Principals of heat transfer Frank Kreith, Raj M.

8.4 Log Mean Temperature Difference 503 1.0 Z= 0.2 0.1 0.9 0.30.40.50.6 0.7 0.8 0.8 F 0.9 1.0 0.7 1.2 1.4 1.6 1.8 2.0 2.5 3.0 4.0 6.0 8.0 10.0 15.0 Z = 20.0 0.6 0.5 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0 P = (Tt, out – Tt, in) /(Ts, in – Tt, in) Ts, in Tt, out Tt, in Ts, out FIGURE 8.15 Correction factor to counterflow LMTD for heat exchanger with two shell passes and a multiple of two tube passes. Source: Courtesy of the Tubular Exchanger Manufacturers Association. 1.0 Ts, in 0.9 0.8 0.2 Tt, in Tt, out F 0.4 0.6 0.7 0.8 1.0 0.6 1.5 2.0 3.0 Z = 4.0 0.5 Ts, out 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 P = Tt, out – Tt, in Ts, in – Tt, in FIGURE 8.16 Correction factor to counterflow LMTD for cross-flow heat exchangers with the fluid on the shell side mixed, the other fluid unmixed, and one tube pass. Source: Extracted from Bowman, Mueller, and Nagel [17], with permission of the publishers, the American Society of Mechanical Engineers. Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

504 Chapter 8 Heat Exchangers Text not available due to copyright restrictions EXAMPLE 8.1 Determine the heat transfer surface area required for a heat exchanger constructed from a 0.0254-m-OD tube to cool 6.93 kg/s of a 95% ethyl alcohol solution (cp ϭ 3810 J/kg K) from 65.6°C to 39.4°C, using 6.30 kg/s of water available at 10°C. Assume that the overall coefficient of heat transfer based on the outer-tube area is 568 W/m2 K and consider each of the following arrangements: (a) Parallel-flow tube and shell (b) Counterflow tube and shell (c) Counterflow exchanger with 2 shell passes and 72 tube passes, the alcohol flow- ing through the shell and the water flowing through the tubes (d) Cross-flow, with one tube pass and one shell pass, shell-side fluid mixed SOLUTION The outlet temperature of the water for any of the four arrangements can be obtained from an overall energy balance, assuming that the heat loss to the atmosphere is neg- ligible. Writing the energy balance as m# hcph(Th, in - Th, out) = m# ccpc(Tc, out - Tc, in) and substituting the data in this equation, we obtain (6.93)(3810)(65.6 Ϫ 39.4) ϭ (6.30)(4187)(Tc,out Ϫ 10) Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

8.4 Log Mean Temperature Difference 505 from which the outlet temperature of the water is found to be 36.2°C. The rate of heat flow from the alcohol to the water is q = m# hcph(Th,in - Th,out) = (6.93 kg/s)(3810 J/kg K)(65.6 - 39.4)(K) = 691,800 W (a) From Eq. (8.18) the LMTD for parallel flow is LMTD = ¢Ta - ¢Tb = 55.6 - 3.2 = 18.4 °C ln(¢Ta> ¢Tb) ln(55.6>3.2) From Eq. (8.16) the heat transfer surface area is A= q = (691,800 W) = 66.2 m2 (U )(LMTD) (568 W/m2 K)(18.4 K) The 830-m length of the exchanger for a 0.0254-m-OD tube would be too great to be practical. (b) For the counterflow arrangement, the appropriate mean temperature differ- ence is 65.6 Ϫ 36.2 ϭ 29.4°C, because m# ccpc = m# hcph. The required area is q = 691,800 = 41.4 m2 A= (U )(LMTD) (568)(29.4) which is about 40% less than the area necessary for parallel flow. (c) For the two-shell-pass counterflow arrangement, we determine the appro- priate mean temperature difference by applying the correction factor found from Fig. 8.15 to the mean temperature for counterflow: Tc,out - Tc,in 36.2 - 10 P = Th,in - Tc,in = 65.6 - 10 = 0.47 and the heat capacity rate ratio is m# tcpt Z = m# scps = 1 From the chart of Fig. 8.15, F ϭ 0.97 and the heat transfer area is A = 41.4 = 42.7 m2 0.97 Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

506 Chapter 8 Heat Exchangers The length of the exchanger for 72, 0.0254-m-OD tubes in parallel would be L= A/72 = 42.7/72 = 7.4 m pD p(0.0254) This length is not unreasonable, but if it is desirable to shorten the exchanger, more tubes could be used. (d) For the cross-flow arrangement (Fig. 8.4), the correction factor is found from the chart of Fig. 8.16 to be 0.88. The required surface area is thus 47.0 m2, about 10% larger than that for the exchanger in part (c). 8.5 Heat Exchanger Effectiveness In the thermal analysis of the various types of heat exchangers presented in the pre- ceding section, we used [Eq. (8.17)] expressed as q ϭ UA ⌬Tmean This form is convenient when all the terminal temperatures necessary for the evaluation of the appropriate mean temperature are known, and Eq. (8.17) is widely employed in the design of heat exchangers to given specifications. There are, however, numerous occasions when the performance of a heat exchanger (i.e., U) is known or can at least be estimated but the temperatures of the fluids leaving the exchanger are not known. This type of problem is encountered in the selection of a heat exchanger or when the unit has been tested at one flow rate, but service conditions require different flow rates for one or both fluids. In heat exchanger design texts and handbooks, this type of prob- lem is also referred to as a rating problem, where the outlet temperatures or the total heat load needs to be determined, given the size (A) and the convective performance (U) of the unit. The outlet temperatures and the rate of heat flow can be found only by a rather tedious trial-and-error procedure if the charts presented in the preceding section are used. In such cases it is desirable to circumvent entirely any reference to the loga- rithmic or any other mean temperature difference. A method that accomplishes this has been proposed by Nusselt [18] and Ten Broeck [19]. To obtain an equation for the rate of heat transfer that does not involve any of the outlet temperatures, we introduce the heat exchanger effectiveness Ᏹ.The heat exchanger effectiveness is defined as the ratio of the actual rate of heat transfer in a given heat exchanger to the maximum possible rate of heat exchange. The latter would be obtained in a counterflow heat exchanger of infinite heat transfer area. In this type of unit, if there are no external heat losses, the outlet temperature of the colder fluid equals the inlet temperature of the warmer fluid when m# ccpc 6 m# hcph; when m# hcph 6 m# ccpc, the outlet temperature of the warmer fluid equals the inlet temperature of the colder one. In other words, the effectiveness compares the actual heat transfer rate to the maximum rate whose only limit is the Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

8.5 Heat Exchanger Effectiveness 507 second law of thermodynamics. Depending on which of the heat capacity rates is smaller, the effectiveness is Ch(Th,in - Th,out) (8.22a) Ᏹ = Cmin (Th,in - Tc,in) or Cc(Tc,out - Tc,in) (8.22b) Ᏹ= Cmin (Th,in - Tc,in) where Cmin is the smaller of the m# hcph and m# ccpc magnitudes. It may be noted that the denominator in Eq. (8.22) is the thermodynamically maximum heat transfer possible between the hot and cold fluids flowing through the heat exchanger, given their respective inlet temperature and mass flow rate, or the maximum available energy. The numerator is the actual heat transfer accom- plished in the unit, and hence the effectiveness Ᏹ represents a thermodynamic performance of the heat exchanger. Once the effectiveness of a heat exchanger is known, the rate of heat transfer can be determined directly from the equation q = Ᏹ Cmin (Th,in - Tc,in) (8.23) since Ᏹ Cmin (Th,in - Tc,in) = Ch(Th,in - Th,out) = Cc(Tc,out - Tc,in) Equation (8.23) is the basic relation in this analysis because it expresses the rate of heat transfer in terms of the effectiveness, the smaller heat capacity rate, and the dif- ference between the inlet temperatures. It replaces Eq. (8.17) in the LMTD analysis but does not involve the outlet temperatures. Equation (8.23) is, of course, also suit- able for design purposes and can be used instead of Eq. (8.17). We shall illustrate the method of deriving an expression for the effectiveness of a heat exchanger by applying it to a parallel-flow arrangement. The effectiveness can be introduced into Eq. (8.13) by replacing (Tc,in Ϫ Tc,out)>(Th,in Ϫ Tc,in) by the effectiveness relation from Eq. (8.22b). We obtain ln c1 - Ᏹ a Cmin + Cmin b d = - a 1 + 1 bUA Ch Cc Cc Ch or 1 - Ᏹ a Cmin + Cmin b = e-(1>Cc + 1>Ch)UA Ch Cc Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

508 Chapter 8 Heat Exchangers Solving for Ᏹ yields Ᏹ= 1 - e-[1 + (Ch>Cc)]UA>Ch (8.24) (8.25a) 1Cmin >Ch2 + 1Cmin >Cc2 (8.25b) When Ch is less than Cc, the effectiveness becomes 1 - e-[1 + (Ch >Cc)]UA>Ch Ᏹ = 1 + 1Ch>Cc2 and when Cc Ͻ Ch, we obtain 1 - e-[1 + (Cc>Ch)]UA>Cc Ᏹ = 1 + 1Cc >Ch2 The effectiveness for both cases can therefore be written in the form Ᏹ = 1 - e-[1 + (Cmin >Cmax )]UA>Cmin (8.26) 1 + 1Cmin >Cmax 2 The foregoing derivation illustrates how the effectiveness for a given flow arrangement can be expressed in terms of two dimensionless parameters, the heat Parallel Flow Exchanger Performance Counterflow Exchanger Performance Hot fluid (m.c)h = Ch Hot fluid (m.c)h = Ch Cold fluid (m.c)c = Cc Cold fluid (m.c)c = Cc Heat transfer surface Heat transfer surface 100 100 Cmin/Cmax = 0 Cmin/Cmax = 0 0.25 0.50 0.75 80 0.25 80 1.00 0.50 60 60 0.75 40 1.00 40 20 20 0 012345 0 Number of transfer units, NTU = AU/Cmin 012345 Number of transfer units, NTU = AU/Cmin FIGURE 8.19 Heat exchanger effective- ness for counterflow. FIGURE 8.18 Heat exchanger effective- ness for parallel flow. Source: With permission from Kays and London [10]. Source: With permission from Kays and London [10]. Effectiveness, ℰ (%)v Effectiveness, ℰ (%) Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

1–2 Parallel-Counterflow 8.5 Heat Exchanger Effectiveness 509 Exchanger Performance Cross-Flow Exchanger with Fluids Unmixed Shell fluid (m.c)s = Cs (m.c)c Cold fluid Tube fluid (m.c)t = Ct (m.c)h Hot fluid One shell pass, 2, 4, 6, etc., tube passes 100 100 Cmin/Cmax = 0 Cmin/Cmax = 0 0.25 0.25 0.50 80 0.50 80 0.75 Effectiveness, ℰ (%) 1.00 Effectiveness, ℰ (%) 0.75 60 60 1.00 40 40 20 20 0 0 012345 012345 Number of transfer units, NTU = AU/Cmin Number of transfer units, NTU = AU/Cmin FIGURE 8.21 Heat exchanger effective- FIGURE 8.20 Heat exchanger effective- ness for cross-flow with both fluids ness for shell-and-tube heat exchanger unmixed. with one well-baffled shell pass and two (or a multiple of two) tube passes. Source: With permission from Kays and London [10]. Source: With permission from Kays and London [10]. capacity rate ratio Cmin/Cmax and the ratio of the overall conductance to the smaller heat capacity rate, UA/Cmin. The latter of the two parameters is called the number of heat transfer units or NTU. The number of heat transfer units is a measure of the heat transfer size of the exchanger. The larger the value of NTU, the closer the heat exchanger approaches its thermodynamic limit. By analyses that, in principle, are similar to the one presented here for parallel flow, effective- ness can be evaluated for most flow arrangements of practical interest. The results have been put together by Kays and London [10] into convenient graphs from which the effectiveness can be determined for given values of NTU and Cmin/Cmax. The effectiveness curves for some common flow arrangements are shown in Figs. 8.18–8.22. The abscissas of these figures are the NTUs of the heat exchangers. The constant parameter for each curve is the heat capacity rate ratio Cmin/Cmax, and the effectiveness is read on the ordinate. Note that for an evapo- rator or condenser, Cmin/Cmax ϭ 0, because if one fluid remains at constant tem- perature throughout the exchanger, its effective specific heat and thus its heat capacity rate are by definition equal to infinity. Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

510 Chapter 8 Heat Exchangers Cross-Flow Exchanger with One Fluid Unmixed Mixed fluid Effectiveness, ℰ (%) 100 Unmixed fluid Cmixed = 0 Cunmixed 0 0.25 0.50 4 80 0.75 2 60 1.33 40 1.00 20 Cmixed = 1 Cunmixed 0 012345 Number of transfer units, NTU = AU/Cmin FIGURE 8.22 Heat exchanger effective- ness for cross-flow with one fluid mixed and the other unmixed. When Cmixed/ Cunmixed Ͼ 1, NTU is based on Cunmixed. Source: With permission from W. M. Kays and A. L. London [10]. EXAMPLE 8.2 From a performance test on a well-baffled single-shell, two-tube-pass heat exchanger, the following data are available: oil (cp ϭ 2100 J/kg K) in turbulent flow inside the tubes entered at 340 K at the rate of 1.00 kg/s and left at 310 K; water flowing on the shell side entered at 290 K and left at 300 K. A change in serv- ice conditions requires the cooling of a similar oil from an initial temperature of 370 K but at three-fourths of the flow rate used in the performance test. Estimate the outlet temperature of the oil for the same water flow rate and inlet temperature as before. SOLUTION The test data can be used to determine the heat capacity rate of the water and the overall conductance of the exchanger. The heat capacity rate of the water is, from Eq. (8.14), Th,in - Th,out 340 - 310 Cc = Ch Tc,out - Tc,in = (1.00 Kg/s)(2100 J/kg K) 300 - 290 = 6300 W/K Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

8.5 Heat Exchanger Effectiveness 511 and the temperature ratio P is, from Eq. (8.20), Tt,out - Tt,in 340 - 310 P = Ts,in - Tt,in = 340 - 290 = 0.6 Z = 300 - 290 = 0.33 340 - 310 From Fig. 8.14, F ϭ 0.94 and the mean temperature difference is (340 - 300) - (310 - 290) ¢Tmean = (F )(LMTD) = (0.94) ln[(340 - 300)>(310 - 290)] = 27.1 K From Eq. (8.17) the overall conductance is UA = q (1.00 kg/s)(2100 J/kg K)(340 - 310)(K) = 2325 W/K ¢Tmean = (27.1 K) Since the thermal resistance on the oil side is controlling, a decrease in velocity to 75% of the original value will increase the thermal resistance by roughly the veloc- ity ratio raised to the 0.8 power. This can be verified by reference to Eq. (6.62). Under the new conditions, the conductance, the NTU, and the heat capacity rate ratio will therefore be approximately UA M (2325)(0.75)0.8 = 1850 W/K NTU = UA (1850 W/K) = 1.17 = Coil (0.75)(1.00 kg/s)(2100 J/kg K) and Coil Cmin (0.75)(1.00 kg/s)(2100 J/kg K) = 0.25 Cwater Cmax = = (6300 W/K) From Fig. 8.20 the effectiveness is equal to 0.61. Hence from the definition of Ᏹ in Eq. (8.22a), the oil outlet temperature is Toil out ϭ Toil in Ϫ Ᏹ ⌬Tmax ϭ 370 Ϫ [0.61(370 Ϫ 290)] ϭ 321.2 K The next example illustrates a more complex problem. EXAMPLE 8.3 A flat-plate-type heater (Fig. 8.23) is to be used to heat air with the hot exhaust gases from a turbine. The required airflow rate is 0.75 kg/s, entering at 290 K; the hot gases are available at a temperature of 1150 K and a mass flow rate of 0.60 kg/s. Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

512 Chapter 8 Heat Exchangers A Gas out A Air in, Air 290 K 19 air passages 0.178 m 18 gas passages 0.343 m Metal thickness = 0.762 mm Gas Gas in 6.71 mm 1150 K 8.23 mm 0.3048 m Enlarged Portion of Section A–A Air out FIGURE 8.23 Flat-plate-type heater. Determine the temperature of the air leaving the heat exchanger for the parameters listed below. Pa ϭ wetted perimeter on air side, 0.703 m Pg ϭ wetted perimeter on gas side, 0.416 m Ag ϭ cross-sectional area of gas passage (per passage), 1.6 ϫ 10Ϫ3 m2 Aa ϭ cross-sectional area of air passage (per passage), 2.275 ϫ 10Ϫ3 m2 A ϭ heat transfer surface area, 2.52 m2 SOLUTION Inspection of Fig. 8.23 shows that the unit is of the cross-flow type, with both fluids unmixed. As a first approximation, the end effects will be neglected. The flow sys- tems for the air and gas streams are similar to flow in straight ducts having the fol- lowing dimensions: La ϭ length of air duct, 0.178 m DHa ϭ hydraulic diameter of air duct, 4Aa = 0.0129 m Pa Lg ϭ length of gas duct, 0.343 m 4Ag DHg ϭ hydraulic diameter of gas duct, Pg = 0.0154 m A ϭ heat transfer surface area, 2.52 m2 The heat transfer coefficients can be evaluated from Eq. (6.63) for flow in ducts (La>DHa ϭ 13.8, Lg>DHg ϭ 22.3). A difficulty arises, however, because the temper- atures of both fluids vary along the duct. It is therefore necessary to estimate an aver- age bulk temperature and refine the calculations after the outlet and wall temperatures have been found. Selecting the average air-side bulk temperature to be Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.

8.5 Heat Exchanger Effectiveness 513 573 K and the average gas-side bulk temperature to be 973 K, the properties at those temperatures are, from Appendix 2, Table 28 (assuming that the properties of the gas can be approximated by those of air): ␮air ϭ 2.93 ϫ 10Ϫ5 N s/m2 ␮gas ϭ 4.085 ϫ 10Ϫ5 N s/m2 Prair ϭ 0.71 Prgas ϭ 0.73 kair ϭ 0.0429 W/m K kgas ϭ 0.0623 W/m K cpair ϭ 1047 J/kg K cpgas ϭ 1101 J/kg K The mass flow rates per unit area are a m# b = (0.75 kg/s) = 17.35 kg/m2 s A air (19)(2.275 * 10-3 m2) a m# b = (0.60 kg/s) = 20.83 kg/m2 s A gas (18)(1.600 * 10-3 m2) The Reynolds numbers are Reair = (m# /A)airDHa = (17.35 kg/m2 s)(0.0129 m) = 7640 ma (2.93 * 10-5 kg/m s) (m# /A)gasDHg (20.83 kg/m2 s)(0.0154 m) Regas = mg = (4.085 * 10-5 kg/m s) = 7850 Using Eq. (6.63), the average heat transfer coefficients are qhair = 0.023 ka Re0ai.r8 Pr 0.4 DHa = 0.023 0.0429 (7640)0.8(0.71)0.4 0.0129 = 85.2 W/m2 K Since La/DHa ϭ 13.8, we must correct this heat transfer coefficient for entrance effects, per Eq. (6.68). The correction factor is 1.377, so the corrected heat transfer coefficient is (1.377)(85.2) = 117 W/m2 K = qhair. hqgas = (0.023) 0.0623 (7850)0.8(0.73)0.4 0.0154 = 107.1 W/m2 K Since Lg/DHg ϭ 22.3, we must correct this heat transfer coefficient for entrance effects, per Eq. (6.69). The correction factor is 1 ϩ 6(DHg/Lg) ϭ 1.27, so the cor- rected heat transfer coefficient is (1.27)(107.1) = 136 W/m2 K = hqgas. The thermal resistance of the metal wall is negligible, therefore the overall con- ductance is UA = 1 = 1 11 11 qhaA + qhgA (117 W/m2 K)(2.52 m2) + (136 W/m2 K)(2.52 m2) = 158 W/K Copyright 2011 Cengage Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.