PROBLEM 11. KNOWN: Type-30 stainless tube with prescribed inner and outer diameters used in a cross-flow heat exchanger. Prescribed fouling factors and internal water flow conditions. FIND: (a) Overall coefficient based upon the outer surface, U o, with air at T o =15 and velocity V o = 0 m/s in cross-flow; compare thermal resistances due to convection, tube wall conduction and fouling; (b) Overall coefficient, U o, with water (rather than air) at T o = 15 and velocity V o = 1 m/s in crossflow; compare thermal resistances due to convection, tube wall conduction and fouling; (c) For the water-air conditions of part (a), compute and plot U o as a function of the air cross-flow velocity for 5 V o 30 m/s for water mean velocities of u m,i = 0., 0.5 and 1.0 m/s; and (d) For the water-water conditions of part (b), compute and plot U o as a function of the water mean velocity for 0.5 u m,i.5 m/s for air cross-flow velocities of V o = 1, 3 and 8 m/s. SHEMATI: ASSUMPTIONS: (1) Steady-state conditions, () Fully developed internal flow, PROPERTIES: Table A.1, Stainless steel, AISI 30 (300 K): k w = 15.1 W/m K; Table A.6, Water ( T m,i = 348 K): ρ i = 974.8 kg/m 3, µ i = 3.746 10-4 N s/m, k i = 0.668 W/m K, Pr i =.354; Table A.4, Air (assume T f,o = 315K, 1 atm): k o = 0.0737 W/m K, ν o = 17.35 10-6 m /s, Pr o = 0.705. ANALYSIS: (a) For the water-air condition, the overall coefficient, Eq. 11.1, based upon the outer area can be expressed as the sum of the thermal resistances due to convection (cv), tube wall conduction (w) and fouling (f): 1UoAo = Rtot = Rcv,i + Rf,i + Rw + Rf,o + Rcv,o Rcv,i = 1 hi Ai Rcv,o = 1 ho Ao Rf,i = R f,i Ai Rf,o = R f,o Ao and from Eq. 3.8, ( ) ( π ) Rw = ln Do Di Lkw The convection coefficients can be estimated from appropriate correlations. ontinued...
PROBLEM 11. (ont.) Estimating h i : For internal flow, characterize the flow evaluating thermophysical properties at T m,i with um,idi 0.5 m s 0.0m ReD,i = = = 8,65 νi 3.746 10 4 N s m 974.8 kg m 3 For the turbulent flow, use the Dittus-Boelter correlation, Eq. 8.60, Nu 0.8 0.4 D,i = 0.03Re D,i Pr i 0.8 0.4 NuD,i = 0.03( 8, 65) (.354) = 119.1 h i = NuD,i ki Di = 119.1 0.668 W m K 0.0m = 3616 W m K Estimating h o : For external flow, characterize the flow with VD o o 0 m s 0.07m ReD,o = = = 31,14 νo 17.35 10 6 m s evaluating thermophysical properties at T f,o = (T s,o + T o )/ when the surface temperature is determined from the thermal circuit analysis result, ( ) = ( ) Tm,i To Rtot Ts,o To Rcv,o Assume T f,o = 315 K, and check later. Using the hurchill-bernstein correlation, Eq. 7.57, find 4/5 0.6 Re 1/ 1/3 5/8 D,o Pro ReD,o NuD,o = 0.3+ 1 /3 1/4 + 8, 000 1+ ( 0.4 Pr o ) 1/ 1/3 5/8 4/5 0.6( 31,14) ( 0.705) 31,14 NuD,o = 0.3+ 1 + /3 1/4 8, 000 1+ ( 0.4 0.705) NuD,o = 10.6 ho = NuD,o ko Do = 10.6 0.0737 W m K 0.07m = 104.0 W m K Using the above values for h i, and h o, and other prescribed values, the thermal resistances and overall coefficient can be evaluated and are tabulated below. R cv,i R f,i R w R f,o R cv,o U o R tot (K/W) (K/W) (K/W) (K/W) (K/W) (W/m K) (K/W) 0.00436 0.00578 0.0016 0.0036 0.1134 9.1 0.18 The major thermal resistance is due to outside (air) convection, accounting for 89% of the total resistance. The other thermal resistances are of similar magnitude, nearly 50 times smaller than R cv,o. (b) For the water-water condition, the method of analysis follows that of part (a). For the internal flow, the estimated convection coefficient is the same as part (a). For an assumed outer film coefficient, T f,o= 9 K, the convection correlation for the outer water flow condition V o = 1 m/s and T o = 15, find
PROBLEM 11. (ont.) Re D,o = 6, 60 NuD,o = 0.6 ho = 4914 W m K The thermal resistances and overall coefficient are tabulated below. R cv,i R f,i R w R f,o R cv,o R tot U o (K/W) (K/W) (K/W) (K/W) (K/W) (K/W) (W/m K) 0.00436 0.00579 0.0016 0.0036 0.0040 0.0171 691 Note that the thermal resistances are of similar magnitude. In contrast with the results for the water-air condition of part (a), the thermal resistance of the outside convection process, R cv,o, is nearly 50 times smaller. The overall coefficient for the water-water condition is 7.5 times greater than that for the waterair condition. (c) For the water-air condition, using the IHT workspace with the analysis of part (a), U o was calculated as a function of the air cross-flow velocity for selected mean water velocities. Water (i) - air (o) condition Uo (W/m^.K) 10 100 80 60 40 5 10 15 0 5 30 Air velocity, Vo (m/s) Water mean velocity, umi = 0. m/s umi = 0.5 m/s umi = 1.0 m/s The effect of increasing the cross-flow air velocity is to increase U o since the R cv,o is the dominant thermal resistance for the system. While increasing the water mean velocity will increase h i, because R cv,i R cv,o, this increase has only a small effect on U o. (d) For the water-water condition, using the IHT workplace with the analysis of part (b), U o was calculated as a function of the mean water velocity for selected air cross-flow velocities. Water (i) - water (o) condition Uo (W/m^.K) 1000 900 800 700 600 0.5 1 1.5.5 Water mean velocity, umi (m/s) Air velocity, Vo = 1 m/s Vo = 3 m/s Vo = 8 m/s Because the thermal resistances for the convection processes, R cv,i and R cv,o, are of similar magnitude according to the results of part (b), we expect to see U o significantly increase with increasing water mean velocity and air cross-flow velocity.
PROBLEM 11.16 KNOWN: Inner tube diameter (D = 0.0 m) and fluid inlet and outlet temperatures corresponding to design conditions for a concentric tube heat exchanger. Overall heat transfer coefficient (U = 500 W/m K) and desired heat rate (q = 3000 W). old fluid outlet temperature after three years of operation. FIND: (a) Required heat exchanger length, (b) Heat rate, hot fluid outlet temperature, overall heat transfer coefficient, and fouling factor after three years. SHEMATI: ASSUMPTIONS: (1) Negligible heat loss to the surroundings and kinetic and potential energy changes, () Negligible tube wall conduction resistance, (3) onstant properties. ANALYSIS: (a) The tube length needed to achieve the prescribed conditions may be obtained from Eqs. 11.14 and 11.15 where T 1 = T h,i - T c,o = 80 and T = T h,o - T c,i = 10. Hence, T 1m = (10-80) /ln(10/80) = 98.7 and q 3000 W L = 0.968 m ( π D) U T = 1m ( 0.0 m) = π 500 W m K 98.7 (b) With q = c (T c,o - T c,i ), the following ratio may be formed in terms of the design and 3 year conditions. Hence, ( ) ( ) c Tc,o T q c,i 60 = = = 1.333 q3 c Tc,o T c,i 3 45 q3 = q 1.33 = 3000 W 1.333 = 50 W Having determined the ratio of heat rates, it follows that q h ( Th,i T h,o ) 0 = = = 1.333 q3 h ( Th,i T h,o ) 160 T 3 h,o(3) Hence, Th,o(3) = 160 0 1.333 = 145 With ( ) ( ) Tlm,3 = 15 95 ln 15 95 = 109.3, q3 50 W U 3 = 338 W m K π DL T = = 1m,3 π 0.0 m 0.968m 109.3 ( ) ( ) ( ) ontinued...
1 PROBLEM 11.16 (ont.) 1 U3 = 1 hi + 1 ho + R f,c, With U= ( 1 hi) + ( 1 ho) and ( ) ( ) 1 1 1 1 R 4 f,c = = m K W 9.59 10 = m K W U3 U 338 500 OMMENTS: Over time fouling will always contribute to a degradation of heat exchanger performance. In practice it is desirable to remove fluid contaminants and to implement a regular maintenance (cleaning) procedure.
PROBLEM 11.0 KNOWN: ounterflow concentric tube heat exchanger. FIND: (a) Total heat transfer rate and outlet temperature of the water and (b) Required length. SHEMATI: ASSUMPTIONS: (1) Negligible heat loss to surroundings, () Negligible kinetic and potential energy changes, (3) Negligible thermal resistance due to tube wall thickness. PROPERTIES: (given): ρ (kg/m 3 ) c p (J/kg K) ν (m /s) k (W/m K) Pr Water 1000 400 7 10-7 0.64 4.7 Oil 800 1900 1 10-5 0.134 140 ANALYSIS: (a) With the outlet temperature, T c,o = 60, from an overall energy balance on the hot (oil) fluid, find q = m& hch Th,i Th,o = 0.1kg/s 1900J/kgK100 60 = 7600 W. ( ) ( ) From an energy balance on the cold (water) fluid, find Tc,o = Tc,i + q/m& ccc = 30 + 7600W/0.1kg/s 400J/kg K = 48.1. (b) Using the LMTD method, the length of the F heat exchanger follows from q = UA Tlm,F = U( πdl) Tlm,F L= q/u( πd) Tlm,F where ( ) ( ) ( ) T1 T 60 30 100 48.1 Tlm,F = = = 40.0 ln( T/ 1 T) ln 30/51.9 L = 7600W/60W/m K 0.05m 40.0 = 40.3m. ( π ) OMMENTS: Using the ε-ntu method, find min = h = 190 W/K and max = c = 40 W/K. Hence ( ) ( ) qmax = min Th,i Tc,i = 190 W/K100 30 K = 13,300 W and ε=q/q max = 0.571. With r = min / max = 0.45 and using Eq. 11.30b, UA 1 ε 1 1 0.571 1 NTU = = ln = ln = 1.00 min r 1 εr 1 0.45 1 0.571 0.45 1 so that with A = πdl, find L = 40.3 m.
PROBLEM 11.1 KNOWN: ounterflow, concentric tube heat exchanger undergoing test after service for an extended period of time; surface area of 5 m and design value for the overall heat transfer coefficient of U d = 38 W/m K. FIND: Fouling factor, if any, based upon the test results of engine oil flowing at 0.1 kg/s cooled from 110 to 66 by water supplied at 5 and a flow rate of 0. kg/s. SHEMATI: ASSUMPTIONS: (1) Steady-state conditions, () Negligible losses to the surroundings and kinetic and potential energy changes, (3) onstant properties. PROPERTIES: Table A-5, Engine oil ( T h = 361 K): c = 166 J/kg K; Table A-6, Water ( c = c,o = ) T 304 K, assuming T 36 : c = 4178 J/kg K. ANALYSIS: For the F conditions shown in the Schematic, find the heat rate, q, from an energy balance on the hot fluid (oil); the cold fluid outlet temperature, T c,o, from an energy balance on the cold fluid (water); the overall coefficient U from the rate equation; and a fouling factor, R, by comparison with the design value, U d. Energy balance on hot fluid q = m hc h 3Th,i T h,o8= 01. kg / s 166 J / kg K 1110 666K = 9530 W Energy balance on the cold fluid q = m cc c 3Tc,o Tc,i8, find Tc,o = 36. 4 Rate equation q = UA T " n,f 3Th,i Tc,o8 3Th,o 1110 36. 46 166 56 T n,f = = = 557. " " n3th,i Tc,o8/ 3Th,o Tc,i8 " n 73.6 / 41.0 9530 W = U 5 m 55. 7 U = 34. W / m K Overall resistance including fouling factor U= 1/ 1/Ud + R f 34. W/ m K = 1/ 1/38 W/ m K+ R f R f = 0. 009 m K/W