027.4 n 56.827.4 π −−− =∆=⋅ − − l l q1335W.= (c) Applying an energy balance to a control volume about the paraffin, inst EE =∆ 2 sfsf qtVhLWHD/4hρρπ  ⋅==−   ( ) ( ) 3 22 5 770kg/m3m t0.25m0.025m2.4410J/kg 1335W4 π×  =−×   4 t2.61810s7.27h.=×= < COMMENTS: (1) The value of o h is overestimated by assuming an infinite quiescent medium. The fact that the paraffin is enclosed will increase the resistance due to free convection and hence decrease q and increase t. (2) Using 2 o h151W/mK =⋅ results in 2 m,o U136W/mK,T57.6C, =⋅=° q = 1009 W and t = 9.62 h. PROBLEM 9.58 KNOWN: A long uninsulated steam line with a diameter of 89 mm and surface emissivity of 0.8 transports steam at 200 ° C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20 ° C. FIND: (a) The heat loss per unit length for a calm day when the ambient air temperature is 20 ° C; (b) The heat loss on a breezy day when the wind speed is 8 m/s; and (c) For the conditions of part (a), calculate the heat loss with 20-mm thickness of insulation (k = 0.08 W/m ⋅ K). Would the heat loss change significantly with an appreciable wind speed? SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Calm day corresponds to quiescent ambient conditions, (3) Breeze is in crossflow over the steam line, (4) Atmospheric air and large surroundings are at the same temperature; and (5) Emissivity of the insulation surface is 0.8. PROPERTIES: Table A-4, Air (T f = (T s + T ∞ )/2 = 383 K, 1 atm): ν = 2.454 × 10 -5 m 2 /s, k = 0.03251 W/m ⋅ K, α = 3.544 × 10 -5 m 2 /s, Pr = 0.693. ANALYSIS: (a) The heat loss per unit length from the pipe by convection and radiation exchange with the surroundings is bcvrad qq q ′′ ′ =+ () ( ) 44 bDbs,b b b b s,b qhPT T PT T P D εσ π ∞∞ ′ =−+− = (1,2) where D b is the diameter of the bare pipe. Using the Churchill-Chu correlation, Eq. 9.34, for free convection from a horizontal cylinder, estimate D h () D 2 1/6 b D 8/27 9/16 0.387 Ra hD Nu 0.60 k 1 0.559/ Pr    ==+    +     (3) where properties are evaluated at the film temperature, T f = (T s + T ∞ )/2 and () 3 s b D gTTD Ra β να ∞ − = (4) Substituting numerical values, find for the bare steam line D Ra D Nu D h (W/m 2 ⋅ K) () cv qW/m ′ () rad qW/m ′ () b qW/m ′ 3.73 × 10 6 21.1 7.71 388 541 929 < Continued … PROBLEM 9.58 (Cont.) (b) For forced convection conditions with V = 8 m/s, use the Churchill-Bernstein correlation, Eq. 7.56, () D 4/5 1/2 1/3 5/8 Db D D 1/4 2/3 0.62 Re Pr hD Re Nu 0.3 1 k 282,000 1 0.4/ Pr    ==+ +      +   where Re D = VD/ ν . Substituting numerical values, find D Re D Nu D,b h (W/m 2 ⋅ K) () cv qW/m ′ () rad qW/m ′ () b qW/m ′ 2.17 × 10 4 82.5 30.1 1517 541 2058 < (c) With 20-mm thickness insulation, and for the calm-day condition, the heat loss per unit length is () ins s,o tot qTT/R ∞ ′′ =− (1) [] 1 t ins cv rad RR 1/R 1/R − ′′ ′ ′ =+ + (2) where the thermal resistance of the insulation from Eq. 3.28 is () [] ins o b RnD/D/2k π ′ =  (3) and the convection and radiation thermal resistances are () cv D,o o R1/hD π ′ = (4) () () ( ) 22 rad rad o rad,o s,o s,o R1/hD h TTTT πεσ ∞∞ ′ ==++ (5,6) The outer surface temperature on the insulation, T s,o , can be determined by an energy balance on the surface node of the thermal circuit. [] s,b s,o s,o 1 ins cv rad TT TT R 1/R 1/R ∞ − −− = ′ ′′ + Substituting numerical values with D b,o = 129 mm, find the following results. 2 ins D,o R 0.7384 m K / W h 5.30 W / m K ′ =⋅ = ⋅ 2 cv rad R 0.4655 K / W h 5.65 W / m K ′ ==⋅ rad ins R 0.4371 K / W q 187 W /m ′′ == < s,o T 62.1 C =° Continued … PROBLEM 9.58 (Cont.) COMMENTS: (1) For the calm-day conditions, the heat loss by radiation exchange is 58% of the total loss. Using a reflective shield (say, ε = 0.1) on the outer surface could reduce the heat loss by 50%. (2) The effect of a 8-m/s breeze over the steam line is to increase the heat loss by more than a factor of two above that for a calm day. The heat loss by radiation exchange is approximately 25% of the total loss. (3) The effect of the 20-mm thickness insulation is to reduce the heat loss to 20% the rate by free convection or to 9% the rate on the breezy day. From the results of part (c), note that the insulation resistance is nearly 3 times that due to the combination of the convection and radiation process thermal resistances. The effect of increased wind speed is to reduce cv R, ′ but since ins R ′ is the dominant resistance, the effect will not be very significant. (4) Comparing the free convection coefficients for part (a), D b = 89 mm with T s,b = 200 ° C, and part (b), D b,o = 129 mm with T s,o = 62.1 ° C, it follows that D,o h is less than D,b h since, for the former, the steam line diameter is larger and the diameter smaller. (5) The convection correlation models in IHT are especially useful for applications such as the present one to eliminate the tediousness of evaluating properties and performing the calculations. However, it is essential that you have experiences in hand calculations with the correlations before using the software. PROBLEM 9.59 KNOWN: Horizontal tube, 12.5mm diameter, with surface temperature 240°C located in room with an air temperature 20°C. FIND: Heat transfer rate per unit length of tube due to convection. SCHEMATIC: ASSUMPTIONS: (1) Ambient air is quiescent, (2) Surface radiation effects are not considered. PROPERTIES: Table A-4, Air (T f = 400K, 1 atm): ν = 26.41 × 10 -6 m 2 /s, k= 0.0338 W/m⋅K, α = 38.3 × 10 -6 m 2 /s, Pr = 0.690, β = 1/T f = 2.5 × 10 -3 K -1 . ANALYSIS: The heat rate from the tube, per unit length of the tube, is ( ) s qhDTTπ ∞ ′ =− where h can be estimated from the correlation, Eq. 9.34, ( ) D 2 1/6 D 8/27 9/16 0.387Ra Nu0.60. 10.559/Pr    =+    +     From Eq. 9.25, ( ) ( ) ( ) 3 2313 3 s D 6262 9.8m/s2.510K24020K12.510m gTTD Ra10,410. 26.4110m/s38.310m/s β να −−− ∞ −− ××−×× − === ××× Hence, ( ) ( ) D 2 1/6 8/27 9/16 0.38710,410 Nu0.604.40 10.559/0.690    =+=    +     D 2 3 k0.0338W/mK hNu4.4011.9W/mK. D 12.510m − ⋅ ==×=⋅ × The heat rate is ( ) ( ) 23 q11.9W/mK12.510m24020K103W/m.π − ′ =⋅××−= < COMMENTS: Heat loss rate by radiation, assuming an emissivity of 1.0 for the surface, is ( ) ( ) ( ) ( ) 44 44384 rads 24 W qPTT112.510m5.671024027320273K mK εσπ −− ∞ ′ =−=××××+−+ ⋅    rad q138W/m. ′ = Note that P = π D. Note also this estimate assumes the surroundings are at ambient air temperature. In this instance, radconv qq. ′′ > [...]... KNOWN: Insulated steam tube exposed to atmospheric air and surroundings at 25°C FIND: (a) Heat transfer rate by free convection to the room, per unit length of the tube; effect on quality, x, at outlet of 30 m length of tube; (b) Effect of radiation on heat transfer and quality of outlet flow; (c) Effect of emissivity and insulation thickness on heat rate SCHEMATIC: ASSUMPTIONS: (1) Ambient air is quiescent,... 0.387Ra D,i  k  0.0263     ho = 0.60 +  = 0.15 0.60 + 8 / 27 8 / 27  Di 1 + ( 0.559 Pr )9 /16  1 + (0.559 0 .70 7 )9 /16              2 Continued PROBLEM 9 .72 (Cont.) 1/ 6  2  h o = 0. 175 0.60 + 2.64 Ts,i − T∞    ( Hence ( ) { ) ( 12.2 343 − Ts,i = 0. 175 0.60 + 2.64 Ts,i − 273 4 )1/ 6 } (Ts,i − 273 ) + 0.5 × 5. 67 × 10−8 Ts,i − ( 273 )4    2 Ts,i ≈ 314 .7 K ≈ 41 .7 C... (3) Ambient air extensive and quiescent PROPERTIES: Table A-4, Air (Tf, 1 atm): β = 1/Tf and Ts (°C) 30 55 80 Tf (K) 302 314 3 27 6 2 ν × 10 m /s 6 2 α × 10 m /s 16.09 17. 30 18.61 3 k × 10 W/m⋅K 22.8 24.6 26.5 Pr 26.5 27. 3 28.3 0 .70 7 0 .70 5 0 .70 3 ANALYSIS: For the cylindrical sensor, using Eqs 9.25 and 9.34, gβ∆TD3 Ra D = να     0.387Ra1/6 hD D   D Nu D = = 0.60 + 8/ 17  k   1 + ( 0.559/Pr... convection and radiation coefficients are h = 9 .75 and hr = 24.5 W/m2⋅K, respectively COMMENTS: Since h and hr decrease with increasing time, the maximum allowable conveyor time is underestimated by the result of part (c) PROBLEM 9 .72 KNOWN: Velocity and temperature of air flowing through a duct of prescribed diameter Temperature of duct surroundings Thickness, thermal conductivity and emissivity of applied... 0.12 and 0.36, respectively Compare these values with those for convection PROBLEM 9. 67 KNOWN: Diameter, emissivity, and power dissipation of cylindrical heater Temperature of ambient air and surroundings FIND: Steady-state temperature of heater and time required to come within 10°C of this temperature SCHEMATIC: ASSUMPTIONS: (1) Air is quiescent, (2) Duct wall forms large surroundings about heater,... a factor of approximately 3 Convection and radiation heat rates at the outer surface are comparable COMMENTS: Over the range of insulation thickness, Ts,o decreases from 45°C to 20°C, while h D o and hr decrease from 6.9 to 3.5 W/m2⋅K and from 3.8 to 3.3 W/m2⋅K, respectively PROBLEM 9 .70 KNOWN: A billet of stainless steel AISI 316 with a diameter of 150 mm and length 500 mm emerges from a heat treatment... be the dominant mode, and would reduce the estimate for Ts Generally such heaters could not withstand operating temperatures above 1000°C and safe operation in air is not possible PROBLEM 9.63 KNOWN: Motor shaft of 20-mm diameter operating in ambient air at T∞ = 27 C with surface temperature Ts ≤ 87 C FIND: Convection coefficients and/ or heat removal rates for different heat transfer processes: (a)... ) = 5 1 3 W / m < < The radiation effect accounts for 16 and 35%, respectively, of the heat rate (b) Cross-wind condition With a cross-wind, find Re D = VD 1 0 m / s × 0.30m = = 1.5 87 ×10 5 −6 m 2 / s ν 18.9 × 10 and using the Hilpert correlation where C = 0.0 27 and m = 0.805 from Table 7. 2, ( Nu D = CRem Pr1/3 = 0.0 27 1.5 87 ×105 D ) 0.805 ( 0 .70 3)1/3 = 368.9 hD = NuD ⋅ k / D = 368.9 × 0.0285W/m ⋅ K/0.300m... bar): hf = 566 kJ/kg, Tsat = 416 K, hg = 272 7 kJ/kg, hfg = 2160 kJ/kg, vg = 0. 476 × 103 m3/kg; Table A.3, magnesia, 85% (310 K): km = 0.051 W/m⋅K; Table A.4, air (assume Ts = 60°C, Tf = (60 + 25)°C/2 = 315 K, 1 atm): ν = 17. 4 × 10-6 m2/s, k = 0.0 274 W/m⋅K, α = 24 .7 × 10-6 m2/s, Pr = 0 .70 5, Tf = 1/315 K = 3. 17 × 10-3 K-1 ANALYSIS: (a) The heat rate per unit length of the tube (see sketch) is given as, T... × 10−6 m 2 s     0.387Ra1/ 6   D Nu D = 0.60 + 8 / 27    1 + ( 0.559 Pr )9 /16          = 27, 981 2 2   1/ 6   0.3 87 ( 27, 981)   Nu D = 0.60 +  = 5.61 9 /16 8 / 27   1 + ( 0.559 0 .70 28 )         h D,fc = Nu D k D = 5.61× 0.02852 W m ⋅ K 0.020m = 8.00 W m 2⋅ K q′ = 8.00 W m 2⋅ K (π × 0.020m )( 87 − 27 ) C = 30.2 W m fc  < Mixed free and forced convection effects . significant if ( ) 0.1 37 3 D D Re 4 .7 Gr Pr < where Gr D = Ra D /Pr, find using results from above and in part (a) for ω = 5000 rpm, () 0.1 37 3 11, 076 ? ? 4 .7 27, 981 0 .70 28 0 .70 18 383  <=   We. 330K 87 27 K 0.020m Ra 27, 981 18.91 10 m s 26.94 10 m s −− − == ××× () 2 1/6 D D 8/ 27 9/16 0.387Ra Nu 0.60 1 0.559 Pr =+ +           () () 2 1/6 D 8/ 27 9/16 0.3 87 27, 981 Nu. Diameter and outer surface temperature of steam pipe. Diameter, thermal conductivity, and emissivity of insulation. Temperature of ambient air and surroundings. FIND: Effect of insulation thickness and