Based on the nature of the heat exchange process

Based on the direction of fluid flow

Based on the mechanical design

Based on the physical state of working fluid

Based on the compactness

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		CHAPTER

Heat exchangers		7
Heat exchangers

7.1 Introduction

A heat exchanger is a device that facilitates the process of heat exchange between two fluids that are at different temperatures. Heat exchangers are used in many engineering applications, such as refrigeration, heating and airconditioning systems, power plants, chemical processing systems, food processing systems, automobile radiators, and waste heat recovery units. Air preheaters, economizers, evaporators, superheaters, condensers, and cooling towers used in a power plant are a few examples of heat exchangers.

This chapter presents the parameters that influence the performance of a heat exchanger and discusses the approaches for the design of a heat exchanger or the prediction of the performance of an already existing heat exchanger.

7.2 Classification of heat exchangers

Heat exchangers can be classified based on different criteria, as listed below.

1.Direct contact–type heat exchanger

In this type, both fluids are from the same substance and a schematic is shown in Fig. 7.1A. For example, the hot fluid is water vapor, and the cold fluid is water.

The limitation of a direct contact–type heat exchanger is that both the fluids must be of the same substance, like hot water and cold water, steam and water, etc.

2.Regenerator type of heat exchanger

In this type of heat exchanger, the hot and cold fluids flow through the heat exchanger alternately and a schematic is shown in Fig. 7.1B. When the hot fluid flows through the heat exchanger, heat is transferred from the hot fluid to the heat exchanger wall (matrix). The hot fluid is then stopped, and the cold fluid is sent so that the heat is transferred from the heat exchanger wall (matrix) to the cold fluid. This is called a fixed matrix regenerator. A rotary regenerator employs a matrix in the form of a wheel that rotates continuously through the counterflowing streams of the hot and cold fluids.

Heat Transfer Engineering. http://dx.doi.org/10.1016/B978-0-12-818503-2.00007-1

199

200 CHAPTER 7 Heat exchangers

FIGURE 7.1

Schematic representation of (A) direct contact–type heat exchanger, (B) regenerator type of heat exchanger, and (C) recuperator type of heat exchanger.

FIGURE 7.2

(A) Parallel-flow heat exchanger and (B) counterflow heat exchanger.

3.Recuperator type of heat exchanger

In this type of heat exchanger, both the hot and cold fluids flow through the heat exchanger simultaneously and are separated by a thin wall, as shown in Fig. 7.1C.

1.Parallel flow

In a parallel-flow heat exchanger, both the hot and cold fluids move in the same direction, as shown in Fig. 7.2A.

2.Counterflow

In a counterflow heat exchanger, the hot and cold fluids move in opposite directions, as shown in Fig. 7.2B.

3.Cross flow

In a cross-flow heat exchanger, the hot and cold fluids move in perpendicular directions, as shown in Figs. 7.3A and B.

7.2 Classification of heat exchangers 201

FIGURE 7.3

Cross-flow heat exchanger with (A) both fluids unmixed and (B) one fluid mixed and the other unmixed.

The difference between the heat exchangers shown in Figs. 7.3A and B is that in the former, both fluids are unmixed while in the latter, the fluid flowing outside the tubes is mixed. It is intuitively apparent that the thermal performance of the two will be different.

1.Concentric tube heat exchanger

Also called a double pipe heat exchanger, it is one wherein one fluid flows through the inner tube, and the other fluid flows through the annulus, as shown in Figs. 7.2A and B.

2.Shell and tube heat exchanger

In this type of heat exchanger, one of the fluids flows through a number of tubes stacked in a shell, and the other fluid flows outside the tubes. Depending on the requirement, there can be multiple tube or shell passes. Flow conditions in a shell and tube heat exchanger are neither parallel flow nor counter flow. Fluid flow outside the tubes is directed by separators known as baffles placed in the shell, as shown in Fig. 7.4. A concentric tube heat exchanger is the simplest form of a shell and tube heat exchanger.

3.Multipass heat exchanger

Shell and tube heat exchangers and crossflow heat exchangers can be of multipass type to enhance their heat transfer capability. Multiple tube passes or shell passes

202 CHAPTER 7 Heat exchangers

FIGURE 7.4

Shell and tube heat exchanger.

FIGURE 7.5

Temperature profiles of the hot and cold fluids in (A) a condenser and (B) an evaporator.

are chosen based on the velocity consideration, the total heat transfer area requirement, and the space (the heat exchanger length) constraints.

1.Condenser

The hot fluid condenses as the heat is transferred to the cold fluid. The temperature of the condensing (hot) fluid remains constant, as shown in Fig. 7.5A.

2.Evaporator

The cold fluid evaporates due to the heat transfer from the hot fluid. The temperature of the evaporating (cold) fluid remains constant, as shown in Fig. 7.5B.

It may be noted that the convection heat transfer coefficients associated with condensation and evaporation are very high compared to single-phase heat transfer coefficients.

= 10.45 m2

U × F × LMTD

1.8 × 0.65 × 20.6

Compact heat exchangers pack a large amount of heat transfer surface area (≥400 m2/m3) per unit volume of the heat exchanger. Gas flow is normally associated with poor heat

7.3 Heat exchanger analysis 203

FIGURE 7.6

Compact heat exchangers: (A, B, C) fin-type heat exchanger and (D) plate-fin heat exchanger.

transfer coefficients, so compact heat exchangers are employed when the heat transfer is between two gases or between a gas and a liquid. Fin-tube heat exchangers and platefin heat exchangers, as shown in Fig. 7.6, are examples of compact heat exchangers.

In this chapter, the heat transfer analysis is presented only for the recuperative type of heat exchangers, which are extensively used in engineering applications.

7.3 Heat exchanger analysis

The heat transfer analysis of a heat exchanger involves relating the total heat transfer rate to variables like inlet and outlet temperatures of the hot and cold fluids, the overall heat transfer coefficient, and the overall heat transfer surface area. The analysis is essentially based on the energy balance between the heat gained by the cold fluid, the heat lost by the hot fluid, and the heat transferred through the wall that separates the two fluids. Let m, cp, and T denote the mass flow rate, the specific heat, and the temperature of a fluid, respectively. The subscripts h and c indicate the hot and cold fluids, respectively. Neglecting the heat transfer between the heat exchanger and its surroundings (assuming that heat transfer is only between the hot and cold fluids flowing through the heat exchanger) and neglecting the changes in kinetic and potential energies of the fluids and assuming a steady flow, the energy balance for a heat exchanger shown in Fig. 7.7 can be written as

mh cph (Thi − Tho ) = mc cpc (Tco − Tci )

(7.1)

204 CHAPTER 7 Heat exchangers

FIGURE 7.7

Schematic representation of a heat exchanger.

The subscripts i and o in Eq. (7.1) indicate the inlet and the outlet, respectively. From Eq. (7.1), one can calculate the unknown temperature. However, it is not possible to determine the size of the heat exchanger required. Therefore, it is necessary to analyze the heat exchanger using the heat transfer coefficients.

		(7.2)
mh cph (Thi − Tho ) = mccpc (Tco − Tci ) = UA( Tmean )

U is the overall heat transfer coefficient in W/m2 K; ∆Tmean is a certain mean temperature difference.

Let us consider the heat exchanger shown in Fig. 7.8.

The heat transfer from the hot fluid to the cold fluid is essentially one dimensional. Using the thermal resistance network,

Rtotal

= Rconv,h + Rcond

+ Rconv,c

(7.3)

where Rconv is convective resistance and Rcond is conductive resistance.

(7.4)

hh A1

2π kL

hc A2

Here, r1 and r2 are the inner and outer radii of the heat exchanger tube, respectively, and A1 and A2 are the inner and outer surface areas of the tube, respectively. Surface area A in Eq. (7.4) can be A1, in which case U is U1, or A2, in which case U is U2.

FIGURE 7.8

Schematic representation of a heat exchanger for heat transfer analysis.

7.4 The LMTD method 205

For a thin wall tube, r2 ≈ r1 , hence Rcond = 0. Eq. (7.4) reduces to

1	=	1	+	1	(7.5)
U	=	h	+	h	(7.5)
U		h		h
		h		c

[Please note that here we have used the idea A1 = A2 in Eq. (7.4), consequent upon

r2 = r1]

Therefore, the overall heat transfer coefficient is purely dependent on the heat transfer coefficients of only the hot and cold fluids and does not depend on the direction of fluid flow. ∆Tmean is an important parameter, and its relationship with the terminal temperatures of the hot and cold fluids needs to be established.

Let Rscaling be the resistance offered by the wall due to the formation of scales (fouling).

					r
				ln		2
				ln
1	=	1	+		r1		+	1	+ R	(7.6)
	=		+				+		+ R	(7.6)
UA hh A1				2π kL hc A2					scaling
UA hh A1				2π kL hc A2

Please note that Eq. (7.6) is the most general one while Eq. (7.5) is for a thin walled heat exchanger with negligible fouling resistance. Eq. (7.6) suggests that the overall heat transfer coefficient can be improved by increasing the heat transfer surface area on the side that has a lower heat transfer coefficient, and engineering efforts to improve heat exchanger performance outcomes need to be in this direction. For example, it is advisable and advantageous to add fins on the side that has a lower heat transfer coefficient, as opposed to the side with the higher coefficient, as the controlling resistance, as aforesaid, is associated with the lower heat transfer coefficient fluid or stream. Stated simply, for an air-liquid exchanger, fins or any other augmentation will be invariably done on the air side that is associated with lower heat transfer coefficients.

7.4 The LMTD method

This section presents the LMTD method of heat exchanger analysis for different types of commonly used heat exchangers.

7.4.1 The parallel-flow heat exchanger

Consider a parallel-flow heat exchanger that is insulated from its surroundings, whose overall heat transfer coefficient and the specific heat are constant. The changes in the potential and kinetic energies are assumed to be negligible. Temperature profiles for a parallel-flow heat exchanger are shown in Fig. 7.9.

For a parallel-flow heat exchanger, the difference between the temperatures of the hot and cold fluids at a section shown in the Figure is

T = Th − Tc

(7.7)

206 CHAPTER 7 Heat exchangers

FIGURE 7.9

Schematic representation of temperature variation along the length for a parallel-flow type of heat exchanger.

d( T ) = dTh − dTc

From energy balance,

dQ = −mhcphdTh = mccpcdTc = UdA T

In the above equation, the negative sign is for the heat lost.

d( T ) = −dQ − dQ

mhcph mccpc

		1		1
d(	T ) = −UdA T		+

	mhcph			mccpc

Separating the variables and integrating on both sides, we have

outlet d(

T )

= −

UdA

∫inlet

∫0

mh cph

mc cpc

outlet

ln( T )

= −UA

inlet

Thi − Tho

Tco − Tci

(7.8)

(7.9)

(7.10)

(7.11)

(7.12)

(7.13)

[Note that in Eq. (7.13), we have used the relation Q = mhcph (Thi − Tho ) = mccpc (Tco − Tci )]

	−UA	(7.14)
ln( To ) − ln( Ti ) =	Q ((Thi − Tci ) − (Tho − Tco ))

7.4 The LMTD method 207

(

T − T

)

Q = UA

(

Ti −

To )

Q = UA(LMTD)

LMTD is a logarithmic mean temperature difference.


LMTD =	Ti −	To	=	θ1 −θ2

		Ti			θ1
	ln			ln	θ2

		To

(7.15)

(7.16)

(7.17)

(7.18)

7.4.2 The counterflow heat exchanger

The assumptions for a counterflow heat exchanger are the same as those for the parallel-flow heat exchanger considered previously.

The temperature profiles for a counterflow heat exchanger are shown in Fig. 7.10.

		(7.19)
dQ = −mhcphdTh = −mcpcdTc = UdA T

The negative sign indicates that the value of the temperature reduces along the positive x direction.

T = Th − Tc

(7.20)

FIGURE 7.10

Schematic representation of temperature variation along the length for a counterflow type of heat exchanger.

208 CHAPTER 7 Heat exchangers

T ) = dTh − dTc

T ) =

−dQ

other side d(

T )

−1

∫one side

∫0

U dA

mh cph

mc cpc

other side

−1

ln(

T )

= UA

one side

Tci

− Tco

Tho − Thi

−

((Tho − Thi )

− (Tci − Tco ))

Tho

− Tci

Q = UA

(Thi − Tco ) − (Tho − Tci )

T − T

Tho − Tci

Q = UA(LMTD)

LMTD =

Tone side −

Tother side

θ1

−θ2

Tone side

θ1

θ2

other side

(7.21)

(7.22)

(7.23)

(7.24)

(7.25)

(7.26)

(7.27)

(7.28)

It is to be noted that while it is possible for Tco to be greater than Tho in a counterflow heat exchanger, this is never possible in a parallel-flow heat exchanger. In a parallel-flow heat exchanger, Tco →Tho (see Fig. 7.9. This condition happens when x → ∞).

7.4.3 Heat exchangers with phase change

Let us consider a condenser for which the temperature profiles of the hot and cold fluids are as shown in Fig. 7.11.

From Fig. 7.11A,

θ1 = Th − Tci

(7.29)

7.4 The LMTD method 209

FIGURE 7.11

Schematic representation of temperature variation along the length for a condenser: (A) parallel-flow configuration and (B) counterflow configuration.

θ2	= Th − Tco	(7.30)
Similarly from Fig. 7.11B,
θ1 = Th − Tci		(7.31)
θ2	= Th − Tco	(7.32)
Therefore, for a condenser,
(LMTD)parallel	= (LMTD)counter	(7.33)

Hence, the performance of parallel-flow and counterflow types of condensing heat exchanger remains the same, and the same holds true for an evaporator.

7.4.4 When is LMTD not applicable?

For a counterflow heat exchanger, the energy balance equation is given by

		(7.34)
mh cph (Thi − Tho ) = mc cpc (Tco − Tci )

If the heat capacities are equal, then Eq. (7.34) is reduced to

Thi − Tho	= Tco − Tci	(7.35)
Thi − Tco	= Tho − Tci	(7.36)
θ1	= θ2	(7.37)

Equation 7.28 becomes

LMTD =	0	(7.38)
	0


210	CHAPTER 7 Heat exchangers
	The above equation is in indeterminate form. Therefore, L’Hospital’s rule needs
	to be applied to Eq. (7.28).
	LMTD = lim		θ1 − θ2		(7.39)
	LMTD = lim			θ1	(7.39)
	θ1→θ2			θ1
			ln
			ln	θ2
				θ2
	Let θ1 = x .
	θ2	θ2 (x		−1)
	LMTD = lim	θ2 (x		−1)	(7.40)
	x→1	ln(x)
	After applying L’Hospital’s rule,
	LMTD = θ2 = θ1				(7.41)
	Therefore, in the case of a counterflow heat exchanger, if the heat capacities are
	equal, then LMTD method is not applicable. LMTD remains constant along the heat
	exchanger length and is equal to the terminal temperature difference (θ1 or θ2).
	The concepts and ideas introduced thus far in Section 7.4 is traditionally known
	as the LMTD analysis or LMTD method.
	In the case of a counterflow heat exchanger, by increasing the length of the heat
	exchanger to infinity, the maximum possible exit temperature of the cold fluid and
	the minimum possible exit temperature of the hot fluid can be obtained.
	Tco, max = Thi				(7.42)
	Tho, min = Tci				(7.43)
	In the case of an infinitely long parallel-flow heat exchanger,
	Tco, max = Tho, min = T				(7.44)
	From the energy balance equation,
	Ch (Thi − T ) = Cc (T − Tci )				(7.45)
	From the above equation, the exit temperature (T) of the fluid(s) can be obtained.

7.4.5 Shell and tube heat exchanger

The flow configuration in a shell and tube heat exchanger is neither parallel nor counter, and hence the LMTD will be different from that of a parallel-flow or counterflow heat exchanger. LMTD analysis can still be used for a shell and tube heat exchanger making use of a correction factor F to account for the deviation from the counterflow configuration. The expression for the heat transfer rate is

Q = UA(LMTD)counterflow F

(7.46)

7.4 The LMTD method 211

FIGURE 7.12

Correction factor for a one-shell pass and 2, 4, 6. (any multiple of 2) -tube pass shell and tube heat exchanger (Bowman et al., 1940).

The correction factor F can be obtained from the charts shown in Figs. 7.12 and 7.13 for one shell pass and two shell passes, respectively. These charts were originally developed by Bowman et al. (1940).

7.4.6 Cross-flow heat exchanger

The hot and cold fluids flow perpendicular to each other in a cross-flow heat exchanger and therefore the flow configuration is neither parallel nor counter. The LMTD approach used for a cross-flow heat exchanger is similar to that for a shell and tube heat exchanger.

Q = UA(LMTD)counterflow F

(7.47)

The correction factor charts are available, as shown in Figs. 7.14 and 7.15, for two different configurations—one fluid mixed and the other unmixed, and both fluids unmixed, respectively.

For analyzing a heat exchanger using the LMTD method, the outlet temperature of at least one of the fluids must be known. Thus, the sizing of the heat exchanger can be done based on the overall heat transfer coefficient and the LMTD. However, in many cases, the size of a heat exchanger is specified, but the outlet temperatures are

212 CHAPTER 7 Heat exchangers

FIGURE 7.13

Correction factor for a two-shell pass and 4, 8, 12...(any multiple of 4) -tube pass shell and tube heat exchanger (Bowman et al., 1940).

unknown. Therefore, its performance needs to be predicted. In such cases, another approach called the effectiveness-NTU method is used. The NTU refers to the number of transfer units and is directly related to the size of the heat exchanger.

7.5 The effectiveness-NTU method

The heat exchanger temperature effectiveness, or simply the effectiveness, ε, is the ratio of actual heat transfer to the maximum possible heat transfer.


	ε =			Qact	(7.48)
	ε =		Q
				max
ε =	Ch (Thi − Tho )	=		Cc (Tco − Tci )	(7.49)
ε =	Cmin (Thi − Tci )	=		Cmin (Thi − Tci )

7.5 The effectiveness-NTU method 213

FIGURE 7.14

Correction factor for a cross-flow heat exchanger with one fluid mixed and the other unmixed (Bowman et al., 1940).

Qmax = Cmin Tmax

(7.50)

For any type of heat exchanger, Tmax = Thi − Tci .

In Eq. (7.50), we have considered Cmin and not Cmax, as the fluid with minimum heat capacity can experience the maximum change in temperature.

The NTU is given by

NTU =	UA		(7.51)
	Cmin

Cmin = (mc		p )min	(7.52)
Cmax = (mc		p )max	(7.53)

The NTU indicates the size of the heat exchanger, as already mentioned. The capacity ratio of a heat exchanger, C, is defined as

214 CHAPTER 7 Heat exchangers

FIGURE 7.15

Correction factor for a cross-flow heat exchanger with both fluids unmixed (Bowman et al., 1940).

C =	Cmin	(7.54)
	Cmax

C varies between 0 and 1. C = 0 corresponds to a heat exchanger with one of the fluids undergoing phase change, condensation or evaporation, and C = 1 corresponds to a heat exchanger with equal heat capacity rates.

7.5.1 Effectiveness of a parallel-flow heat exchanger

The temperature profiles for the hot and cold fluids in a parallel-flow heat exchanger were shown in Fig. 7.9. Please refer to the same figure.

Let Cc = Cmin

T = Th − Tc	(7.55)
d( T ) = dTh − dTc	(7.56)

7.5 The effectiveness-NTU method 215

From energy balance,

dQ = −mhcphdTh = mccpcdTc = UdA T

In the above equation, the negative sign is for the heat lost by the hot fluid.

d( T ) = −dQ − dQ

mhcph mccpc

		1		1
d(	T ) = −UdA T		+

	mhcph			mccpc

Separating the variables and integrating on both sides

outlet

T )

∫inlet

= −

∫0

U dA

mh cph

outlet

ln(

T )

= −UA

inlet

= −UA

+ 1

= −NTU(1 + C)

− T

= −NTU(1 + C)

Thi − Tci

+	1


	mc cpc

Here NTU = UA , and C = Cc

Cc Ch

From energy balance,

mh cph (Thi − Tho ) = mc cpc (Tco − Tci )

Ch (Thi − Tho ) = Cc (Tco − Tci )

Tho = Thi − C(Tco − Tci )

(7.57)

(7.58)

(7.59)

(7.60)

(7.61)

(7.62)

(7.63)

(7.64)

(7.65)

(7.66)

(7.67)

216CHAPTER 7 Heat exchangers

From Eq. (7.64),

		Thi − C(Tco − Tci ) − Tco					= −NTU(1 + C)			(7.68)
	ln
				Thi − Tci

Thi − C(Tco − Tci ) − Tco + Tci − Tci										(7.69)
ln								= −NTU	(1 + C)
				Thi − Tci

					Tco − Tci
	ln	Thi − Tci		− (1 + C)				= −NTU(1	+ C)	(7.70)

	Thi − Tci				Thi	− Tci
			ln(1 − ε(1 + C)) = −NTU(1 + C)							(7.71)
			1 − ε(1 + C) = exp(−NTU(1 + C))							(7.72)
			ε = 1 − exp(−NTU(1 + C))							(7.73)
					(1 + C)

7.5.2 Effectiveness of a counterflow heat exchanger

An expression for the effectiveness of a counterflow heat exchanger can be derived in a way similar to that of a parallel-flow heat exchanger. The temperature profiles of fluids for a counterflow heat exchanger were shown in Fig. 7.10. Please refer to the same figure.

From energy balance,

dQ = −mhcphdTh = −mcpcdTc = UdA T

(7.74)

The negative sign indicates that the value of temperature reduces along the positive x direction.

T = Th

− Tc

(7.75)

T ) = dTh − dTc

(7.76)

T ) =

−dQ

(7.77)

other si de d(

T )

−1

U dA

(7.78)

∫one si de

∫0

mh cph

mc cpc

7.5 The effectiveness-NTU method 217

ln( T )		other side
		one side
T		− T
ln	ho	ci

Thi		− Tco

Thi − Tco

Tho − Tci

=UA −1 + 1

Ch Cc

=UA −1 + 1

Ch Cc

= exp(−NTU(1 − C))

LHS:

− T

Thi − Tco

Thi − Tci

− T

Thi

− Tci

We know that

ε =

Ch (Thi − Tho )

Cc (Tco − Tci )

min

(T − T )

min

(T − T )

In this derivation, let Ch = Cmax , Cc

= Cmin

Therefore,

ε =

Cmax (Thi

− Tho )

Cmin (Tco − Tci )

min

− T )

min

(T − T )

From Eq. (7.84),

Tho = Thi − εC(Thi − Tci )

Tco = Tci + ε(Thi − Tci )

Substituting Tho and Tco

in Eq. (7.82),

Thi − (Tci + ε(Thi − Tci ))

Thi

− Tci

1 − ε

(Thi − εC(Thi − Tci ))

− Tci

− Cε

Thi

− Tci

From Eq. (7.81),

1− ε

= exp(−NTU(1− C))

1− Cε

(7.79)

(7.80)

(7.81)

(7.82)

(7.83)

(7.84)

(7.85)

(7.86)

(7.87)

(7.88)

218 CHAPTER 7 Heat exchangers

ε = 1− exp(−NTU(1− C))

1− C exp(−NTU(1− C))

(7.89)

Similar expressions for the effectiveness of other types of commonly used heat exchangers were developed by Kays and London (1984) and are presented in Table 7.1. For a shell and tube heat exchanger with n shell passes, ε would be determined based on the (NTU)1 of a single shell and the corresponding effectiveness (ε1), assuming that the total NTU is equally shared between shell passes, NTU = n(NTU)1. The effectiveness can also be determined from the effectiveness-NTU charts, as shown in Figs. 7.16–7.21.

Table 7.1 Heat exchanger effectiveness relations (Kays and London, 1984).

Flow configuration

Concentric tube

Parallel flow

Counterflow

Shell and tube

One shell pass (2, 4...tube passes)

n shell passes (2n, 4n... tube passes)

Cross-flow (single pass)

Both fluids unmixed

Cmax (mixed), Cmin (unmixed)

Cmax (unmixed), Cmin (mixed)

All exchangers (C = 0)

Relation

ε =

1− exp(−NTU(1+ C))

(1+ C)

ε =

1− exp(−NTU(1− C))

(C < 1)

1− C exp(−NTU(1− C))

ε =

NTU

(C = 1)

1+ NTU

)

1+ exp(−(NTU) (1+ C2 )1/ 2 ) −1

ε1 = 2 1+ C + (1+ C

2 1/ 2

1− exp(−(NTU)1(1+ C )

)

1− ε C n

−1

ε =

− 1

− C

1− ε1

1−

ε1

0.22

{exp(−C(NTU)0.78 )−1}

ε =

1− exp NTUC

ε = 1− exp{−C(1− exp(−NTU))}

ε =

1− exp(−C(NTU))

1− exp

−C

ε = 1− exp(−NTU)

7.5 The effectiveness-NTU method 219

FIGURE 7.16

Effectiveness of a parallel-flow heat exchanger.

7.5.3 Comparison between parallel-flow and counterflow heat exchangers

Having seen heat exchangers with different configurations, a comparison between two simple configurations – parallel-flow and counterflow heat exchangers, is warranted. The expressions for the effectiveness of parallel-flow and counterflow configurations, as derived before, are,

ε p =	1− exp(−NTU(1+ C))	(7.90)
	(1+ C)
εc =	1 − exp(−NTU(1 − C))	(7.91)
εc =	1 − Cexp(−NTU(1 − C))	(7.91)
	1 − Cexp(−NTU(1 − C))

Figures 7.16 and 7.17 show the effectiveness-NTU relations in graphical form for the parallel-flow and counterflow configurations, respectively. It can be seen that the effectiveness of the counterflow heat exchanger is higher than that of the parallelflow heat exchanger for the same NTU and heat capacity ratio. This is intuitively apparent as the stream to stream temperature difference, which holds the key to the superior thermal performance of a heat exchanger, is more or less preserved in a

220 CHAPTER 7 Heat exchangers

FIGURE 7.17

Effectiveness of a counterflow heat exchanger.

FIGURE 7.18

Effectiveness of a one-shell pass and 2, 4, 6...(any multiple of 2) -tube pass shell and tube heat exchanger.

7.5 The effectiveness-NTU method 221

FIGURE 7.19

Effectiveness of a two-shell pass and 4, 8, 12 . . . (any multiple of 4) -tube pass shell and tube heat exchanger.

FIGURE 7.20

Effectiveness of a cross-flow heat exchanger with both fluids unmixed.

222 CHAPTER 7 Heat exchangers

FIGURE 7.21

Effectiveness of a cross-flow heat exchanger with one fluid mixed and the other fluid unmixed.

counterflow heat exchanger from the entry to the exit, while there is a continuous decline in the stream to stream temperature difference between the hot and cold fluids from the entry to the exit in a parallel flow heat exchanger.

For a condenser,

Qh = Ch (Th1 − Th2 )

Since the hot fluid condenses,

Ch → ∞

C = Cc = 0

If C = 0

ε p = 1− exp(−NTU)

εc = 1− exp(−NTU)

(7.92)

(7.93)

(7.94)

(7.95)

(7.96)

Therefore, for a condenser, the effectiveness is the same for both the parallel-flow and counterflow configurations. Eqs. (7.95) and (7.96) are true for an evaporator as well. If C = 1 for a parallel-flow heat exchanger,

ε p =	1− exp(−2NTU)	(7.97)
	2

7.6 Comparison between the LMTD and effectiveness-NTU methods 223

If C = 1 for a counterflow heat exchanger,
εc =	0	= indeterminate form			(7.98)
εc =	0	= indeterminate form

Applying L’Hospital’s rule,
	εc	=		NTU	(7.99)

			1+ NTU
			1+ NTU

Figures 7.18–7.21 show the effectiveness-NTU relations in graphical form for other configurations - a one-shell pass shell and tube, a two-shell pass shell and tube, a cross-flow with both fluids unmixed, and a cross-flow with one fluid mixed and the other fluid unmixed, respectively.

7.6 Comparison between the LMTD and effectiveness-NTU methods

1.If the exit temperature of at least one of the fluids is known, both the methods can be used for analyzing the heat exchanger. Out of the two, LMTD is easier and faster as compared to the effectiveness-NTU method.

2.If the exit temperatures of both fluids are unknown, the only method that can be used is the effectiveness-NTU method.

Example 7.1 A 1 kg/s of hot water stream is cooled from 90 °C to 60 °C in a par- allel-flow heat exchanger in which the cooling agent is 2 kg/s of water with inlet temperature of 40 °C. The overall heat transfer coefficient U is 1000 W/m2 K.

Using the LMTD method, determine the required area of the heat exchanger. Take cph = cpc = 4182 J/kg K .

Solution:

mhcph (Thi − Tho ) = mccpc (Tco − Tci )

1× 4182(90 − 60) = 2 × 4182(Tco − 40)

Tco = 55	°C
Ti = 90	− 40 = 50 °C
To = 60	− 55 = 5 °C

LMTD =	50 − 5		= 19.54 °C
	ln	50

		5

224 CHAPTER 7 Heat exchangers

Q = UA(LMTD)

1× 4182 × (90 − 60)= 1000 × A × 19.54

Ap = 6.42 m2

Example 7.2 Revisit the above problem and solve it for a counter-flow heat exchanger.

Solution:

In the case of a counterflow–type heat exchanger,

Toneside = 90 − 55 = 35 °C

Totherside = 60 − 40 = 20 °C

LMTD =	35 − 20		= 26.80 °C
	ln	35

		20

Q = UA(LMTD)

1× 4182 × (90 − 60) = 1000 × A × 26.80

Ac = 4.68 m2

From the above two examples it is clear that LMTDparallel < LMTDcounter . Therefore the performance of the counterflow heat exchanger is better than the par-

allel-flow type.

Example 7.3 Revisit the above problem and solve it for a cross-flow heat exchanger with one stream mixed and the other unmixed.

Solution:

For a cross-flow heat exchanger,

Q = UA(LMTD)counter F

Here F is the correction factor obtained from charts, and the procedure is as follows:

P = t2 − t1 T1 − t1

P = 55 − 40 = 0.3 90 − 40

7.6 Comparison between the LMTD and effectiveness-NTU methods 225

R = T1 − T2 t2 − t1

R = 90 − 60 = 2 55 − 40

Using P and R from Fig. 7.14 we get F = 0.9.

1× 4182 × (90 − 60) = 1000 × A × 26.80 × 0.9

Across = 5.20 m2

Example 7.4 Revisit the above problem and assume that the outlet temperature of the hot fluid is not provided. Determine the outlet temperature of the hot fluid using the effectiveness-NTU method for a parallel-flow heat exchanger with A = 6.42 m2.

Solution:

mhcph = 1× 4182 = 4182 W/K

mccpc = 2 × 4182 = 8364 W/K

From the above

Cmin = 1× 4182 = 4182 W/K

Cmax = 2 × 4182 = 8364 W/K

C =	Cmin		= 0.5
	Cmax

NTU =	UA	= 1000 × 6.42		= 1.535

	Cmin		4182
ε p =	1 − exp(−NTU(1 + C))
			(1 + C)
ε p =	1− exp(−1.535(1.5)) = 0.6
			(1.5)

Q= ε pQmax

=0.6 × Cmin × (Thi − Tci )

=0.6 × 4182 × (90 − 40)

=125460 W


226	CHAPTER 7 Heat exchangers
	Tho = Thi −			Q
	Tho = Thi −			mhcph
				mhcph
		= 90 −	125460
			1× 4182
		= 60 °C.
	This is the same as what we started out with in Example 7.1.
	Example 7.5 A cardiac-pulmonary bypass procedure uses a cross-flow heat exchanger
	(both fluids unmixed) to cool the blood flowing at 5000 ml/min from a body tempera-
	ture of 37 °C to 25 °C so as to induce hypothermia, which slows down metabolism and
	reduces oxygen demand. Iced water at 0 °C is used as the coolant, and its outlet temper-
	ature is 15 °C. The density and specific heat of blood are 1050 kg/m3 and 3740 J/kg K,
	respectively, and that of water are 1000 kg/m3 and 4198 J/kg K, respectively.
	The overall heat transfer coefficient is 750 W/m2K.
	a. Calculate the heat transfer rate for the heat exchanger.
	b. Determine the coolant flow rate.
	c. Calculate the heat transfer area of the heat exchanger.
	Solution:
	mh =	5 ×10−3		×1050 = 0.0875 kg/s
	mh =	60		×1050 = 0.0875 kg/s
		60

From thermodynamics

0.0875 × 3740(37 − 25) = mw × 4198(15 − 0)

mw =	0.0623 kg/s
Q =	0.0875 × 3740(37 − 25) = 3.927 kW
θ1 =	37 −15 = 22 °C
θ2 =	25 − 0 = 25 °C
LMTD =	θ1 −θ2

	ln(	θ1	)

	θ2
LMTD =	22 − 25 = 23.46
	22
	ln
	25

7.6 Comparison between the LMTD and effectiveness-NTU methods 227

Q = UA(LMTD)F

In the above equation, F is the correction factor, which can be obtained from Fig. 7.15.

P =	t2 − t1	=	18 − 0	= 0.4
	T1 − t1		37 − 0

R = 37 − 25 = 0.8 15 − 0

From the figure, F = 0.95

3927 = 750 × A × (23.46) × 0.95

A = 0.234 m2

Example 7.6 In a chemical processing plant, 10800 kg/h of water (cp = 4.2 kJ/kg K) is to be heated from 42 °C to 62 °C with 7200 kg/h of hot water (cp = 4.2 kJ/kg K) available at 88 °C. A shell and tube heat exchanger is proposed with cold water flowing through the tubes of 19 mm diameter at an average velocity of 0.6 m/s and hot water flowing through the shell. The space constraint limits the maximum length of the tube to 2.5 m. Assuming the overall heat transfer coefficient as 1800 W/m2 K and one shell pass, determine the number of tube passes, the total number of tubes, and the length of the tubes.

Solution:

Since the number of tube passes is not known, it can be assumed to be one. The total heat transfer rate is


Qc = mccpc Tc =	10800 × 4.2 × (62 − 42) = 252 kW
	3600
	Qh = Qc
The outlet temperature of the hot fluid is
Tho = Thi −			Qh
Tho = Thi −			mhcph
	252
=	88 −
=	88 −	7200		× 4.2
	3600

= 58 °C

228CHAPTER 7 Heat exchangers

LMTD for the counterflow configuration is

LMTD =	θ1 −θ2
LMTD =		θ1
		θ1
	ln
where,		θ2
where,
θ1 = Thi − Tco = 88 − 62 = 26 °C
θ2 = Tho − Tci = 58 − 42 = 16 °C
LMTD =	26 − 16		= 20.6 °C
LMTD =		26	= 20.6 °C
		26
	ln
		16

Heat transfer area is
A =	Q	=		252	= 6.8 m2

	U × LMTD		1.8	× 20.6
	U × LMTD		1.8	× 20.6

The mass flow rate of cold water is

m = ρu Nπ D2 c 4

where N is the number of tubes.

N =	mc
		π D2

	ρu	4

Length of each tube can be calculated as

=	10800 / 3600

	1000 × 6 ×	π × 0.0192
	1000 × 6 ×	π × 0.0192
		4

= 17.63 ≈ 18

L =	A	=		6.8	= 6.33m
	Nπ D		18 × π × 0.019

Since the length is larger than the maximum allowable length of 2.5 m, we need to use more than one tube pass. Next, a two-tube pass is tried.

LMTD correction factor needs to be determined from Fig. 7.12.

P =	Tco − Tci	=	62	− 42	= 0.434
	Tho − Tci		88	− 42

7.7 Other considerations in the design of a heat exchanger 229

R =	Thi − Tho	=	88 − 58	= 1.5
			62 − 42
	Tco − Tci

From Fig. 7.12, F ≈ 0.65.

Therefore the total area required is,

The number of tubes per pass will still be 18, as the velocity has to be the same. The length of the tube per pass is

L =	A	=	10.45		= 4.86 m

	2Nπ D		2 ×18 × π ×	0.019

This length is greater than the maximum allowable length of 2.5 m. Hence a twotube pass is not sufficient. Next, a four-tube pass can be tried.

From Fig. 7.12, F factor is the same for a four-tube pass. Therefore, the area required remains the same (10.45 m2).

The length of the tube per pass is

L =	A	=	10.45	= 2.43m

	4Nπ D		4 ×18 × π × 0.019

This length is acceptable as it is within the permitted length of 2.5 m.

Therefore, the final parameters are as follows:

Number of tube passes = 4

Length of tube per pass = 2.43 m

Number of tubes per pass = 18

Total number of tubes = 72

Heat duty = 252 kW

For elaborate illustrations and more examples on heat exchangers, readers can refer to Lienhard IV and Lienhard V (2020), Bergman et al. (2011), Cengel (2003), and Bejan (1993).

7.7 Other considerations in the design of a heat exchanger

In addition to the heat transfer, the factors that influence the design or selection of a heat exchanger are the incurred pressure drop (or pumping power) as the fluids are forced through the heat exchanger; the construction material and its strength; the size and weight considerations; and the total cost taking into account the material, fabrication, and space requirements. These are beyond the scope of this book.

Problems

7.1.A thin-walled, double-pipe (concentric tube) heat exchanger has air flowing through the annulus and water flowing through the inner tube. The air side

230 CHAPTER 7 Heat exchangers

and water side heat transfer coefficients are 80 W/m2 K and 4000 W/m2 K, respectively. Determine the overall heat transfer coefficient. To enhance heat transfer, in one case, more turbulence is created on the water side, leading to an increase in the water side heat transfer coefficient by 40%, and in the other case, more turbulence is created on the air side, resulting in an increase in the air side heat transfer coefficient by 40%. Determine the increase in the overall heat transfer coefficients for both cases and comment on the results.

7.2.A heat recovery unit involves heat transfer from hot gases flowing over an aluminum tube (k = 205 W/m K) to water flowing through the tube. The inner and outer diameters of the tube are 19 mm and 25 mm, respectively. The outer and inner heat transfer coefficients are 100 W/m2 K and 5000 W/m2 K, respectively. Determine the overall heat transfer coefficient based on the inner surface area of the tube. In order to enhance the heat transfer, fins are proposed on the outer side. The scheme consists of 10 straight fins of rectangular profile provided longitudinally (evenly spaced around the circumference) over the outer surface of the tube. The fin thickness and length are 3 mm and 20 mm, respectively. Determine the enhanced overall heat transfer coefficient based on the inner surface area of the tube.

7.3.The operating condition for a certain double-pipe (concentric tube) heat exchanger is such that the LMTD is the same for both the parallel-flow and counterflow configurations. The effectiveness of the heat exchanger is 60%. Determine the required increase in the length of the heat exchanger if the effectiveness of the heat exchanger has to be increased to 80%.

7.4.A double-pipe (concentric tube) counterflow heat exchanger is designed to transfer heat from oil flowing through the annulus to water flowing through the inner tube. The inlet and outlet temperatures of water are 30 °C and 70 °C, respectively, and that of oil are 130 °C and 110 °C, respectively. The inner tube has a diameter of 25 mm, and the overall heat transfer coefficient is 400 W/m2 K.

a.Determine the length of the heat exchanger for a total heat transfer rate of 2500 W.

b.Due to deterioration in the performance caused by scaling on the water side of the heat exchanger, the water outlet temperature decreases to 60 °C for the same inlet temperatures and fluid flow rates. Determine the overall heat transfer coefficient, the scaling (fouling) factor on the water side, the heat transfer rate, and the oil outlet temperature.

7.5.A one-shell and two-tube pass condenser of a steam power plant consists of brass tubes (k = 111 W/m K) of 19 mm outer diameter and 16 mm inner diameter. The condensate heat transfer coefficient on the outer surfaces of the tubes is 12500 W/m2 K. Cooling water from a lake at 20 °C is pumped at 1000 kg/s through 2000 brass tubes per tube pass.

a.For a steam condensation rate of 50 kg/s at 52 °C (latent heat of condensation,

hfg = 2377 kJ/kg), determine the inner heat transfer coefficient (based on the appropriate Nusselt number correlation), the overall heat transfer

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