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HEAT EXCHANGERS Tutor: Nazaruddin Sinaga email: [email protected] Phone 024-76480655 Powerpoint Templates Page 1 Contents • • • • Types of Heat Exchangers The Overall Heat Transfer Coefficient Analysis of Heat Exchangers The Log Mean Temperature Difference Method • The Effectiveness–NTU Method Objectives • Perform an energy balance over the heat exchanger • Identify key variables affecting heat transfer and quantify the nature of any significant effects • Develop appropriate models to describe heat transfer characteristics Introduction to Heat Exchangers What Are Heat Exchangers? Heat exchangers are units designed to transfer heat from a hot flowing stream to a cold flowing stream. Why Use Heat Exchangers? Heat exchangers and heat recovery is often used to improve process efficiency. Types of Heat Exchangers There are three broad categories: • The recuperator, or through-the-wall non storing exchanger • The direct contact non storing exchanger • The regenerator, accumulator, or heat storage exchanger Recuperators Direct Contact Regenerators Heat Transfer Within a Heat Exchanger Heat transfer within a heat exchanger typically involves a combination of conduction and convection The overall heat transfer coefficient U accounts for the overall resistance to heat transfer from convection and conduction 1 1 1 x UA h1 A1 h2 A2 kAmean The schematic of a shell-and-tube heat exchanger (one-shell pass and one-tube pass). A plate-and-frame liquid-to-liquid heat exchanger Thermal resistance network associated with heat transfer in a double-pipe heat exchanger. The two heat transfer surface areas associated with a double-pipe heat exchanger (for thin tubes, Di ≈ Do and thus Ai ≈ Ao). The thermal resistance of the tube wall : The total thermal resistance : The rate of heat transfer between the two fluids as : U is the overall heat transfer coefficient [ W/m2.oC] When the wall thickness of the tube is small and the thermal conductivity of the tube material is high, as is usually the case, the thermal resistance of the tube is negligible (Rwall ≈ 0) and the inner and outer surfaces of the tube are almost identical (Ai ≈ Ao ≈ As). Equation for the overall heat transfer coefficient simplifies to: The overall heat transfer coefficient U dominated by the smaller convection coefficient, since the inverse of a large number is small. When one of the convection coefficients is much smaller than the other (say, hi << ho), we have 1/hi >>1/ho, and thus U ≈ hi. Therefore, the smaller heat transfer coefficient creates a bottleneck on the path of heat flow and seriously impedes heat transfer. This situation arises frequently when one of the fluids is a gas and the other is a liquid. In such cases, fins are commonly used on the gas side to enhance the product UAs and thus the heat transfer on that side. Type of heat exchanger Gas-to-gas Water-to-air in finned tubes (water in tubes) Water-to-air in finned tubes (water in tubes) Steam-to-air in finned tubes (steam in tubes) Steam-to-air in finned tubes (steam in tubes) Steam-to-heavy fuel oil Water-to-oil Steam-to-light fuel oil Alcohol condensers (water cooled) Freon condenser (water cooled) Water-to-gasoline or kerosene Ammonia condenser (water cooled) Water-to-water Feedwater heaters Steam condenser U, W/m2..°C 10–40 30–60† 300–850‡ 30–300† 400–4000‡ 50–200 100–350 200–400 250–700 300–1000 300–1000 800–1400 850–1700 1000–8500 1000–6000 *Multiply the listed values by 0.176 to convert them to Btu/h ·ft2 ·°F. †Based on air-side surface area. ‡Based on water- or steam-side surface area. When the tube is finned on one side to enhance heat transfer, the total heat transfer surface area on the finned side becomes For short fins of high thermal conductivity, we can use this total area in the convection resistance relation Rconv =1/hAs since the fins in this case will be very nearly isothermal. Otherwise, we should determine the effective surface area As from Fouling Factor • The performance of heat exchangers usually deteriorates with time as a result of accumulation of deposits on heat transfer surfaces. • The layer of deposits represents additional resistance to heat transfer and causes the rate of heat transfer in a heat exchanger to decrease. • The net effect of these accumulations on heat transfer is represented by a fouling factor Rf , which is a measure of the thermal resistance introduced by fouling. • The most common type of fouling is the precipitation of solid deposits in a fluid on the heat transfer surfaces. • Another form of fouling, which is common in the chemical process industry, is corrosion and other chemical fouling. • Heat exchangers may also be fouled by the growth of algae in warm fluids (biological fouling) Precipitation fouling of ash particles on superheater tubes • The fouling factor depends on the operating temperature and the velocity of the fluids, as well as the length of service. • Fouling increases with increasing temperature and decreasing velocity. • For an unfinned shell-and-tube heat exchanger : Rf, i and Rf, o are the fouling factors Representative fouling factors (thermal resistance due to fouling for a unit surface area) (Source: Tubular Exchange Manufacturers Association.) EXAMPLE : Overall Heat Transfer Coefficient of a Heat Exchanger Hot oil is to be cooled in a double-tube counter-flow heat exchanger. The copper inner tubes have a diameter of 2 cm and negligible thickness. The inner diameter of the outer tube (the shell) is 3 cm. Water flows through the tube at a rate of 0.5 kg/s, and the oil through the shell at a rate of 0.8 kg/s. Taking the average temperatures of the water and the oil to be 45°C and 80°C, respectively, determine the overall heat transfer coefficient of this heat exchanger. SOLUTION Hot oil is cooled by water in a double-tube counter-flow heat exchanger. The overall heat transfer coefficient is to be determined. Assumptions 1. The thermal resistance of the inner tube is negligible since the tube material is highly conductive and its thickness is negligible. 2. Both the oil and water flow are fully developed. 3. Properties of the oil and water are constant. • The properties of water at 45°C are • The properties of oil at 80°C are • The hydraulic diameter for a circular tube is the diameter of the tube itself, Dh = D= 0.02 m. • The mean velocity of water in the tube and the Reynolds number are • Therefore, the flow of water is turbulent. • Assuming the flow to be fully developed, the Nusselt number can be determined from • Now we repeat the analysis above for oil. • The properties of oil at 80°C are : • The hydraulic diameter for the annular space is The mean velocity and the Reynolds number in this case are which is less than 4000. Therefore, the flow of oil is laminar. Nusselt number for fully developed laminar flow in a circular annulus with one surface insulated and the other isothermal (Kays and Perkins, Ref. 8.) • Assuming fully developed flow, the Nusselt number on the tube side of the annular space Nui corresponding to Di /Do = 0.02/0.03 = 0.667 can be determined from the table by interpolation, we find : Nu = 5.45 • and • Then the overall heat transfer coefficient for this heat exchanger becomes Discussion • Note that U ≈ ho in this case, since hi >> ho. • This confirms our earlier statement that the overall heat transfer coefficient in a heat exchanger is dominated by the smaller heat transfer coefficient when the difference between the two values is large. • To improve the overall heat transfer coefficient and thus the heat transfer in this heat exchanger, we must use some enhancement techniques on the oil side, such as a finned surface. DIMENSIONLESS ANALYSIS TO CHARACTERIZE A HEAT EXCHANGER Nu f(Re, Pr, L/D, μ /μ ) b o h.D k •Further Simplification: v.D. C p . k Nu a.Re b .Pr c Nu Can be obtained from 2 set of experiments One set, run for constant Pr And second set, run for constant Re q k D h A(Tw T ) •Empirical Correlation • For laminar flow Nu = 1.62 (Re*Pr*L/D) • For turbulent flow b Nu 0.026. Re . P r . o 0.8 Good to predict within 20% Conditions: L/D > 10 0.6 < Pr < 16,700 Re > 20,000 1/ 3 0.14 Compact Heat Exchangers • Analysis based on NTU method • Convection (and friction) coefficients have been determined for selected HX cores by Kays and London. Proprietary data have been obtained by manufacturers of many other core configurations. • Results for a circular tube-continuous fin HX core: jh St Pr 2 / 3 St h / Gc p G Vmax Results for a circular tube-continuous fin HX core jh St Pr 2 / 3 St h / Gc p G Vmax General Considerations • Computational Features/Limitations of the LMTD Method: The LMTD method may be applied to design problems for which the fluid flow rates and inlet temperatures, as well as a desired outlet temperature, are prescribed. For a specified HX type, the required size (surface area), as well as the other outlet temperature, are readily determined. If the LMTD method is used in performance calculations for which both outlet temperatures must be determined from knowledge of the inlet temperatures, the solution procedure is iterative. For both design and performance calculations, the effectiveness-NTU method may be used without iteration. LMTD Method Q = U As Tlm The procedure to be followed by the selection process is: 1. Select the type of heat exchanger suitable for the application. 2. Determine any unknown inlet or outlet temperature and the heat transfer rate using an energy balance. 3. Calculate the log mean temperature difference Tlm and the correction factor F, if necessary. 4. Obtain (select or calculate) the value of the overall heat transfer coefficient U. 5. Calculate the heat transfer surface area As . CONCURRENT FLOW 1 2 ∆T1 ∆T2 ∆A A T10 T1 T4 T5 T2 T6 T3 T9 T8 T7 P ara llel Fl ow Calculating U using Log Mean Temperature Hot Stream : h.Cph.dTh dqh m Cold Stream: c .C pc .dTc dqc m dq dqhot dqcold dq U .T .dA T2 T1 d (T ) dTh dTc d (T ) d (Th Tc ) T Th Tc dqh dqc d (T ) m .C h m .C c c p h p 1 1 d (T ) U .T .dA. m .C h m .C c c p h p 1 d (T ) 1 U . h c T m . C m . C c p h p A2 . dA A1 T2 U . A. Th Tc U . A Thin Thout Tcin Tcout ln q q T1 h.Cph.Th qm Log Mean Temperature Difference T T 2 1 q U . A T 2 ln T 1 Calculating U using Log Mean Temperature T2 U . A. U . A in out in out ln T T T T T T h c h h c c T q q 1 T2 ln T1 U .A Thin Tcin Thout Tcout q T2 ln T1 U .A U .A T1 T2 T2 T1 q q T T 2 1 q U .A T 2 ln T 1 Log Mean Temperature Difference Log Mean Temperature Evaluation CONCURRENT FLOW 1 T1 Thin Tcin T3 T7 2 T2 Thout Tcout T6 T10 ∆ T1 ∆ T2 TLn ∆A A U T 10 T1 T4 T5 T2 T6 T3 T9 T8 T7 P ara ll el Fl ow T2 T1 T2 ln T1 h .C ph .T3 T6 m A.TLn c .C pc .T7 T10 m A.TLn Log Mean Temperature Evaluation COUNTER CURRENT FLOW 1 T1 2 T3 T4 T6 T1 T6 Wall T7 T2 T8 T9 T10 A T 10 T1 T4 T2 T5 T3 T6 T7 T8 Co un t er - C u r re n t F low T9 T2 T1 Thin Tcout T3 T7 T2 Thout Tcin T6 T10 q hh Ai Tlm (T T1 ) (T6 T2 ) Tlm 3 (T T1) ln 3 (T6 T2 ) 1 2 T3 T4 T1 T6 T6 Wall T2 T7 T8 T9 T10 q hc Ao Tlm Tlm (T1 T7 ) (T2 T10 ) (T1 T7 ) ln (T2 T10 ) A The Effectiveness – NTU Method The Effectiveness – NTU Method In an attempt to eliminate the iterations from the solution of such problems, Kays and London came up with a method in 1955 called the effectiveness–NTU method, which greatly simplified heat exchanger analysis. This method is based on a dimensionless parameter called the heat transfer effectiveness, defined as The actual heat transfer rate in a heat exchanger can be determined from an energy balance on the hot or cold fluids and can be expressed as Definitions • Heat exchanger effectiveness : q qmax 0 1 • Maximum possible heat rate : qmax Cmin Th,i Tc,i Cmin Ch if Ch Cc or Cc if Cc Ch Will the fluid characterized by Cmin or Cmax experience the largest possible temperature change in transit through the HX? Why is Cmin and not Cmax used in the definition of qmax? For a parallel-flow heat exchanger can be rearranged as Effectiveness relations of the heat exchangers typically involve the dimensionless group UAs /Cmin. This quantity is called the number of transfer units NTU and is expressed as In heat exchanger analysis, it is also convenient to define another dimensionless quantity called the capacity ratio c as Example A counter-flow double-pipe heat exchanger is to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be accomplished by geothermal water available at 160°C at a mass flow rate of 2 kg/s. The inner tube is thin-walled and has a diameter of 1.5 cm. If the overall heat transfer coefficient of the heat exchanger is 640 W/m2 .°C, determine the length of the heat exchanger required to achieve the desired heating. Assumptions 1. Steady operating conditions exist. 2. The heat exchanger is well insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3. Changes in the kinetic and potential energies of fluid streams are negligible. 4. There is no fouling. 5. Fluid properties are constant. Problem 11.28: Use of twin-tube (brazed) heat exchanger to heat air by extracting energy from a hot water supply. KNOWN: Counterflow heat exchanger formed by two brazed tubes with prescribed hot and cold fluid inlet temperatures and flow rates. FIND: Outlet temperature of the air. SCHEMATIC: Problem: Twin-Tube Heat Exchanger (cont.) ASSUMPTIONS: (1) Negligible loss/gain from tubes to surroundings, (2) Negligible changes in kinetic and potential energy, (3) Flow in tubes is fully developed since L/D h = 40 m/0.030m = 1333. PROPERTIES: Table A-6, Water ( Th = 335 K): ch = cp,h = 4186 J/kgK, = 453 10-6 Ns/m2, k = 0.656 W/mK, Pr = 2.88; Table A-4, Air (300 K): cc = cp,c = 1007 J/kgK, = 184.6 10-7 Ns/m2, k = 0.0263 W/mK, Pr = 0.707; Table A-1, Nickel ( T = (23 + 85)C/2 = 327 K): k = 88 W/mK. ANALYSIS: Using the NTU - method, from Eq. 11.30a, 1 exp NTU 1 Cr 1 Cr exp NTU 1 Cr NTU UA / Cmin Cr Cmin / Cmax . (1,2,3) and the outlet temperature is determined from the expression Cc Tc,o Tc,i / Cmin Th,i Tc,i . (4) From Eq. 11.1, the overall heat transfer coefficient is 1 1 1 1 UA o hA h Kt L o hA c (5) Since circumferential conduction may be significant in the tube walls, o needs to be evaluated for each of the tubes. Problem: Twin-Tube Heat Exchanger (cont.) The convection coefficients are obtained as follows: 4mh 4 0.04 kg / s Water-side: ReD 11, 243. D 0.010m 453 106 N s / m2 The flow is turbulent, and since fully developed, the Dittus-Boelter correlation may be used, 0.3 0.023 11, 243 Nu h h h D / k 0.023Re0.8 D Pr 0.8 2.880.3 54.99 h h 54.99 0.656 W / m K / 0.01m 3, 607 W / m2 K. ReD Air-side: 4mc 4 0.120 kg / s 275,890. D 0.030m 184.6 107 N s / m2 The flow is turbulent and, since fully developed, 0.4 0.023 275,890 Nu c h c D / K 0.023Re0.8 D Pr 0.8 0.707 0.4 450.9 h c 450.9 0.0263 W / m K / 0.030m 395.3 W / m 2 K. Water-side temperature effectiveness: A h Dh L 0.010m 40m 1.257 m 2 o,h f ,h tanh mLh / mLh m 3607 W / m2 K / 88 W / m K 0.002m m hh P / kA 1/ 2 1/ 2 143.2 m 1 h h / kt 1/ 2 Problem: Twin-Tube Heat Exchanger (cont.) With Lh = 0.5 Dh, o,h = tanh(143.2 m-1 0.5 0.010m)/143.2 m-1 0.5 0.010 m = 0.435. Air-side temperature effectiveness: Ac = DcL = (0.030m)40m = 3.770 m2 o,c f ,c tanh mLc / mLc m 395.3 W / m 2 K / 88 W / m K 0.002m 1/ 2 47.39 m 1 With Lc = 0.5Dc, o,c = tanh(47.39 m-1 0.5 0.030m)/47.39 m-1 0.5 0.030m = 0.438. Hence, from Eq. (5) the UA product is 1 UA 1 2 0.435 3607 W / m K 1.257 m 2 1 100 W / m K 40m UA 5.070 104 2.50 104 1.533 103 With 1 1 2 0.438 395.3 W / m K 3.770 m 2 W / K 437 W / K. Ch mh ch 0.040 kg / s 4186 J / kg K 167.4 W / K C max Cr Cmin / Cmax 0.722 Cc mccc 0.120 kg / s 1007 J / kg K 120.8 W / K C min NTU 437 W / K UA 3.62 Cmin 120.8 W / K Problem: Twin-Tube Heat Exchanger (cont.) and from Eq. (1) the effectiveness is 1 exp 3.62 1 0.722 1 0.722 exp 3.62 1 0.722 0.862 Hence, from Eq. (4), with Cmin = Cc, 0.862 Cc Tc,o 23C Cc 85 23 C Tc,o 76.4C < COMMENTS: (1) Using the overall energy balance, the water outlet temperature is Th,o Th,i Cc / Ch Tc,o Tc,i 85C 0.722 76.4 23 C 46.4C. (2) To initially evaluate the properties, we assumed that Th 335 K and Tc 300 K. From the calculated values of Th,o and Tc,o, more appropriate estimates of Th and Tc are 338 K and 322 K, respectively. We conclude that proper thermophysical properties were used for water but that the estimates could be improved for air. Problem: Heat Transfer Enhancement Problem 11.65: Use of fluted spheres and solid spheres to enhance the performance of a concentric tube, water/glycol heat exchanger. KNOWN: Flow rates and inlet temperatures of water and glycol in counterflow heat exchanger. Desired glycol outlet temperature. Heat exchanger diameter and overall heat transfer coefficient without and with spherical inserts. FIND: (a) Required length without spheres, (b) Required length with spheres, (c) Explanation for reduction in fouling and pump power associated with using spheres. SCHEMATIC: Th,i = 100oC Th,o = 40oC . mh = 0.5 kg/s L Tc,i = 15oC . mc = 0.5 kg/s Di = 0.075 m Problem: Heat Transfer Enhancement (cont.) ASSUMPTIONS: (1) Negligible kinetic energy, potential energy and flow work changes, (2) Negligible heat loss to surroundings, (3) Constant properties, (4) Negligible tube wall thickness. PROPERTIES: Table A-5, Ethylene glycol Th 70C : cp,h = 2606 J/kgK; Table A-6, Water Tc 35C : cp,c = 4178 J/kgK. ANALYSIS: (a) With Ch = Cmin = 1303 W/K and Cc = Cmax = 2089 W/K, Cr = 0.624. With actual and maximum possible heat rates of q Ch Th,i Th,o 1303 W / K 100 40 C 78,180 W q max Cmin Th,i Tc,i 1303 W / K 100 15 C 110, 755 W the effectiveness is = q/qmax = 0.706. From Eq. 11.30b, 1 1 0.294 NTU ln 2.66 ln 1.71 Cr 1 Cr 1 0.559 Hence, with A = DL and NTU = UA/Cmin, L 1303 W / K 1.71 Cmin NTU 9.46m 2 Di U 0.075m 1000 W / m K (b) Since mc, mh, Th,i, Th,o and Tc,i are unchanged, Cr, and NTU are unchanged. Hence, with U = 2000 W/m2K, L 4.73m < Problem: Heat Transfer Enhancement (cont.) (c) Because the spheres induce mixing of the flows, the potential for contaminant build-up on the surfaces, and hence fouling, is reduced. Although the obstruction to flow imposed by the spheres acts to increase the pressure drop, the reduction in the heat exchanger length reduces the pressure drop. The second effect may exceed that of the first, thereby reducing pump power requirements. COMMENTS: The water outlet temperature is Tc,o = T c,i + q/Cc = 15C + 78,180 W/2089 W/K = 52.4C. The mean temperature Tc 33.7C is close to that used to evaluate the specific heat of water. 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