PROBLEM 8.28 KNOWN: Inlet temperature, flow rate and properties of hot fluid. Initial temperature, volume and properties of pharmaceutical. Heat transfer coefficient at outer surface and dimensions of coil. FIND: (a) Expressions for T c (t) and T h,o (t), (b) Plots of T c (t) and T h,o (t) for prescribed conditions. Effect of flow rate on time for pharmaceutical to reach a prescribed temperature. SCHEMATIC: ASSUMPTIONS: (1) Constant properties, (2) Negligible heat loss from vessel to surroundings, (3) Pharmaceutical is isothermal, (4) Negligible work due to stirring, (5) Negligible thermal energy generation (or absorption) due to chemical reactions associated with the batch process, (6) Negligible kinetic energy, potential energy and flow work changes for the hot fluid, (7) Negligible tube wall conduction resistance. ANALYSIS: (a) Performing an energy balance for a control surface about the stirred liquid, it follows that () () cc c c v,c c c c v,c dU dT d cT Vc qt dt dt dt ρρ =∀ = = (1) where, () () hp,h h,i h,o qt m c T T =−  (2) or, () sm qt UA T =∆ (3a) where ()( )() h,i c h,o c h,i h,o m h,ic h,ic h,o c h,o c TT T T TT T TT TT nn TT TT −− − − ∆= =  −−   −−   (3b) Substituting (3b) into (3a) and equating to (2), () () h,i h,o hp,h h,i h,o s h,i c h,o c TT mc T T UA TT n TT − −=  −   −    Hence, h,i c s h,o c h p,h TT UA n TTmc  − =   −    or, () () () h,o c h,i c s h p,h T t T T T exp UA /m c =+ − −  (4) < Substituting Eqs. (2) and (4) into Eq. (1), Continued … PROBLEM 8.28 (Cont.) () () c c c v,c h p,h h,i c h,i c s h p,h dT c m c T T T T exp UA / m c dt ρ  ∀= −−−−   () () hp,h h,i c c s h p,h ccv,c mc T T dT 1 exp UA / m c dt c ρ −  =−−  ∀   () () () c c,i Tt t hp,h c s h p,h To ccv,c ch,i mc dT 1 exp UA / m c dt c TT ρ  −=−−  ∀ − ∫∫   () hp,h ch,i s h p,h c,i h,i c c v,c mc TT n1expUA/mct TT Vc ρ  −  −=−−    −     () () () hp,h hp,h ch,ih,ic,i ccv,c mc 1exp UA/mc t Tt T T T exp c ρ   −−   =− − −  ∀    (5) < Eq. (5) may be used to determine T c (t) and the result used with (4) to determine T h,o (t). (b) To evaluate the temperature histories, the overall heat transfer coefficient, () 1 11 oi Uh h , − −− =+ must first be determined. With () 2 D Re 4m / D 4 2.4 kg /s / 0.05m 0.002 N s / m 30,600, π µ π ==× ⋅= the flow is turbulent and ()() 4/5 0.3 2 iD k 0.260W / m K h Nu 0.023 30,600 20 1140W /m K D 0.05m ⋅  == = ⋅   Hence, ()() 1 22 11 U 1000 1140 W / m K 532W / m K. − −− =+ ⋅= ⋅   As shown below, the temperature of the pharmaceuticals increases with time due to heat transfer from the hot fluid, approaching the inlet temperature of the hot fluid (and its maximum possible temperature of 200 ° C) at t = 3600s. Continued … 0 400 800 12001600 2000 2400 280032003600 Tim e(s) 20 40 60 80 100 120 140 160 180 200 Temperature(C) Pharmaceutical, Tc H ot fluid , Th PROBLEM 8.28 (Cont.) With increasing T c , the rate of heat transfer from the hot fluid decreases (from 4.49 × 10 5 W at t = 0 to 6760 W at 3600s), in which case T h,o increases (from 125.2 ° C at t = 0 to 198.9 ° C at 3600s). The time required for the pharmaceuticals to reach a temperature of T c = 160 ° C is c t 1266s = < With increasing h m, the overall heat transfer coefficient increases due to increasing h i and the hot fluid maintains a higher temperature as it flows through the tube. Both effects enhance heat transfer to the pharmaceutical, thereby reducing the time to reach 160 ° C from 2178s for h m1kg/s = to 906s at 5 kg/s. For 2 hDi 1 m 5kg /s, 12,700 Re 63, 700 and 565 h 2050W / m K. ≤≤ ≤ ≤ ≤≤ ⋅ COMMENTS: Although design changes involving the length and diameter of the coil can be used to alter the heating rate, process control parameters are limited to h,i h Tandm. 1 2 3 4 5 Mass flowrate, m doth (kg/s) 800 1000 1200 1400 1600 1800 2000 2200 Tim e, tc(s ) PROBLEM 8.29 KNOWN: Tubing with glycerin welded to transformer lateral surface to remove dissipated power. Maximum allowable temperature rise of coolant is 6 ° C. FIND: (a) Required coolant rate m , tube length L and lateral spacing S between turns, and (b) Effect of flowrate on outlet temperature and maximum power. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) All heat dissipated by transformer transferred to glycerin, (3) Fully developed flow (part a), (4) Negligible kinetic and potential energy changes, (5) Negligible tube wall thermal resistance. PROPERTIES: Table A.5, Glycerin ( m T ≈ 300 K): ρ = 1259.9 kg/m 3 , c p = 2427 J/kg ⋅ K, µ = 79.9 × 10 - 2 N ⋅ s/m 2 , k = 286 × 10 -3 W/m ⋅ K, Pr = 6780. ANALYSIS: (a) From an overall energy balance assuming the maximum temperature rise of the glycerin coolant is 6 ° C, find the flow rate as () pm,o m,i qmcT T =− () () 2 pm,o m,i m q c T T 1000W 2427J kg K 6K 6.87 10 kg s − =−= ⋅=× < From Eq. 8.43, the length of tubing can be determined, () sm,o p sm,i TT exp PLh mc TT − =− −  where P = π D. For the tube flow, find 2 D 22 4m 4 6.87 10 kg s Re 5.47 D 0.020m 79.9 10 N s m πµ π − − ×× == = ×××⋅  which implies laminar flow, and if fully developed, D hD Nu 3.66 k == 3 2 3.66 286 10 W m K h 52.3W m K 0.020m − ×× ⋅ ==⋅ () () () ( ) ( ) 22 47 30 C exp 0.020m 52.3W m K L 6.87 10 kg s 2427J kg K 47 24 C π − − =− × ⋅× × × ⋅ −    L = 15.3 m. < The number of turns of the tubing, N, is N = L/( π D) = (15.3 m)/ π (0.3 m) = 16.2 and hence the spacing S will be S = H/N = 500 mm/16.2 = 30.8 mm. < Continued PROBLEM 8.29 (Cont.) (b) Parametric calculations were performed using the IHT Correlations Toolpad based on Eq. 8.56 (a thermal entry length condition), and the following results were obtained. 0.05 0.09 0.13 0.17 0.21 0.25 Mass flowrate, mdot(kg/s) 1000 1400 1800 2200 2600 3000 Heat rate, q(W) 0.05 0.09 0.13 0.17 0.21 0.25 Mass flowrate, mdot(kg/s) 25 27 29 31 33 35 Outlet temperature, Tmo(C) With T s maintained at 47 ° C, the maximum allowable transformer power (heat rate) and glycerin outlet temperature increase and decrease, respectively, with increasing m . The increase in q is due to an increase in D Nu (and hence h ) with increasing Re D . The value of D Nu increased from 5.3 to 9.4 with increasing m from 0.05 to 0.25 kg/s. COMMENTS: Since () 1 D D Gz L D Re Pr − = = (15.3 m/0.02 m)/(5.47 × 6780) = 0.0206 < 0.05, entrance length effects are significant, and Eq. 8.56 should be used to determine D Nu . PROBLEM 8.30 KNOWN: Diameter and length of copper tubing. Temperature of collector plate to which tubing is soldered. Water inlet temperature and flow rate. FIND: (a) Water outlet temperature and heat rate, (b) Variation of outlet temperature and heat rate with flow rate. Variation of water temperature along tube for the smallest and largest flowrates. SCHEMATIC: ASSUMPTIONS: (1) Straight tube with smooth surface, (2) Negligible kinetic/potential energy and flow work changes, (3) Negligible thermal resistance between plate and tube inner surface, (4) Re D,c = 2300. PROPERTIES: Table A.6, water (assume m T = (T m,i + T s )/2 = 47.5 ° C = 320.5 K): ρ = 986 kg/m 3 , c p = 4180 J/kg ⋅ K, µ = 577 × 10 -6 N ⋅ s/m 2 , k = 0.640 W/m ⋅ K, Pr = 3.77. Table A.6, water (T s = 343 K): µ s = 400 × 10 -6 N ⋅ s/m 2 . ANALYSIS: (a) For m = 0.01 kg/s, Re D = 4 mD π µ  = 4(0.01 kg/s)/ π (0.01 m)577 × 10 -6 N ⋅ s/m 2 = 2200, in which case the flow may be assumed to be laminar. With fd,t xD ≈ 0.05Re D Pr = 0.05(2200)(3.77) = 415 and L/D = 800, the flow is fully developed over approximately 50% of the tube length. With ()() 1/3 0.14 Ds Re Pr L D µµ   = 2.30, Eq. 8.57 may therefore be used to compute the average convection coefficient 0.14 1/3 D D s Re Pr Nu 1.86 4.27 LD µ µ   ==     () () 2 D h k D Nu 4.27 0.640W m K 0.01m 273W m K == ⋅=⋅ From Eq. 8.42b, 2 sm,o s m,i p TT DL 0.01m 8m 273W m K exp h exp T T mc 0.01kg s 4180J kg K ππ   − ××× ⋅  =− =−    −×⋅    () m,o s s m,i T T 0.194 T T 70 C 8.7 C 61.3 C =− − = − = < Hence, () ()() pm,o m,i q mc T T 0.01kg s 4186J kg K 36.3K 1519W =−= ⋅ =  < (b) The IHT Correlations, Rate Equations and Properties Tool Pads were used to determine the parametric variations. The effect of m  was considered in two steps, the first corresponding to m  < 0.011 kg/s (Re D < 2300) and the second for m  > 0.011 kg/s (Re D > 2300). In the first case, Eq. 8.57 was used to determine h , while in the second Eq. 8.60 was used. The effects of m  are as follows. Continued PROBLEM 8.30 (Cont.) 0.005 0.006 0.007 0.008 0.009 0.01 0.011 Mass flowrate, mdot(kg/s) 60 61 62 63 64 65 66 67 Outlet temperature, Tmo(C) Laminar flow (ReD < 2300) 0.01 0.02 0.03 0.04 0.05 Mass flowrate, mdot(kg/s) 69 69.2 69.4 69.6 69.8 70 Outlet temperature, Tmo(C) Turbulent flow (ReD>2300) 0.005 0.006 0.007 0.008 0.009 0.01 0.011 Mass flowrate, mdot(kg/s) 800 900 1000 1100 1200 1300 1400 1500 1600 1700 Heat rate, q(W) Laminar flow (ReD < 2300) 0.01 0.02 0.03 0.04 0.05 Mass flowrate, mdot(kg/s) 1500 3500 5500 7500 9500 Heat rate, q(W) Turbulent flow (ReD>2300) The outlet temperature decreases with increasing m  , although the effect is more pronounced for laminar flow. If q were independent of m , (T m,o - T m,i ) would decrease inversely with increasing m . In turbulent flow, however, the convection coefficient, and hence the heat rate, increases approximately as 0.8 m  , thereby attenuating the foregoing effect. In laminar flow, q ~ 0.5 m  and this attenuation is not as pronounced. The temperature distributions were computed from Eq. 8.43, with h assumed to be independent of x. For laminar flow ( m  = 0.005 kg/s), h was based on the entire tube length (L = 8 m) and computed from Eq. 8.57, while for turbulent flow ( m  = 0.05 kg/s) it was assumed to correspond to the value for fully developed flow and computed from Eq. 8.60. The corresponding temperature distributions are as follows. Continued PROBLEM 8.30 (Cont.) 0 2 4 6 8 Axial location, x(m) 20 30 40 50 60 70 Mean temperature, Tm(C) mdot = 0.005 kg/s 0 2 4 6 8 Axial location, x(m) 20 30 40 50 60 70 Mean temperature, Tm(C) mdot = 0.05 kg/s The more pronounced increase for turbulent flow is due to the much larger value of h (4300 W/m 2 ⋅ K for m = 0.05 kg/s relative to 217 W/m 2 ⋅ K for m  = 0.05 kg/s). PROBLEM 8.31 KNOWN: Diameter and surface temperature of ten tubes in an ice bath. Inlet temperature and flowrate per tube. Volume ( ∀ ) of container and initial volume fraction, f v,i , of ice. FIND: (a) Tube length required to achieve a prescribed air outlet temperature T m,o and time to completely melt the ice, (b) Effect of mass flowrate on T m,o and suitable design and operating conditions. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) Negligible kinetic/potential energy and flow work changes, (3) Constant properties, (4) Fully developed flow throughout each tube, (5) Negligible tube wall thermal resistance. PROPERTIES: Table A.4, air (assume m T = 292 K): c p = 1007 J/kg ⋅ K, µ = 180.6 × 10 -7 N ⋅ s/m 2 , k = 0.0257 W/m ⋅ K, Pr = 0.709; Ice: ρ = 920 kg/m 3 , h sf = 3.34 × 10 5 J/kg. ANALYSIS: (a) With Re D = 4 m  / π D µ = 4(0.01 kg/s)/ π (0.05 m)180.6 × 10 -7 N ⋅ s/m 2 = 14,100 for m  = 0.01 kg/s, the flow is turbulent, and from Eq. 8.60, ()() 0.8 0.3 0.8 0.3 D D D Nu Nu 0.023Re Pr 0.023 14,100 0.709 43.3 == = = () () 2 D h Nu k D 43.3 0.0257W m K 0.05m 22.2W m K == ⋅=⋅ With T m,o = 14 ° C, the tube length may be obtained from Eq. 8.42b, () ( ) () 2 sm,o s m,i p 0.05m 22.2W m K L TT 14 DLh exp exp T T 24 mc 0.01kg s 1007J kg K π π  ⋅  − −  ==− =−    −− ⋅     L = 1.56 m < The time required to completely melt the ice may be obtained from an energy balance of the form, () ( ) v,i sf qt f h ρ −=∀ where () () () p m,i m,o q Nmc T T 10 0.01kg s 1007J kg K 10K 1007W =−= ⋅=  . Hence, ( ) ( ) 335 6 0.8 10m 920kg m 3.34 10 J kg t 2.44 10 s 28.3days 1007W × ==×= < (b) Using the appropriate IHT Correlations and Properties Tool Pads, the following results were obtained. Continued [...]... turbulent and L/D = 31, the assumption of fully developed flow throughout a tube is marginal and the foregoing analysis overestimates the discharge temperature PROBLEM 8.32 KNOWN: Thermal conductivity and inner and outer diameters of plastic pipe Volumetric flow rate and inlet and outlet temperatures of air flow through pipe Convection coefficient and temperature of water FIND: Pipe length and fan power... f = ( 0.790 * ln (ReD) - 1 .64 ) ^ -2 PROBLEM 8.34 KNOWN: Flow rate and inlet temperature of water passing through a tube of prescribed length, diameter and surface temperature FIND: (a) Outlet water temperature and rate of heat transfer to water for prescribed conditions, and (b) Compute and plot the required tube length L to achieve Tm,o found in part (a) as a function of the surface temperature for... twice xfd,t, the average heat transfer coefficient is larger as we would expect, but amounts to only a 10% increase PROBLEM 8.39 KNOWN: Flow rate and temperature of atmospheric air entering a duct of prescribed diameter, length and surface temperature FIND: (a) Air outlet temperature and duct heat loss for the prescribed conditions and (b) Calculate and plot q and ∆p for the range of diameters, 0.1 ≤ D... 0.0 261 W/m⋅K, Pr = 0.71 ANALYSIS: For the external and internal flows, Re D = VD u mD 30 m/s × 0.05 m = = = 9.55 ×10 4 -6m 2 / s ν ν 15.71×10 From the Zhukauskas relation for the external flow, with C = 0. 26 and m = 0 .6, ( NuD = C Re m Pr n ( Pr/Prs )1/4 = 0. 26 9.55 × 104 D ) 0 .6 ( 0.71)0.37 (1)1 / 4 = 223 Hence, the convection coefficient and heat flux are h= k 0.0 261 W/m ⋅ K Nu D = × 223 = 1 16. 4... KNOWN: Diameters and thermal conductivity of steel pipe Temperature and velocity of water flow in pipe Temperature and velocity of air in cross flow over pipe Cost of producing hot water FIND: Daily cost of heat loss per unit length of pipe SCHEMATIC: ASSUMPTIONS: (1) Steady state, (2) Constant properties, (3) Negligible radiation from outer surface, (4) Fully-developed flow in pipe -6 2 PROPERTIES:... negligible error The implication is that the temperature of the pipe’s inner surface closely approximates that of the water If R ′ cnv,w is neglected, the heat loss is q ′ = 3 46 W / m PROBLEM 8. 36 KNOWN: Inner and outer diameter of a steel pipe insulated on the outside and experiencing uniform heat generation Flow rate and inlet temperature of water flowing through the pipe FIND: (a) Pipe length required... right-hand side the product P⋅ h will be independent of D Hence, Tm,o will  depend only on m This is, of course, a consequence of the laminar flow condition and will not be the same for turbulent flow PROBLEM 8. 46 KNOWN: Gas-cooled nuclear reactor tube of 20 mm diameter and 780 mm length with helium heated -3 from 60 0 K to 1000 K at 8 × 10 kg/s FIND: (a) Uniform tube wall temperature required to heat. .. approximated as a tube of prescribed diameter and length maintained at a known surface temperature Air inlet temperature and flowrate FIND: (a) Outlet temperature of the air coolant for the prescribed conditions and (b) Compute and plot  the air outlet temperature Tm,o as a function of flow rate, 0.1 ≤ m ≤ 0 .6 kg/h Compare this result with those for vanes having passage diameters of 2 and 4 mm SCHEMATIC:... 8.43 KNOWN: Surface temperature and diameter of a tube Velocity and temperature of air in cross flow Velocity and temperature of air in fully developed internal flow FIND: Convection heat flux associated with the external and internal flows SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform cylinder surface temperature, (3) Fully developed internal flow -6 PROPERTIES: Table A-4, Air (298K):... initial estimate of 325 K is reasonable and iteration is not necessary (2) For a steam flow rate of 0.01 kg/s, approximately 10% of the outflow would be in the form of saturated liquid, (3) With L/Di = 100, it is reasonable to assume fully developed flow throughout the tube PROBLEM 8.41 KNOWN: Duct diameter and length Thermal conductivity of insulation Gas inlet temperature and velocity and minimum allowable . second Eq. 8 .60 was used. The effects of m  are as follows. Continued PROBLEM 8.30 (Cont.) 0.005 0.0 06 0.007 0.008 0.009 0.01 0.011 Mass flowrate, mdot(kg/s) 60 61 62 63 64 65 66 67 Outlet temperature,. 0.01 0.02 0.03 0.04 0.05 Mass flowrate, mdot(kg/s) 69 69 .2 69 .4 69 .6 69.8 70 Outlet temperature, Tmo(C) Turbulent flow (ReD>2300) 0.005 0.0 06 0.007 0.008 0.009 0.01 0.011 Mass flowrate, mdot(kg/s) 800 900 1000 1100 1200 1300 1400 1500 160 0 1700 Heat. heat loss is q 3 46 W / m. ′ = PROBLEM 8. 36 KNOWN: Inner and outer diameter of a steel pipe insulated on the outside and experiencing uniform heat generation. Flow rate and inlet temperature of