Chapter 12_Heat Exchangers

10/1/2016

Chapter 12. Heat Exchangers Fundamentals of Heat and Mass Transfer

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Fundamentals of Heat and Mass Transfer

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Chapter 12 Heat Exchangers 12.1 Introduction ‘Heat exchanger’ is one of the most commonly used process equipments in industry and research. Function of a heat exchanger is to transfer energy; this transfer of energy may occur to a single fluid (as in the case of a boiler where heat is transferred to water) or between two fluids that are at different temperatures (as in the case of an automobile radiator where heat is transferred from hot water to air). In some cases, there are more than two streams of fluid exchanging heat in a heat exchanger. Heat exchangers of several designs in a variety of sizes varying from ‘miniature’ to ‘huge’ (with heat transfer areas of the order of 5000 to 10,000 sq. metres) have been developed over the years. Some typical examples of heat exchanger applications are: 1. Thermal power plants (boilers, superheaters, steam condensers, etc.) 2. Refrigeration and airconditioning (evaporators, condensers, coolers) 3. Automobile industry (radiators, all engine cooling and fuel cooling arrangements) 4. Chemical process industry (variety of heat exchangers between different types of fluids, in cumbustors and reactors) 5. Cryogenic industry (condenserreboilers used in distillation columns, evaporators to produce gas from cryogenic liquids, etc.) 6. Research (‘regenerators’ used in Stirling engines, special ceramic heat exchangers used in ultralow temperature devices, superconducting magnet systems, etc.).

12.2 Types of Heat Exchangers Heat exchangers may be classified in several ways: 1. according to heat exchange process 2. according to relative direction of flow of hot and cold fluids 3. according to constructional features, compactness, etc. 4. according to the state of the fluid in the heat exchanger. (i) Classification according to heat exchange process Heat exchangers may be of ‘direct contact type’ or of ‘indirect contact type’. In direct contact type, two immiscible fluids come in direct contact with each other and exchange heat, e.g. air and water exchanging heat in a cooling tower. Indirect contact type can be further classified as ‘recuperators’ and ‘regenerators’. Recuperators are most commonly used; here, the hot and cold fluids are separated from each other by a solid wall and heat is transferred from one fluid to the other across this wall. In regenerators, also called ‘periodic flow heat exchangers’, hot and cold streams alternately flow through a solid matrix (made of solid particles or wire mesh screens); during the ‘hot blow’, the matrix stores the heat given up by the hot stream and during ‘cold blow’, the stored heat is given up by the solid matrix to the cold stream. Sometimes, the solid matrix is made to rotate across fluid passages arranged side by side, so that the heat exchange process is ‘continuous’. (ii) Classification according to relative direction of hot and cold fluids If the hot and cold fluids flow parallel to each other, it is known as ‘parallel flow’ heat exchanger; if the two fluids flow opposite to each other, it is of ‘counterflow’ type. If the fluids flow perpendicular to each other, then, we have ‘cross flow’ type of heat exchanger. These three types of heat exchangers are shown schematically in Fig. 12.1.

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FIGURE 12.1(a) Parallel flow heat exchanger

FIGURE 12.1(b) Counterflow flow heat exchanger Further, when a fluid is constrained to flow within a channel (such as a tube), the fluid is said to be ‘unmixed’; otherwise, it is ‘mixed’. In Fig. 12.1 (c), hot fluid is unmixed since it flows constrained within the tubes, whereas the cold fluid is perfectly mixed as it flows through the heat exchanger. In Fig. 12.1 (d), both the cold and hot fluids are constrained to flow within the tubes and therefore, both the fluids are unmixed. (iii) Classification according to constructional features Basically, there are three types: (a) concentric tubes type (b) shell and tube type, and (c) compact heat exchangers. In concentric tubes type of heat exchanger, one tube is located inside another; one fluid flows through the inside tube and the other fluid flows in the annular space between the tubes. Fluids may flow parallel to each other as shown in Fig. 12.1 (a), or they may flow in opposite directions, as shown in Fig. 12.1 (b). Shell and tube type of heat exchanger is very popular in industry because of its reliability and high heat transfer effectiveness. Here, one of the fluids flows within a bundle of tubes placed within a shell. And, the other fluid flows through the shell over the surfaces of the tubes. Suitable baffles are provided within the shell to make the shell fluid change directions and provide good turbulence, so that heat transfer coefficient is increased. Fig. 12.2 shows a schematic diagram of a typical shell and tube heat exchanger. Fig. 12.2 is an example of two tube pass and one shell pass heat exchanger, i.e. flow passes through the tubes twice in opposite directions, and shell fluid passes through the shell once. Other flow arrangements are also used, such as: one shell pass + two, four or six tube passes; two shell passes and four, eight, twelve, etc. tube passes. Compact heat exchangers are special purpose heat exchangers which provide very high surface area per cubic metre of volume, known 2

3

as ‘area density’. According to usually accepted norms, a ‘compact heat exchanger’ has an area density of 700 m /m or more. These are generally used for gases, since usually gas side heat transfer coefficient is small and therefore, it is needed to provide larger areas. Compact heat exchangers are of platefin type or tubefin type. A typical example of a platefin type of compact heat exchanger is shown in Fig. 12.3.



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FIGURE 12.1(c) Crossflow heat exchanger, cold fluid ‘mixed’, hot fluid ‘unmixed’

FIGURE 12.1(d) Crossflow heat exchanger, both cold and hot fluids ‘unmixed’

FIGURE 12.2 Diagram of a typical (fixed tube sheet) shellandtube heat exchanger

FIGURE 12.3 Section of a platefin heat exchanger (iv) Classification according to state of the fluid In all the types of heat exchangers discussed above, both the fluids changed their temperature along the length of heat exchanger. But, this need not be the case always. A heat exchanger may be used to condense a fluid in which case the condensing fluid will be at a constant temperature throughout the length of the heat exchanger, while the other (cold) fluid will increase in temperature as it passes through the heat exchanger, absorbing the latent heat of condensation released by the condensing fluid. Such a heat exchanger is called a ‘Condenser’. If, on the other hand, one of the fluids evaporates in a heat exchanger, temperature of this fluid will remain constant throughout the length of heat exchanger, whereas the temperature of the other fluid, which supplies the latent heat of evaporation to the evaporating fluid, goes on decreasing along the length of the heat exchanger. Such a heat exchanger is called an ‘Evaporator’. Enjoy Safari? Subscribe Today


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It is interesting to compare the surface areatovolume ratios of different types of heat exchangers. See Table 12.1: TABLE 12.1 Surface areatovolume ratios of different heat exchangers

Type of HX

Hydraulic diameter (mm)

Surfacearea/Volume, 2

3

(m /m )

Plain tube, shellandtube

40 to 6

60 – 600

Plate heat exchangers

20 to 10

180 – 350

Strip fin and louvred fin

heat exchangers

10 to 0.5

350 – 7100

Automotive radiators

5 to 2.5

710 – 1500

Cryogenicheat exchangers

3.7 to 1.7

1000 – 2500

Gas turbine rotary regenerators

1.2 to 0.5

3000 – 7100

Matrix types, wire screen, sphere

2.5 to 0.2

1500 – 18000

bed, corrugated sheets

Human lungs

0.2 to 0.15

18000 – 25000

In Table 12.1, hydraulic diameter of the flow passage is also given; note that smaller the hydraulic diameter, larger is the ratio of surface areatovolume. Note that the human lungs have the largest of all surface areatovolume ratios.

12.3 Overall Heat Transfer Coeⴠ〠icient ‘Overall heat transfer coefficient’, was first introduced in Chapter 4. So, the reader may please refer back to Chapter 4 to refresh memory. In most of the practical cases of heat exchangers, temperature of the hot fluid (Ta) and that of the cold fluid (Tb ) are known; then we would like to have the heat transfer given by a simple relation of the form Q = U · A · (Ta − Tb ) = U · A · ΔT …(4.21) where, Q is the heat transfer rate (W), A is the area of heat transfer perpendicular to the direction of heat transfer, and (Ta – Tb ) = ∆T is the overall temperature difference between the temperature of hot fluid (Ta) and that of the cold fluid (Tb ). In a normally used recuperative type of heat exchanger, the hot and cold fluids are separated by a solid wall. This may be a flat type of wall (as in the case of platefin type of heat exchangers), or, more often, a cylindrical wall (as in the case of a tubeintube type of heat exchangers). See Fig. 12.4. Recall from Chapter 4 that, in general, the overall heat transfer coefficient is related to the total thermal resistance of the system, as follows:

Therefore, the task of finding the overall heat transfer coefficient reduces to finding out the total thermal resistance of the system. For plane wall:



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Remember that for a plane wall, thermal resistance is L/(k.A), and convective resistance is 1/(h.A), and since the resistances are in series, we get:

FIGURE 12.4 Heat exchanger walls

Now, if the thermal resistance of the wall is negligible compared to other resistances, we get:

For cylindrical wall: Remember that for a cylindrical wall, thermal resistance is:

and, convective resistance is 1/(h.A) and the resistances are in series. However, the area to be considered has to be specified since the inner surface area and the outer surface area of the cylinder are different. Now, we have, the general relation for U:

We can also write:

Therefore, referred to outer surface area, U becomes:

Now, for a cylindrical system, we have: Ai = 2 · π · ri · L and, Ao = 2 · π · ro · L



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Then,

i.e.

Similarly, referred to inner surface area, U becomes:

and,

i.e.

Again, if the thermal resistance of the wall is negligible compared to other resistances, (i.e. high value of thermal conductivity, k), or, wall thickness of the tube is very small (i.e. (ri/ro) ≈ 1), we get:

Note that Eq. 12.8 is the same as Eq. 12.2. For many practical situations, this simple equation gives a quick estimate of overall heat transfer coefficient, U. Observe from Eq. 12.2 or 12.8 that the value of U is controlled by the smaller of the two heat transfer coefficients, hi and ho. Therefore, aim of the designer should be to focus on the smaller of the two heat transfer coefficients and improve it, if possible. For example, in a gastoliquid heat exchanger, heat transfer coefficient is generally smaller on the gas side, and, therefore, the gas side heat transfer coefficient controls the final value of overall heat transfer coefficient. So, one tries to improve the heat transfer coefficient on the gas side by providing fins on the gas side surface. If fins are provided on a particular surface, then the total heat transfer area on that surface is: Atotal = Afin + Aunfinned …(12.9) where, Afin is the surface area of the fins and Aunfinned is the area of the unfinned portion of the tube. For short fins of a material of high thermal conductivity, since there is practically no temperature drop along the length we can use the value of total area as given by Eq. 12.9 to calculate the convection resistance on the finned surface. However, for long fins where there is a temperature drop along the length of fin, we should use the total or effective area, given by: Atotal = Aunfinned + ηfin · Afin …(12.10) where, ηfin is the ‘fin efficiency’. Sometimes, an overall surface efficiency’ ηo is used. η0 is defined as: ηo · Atotal = Aunfinned + ηfin · Afin



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i.e. ηo tells us how much of the total surface area is really effective in transferring heat. Then, since the effective surface area is also equal to the unfinned area plus the effective area of fin, we can get an expression for overall surface efficiency as follows:

Then, while determining U, we should use ηo.Atotal for the finned surface, whether it is inner surface area, outer surface area or both. For example, if the outer surface of the tube is finned (which is usually the case), with a fin efficiency of ηfin, we write, neglecting the thermal resistance of tube material:

Instead, if the total (i.e. unfinned + finned) surface area and the overall surface efficiency (ηo) is given for the outer surface, Eq. 12.12 can be written as:

Typical values of overall heat transfer coefficients are given in Table 12.2: TABLE 12.2 Typical values of overall heat transfer coefficients



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2

Type of HX

U (W/(m C)

Watertowater

850 – 1700

Watertooil

100 – 350

Watertogasoline or kerosene

300 – 1000

Feed water heaters

1000 – 8500

Steamtolight fuel oil

200 – 400

Steamtoheavy fuel oil

50 – 200

Steam condenser

1000 – 6000

Freon condenser (water cooled)

300 – 1000

Ammonia condenser (water cooled) 800 – 1400

Alcohol condenser (water cooled)

250 – 700

Gastogas

10 – 40

Watertoair in finned tubes

30 – 60 (based on

(water in tubes)

water side surface area)

Steamtoair in finned tubes

400 – 4000 (based on

(steam in tubes)

steam side surface area)

Fouling factors Note that above analysis was for clean heat transfer surfaces. However, with passage of time, the surfaces become ‘dirty’ because of scaling, deposits, corrosion, etc. This results in a reduction in heat transfer coefficient since the scale offers a thermal resistance to heat transfer. Fouling may be categorized as follows: 1. due to scaling or precipitation 2. due to deposits of finely divided particulates 3. due to chemical reaction 4. due to corrosion 5. due to attachments of algae or other biological materials 6. due to crystallization on the surface by subcooling. Effect of fouling is accounted for by a term called, ‘Fouling factor’, (or, ‘dirt factor’), defined as:

Rf is zero for a new heat exchanger. Rf for a fouled heat exchanger cannot be ‘calculated’ theoretically, but has to be determined experimentally by finding out the heat transfer coefficients for a ‘clean’ heat exchanger and a ‘dirty’ heat exchanger of identical design, operating under identical conditions. While taking into account the effect of fouling, the ‘fouling resistance’ (= Rf/area) should be added to the other thermal resistances. For example, for a tube, we can write:



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where, Rfi and Rfo are the fouling factors for the inside and outside surfaces, respectively, and L is the length of tube. From Eq. 12.15, Ui or Uo can easily be calculated. Fouling factor depends on flow velocity and operating temperature; fouling increases with decreasing velocity and increasing temperature. Based on experience, Tubular Exchanger Manufacturers’ Association (TEMA) have given suggested values of fouling factors. Some of these values are given in Table 12.3: TABLE 12.3 Fouling factors for industrial fluids (TEMA, 1988)



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2

Fluid

R (m C/W) f

LIQUIDS:

Fuel oil

0.00088

Quench oil

0.0007

Transformer oil

0.00018

Hydraulic fluid

0.000238

Molten salts

0.000119

Industrial organic heat transfer media

0.000119

Refrigerant liquids

0.00018

Caustic solutions

0.000476

Vegetable oils

0.000715

Gasoline, naptha, light distillates, kerosene

0.000238

Light gas oil

0.000476

Heavy gas oil

0.000715

GASES & VAPOURS:

Solvent vapours

0.000238

Acid gases

0.000238

Natural gas

0.000238

Air

0.000119 – 0.000238

Flue gases

0.000238 – 0.000715

Steam (sat., oil free)

0.000119 – 0.000357

WATER:

River water, sea water, distilled water, boiler feed water:

Below 50 deg.C

Above 50 deg.C

0.0001

0.0002



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Example 12.1. Water at a mean temperature of Tm = 90°C and a mean velocity of um = 0.10 m/s flows inside a 2.5 cm ID, thin walled copper tube. Outer surface of the tube dissipates heat to atmospheric air at Ta = 20°C, by free convection. Calculate the tube wall temperature, overall heat transfer coefficient and heat loss per metre length of tube. Use following simplified expression for air to determine heat transfer coefficient by free convection:

Solution. Data Tm : = 90°C um : = 0.1 m/s D := 0.025 m Ta := 20°C Properties of water at mean temperature of 90°C: 3

−3

ρ := 965.3 kg/m k := 0.675 W/(mC) μ := 0.315 × 10 kg/(ms) Pr := 2.22 We need to calculate the heat transfer coefficients for the inner and outer surfaces: For the water side (i.e. inner surface):

3

i.e. Re = 7.66 × 10 > 4000 (therefore, turbulent) Using Ditus–Boelter equation to determine heat transfer coefficient for inside surface: Nu : = 0.023 · Re

0.8

· Pr

0.3

i.e. Nu = 37.417 (Nusselts number)

Therefore,

3

2

i.e. hi = 1.01 × 10 W/(m C) (inside surface heat transfer coefficient) For the air side (i.e. outer surface): Approximate, value of film temperature for air:

Properties of air at film temperature of 55°C: 3

−5

ρ : = 1.076 kg/m k := 0.0283 W/(mC) μ : = 1.99 × 10 kg/(ms) Pr := 0.708 Then, free convection heat transfer coefficient for outer surface is given by:

i.e. ho = 3.32 · (Ts − 20)

0.25

…(a)

However, Ts is not known, Applying overall energy balance, with Ai = Ao for thinwalled tube: hi · (Ti − Ts) = ho · (Ts − Ta)


Substituting for hi and ho

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1010 · (90 − Ts) = [3.32·(Ts − 20)

0.25

i.e. 1010 · (90 − Ts) = [3.32·(Ts − 20)

] · (Ts − 20)

1.25

]

This equation may now be solved for Ts by trial and error. But, with Mathcad, it is easily solved using solve block. Start with a guess value of Ts, then, after typing ‘Given’ write down the constraint, and then, typing ‘Find (Ts)’ gives the value of Ts immediately: Ts := 40°C (guess value) Given

Then, ho is calculated from Eq. a: ho : = 3.32 · (Ts − 20)

0.25

…(a)

2

i.e. ho = 9.58 W/(m C) (outside surface heat transfer coefficient) and, overall heat heat transfer coefficient, U:

Note that overall heat transfer coefficient is nearly equal to ho. As commented earlier, since hi >> ho, overall heat transfer coefficient is conrolled by ho. Heat loss per metre length of tube: Q : = U × (π × D × 1) × (Ts − Ta) W/m i.e. Q = 51.686 W/m. Example 12.2. In Exaple 12.1, if we desire to increase the value of overall heat transfer coefficient U, the obvious choice is to focus on the airside, since the air side heat transfer coefficient is the lower of the inside and outside heat transfer coefficients. Let us increase the area on the air side by providing 8 numbers of radial fins of rectangular cross section, 2 mm thick and 20 mm height. Material of fins is the same as that of the tube, i.e. copper (k = 380 W/(mK)). Then, determine the overall heat transfer coefficient and the rate of heat transfer. Solution. Data:

Overall heat transfer coefficient Ui referred to the inside surface: Neglecting the thermal resistance of tube wall, we write:

First term in the denominator in RHS is the thermal resistance due to film coefficient on the inside, and the second term is the thermal resistance of the film coefficient on the outside. Unfinned surface (or the base surface) on the outside is at the wall temperature and is fully effective for heat transfer whereas the finned surface is not fully effective because of temperature drop along the length of fins; therefore, effective area of fins is obtained by


multiplying the total area of fins by the fin effectiveness, ηfin.


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Therefore, we need to find out the fin efficiency. Fin efficiency: For a rectangular fin with adiabatic tip, the fin efficiency is given by:

where,

(fin parameter)

Then, from Eq. b, we get:

Areas:

Therefore, overall heat transfer coefficient U referred to the inside surface: We have:

2

Note: compare this to the earlier U value of 9.49 W/(m C); there is great improvement in value of U by providing fins. Heat loss per metre length of tube: Q := Ui × (π × D × l) × (Ts − Ta) W/m i.e. Q = 221.34 W/m. Note: Compare this to the earlier Q value of 51.686 W/m; this substantial improvement in value of Q is the result of providing fins. Also, note that in the above analysis, we assumed that the outside heat transfer coefficient ho is the same for the unfinned surface as well as for the finned surfaces. Example 12.3. A shell and tube counterflow heat exchanger uses copper tubes (k = 380 W/(mC)), 20 mm ID and 23 mm OD. Inside and outside film coefficients are 5000 and 1500 W/(m C), respectively. Fouling factors on the inside and outside may be taken 2

as 0.0004 and 0.001 m C/W respectively. Calculate the overall heat transfer coefficient based on: (i) outside surface, and (ii) inside surface. Solution.



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Data:

Heat loss per metre length of tube:

Overall heat transfer coefficient: We have:

In the denominator of RHS of Eq. 12.15 above, we have the various thermal resistances, as follows: −3

first term → convective film resistance on the inside surface = 3.1831 × 10 C/W –3

second term → fouling resistance on the inside surface = 6.3662 10 C/W −5

third term → conductive resistance of the tube wall = 5.854 × 10 C/W −3

fourth term → convective film resistance on the outside surface = 9.2264 × 10 C/W fifth term → fouling resistance on the outside surface = 0.01384 C/W. Note the relative magnitude of fouling resistances, as compared to other resistances. As expected, conductive resistance of the tube wall (of copper, which is a good conductor) is the smallest of all. Calculating the RHS, we get:

Comments: ‘Fouling’ affects the value of overall heat transfer coefficient and therefore, the size (or area) of the heat exchanger adversely. If the fouling resistances were not included, we should have obtained the following values for the overall heat transfer coefficient: RHS of Eq. 12.15, deleting the fouling resistances, will become:



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i.e. Uo and Ui with no fouling are about 2.62 times the corresponding values when fouling resistances are included. Therefore, it is advisable to include the effect of fouling, if practicable, at the design stage.

12.4 The LMTD Method for Heat Exchanger Analysis Basically, a complete design of a heat exchanger is a huge topic which involves an analysis of: 1. Thermal aspects (i.e. temperatures of fluids at inlet/exit, rate of heat transfer, etc., offdesign performance, etc.) 2. Hydrodynamic aspects (i.e. pressure drops in the flow channels) 3. Structural aspects (mechanical design and structural design). However, here we shall consider only the thermal analysis aspects.

12.4.1 Parallel Flow Heat Exchanger Consider a double pipe, parallel flow heat exchanger, in which a hot fluid and a cold fluid flow parallel to each other, separated by a solid wall. Hot fluid enters at a temperature of Th1 and leaves the heat exchanger at a temperature of Th2 ; cold fluid enters the heat exchanger at a temperature of Tc1 and leaves at a temperature of Tc2 . This situation is shown in Fig. 12.5.

FIGURE 12.5 Parallel flow heat exchanger We desire to get an expression for the rate of heat transfer in this heat exchanger in the following form: Q = U · A · ΔTm …(12.16) where, U = overall heat transfer coefficient A = area for heat transfer (should be the same area on which U is based), and ∆Tm = a mean temperature difference between the fluids. Now, we make the following assumptions: 1. U is considered as a constant throughout the length (or area) of the heat exchanger 2. Properties of fluids (such as specific heat) are also considered to be constant with temperature 3. Heat exchange takes place only between the two fluids and there is no loss of heat to the surroundings, i.e. perfect insulation of heat exchanger is assumed 4. Changes in potential and kinetic energy are negligible 5. Temperatures of both the fluids remain constant (equal to their bulk temperatures) over a given cross section of the heat exchanger. Area ‘A’ is constant for a given heat exchanger. However, we see from Fig. 12.5 that the temperature of the two fluids vary along the length (or area) of the heat exchanger, i.e. the temperature difference between the hot and cold fluids is not a constant along the length



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of the heat exchanger, but varies along the length. Our aim is to find out the appropriate ‘mean temperature difference (∆Tm)’ between the hot and cold fluids, so that Eq. 12.16 can be applied. We proceed as follows: Consider an elemental area dA of the heat exchanger. Then, by applying the First law, we can write: Heat given up by the hot fluid = heat received by the cold fluid. i.e. dQ = − mh · Cph · dTh = mc · Cpc · dTc …(12.17) Here, the temperature of hot fluid decreases as the length increases. So, a negative sign is put in front of mh.Cph.dTh, so that the heat transferred is a positive quantity. Now, dQ for the elemental area dA, can also be expressed as: dQ = U · (Th − Tc ) · dA … (12.18) Now, from Eq. 12.17, we have:

and,

where, mh and mc are the mass flow rates, and Cph and Cpc are the specific heats of hot and cold fluids, respectively. Therefore,

Substituting for dQ from Eq. 12.18, we get:

Integrating Eq. 12.20 between the inlet and exit of the heat exchanger (i.e. between conditions 1 and 2):

Now, considering the total heat transfer rate for the entire heat exchanger, we have:

and,

Substituting in Eq. 12.21:



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Now, comparing Eq. 12.22 and Eq. 12.16, we observe that:

Since this mean temperature difference varies in a logarithmic manner, it is called ‘Logarithmic Mean Temperature Difference’ or, simply LMTD. So, we write:

Now, note that (Th2 – Tc2 ) is the temperature difference at the exit and (Th1 – Tc1) is the temperature difference at the inlet of the heat exchanger. If we denote the temperature differences at the inlet and exit of the heat exchanger by ∆T1 and ∆T2 , respectively, we can write:

We can state Eq. 12.25 in words as follows: LMTD is equal to the ratio of the difference between the greater and lower of the temperature differences at the two ends to the natural logarithm of the ratio between those temperature differences. Equation for LMTD is easily remembered as follows:

where,

GTD = ‘greater (of the two) temperature difference’, and

LTD = ‘lower temperature difference’.

12.4.2 Counter-flow Heat Exchanger Again, consider a double pipe, counterflow heat exchanger, in which a hot fluid and a cold fluid flow in directions opposite to each other, separated by a solid wall. Hot fluid enters at a temperature of Th2 and leaves the heat exchanger at a temperature of Th2 ; cold fluid enters the heat exchanger at a temperature of Tc1 and leaves at a temperature of Tc2 . This situation is shown in Fig. 12.6. We desire to get an expression for the rate of heat transfer in this heat exchanger in the following form: Q = U · A · ΔTm …(12.16)

where,



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U = overall heat transfer coefficient

A = area for heat transfer (should be the same area on which U is based), and

∆Tm = a mean temperature difference between the fluids.

We see from Fig. 12.6 that the temperatures of the two fluids vary along the length (or area) of the heat exchanger, i.e. the temperature difference between the hot and cold fluids is not a constant along the length of the heat exchanger, but varies along the length. Our aim is to find out the appropriate ‘mean temperature difference (∆Tm)’ between the hot and cold fluids, so that Eq. 12.16 can be applied. We proceed as follows, with the same assumptions as made for the analysis of parallel flow heat exchanger: Consider an elemental area dA of the heat exchanger. Then, by applying the First law, we can write: Heat given up by the hot fluid = heat received by the cold fluid.

FIGURE 12.6 Counterflow heat exchanger i.e. dQ = − mh · Cph · dTh = − mc · Cpc · dTc …(12.27) Here, the temperatures of both hot and cold fluids decrease as the length increases. So, negative sign is put in front of mh.Cph.dTh and mc .Cpc .dTc so that the heat transferred is a positive quantity. Now, dQ for the elemental area dA, can also be expressed as: dQ = U ·(Th − Tc ) · dA …(12.28) Now, from Eq. 12.27, we have:

and,

where, mh and mc are the mass flow rates, and Cph and Cpc are the specific heats of hot and cold fluids, respectively. Therefore,



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Substituting for dQ from Eq. 12.28, we get:

Integrating Eq. 12.30 between the inlet and exit of the heat exchanger (i.e. between conditions 1 and 2):

Now, considering the total heat transfer rate for the entire heat exchanger, we have:

and,

Substituting in Eq. 12.31:

Now, comparing Eq. 12.32 and Eq.12.16, we observe that:

Note that this mean temperature difference varies in a logarithmic manner; so, it is called ‘Logarithmic Mean Temperature Difference’ or, simply LMTD. So, we write:

Now, note that (Th2 – Tc2 ) is the temperature difference at the exit and (Th1 – Tc2 ) is the temperature difference at the inlet of the heat exchanger. If we denote the temperature differences at the inlet and exit of the heat exchanger by ∆T1 and ∆T2 , respectively, we can write:

Note that the LMTD expressions for the parallel flow and the counterflow heat exchangers (i.e. Eqs. 12.25 and 12.35) are the same.



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Again, equation for LMTD is easily remembered as follows:

where,

GTD = ‘greater (of the two) temperature difference’, and

LTD = ‘lower temperature difference’

Comments: (i) When ∆T1 = ∆T2 : This is a special case, which can occur sometimes in the case of a counterflow heat exchanger. Then, Eq. 12.35 reduces to a form 0/0, which is indeterminate. However, from physical considerations, ∆T1 = ∆T2 means that the temperature difference between the hot and cold fluids is equal throughout the heat exchanger. Therefore, obviously, the mean temperature difference between the two fluids is ∆T1 = ∆T2 . (This can be proved mathematically also, by applying L’Hospital’s rule). (ii) LMTD for a counterflow heat exchanger is always greater than that for a parallel flow heat exchanger. This means that to transfer the same amount of heat, counterflow unit will require a smaller heat transfer surface as compared to a parallel flow unit. This is the reason why a counterflow heat exchanger is usually preferred. (iii) LMTD can easily be calculated when all the end temperatures of the fluids are known. Then, immediately, the heat transfer rate is determined from the Eq. 12.16, i.e. Q = U.A. (LMTD). Therefore, calculation of LMTD is an important step in the design of a heat exchanger. To facilitate quick calculation of LMTD, when both the end temperatures are known, following graph (Fig. 12.7) is provided. Here, (LMTD/GTD) is plotted against the ratio (LTD/GTD), where GTD = greater of the two end temperature differences, and LTD = lower of the two end temperature differences. First, calculate the ratio (LTD/GTD), and then, read (either from the graph or Table 12.4) the value of LMTD/GTD. Next, multiply this value by GTD to get LMTD.

FIGURE 12.7 LMTD and AMTD for parallel and counterflow HX In the same graph, the value of (AMTD/GTD) is also plotted, for comparison. Here, AMTD is the arithmetic mean temperature difference; AMTD = (∆T1 + ∆T2 )/2. It may be noted that for values beyond about ∆T2 /∆T1 = 0.7, AMTD and LMTD are almost the same, i.e. when ∆T2 / ∆T1 > 0.7, it would suffice to use AMTD (which is easier to calculate) instead of LMTD. However, for lower values of ∆T2 /∆T1, LMTD has to be used. Graph for LMTD shown above is also represented in tabular form (for better accuracy) in Table 12.4, (iv) One term occurring in the derivation of LMTD shown above, is the product of mass flow rate and the specific heat of a fluid, i.e. C = m.Cp. Here, C is known as ‘heat capacity rate’ or, simply ‘capacity rate’ of that particular fluid. Thus, the capacity rates for hot and cold fluids are: Ch = mh · Cph W/C ((12.36)…capacity rate for hot fluid) Cc = mc · Cpc W/C ((12.37)…capacity rate for cold fluid) Then, the heat transfer rate is given by: Q = Ch · (Th1 − Th ) W ((12.38)…for hot fluid) 2

Q = Cc · (Tc − Tc ) W ((12.38)…for cold fluid) 2

1

i.e. to transfer a given amount of heat, higher the heat capacity rate of a fluid, lower will be the temperature rise (or fall) of that particular fluid. If the heat capacity rates of both the hot and cold fluids are equal, then, the total temperature drop of the hot fluid will be equal to the total temperature rise of the cold fluid. See Fig. 12.8 (a). (v) When a fluid is condensing or boiling, its temperature is essentially constant, i.e. Th1 = Th2 for a condensing fluid and Tc1 = Tc2 for a boiling liquid. In other words, ∆T for the condensing or boiling fluid is zero. But, since a finite amount of heat is transferred, (= m.hfg), we say that capacity rate of a condensing or boiling fluid tends to infinity. Temperature profiles for fluids in a heat exchanger when one of the fluids is condensing or boiling are shown in Fig. 12.8 (b) and (c), respectively. LMTD for both these cases is

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Example 12.4. Furnace oil, flowing at a rate of 4000 kg/h, is heated from 10 to 20°C by hot water flowing at 75°C, with a velocity of 0.8 m/s, through a copper pipe 2.15 cm OD, 1.88 cm ID. Oil flows through annulus between copper and steel pipe of 3.35 cm OD and 3 cm ID. Find the length of counterflow heat exchanger. Fluid properties are given. Use Dittus–Boelter equation Nu = 0.023.Re0.8.Pr0.4. Solution.

TABLE 12.4 LMTD/GTD for parallel flow and counterflow HX LMTD = Log mean temperature difference GTD = Greater of two end temperature differences LTD = Lower of two end temperature differences

FIGURE 12.8 (a) Both fluids have same capacity rates

FIGURE 12.8 (b) One of the fluids condensing (Ch ⇒ ∞)

Figure 12.8 (c) One of the fluids boiling (Ch ⇒ ∞)

FIGURE Example 12.4 Counterflow heat exchanger Fluid properties:

Property

Water

Cp (kJ/kg K)

k (W/mK) 2

4.187

1.884

0.657

0.138

4.187 × 10

v (m /s)

Oil

−7

7.43 × 10

3

982

ρ (kg/m )

−6

854

−2

Dhi : = 1.88 × 10 m (inside diameter of tube for hot fluid flow) −2

Dho : = 2.15 × 10 m (outside diameter of tube for hot fluid flow) −2

Dci : = 3 × 10 m (inside diameter of tube for cold fluid flow) −2

Dco : = 3.35 × 10 m (outside diameter of tube for cold fluid flow)

Total heat transferred: Q := mc × cpc × (Tc2 − Tc1) W 4

i.e. Q = 2.093 × 10 W Inside heat transfer coefficient:



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Therefore,

We have:

3

2

i.e. hh = 5.212 × 10 W/(m C) (heat transfer coefficient for hot (inside) fluid) Outside heat transfer coefficient: Equivalent diameter of annulus: Deq = 4 × (area of cross section/wetted perimeter)

Therefore,

Also,

From DittusBoelter’s equation

Therfore,

Overall heat transfer coefficient U: We have: U · A = 1/(Total thermal resistance)

Therefore,

Now, calculate LMTD: Exit temperature of hot fluid:

Therefore,

Alternatively: We can calculate LMTD quickly by using graph of Fig. 12.7 or from Table 12.4: We have: ∆T2 /∆T1 = 42.074/55 = 0.765. From the table, we read against ∆T2 /∆T1 = 0.765, a value of LMTD/∆T1 = 0.8775. Then, LMTD = 0.8775 × 55 = 48.262°C. Length of HX required: Heat exchange area required:

Therefore,



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FIGURE Example 12.5 Heat exchanger with one of the fluids condensing (Ch ⇒ ∞) Example 12.5. In a shellandtube heat exchanger, tubes are 4 m long, 3.1 cm OD, 2.7 cm ID. Water is heated from 22°C to 45°C by condensing steam at 100°C on the outside of tubes. Water flow rate through the tubes is 10 kg/s. Heat transfer coefficient on steam 2

2

side is 5500 W/(m K) and on waterside, 850 W/(m K). Neglecting all other resistances, find the number of tubes. Solution. Data:

Alternatively: We can calculate LMTD quickly by using graph of Fig. 12.7 or from Table 12.4. We have: ∆T2 /∆T1 = 55/78 = 0.705. From the table, we read against ∆T2 /∆T1 = 0.705, value of LMTD/∆T1 = 0.844. Then, LMTD = 0.844 × 78 = 65.832°C….same value as obtained above by calculation. Heat transfer:

Number of tubes required:

Therefore,

i.e. Number of tubes required is, say, 58. Example 12.6. In a double pipe counterflow heat exchanger, 10,000 kg/h of oil (Cp = 2.095 kJ/kgK) is cooled from 80°C to 50°C 2

by 8000 kg/h of water entering at 25°C. Determine the area of heat exchanger for an overall U = 300 W/(m K). Take Cp for water as 4.18 kJ/kgK. (M.U. 1997) Solution. Data:

FIGURE Example 12.6 Counterflow heat exchanger

To calculate LMTD: We have:

Alternatively: We can calculate LMTD quickly by using graph of Fig. 12.7 or from Table 12.4. We have: ∆T2 /∆T1 = 55/78 = 0.705. From the table, we read against ∆T2 /∆T1 = 0.705, value of LMTD/∆T1 = 0.844. Then, LMTD = 0.844 × 78 = 65.832°C….same value as obtained above by calculation. Area required:



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12.5 Correction Factors for Multi-pass and Cross-flow Heat Exchangers LMTD relations derived above are applicable to parallel flow and counterflow heat exchangers only. But, in practice, crossflow heat exchangers (e.g. automobile radiators) and shellandtube heat exchangers, with more than one pass in shell side and/or tube side, are also used. In such cases, the flow situation is complex and the analytic relations for mean temperature difference are very complicated. Then, first, LMTD is calculated as if for a counterflow heat exchanger with the inlet and exit temperatures for the two fluids as per the actual data, and next, a ‘correction factor (F)’ is applied to the calculated LMTD to get the mean temperature difference between the fluids. Now, heat transfer rate is calculated as: Q = U · A · (F · LMTD) W …(12.39) where, A is the area of heat transfer, U is the overall heat transfer coefficient referred to that area, and F is the correction factor. Note again that LMTD is calculated as if for a counterflow heat exchanger, taking the inlet and exit temperatures of the two fluids the same as for the actual heat exchanger. Values of correction factor (F) for a few selected heat exchangers are given in graphical representation in Fig. 12.9. F varies from 0 to 1. In these graphs, correction factor F is plotted as function of two parameters, i.e. P and R, defined as:

FIGURE 12.9(a) One shell pass and 2,4,6, etc. (any multiple of 2), tube passes

FIGURE 12.9(b) Two shell passes and 4,8,12, etc. (any multiple of 4), tube passes

FIGURE 12.9(c) Single pass crossflow with both fluids unmixed

FIGURE 12.9(d) Single pass crossflow with one fluid mixed and the other unmixed

where, C is the capacity rate = m.Cp. Also, for a shellandtube heat exchanger, T and t represent the temperatures of fluids flowing through the shell and tube sides, respectively. And, subscripts 1 and 2 refer to the inlet and exit, respectively. It makes no difference whether hot or cold fluid flows through the shell or the tube. Values of P vary from 0 to 1 and it is equal to the ratio of the temperature change of the tube side fluid to the maximum temperature difference between the two fluids; thus, P represents the thermal effectiveness of the tube side fluid; values of R vary from 0 to When R = 0, it means that the fluid on the shell side is undergoing a phase change (i.e. boiling or condensation, which occurs at a practically constant temperature, Tsat), and when R = ∞, the tube side fluid is undergoing a phase change. Observe from the graphs that, when R = 0 or ∞, the correction factor F is equal to 1. Therefore, for a condenser or boiler, F = 1, irrespective of the configuration of the heat exchanger. Note: To apply the correction factor F from these graphs, it is necessary that the end temperatures of both the fluids must be known. Example 12.7. A one shell pass, two tube pass heat exchanger, with flow arrangement similar to that shown in Fig. 12.9 (a), has water flowing through the tubes and engine oil flowing on the shell side. Water flow rate is 1.2 kg/s and its temperatures at inlet and 2

exit are 25°C and 75°C, respectively. Engine oil enters at 110°C and leaves at 75°C. Overall U = 300 W/(m K). Take Cp for water as 4.18 kJ/(kgK) and calculate the heat transfer area required.

FIGURE Example 12.7 Counterflow heat exchanger Solution. Data: mc : = 1.2 kg/s Cpc := 4180 J/kgK 2

U : = 300 W/(m K) T1 := 110°C (hot fluid, inlet) T2 : = 75°C (hot fluid, exit) t1 : = 25°C (cold fluid, inlet)



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t2 : = 75°C (cold fluid, exit) Therefore, total heat load: Q := mc · Cpc · (t2 − t1) W 5

i.e. Q = 2.508 × 10 W Since this is a multipass HX, LMTD must be calculated as for a counterflow HX, and, then a correction factor applied from Fig. 12.9 (a): To calculate LMTD: ∆T1 := T1 – t2 i.e. ∆T1 = 35°C and, ΔT2 := T2 − t1 i.e. ΔT2 = 50°C Therefore,

Correction factor, F: We have:

Then, from Fig. 12.9 (a): F = 0.8 (correction factor) And, the corrected temperature difference becomes: ΔT := 0.8 ° LMTD i.e. ΔT = 33.644°C (actual mean temperature) Therefore, heat transfer area:

Example 12.8. In a shell and tube HX, 50 kg/min of furnace oil is heated from 10 to 90°C. Steam at 120°C flows through the shell and oil flows inside the tube. Tube size: 1.65 cm ID and 1.9 cm OD. Heat transfer coefficient on oil and steam sides are: 85 and 7420 2

W/(m K), respectively. Find the number of passes and number of tubes in each pass if the length of each tube is limited to 2.85 m. Velocity of oil is limited to 5 cm/s. Density and specific heat of oil are 1970 J/(kg.K), respectively. (M.U. 1994) Solution. Data:

Total heat transferred:

Number of tubes required: Total crosssectional area for flow:

Crosssectional area of each tube:

Therefore,



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FIGURE Example 12.8 Heat exchanger with one of the fluids condensing (Ch ⇒ ∞) Overall U, based on outer surface area: Total thermal resistance:

Then,

Therefore, heat transfer area required:

But, length is limited to 2.85 m. So, use 2 tube passes. Then, it becomes a shellandtube HX with two tube passes. So, it appears at first sight that correction factor (F) has to be obtained from Fig. 12.9; but, observe that one of the fluids is condensing. So, F = 1, irrespective of HX configuration. i.e. F = 1 Therefore, Aht remains same. Then,

12.6 The Eⴠ〠ectiveness-NTU Method for Heat Exchanger Analysis LMTD can readily be determined when all the four end temperatures are either given, or can easily be calculated. Then, the area required, A (i.e. the size of the HX) is easily found out by applying the equation: Q = U.A.(LMTD). In other words, LMTD method is very convenient to use for sizing problems, when all the end temperatures are known. However, there are certain problems where only the inlet temperatures of both the fluids are specified, along with the flow rates and the overall heat transfer coefficients, and the heat transfer rate and the exit temperatures of the fluids are to be calculated. Solution of such rating problems by the LMTD method would require tedious iterations. However, the EffectivenessNTU method, developed by Kays and London in 1955, overcomes this problem and makes the solution straight forward. EffectivenessNTU method is also useful in solving heat exchanger problems, where off design conditions exist; i.e. for example, the heat exchanger might have been designed for some particular flow rates of fluids; now, to find out what happens to the performance if flow rate of one of the fluids is reduced to, say, 75 % of the design flow rate, and so on. The effectivenessNTU method is not an altogether new method; fundamental equations are the same as used in the LMTD method, but the different variables are arranged rather differently. Before we develop the EffectivenessNTU relations for different types of heat exchangers, let us define a few quantities: Effectiveness of a heat exchanger (ε):

where,

Q = actual heat transferred in the heat exhanger

Qmax = maximum possible heat transfer in the heat exchanger

Now, actual heat transfer rate in a heat exchanger is given by: Q = mh · Cph · (Th1 − Th2 ) = Ch · (Th1 − Th2 ) and, Q = mc · Cpc ·(Tc2 − Tc1) = Cc · (Tc2 − Tc1)



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where, Ch = capacity rate of the hot fluid, and Cc = capacity rate of the cold fluid Now, Ch may be equal to Cc or less than Cc or greater than Cc . If Ch < Cc , we designate Ch as Cmin; Instead, if Ch > Cc , we designate Cc as Cmin. And, in each case, capacity rate of the other fluid is designated as Cmax. Capacity Ratio (C): Capacity ratio is defined as:

Number of Transfer Units (NTU): Number of Transfer Units (which is a dimensionless number), is defined as:

where, U is the overall heat transfer coefficient and A is the corresponding heat transfer area. For given value of A and flow conditions, NTU is a measure of the area (i.e. size) of the heat exchanger. Larger the NTU, larger the size of the heat exchanger. Maximum possible heat transfer in a heat exchanger (Qmax): Now, consider a heat exchanger where the hot fluid is cooled from a temperature of Th1 to Th2 and the cold fluid heated from Tc1 to Tc2 . So, the maximum temperature differential in the heat exchanger is (Th1 – Tc1). Now, if the heat exchanger had an infinite area, the hot fluid will be cooled from Th1 to Tc1 or the cold fluid may be heated from Tc1 to Tc1. However, which fluid will experience the maximum temperature differential (Th1 – Tc1) will depend upon which fluid has the minimum capacity rate. If hot fluid has the minimum capacity rate, we can write: Qmax = Ch · (Th1 − Tc1) (if Ch is minimum capacity rate, Cmin) Instead, if cold fluid has the minimum capacity rate, we write: Qmax = Cc · (Th − Tc1) (if Cc is minimum capacity rate, Cmin.) 1

Or, more generally, we write: Qmax = Cmin · (Th1 − Tc ) …(12.45) 1

These situations are represented graphically in Fig. 12.9:

FIGURE 12.9(a) Cold fluid has minimum capacity rate

FIGURE 12.9(b) Hot fluid has minimum capacity rate Therefore, we can write for effectiveness:

Now, if hot fluid is the ‘minimum fluid’ (i.e. Ch < Cc ), we get from Eq. 12.46:

And, if cold fluid is the ‘minimum fluid’ (i.e. Cc < Ch), we get from Eq. 12.46:



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i.e. by suitably choosing the fluid, the effectiveness of a heat exchanger can be expressed as a ratio of temperatures (or, as a temperature effectiveness). If Ch – Cc , obviously, both the fluids will experience the maximum possible temperature differential, if the heat exchanger had an infinite area. Now, for any heat exchanger, effectiveness can be expressed as a function of the NTU and capacity ratio, Cmin/Cmin, i.e.

We shall derive below eNTU relation for a parallel flow HX.

12.6.1 E��ectiveness-NTU Relation for a Parallel-flow Heat Exchanger Consider the parallelflow heat exchanger shown in Fig. 12.5. Assumptions for this derivation remain the same as for the LMTD method. Continuing from Eq. 12.21:

Now, out of the two fluids, one is the ‘minimum’ fluid and the other is the ‘maximum’ fluid. Whichever may be the minimum fluid, we can write Eq. 12.21 as:

Now, substituting for Th2 and Tc2 from Eq. 12.46, we get:

Now, assuming Ch > Cc , i.e. cold fluid as the ‘minimum fluid‘, we have: Cmin = Cc and Cmax = Ch. Therefore,

Eq. 12.50 is the desired expression for effectiveness of a parallel flow heat exchanger. Note that the same result would be obtained, if we assume the hot fluid as the ‘minimum’ fluid. Eq. 12.50 is concisely expressed as:

where, N = NTU and, Special cases: (i) For a condenser or boiler i.e. one of the fluids undergoes a phase change. Therefore, Cmax → ∞ i.e. Capacity ratio, C = 0. Then effectiveness relation (for all heat exchangers) reduces to: ε = 1 − exp(− NTU) …(12.52) (ii) When C = 1, i.e. Cmin = Cmax This is the case of a typical, gas turbine regenerator. In this case,

12.6.2 E��ectiveness-NTU Relation for a Counter-flow Heat Exchanger Again, consider the counterflow heat exchanger shown in Fig. 12.6. Assumptions for this derivation remain the same as for the LMTD method. Continuing from Eq. 12.31:



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This can be written as:

Assuming hot fluid as the ‘minimum’ fluid, Cmin = Ch and, Cmax = Cc we have:

Now, substituting for Th2 and Tc2 from Eq. 12.46, we get:

Now, put Cmin = Ch, C = Cmin/Cmax, and N = NTU, in Eq. 12.55:

Instead of assuming that the hot fluid is the minimum fluid, if we assume that the cold fluid is the ‘minimum’ fluid, then also the same relation (namely, Eq. 12.56), will result. Eq. 12.56 is the desired expression for the effectiveness of the counterflow heat exchanger. Special cases: 1. For a condenser or boiler i.e. one of the fluids undergoes a phase change. Therefore, Cmax → ∞ i.e. Capacity ratio, C = 0. Then effectiveness relation (for all heat exchangers) reduces to: ε = 1 − exp(−NTU) …(12.57) 2. When C = 1, i.e. Cmin = Cmax This is the case of a typical, gas turbine regenerator. In this case, relation for ε reduces to the indeterminate form, 0/0. Then, apply the L’Hospital’s rule to evaluate e. i.e. differentiate the numerator and denominator w.r.t. C and taking the limit C → 1, we get:

EffectivenessNTU relations and the corresponding graphical representations for several types of heat exchangers are given by Kays and London. Table 12.5 gives the Effectiveness relations for a few types of heat exchangers; and Table 12.6 gives the NTU relations: TABLE 12.5 Effectiveness relations for heat exchangers [N = NTU = U.A/Cmin, C = Cmin/Cmax]



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Flow geometry

Relation

Double pipe: parallelflow

Double pipe: counterflow

Counterflow, C = 1

Crossflow: (single pass) both fluids un mixed

Crossflow: (single pass) both fluids mixed

Crossflow: (single pass) Cmax mixed, Cmin unmixed

Crossflow: (single pass) Cmax unmixed, Cmin mixed

Shell and tube:

One shell pass, 2, 4, 6 tube passes

Multiple shell passes, 2n, 4n, 6n tube passes (εp = effectiveness of each shell pass, n = number. of shell passes)

Special case for C = 1

All exchangers, with C = 0 (Condensers

ε = 1 − e

−N

and Evaporators)

TABLE 12.6 NTU relations for heat exchangers [N = NTU = U.A/Cmin, C = Cmin/Cmax, ε = effectiveness]



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Flow geometry

Relation

Double pipe: parallelflow

Double pipe: counterflow, for C = 1

Counterflow, C = 1

Crossflow: Cmax mixed, Cmin unmixed

Crossflow: Cmax unmixed, Cmin mixed

Shell and tube:

One shell pass, 2, 4, 6 tube passes

All exchangers, with C = 0 (Condensers and

N = −ln(1 − ε)

Evaporators)

NTUEffectiveness graphs: NTUEffectiveness relations are also represented in graphical form and these are quite instructive. However, it is a bit difficult to read these graphs accurately; so, analytical relations may be used wherever possible. NTUEffectiveness relations for parallelflow and counterflow heat exchangers are shown graphically in Fig. 12.10 and 12.11, respectively. In these figures, effectiveness values are plotted against NTU for different values of capacity ratio, C. For convenience and accuracy in reading, effectiveness values for the parallel flow and counterflow heat exchangers are given in Tabular form, in Table 12.7 and 12.8:

FIGURE 12.10 NTU Vs. effectiveness for parallelflow heat exchangers

FIGURE 12.11 NTU Vs. effectiveness for counterflow heat exchangers NTUeffectiveness graphs for some other types of heat exchangers, are given by Kays and London, and are reproduced below:

FIGURE 12.12 Crossflow heat exchanger with both fluids unmixed

FIGURE 12.13 Crossflow heat exchanger with one fluid mixed and the other unmixed



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Note: In Fig. 12.13, the dashed lines are for the case of Cmin unmixed and Cmax mixed. And, the solid lines are for the case of Cmin mixed and Cmax, unmixed. From the NTU—Effectiveness graphs, following important points may be observed: (i) For a given value of capacity ratio, C, the effectiveness increases with NTU. Value of effectiveness varies from 0 to 1. (ii) Initially, effectiveness increases rather rapidly as NTU increases (up to a value of NTU = about 1.5) and then, slowly for larger values of NTU. Remember that NTU is a measure of the size (i.e. heat exchange area, A) of the heat exchanger; so, we can conclude that increasing the size of the heat exchanger beyond about NTU = 3, cannot be economically justified, since there will not be any corresponding increase in effectiveness. TABLE 12.7 NTU Vs. effectiveness for parallelflow HX

TABLE 12.8 NTU Vs. effectiveness for counterflow HX

FIGURE 12.14 Shell and tube heat exchanger, with one shell pass and 2, 4, 6 tube passes

FIGURE 12.15 Shell and tube heat exchanger, with two shell passes and 4, 8, 12 tube passes (iii) At a given value of NTU, effectiveness is maximum for C = 0, (i.e. for a condenser or evaporator), and decreases as C increases. (iv) For NTU less than about 0.3, effectiveness is independent of capacity ratio, C. (v) For given NTU and C, a counterflow heat exchanger has highest effectiveness and a parallel, flow heat exchanger has the lowest effectiveness. (vi) When C = 1 (i.e. capacity rates of both the fluids are equal, as in the case of a typical regenerator), maximum effectiveness of a parallelflow heat exchanger is 50% only, whereas there is no such limitation for a counterflow HX. Therefore, for such applications, obviously, the counterflow arrangement is preferred. Example 12.9. Consider a heat exchanger for cooling oil which enters at 180°C, and cooling water enters at 25°C. Mass flow rates of 2

oil and water are: 2.5 and 1.2 kg/s, respectively. Area for heat transfer = 16 m . Specific heat data for oil and water and overall U are 2

given: Cpoil = 1900 J/kgK; Cpwater = 4184 J/kgK; U = 285 W/m K. Calculate outlet temperatures of oil and water for parallel and counterflow HX. (M.U. 1995) Solution. Here, the outlet temperatures of both the fluids are not known. Use of LMTD method would require an iterative solution. i.e. to start with, assume outlet temperature of, say, hot fluid, Th2 and calculate the exit temperature of cold fluid, Tc2 and then, the LMTD; then, calculate the heat transfer rate Q. From Q and capacity rates, recalculate Th2 , and compare this value with the initially assumed value; if they do not match, say, within 0.5 deg.C, repeat the iterative But, as will be shown below, EffectivenessNTU method, offers a direct, straightforward solution: Data:

Capacity rates:

i.e. C = 0.946 (capacity ratio) Number of Transfer Units:

Case (i): Parallelflow HX: For parallelflow HX, we have the effectiveness relation:



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Then, since hot fluid is the ‘minimum’ fluid, we have:

and, Tc2 is obtained from heat balance:

Case (ii): Counterflow HX: For counterflow HX, we have:

Then, again, since hot fluid is the ‘minimum’ fluid, we have:

And, Tc2 is obtained from:

Note: In this problem, it is difficult to read accurately the ε values from the graphs, for the given values of NTU and C. It is suggested that the analytical relations may be used to get accurate results. (b) In the above Example, suppose that the flow rate of water is increased to 2 kg/s. Calculate the new outlet temperatures of oil and water for parallel and counterflow HX. Rest of the data remain the same. Now, the heat exchanger is operated at an offdesign condition, i.e. the water flow is changed from 1.2 kg/s to 2 kg/s. Then, eNTU method is convenient to use to find out the exit temperatures of both the fluids. mc := 2 kg/s (mass flow rate of cold fluid (water)) Note that still, hot fluid is the ‘minimum’ fluid and NTU remains the same, but C changes: Capacity rates: Ch := mh = Cph

Therefore, Capacity ratio:

Case (i): Parallelflow HX: We have the effectiveness relation:

Compare this ε with ε = 0.435 obtained earlier. Then, since hot fluid is the ‘minimum’ fluid, we have:

And, Tc2 is obtained from heat balance:

Case (ii): Counterflow HX: For Counterflow HX, we have:



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Compare this ∞ with ∞ = 0.496 obtained earlier. Now, again, since hot fluid is the ‘minimum’ fluid, we have:

And, Tc2 is obtained from:

Note that as a result of increasing the cold fluid (water) flow rate, the new exit temperature of both the hot and cold fluids are lower, for both the parallel and counterflow cases. Example 12.10. A steam condenser, condensing at 70°C has to have a capacity of 100 kW. Water at 20°C is used and the outlet 2

water temperature is limited to 45°C. If the overall heat transfer coefficient is 3100 W/m K, determine the area required. (b) If the inlet water temperature is increased to 30°C, determine the increased flow rate of water to maintain the same outlet temperature. (M.U. 1998) Solution. This problem can be solved by LMTD method, too. But, in part (b), since the heat exchanger is operated at an offdesign condition, we shall adopt the εNTU method. Data:

Therefore, effectiveness: Since steam is condensing, it is the ‘maximum’ fluid. So, we can write:

Now, for a condenser, we have, from Table 12.6:

Case (b): If Tc1 is increased to 30°C, and Tc2 maintained at 45°C, what is the increased flow rate?

Compare this value with Cmin = 4000 obtained earlier. Therefore, increased flow rate:

Example 12.11. Hot oil at a temperature of 180°C enters a shell and tube HX and is cooled by water entering at 25°C. There is one 2

shell pass and 6 tube passes in the HX and the overall heat transfer coefficient is 350 W/(m K). Tube is thinwalled, 15 mm ID and length per pass is 5 m. Water flow rate is 0.3 kg/s and oil flow rate is 0.4 kg/s. Determine the outlet temperatures of oil and water and also the heat transfer rate in the HX. Given: specific heat of oil = 1900 J/(kgK) and specific heat of water = 4184 J/(kgK) Solution. Since the exit temperatures of both the fluids are not known, we shall use ε – NTU method. Data:

FIGURE Example 12.11 Shell and tube heat exchanger with one shell pass and 6 tube passes Therefore, oil is the ‘minimum’ fluid.

Therefore, capacity ratio:

Number of Transfer Units:



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Effectiveness: This is a shell and tube HX with one shell pass and 6 tube passes. So, 0.605 and NTU = 0.651, from Fig. 12.14. Since it is difficult to read from the graph accurately, let us calculate ε from analytical relation given in Table 12.5: N := NTU (notation in following equation)

Outlet temperatures of hot and cold fluids:

Example 12.12. A feed water heater heats water entering at a temperature of 25°C, at a rate of 3 kg/s. Heating is due to steam condensing at 117°C. When the feed water heater was new (i.e. ‘clean’ condition), the exit temperature of water was 85°C. After prolonged operation, for the same flow rates and inlet conditions, it was observed that the outlet temperature was 75°C. Determine the 2

value of fouling factor. Given: area of heat exchange = 5.5 m . Solution. Fouling resistance, Rf is calculated from the relation:

Also, since the steam is condensing, it is the ‘maximum’ fluid, and the water is the ‘minimum fluid. Data:

Since this is a condenser, water is the ‘minimum’ fluid, and capacity rate of condensing steam is ∞ i.e. Cmin := Cc Therefore, capacity ratio:

Effectiveness: Remembering that water is the minimum fluid, effectiveness is given by:

and, from Table 12.6, NTU of the condenser is given by:

But, by definition of NTU:

After prolonged operation: Tc2 := 75°C (exit temperature of water for ‘dirty’ HX.) Therefore, effectiveness of dirty HX:

Therefore, NTU of condenser:

Therefore, Fouling factor:



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Example 12.13. Oil at 100°C (Cp = 3.6 kJ/kgK) flows at a rate of 30,000 kg/h and enters into a parallelflow HX. Cooling water 2

2

(Cp = 4.2 kJ/kg.K) enters the HX at 10°C at the rate of 50,000 kg/h. The heat transfer area is 10 m and U = 1000 W/(m K). Calculate the following: (i) outlet temperature of oil and water (ii) maximum possible outlet temperature of water. Solution. Exit temperature of both the fluids are not known; therefore, NTU method is to be used: Data

Capacity rates:

Therefore, oil is the ‘minimum’ fluid.

Therefore, Capacity ratio:

Effectiveness: For parallel flow HX, we have:

FIGURE 12.9 Example 12.13 Parallelflow heat exchanger i.e. ε = 0.262 (effectiveness of parallel flow HX with NTU = 0.333 and C = 0.514.) Note: We can use the graph of Fig. 12.10 or Table 12.7, but using the analytical relation is more accurate. Outlet temperature of hot and cold fluids:

And, from heat balance:

(b) Maximum possible outlet temperature of water: For a very long parallelflow HX, the outlet temperatures of hot and cold fluids would be the same: i.e. Th2 = Tc2 Therefore, writing the heat balance:

12.7 The Operating-line/Equilibrium-line Method NTU∞ method can be represented graphically in another way. Refer to Fig. l2.l6. Here, the xaxis represents the cold fluid temperature and the yaxis, the hot fluid temperature. Now, if we plot the entrance and exit temperatures of a heat exchanger on these axes, we see that the operating range of the HX is represented by a single line; this line is called ‘the operating line’. On the same graph, a line drawn at 45 deg. is called ‘the equilibrium line’. For equilibrium line, Th = Tc . Thermodynamically, it is impossible for the operating line of a heat exchanger to drop below the equilibrium line, since, if it does, it would mean a violation of the second law. Slope of the operating line for the counterflow HX is: (Cc / Cc ) = (Th1–Th2 )/(Tc2 –Tc1). And, the slope of the operating line for the parallelflow HX is: – (Cc /Ch), i.e. negative slope. For a condenser, operating line is horizontal with Th = constant and (Cc /Ch) = 0, and for an evaporator, the operating line is a vertical line with Tc = constant, and (Cc /Ch) ∞ Advantage of this method of representation is that the effectiveness of the heat exchanger can now be shown geometrically as a ratio of two lengths. For example, for the counterflow HX shown in Fig. l2.l6 (a), we have: Cc > Ch and the effectiveness is equal to δ/∆. For constant specific heats of fluids, the operating line is a straight line. Variation in specific heats of fluids is also shown easily in these graphs: As shown in Fig. l2.l6 (d), if the operating line curves upwards, i.e. the slope increases as the temperature increases, it



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means that Cp of cold fluid increases with temperature (or, the Cp of hot fluid decreases with temperature). Similarly, if the operating line curves downwards, it means that Cp of cold fluid decreases with temperature (or, the Cp of hot fluid increases with temperature). Effectiveness of a parallelflow heat exchanger: Operatingline/Equilibrium line method can be used to determine the effectiveness of a heat exchanger. Let us illustrate this briefly with reference to a parallelflow HX: Refer to Fig. l2.l7. Line 1–2 is the operating line for the parallelflow HX. We see from the figure that Cc < Ch, since the slope of the operating line = – Cc /Ch. And, Capacity ratio, C = Cc /Cc . From the Fig l2.17: Th1 − Tc1 = Δ Tc2 − Tc1 = a1

FIGURE 12.16 Operating line and equilibrium lines for heat exchangers

Therefore, For parallelflow HX, we have:

Now, from the Fig. 12.17 we have: Th2 − Tc2 = b1 Th − Tc1 = Δ 1

We also see from the Fig. 12.17:

FIGURE 12.17 Parallelflow heat exchanger b1 = b2 − (Th1 − Th2 ) But, b2 = Δ − a1 (from the Fig. 12.17, since equilibrium line is at 45 deg. to horizontal.) Therefore, b1 = (Δ − a1) − (Th1 − Th2 )

Substituting in Eq. A:

Using Eq. B:

Eq. C is the desired equation for the effectiveness of the parallelflow HX. This is the same as the equation derived earlier for parallelflow HX, i.e. Eq. 12.51. While deriving Eq. C, it was assumed that the cold fluid was the ‘minimum’ fluid; if we assume that the hot fluid is the minimum fluid, then also, the same result would be obtained.

12.8 Compact Heat Exchangers 2

3

Heat exchangers with an area density greater than about 700 m /m are classified as ‘compact heat exchangers’. Generally, they are used for gases. Compact heat exchangers are, typically, of three types: 1. array of finned circular tubes



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2. array of platefin matrix, and 3. array of finned flattube matrix. Heat transfer and pressure drops for these compact heat exchangers are determined experimentally and are supplied by manufacturers as their proprietary data. As an example, a platefin type of heat exchanger matrix, manufactured by MarstonExcelsior Ltd., is shown in Fig. 12.18. As shown in the Fig.12.18, a single element consists of two plates in between which is sandwiched a corrugated sheet. The two edges are sealed. Dip brazing technique is used to build a complete heat exchanger block from individual elements. Multiflow configurations are possible, and the generally used corrugations types are: plain (P), plainperforated (R), serrated (S) and herringbone (H). Table 12.9 gives the geometrical data for some typical corrugations.

FIGURE 12.18 Platefin heat exchangers for cryogenic service (MarstonExcelsior Ltd. TABLE 12.9 Geometrical data for typical corrugations (Marston Excelsior Ltd.)

In the above table, a = free flow area per metre width of corrugation A1 = (primary surface area per metre width) x (metre length of corrugation) A2 = (secondary surface area per metre width) x (metre length of corrugation) Dh = hydraulic mean diameter i.e.

Kays and London have studied a large number of compact heat exchanger matrices and presented their experimental results in the form of generalised graphs. Heat transfer data is plotted as St.Pr

2/3

against Re, where, St = Stanton number = h/(G.Cp), Pr = Prandtl 2

number = μ.Cp/k, and Re = G.Dh/μ, G = mass velocity (= mass flow rate/Area of cross section), kg/(sm .) In the same graphs, friction factor, f, is also plotted against Re. As an example, heat transfer and friction factor characteristics for a particular tubefin matrix are shown in Fig.12.19. Pressure drop in platefin heat exchangers: Total pressure drop for the fluid flowing across the heat exchanger is given by:

FIGURE 12.19 Heat transfer and friction factor for platefinned circular tube matrix (Trane Company) where

Kc and Ke = flow contraction and expansion coefficients, respectively ρi and ρo = density at inlet and exit, respectively

In Eq. 12.59, on the RHS, the first term inside the square brackets represents the entrance contraction effect, second term—the flow acceleration, the third term—core friction and the fourth term—the exit expansion effect. Core friction drop is generally 90% of the total pressure drop. For liquids, entrance and exit losses are negligible. Values of Kc and Ke are given in Kays and London. Pressure drop for finnedtube exchangers: Entrance and exit effects are included in the friction factor; therefore, Kc = Ke = 0. Then, total pressure drop across the tube bank is:

Here, first term on the RHS is the flow acceleration effect, and the second term is the core friction.



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Fig. 12.20 shows a platefin exchanger for an ethylene plant and Fig. 12.21 shows another platefin exchanger for an air liquefier. Regenerators: Regenerators are extensively used in blast furnace stoves, open hearth furnaces, coke manufacture, glass production, for air pre heating in power plants, in gas turbine systems and in cryogenic plants, in Stirling cycle air (or helium) liquefiers, in cryogenic mini coolers used for cooling infrared detectors, etc. In a regenerator, hot and cold fluids flow alternately through the regenerator matrix. The matrix may be sandlime bricks, metal packings, wire screen mesh, lead balls, etc., depending upon application. During the ‘hot blow’ hot fluid flows through the matrix and the matrix absorbs the heat from the fluid; during the ‘cold blow’, cold fluid flows through the matrix and the matrix gives up the absorbed heat to the cold fluid, thus heating the fluid. Thus, suitable valving is necessary to alternately switch the hot and cold fluids through the regenerator. In a valved type of exchanger, generally, two identical matrices are provided such that when one matrix is being heated, the other is being cooled. Alternately, regenerator may be of rotary type, where a porous matrix is rotated around its axis cutting the hot and cold fluid lines, thus transferring heat from the hot to the cold fluid.

FIGURE 12.20 A heat exchanger assembly with associated pipe work for an ethylene plant (Marston Excelsior Ltd.) Analysis of a periodic flow HX is complicated since the matrix and gas temperatures vary with both position and time. A rough outline of the analysis is given below: Refer to Fig. 12.22, which shows a regenerator diagrammatically. Hot fluid flows through the matrix during the hot blow and heats the matrix. Then, the flow is switched to effect the cold blow and the cold fluid flows through the matrix and gets heated up. Thus, in effect, heat is transferred from the hot fluid to the cold fluid. We are interested in the gas and matrix temperatures at any location and at any time. These are obtained by writing an energy balance for an element of width dx, shown in the Fig. 12.22. Following notations are used:

FIGURE 12.21 Air liquefier for a 400 Ton/day Oxygen plant using two, 763 mm × 763 mm blocks in parallel (Marston Excelsior Ltd.)

FIGURE 12.22 Periodicflow heat exchanger (Regenerator)

Ms

= Mass of solid (matrix filling) per unit length, kg/m

M

= Mass flow rate of gas, kg/s

Cps

= specific heat of solid, J/(kgK)

Cpg

= specific heat of gas, J/(kgK)

V

= free volume per unit length

A

= heat transfer area per unit length

ρ

= density of gas, kg/m

L

= length of matrix column

h

= convective heat transfer coefficient between the gas and the

3

matrix



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t

= solid temperature at a given location x

tg

= gas temperature at loc

Writing the heat balance: Heat transferred by convection between the gas and the solid = Heat stored in the solid i.e.

Now, the heat transferred by convection is also equal to the heat stored in the gas contained in length dx plus the increase in the enthalpy of the gas as it passes through the element dx. i.e.

Above equations are simplified as:

In most of the practical situations, the term (ρ.V/M) is very small and is neglected. Then, making following substitutions

the resulting equations are solved with the following boundary conditions:

The results are presented usually in graphical form and the nature of graphs is shown in Fig. 12.23 (a) and (b).

FIGURE 12.23 Gas and solid temperature charts for a regenerator In these graphs, tgo is the initial temperature of the gas, and to is the initial temperature of the solid. Fig. 12.23 (a) presents the dimensionless gas temperature at any location as a function of ξ and η and Fig. 12.23 (b) shows the dimensionless solid temperature as a function of ξ and η. EffectivenessNTU relations for regenerator: Effectiveness of a regenerator is presented as a function of three dimensionless parameters, as follows:

where, NTUmod = modified NTU, given by:

and, matrix capacity rate is equal to matrix mass rate times the specific heat of the solid. For the rotary type of regenerator,

For the valved type of regenerator, total mass of both the identical matrices is used, multiplied by valve cycles/s, where period is the interval between ‘valveontoofftoon’. Kays and London have presented εNTUmod graphs for different Cr /Cmin ratios (ranging from 1 to infinity), for given Cmin/Cmax ratios (ranging from 0.5 to 1). Table 12.10 is a sample table showing ε values for Cmin/Cmax = 1. Fig. 12.24 presents this table in graphical form.



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TABLE 12.10 Effectiveness of periodic flow HX (Cmin/Cmax = 1)

FIGURE 12.24 Effectiveness of a periodicflow HX (regenerator) for (Cmin/Cmax = 1) Higher NTUmod ranges (of the order of 100 or more) are generally applicable to regenerators used in cryogenic applications; in such cases, since the effectiveness approaches unity asymptotically, for better clarity, graphs are plotted with (1 – ε) vs. NTUmod. (See Appendix at the end of chapter). To calculate NTUmod we need the heat transfer coefficients on the cold and hot fluid sides. We also need the heat transfer area. Heat transfer characteristics in terms of Colburn j factor vs. Reynolds number, and friction factor vs. Reynolds number are presented for many types of matrices by Kays and London. They also provide physical data such as hydraulic diameter, heat transfer area, porosity, etc., for those matrices. As an example, heat transfer characteristics and friction factor data for a randomly stacked, wire screen matrix (used typically, in gas turbine regenerators) are shown in Fig. 12.25 and Fig. 12.26, respectively. Advantages of regenerators: 2

3

1. High surface density, of the order of 3000 m /m (for a 24 mesh screen matrix, typically used in gas turbine regenerators), can be packed into a given volume 2. Tends to be ‘selfcleaning’ because of periodic flow reversals 3. Cheaper on per unit heat transfer area basis. Disadvantages: 1. Some mixing of hot and cold fluids is unavoidable 2. Sealing between the fluids presents some problem if the pressure differential is large.

FIGURE 12.25 Colburn j factor vs. Reynolds number for a randomly sacked wire screen matrix

FIGURE 12.26 Friction factor vs. Reynolds number for a randomly stacked wire screen matrix

12.9 Hydro-mechanical Design of Heat Exchangers So far, we studied thermal design aspects for a heat exchanger. But, from a practical point of view, the pressure drop that occurs when the fluid passes through the heat exchangers and the pumping power required to effect this flow, are also important. We shall only briefly mention about this aspect. Obviously, flow of fluid through heat exchanger passages involves pressure drop. And, higher the viscosity of the fluid, higher the pressure drop. Total pressure drop in a heat exchanger section is calculated by summing up the following individual pressure drops: 1. Pressure drops in straight passages and pipe bends, ∆Pf 2. Pressure drops due to ‘end effects’, i.e. due to flow contraction and expansion at the ends, ∆Pe 3. Pressure drops due to flow acceleration (in cases of gases in nonisothermal flow), ∆Pa, and 4. Pressure drop due to self draught (due to buoyant forces) as a result of change in elevation of flow channels, ∆Ps (a) Pressure drops in straight passages and bends These are determined by Darcy formula, as explained in the chapter on convection.

3

In Eq. 12.68, ρ is density of fluid (kg/m ) and V is mean velocity of flow (m/s). The friction factor, fD is determined depending on the Reynolds number, as explained in the chapter on Forced convection. Effect of bends and valves in the flow lines is generally accounted for by adding an ‘equivalent length’ for each bend or valve, to the straight length. Equivalent lengths of a few fittings are shown in Table 12.11: (b) Pressure drops due to contractions and expansions ∆ Pe is a function of area ratio A1/A2 , where A1 is the smaller area.

where, V refers to velocity at smaller cross section. Values of fcont and fexpn are given in Table 12.12; these are shown graphically in Fig. 12.27. (c) Pressure drops due to flow acceleration This pressure drop, in a channel of constant crosssection, is equal to twice the difference in velocity heads, i.e.



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TABLE 12.11 Equivalent lengths of fittings

Fitting

45° Elbow

90° Elbow (standard radius)

90° Elbow (medium radius)

90° Elbow (long sweep)

90° Square elbow

180° Close return bend

Swing check valve, open

Tee (as eL, entering run)

Tee (as eL, entering branch)

Couplings, unions

Gate valve, open

Gate valve, 1/4 closed



Globe valve, open

Angle valve, open

L /D (for turbulent flow only) e

15

31

26

20

65

75

77

65

90

Negligible

7

40

190

840

340

170

TABLE 12.12 Sudden contraction and expansion coefficients for a tube



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A /A 1

2

f

cont

f

expn

0

1

0.5

0.1

0.8

0.45

0.2

0.65

0.42

0.3

0.5

0.38

0.4

0.37

0.34

0.5

0.22

0.28

0.6

0.14

0.23

0.7

0.1

0.2

0.8

0.03

0.12

0.9

0.01

0.07

1

0

0

FIGURE 12.27 Sudden conraction and expansion coefficients for a tube where, subscripts 1 and 2 refer to inlet and outlet, respectively. (d) Pressure drop due to selfdraught If the height of the vertical channel (flue) is ‘h’, ρ0 = density of cold fluid (say, air) and p = density of hot fluid (say, flue gas), then pressure drop due to selfdraught is given by:

where, ‘g’ is the acceleration due to gravity. ∆Ps is positive for the descending fluid and negative, if the fluid is ascending through the channel. ∆Ps is zero if the heat exchanger is not exposed to ambient air, but is connected in a closed system. Then, the total pressure drop is given by the summation of all these pressure drops:

Power required to originate fluid flow: Once the total pressure drop in the system is determined, the power required to circulate the fluid through the system is easily calculated:

3

where,

Flow = volumetric flow rate, m /s,

M = mass flow rate of fluid, kg/s

∆Pt = total pressure drop, N/m2, and


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3

ρ = density of liquid or gas, kg/m

η = efficiency of pump or fan.

12.10 Summary Heat exchanger is one of the important pieces of process equipment, used extensively in research as well as industrial applications. Heat exchangers may be of recuperative, regenerative or direct contact type. In this chapter, we focussed on the thermal design aspects of heat exchangers. First, the method of calculating the overall heat transfer coefficient was explained. Inclusion of fouling resistance is an important aspect of design and this was discussed next. Calculation of logarithmic mean temperature difference, LMTD, between the two fluid streams exchanging heat, is an important step in the design. Procedure of calculating the LMTD for parallel and counterflow heat exchangers was explained; for more complicated type of exchangers, such as cross flow or multipass shellandtube heat exchangers, mean temperature difference is calculated by multiplying the LMTD of a counterflow HX by a correction factor. Correction factor graphs have been given for a few types of heat exchangers. Problems in heat exchanger are mainly of two types: (i) design problems where one has to calculate the area of the HX, and (ii) performance problems where one has to calculate the outlet temperatures of both the fluids, given the inlet temperatures. LMTD approach is suitable for the first type of problems, whereas for the second type of problems, eNTU approach is recommended, since in this case LMTD approach would require a laborious iterative procedure. εNTU relations and graphs for a few important cases have been given. Further, operatingline/equilibriumline method was also briefly explained. Compact heat exchangers and regenerators are also used in a variety of applications. Brief mention has been made about these; however, their design is rather more involved and use of proprietary technical information from the suppliers’ catalogues will be required. Finally, calculation of pressure drops and the necessary pumping power in a heat exchanger, has been explained. Selection of heat exchangers for a particular application is a serious task for the engineer and the following aspects must be borne in mind while selecting a heat exchanger: 1. required heat transfer rate 2. necessary pumping power 3. type of heat exchanger most suitable, depending upon the process 4. materials of construction and fabrication and testing procedures, with due consideration to operating temperatures and pressures 5. size and weight, depending upon application 6. ease of maintenance and servicing 7. safety and reliability aspects, and 8. cost.

Questions 1. How are heat exchangers classified? Discuss briefly different types of heat exchangers. Why is counterflow HX better than parallelflow HX? [M.U.] 2. Draw temperature vs. length profiles for: (i) Condenser (ii) Evaporator (iii) Counterflow HX with Ch = Cc [M.U.] 3. What is overall heat transfer coefficient? What is its importance? Derive an expression for overall heat transfer coefficient for a tubular HX based on inner surface area. [M.U.] 4. Explain the terms: Fouling factor, Effectiveness, NTU and LMTD. [M.U.] 5. Write short notes on correction factor charts for crossflow heat exchangers.



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[M.U.] 6. Starting from fundamentals, derive an expression for the mean temperature difference for counterflow HX in terms of inlet and outlet temperatures of hot and cold fluids. [M.U.] 7. Derive an expression for the LMTD of a parallelflow HX. State clearly the assumptions. [M.U.] 8. Derive an expression for the effectiveness of a counterflow HX when capacity rate of hot fluid is more than that of cold fluid. Hence show that effectiveness of a condenser is given by: ε = 1 − exp(−NTU) …[M.U.] 9. Starting from basics, derive an equation for the effectiveness of a parallelflow HX in terms of NTU and capacity ratio. Also, show that when capacity ratio is 1, effectiveness is given by: ε = (½).{1 − exp(− 2.NTU)} …[M.U.] 10. Prove that for a counterflow HX, when Cmin/Cmax = 1, ε = NTU/(1 + NTU). …[M.U.] 11. Compare LMTD and εNTU methods of solving heat exchanger problems. 12. Using the operatingline/equilibriumline method, derive an expression for the effectiveness of a counterflow HX. Assume Cc > Ch. 13. Write a short note on compact heat exchangers and regenerators.

Problems 1. A copper pipe (k = 350 W/mK) of 17.5 mm ID and 20 mm OD conveys water and the oil flows through the annular passage 2

between this pipe and a steel pipe. On the water side, the film coefficient is fouling factor is 0.00034 m K/W. The 2

corresponding values for the oil side are m K/W. Calculate the overall heat transfer coefficient between the water and oil, based on outside surface area of inner pipe. 2. In a shell and tube counterflow HX, water flows through a copper tube (20 mm ID, 23 mm OD), while oil flows through the shell. Water enters at 20°C and comes out at 30°C while oil enters at 75°C and comes out at 60°C. The water and oil side film 2

coefficients. are: 4500 and 1250 W/(m K), respectively. Thermal conductivity of tube wall is 355 W/(mK). Fouling factors on 2

water and oil sides are: 0.0004 and 0.001 m K/W, respectively. If the length of tube is 2.4 m, calculate the overall heat transfer coefficient and rate of heat transfer. [M.U.] 3. Saturated seam at 120°C is condensing on the outer surface of a single pass HX. The overall heat transfer coefficient is 1600 2

W/(m K). Determine the surface area of the HX required to heat 2000 kg/h of water from 20°C to 90°C. Also, determine the rate of condensation of steam in kg/h. Assume latent heat of steam to be 2195 kJ/kg. [M.U.] 4. A HX is required to cool 55,000 kg/h of alcohol from 66°C to 40°C using 40000 kg/h of water entering at 5°C. Calculate (i) the exit temperature of water (ii) heat transfer (iii) surface area required for: (a) parallelflow type (b) counterflow type of HX. 2

Take overall heat transfer coefficient U = 580 W/(m K). Cp (alcohol) = 3760 J/(kgK) and Cp (water) = 4180 J/(kgK). 5. In a counterflow double pipe HX, water flow rate is 1300 kg/h. and it enters at 15°C. It is heated by oil, Cp = 2 kJ/kgK; oil flow 2

2

rate is 550 kg/h. Oil inlet temperature is 95°C. Overall U = 800 W/m K. Surface area of HX: 1.34 m . Table of NTUe is given as follows:



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Capacity ratio, R

NTU

Effectiveness

0.202

3

0.93

0.202

4

0.96

Find out ε, NTU and outlet temperatures. [M.U.] 6. In a gas turbine installation, a counterflow HX, has hot exhaust gas outlet at 330°C, and air outlet at 460°C. For each element of HX, (dQ/(dT.dX) is uniform and is equal to C and C.L = 52.3 kW/K. Capacity rate for hot fluid = 21.76 kW/K and for cold fluid = 19.04 kW/K. Temperature variation along the length is linear for both fluids. Calculate temperatures at entry. [M.U.] 7. A one shell, 2tube pass steam condenser, has 2000 tubes of 20 mm diameter, with cooling water entry at 20°C, flow rate 3000 2

8

kg/s; U = 6890 W/m K. Total heat to be transferred, Q = 2.331 × 10 W. Steam condenses at 50°C. Determine tube length per pass using NTU method. Given that at 0.6 and 0.64 effectiveness, NTU is 0.78 and 0.82. [M.U.] 8. A water preheater of ID:3.2 cm, OD:3.52 cm, is heated by steam at 180°C. Water flows through pipe at a velocity of 1.2 m/s. h’ 2

on steam side:11,000 W/m K; water is heated from 25°C to 95°C. k of pipe material: 59 W/mK. Properties of water at 60°C are −4

given. Calculate the length required. Use appropriate empirical relation. Data: μ = 4.62 × 10 kg/ms; k = 0.653 W/mK; Cp = 4200 J/kgK. [M.U.] 9. Consider a HX for cooling oil entering at 180°C, by water entering at 25°C; mass flow rates of oil and water are: 2.5 and 1.2 2

kg/s, respectively. Area: 16 m . Specific heat data for oil and water and overall U are given: Data: Cp of oil = 1900 J/kgK; Cp of 2

water = 4184 J/kgK; U = 285 W/m K. Calculate outlet temperature of oil and water for parallel and counterflow HX. [M.U.] 10. In a shellandtube HX, 50 kg/min of furnace oil is heated from 10°C to 90°C. Steam at 120°C flows through the Shell and oil flows inside the tube. Tube size: 1.65 cm ID and 1.9 cm OD. Film coefficients on oil and steam sides are: 85 and 7420 2

W/(m K). Find the number of passes and number of tubes in each pass if the length of each tube is limited to 2.85 m. Velocity of oil is limited to 5 cm/s. Density and specific heat of oil are 1970 J/(kgK), respectively. [M.U.] 11. Water at a rate of 4080 kg/h is heated from 35°C to 75°C by an oil of Cp = 1.9 kJ/(kgK). The HX is of counterflow, double pipe design. The oil enters at 110°C and leaves at 75°C. Determine: (i) mass flow rate of oil (ii) area of HX necessary to handle this 2

load, if overall heat transfer coefficient, U = 320 W/(m K). [M.U.] 12. A steam condenser, condensing at 70°C has to have a capacity of 100 kW. Water at 20°C is used and the outlet water 2

temperature is limited to 45°C. If the overall heat transfer coefficient is 3100 W/(m K), determine the area required. If the inlet water temperature is increased to 30°C, determine the increased flow rate of water to maintain the same outlet temperature. [M.U.] 13. Water enters a counterflow, double pipe HX at 38°C flowing at a rate of 0.76 kg/s. It is heated by oil (Cp = 1.88 kJ/kgK) 2

flowing at a rate of 0.152 kg/s from an inlet temperature of 116°C. For an area of 1.3 m Enjoy Safari? Subscribe Today and an overall heat transfer coefficient

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2

of 340 W/(m K), determine the total heat transfer rate. Take Cp for water = 4.170 kJ/ kgK. Given: expression for effectiveness of double pipe, countercurrent HX:

14. A refrigerator is designed to cool 250 kg/h of hot liquid (Cp = 3350 J/kgK) at 120°C using a parallelflow arrangement. 1000 2

2

kg/h of cooling water is available at a temperature of 10°C. If overall U = 1160 W/(m K) and the surface area of HX is 0.25 m , calculate the outlet temperatures of the fluids and also the effectiveness of the HX. [M.U.] 15. A steam condenser, condensing at 100°C has a capacity of 150 kW. Water at 25°C is used and the outlet water temperature is 2

35°C. If the overall heat transfer coefficient is 3000 W/(m K), determine the area required for the HX. [M.U.] 16. A parallelflow HX has hot and cold water streams running through it and has following data: mh = 10 kg/min, mc = 25 kg/min, Cph = Cpc = 4.18 kJ/(kgK), Th1 = 70°C, Th2 = 50°C, Tc1 = 25°C. Individual heat transfer coefficients on both sides are area of HX (ii) exit temperatures of hot and cold fluids, if hot water flow rate is doubled. [M.U.] 17. Steam at atmospheric pressure enters the shell of a surface condenser in which water flows through a bundle of tubes of diameter = 25 mm at a rate of 0.05 kg/s. The inlet and outlet temperatures of water are 15°C and 70°C, respectively. 2

Condensation of steam takes place on the outside surface of the tube. If U = 230 W/(m K), using NTU method, find: (i) effectiveness of the HX (ii) length of tube required, and (iii) rate of steam condensation. [M.U.] 18. An economiser in a boiler has water flowing inside the pipes and hot gases on the outside, flowing across the pipes. The flow rate of the gases is 2000 tonne/h and they are cooled from 390°C to 200°C and their specific heat is 1005 J/(kgC). Water is 2

heated under high pressure from 100°C to 220°C. Assuming an overall heat transfer coefficient of 35 W/(m C), determine the area required. The correction factor, F = 0.8. [M.U.] 19. Hot oil is being cooled from 200°C to 130°C in a parallelflow HX by water entering a 25°C and exiting at 60°C. Determine the outlet temperatures of both the streams if the HX is made counterflow. [M.U.] 20. A shell and tube HX with two shell passes and 8 tube passes has ethyl alcohol (Cp = 2670 J/kgC) flowing inside the tubes, and water (Cp = 4190 J/kgC) flows through the shell. Ethyl alcohol enters at 25°C and leaves at 75°C with a flow rate of 2 kg/s 2

whereas water enters at 95°C and leaves at 45°C. Overall heat transfer coefficient U = 850 W/(m K). Determine the surface area required for the HX. Appendix

In this Appendix to Chapter 12, some more information on compact heat exchangers and regenerators is given. Example A12.1. Air at 2 atm and 400 K flows at a rate of 5 kg/s, across a finned circular tube matrix shown in Fig. 12.19. Dimensions of the heat exchanger matrix are: 1 m (W) × 0.6 m (Deep) × 0.5 m (H), as shown in Fig. A12.1. Find: (a) the heat transfer coefficient (b) the friction factor, and (c) ratio of core friction pressure drop to the inlet pressure. Solution. Data:

Physical properties of air at 2 atm and 400 K:



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FIGURE A12.1 Configuration of compact heat exchanger for Example A12.1

From Fig. 12.19, we have:

(a) And, heat transfer coefficient:

(b) Friction factor: From Fig. 12.19, for Re = 2972, we get: f = 0.024 (friction factor) (c) Pressure drop:

i.e. frictional pressure drop is 0.78% of the inlet pressure. Fig. A12.2, A12.3, and A12.4 show data for three more compact heat exchanger matrices, from Kays and London. Fig. A12.5 shows data for crossed rod matrices, random stacking (d = 0.375 in.) used in regenerators (Kays and London): Fig. A12.6 gives data for an infinite, randomly stacked sphere matrix, with porosity varying from 0.37 to 0.39. (Kays and London) Data of Fig. A12.6 is given in tabular form in Table A12.1. Analysis of regenerators by NTU method: As explained earlier, in a regenerator, the same space is alternately occupied by the hot and cold fluids; regenerator matrix stores the ‘heat’ during the flow of hot fluid (i.e. during ‘hot blow’) and rejects this heat to the cold fluid during the flow of cold fluid through the regenerator matrix (i.e. during ‘cold blow’). Temperatures of the gas as well as of the matrix solid are functions of both position and time. After sufficiently long time, some sort of steady state is reached, and the same temperature distribution is repeated in each cycle of operation. An NTU analysis, similar to the one done for heat exchangers, proceeds as follows, with the assumption that the same mass flow rate of gas is maintained during both the hot and cold blows. (Ref: Cryogenic Systems by R.F. Barron).

FIGURE A12.2 Heat transfer coefficient and friction factor data for finned circular tube matrix (surface CF 8.75/8J)

FIGURE A12.3 Heat transfer coefficient and friction factor data for finned circular tube matrix (surface CF–7.0–5/8J) Following notations are used:

ms

= mass of solid (matrix filling) in regenerator, kg

m

= mass flow rate of gas through regenerator, kg/s

cs

= specific heat of solid, J/(kgK)

cp

= specific heat of gas, J/(kgK)



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L

= length of regenerator, m

Ah

= total heat transfer area of solid material in regenerator, m2

3

ρ

= density of gas, kg/m

T

= temperature of gas at location x and time τ

Ts

= temperature of solid material at location x and time τ

FIGURE A12.4 Heat transfer coefficient and friction factor data for finned flat tube matrix (surface 9.–0.7375)

FIGURE A12.5 Crossed rod matrix, random stacking (rod diameter d = 0.375 in.) 2

hc = heat transfer coefficient between the solid material and gas, W/(m /C). Now, applying the First law to a differential element of gas: heat transferred to or from the gas = change in enthalpy of gas in a length dx, as it flows through the regenerator i.e.

Here, change in energy stored in the gas within the differential element is neglected. (This is true for cryogenic regenerators.) Again, applying the First law to a differential element of solid material, we have: heat transferred to or from the solid material = change in energy stored within the solid material

FIGURE A12.6 Data for an infinite, randomly stacked sphere matrix with porosity varying from 0.37 to 0.39 TABLE A12.1 Heat transfer and friction data for sphere bed matrices (Kays and London) (Random packing, p = 0.37 to 0.39)



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2/3

Reynolds number, N

N .N

R

st

f

pr

50000

0.0089

0.30

20000

0.0118

0.34

10000

0.0144

0.37

5000

0.0178

0.41

2000

0.023

0.47

10000

0.029

0.52

500

0.0355

0.59

200

0.046

0.80

100

0.056

1.10

50

0.069

1.65

20

0.091

3.0

10

0.112

5.2

Solving Eqs. 1 and 2 , we get the partial differential equation for the temperature of gas flowing through the regenerator:

From Eq. (3), we observe that two important dimensionless quantities are involved in the analysis of a regenerator, i.e.

Ntu

= hc Ah/(m cp) = Number of heat transfer units, and

Fn

= hc Ah/(ms cs f) = Frequency number, where f = 1/P = frequency of switching the hot and cold stream, P = heating or cooling period.

f.τ

= dimensionless time.

FIGURE A12.7 NTU analysis of a regenerator

FIGURE A12.8 Gas temperature distribution in a counterflow regenerator



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Eq. 3 has been solved numerically for temperature distribution of gas, by Hausen for steady state cyclic operation of a regenerator. Results for a particular case of Ntu = 10 are shown in Fig. A 12.8, as an example. Once the temperature distribution is known, actual energy transferred is calculated as:

Maximum possible heat transfer in the regenerator occurs when the gas is heated from Tco (i.e. temperature of cold gas entering at x = 0) to the temperature ThL (i.e. temperature of hot gas entering the regenerator at x = L). We get:

Then, regenerator effectiveness e, is given by:

Hausen’s numerical solution for the effectiveness of a regenerator as a function of frequency number and number of transfer units, is shown in Fig. A 12.9. It is clear from Fig. A12.9 that for a large effectiveness, we need a small frequency number (F n) and a large number of transfer units (Nu), i.e. for a large effectiveness of a regenerator, the requirements are: 1. large heat transfer coefficient, hc 2. small gas mass flow rate, m, or small gas capacity rate m.cp 3. large product of regenerator matrix mass and its specific heat, ms.cs 4. large frequency, f. 2

Example A12.2. Following data are given for a regenerator: heat transfer coefficient = 640 W/(m K), heat transfer area per unit 2

length = 100 m /m, mass of matrix solid per unit length = 8 kg/m, specific heat of matrix material = 800 J/(kgK), frequency of operation = 60 cpm (= 1 cycle/s), mass flow rate of gas through regenerator = 0.013 kg/s and specific heat of gas = 5200 J/(kgK). Desired effectiveness of this regenerator, operating in a counterflow mode, is 0.95. Determine the length of the regenerator required. Solution. Data:

FIGURE A12.9 Effectiveness of a counterflow regenerator

Number of heat transfer units: For a F n = 10 and ε = 0.95, get the value of Ntu from Fig. A 12.9:

Therefore, heat transfer area required:

Therefore, regenerator length required:

Regenerator ineffectiveness (1 – ε) vs. NTU graphs for cryogenic regenerators: As stated in the text, cryogenic regenerators, generally, have large values of modified NTU (i.e. NTUmod), of the order of 100 or more. It may be observed that in the usual εNTU graphs, the value of ε approaches unity asymptotically; so, for cryogenic heat exchangers, it is more instructive and convenient to draw regenerator ineffectiveness (1 – ε) against NTUmod in loglog coordinates. Two sample graphs, one for Cmin/Cmax = 1, and the other for Cmin/Cmax = 0.95, are shown in Fig. A12.10 and Fig. A12.11, respectively. (Ref: Compact Heat Exchangers by Kays and London).



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FIGURE A12.10 Regenerator ineffectiveness as a function of Ntu0 and matrix capacity rate ratio (Cmin/Cmax = 1)

FIGURE A12.11 Regenerator ineffectiveness as a function of Ntu0 and matrix capacity rate ratio (Cmin/Cmax = 0.95)

Example 12.9. Example 12.9. Calculate the surface area required for a HX to cool 55,000 kg/h of alcohol from 66°C to 40°C using 40,000 kg/h of water entering at 5°C, for the following arrangements: (i) counterflow, tube & shell (ii) Parallel flow, tube & shell (iii) Reversed current HX with 2 shell passes and 12 tube passes with alcohol flow in the shell. Assume LMTD correction factor as 0.96 (iv) cross flow with one tube pass, with shell side fluid assumed to be mixed, with LMTD correction factor as 0.91. Assume U based on outside 2

area of tubes as 570 W/(m K). Cp of alcohol is 3.8 kJ/(kgK) and for water 4.187 kJ/(kgK). Solution. **** Figure Example 12.9 Counterflow heat exchanger Data: Mh = **** i.e. mh = 15.278 kg/s…mass flow rate of hot fluid (Alcohol) mc = **** i.e. mc = 11.111 kg/s…mass flow rate of cold fluid (Water) Th1 = 66°C…inlet temperature of hot fluid Th2 = 40°C…exit temperature of hot fluid U = 570 W/(m2K)…overall heat transfer coefficient Tc1 = 5°C…inlet temperature of cold fluid Cph = 3.8 kJ/(kgK)…specific heat of hot fluid Cpc = 4.187 kJ/(kgK)…specific heat of cold fluid F1 = 0.96 …Correction factor for case (iii) F2 = 0.91 …Correction factor for case (iv) Case (i): Counterflow HX Total heat transferred: Q = mh Cph (Th1 – Th2) i.e. Q = 1.509 × 103 kW Exit temperature of cold fluid: Tc2 = Tc1 + **** i.e. Tc2 = 37.446°C To calculate LMTD for counterflow HX: Now, d*T1 = Th1 – Tc2 i.e. d*T1 = 28.554°C…temperature difference at inlet and, d*T2 = Th2 – Tc1 i.e. d*T2 = 35°C…temperature difference at exit and, LMTD1 = **** i.e. LMTD1 = 31.668°C…Log Mean Temperature Difference Enjoy Safari? Subscribe Today


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Therefore, area required: A1 = **** i.e. A1 = 83.622 m2…area required for a counterflow HX. Case (ii): Parallel flow Hx: Total heat transferred: Q = mh Cph (Th1 – Th2) i.e. Q = 1.509 ×* 103 kW Exit temperature of cold fluid: Tc2 = Tc1 + **** Tc2 = 37.446°C To calculate LMTD: Now, d*T1 = Th1 – Tc1 i.e. d*T1 = 61°C…temperature difference at inlet and, d*T2 = Th2 – Tc2 i.e. d*T2 = 2.554°C…temperature difference at exit and, LMTD2 = **** i.e. LMTD2 = 18.419°C…Log Mean Temperature Difference Therefore, area required: A2 = **** i.e. A2 = 143.771 m2…area required for a parallel flow HX Case (iii): Reversed current HX: A3 = **** …including the correction factor F1 A3 = 87.107 m2…area required for a reversed current HX. Case (iv): Cross flow HX: A4 = **** …including the correction factor F2 i.e. A4 = 91.893 m2…area required for a cross flow HX.

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