8. Heat Exchangers - Heat and M= ass Transfer, 2nd Edition

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8 Heat Exchangers

KEY CONCEPTS

The transfer of thermal energy between fluids is one of the mo= st important and frequently used processes in engineering.
The transfer of heat is usually accomplished by means of a device k= nown as a heat exchanger.
Common applications of heat exchangers is the power plants with= boilers, fan coolers, cooling water heat exchangers, and condensers.

= 8.1 CLASSIFICATION OF HEAT EXCHANGERS

Heat exchanger is a device, w= hose primary purpose is the transfer of heat energy between two fluids. Hea= t exchangers are classified into regenerators, open-type exchangers, and cl= osed-type exchangers or recuperators. Regenerators are heat exchangers in w= hich hot and cold fluids flow alternately through the same space with as li= ttle physical mixing between the two streams as possible. Open-type heat ex= changers are devices wherein physical mixing of the two fluid streams actua= lly occurs. Hot and cold fluids enter open-type heat exchangers and leave a= s a single stream. The recuperator is a type in which the hot and cold flui= d streams do not come into direct contact with each other but are separated= by a tube wall or a surface. Heat exchangers are also classified according= to the relative directions of the two streams: parallel flow if the fluids= flow in the same direction, counter flow if the fluids flow in opposite di= rections, and cross flow if the two fluids flow at right angles to one anot= her. Different types of heat exchangers are shown in Fig. 8.1.

Fi= gure 8.1 Indirect Contact Heat Exchangers and Direct Contact Heat = Exchangers

= 8.2 OVERALL HEAT TRANSFER COEFFICIENT OF HEAT EX= CHANGERS

The concept of overall heat t= ransfer coefficient is mainly applied to composite systems. In heat exchang= ers, the heat transfer takes place between the fluids through solid walls. = In this process, convection and conduction heat transfers are taking place.= In order to combine these conduction and convection thermal resistances in= the path of heat flow from the hot to the cold fluid, overall heat transfe= r coefficient is introduced.

In a heat exchanger, two flu= ids are separated by a wall as shown in

Fi= gure 8.2 Heat I ransfer Across Wall

KEY CONCEPTS

The basic design of a heat exchanger no= rmally has two fluids of different temperatures separated by some conductin= g medium.
The most common design has one fluid flowing through metal tubes an= d the other fluid flowing around the tubes.
On either side of the tube, heat is transferred by convection.
<= /li>
Heat is transferred through the tube wall by conduction.
Heat exchangers may be divided into serveral categories or classifi= cations.
In the most commonly used type of heat exchanger, two fluids of dif= ferent temperature flow in spaces separated by a tube wall.
This type is referred to as an =E2=80=98ordinary heat exchanger=E2= =80=99, as compared to the other two types classified as =E2=80=98regenerat= ors=E2=80=99 and =E2=80=98cooling towers=E2=80=99.
An ordinary heat exchanger is single-phase or two-phase.
In a single-phase heat exchanger, both of the fluids (cooled and he= ated) remain in their initial gaseous or liquid states.
In two-phase exchangers, either of the fluids may change its phase = during the heat exchange process.
The steam generator and main condenser of power plants are of the t= wo-phase, ordinary heat exchanger classification.
Single-phase heat exchangers are usually of the tube-and-shell type= ; that is, the exchanger consists of a set of tubes in a container called a= shell.
At the ends of the heat exchanger, the tube-side fluid is separated= from the shell-side fluid by a tube sheet.
The design of two-phase exchangers is essentially the same as t= hat of single-phase exchangers.

The heat transfer can be put= in terms of overall heat transfer coefficient (U) and overall tem= perature difference as

Q =3D UA (T_h =E2=88= =92 T_c). (= 8.2)

From Equations (8.1) and (8= .2), we can write

If the fluids are separated = by a tube wall in a heat exchanger as shown in Fig. 8.3, the total resistance is given by

Fi= gure 8.3 lube Wall

But

and U changes with area for the same heat flow rate.

Then the overall heat transf= er coefficients based on either the inside or outside surface area of the t= ube are

and

If the wall thickness is sma= ll and its thermal resistance is negligible, then

Some approximate range of ov= erall heat transfer coefficients is given in Table 8.1 for quick calculations.

Table 8.1 Overall heat transfer coefficients = for heat exchangers

Types of heat exchanger U (W/m^2K)

Gas-to-gas
10=E2=80=9340

Water-to-air in finned tubes (water in t= ubes)
30=E2=80=9360

Water-to-oil
100=E2=80=93350

Water-to-gasoline or kerosene
300=E2=80=931000

Water-to-water
850=E2=80=931700

Feedwater heaters
1000=E2=80=938500

steam-to-air in finned tubes (steam in t= ubes)
30=E2=80=93300

steam-to-light fuel oil
200=E2=80=93400

steam-to-heavy fuel oil
50=E2=80=93200

steam condenser
1000=E2=80=936000

Freon condenser (water-cooled)
300=E2=80=931000

Ammonia condenser (water-cooled)
800=E2=80=931400

Alcohol condensers (water-cooled)
250=E2=80=93700

= 8.3 FOULING FACTORS

In heat exchangers, during op= eration with fluids, a deposit gradually builds up on the surface of the tu= bes which affects the heat transfer area. The deposits are rust, boiler sca= le, silt, coke, etc. The deposits are known as scaling and they increase th= e thermal resistance to heat flow. The effect of scaling (fouling) is intro= duced in the design of heat exchanger by means of the fouling factor.

If there is deposit formatio= n on the inside and outside surfaces of the tubes, then the thermal resista= nce (R) is given as

where F_i and F_o are the fouling = factors at the inside and outside surfaces of the tube. The overall heat tr= ansfer coefficient is determined as discussed earlier based on inside or ou= tside surfaces.

= 8.4 LOG MEAN TEMPERATURE DIFFERENCE (LMTD)

The temperatures of hot and c= old fluids in a heat exchanger are generally not constant and vary from ent= ry to exit due to transfer of heat from the hotter to the colder fluid. Due= to the variation of temperature difference at various sections, the rate o= f heat flow varies along the exchanger even for a constant thermal resistan= ce. Neglecting heat loss to the surroundings and potential and kinetic ener= gy change, the general heat transfer equation for a heat exchanger is given= as

Heat lost by the hot fluid (= Q_h) =3D Heat transfer rate in heat exch= anger =3D Heat gained by the cold fluid (Q_c<= /em>)

m_hC_ph(T<= sub class=3D"EBsub">hi =E2=88=92 T_ho) =3D UA LMTD =3D m_cC_pc(T_co =E2=88= =92 T_ci)     (= 8.9)

where = the variables are defined in the box given below.

m_h =E2=80=93 mass flow rate of hot fluid (kg/s)

C_ph<=
/sub> =E2=80=93 specific heat of hot fluid (kJ/kgK)

T_hi<=
/sub> =E2=80=93 hot fluid inlet temperature (K)

T_ho<=
/sub> =E2=80=93 hot fluid outlet temperature (K)

U =E2=80=93 overall = heat transfer coefficient (kW/m²K)

A =E2=80=93 heat tra= nsfer surface area of hat exchanger (m²)

LMTD =E2=80=93 log mean tempe= rature difference of hot and cold fluids (K)

m_c =E2=80=93 mass flow rate of cold fluid (kg/s)

C_pc<=
/sub> =E2=80=93 specific heat of cold fluid (kJ/kgK)

T_co<=
/sub> =E2=80=93 cold fluid outlet temperature (K)

T_ci<=
/sub> =E2=80=93 cold fluid inlet temperature (K)

8.4.1 = LMTD for Parallel-flow Heat Exchangers

Consider a parallel-flow heat= exchanger as shown in Fig. 8.4. T= he temperature profiles are plotted with respect to the area covered along = the flow direction in the heat exchanger.

Fi= gure 8.4 Temperature Profiles in Parallel-Flow Heat Exchanger

Assumptions<= /p>

The flow conditions are steady an= d no phase change takes place in the fluids.

Heat loss to the surroundings, changes= in potential and kinetic energies are negligible.

The overall heat transfer coeffici= ent, specific heats, and mass flow rates of fluids are constants.

Consider an elementary area = dA of the heat exchanger. The rate of heat transfer dQ fr= om the hotter to colder fluid in this area is given by

Since T_h and T_c are assumed to = be constant over the elementary area dA, due to transfer of heat, = the hot fluid is cooled by dT_h, whereas= the cold fluid is heated up by dT_c. Th= en dQ can be written as

dQ =3D =E2=88=92mh=C2=B7C_ph =C2=B7 dT_h =3D m_c C_pc dT_c     (8.11)

i.e.,

and

where C_h and C_c are the heat cap= acity of hot and cold fluids, respectively.

i.e., C_h =3D m_=
h C_ph   and  = ; C_c =3D m_cC_pc

From Equation (8.10), Equ= ation (8.12) becomes

Integ= rating the above equation between inlet and outlet conditions, we get

i.e.,

Now considering the changes = in inlet to outlet temperatures of fluids, we can write as

Q =3D C_h (T_hi =E2=88=92 T_ho) =3D = C_c (T_co = =E2=88=92 T_ci)

Substituting the values of <= em>C_h and C_c= into Equation (8.13), we get=

where (logarithmic mean tempe= rature difference)

8.4.2 = LMTD for Counter-flow Heat Exchangers

The temperature variations of= hot and cold fluids in a counter-flow heat exchanger are shown in Fig. 8.5. Always the inlet condition is r= eferred from the hot fluid entry. The temperature differences can be writte= n as

Fi= gure 8.5 Temperature Profiles in Counter-Flow Heat Exchanger

=CE=B8_i =3D T_hi =E2=88=92 T_co

   =CE=B8_o =3D T_ho =E2= =88=92 T_ci =C2=B7

where =CE=B8_i =3D =CE=B8_o, LMTD = becomes indeterminate. In such a condition, as per L=E2=80=99 Hospital's ru= le,

LMTD =3D = =CE=B8_i =3D =CE=B8_=
o.     (8.15)

For the same inlet and outle= t temperatures, LMTD for the counter flow is higher than that for the paral= lel flow. Therefore, more heat transfer takes place for the same values of = U and A.

In counter flow, T_co can be more than T_ho<=
/sub> but not so in parallel flow. The arithmetic and logarithmic mean= s are equal when =CE=B8₀ =3D =CE=B8_i. If =CE=B8₀ = =E2=89=A0 =CE=B8_i, then the LMTD's alwa= ys less than the arithmetic mean.

The temperature distribution= for single-pass condenser and evaporator are shown in Fig. 8.6.

Fi= gure 8.6 Temperature Profiles for Condenser and Evaporator

8.4.3 = LMTD for Cross-flow Heat Exchangers

The LMTD calculation is more = complicated for multipass and cross-flow arrangements compared to single-pa= ss, parallel, and counter-flow heat exchangers and is also difficult to tre= at analytically. Correction factors are to be multiplied with the LMTD calc= ulated on the basis of counter flow to estimate the LMTD of multipass and c= ross-flow heat exchangers as shown in = Fig. 8.7.

Fi= gure 8.7 Cross-Flow Heat Exchanger

For multipass and cross-flow= types,

Q =3D FU A(LMTD)_{counter flow}.     (8.16)

where F is the corre= ction factor which can be taken from the charts shown in Figs. 8.8=E2=88=928.11.

Fi= gure 8.8 Correction Factor to the Counter-Flow LMT D for a Heat Ex= changer with One Shell Pass and 2, 4, 6, or Any Multiple of 2 Tube Passes

Fi= gure 8.9 Correction Factor to the Counter-Flow LMTD for a Heat Exc= hanger with Two Shell Passes and 4, 8, 12, =E2=80=A6 Tube Passes

F= igure 8.10 Correction Factor to the Counter-Flow LMTD for a Cross-= Flow Heat Exchanger with Shell Side Fluid Mixed and Tube Fluid Unmixed, Hav= ing One Tube Pass

F= igure 8.11 Correction Factor to the Counter-Flow LMT D for a Cross= -Flow Heat Exchanger with Both Fluids Unmixed and One Tube Pass

KEY= CONCEPTS

Many applications involving convective heat transfer take plac= e within pipes, tubes, or some similar cylindrical device.

In such circumstances, the surface area of heat transfer normally g= iven in the convection equation varies as heat passes through the cylinder.=

In addition, the temperature difference existing between the inside= and the outside of the pipe, as well as the temperature differences along = the pipe, necessitates the use of some average temperature value in order t= o analyze the problem.

This average temperature difference is called the log mean temperat= ure difference (LMTD), described eariler.

It is the temperature difference at one end of the heat exchanger m= inus the temperature difference at the other end of the heat exchanger, div= ided by the natural logarithm of the ratio of these two temperature differe= nces.

The above definition for LMTD involves two important assumption= s: (1) the fluid specific heats do not vary significantly with temperature,= and (2) the convection heat transfer coefficients are relatively constant = throughout the heat exchanger.

= 8.5 The NTU Method

To obtain an equation for the= rate of heat transfer without involving any of the outlet temperatures of = the fluids, heat exchanger effectiveness (=CE=B5) is introduced. T= he term effectiveness (=CE=B5) is defined as the ratio of the actu= al heat transfer in a heat exchanger to the maximum possible heat transfer.= Maximum possible heat transfer would be obtained in a counter-flow heat ex= changer of infinite heat transfer area without heat loss. In general, one f= luid undergoes a greater total temperature change than the other. The fluid= experiencing the larger change in temperature is the one having the smalle= r capacity coefficient, which is designated as C_min. The outlet temperature (T_co) of the cold fluid equals the inlet temperature (T_hi) of the hot fluid when C_c =E2=89=AA C_h, i.e., C_c =3D C_min =3D m_cC_pc. If C_h =E2=89=AA C_c, then T_h_o =3D T_ci^=C2=B7.

Effectiveness,

where C_min is the smaller value of m_h= c_ph and m_cc<= sub class=3D"EBsub">pc.

If the effectiveness is know= n, the rate of heat transfer is

Q =3D =CE=B5 C_min (T_hi =E2=88=92 T_ci).     (8.17)

Substituting the value of =CE=B5 from Equation (8.17)= , we get

Q =3D C_h (T_hi =E2=88=92 T_ho) =3D = C_c (T_co<= /em> =E2=88=92 T_ci)   &n= bsp; (8.18)

Equation (8.18) is more suitable for the design of heat exchan= gers than an equation using LMTD.

8.5.1 = Heat Exchanger Effectiveness of Various Types of Heat Exchangers

8.5.1.1 Parallel-flow heat excha= nger

Equation (8.13) can be written as

In order to eliminate the te= mperatures in the above equation, we can write from Equation (8.17):

By substituting the values o= f T_ho and T_co<=
/sub> in Equation (8.19), we = get

If C_h is C_min and C_c is C_max, the = effectiveness becomes

and if C_c is C_min. and Ch is C_max, the effectiveness becomes

Equations (8.22) and (8.23) = are same by taking C_min as C_h or C_c. The r= atio C_min/C_{max is called the heat capacity rate (C_h=
) and the ratio UA/C_min is cal=
led the number of heat transfer units (NTU). NTU is a measure of heat excha=
nger size. It is defined as the ratio between the heat capacity of exchange=
r and the heat capacity of flow. The effectiveness for various heat exchang=
ers can also be directly obtained from the charts shown in Figs. 8.12=E2=88=928.17.}

F= igure 8.12 Heat Exchanger Effectiveness for Parallel Flow

F= igure 8.13 Heat Exchanger Effectiveness for Counter Flow

F= igure 8.14 Heat Exchanger Effectiveness for Shell-and-Tube Heat Ex= changer with One Well-Baffled Shell Pass and Two (or a Multiple of Two) Tub= e Passes

F= igure 8.15 Heat Exchanger Effectiveness for Cross Flow with Both F= luid Unmixed

1, NTU is Based on Cunmixed" width=3D"273" height=3D"393" da= ta-mfp-src=3D"/library/view/heat-and-mass/9789332515680/images/pg265_img01.= png">

F= igure 8.16 Heat Exchanger Effectiveness for Cross Flow with One Fl= uid Mixed and the Other Unmixed. When C_{mixed<=
/sub>/ C_unmixed > 1, NTU is Based on=
C_unmixed}

F= igure 8.17 Heat Exchanger Effectiveness for Shell-and-Tube Heat Ex= changer with Two Shell Passes and 4, 8, 12, =E2=80=A6 Tube Passes

8.5.1.2 Counter-flow heat exchan= ger

8.5.1.3 Cross-flow heat exchange= r: both fluids unmixed

where =E2=80=98n=E2=80=99= is the number of tubes passes.

8.5.1.4 Cross-flow heat exchange= r: both fluids mixed

8.5.1.5 C= ross-flow type with C_max mixed and = C_min. unmixed

8.5.1.6 Cross-flow type with C_max unmixed and C_min. mixed

8.5.1.7 Shell-and-tube type: one= shell pass and 2, 4, 6 =E2=80=A6 tube passes

All exchangers with

=CE=B5 =3D 1 =E2=88=92 e^{=E2=88=92N.     (8.30)}

= 8.6 HEAT TRANSFER ENHANCEMENT

In a heat exchanger, the rate= of heat transfer depends on the overall heat transfer coefficient and heat= transfer area. The overall heat transfer coefficient can be increased eith= er by increasing the convection heat transfer coefficients (natural to forc= ed convection) or by increasing the heat transfer area by adding fins. The = overall heat transfer coefficient varies from 10 W/m^{2<=
/sup>K (gas-to-gas heat exchanger) to about 10,000 W/ m²K for phase change heat exchanges. Heat transfer coefficients depen=
d on the velocity of fluid. In the case of natural convection, the fluid ve=
locity is very low and the flow is laminar, the heat transfer coefficients =
are usually very low. This happens in gas-to-gas heat exchangers and fins a=
re used to increase the area and hence rate of heat transfer. The other met=
hods used to increase the rate of heat transfer are by making the surface r=
ough, twisting the fluid flow, and introducing acceptable level of vibratio=
n. Increase in heat transfer is also achieved by making the flow turbulent =
by introducing twisted tapes inside the tubes.}

In shell-and-tube heat excha= nger, baffles are commonly used in the shell side to increase the fluid sta= y time inside the heat exchanger. Multipass methods are also used to increa= se the heat transfer rate. In one-shell pass and two tube pass heat exchang= ers, all the tubes make one U-turn inside the shell. Similarly, two-shell p= ass and four tube pass heat exchangers are also made. Another type of heat = exchanger that is found in many industrial applications is the plate-type h= eat exchanger. In this case, the hot and cold fluids flow in alternate pass= ages and the heat transfer capacity is enhanced by the addition of more pla= tes in series. Figure 8.18 shows = the multipass and plate heat exchangers.

F= igure 8.18 Multipass Heat Exchangers

Estimation of LMTD in multip= ass heat exchangers is very difficult and hence charts are used to find LMT= D rather than mathematical expressions. Usually, in such situations, the fl= ow arrangement is considered to be of the single-pass type and LMTD is calc= ulated, and then this LMTD is multiplied by a correction factor obtained fr= om the charts of Bowman and Tubular Exchanger Manufacturers=E2=80=99 Associ= ation (TEMA).

= 8.7 COMPACT HEAT EXCHANGERS

In any heat exchanger, the ra= tio of the surface area for heat transfer to its volume is called area dens= ity. Heat exchangers with area density larger than 700 m²/m³ are called compact heat exchangers.= Car radiator with an area density of more than 1000= m²/m³ is a compact h= eat exchanger. Compact heat exchangers are produced by using corrugated fin= s or closely spaced thin plates.

= 8.8 CONCLUSIONS

Heat exchanger is a device us= ed to promote the transfer of heat between two or more fluids at different = temperatures. Hot fluid of a heat exchanger is the one that gives the heat = energy to another fluid during the phase change from vapor to liquid state = or without phase change. Cold fluid of a heat exchanger is the one that rec= eives the heat energy from another fluid for phase change or temperature ri= se. Area density is defined as the ratio of the heat transfer area of a hea= t exchanger to its volume. Area density for car radiator is 1000 m²/m³ and for human lung is 20,= 000 m²/m³.

Compact heat exchangers have= heat transfer area densities greater than 700 m^{2/m³ and are essential in application where the =
size and weight of the heat exchanger is an important design constraint.}
In recuperators, the hot and= cold fluids are separated by a wall and the heat is transferred by a combi= nation of convection to and from the wall and conduction through wall. In r= egenerators, the hot and cold fluids occupy alternatively the same space in= the exchanger core. The exchanger serves as a heat storage device that is = periodically heated by the warmer of the two fluids and then transfers the = heat to the cold fluid. In direct contact heat exchanger, the hot and cold = fluids contact each other directly. In condenser, the hot fluid remains at = constant temperature, from vapor to liquid phase and the cold fluid tempera= ture increases. In evaporator, the cold fluid remains at constant temperatu= re, from liquid to vapor phase and the hot fluid temperature decreases.

According to the direction o= f flow of fluids, the heat exchangers are classified as parallel flow, coun= ter flow, and the cross flow. In the parallel-flow arrangement. the hot and= cold fluids enter the exchanger from the same side, flow in the same direc= tion, and leaves from the same side. In the counter-flow arrangement, the h= ot and cold fluids enter the exchanger from the opposite ends, travel in op= posite direction, and leaves from the same ends. In the cross-flow arrangem= ent, the hot and cold fluids flow at right angles to each other.

Heat exchangers are also cla= ssified based on the design and construction as (a) concentric tubes type, = (b) shell-and-tube type, (c) multiple shell-and-tube passes type, and (d) c= ompact exchangers.

Fouling is a phenomenon of d= ust formation and deposition of fluid impurities inside heat exchangers tub= es during normal operations. The resistance to heat flow due to fouling is = fouling resistance. This resistance should be taken into account in designi= ng the heat exchanger.

LMTD in a heat exchanger is = the logarithmic average of temperatures change from inlet to outlet. In ord= er to calculate heat transfer between the fluids, the temperature differenc= e is written in terms of LMTD. The arithmetic mean temperature difference (= AMTD) is used instead of LMTD as an approximate method for low-temperature = differences in a heat exchanger. It is the average of temperatures differen= ce between the fluids at inlet and outlet of the heat exchanger.

The assumptions made in the = determination of LMTD in a heat exchanger are (a) overall heat transfer coe= fficient is constant; (b) specific capacity of fluids are constant; (c) the= re is no heat loss to surroundings; (d) the conditions are steady; and (e) = the kinetic and potential energy are negligible. LMTD expressions are simpl= e for parallel-flow and counter-flow heat exchangers. But for cross flow an= d multiple shell and tube heat exchangers, it is complicated because of the= complex flow nature and in such cases, LMTD is written as F(LMTD)= _{counter=E2=80=99} where F is the correct= ion faction, which depends on the geometry of the exchanger and temperature= of fluids.

NTU is defined as the ratio = between UA and C_min where = U is the overall heat transfer coefficient, A is the area of = the heat exchanger, and C_min is the min= imum heat capacity of the fluids. It is the measure of heat transfer surfac= e area. Larger the NTU, larger the size of the heat exchanger. Effectivenes= s of a heat exchanger is defined as the ratio of the actual heat transfer o= f the fluids to maximum possible heat transfer. Duri= ng boiling and condensation, the phase change takes place and one fluid rem= ains at constant temperature throughout the exchanger. The heat capacity of= phase change fluid is infinity. In a gas turbine recuperator, both the exh= aust gas and compressed air have equal thermal capacities and also the capa= city ratio becomes very close to unity. If the hot and cold fluids outlet t= emperatures are not known, the effectiveness NTU method is preferred becaus= e LMTD method requires tedious iteration.

= EXAMPLE PROBLEMS

8.1. A counter-flo= w double-pipe heat exchanger is used to heat water from 20=C2=B0C (T_ci) to 40=C2=B0C (T_co) using= a hot oil which enters the exchanger at 180=C2=B0C (T_=
hj) and leaves at 140=C2=B0C (T_ho). The fl= ow rate of water is 3.0 kg/s (m_c) and the overal= l heat transfer coefficient is 130 W/m²C (U). As= sume that the specific heat of oil is 2.1 kJ/kgK. Suppose the water flow ra= te is reduced to half, what would be the new oil flow rate necessary to mai= ntain the same outlet water temperature.

Case I:

Heat lost by oil =3D Heat gained by= the water

m_h C= _ph(T_hi =E2= =88=92 T_ho) =3D m_c= C_pc(T_{co =E2=88=92 T_ci} )

Mass flow rate of oil,

Case II:

The mass flow rate of oi= l if water flow rate is halved is given by the equation:

8.2. A heat exchan= ger, which is designed to cool 10 kg/s of air (m_h) from 300=C2=B0C (T=E2=80=B2_{hi) to 80=C2=B0C (T_ho)}, contains 400 tu= bes, each having a diameter of 40 mm (d). The inlet and outlet tem= peratures of cooling water are 30=C2=B0C (T_ci) a= nd 90=C2=B0C (T_co), respectively. The i= nside and outside heat transfer coefficients (h) on the sides are 200 and 6= 50 W/m²K. Determine (i) the mass flow rate of wa= ter, (ii) the type of exchanger, (iii) the heat transfer = rate and (iv) the length of the heat exchanger.

Mass flow rate of water (mc)

Heat gained by water =3D Heat lost = by air

m_c C= _pc (T_co =E2= =88=92 T_ci) =3D m_h= C _ph (T_{hi =E2=88=92 T_ho})

Type of exchanger
Since T_co > T_ho, the analysis for a co= unter-flow heat exchanger is applicable.

Heat transfer rate (Q)

Q =3D m_{hC_ph}(T_hi = =E2=88=92 T_ho) =3D 10 =C3=97 1 =C3=97 (300 = =E2=88=92 80) =3D 2100 kW

Length of e= xchanger (L)
Heat transfer rate,

Q =3D UA =C3=97 LMTD =3D U=CF=80dLh =C3= =97 LMTD

Therefore,

Here

8.3. Oil at temper= ature 40=C2=B0C (T_ci) enters a tube at a rate of= 3.0 kg/min (m_c). The tube is maintained at an u= niform temperature of 100=C2=B0C (T_w) due to con= densation of steam outside the tube. The inner diameter of the tube is 25 m= m (d_i.). Determine the length of tube required t= o maintain the outlet temperature of the oil (T_co) at 80=C2=B0C. The overall heat transfer coefficient (U) is 21 W= /m²K and C_p,_oil =3D 2050 J/kgK.

Heat transfer rate,

Also,

Therefore, 4100 =3D 21 =C3=97 A= =C3=97 36.4

=E2=87=92 A =3D 5.36 m2

Length of tube,

8.4. A heat exchanger is to be designed to condense a vapour at the ra= te of 8.5 kg/s (m_h) which is available at its sa= turation temperature of 80=C2=B0C (T_h) The cooli= ng liquid at 20=C2=B0C (T_ci) and a flow rate of = 62 kg/s (m_c) is used to remove the heat. The ove= rall heat transfer coefficient (U) is 500 W/m²K.= The latent heat of condensation of the vapour (h_{fg) is 550 kJ/kg. Determine (i) the number of tubes required if the outer =
diameter is 30 mm (d_o)}, inner diameter = (d_i), is 25 mm and length (l) is 5 m of the tube= s that are available and (ii) the number of tube passes, if the co= oling liquid velocity should not exceed 1.8 m/s (v_{c). (Assume that C_pc of the liquid is 4.2 kJ/k=
gK and =CF=81 =3D 1000 kg/m³.)}

The number of tubes required (N)

Heat of condensation =3D Heat gaine= d by liquid

m_h= =C3=97h_fg =3D m_c=C3= =97C_pc(T_co= =E2=88=92T_ci)

8.5 =C3=97 550 =3D 62 =C3=97 4.2 = =C3=97 (T_co =E2=88=92 20)

Therefore,

Area,

The number of tube passes (P)
Mass flow rate of the liqui= d,

Here the number of tubes in e= ach pass =3D N_p

Therefore,

Number of passes =3D that is 6 passes.

8.5. Light lubricating oil (C_ph =3D 2090 J/= kgK) is cooled by allowing it to exchange energy with water in a small heat= exchanger. The oil enters and leaves the heat exchanger at 375 K (T_hi) and 350 K (T_ho), respectively, and flows at a rate of 0.5 kg/s (m_h). Water at 280 K (T_ci) is available= in sufficient quantity to allow 0.201 kg/s (m_c)= to be used for cooling purposes. Determine the required heat transfer area= for (i) counter-flow and (ii) parallel-flow operation. The overall heat tr= ansfer coefficient (U) may be taken as 250 W/m²K= in both cases.

Heat loss by hot fluid,<= /p>

Heat gained by cold flui= d,

Q =3D m_c C_pc (T_co =E2=88=92 T_ci) =3D 26125W

This result applies to b= oth parallel-flow and counter-flow operations. For the counter-flow configu= ration, LMTD is calculated as

Performing similar calcu= lations for the parallel-flow situation,

The area required is low= er for the counter-flow compared with the parallel-flow arrangement by appr= oximately 7%.

8.6. The condensat= ion temperature (T_h) of a vapour in a condenser = is 120=C2=B0C. The inlet and outlet temperatures of cooling water flowing a= t the rate of 1.0 kg/s through the tubes are 30=C2=B0C (T_ci) and 65=C2=B0C (T_co), respectively. The tube diameter (d) is 40 mm, length (L) = is 8 m and the tubes are 10 in number (N). Determine the (i) the overall he= at transfer coefficient (U), (ii) the number of transfer units (NT= U), (iii) the effectiveness of the condenser (=CE=B5) and (iv) the= rate of condensation (m_h) if the latent heat of= condensation (h_fg) is 2200 kJ/kg.

Overall heat transfer coefficient= (U)
Heat gained by cooling water<= /p>

Area, A =3D =CF=80dLN =3D= =CF=80 =C3=97 0.04 =C3=97 10 =C3=97 10 =3D 12.57 m²

Therefore,

The number of transfer units (NTU)
In a condenser, Cmin refers to the cooling water.

Therefore, C_min =3D m_c =3D 1.0 =C3=97 4.2 =3D 4.2 W/K

The effectiveness of the condenser, =CE=B5

It can also be calculated as<= /p>

Rate of condensation (m_h)
Heat of condensation, Q = =3D m_h, h_fg.

Therefore,

8.7. Oil at a temperature of 500=C2=B0C (T_{hi) is used to heat the water at 50=C2=B0C (T_ci).=
The specific heat of oil (C_ph) is 2.5 kJ/kgK an=
d the flow rate of oil (m_h) is 20 kg/s. The flow=
rate of water (m_c) is 25 kg/s. If the overall h=
eat transfer coefficient}, U, and area, A, <= em>are 1.5 W/m²K and 50 m^{2<=
/sup>, respectively, calculate the oil and water outlet t=
emperatures for counter-flow and parallel-flow arrangements (i) by calculat=
ion and (ii) by charts.}

By calculation, the outlet temper= ature of fluids (T_ho and T_co)

Counter-flow arrangement

Heat capacity of hot fluid,

Heat capacity of cold fluid,<= /p>

Here

C_c =3D C_max = and C_h =3D C_min

Therefore,

Effectiveness,

Here

Also

Also

Parall= el-flow arrangement

Effectiveness,

Here R =3D 0.48 and = NTU =3D 0.8

Also

Therefore,

T_ho = =3D 288.5=C2=B0 C and T_co =3D 150.= 7=C2=B0C

Using charts

Counter-flow arrangement

From the chart in Fig. 8.13, for

Then the values of T_h_o =3D 266=C2=B0C and= T_co =3D 161.4=C2=B0C

Parallel-flow arrangement

From the chart in Fig. 8.12, =CE=B5 =3D 0.48

Then T_c_o =3D 284=C2=B0C and T_co =3D 152.9=C2=B0C.

8.8. In a shell-an= d-tube heat exchanger, water flowing at a rate of 10 kg/s (m_h) through 50 double pass tubes heats air that flows t= hrough the shell side. The length of the tube is 6.7 m and its outer diamet= er is 26 mm (d_o). The heat transfer coefficients= of the water (h_i) and air (h_o) sides are 4 70 W/m²K and 210 W/m²K, respectively. Air enters the shell at a = temperature (T_ci) of 15=C2=B0C and a flow rate o= f 1.6 kg/s (m_c). The inlet temperature of water = (T_hj) is 75=C2=B0C. Determine the effectiveness = of the heat exchanger.

Effectiveness of the heat exchang= er (=CE=B5)
To use the graph, we have to = calculate

Heat capacity of hot fluid,

C_h = =3D m_hC_ph=3D 10 = =C3=97 4.2 =3D 42 kW/K

Heat c= apacity of cold fluid,

C_c = =3D m_cC_pc =3D 1.= 6 =C3=97 1 =3D 1.6 kW/K

C_c = =3D C_min and C_{h =3D C}_max

Here A =3D =CF=80d_oL =3D =CF=80 =C3=97 0.026 =C3=97 6.7 =3D 0.547 m= ²

Therefore,

From the chart in Fig. 8.14, effectiveness =3D 0.1.

8.9. In a shell-an= d-tube heat exchanger, saturated steam at 120=C2=B0C (T_h) is condensing on the shell side. The cooling water ente= rs the tubes at 25=C2=B0C (T_ci) and leaves at 75= =C2=B0C (T_co). Determine the LMTD for the follow= ing arrangement and discuss (i) parallel-flow (ii) counter-flow and (iii) c= ross-flow arrangement if shell side flow is mixed and other fluid unmixed.<= /em>

Parallel-flow arrangement

Counter-flow arrangement

Cross-flow = arrangement

(LMTD)_{cross_-_fl0W =3D F (LMTD)_counter-flow}

For F, we have to ca= lculate R₁ and R₂

From the chart in Fig. 8.10, F =3D 1;

LMTD_cross-flow= =3D LMTD_cross-flow

In a condenser, phase change = takes place at constant temperature, therefore, for all arrangements the LM= TD is same and the heat transfer is also the same.

8.10. In a cross-f= low heat exchanger, water at 5=C2=B0C (T_ci) ente= rs at the rate of 1.5 kg/s (m_c) to cool 1 kg/s o= f air (m_h) that is initially at 40=C2=B0C (Thi). The overall heat transfer coefficient (U) is 15= 0 W/m²K and area (A) is 25 m². Determine the outlet temperature of air (T_ho=) and water (T_co). (Assume that C_ph for air =3D 1.0 kJ/ kgK and C_{pc for water is 4.2 kJ/kgK, both fluids unmixed.)}

In order to find the out= let temperature, we have to calculate C_{min/C_max and NTU in order to read =
=CE=B5.}

Heat capacity of cold fl= uid, C_c =3D m_cCpc

=3D 1.5 =C3=97 4.2 =3D 6.3kW/K

Heat capacity of hot flu= id, C_h =3D m_hCph

=3D 1 =C3=97 1 =3D 1 kW/K

Here C_h=3D C_min and Cc=3D C_max=

and

From the chart in Fig. 8.17, =CE=B5 =3D 0.93

But the actual heat tran= sfer,

Q =3D =CE=B5Q_max =3D =CE=B5C_min(T<= sub class=3D"EBsub">hi =E2=88=92 T_ci)=             &nb= sp;

=3D 0.93 =C3=97 1 =C3=97 (40 =E2=88= =92 5) =3D 32.55 kW

Also Q =3D C_c(T_co =E2=88=92 Tci) =3D C_h= (T_hi =E2=88=92 T_ho)

Therefore, and

8.11. A cross-flow= heat exchanger is used to heat unmixed oil in the tubes from 20=C2=B0C (T<= sub class=3D"EBsub">ci) to 90=C2=B0C (T_o).= The specific heat (C_pc) of oil is 2.0 kJ/kgK. S= team (mixed) enters the exchanger with temperature 140=C2=B0C (T_hi) and leaves at 115=C2=B0C (T_{ho) with a mass flow rate of 5.0 kg/s (m_h). The =
overall heat transfer coefficient (U) is 300 W/m^{2K and C_p for steam is 2.0 kJ/kgK (C_ph). Calculate the surface area of the heat exchanger. Als=
o determine the heat transfer rate if the flow rate of oil (m_h) is reduced to half while the steam flow rate and U remain t=
he same. If the tube diameter (d) is 5 cm and length (l) is 2.5 m}}, find the number of tubes for oil flow.

Heat transfer rate Q =3D m_hCph(T_hi=E2=88=92= T_ho) =3D m_cC= _pc(T_co=E2=88= =92T_ci) =3D UAF LMTD_counter-flow

To determine F we ha= ve to know R₂ and R₁

From the chart in Fig. 8.10, F =3D 0.97

Heat transfer,

Q =3D m_h C_ph (T_hi =E2=88=92 T_ho) &nb= sp;            =

=3D 5 =C3=97 2.0(140=E2=88=92115) = =3D 250kW

Mass flow of oil,

In this case outlet temperatures of oi= l and steam are changing due to the reduced flow of oil. Using the NTU meth= od, the temperatures are calculated.

To know =CE=B5, we have to de= termine C_minC_max and NTU

Therefore, heat capacity of s= team,

C_h = =3D m_hC_ph =3D 5 = =C3=97 2 =3D 10 kW/K

Heat c= apacity of oil, C_c =3D m_cC_pc

In this, oil flow m_c is reduced to half

C_c =3D 0.895 =C3=97 2 =3D 1.79kW/K

Here C_c =3D C_mi_n and C_h =3D C_max (mixed fluid)

Therefore,

and

Then using the formula, effec= tiveness, =CE=B5, for C_min(oil, unmixed= ) and C_max (steam, mixed) is given by

=CE=B5 can also be t= aken from the chart in Fig. 8.16<= /p>
We know that

That is Q_actual =3D =CE=B5Qmax =3D C_c (T_co =E2=88=92 T_ci) =3D C_h (T_h=
i =E2=88=92 T_ho)

Similiarly,

Therefore,

Here a reduction of 50% in th= e oil flow rate reduces the heat flow by 33%.

Here the su= rface area,

8.12. Calculate th= e overall heat transfer coefficients (U_i and Uo) on the inner and outer surfaces of a brass tube= of inner diameter 3 cm (d_i) and outer diameter = 4 cm (d_o). The thermal conductivity (K) of brass= is 110 W/mK. The inside (h_i) and outside (ho) heat transfer coefficients are 1200 W/m²K and 2000 W/m²K, res= pectively. The fouling factors for the inner (F_i= ) and outer (F_o) surfaces are 0.00018 m²C/W. Also, calculate U_{i and U_o}, neglecting tube resistance = and fouling resistance.

Overall heat transfer co= efficient for the inner surface, U_i

But Q =3D A_iU_i LMTD =3D A_oU_o LMTD

Therefore, A_iU_i =3D A_oUc

By neglecting tube resis= tance and fouling resistance, U_i =3D U_o =3D U.

8.13. A shell-and-= tube heat exchanger has one shell pass and four tube passes. The fluid in t= he tubes enters at 200=C2=B0C (T_hi) and leaves a= t 100=C2=B0C (T_ho). The temperature of the fluid= entering the shell is 20=C2=B0C (T_ci) and as it leaves the shell is (T_co= ) 90=C2=B0C. The overall heat transfer coefficient (U) based on a surface a= rea of 12 m² (A) is 300 W/m^=
2K. Calculate the heat transfer rate between the fluids.

Heat transfer rate, = Q =3D UAF LMTD_counter-flow

To find the correction f= actor, F, from the chart, the temperature ratios required are

From the chart in Fig. 8.8, F=3D 0.85

Therefore, Q =3D 300 =C3= =97 12 =C3=97 0.85 =C3=97 94.21 =3D 288.283 kW.

8.14. A tube of ou= ter diameter 0.025 m (d_o) is used to cool 7.0 kg= /s (m_h) of a solution from 70=C2=B0C (T_hi) to 40=C2=B0C (T_ho) using 6= .7 kg/s (m_c) of water available at 10=C2=B0C (T<= sub class=3D"EBsub">ci). The overall heat transfer coefficient on the= outer surface of the tube (U_o) is 600 W/m²K and the specific heat of the solution is 3.81 kJ/kg= K (C_ph). Calculate the heat transfer surface are= a required for (i) parallel-flow, (ii) counter-flow arrangement, (iii) with 2 shell passes and 100 tube passes (n) (water is flowing= inside the tubes) and (iv) a cross-flow exchanger (both fluids unmixed)

Parallel-flow arrangement

In order to determine LMTD, w= e have to calculate outlet temperature of water, (T_co).

Using the energy balance equ= ation

Heat gained by water =3D Hea= t lost by solution

m_cc_pc (T_co =E2=88=92 T_ci) =3D m_h c_ph (T_hi =E2=88=92 T= _hi)

LMTD for parallel-flow arrang= ement

Rate = of heat flow from solution to water,

Q =3D m_hC_ph(T_=
hi=E2=88=92 T_h0)   &nb= sp;            =

      = ;      =3D 7 =C3=97 3.81 =C3=97 (70 =E2=80=93= 40) =3D 800.1 kW

Heat transfer area, A, required =3D Q/U_o LMTD =3D= 800.1 =C3=97 10³/600 =C3=97 15.7 =3D 84.94 m2

Length of the exchanger,

It is not preferable in pract= ice to build a heat exchanger of this length.

Counter-flow arrangement

Heat transfer area,

Length of the exchanger, = L =3D 45.23/=CF=80 =C3=97 0.025 =3D 575.94 m

In this case, 46.75% of the= length is reduced. Still it is difficult to build a heat exchanger of this= length.

Two shell and 100 tube passes
Heat transfer,

Q =3D FAU₀₌ LMTD_counter-fl0W

For F, we have to calculate R₂ and R₁

From the chart in Fig. 8.9, F =3D 0.96

Length of exchanger,

This is a reasonable length f= or a heat exchanger.

Cross-flow exchanger
Heat transfer,

Q =3D FAU₀₌ LMTD_Counter-fl0W

From the chart for the above = values of R₁ and R₂, F is 0.87

This area is more than that o= f the multipass shell-and-tube exchanger.

8.15. Air is suppl= ied to a furnace after preheating with exhaust gas from a power plant in a = cross-flow heat exchanger. The air enters at 70=C2=B0C (T_ci), leaves at 250=C2=B0C (T_co= ) and has a mass flow rate of 10 kg/s (m_c). The = exhaust gas enters the exchanger at 450=C2=B0C (T_{hi) and leaves at 200=C2=B0C (T_ho). The specifi=
c heats of gas (C_ph) and air (C_pc) are 1.2 and 1.0 kJ/kgK}, respectively. The overall he= at transfer coefficient (U) is 150 W/m²K. Find t= he surface area of the heat exchanger when (i) the air is unmixed (ii) both= air and gas are unmixed.

Air unmixed and gas mixed

Heat transferred to air,

Q =3D m_{cC_pc (T_co =E2=80=
=93 T_ci})      = ;

      = ;     =3D 10 =C3=97 1 =C3=97 (250 =E2=80=93 70) =3D 180= 0 kW

Again Q =3D FAU_o LMTD_counter-flow

To find F we have to= calculate R₂ and R₁

Heat lost by the exhaust ga= s =3D Heat gained by the air

m_hC<= sub class=3D"EBsub">ph (T_hi=E2=88= =92 T_ho) =3D m_{cC_pc}(T_co = =E2=88=92 T_ci)

Then <= em>F, from the chart in Fig. 8.1= 0, is 0.65

Both air and gas are unmixed

For the same values of R<= /em>₁ and R₂= , from the chart in Fig. 8.11, F =3D 0.82,

8.16. An economize= r is to be purchased for a power plant. The unit is to be large enough to h= eat 8 kg/s (m) of water from 30=C2=B0C (T _ci) to= 160=C2=B0C (T_co). Flue gas flowing at the rate = of 25 kg/s (m_h) at (T_{hi) 425=C2=B0C is available. The specific heat of flue gas is 1.0 kJ/kgK (C=
_ph). Calculate the outlet temperature of flue ga=
s}, the effectiveness and the heat transfer area required for a cou= nter-flow arrangement. (Assume U =3D 60 W/m²K.)<= /em>

Heat loss of flue gas = =3D Heat gained by water

m_h C_ph (= T_hi =E2=88=92 T_ho<=
/sub>) =3D m_c C_pc<=
/sub> (T_co =E2=88=92 T_ci)

Heat transfer rate, = Q =3D UA LMTD

Effectiveness,

H= ere

C_h = =3D m_hC_ph =3D 25= =C3=97 l =3D 25kW/=C2=B0C

C_c = =3D m_cC_pc =3D 8 = =C3=97 4.2 =3D 33.6 kW/=C2=B0C

8.17. Hot air is c= ooled from 66=C2=B0C (T_hi) to (T_ho)38=C2=B0C, while flowing at the ra= te of 1.25 kg/s (m_h) in a simple adiabatic count= er-flow heat exchanger, using cold air which enters at 15.5=C2=B0C= (T_ci) and flows at a rate of 1.6 kg/s (m_c). It is planned to lengthen the heat exchanger to att= ain the hot air temperature 26.5=C2=B0C (T=E2=80=B2_ho) using same cold air. Determine the ratio of the length of t= he new heat exchanger to the length of the original. (Assume that the speci= fic heat of air is constant.)

Case I:

Heat lost by hot air =3D= Heat gained by cold air

m_hC<= sub class=3D"EBsub">ph(T_hi =E2=88= =92 T_ho) =3D m_{cC_pc}(T_co = =E2=88=92 T_ci)

Rate of heat transfer,

Therefore,

Case II:

A= lso

8.18. It is desire= d to heat 230 kg/h of water (m_c) from 35=C2=B0C = (T_ci) to 93=C2=B0C (T_{co) with oil having an initial temperature of 175=C2=B0C (T_hi). The mass flow rate of oil (m_h) i=
s also 230 kg/h. C_p of oil =3D 2.1 kJ/kgK. Two d=
ouble-pipe heat exchangers are available such that}

U =3D 570 W/m²C, A =3D 0.47 m^=
2 (Exchanger 1);

U =3D 370 W= /m²C, A =3D 0.94 m2 (Exchanger 2).

Which exchanger shou= ld be selected?

Rate of heat transfer,

The outlet temperature o= f oil,

Since the outlet tempera= ture of oil is less than the outlet temperature of water, we have to use a = counterflow arrangement

Therefore,

H= eat exchanger 1

Q =3D UA LMTD  &= nbsp;           &nbs= p;

      = ;             = =3D 570 =C3=97 0.47 =C3=97 47.15 =3D 12.633 kW

Heat exchanger 2

Q =3D 370 =C3=97 0.94 =C3= =97 47.15 =3D 16.4 kW

Therefore, Heat exchange= r 2 is to be selected.

8.19. A feed water= heater in a power plant uses a parallel-flow exchanger with condensing ste= am in one shell pass at 120=C2=B0C (T_h). Water e= nters the tubes at 30=C2=B0C (T_ci) and makes fou= r passes to produce an overall heat transfer coefficient of 2000 W/m²K (U). The flow rate of water is 2.5 kg/s (m_c) and the water exit temperature is 100=C2=B0C (T_co). Calculate the area of the exchanger. Also find the = exit temperature of water if the heat exchanger has been in service a long = time and has a fouling factor of 0.0002 m²K/W (R= _f).

Heat transfer rate,

Q =3D m_{cC_pc} (T_co= =E2=88=92 T_ci)    &nbs= p;            &= nbsp;

      = ;       =3D 2.5 =C3=97 4.2 =C3=97 (100 = =E2=88=92 30) =3D 735 kW

Area required,

We know that

On trial and error,

T_co<= /em> =3D 74.5=C2=B0C.

8.20. Determine the required heat transfer area for the conditions spe= cified in Problem S8.5 using charts and the equations of effectiveness (NTU= ) method if the configurations are

counter-flow;

parallel-flow;

cross-flow, water-mixed;<= /em>

and shell-and-tube with four t= ube-side passes.

It is first necessary to= determine the capacity coefficients for the oil and water,

C_{oil<=
/sub> =3D (mc_p)_oil= =3D (0.5 kg/s) =C3=97 (2090 J/kgK) =3D 1045 J/sK}

and

C_wate=
r =3D (mc_p)_w= =3D (0.201 kg/s) =C3=97 (4177 J/kgK) =3D 841.2 J/sK

thus water is the minimu= m fluid. From the equation, the effectiveness is evaluated as

By using the appropriate= chart in the appropriate NTU values, in turn, the required area may be eva= luated for each heat exchanger configuration:

Counter-flow (chart in Fig. 8.13)

NTU =3D 0.47

Parallel-flow (chart in Fig. 8.12)

NTU =3D 0.50

Cross-flow, water-mixed (chart in Fig. 8.16)

NTU =3D 0.48

Shell-and-tube, four tube-side pas= ses

NTU =3D 0.49

These results are compar= able to those obtained earlier, with some possible inaccuracies involved in= reading the chart.

8.21. A room wall is made up of wood (thickness 15 mm and K_w =3D 0.2 W/mK, foam (thickness 80 mm and K_f =3D 0.02 W/mK and brick (thickness 225 mm and K_i. =3D 1 W/mK) from inside to outside. The room air i= s at a temperature of 270 K with a convection heat transfer coefficient of = 30 W/m²K. The outside air at 300 K and the outsi= de convection heat transfer coefficient is 15 W/m^{2K. If the wall area is 100 m²}, find t= he overall heat transfer coefficient and heat gain to the room. If the thic= kness of the foam layer is doubled, find the change in heat gain. = If the wood thickness is doubled, what is the change in heat gain?=

Overall heat transfer co= efficient,

Heat gain,

Q =3D UA(T_o =E2=88=92 T_i)  = ;            &n= bsp;

      = ;            &n= bsp;=3D 1.25 =C3=97 100 =C3=97 (300 =E2=88=92 270) =3D 3750 W

If the foam thickness is doubled,

If the wood thickness is doubled,

Heat gain =3D Q =3D = 1.143 =C3=97 100 =C3=97 (300 =E2=88=92 270) =3D 3429 W.

8.22. A heat excha= nger is to be designed to cool lubricant oil from 90=C2=B0C (T_oi) to 60=C2=B0C (T_oo) flowing at = the rate of 980 kg/hr (m_ow). Water is to be used= as cooling medium at the same mass flow rate (m_{w =3D m_o) and its available temperature is 30=C2=
=B0C (T). Assuming an overall heat transfer coefficient as 30 W/m²K find the surface area required for (i) parallel-flow an=
d (ii) counter-flow arrangements. Assume that specific heats of water (Cpw}) and oil (C_po) are 4.= 2 and 2 kJ/kgK, respectively.

Heat lost by oil =3D Hea= t gained by water

m_oC<= sub class=3D"EBsub">po(T_oi =E2=88= =92 T_oo) =3D m_{wC_pw}(T_wo = =E2=88=92 T_wi)

Therefore,           &nbs= p;

Parallel-f= low arrangement

Counter-flow arrangement

8.23. A heat excha= nger is to be designed to evaporate water continuously at 100=C2=B0C (Tw). The latent heat (LH) of vaporization is 2255 kJ/= kg. Hot air at the rate of 600 kg/hr (m_a) enters= at a temperature of 200=C2=B0C (T_ai) and leaves= at 125=C2=B0C (T_ao). Assuming an overall heat t= ransfer coefficient of 50 W/m²K and that C_p of air is 1 kJ/kgK, find the surface area o= f the heat exchanger and rate of evaporation.

Heat supplied by air =3D= Heat transfer rate in heat exchanger =3D Rate of evaporation

m_aC<= sub class=3D"EBsub">pa (T_ai =E2= =88=92 T_ao) =3D UA LMTD =3D m<= sub class=3D"EBsub">w =C3=97 LH

S= urface area,

Rate of evaporation,

8.24. A condenser = in a thermal power plant is one shell pass and two tube passes type and has= (N) 20, 000 tubes per pass to transfer the heat rate of (Q) 3 =C3=97 109 W. The tube diameter (D) is 30 mm and steam condenses on t= he outer surface with a convection heat transfer coefficient (h_o) of 15, 000 W/m². Co= oling water enters the condenser tubes at the rate (m_c=) of 4 =C3=97 104 kg/s with a inlet temperature (T_h) of 30=C2=B0C while the steam condenses at a temperatur= e (T_h) of 60=C2=B0C. Find the = temperature of the cooling water at the outlet and length of tube per pass.=

The properties of water = are

C_pc=3D 4.2 kJ/kg; m =3D 855 =C3=97 10^-6 NS/m²; k =3D 0.613 W/mK; = Pr =3D 5.83

Cooling water outlet tem= perature,

Rate of heat transfer, <= em>Q =3D UAF (LMTD)

where A =3D N =C3=97 2L =C3=97 pD and

For water flowing inside= the tube

Reynolds number, Re

Flow inside the tube is = turbulent. The heat transfer correlation is

Heat transfer coefficien= t,

O= verall heat transfer coefficient,

The correction factor (<= em>F) from the shell and tube exchanger chart for P and R=

and R =3D 0. Th= en F is 1.

8.25. The bed of a= fluidized bed combustor is to be maintained at a temperature (T_b) of 880=C2=B0C. The heat transfer coefficients between b= ed to surface (h_b) and tube surface to liquid (h= _t,) are 360 W/m²K and= 6500 W/m²K, respectively. A tube of 30 mm diame= ter (D) is immersed in the bed at rate of 40 kW is extracted. Liquid enters= the tube (T_I) at 40=C2=B0C and leaves (T_a) at 90=C2=B0C. Estimate the length of the immersed tu= be. Neglect the thermal resistance of the tube wall.

The overall heat transfe= r coefficient,

Log mean temperatute dif= ference

8.26. Liquid benzene is to be supplied at the rate of 35 m³ (V) at a temperature (T) of 160=C2=B0C for an industrial= process. The liquid is stored at a temperature (T_{i) of 0=C2=B0C and circulated (m) at the rate of 5.0 kg/s to the tube sid=
e of a one shell and two tube pass heat exchanger. A gas at (m_g) 2.5 kg/s and (T_gi) 210=C2=B0C p=
asses through the shell side of the exchanger. Estimate the time taken to h=
eat the benzene using this external heat exchanger if U =3D 250 W/m²K and A =3D 50 m².}

The properties of benzen= e are

=CF=81 =3D 890 = kg/m³; C_pl = =3D 2 kJ/kgK; C_pg =3D 2.4 kJ/kgK.

For a mixed batch liquid= / external shell and tube exchanger

where the mass of benzen= e in the tank,

W =3D volume =C3=97 densit= y            &n= bsp;

=3D 35 =C3=97 890 =3D 31150 kg/m3

and the factor

where F is the = correction factor.

The factor F is= 0.68 (from the chart)

Time, t =3D 22,= 758 s.

= REVIEW QUESTIONS

What is a heat exchanger?

Classify the types of heat exchangers.=

Which type of heat exchanger is more e= ffective? Why?

What is log mean temperature differenc= e (LMTD)?

Derive an expression for LMTD in (i) p= arallel-flow and (ii) counter-flow heat exchangers.

Why are the directions of fluid flows = immaterial when one fluid undergoes a phase change?

When is the NTU method useful in heat = exchanger analysis?

What does the term effectiveness mean = in heat exchanger analysis?

Derive an expression for the effective= ness of (i) a parallel-flow and (ii) a counter-flow heat exchanger.

How will you estimate the effectiv= eness of a heat exchanger when one fluid undergoes a phase change?

OBJECTIVE QUESTIONS

The most effective heat exchanger= is

the parallel-flow type

the counter-flow type

the cross-flow type

all of the above

In a parallel-flow heat exchanger, bot= h hot and cold fluid will

enter at one side and leave at th= e other side parallel to each other

enter from opposite sides parallel to = each other

enter at perpendicular directions to e= ach other

all of the above

In a counter-flow heat exchanger, if t= he entry and exit hot fluid temperatures are 80=C2=B0C and 40=C2=B0C and th= at of the cold fluid are 30=C2=B0C and 50=C2=B0C, the LMTD is

1.82=C2=B0C

18.2=C2=B0C

0.182=C2=B0C

182=C2=B0C

In an evaporator, water evaporates at = 100=C2=B0C taking heat from hot air entering at 260=C2=B0C and leaving at 1= 50=C2=B0C. The LMTD is

9.45=C2=B0C

95.4=C2=B0C

94.5=C2=B0C

0.954=C2=B0C

In a heat exchanger, oil flowing at th= e rate of 1000 kg/hr is to be cooled from 80=C2=B0C to 40=C2=B0C using wate= r available at 30=C2=B0C at the same mass flow rate. Specific heats of oil = and water are 2.09 and 4.18 kJ/kg K. The exit temperature of water will be

45=C2=B0C

55=C2=B0C

65=C2=B0C

50=C2=B0C

Heat exchanger effectiveness is define= d as the ratio of actual heat transfer to

the maximum possible heat transfe= r

the minimum possible heat transfer
the average heat transfer

the area of heat transfer

The term NTU is related to

a heat exchanger

UA/C_min= .

the effectiveness

all of the above

LMTD of a cross-flow heat exchanger is

higher than that of a parallel-fl= ow heat exchanger

higher than that of a counter-flow hea= t exchanger

zero

both (a) and (b)

Fouling factor in a heat exchanger

increases the resistance for heat= transfer

decreases the resistance for heat tran= sfer

keeps the resistance for heat transfer= constant

all of the above

In a heat exchanger, if the value = of the overall heat transfer coefficient is 200 W/m^{2 K, the area is 10 m² and C_min. is 2500 W/K, then the value of NTU is

1.25
0.8
125
0.08}

Answers:

c

a

b

c

d

a

d

d

a

b

= EXERCISE PROBLEMS

In a heat exchanger, a pipe of ou= ter diameter 60 mm and inner diameter 52.5 mm with thermal conductivity 0.0= 54 kW/mK is used to heat water. If the inside and outside heat transfer coe= fficients of the pipe are 1.960 kW/m²K and 0.007= 9 kW/m²K, estimate the overall heat transfer coe= fficient. (Ans. 0.0079 kW/m² K)=

In a counter-flow heat exchanger, wate= r flowing at the rate of 4080 kg/hr is heated from 310 K to 350 K using an = oil which enters at 385 K and leaves at 350 K. Assuming an overall heat tra= nsfer coefficient of 0.319 kW/m²=C2=B0C for the = heat exchanger, specific heats of oil and water as 1.895 kJ/kg=C2=B0C and 4= .18 kJ/kg=C2=B0C, find the surface area of the heat exchanger. (Ans= . 15.8m²)

Glycol flowing at the rate of 30 kg/ m= in is to be cooled from 375 K to 335 K in a counter-flow heat exchanger usi= ng water available at 285 K with a flow rate of 1249.9kg/hr. Assuming an ov= erall heat transfer coefficient of (2624 =C3=97 4.18/3600) kW/m²K find the surface area of the heat exchanger. Specific h= eats of glycol and water may be assumed to be (0.632 =C3=97 4.18) and 4.18 = kJ/kgK. (Ans. 0.34 m²)

A refrigerant is condensed at 40=C2=B0= C and (67500 =C3=97 4.18/3600) kW of heat is removed using water which ente= rs at 30=C2=B0C and leaves at 35=C2=B0C in a condenser having an overall co= efficient of (848 =C3=97 4.18/3600) kW/m² K. Fin= d the area required for heat transfer. (Ans. 11 m²)

Oil is cooled in a tubular heat exchan= ger using water. The inside tube diameter is 6 cm and tube thickness is 0.1= 95 cm. The inner tube has a thermal conductivity of 0.0159 kW/ mK and has a= n inside convection heat transfer coefficient of 0.64 kW/m²K. The convection heat transfer coefficient on the outer surface= of the tube is 0.022 kW/m² K. If the inside and= outside fouling factors are 0.0001 and 0.0004 m^{2 K/W, find the overall heat transfer coefficient. (Ans. 2=
2 W/m² K)}

Water with a flow rate of 18 kg/ min, = available at 285 K, is used to cool oil from 365 K to 325 K in a heat excha= nger. The mass flow rate of oil is 9 kg/min and its specific heat is 2.13 k= J/kgK. If the overall heat transfer coefficient is 22 W/m² K, find the surface area required if the heat exchanger is of (i= ) the parallel-flow type and (ii) the counter-flow type. (Ans. 11.5 m², 11 m²)<= /li>
If the parallel-flow heat exchanger of= Problem E8.6 is used to cool oil, with the outlet temperature of the oil b= egin unknown, using the NTU method estimate the oil outlet temperature. (Ans. 325 K)

A counter-flow heat exchanger is used = to cool compressed air from 525 K to 315 K using water available at 295 K. = If the overall heat transfer coefficient is 0.1 kW/m^{2<=
/sup> K and the mass flow rates of water and compressed air are 300 kg/min =
and 60 kg/min, respectively, find the area required. (Ans.=
25 m²)}

Hot water is cooled from 330 K to 315 = K using a coolant which enters a one shell pass and two tube pass heat exch= anger at 280 K and leaves at 305 K. If the hot water flows through the shel= l side at the rate of 15 kg/ min and the overall heat transfer coefficient = is 0.899 kW/m² K, find the area required for hea= t transfer. (Ans. 0.65 m²)

Water is heated in a cross-flow heat e= xchanger from 315 K to 355 K when it is used as a coolant to cool hot oil w= hich enters the heat exchanger at 375 K. If the flow rates of oil and water= are 156 kg/ min and 60 kg/ min, respectively, find the overall heat transf= er coefficient. The area available for heat transfer is 19.95 m². Assume the specific heat of oil as 1.895 kJ/kg K. (Ans. 450 W/m² K)

Water flowing at the rate of 22.5kg/ m= in is cooled from 355 K to 320 K in a shell-and-tube heat exchanger by a ga= s entering at 295 K. The overall heat transfer coefficient is 0.16 kW/m2 K and the area available for heat transfer is 9.99= 5 m². If the water flows through the tube side o= f the two shell pass and eight tube pass arrangement, find the flow rate of= gas and its exit temperature. (Ans. 72 kg/min, 300 K)
A steam power plant uses a shell and t= ube condenser to condense steam at 325 K. Water is used as a coolant and it= passes through the tubes at the rate of 1800 tons/min and removes heat at = the rate of 2000 MW. If the water enters at 295 K in a single shell and two= tube pass arrangement with the lengths and diameters of the tubes in one p= ass being 4.5 m and 2.5 cm, find the number of tubes required. Assume the o= verall heat transfer coefficient to be 4.5 kW/m²= K.
(Ans. 30,000= )

An evaporat= or with a heat transfer area of 3.5 m² and an ov= erall heat transfer coefficient of 0.0459 kW/m² = K is used evaporate water at 375 K using hot gas entering and leaving the h= eat exchanger at temperatures of 535 K and 425 K, respectively. Assuming th= at the specific heat of the gas is 1 kJ/kg K, find the mass flow rate of th= e gas required. (Ans. 8.33 kg/ min)

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Types of heat exchanger	U (W/m^2K)
Gas-to-gas	10=E2=80=9340
Water-to-air in finned tubes (water in t= ubes)	30=E2=80=9360
Water-to-oil	100=E2=80=93350
Water-to-gasoline or kerosene	300=E2=80=931000
Water-to-water	850=E2=80=931700
Feedwater heaters	1000=E2=80=938500
steam-to-air in finned tubes (steam in t= ubes)	30=E2=80=93300
steam-to-light fuel oil	200=E2=80=93400
steam-to-heavy fuel oil	50=E2=80=93200
steam condenser	1000=E2=80=936000
Freon condenser (water-cooled)	300=E2=80=931000
Ammonia condenser (water-cooled)	800=E2=80=931400
Alcohol condensers (water-cooled)	250=E2=80=93700

=20 =20 =20 =20 Previous Chapter
7. Condensation =20 =20

=20 =20 =20 =20 Next Chapter
9. Radiation Heat Tr= ansfer
=20 =20
=20

= REVIEW QUESTIONS

OBJECTIVE QUESTIONS

Answers:

= EXERCISE PROBLEMS

Table= of Contents for
=20 Heat and Mass Transfer, 2nd Edition =20

8

Heat Exchangers

KEY CONCEPTS

= 8.1 CLASSIFICATION OF HEAT EXCHANGERS

= 8.2 OVERALL HEAT TRANSFER COEFFICIENT OF HEAT EX= CHANGERS

KEY CONCEPTS

= 8.3 FOULING FACTORS

= 8.4 LOG MEAN TEMPERATURE DIFFERENCE (LMTD)

8.4.1 = LMTD for Parallel-flow Heat Exchangers

8.4.2 = LMTD for Counter-flow Heat Exchangers

8.4.3 = LMTD for Cross-flow Heat Exchangers

KEY= CONCEPTS

= 8.5 The NTU Method

8.5.1 = Heat Exchanger Effectiveness of Various Types of Heat Exchangers

8.5.1.1 Parallel-flow heat excha= nger

8.5.1.2 Counter-flow heat exchan= ger

8.5.1.3 Cross-flow heat exchange= r: both fluids unmixed

8.5.1.4 Cross-flow heat exchange= r: both fluids mixed

8.5.1.5 C= ross-flow type with C_max mixed and = C_min. unmixed

8.5.1.6 Cross-flow type with C_max unmixed and C_min. mixed

8.5.1.7 Shell-and-tube type: one= shell pass and 2, 4, 6 =E2=80=A6 tube passes

= 8.6 HEAT TRANSFER ENHANCEMENT

= 8.7 COMPACT HEAT EXCHANGERS

= 8.8 CONCLUSIONS

= EXAMPLE PROBLEMS

Table= of Contents for =20 Heat and Mass Transfer, 2nd Edition =20

8

Heat Exchangers

KEY CONCEPTS

= 8.1 CLASSIFICATION OF HEAT EXCHANGERS

= 8.2 OVERALL HEAT TRANSFER COEFFICIENT OF HEAT EX= CHANGERS

KEY CONCEPTS

= 8.3 FOULING FACTORS

= 8.4 LOG MEAN TEMPERATURE DIFFERENCE (LMTD)

8.4.1 = LMTD for Parallel-flow Heat Exchangers

8.4.2 = LMTD for Counter-flow Heat Exchangers

8.4.3 = LMTD for Cross-flow Heat Exchangers

KEY= CONCEPTS

= 8.5 The NTU Method

8.5.1 = Heat Exchanger Effectiveness of Various Types of Heat Exchangers

8.5.1.1 Parallel-flow heat excha= nger

8.5.1.2 Counter-flow heat exchan= ger

8.5.1.3 Cross-flow heat exchange= r: both fluids unmixed

8.5.1.4 Cross-flow heat exchange= r: both fluids mixed

8.5.1.5 C= ross-flow type with Cmax mixed and = Cmin. unmixed

8.5.1.6 Cross-flow type with Cmax unmixed and Cmin. mixed

8.5.1.7 Shell-and-tube type: one= shell pass and 2, 4, 6 =E2=80=A6 tube passes

= 8.6 HEAT TRANSFER ENHANCEMENT

= 8.7 COMPACT HEAT EXCHANGERS

= 8.8 CONCLUSIONS

= EXAMPLE PROBLEMS

= REVIEW QUESTIONS

OBJECTIVE QUESTIONS

Answers:

= EXERCISE PROBLEMS

Highlight

Highlight

Table= of Contents for
=20 Heat and Mass Transfer, 2nd Edition =20

8.5.1.5 C= ross-flow type with C_max mixed and = C_min. unmixed

8.5.1.6 Cross-flow type with C_max unmixed and C_min. mixed