# HEAT EXCHANGERS AND HEAT-RECOVERY UNITS General Theory of Heat Exchangers

A heat exchanger is a device that transfers heat from one medium to an­other; the medium may be a solid, liquid, or gas. Some of the most complex engineering design problems relate to heat exchangers.

Heat exchangers are divided into the following types.

1. Recuperator: A wall separating the flowing fluids is the most commonly encountered problem.

2. Regenerator: The hot and cold fluids pass alternately through a space containing solid areas/particles, which provide alternately a heat sink and a heat source. An example of the rotating-type matrix is a cooling tower.

The direction of flow is important, as it has a pronounced effect on the ef­ficiency of a heat exchanger. The flows may be in the same direction (parallel flow, cocurrent), in the opposite direction (counterflow), or at right angles to each other (cross-flow). The flow may be either single-pass or multipass; the latter method reduces the length of the pass.

A schematic diagram of a parallel-flow (cocurrent) heat exchanger is shown in Fig. 9.4.

Hot fluid: Thi> Tho Cold fluid: Tct < Tco The hot and cold fluids are denoted by the subscripts h and c and the fluid in

Let and outlet by i and o, respectively.

In the case of Fig. 9.4, Thi = Thl. The relative temperature differences are defined by (9.1) (9.2)

 L 2

Thi> cpb™h The relative temperature differences can be calculated by dividing the temper­ature difference of the hot or cold side by the maximum temperature differ­ence, Tj, t — Tci, that occurs in the heat exchanger.

There are two relative temperature changes for a heat exchanger. 1 he greatest of them is the heat exchanger effectiveness, e:

E = ma x{8Tc,8Th} (9.3)

The maximum temperature difference that takes place in a heat exchanger is Thi — Tci. A higher temperature difference cannot occur due to the second law of thermodynamics. The maximum theoretical heat transfer rate in a heat exchanger is

‘Amax — (~‘2 (1 hr~ ^ ! ^ &max ~ ^j( ? hr~

Where

C = qmcp (9.4)

Is the heat capacity rate, qm is the mass flow through the heat exchanger, and cp is the specific heat capacity.

Assuming that the maximum possible temperature difference is on the fluid side with the higher heat capacity rate, then

</w = C2(Thl — T,,-) .

The heat exchanger balance is

Cif T/;, — TЈ/) = C| AT, ,

Where ATX is the temperature difference of the smallest heat capacity rate fluid in the heat exchanger. Then

AT, =&(Tbi-Tci) > Tbt-Tct.

This equation, which gives a higher temperature difference than Tt(-T,,, cannot occur. Then

^max = C, (Th,-Ta) , where C, is the smaller heat capacity rate. The actual power is

<f> = C1ATl =C2AT2,

From which

<j> . QATt = AT’! C2A T2 A Tz

«/»max ei(Tfcl-TCf.) (Thl-Ta) Ci(Thi-Tci) Thi-Tcr

From this it follows that

E — & — Temperature difference of smaller heat capacity fluid m ^

Tfrmax initial temperature difference ‘

The heat exchanger effectiveness shows how close the heat exchanger is oper­ating to the maximum heat transfer performance. Equation (9.5) is valid for any type of heat exchanger.

In a counterflow or countercurrent heat exchanger, hot and cold fluids en­ter the exchanger from opposite sides. The counterflow heat exchanger in Fig.

9.5 serves as a reference for all other heat exchanger configurations. A de­tailed analysis of this type of heat exchanger is therefore necessary.

The heat is transferred by convection and conduction from the hot to the colder fluid through an infinitesimal surface area dA. The temperature of the hot fluid reduces by an amount dl , and the temperature of the cold fluid in­creases by an amount dTc. CD

T T,

 DA

 4—- ►

* A T T

T

 DTh-dTc = d(Th-Tt:) = -L—L ^b ^c. Equation (9.9), after integrating,
 G dAiTfo — TL. ).
 The heat balance gives G" dA(Th — Tc) = Ch dTh = C,: dTc,

 (9.6!

 Where G" is the conductance per unit of surface area of the separating wall, which for a plane wall can be written as

 _4 = i + ± + s, G" ac ai, А ’

 (9.7)

 Where ac and ah are the cold — and hot-side convective heat transfer coefficients, respectively, 5 is the wall thickness, and А is the wall thermal conductivity. For a thick circular tube, the conductance per unit tube length is

 (9.8)

 G’ 1ldoao 771ijOLj

 2 7TА

 Where a0 is the outer convective heat transfer coefficient, a, is the inner con­vective heat transfer coefficient, dt is the inner tube diameter, and d0 is the outer tube diameter. Equation (9.6) gives (9.9)

 ■d(Th-Tc) Th-Ti: ‘
 J__ JL YCh Ct-
 G" J dA, O (9.10)

 Gives

 AT, AT!
 In
 / Ch C, j
 JL_ A. Ch Cc
 G" A = G  (9.11)

 AT, r„ Sr; “ “p &
 YC), (‘■,)
 Using Fig. 9.5 and Eq. (9.12) gives Tb,-Tc j + Tci-Tco Th, — Tci + Tho — Thi For the case of Q, < Cc, the heat exchanger effectiveness is _ Tht-Tho Th,-Tcr And the heat balance is Cf,(Tbj-Tho) = Cc( Tco-Tci). This then is written as Teo — Tci = ^Th, — Th0) = R(Tht — Th0),
 Gi-1-4- Ch Cc
 Exp
 Or (9.12) (9.13)

 (9.14) (9.15) (9.16)

 Where R = Cb/Cr < 1 •

 Pf(l-R) = exp(-z(l — R)), (9.18) 1~Re = exp

1 — Ђ

From which

1 — e

1 — Re

/

^ (9.1“)

Where z = G/Cc is the dimensionless conductance.

Taking the case C( < Ch, it can then be shown that we also get

C_ 1 — exp(-z(l — R)) {9A9)

1 — R exp(-z(l — R))’ ‘

Which is the same as Eq. (9.18).

This heat exchanger effectiveness is one of the important parameters that describes the performance of a counterflow heat exchanger.

In Eq. (9.19) R = Cmm/Cmax < ] is between the minimum and maxi­mum heat capacity rates, z = G/C^ is the heat conductance divided by the

Minimum heat capacity rate. In heat transfer literature, it is also denoted by z = NTU (number of heat transfer units).

Solving Eq. (9.19) for the dimensionless conductance z gives

 1 In
 1 — R
 1 — Re’ 1 — Ђ (9.20)

If the cold and hot fluids heat capacity rates are equal, then R = 1. Equation

(9.22) gives an indefinite value, and this equation cannot be used directly. Using l’Hopital’s rule as R —> 1 gives

Lim-4r(l — exp(-z( 1 — R)))

Lime =

TOC o "1-5" h z, im^(l ~R exp(-z(l — R))) (9 21)

Lim(-z exp(-z(l — R)))

Lim(~ Rz exp(-z(l — R)) — exp(-z(l — R))) 1 + z’

From which

(9.22)

1 — e

In Fig. 9.6 the temperature profiles in a counterflow heat exchanger are shown when Ch > C.

1.3.1.3 Logarithmic Mean Temperature Difference

The rate of temperature drop of a fluid as it flows along the length of a heat exchanger is not constant. In order to take account of this nonlin­ear relationship, the logarithmic mean temperature difference (LMTD) is used. If the inlet and outlet temperatures do not differ widely, an arith­metic mean can be used, because the relationship is considered to be lin­ear.

In order to calculate the heat transfer from the hot to the cold fluid, the heat exchanger conductance and the temperature of the fluids at both sides of

 T„,-Tri = *0

 FIGURE 9.6 Counterflow heat exchanger temperature profiles when Ch > Cc — 0 — Thj~ Ta. The heat exchanger must be known. The mass flow measurement is often diffi­cult to determine; however, the temperatures are easily measured.

A temperature difference is defined that satisfies the following equation:

<p = G AT, (9.2.3)

Where <j!) is the heat transfer rate in the heat exchanger and G the conductance. For counterflow Eq. (9.12) gives

 / J___i _ A. ( A Cc, AT Ch (9.24) 1 AT2 ^ at7 =

On the other hand, Eq. (9.15) gives

4> ~ Ch(Thi — Tho) = Ccfl co — Tci) ,

Giving

,_AT, I — r, ^ T

ДТ = ЛІ2_Д7’, = ДГ)п _ (9.27)

LnAT,

From Fig. 9.6 for a countercurrent heat exchanger,

AT, — Tho — Ta

And

AT2 = Thl — 7 co _

For a countercurrent heat exchanger, the logarithmic temperature difference is then

AT| = (^bi~ Tco) — (Ij,0 — Tcj) ^

 B

Thi — Tco

T—— T~

* ho ~ * ci

This is the logarithmic temperature difference for the counterflow heat ex­changer.

The logarithmic mean temperature difference is defined when AT2 ^ AT) Consider the case where AT2 = AT,. The logarithmic temperature differ­ence is obtained by applying l’FIopital’s rule as AT2 —> AT,, giving

AT — AT ^’m dAT ^ ~ ATi)

 , AT7 AT, Lim AT, n = lim 2 , , = “A1! ——————- = AT?. (9.29)

D

A7i dAT1

The logarithmic mean temperature difference is the same as the temperature difference at the entrance and exit of the heat exchanger, i. e., AT, = A T2 = ATln

For a counterflow heat exchanger, when AT2 — AT,,

Thi-TC0 = Th0-Tci (9.30)

Or

Thi-Tco = Tco-Tci . (9.31)

Then from Eq. (9.15), Ch = Cc when AT = AT2 = ATln

Example I

A brine solution enters a counterflow heat exchanger at T(( = 31.7 °С, and air enters at Tai = 24.4 °С. The measured outlet temperatures of the brine solution and air are Tio = 27.2 °С and Tao = 30 °С, respectively.

The mass flow rate of the hot fluid is qml = 0.382 kg s-1 and of the cold fluid, qma = 0.9 kg s-1. The specific heat capacity for brine is срі = 3.12 kj kg-1 K-1 and for air, cpa = 1.007 kj kg-1 K_1.

A. Calculate the total conductance of the heat exchanger.

B. If the liquid mass flow rate is reduced to qml = 0.3 kg s_1, calculate the new outlet temperatures of the brine and air.

Solution

A. The total heat conductance can be calculated from Eqs. (9.23) and (9,28)

As T/„ = T* Tho = Tlo% Ta = Xu, and Tc0 = T.;0

Q _—————— di—————- 1„ hi—Luj

Tot (T, i — TJO) — (Tl0 — TjnTt0 — Td1 ’

Where (f) = q„wCpa(Tao-Tai) = 5.1 kW is the heat flow to the air, which gives Gtot = 2310 W K-1, 4) = qmicpi(Tii — T,0) = 5.36 kW is the heat flow from the liquid, which gives Gtot = 2430 W K-1. The difference is due to er­rors in flow and temperature measurements. The difference between the heat flow in the two cases, 5%, is satisfactory. Then an average estimation can be obtained as Gtot = 2370 W K-1.

B. The heat capacity rates of liquid and air are calculated by, respectively,

Q = qmicpi = 0.3 x 3120 = 936 W

And

= 4maCPa = °-9 X 1007 = 906 W K~

Therefore,

0/ <

The maximum heat transfer in the heat exchanger is

<*W = Cmin(T,(- Tai) = 906 x (31.7-24.4) = 6.61 kW.

The total heat conductance is considered constant and not influenced by the mass flow rate of liquid.

Then the number of heat transfer units is

? = ^tot = 2370 j Z c 906

^min

Calculating the heat capacity rates ratio gives

R = = 906 = 0_968

C 936

Vjmax

From Eq. (9.19),

= 1 — exp(-z(l — R)) = 1 -exp(-2.62(l -0.968)) = „

L-Rexp(-z(l-R)) 1 -0.968 exp(-2.62(l -0.968))

The actual heat transfer is calculated from

</> = e4w = 0.73 x 6.61 = 4.84 kW.

The exit temperature of the liquid is calculated from

4> = C/(T/; — T;0) ,

Which gives

T, o = Ti, — Ј = 31.7 — ^ = 26.5 °C.

<t> = Ca(Tao-Tai),

Which gives

T =T +iЈ. = 24 4 + 4840 = 29 7 °C

1 ao 1 at ^ ^s ’ 906

A reduction in liquid flow reduces the outlet temperature of the liquid and

Air.