11.1 SECTION 11 HEAT TRANSFER AND HEAT EXCHANGE Selecting Type of Heat Exchanger for a Specific Application 11.1 Shell-and-Tube Heat Exchanger Size 11.4 Heat Exchanger Actual Temperature Difference 11.6 Fouling Factors in Heat-Exchanger Sizing and Selection 11.8 Heat Transfer in Barometric and Jet Condensers 11.10 Selection of a Finned-Tube Heat Exchanger 11.12 Spiral-Type Heating-Coil Selection 11.15 Sizing Electric Heaters for Industrial Use 11.16 Economizer Heat Transfer Coefficient 11.19 Boiler-Tube Steam-Generating Capacity 11.20 Shell-and-Tube Heat Exchanger Design Analysis 11.21 Designing Spiral-Plate Heat Exchangers 11.37 Spiral-Tube Heat Exchanger Design 11.46 Heat-Transfer Design for Internal Steam Tracing of Pipelines 11.54 Designing Heat-Transfer Surfaces for External Heat Tracing of Pipelines 11.62 Air-Cooled Heat Exchangers: Preliminary Selection 11.67 Heat Exchangers: Quick Design and Evaluation 11.70 SELECTING TYPE OF HEAT EXCHANGER FOR A SPECIFIC APPLICATION Determine the type of heat exchanger to use for each of the following applications: (1) heating oil with steam; (2) cooling internal combustion engine liquid coolant; (3) evaporating a hot liquid. For each heater chosen, specify the typical pressure range for which the heater is usually built and the typical range of the overall coefficient of heat transfer U. Calculation Procedure: 1. Determine the heat-transfer process involved In a heat exchanger, one or more of four processes may occur: heating, cooling, boiling, or condensing. Table 1 lists each of these four processes and shows the usual heat-transfer fluids involved. Thus, the heat exchangers being considered here involve (a) oil heater—heating—vapor-liquid; (b) internal-combustion engine coolant—cooling—gas-liquid; (c) hot-liquid evaporation—boiling—liquid-liquid. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 11.2 TABLE 1 Heat-Exchanger Selection Guide* Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. HEAT TRANSFER AND HEAT EXCHANGE 11.3 TABLE 1 (Continued ) Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. HEAT TRANSFER AND HEAT EXCHANGE 11.4 PLANT AND FACILITIES ENGINEERING 2. Specify the heater action and the usual type selected Using the same identifying letters for the heaters being selected, Table 1 shows the action and usual type of heater chosen. Thus, 3. Specify the usual pressure range and typical U Using the same identifying letters for the heaters being selected, Table 1 shows the action and usual type of heater chosen. Thus, 4. Select the heater for each service Where the heat-transfer conditions are normal for the type of service met, the type of heater listed in step 2 can be safely used. When the heat-transfer conditions are unusual, a special type of heater may be needed. To select such a heater, study the data in Table 1 and make a tentative selection. Check the selection by using the methods given in the following calculation procedures in this section. Related Calculations. Use Table 1 as a general guide to heat-exchanger selec- tion in any industry—petroleum, chemical, power, marine, textile, lumber, etc. Once the general type of heater and its typical U value are known, compute the required size, using the procedure given later in this section. SHELL-AND-TUBE HEAT EXCHANGER SIZE What is the required heat-transfer area for a parallel-flow shell-and-tube heat ex- changer used to heat oil if the entering oil temperature is 60 Њ F (15.6 Њ C), the leaving oil temperature is 120 Њ F (48.9 Њ C), and the heating medium is steam at 200 lb /in 2 (abs) (1378.8 kPa)? There is no subcooling of condensate in the heat exchanger. The overall coefficient of heat transfer U ϭ 25 Btu / (h ⅐ Њ F ⅐ ft 2 ) [141.9 W / (m 2 ⅐ Њ C)]. How much heating steam is required if the oil flow rate through the heater is 100 gal/min (6.3 L / s), the specific gravity of the oil is 0.9, and the specific heat of the oil is 0.5 Btu/(lb ⅐ Њ F) [2.84 W / (m 2 ⅐ Њ C)]? Calculation Procedure: 1. Compute the heat-transfer rate of the heater With a flow rate of 100 gal / min (6.3 L/s) or (100 gal/ min)(60 min/ h) ϭ 6000 gal/h (22,710 L/h), the weight flow rate of the oil, using the weight of water of Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. HEAT TRANSFER AND HEAT EXCHANGE HEAT TRANSFER AND HEAT EXCHANGE 11.5 FIGURE 1 Temperature relations in typical parallel-flow and counterflow heat exchang- ers. specific gravity 1.0 as 8.33 lb/gal, is (6000 gal/h) (0.9 specific gravity)(8.33 lb / gal) ϭ 45,000 lb/h (20,250 kg/h), closely. Since the temperature of the oil rises 120 Ϫ 60 ϭ 60 Њ F (33.3 Њ C) during passage through the heat exchanger and the oil has a specific heat of 0.50, find the heat- transfer rate of the heater from the general relation Q ϭ wc ⌬ t, where Q ϭ heat- transfer rate, Btu/ h; w ϭ oil flow rate, lb/h; c ϭ specific heat of the oil, Btu / (lb ⅐ Њ F); ⌬ t ϭ temperature rise of the oil during passage through the heater. Thus, Q ϭ (45,000)(0.5)(60) ϭ 1,350,000 Btu/h (0.4 MW). 2. Compute the heater logarithmic mean temperature difference The logarithmic mean temperature difference (LMTD) is found from LMTD ϭ (G Ϫ L)/ln (G/L), where G ϭ greater terminal temperature difference of the heater, Њ F; L ϭ lower terminal temperature difference of the heater, Њ F; ln ϭ logarithm to the base e. This relation is valid for heat exchangers in which the number of shell passes equals the number of tube passes. In general, for parallel flow of the fluid streams, G ϭ T 1 Ϫ t 1 and L ϭ T 2 Ϫ t 2 , where T 1 ϭ heating fluid inlet temperature, Њ F; T 2 ϭ heating fluid outlet temperature, Њ F; t 1 ϭ heated fluid inlet temperature, Њ F; t 2 ϭ heated fluid outlet temperature, Њ F. Figure 1 shows the maximum and minimum terminal temperature differences for various fluid flow paths. For this parallel-flow exchanger, G ϭ T 1 Ϫ t 1 ϭ 382 Ϫ 60 ϭ 322 Њ F (179 Њ C), where 382 Њ F (194 Њ C) ϭ the temperature of 200-lb/in 2 (abs) (1379-kPa) saturated steam, from a table of steam properties. Also, L ϭ T 2 Ϫ t 2 ϭ 382 Ϫ 120 ϭ 262 Њ F (145.6 Њ C), where the condensate temperature ϭ the saturated steam temperature Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. HEAT TRANSFER AND HEAT EXCHANGE 11.6 PLANT AND FACILITIES ENGINEERING because there is no subcooling of the condensate. Then LMTD ϭ G Ϫ L /ln (G/L) ϭ (322 Ϫ 262)/ln(322 / 262) ϭ 290 Њ F (161 Њ C). 3. Compute the required heat-transfer area Use the relation A ϭ Q/U ϫ LMTD, where A ϭ required heat-transfer area, ft 2 ; U ϭ overall coefficient of heat transfer, Btu /(ft 2 ⅐ h ⅐ Њ F). Thus, A ϭ 1,350,000/ [(25)(290)] ϭ 186.4 ft 2 (17.3 m 2 ), say 200 ft 2 (18.6 m 2 ). 4. Compute the required quantity of heating steam The heat added to the oil ϭ Q ϭ 1,350,000 Btu/ h, from step 1. The enthalpy of vaporization of 200-lb/in 2 (abs) (1379-kPa) saturated steam is, from the steam ta- bles, 843.0 Btu/ lb (1960.8 kJ / kg). Use the relation W ϭ where W ϭ flow Q/h , ƒg rate of heating steam, lb/ h; ϭ enthalpy of vaporization of the heating steam, h ƒg Btu/lb. Hence, W ϭ 1,350,000/843.0 ϭ 1600 lb/h (720 kg/h). Related Calculations. Use this general procedure to find the heat-transfer area, fluid outlet temperature, and required heating-fluid flow rate when true parallel flow or counterflow of the fluids occurs in the heat exchanger. When such a true flow does not exist, use a sitable correction factor, as shown in the next calculation procedure. The procedure described here can be used for heat exchangers in power plants, heating systems, marine propulsion, air-conditioning systems, etc. Any heating or cooling fluid—steam, gas, chilled water, etc.—can be used. To select a heat exchanger by using the results of this calculation procedure, enter the engineering data tables available from manufacturers at the computed heat- transfer area. Read the heater dimensions directly from the table. Be sure to use the next larger heat-transfer area when the exact required area is not available. When there is little movement of the fluid on either side of the heat-transfer area, such as occurs during heat transmission through a building wall, the arithmetic mean (average) temperature difference can be used instead of the LMTD. Use the LMTD when there is rapid movement of the fluids on either side of the heat-transfer area and a rapid change in temperature in one, or both, fluids. When one of the two fluids is partially, but not totally, evaporated or condensed, the true mean tem- perature difference is different from the arithmetic mean and the LMTD. Special methods, such as those presented in Perry—Chemical Engineers’ Handbook, must be used to compute the actual temperature difference under these conditions. When two liquids or gases with constant specific heats are exchanging heat in a heat exchanger, the area between their temperature curves, Fig. 2, is a measure of the total heat being transferred. Figure 2 shows how the temperature curves vary with the amount of heat-transfer area for counterflow and parallel-flow exchangers when the fluid inlet temperatures are kept constant. As Fig. 2 shows, the counterflow arrangement is superior. If enough heating surface is provided, in a counterflow exchanger, the leaving cold-fluid temperature can be raised above the leaving hot-fluid temperature. This cannot be done in a parallel-flow exchanger, where the temperatures can only ap- proach each other regardless of how much surface is used. The counterflow ar- rangement transfers more heat for given conditions and usually proves more eco- nomical to use. HEAT-EXCHANGER ACTUAL TEMPERATURE DIFFERENCE A counterflow shell-and-tube heat exchanger has one shell pass for the heating fluid and two shell passes for the fluid being heated. What is the actual LMTD for this Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. HEAT TRANSFER AND HEAT EXCHANGE HEAT TRANSFER AND HEAT EXCHANGE 11.7 FIGURE 2 For certain conditions, the area between the temperature curves measures the amount of heat being transferred. exchanger if T 1 ϭ 300 Њ F (148.9 Њ C), T 2 ϭ 250 Њ F (121 Њ C), t 1 ϭ 100 Њ F (37.8 Њ C), and t 2 ϭ 230 Њ F (110 Њ C)? Calculation Procedure: 1. Determine how the LMTD should be computed When the numbers of shell and tube passes are unequal, true counterflow does not exist in the heat exchanger. To allow for this deviation from true counterflow, a correction factor must be applied to the logarithmic mean temperature difference (LMTD). Figure 3 gives the correction factor to use. 2. Compute the variables for the correction factor The two variables that determine the correction factor are shown in Fig. 3 as P ϭ (t 2 Ϫ t 1 )/(T 1 Ϫ t 1 ) and R ϭ (T 1 Ϫ T 2 )/(t 2 Ϫ t 1 ). Thus, P ϭ (230 Ϫ 100)/(300 Ϫ 100) ϭ 0.65, and R ϭ (300 Ϫ 250)/(230 Ϫ 100) ϭ 0.385. From Fig. 3, the correction factor is F ϭ 0.90 for these values of P and R. 3. Compute the theoretical LMTD Use the relation LMTD ϭ (G Ϫ L)/ln(G/L), where the symbols for counterflow heat exchange are G ϭ T 2 Ϫ t 1 ; L ϭ T 1 Ϫ t 2 ;ln ϭ logarithm to the base e. All temperatures in this equation are expressed in Њ F. Thus, G ϭ 250 Ϫ 100 ϭ 150 Њ F (83.3 Њ C); L ϭ 300 Ϫ 230 ϭ 70 Њ F (38.9 Њ C). Then LMTD ϭ (150 Ϫ 70)/ln (150 / 70) ϭ 105 Њ F (58.3 Њ C). Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. HEAT TRANSFER AND HEAT EXCHANGE 11.8 PLANT AND FACILITIES ENGINEERING FIGURE 3 Correction factors for LMTD when the heater flow path differs from the counterflow. (Power.) 4. Compute the actual LMTD for this exchanger The actual LMTD for this or any other heat exchanger is ϭ LMTD actual ϭ 0.9(105) ϭ 94.5 Њ F (52.5 Њ C). Use the actual LMTD to computeF(LMTD ) computed the required exchanger heat-transfer area. Related Calculations. Once the corrected LMTD is known, compute the re- quired heat-exchanger size in the manner shown in the previous calculation pro- cedure. The method given here is valid for both two- and four-pass shell-and-tube heat exchangers. Figure 4 simplifies the computation of the uncorrected LMTD for temperature differences ranging from 1 to 1000 Њ F( Ϫ 17 to 537.8 Њ C). It gives LMTD with sufficient accuracy for all normal industrial and commercial heat-exchanger applications. Correction-factor charts for three shell passes, six or more tube passes, four shell passes, and eight or more tube passes are published in the Standards of the Tubular Exchanger Manufacturers Association. FOULING FACTORS IN HEAT-EXCHANGER SIZING AND SELECTION A heat exchanger having an overall coefficient of heat transfer of U ϭ 100 Btu / (ft 2 ⅐ h ⅐ Њ F) [567.8 W / (m 2 ⅐ Њ C)] is used to cool lean oil. What effect will the tube fouling have on the value of U for this exchanger? Calculation Procedure: 1. Determine the heat exchange fouling factor Use Table 2 to determine the fouling factor for this exchanger. Thus, the fouling factor for lean oil ϭ 0.0020. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. HEAT TRANSFER AND HEAT EXCHANGE HEAT TRANSFER AND HEAT EXCHANGE 11.9 FIGURE 4 Logarithmic mean temperature for a variety of heat- transfer applications. 2. Determine the actual U for the heat exchanger Enter Fig. 5 at the bottom with the clean heat-transfer coefficient of U ϭ 100 Btu/ (h ⅐ ft 2 ⅐ Њ F) [567.8 W / (m 2 ⅐ Њ C)] and project vertically upward to the 0.002 foul- ing-factor curve. From the intersection with this curve, project horizontally to the left to read the design or actual heat-transfer coefficient as U a ϭ 78 Btu/(h ⅐ ft 2 ⅐ Њ F) [442.9 W/ (m 2 ⅐ Њ C)]. Thus, the fouling of the tubes causes a reduction of the U value of 100 Ϫ 78 ϭ 22 Btu / (h ⅐ ft 2 ⅐ Њ F) [124.9 W /(m 2 ⅐ Њ C)]. This means that the required heat transfer area must be increased by nearly 25 percent to compensate for the reduction in heat transfer caused by fouling. Related Calculations. Table 2 gives fouling factors for a wide variety of ser- vice conditions in applications of many types. Use these factors as described above; or add the fouling factor to the film resistance for the heat exchanger to obtain the Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. HEAT TRANSFER AND HEAT EXCHANGE 11.10 PLANT AND FACILITIES ENGINEERING TABLE 2 Heat-Exchanger Fouling Factors* total resistance to heat transfer. Then U ϭ the reciprocal of the total resistance. Use the actual value U a of the heat-transfer coefficient when sizing a heat exchanger. The method given here is that used by Condenser Service and Engineering Com- pany, Inc. HEAT TRANSFER IN BAROMETRIC AND JET CONDENSERS A counterflow barometric condenser must maintain an exhaust pressure of 2 lb/in 2 (abs) (13.8 kPa) for an industrial process. What condensing-water flow rate is re- quired with a cooling-water inlet temperature of 60 Њ F (15.6 Њ C); of 80 Њ F (26.7 Њ C)? How much air must be removed from this barometric condenser if the steam flow rate is 25,000 lb / h (11,250 kg / h); 250,000 lb / h (112,500 kg / h)? Calculation Procedure: 1. Compute the required unit cooling-water flow rate Use Fig. 6 as a quick guide to the required cooling-water flow rate for counterflow barometric condensers. Thus, entering the bottom of Fig. 6 at 2-lb /in 2 (abs) (13.8- kPa) exhaust pressure and projecting vertically upward to the 60 Њ F (15.6 Њ C) and 80 Њ F (26.7 Њ C) cooling-water inlet temperature curves show that the required flow rate is 52 gal / min (3.2 L / s) and 120 gal/min (7.6 L /s), respectively, per 1000 lb /h (450 kg/h) of steam condensed. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. HEAT TRANSFER AND HEAT EXCHANGE [...]... the length of finned tubing required The total area of a finned tube is the sum of the tube and fin area per unit length The tube area is a function of the tube diameter, whereas the finned area is a function of the number of fins per inch of tube length and the tube diameter Assume that 1-in (2.5-cm) tubes having 4 fins per inch (6.35 mm per fin) are used in this radiator A tube manufacturer’s engineering. .. from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website 11.29 HEAT TRANSFER AND HEAT EXCHANGE 11.30 PLANT AND FACILITIES ENGINEERING TABLE 8 Results of Trial Calculations 1st Trial Number of tubes Shell diameter, in Baffle spacing, in Product of factors:... ϫ 108) ft (hr)2 Film coefficient of heat transfer, Btu / [(hr) (ft2) (ЊF)] Thermal conductivity, Btu / [(hr) (ft2) Total series length of tubes, (Lo NPT ϫ number of shells), ft Length of condensing zone, ft LB Subscripts Length of subcooled zone, ft o Lo Length of shell, ft i M NPT Molecular weight, lb / (lb-mol) Number of tube passes per shell, dimensionless Number of tubes per pass (or in parallel)... tube of these dimensions has 5.8 ft2 of area per linear foot (1.8 m2 / lin m) of tube To compute the linear feet L of finned tubing required, use the relation L ϭ A / (ft2 / ft), or L ϭ 14,650 / 5.8 ϭ 2530 lin ft (771.1 m) of tubing 7 Compute the number of individual tubes required Assume a length for the radiator tubes Typical lengths range between 4 and 20 ft (1.2 and 6.1 m), depending on the size of. .. properties of the fluid, performance or duty of the exchanger, and mechanical design or arrangement of the heat-transfer surface These groups are then multiplied together with a numerical factor to obtain a product that is equal to the fraction of the total driving force—or log mean temperaturedifference (LMTD or ⌬TM )—that is dissipated across each element of resistance in the heat-flow path When the sum of. .. the number of tubes to give a pressure drop of 10 lb / in2 (68.9 kPa) on the tubeside The pressure drop varies inversely as n 1.8 Therefore, 14.3 / 10 ϭ (n / 89)1.8, and n ϭ 109 Each individual product of the factors is then adjusted in accordance with the applicable exponential function of the number of tubes Since the tubeside product is inversely proportional to the 0.2 power of the number of tubes,... is Hm ϭ (weight of solid melted, lb)(heat of fusion of the solid, Btu / lb) Since the heat of fusion of lead is 10 Btu / lb (23.2 kJ / kg), Hm ϭ (600)(10) ϭ 6000 Btu / h, or 6000 / 3412 ϭ 1.752 kWh 3 Compute the heat required to reach the working temperature Use the same relation as in step 1, except that the temperature range is expressed as tw Ϫ tm, where tw ϭ working temperature of the melted solid... versions of such tables whenever available, as necessary BOILER TUBE STEAM-GENERATING CAPACITY A counterflow bank of boiler tubes has a total area of 900 ft2 (83.6 m2) and its overall coefficient of heat transfer is 13 Btu / (h ⅐ ft2 ⅐ ЊF) [73.8 W / (m2 ⅐ K) The boiler tubes generate steam at a pressure of 1000 lb / in2 absolute (6900 kPa) The tube bank is heated by flue gas which enters at a temperature of. .. of 450,000 lb / h (56.7 kg / s) Assume an average specific heat of 0.25 Btu / (lb ⅐ ЊF) [1.05 kJ / (kg ⅐ K)] for the gas and calculate the temperature of the gas that leaves the bank of boiler tubes Also, calculate the rate at which the steam is being generated in the tube bank Calculation Procedure: 1 Find the temperature of steam at 1000 lfƒ / in2 (6900 kPa ) From Table 2, Saturation: Pressures, of. .. (2.3 ϫ 450,000 ϫ 0.25) ϭ 0.0452 The antilog of 0.0452 ϭ 1.11, hence, [(1455.4 / (t2 Ϫ 544.7)] ϭ 1.11, and t2 ϭ (1455.4 / 1.11) ϩ 544.6 ϭ 1850ЊF (1280 K) 4 Find the heat of vaporization of the water From the Steam Tables, the heat of vaporization of the water at 1000 lb / in2 (6900 kPa), hƒg ϭ 649.5 Btu / lb (1511 kJ / kg) 5 Compute the steam-generating rate of the boiler tube bank Heat absorbed by the . reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 11.2 TABLE 1 Heat-Exchanger. to melt a solid is H m ϭ (weight of solid melted, lb)(heat of fusion of the solid, Btu /lb). Since the heat of fusion of lead is 10 Btu/ lb (23.2 kJ/kg),