PROBLEM 8.6 KNOWN: Water, engine oil and NaK flowing in a 20 mm diameter tube, temperature of the fluids. FIND: (a) The mean velocity as well as hydrodynamic and thermal entrance lengths, for a flow rate of 0.01 kg/s and mean temperature of 366 K, (b) The mass flow rate as well as hydrodynamic and thermal entrance lengths for water and oil at a mean velocity of 0.02 m/s at mean temperatures of 300 and 400 K. SCHEMATIC: D = 0.020 m Water, NaK, or Engine oil ASSUMPTIONS: (1) Constant properties. PROPERTIES: Liquid Water T(K) 300 366 400 Table A.6 A.6 A.6 ρ(kg/m3) 997 963 937 μ(N⋅s/m2) 855 × 10-6 303 × 10-6 217 × 10-6 ν(m2/s) - Pr 5.83 1.89 1.34 Oil 300 366 400 A.5 A.5 A.5 884 844 825 48.6 × 10-2 2.12 × 10-2 0.874 × 10-2 - 6400 338 152 NaK 366 A.7 849 - 5.797 × 10-7 0.019 ANALYSIS: (a) The mean velocity is given by & ρA c = 0.01 kg/s/ ⎡ρπ(0.020m)2 /4 ⎤ = 31.8 kg/s ⋅ m 2 / ρ um = m ⎣ ⎦ The Reynolds number is & 4m 4 × 0.01 kg/s 0.636 kg/s ⋅ m Re D = = = πDμ π(0.020 m)μ μ The hydrodynamic entrance length is 0.636 kg/s ⋅ m x fd,h = 0.05Re D D = 0.05 × × (0.020 m) μ = 636 × 10-6 kg/s ⋅ m μ (1) (2) (3) Continued… t with Pr for large and small values of the Prandtl number.6 (Cont.030 1.0063 0. μ is found from the definition μ = νρ = 5.020 m) 2 m3 & = ρA c u m = m ρ = 6. may increase the convection heat transfer coefficients.72 2.02 m (ρ/μ) x fd.h Pr = 400 × 10-9 m3 /s (ρ/μ) Pr (8) < Solving Equations (5). (3) and (4) yields Liquid water engine oil NaK um (m/s) 0.h = 400 × 10-9 m3 /s (ρ/μ) (7) The thermal entrance length is x fd.05ReD DPr = x fd. .3 xfd. (2) Note the variation of xfd.t → 1.t = 0.74 COMMENTS: (1) As the momentum and thermal diffusivities approach similar values (Pr → 1) xfd.033 0. (7) and (8) yields Liquid Water Water Engine Oil Engine Oil T (k) 300 400 300 400 & (kg/s) m 0.h Pr = 636 × 10-6 kg/s ⋅ m Pr μ (4) < Solving Equations (1).t (m) 2.7 × 10-3 xfd. (c) The Reynolds number associated with the oil is very small. Buoyancy forces are likely to be significant and may induce secondary fluid motion which.797 × 10-7 m 2 /s × 849 kg/m3 = 492 × 10-6 N ⋅ s/m 2 (b) The mass flow rate is given by 0.05 × 400 × 10-6 m 2 /s × 0.020 m)μ (5) (6) The hydrodynamic entrance length is x fd.0056 0.28 × 10-6 m3 /s × ρ Re D = = = 400 × 10-6 m 2 /s × (ρ/μ) πDμ π(0.28 × 10-6 ρ 4 s The Reynolds number is & 4m 4 × 6.t (m) 3.025 where.02 m/s × π × (0. for the NaK.65 5.t = x fd.PROBLEM 8.037 xfd.h (m) 2.97 10.h/xfd.038 0.27 × 10-4 37.30 4.0052 xfd. in turn.05Re D D = 0.h/xfd.1 0.464 1.0059 0.72 7.h (m) 0. We will treat buoyancy effects in Chapter 9.1 0.h = 0.) The thermal entrance length is x fd. < COMMENTS: Due to increasing heat loss with increasing Tm.o = 40o C + ( 700 W/m2 ) (1× 3) m2 0. the air outlet temperature is Tm.17 KNOWN: Surface heat flux for air flow through a rectangular channel. PROPERTIES: Table A-4. find q′′ ( w ⋅ x ) Tm ( x ) = Tm.o = Tm. (2) No heat loss through bottom of channel. the net flux q′′o will actually decrease slightly with increasing x. Air (T ≈ 50°C.1 kg/s (1008 J/kg ⋅ K ) Tm. & cp m < (b) Substituting numerical values. SCHEMATIC: ASSUMPTIONS: (1) Ideal gas with negligible viscous dissipation and pressure variation. .8o C.i + o .o = 60. E& in =E& out & c p Tm + q′′o ( w ⋅ dx ) = m & cp ( Tm + d Tm ) m d Tm q′′o ⋅ w = & cp dx m Separating and integrating between the limits of x = 0 and x. (b) Air outlet temperature for prescribed conditions. FIND: (a) Differential equation describing variation in air mean temperature. ANALYSIS: (a) For the differential control volume about the air.PROBLEM 8. 1 atm): cp = 1008 J/kg⋅K.i + o & cp m q′′ ( w ⋅ L ) Tm. (3) Uniform heat flux at top of channel. 032 kg m ⋅ s ) Hence the flow is laminar. k = 138 × 10-3 W/m⋅K.57 Nu D = 3.05/25)398 × 490 = 390.05m )( 0. and (b) Effect of flowrate.95. Hence ⎛ π DL ⎞ Tm.05 m )( 398 )( 490 ) = 486 m .05 m Nu D = 3.04Gz D2/3 .66 + Continued. h = Nu D = 11. cp = 2131 J/kg⋅K. first calculate ReD from Eq. SCHEMATIC: ASSUMPTIONS: (1) Negligible temperature drop across tube wall.05 ( 0. Pr = 490.95 = 33 W m 2 ⋅ K and it follows that D 0.032 kg/m⋅s. PROPERTIES: Table A.i exp ⎜ − h⎟ ⎜ mc & p ⎟ ⎝ ⎠ ( ) To determine h .5 × 10-6 m2/s.o = Ts − Ts − Tm.25 KNOWN: Inlet temperature and flowrate of oil flowing through a tube of prescribed surface temperature and geometry.t ≈ 0. With GzD = (D/L)ReDPr = (0. hence Tm = 80°C = 353 K): ρ = 852 kg/m3.o = 140°C.66 + 0. . FIND: (a) Oil outlet temperature and total heat transfer rate.PROBLEM 8. 8.5 kg s ) & 4m = = 398 π Dμ π ( 0. (2) Incompressible liquid with negligible viscous dissipation. 8.05D ReD Pr = 0. h may be determined from Eq. Since L = 25 m the flow is far from being thermally fully developed.23 the thermal entry length is x fd. μ = ρ⋅ν = 0. ReD = 4 ( 0.. Moreover.138 W m ⋅ K Hence.41b. 8. 1 + 2. Engine oil (assume Tm. Since Pr > 5..6. ν = 37.14 k 0. it follows that 26 = 11. from Eq.5. ANALYSIS: (a) For constant surface temperature the oil outlet temperature may be obtained from Eq.0668Gz D 1 + 0. 8. one would obtain Tm.i = q mc & p .5 2 Mass flowrate. Eq. and q = 15.o.o.o − Tm. The heat rate increases with increasing m ( ) & . The value of Tm. it follows that ( ) & p Tm.05 m )( 25 m ) × 33 W m 2 ⋅ K ⎥ Tm. the increase is not proportional to m & = 0.o − Tm.) ⎡ ⎤ π ( 0. Following such a procedure.o and q was determined by using the appropriate IHT Correlations and Properties Toolpads. h = 32.o = 36. and hence Tm.660 W. Iteration should continue until satisfactory convergence is achieved between the calculated and assumed values of Tm.980 W. COMMENTS: Note that significant error would be introduced by assuming fully developed thermal conditions and Nu D = 3.o = 35°C. m . The properties should be re-evaluated at T = (20 + 35)/2 = 27°C and the calculations repeated.25 (Cont. Tmo(C) 40 32 28 25000 20000 24 20 15000 0.66. 30000 36 Heat rate.4°C. From the overall energy balance.8. to However. (b) The effect of flowrate on Tm. 8.5 kg s × 2131J kg ⋅ K ⎦ ) ( < Tm.8 W/m2⋅K. The small effect of reevaluating the properties is attributed to the compensating effects on ReD (a large decrease) and Pr (a large increase).o = 150o C − 150o C − 20o C exp ⎢ − ⎣ 0.5 1 1.20 & .5 kg s × 2131J kg ⋅ K × ( 35 − 20 )o C q = mc < q = 15. mdot(kg/s) & due to the corresponding increase in ReD and hence h . The flow remains well within the laminar region over the entire range of &. The maximum heat rate corresponds to the maximum flowrate ( m decrease with increasing m kg/s).o has been grossly overestimated in evaluating the properties. ReD = 27.5 1 Mass flowrate.5 2 0. causing Tm. q(W) Outlet temperature.i = 0.PROBLEM 8.34. mdot(kg/s) 1. h = 0. (b) Conditions at the tube exit for reduced tube length.005m = 29. (3) Negligible viscous dissipation. ANALYSIS: (a) We begin by calculating the Reynolds number ReD = 4m& 4 × 135 × 10−6 kg/s = = 1494 π Dμ π × 0. the flow is laminar.h Pr = 0.36. 2) Equating Eqs. (c) Conditions at the tube exit for increased air flow rate. the Nusselt number is NuD = 4.1×10-7 N⋅s/m2.4. Air ( Tm ≈ 400 K. The hydrodynamic and thermal entrance lengths are x fd . air temperature and pressure at the tube inlet.0338 W/m ⋅ K / 0. Surface temperature at the tube exit. Pr = 0.36k / D = 4. (1) and (2) yields Continued… .o − Tm. p = 1 atm): μ = 230. FIND: (a) The heat transfer rate of the problem.o − Tm.47 W/m 2 ⋅ K Two independent expressions for the heat flux may be written based upon application of Newton’s law of cooling at the tube exit and an overall energy balance.o D = 5 mm Ts. SCHEMATIC: Air L=2 m · m = 135 × 10-6 kg/s Tm. k = 0.26 m Therefore.o ) .690. q " = h(Ts .0338 W/m⋅K.i = 100°C Constant wall heat flux Tm.30 KNOWN: Diameter and length of tube.1 × 10−7 N ⋅ s/m 2 Therefore.690 = 0.o = 160°C ASSUMPTIONS: (1) Steady-state conditions. air flow rate.005 m × 1494 = 0. q" = & p (Tm. the local heat transfer coefficient at the tube exit is h = 4.36 × 0. Therefore.005m × 230. PROPERTIES: Table A. For fully-developed laminar flow with constant heat flux conditions. the flow is fully-developed at the tube exit.t = x fd. (2) Constant properties.05DReD = 0.37 m × 0.37 m x fd.05 × 0. cp = 1014 J/kg⋅K.i ) mc π DL (1.PROBLEM 8. 940.) & p mc ⎡ ⎤ Tm.4 = 43.16 W q = mc < (b) If L = 0. and the flow will be turbulent at the tube exit. h = 29. 8.3°C = 7.3.97W/m 2 ⋅ K ⎥ ⎢ × × π 0. Hence.47W/m ⋅ K × 160°C + π × 0. For Ts.i ⎥ π DL ⎣ ⎦ & p ⎡ mc ⎤ ⎢ π DL + h ⎥ ⎣ ⎦ ⎡ ⎤ 135 × 10−6 kg/s × 1014 J/kg ⋅ K 2 × 100°C⎥ ⎢ 29.039. the turbulent flow at the tube exit will also be fully developed.i ) = 135 × 10−6 kg/s × 1014J/kg ⋅ K × 52. h = 29.36) = 292.2 m/ 0. the Reynolds number will increase to ReD = 14.005 m)/(1494 × 0.3/4. NuD ≈ 4. the heat rate is & p (Tm. Specifically. conditions at x = L are not fully developed and the value of the heat transfer coefficient at the tube exit would exceed that of part (a).005 m 2 m ⎣ ⎦ = 152.47 W/m2 × (4.PROBLEM 8.6/4.o to remain the same.6.023×(14.47 W/m2 and could be evaluated using the Dittus-Boelter correlation.6900.7 W/m2⋅K.005 m = 400. The heat transfer coefficient at the tube exit would exceed that of part (a).o + Tm. In part (c). the heat rate associated with either part (b) or part (c) would have to exceed that of part (a).3°C Hence.o = ⎢ hTs . Specifically. Hence. Since L/D = 2 m / 0.o − Tm. .10a. (c) If the flow rate is increased by an order of magnitude.1 W/m2⋅K.47 W/m2 × (43.005 m × 2 m ⎦ =⎣ − 6 ⎡ 135 × 10 kg/s × 1014 J/kg ⋅ K ⎤ + 24. COMMENTS: In part (b).690) = 0. NuD = 0.30 (Cont.2 m. for Gz-1 = (x/D)/(ReDPr) = (0. the local heat transfer coefficient would exceed h = 29.36) = 31.47 W/m2 at the tube exit and could be estimated using Fig. the local heat transfer coefficient would exceed h = 29.940)4/50. FIND: (a) Pipe length required to achieve desired outlet temperature. μ = 803 × 10 N⋅s/m .02m ) ⎤ ⎣⎢ ⎦⎥ < L = 8. r = ro : dT ⎞ q& C =0=− ro + 1 ⎟ dr ⎠r = r 2k ro o C1 = & o2 qr 2k Continued … . (3) Incompressible liquid with negligible viscous dissipation. Flow rate and inlet temperature of water flowing through the pipe.o − Tm. k = 0.04m ) − ( 0. Pr = 5.56. 3.i = q = q& (π /4 ) Do2 − Di2 L m L= ( & cp Tm. (4) One-dimensional radial conduction in pipe wall. (2) Constant properties.37 KNOWN: Inner and outer diameter of a steel pipe insulated on the outside and experiencing uniform heat generation. (b) The maximum wall temperature exists at the pipe exit (x = L) and the insulated surface (r = ro).87m. Stainless steel 316 (T ≈ 400K): k = 15 W/m⋅K. From Eq. Table A-6.i m ( q& (π /4 ) Do2 − Di2 )= ) ( 0. PROPERTIES: Table A-1.617 W/m⋅K.PROBLEM 8. -6 2 Water ( Tm = 303K ) : cp = 4178 J/kg⋅K.o − Tm. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions. the radial temperature distribution in the wall is of the form T (r) = − q& 2 r + C1ln r + C2 . it follows that ( ( ) ) & cp Tm. (5) Outer surface is adiabatic.45. (b) Location and value of maximum pipe temperature. ANALYSIS: (a) Performing an energy balance for a control volume about the inner tube.1 kg/s ) 4178 ( J/kg ⋅ K ) 20o C 2 2 106 W/m3 (π / 4 ) ⎡( 0. 4k Considering the boundary conditions. Hence.6oC.max = − 106 W/m3 ⎡ 2 2 0. 8.7(f / 8)1/ 2 (Pr 2 / 3 − 1) ⎥ ⎣ ⎦ ⎡ 0. f = (0.1) ⎥⎦ where from Eq.617 W/m ⋅ K (0. from the Gnielinski correlation. the inner surface temperature of the wall at the exit.1 kg/s ReD = = = 7928 π Di μ π ( 0. (f / 8)(ReD − 1000) Pr ⎤ ⎢ ⎥ Di ⎢1 + 12. Eq.max = T ( ro ) = − ) ( q& 2 2 q& ro2 ro − ri + ln 4k 2k ro + Ts ri where Ts.02m ) 2 0. 4k i 2k The temperature distribution and the maximum wall temperature (r = ro) are T (r) = − ) ( q& 2 2 q& ro2 r r − ri + ln + Ts 4k 2k ri Tw.7(0.01 COMMENTS: The physical situation corresponds to a uniform surface heat flux. it is also fully developed.033618) (5. < 2 ×15 W/m ⋅ K 0.02m ) h= k ⎡ ) ( and Tw.0336.02m) = 444 >> (xfd/D) ≈ 10.) T ( ri ) = Ts = − r = ri : q& 2 q& ro2 r + ln ri + C2 4k i 2k C2 = q& 2 q& ro2 r − ln ri + Ts .4oC + Tm.04m ) − ( 0.02 m ⎢⎣1 + 12. Hence.o = Ts = 2 4 h Di 4 ×1796 W/m ⋅ K ( 0. the inner surface temperature of the wall at the exit is 2 2 106 W/m3 ⎡( 0.45 ⎤ 2 = ⎢ ⎥ = 1796 W/m ⋅ K 1/2 2/3 0.64)-2 = 0.01m ) ⎤ ( ⎢ ⎥⎦ 4 × 15 W/m ⋅ K ⎣ 106 W/m3 ( 0.87 m/0.37 (Cont. follows from q′′s = ) ( q& (π /4 ) Do2 − Di2 L π Di L = ( q& Do2 − Di2 4 Di ) = h ( Ts − Tm.4o C = 52. In the fully developed region.02m ) ⎤ q& Do2 − Di2 ⎢ ⎣ ⎦⎥ + 40o C = 48. and Tm + ln increases linearly with x. 8.PROBLEM 8. .62.790 ln ReD-1.02m ) 803 × 10−6 N ⋅ s/m 2 the flow is turbulent and. with (L/Di) = (8.45 . With & 4m 4 × 0.o ) where h is the local convection coefficient at the exit.033618)(7928 .02m ) − ( 0. Ts also increases linearly with x.21.02 + 48.1000)5. 4 / Pr) 2 / 3 ⎤ ⎣ ⎢⎣ ⎥⎦ 4/5 5/8⎤ ⎡ ⎛ 4 1/ 2 1/ 3 4 ⎞ 0.0290 W/m⋅K. FIND: Convection heat flux associated with the external and internal flows. Velocity and temperature of air in fully developed internal flow. 000 ⎢ ⎝ ⎠ ⎥⎦ ⎡1 + (0.4 / 0.69 ×104.69 × 104 Nu D = 0.023 Re4/5 D Pr h= ) 4/5 ( 0. which is turbulent.69 × 10 ) 0. air is an ideal gas with negligible viscous dissipation and pressure variations. (2) Uniform cylinder surface temperature. ANALYSIS: For the external and internal flows. the convection coefficient and heat flux are h= k 0.4 = 162 k 0. 000 ⎟ 2 / 3 ⎡1 + (0.05 m = m = = 7. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions.51 × 10 m /s.702.05 m Continued… .0290 W/m ⋅ K Nu D = ×180 = 104 W/m 2 ⋅ K D 0.69 ×10 ⎟ ⎥ = 0.3 + 1+ ⎜ ⎟ ⎥ 1/ 4 282. ( 0.46 KNOWN: Surface temperature and diameter of a tube. o < Using the Dittus-Boelter correlation.702) ⎤ ⎠ ⎥⎦ ⎢⎣ ⎝ ⎢⎣ ⎥⎦ 2 1/ 3 0.60. -6 2 PROPERTIES: Table A-4.4 = 0.023 7.702 ⎢1 + ⎜ 7. ReD = u D 30 m/s × 0. (4) For internal flow. Eq. -6 2 ν ν 19.PROBLEM 8. Pr = 0. Air (336 K): ν = 19.62 Re1/ D Pr Hence. 4/5 5/8 ⎡ ⎛ Re ⎞ ⎤ D ⎢ Nu D = 0.3 + = 180 ⎥ 1/ 4 ⎢ ⎜ 282. Velocity and temperature of air in cross flow. (3) Fully developed internal flow.0290 W/m ⋅ K × 162 = 94 W/m 2 ⋅ K Nu D = D 0. 8.702 )0.62(7.05 m q′′ = h ( Ts − T∞ ) = 104W/m 2 ⋅ K (100 − 25 ) C = 7840 W/m 2 .71× 10 m / s VD From the Churchill-Bernstein relation for the external flow. for the internal flow. k = 0. 46 (Cont. . o < COMMENT: Convection effects associated with the two flow conditions are comparable.) and the heat flux is q′′ = h ( Ts − Tm ) = 94 W/m 2 ⋅ K (100 − 25 ) C = 7040 W/m 2 .PROBLEM 8. (c) Discuss whether the flow is laminar or turbulent for Tm.1 K ⎝ 0.6. the flow is in a fully turbulent condition.41b).1m × 769 × 10−6 N ⋅ s/m 2 Continued… . Pr = 0. SCHEMATIC: Ts = 310 K.635 W/m⋅K. assumed): μ = 769×10-6 N⋅s/m2. PROPERTIES: Table A.1m = 60. liquid water flow rate. ANALYSIS: (a) We begin by calculating the Reynolds number ReD = 4m& 4 × 0.i = 300 K or 500 K L=6m ASSUMPTIONS: (1) Steady-state conditions.PROBLEM 8. liquid water entrance temperatures and tube surface temperatures.0144 / 50. (8.i )exp ⎜ − h⎟ ⎜ mc ⎟ & p ⎝ ⎠ ⎛ ⎞ π × 0. (2) Constant properties in parts (a) and (b).1 m Tm. liquid water ( Tm = 505 K. Ts = 510 K and (b) Tm.635W/m ⋅ K ⎡ 0.4 ⎤ = 0. we conclude that entrance effects are not important.5 × 10−6 N ⋅ s/m 2 Therefore.1 kg/s = = 1655 π Dμ π × 0.855. We may use Dittus-Boelter (Eq.1m ⎛ PL ⎞ Tm.014 π Dμ π × 0.5×10-6 N⋅s/m2. h= k ⎡ 0.i = 500 K.o = Ts − (Ts − Tm. assumed): μ = 115. Liquid water ( Tm = 305 K.i = 300 K. k = 0.48 KNOWN: Diameter and length of circular tube. FIND: Water outlet temperatures for (a) Tm. cp = 4700 J/kg⋅K.i = 300 K.1 kg/s × 4700 J/kg ⋅ K ⎠ < (b) The Reynolds number is ReD = 4m& 4 × 0. Ts = 310 K.4 ⎤ = 235 W/m 2 ⋅ K ⎣ ⎦ ⎣ ⎦ D 0.1 m × 6 m = 510 K − 10 K × exp ⎜ − 235 W/m 2 ⋅ K ⎟ = 506.620 W/m⋅K.023 × 11.1 kg/s = = 11.8550. Ts = 647 K.023ReD4 /5 Pr 0. Since L/D = 6m/0. k = 0.1m × 115. m = 0. 510 K or 647 K Liquid water .60) to determine the average heat transfer coefficient and the mean outlet temperature may be found from Eq.20.1 kg/s D = 0. Pr = 5. (3) Negligible viscous dissipation. cp = 4178 J/kg⋅K. 8. 0 W/m 2 ⋅ K ⎨ 2/3 ⎬ 0.04 [(0.1 kg/s × 4178 J/kg ⋅ K ⎠ < (c) The temperature variations within the water are very large.66 + 2/3 ⎬ D⎪ 1 + 0. With Pr > 5. . h= = k ⎧⎪ 0.1 m × 6 m 51 W/m 2 ⋅ K ⎟ = 302.1 m / 6 m) × 1655 × 5.o = Ts − (Ts − Tm.05 × 0. properties are expected to vary significantly from location to location. average temperatures will be low.) Therefore. The assumption of constant properties under the conditions of part (c) may not be appropriate.05 × D × ReD × Pr = 0. higher temperatures will exist in a greater portion of the liquid and viscosities may drop to very low values. The thermal entrance length is xfd.620W/m ⋅ K ⎧⎪ 0.i )exp ⎜ − h⎟ ⎜ mc ⎟ & p ⎝ ⎠ ⎛ ⎞ π × 0.0 m > L. we expect entrance effects to be significant. and the flow is expected to be laminar.1 K = 310 K − 10 K × exp ⎜ − ⎝ 0. (8.1 m 1 + 0. However.PROBLEM 8.0668(0.66 + = 51.0668( D / L) ReD Pr ⎫⎪ ⎨3. Near the entrance of the tube.20 3. (8. we may use Eq. to estimate the value of h .1 m / 6.t = 0. COMMENTS: Even though entrance effects are important for the laminar flow conditions of part (b).20 = 43.04 [( D / L) ReD Pr ] ⎭⎪ ⎩ ⎫⎪ 0. (8.48 (Cont. Therefore. the flow is laminar. the heat transfer coefficient is small relative to that associated with the turbulent conditions of part (a).1m × 1655 × 5.0 m) × 1655 × 5.20] ⎪⎭ ⎪⎩ Using Eq. Therefore. the flow may trip into turbulent conditions at a location between the tube entrance and the tube exit. as the boundary layer regions grow.57) with Eq. Hence.56) for the Graetz number.41b) ⎛ PL ⎞ Tm. 500 = exp ⎢ − Ts − 600 K ⎢⎣ ⎥⎦ 8 × 10-3 kg/s × 5193 J/kg ⋅ K < Ts = 1400 K.39 × 10 m /s. Hence. & 4m 4 × 8 × 10−3 kg/s ReD = = = 1.020 m ) × 0.7 × 0. q=m Continued … .304 W/m ⋅ K/0. Pr = 0.i ⎝ ⎠ Ts − Tm. ( 0.4 = 38.780 m × 588 W/m2 ⋅ K ⎤ Ts − 1000 K ⎥ = 0.PROBLEM 8. ⎛ PLh ⎞ = exp ⎜ − ⎟ ⎜ m & cp ⎟ Ts − Tm. cp = 1099 J/kg⋅K. ν = 6.o − Tm. fully developed flow.02 m = 588 W/m2 ⋅ K.654. 3 PROPERTIES: Table A-4. μ = 382 × 10 N⋅s/m . k = 57.654 )0. The heat rate with helium coolant is ( ) & c p Tm. Helium ( Tm = 800K.333 × 104 -7 2 π Dμ π × 0. For the circular tube. Air ( Tm = 800K. (2) Ideal gas with negligible viscous dissipation and pressure variation.i = 8 × 10−3 kg/s × 5193 J/kg ⋅ K (1000 − 600 ) K = 16.304 W/m⋅K. 1 atm ) : ρ 3 -3 -6 2 = 0.3 × 10 W/m⋅K.7 h = Nu ⋅ k/D = 38. 1 atm ) : ρ = 0.333 × 104 Nu = 0.4354 kg/m . cp = 5193 J/kg⋅K. the surface temperature is ⎡ π ( 0.93 × 10 m /s. from Eq. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions. 8.62 kW. k = -7 2 -4 2 0. ν = 84. ANALYSIS: (a) For helium and a constant wall temperature.023 Re4/5 D Pr ) 4/5 ( 0. Pr = 0. (b) Outlet temperature and required flow rate to achieve same removal rate and wall temperature if the coolant gas is air.o where P = πD.023 1.41b.4 = 0. FIND: (a) Uniform tube wall temperature required to heat the helium.51 KNOWN: Gas-cooled nuclear reactor tube of 20 mm diameter and 780 mm length with helium -3 heated from 600 K to 1000 K at 8 × 10 kg/s.020 m × 382 ×10 N ⋅ s/m and using the Dittus-Boelter correlation for turbulent. (3) Fully developed conditions.06272 kg/m .709. 321 × 10 -2 5. (6) Eq.041 × 10 -2 5.215 × 10 ( W/m2 ⋅ K ) Tm. (5) 1000 950 900 890 ha -2 3.51 (Cont.4 = 0. we find & a = 5. written in the order they would be used in the iteration.o − Tm. use m m results and iterate.023 Re4/5 D Pr k (3.o Re = &a 4m π Dμ (2) hD 0. (2) to evaluate Tm.i.620 W = m ) (1) ⎡ PLh ⎤ a ⎥ = exp ⎢ − & Ts − Tm.o (K) Eq.5 times that obtained with helium. (1).PROBLEM 8. (3) and (4) to find h and then Eq. the required mass rate is 6.459 × 10-5 ( ha / m ⎥⎦ ⎣ (7) Results of the iterative solution are Trial 1 2 3 4 Tm.i q = 16. compare & from Eq. (7) 407 905 453 899 513 891 527 890 Hence.i ⎣⎢ ma c p ⎦⎥ Ts − Tm.22 ×10−2 kg/s m Tm.o = 890 K. the above relations. COMMENTS: To achieve the same cooling rate with air.4) & are unknown.) (b) For the same heat removal rate (q) and wall temperature (Ts) with air supplied at Tm.o and find where Tm. Using thermophysical properties of air evaluated at Tm = 800K. < . the relevant relations are ( & a c p Tm.781 × 10 -2 4. become &a = m 15.o − 600 (5) & a4/5 ha = 5600m (6) & a )⎤ Tm. An iterative solution is required: assume a value of Tm.o (K) & (kg/s) m (Assumed) Eq.1 Tm.o = 1400 K − 800 K × exp ⎡⎢ −4.o and m & in Eqs.o.