Module7.pdf

Prof. Pradip Dutta

Heat and Mass Transfer

Module 7: Learning objectives •

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One of the most common applications of heat transfer is to design equipment for exchanging heat from one fluid to another. Such devices are generally called Heat Exchangers. Because there are many important applications, heat exchanger research and development has had a long history. Such activity is by no means complete, however as many talented workers continue to seek ways of improving deign and performance. From the view point of energy conservation and space constraint, there has been a steady and substantial increase in research activity. a focal point for this work has been heat transfer enhancement, which includes the search for special heat exchanger surfaces through which enhancement may be achieved. In this chapter we have attempted to develop tools that will allow you to perform approximate heat exchanger calculations. Although we have restricted ourselves to heat exchangers involving separation of hot and cold fluids by stationary walls, there are other important options. for example, evaporative heat exchangers enable direct contact between liquid and gas and because of latent energy effects, large heat transfer rates per unit volume are possible. Also for gas-to-gas heat exchange, use is often made of regenerators in which the same space is alternately occupied by hot and cold gases. In a fixed regenerator such as a packed bed, the hot and cold gases alternately enter a stationary, porous solid in a rotary regenerator, the porous solid is a rotation wheel, which alternately exposes its surfaces to the continuously flowing hot and cold gases.

Indian Institute of Science Bangalore

Prof. Pradip Dutta


MODULE 7

HEAT EXCHANGERS 7.1 What are heat exchangers?

Heat exchangers are practical devices used to transfer energy from one fluid to another. Around the household, we are accustomed to seeing the condensers and evaporators used in air conditioning units. In automobiles we see radiators and oil coolers. In the power industry we see boilers, condensers, economizers, pre-heaters and numerous other heat exchangers. Within the process industry, we find heat exchangers used extensively for a variety of purposes. Because of the wide variety of uses for heat exchangers, their construction may vary widely. We will consider only the more common types here, but the considerations included are common to all types. In industrial drawings it is common to use the abbreviation HX to indicate heat exchangers. We will use this terminology here to shorten the discussions. 7.2 Heat Transfer Considerations The energy flow between hot and cold streams, when viewed from one end of the heat exchanger, will appear as shown in Figure to the right. Heat transfer will occur by convection to the outside of the inner tube, by conduction across the tube and by convection to the cooler fluid from the inside tube surface. Since the heat transfer occurs across the smaller tube, it is this internal surface which controls the heat transfer process. By convention, it is the outer surface, termed Ao, of this central tube which is referred to in describing heat


t

di do

T

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exchanger area. Applying the electrical analogy, an equivalent thermal resistance may be defined for this tube. r ln⎛⎜ o r ⎞⎟ ⎝ i⎠ 1 1 R= + + ho Ao hi Ai 2π ⋅ kl

If we define the heat exchanger coefficient, Uc, as: Uc ≡

1 RAo

Substituting the value of R above this yields: r ro ln⎛⎜ o r ⎞⎟ ⎝ i⎠ A 1 1 = + + o U c ho k hi Ai

Both convective coefficients, ho and hi, can be evaluated from experimentally developed convective correlations. Areas and radii are determined from the geometry of the internal tube. The thermal conductivity, k, corresponds to that for the material of the internal tube. In this fashion each of the terms are generally available for determining Uc and the term is well defined for most heat exchangers. 7.3 Fouling Material deposits on the surfaces of the heat exchanger tube may add further resistances to heat transfer in addition to those listed above. Such deposits are termed fouling and may significantly affect heat exchanger performance. The heat exchanger coefficient, Uc, determined above may be modified to include the fouling factor Rf. 1 1 = + R" Ud Uc

• Scaling is the most common form of fouling and is associated with inverse solubility salts. Examples of such salts are CaCO3, CaSO4, Ca3(PO4)2, CaSiO3, Ca(OH)2, Mg(OH)2, MgSiO3, Na2SO4, LiSO4, and Li2CO3. The characteristic which is termed inverse solubility is that, unlike most inorganic materials, the solubility decreases with temperature. The most important of these compounds is calcium carbonate, CaCO3. Calcium carbonate exists in several forms, but one of the more important is limestone. The material frequently crystallizes in a form closely resembling marble, another form of calcium carbonate. Such


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materials are extremely difficult to remove mechanically and may require acid cleaning. Corrosion fouling is classified as a chemical reaction which involves the heat exchanger tubes. Many metals, copper and aluminum being specific examples, form adherent oxide coatings which serve to passivate the surface and prevent further corrosion. Metal oxides are a type of ceramic and typically exhibit quite low thermal conductivities. Even relative thin coatings of oxides may significantly affect heat exchanger performance and should be included in evaluating overall heat transfer resistance. Chemical reaction fouling involves chemical reactions in the process stream which results in deposition of material on the heat exchanger tubes. When food products are involved this may be termed scorching but a wide range of organic materials are subject to similar problems. This is commonly encountered when chemically sensitive process fluids are heated to temperatures near that for chemical decomposition. Because of the no flow condition at the wall surface and the temperature gradient which exists across this laminar sub–layer, these regions will operate at somewhat higher temperatures than the bulk and are ideally suited to promote favorable conditions for such reactions. Freezing fouling is said to occur when a portion of the hot stream is cooled to near the freezing point for one of its components. This is most notable in refineries where paraffin frequently solidifies from petroleum products at various stages in the refining process, obstructing both flow and heat transfer. Biological fouling is common where untreated water is used as a coolant stream. Problems range from algae or other microbes to barnacles. During the season where such microbes are said to bloom, colonies several millimeters deep may grow across a tube surface virtually overnight, impeding circulation near the tube wall and retarding heat transport. Viewed under a microscope, many of these organisms appear as loosely intertwined fibers– much like the form of fiberglass insulation Traditionally these organisms have been treated which chlorine, but present day concerns on possible contamination to open water bodies has severely restricted the use of oxidizers in open discharge systems. Particulate fouling results from the presence of Brownian sized particles in solution. Under certain conditions such materials display a phenomenon known as thermophoresis in which motion is induced as a result of a temperature gradient. Thermodynamically this is referred to as a cross-coupled phenomenon and may be viewed as being analogous to the Seabeck effect. When such particles accumulate on a heat


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exchanger surface they sometimes fuse, resulting in a buildup having the texture of a sandstone. Like scale these deposits are difficult to remove mechanically. Most of the actual data on fouling factors is tightly held be a few specialty consulting companies. The data which is commonly available is sparse. An example is given below: Fluid

R”, m K/Watt 0.0001 0.0002 0.0002-0.001 0.0009 0.0002 0.0001 2

Seawater and treated boiler feedwater (below 50oC) Seawater and treated boiler feedwater (above 50oC) River water (below 50oC) Fuel Oil Regrigerating liquids Steam (non-oil bearing)

Table: Representative Fouling Factors 1 = Ud

r ln( o r ) 1 1 i + + + R" k ho ⎛⎜ ri ⎞⎟ hi ⋅ ⎝ r ⎠ o

7.4 Basic Heat Exchanger Flow Arrangements Basic flow arrangements are as shown in the Figure below. Parallel and counterflow provide alternative arrangements for certain specialized applications. In parallel flow both the hot and cold streams enter the heat exchanger at the same end and travel to the opposite end in parallel streams. Energy is transferred along the length from the hot to the cold fluid so the outlet temperatures asymptotically approach one another. In counter flow the two streams enter at opposite ends of the heat exchanger and flow in opposite directions. Temperatures within the two streams tend to approach one another in a nearly linearly fashion resulting in a much more uniform heating pattern. Shown below the heat exchangers are representations of the axial temperature profiles for each. Parallel flow results in rapid initial rates of heat exchange but rates rapidly decrease as the temperatures of the two streams approach one another. Counter flow provides for relatively uniform temperature differences and, consequently, lead toward relatively uniform heat rates throughout the length of the unit.


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t1

t2

Temperature

T1

T2

Parallel Flow

T1

t1

T2 t1

t2

T2

Counter Flow

T1 Temperature

T1

t2

T2 t2

t1

Position

Position

Basic Flow Arrangements for Tube in Tube Heat Exchangers.

7.5 Log Mean Temperature Differences Heat flows between the hot and cold streams due to the temperature difference across the tube acting as a driving force. As seen in the Figure below, the difference will vary with axial position within the HX so that one must speak in terms of the effective or integrated average temperature differences.

Counter Flow

T1

Parallel Flow

1

T1

t1

t2

2

t1

2

t2

T2

1

T2

Position

Position

Temperature Differences Between Hot and Cold Process Streams Working from the three heat exchanger equations shown above, after some development it if found that the integrated average temperature difference for either parallel or counter flow may be written as: Δθ = LMTD =

θ1 − θ 2 ⎛θ ⎞ ln⎜ 1 ⎟ ⎝θ2 ⎠

The effective temperature difference calculated from this equation is known as the log mean temperature difference, frequently abbreviated as LMTD, based on the type of mathematical average that it describes. While the equation applies to either parallel or counter flow, it can be shown that Δθeff will always



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be greater in the counter flow arrangement. This can be shown theoretically from Second Law considerations but, for the undergraduate student, it is generally more satisfying to arbitrarily choose a set of temperatures and check the results from the two equations. The only restrictions that we place on the case is that it be physically possible for parallel flow, i.e. θ1 and θ2 must both be positive. Another interesting observation from the above Figure is that counter flow is more appropriate for maximum energy recovery. In a number of industrial applications there will be considerable energy available within a hot waste stream which may be recovered before the stream is discharged. This is done by recovering energy into a fresh cold stream. Note in the Figures shown above that the hot stream may be cooled to t1 for counter flow, but may only be cooled to t2 for parallel flow. Counter flow allows for a greater degree of energy recovery. Similar arguments may be made to show the advantage of counter flow for energy recovery from refrigerated cold streams.

7.6 Applications for Counter and Parallel Flows We have seen two advantages for counter flow, (a) larger effective LMTD and (b) greater potential energy recovery. The advantage of the larger LMTD, as seen from the heat exchanger equation, is that a larger LMTD permits a smaller heat exchanger area, Ao, for a given thermal duty, Q. This would normally be expected to result in smaller, less expensive equipment for a given application. This should not lead to the assumption that counter flow is always a superior. Parallel flows are advantageous (a) where the high initial heating rate may be used to advantage and (b) where the more moderate temperatures developed at the tube walls are required. In heating very viscous fluids, parallel flow provides for rapid initial heating. The quick decrease in viscosity which results may significantly reduce pumping requirements through the heat exchanger. The decrease in viscosity also serves to shorten the distance required for flow to transition from laminar to turbulent, enhancing heat transfer rates. Where the improvements in heat transfer rates compensate for the lower LMTD parallel flow may be used to advantage. A second feature of parallel flow may occur due to the moderation of tube wall temperatures. As an example, consider a case where convective coefficients are approximately equal on both sides of the heat exchanger tube. This will result in the tube wall temperatures being about the average of the two stream temperatures. In the case of counter flow the two extreme hot temperatures are at one end, the two extreme cold temperatures at the other. This produces relatively hot tube wall temperatures at one end and relatively cold temperatures at the other. Temperature sensitive fluids, notably food


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products, pharmaceuticals and biological products, are less likely to be “scorched” or “thermally damaged” in a parallel flow heat exchanger. Chemical reaction fouling may be considered as leading to a thermally damaged process stream. In such cases, counter flow may result in greater fouling rates and, ultimately, lower thermal performance. Other types of fouling are also thermally sensitive. Most notable are scaling, corrosion fouling and freezing fouling. Where control of temperature sensitive fouling is a major concern, parallel flow may be used to advantage. 7.7 Multipass Flow Arrangements In order to increase the surface area for convection relative to the fluid volume, it is common to design for multiple tubes within a single heat exchanger. With multiple tubes it is possible to arrange to flow so that one region will be in parallel and another portion in counter flow. An arrangement where the tube side fluid passes through once in parallel and once in counter flow is shown in the Figure below. Normal terminology would refer to this arrangement as a 12 pass heat exchanger, indicating that the shell side fluid passes through the unit once, the tube side twice. By convention the number of shell side passes is always listed first. The primary reason for using multipass designs is to increase the average tube side fluid velocity in a given arrangement. In a two pass arrangement the fluid flows through only half the tubes and any one point, so that the Reynold’s number is effectively doubled. Increasing the Reynolds’s number results in increased turbulence, increased Nusselt numbers and, finally, in increased convection coefficients. Even though the parallel portion of the flow results in a lower effective ΔT, the increase in overall heat transfer coefficient will frequently compensate so that the overall heat exchanger size will be smaller for a specific service. The improvements achievable with multipass heat exchangers is sufficiently large that they have become much more common in industry than the true parallel or counter flow designs. The LMTD formulas developed earlier are no longer adequate for multipass heat exchangers. Normal practice is to calculate the LMTD for counter flow, LMTDcf, and to apply a correction factor, FT, such that Δθ eff = FT ⋅ LMTDCF


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The correction factors, FT, can be found theoretically and presented in analytical form. The equation given below has been shown to be accurate for any arrangement having 2, 4, 6, .....,2n tube passes per shell pass to within 2%. ⎡ 1− P ⎤ R 2 + 1 ln ⎢ ⎣1 − R ⋅ P ⎥⎦ FT = ⎡2 − P R + 1− R2 + 1 ⎤ ⎥ ( R − 1) ln ⎢⎢ ⎥ 2 ⎢⎣ 2 − P R + 1 + R + 1 ⎥⎦

( (

) )

where the capacity ratio, R, is defined as:

T1 − T2 t 2 − t1

R=

The effectiveness may be given by the equation: 1 − X 1/ N shell P= R − X 1/ N shell

provided that R≠1. In the case that R=1, the effectiveness is given by: P=

N shell

Po − Po ⋅ ( N shell − 1)

where Po =

and X =

t2 − t1 T1 − t1

Po ⋅ R − 1 Po − 1

As an alternative to using the formulas for the correction factors, which can become tedious for non-computerized calculations, charts are available. Several are included in standard texts. Experience has shown that, due to variability in reading charts, considerable error can be introduced into the calculations and the equations are recommended. When charts are used, they should be reproduced at a sufficiently large scale, and considerable care should be used in making interpolations.



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7.8 Limitations of Multipass Arrangements Since the 1-2 heat exchanger uses one parallel pass and one counter current, it follows that the maximum heat recovery for these units should be between that of parallel and counter flow. As a practical limit it is Figure . Temperature Profiles for a 1-2 HX with a important that nowhere in the Temperature Cross. unit should the cold fluid temperature exceed that of the hot fluid. If so, then heat transfer is obviously in the wrong direction. Such a situation can arise in a multipass heat exchanger as seen in Figure 6. This unit represents a cold fluid, located on the tube side of the heat exchanger, making two passes through the unit, the hot fluid, on the shell side, traveling across the unit only once. Here the cold fluid is heated to a temperature slightly above that of the hot fluid near the exit for the two streams. At this axial location, near the left end of the unit, the temperature of the cold fluid in the first pass remains well below that of the hot fluid so that considerable heat transfer occurs. The cold fluid in the second pass is slightly above that of the hot fluid at the same location. The small temperature difference between the second pass cold fluid and the hot stream, indicates that only a small amount of heat will be transferred between these streams. Overall heat will flow from hot to cold fluid, but a portion of the heat transfer surface is being used in a counter productive way. This condition is termed as a temperature cross. In the limit the hot fluid exit temperature could be cooled to the average of the cold fluid inlet and exit temperature. This would, however, be highly inefficient and would require an excessively large surface area. Some engineers advocate that good design should not permit a temperature cross, indicating that the 1-2 should operate with the same heat recovery limit as a true parallel flow. The preferred method of attaining additional heat recovery is to stage heat exchangers in series so that no temperature cross occurs in any unit. An equivalent solution is to put multiple 1–n arrangements within a single shell. A 2–4 unit is the equivalent of 2 1–2 units provided that the total heat transfer area is equal. Similarly a 3–6 unit is the equivalent of 3 1–2 units with equal overall area. Other engineers suggest that a small temperature cross may be acceptable and may provide a less expensive design than the more complex alternatives. If one were to plot the locus of points where the temperature cross occurs for the 1-2 heat exchanger on the temperature correction chart, it would be found to correspond to a relatively narrow range of FT values ranging from about 0.78 to 0.82. Lower values of FT may be taken as an indication that a temperature cross will occur.


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A second consideration is that at lower FT values the slope, dFT/dP, becomes extremely steep. This is an indication that the temperature efficiency, P, is asymptotically approaching its upper limit and the design has no margin to accommodate uncertainties. A good rule of thumb is that the minimum slope of dFT/dP, which is negative, should not fall below -1.5. Instead a 2-4 or even a 3-6 should be selected to provide the needed operational design margin.. Similar restrictions exist for these designs as well. In the limit a counter flow design may be the only suitable selection for high heat recovery applications.

7.9 Effectiveness-NTU Method: In our previous discussions, we have been looking at practical HX designs using the LMTDCF with a Ft correction factor to account for the mixed flow conditions. Now we wish to consider an alternate, more recent approach that is in common use today. This is the effectiveness-NTU method. Effectiveness, ε Consider two counter-flow heat exchangers, one in which the cold fluid has the larger ΔT (smaller m⋅cp) and a second in which the cold fluid has the smaller ΔT (larger m⋅cp): Δt > ΔT M⋅Cp > m⋅cp T2

T1

ΔT > Δt m⋅cp > M⋅Cp

t2 T2

t1

T1

t2

t1

We may see in the first case that, because the cold fluid heat capacity is small, its temperature changes rapidly. If we seek efficient energy recovery, we see that in the limit a HX could be designed in which the cold fluid exit temperature would reach that of the hot fluid inlet. In the second case, the hot fluid temperature changes more rapidly, so that in the limit the hot fluid exit temperature would reach that of the cold fluid inlet.


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The effectiveness is the ratio of the energy recovered in a HX to that recoverable in an ideal HX. ε=

m ⋅ c p ⋅ (t2 − t1 )

m ⋅ c p ⋅ (T1 − t1 )

⋅ C ⋅ (T − T ) M p 1 2 ε= ⋅ Cp ⋅ (T1 − t1 ) M

Δt > ΔT

ΔT > Δt

Canceling identical terms from the numerator and denominator of both terms: ε=

(t − t ) (T − t ) 2

1

1

1

ε=

Δt > ΔT

(T − T ) (T − t ) 1

2

1

1

ΔT > Δt

We see that the numerator, in the two cases, is the temperature change for the stream having the larger temperature change. The denominator is the same in either case: ε=

Δ Tmax

In the LMTD-Ft method an effectiveness was defined:

(T − t ) 1

1

P=

t 2 − t1 T1 − t1

Note that the use of the upper case T in the numerator, in contrast to our normal terminology, does not indicate that the hot fluid temperature change is used here. The max subscript over-rides the normal terminology and indicates that this refers to the side having the larger temperature change. Number of Transfer Units (NTU) Recall that the energy flow in any HX is described by three equations: Q = U⋅A⋅Δθeff Q = -M⋅Cp⋅ΔT Q = m⋅cp⋅Δt

HX equation 1st Law Equation 1st Law Equation

We may generalize the latter two expressions, using ε-NTU terminology as follows: Q = (M⋅Cp)min⋅ΔTmax Q = (M⋅Cp)max⋅ΔTmin


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Again the use of the upper case letters is over-ridden by the use of the subscripts. If we eliminate Q between the HX equation and one of the 1st Law equations, U⋅A⋅Δθeff = (M⋅Cp)min⋅ΔTmax This expression may be made non-dimensional by taking the temperatures to one side and the other terms to the other side: NTU ≡

(

U⋅A ⋅C M p

)

=

Δ Tmax Δ θ eff

min

Physically we see that a HX with a large product U⋅A and a small (M⋅Cp)min should result in a high degree of energy recovery, i.e. should result in a large effectiveness, ε. Capacity Ratio, CR The final non-dimensional ratio needed here is the capacity ratio, defined as follows: CR ≡


( M ⋅ Cp ) min ( M ⋅ Cp ) max

=

Δ Tmin Δ Tmax

In the LMTD-Ft method a capacity ratio was defined: R=

T1 − T2 t 2 − t1

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ε-NTU Relationships In the LMTD-Ft method, we found a general equation which described Ft for all 1-2N, 2-4N, 3-6N, etc. heat exchangers. Another relationship, not given here, is required for cross flow arrangements. In a similar fashion, we may develop a number of functional relationships showing ε = ε (NTU, CR) or, alternatively: NTU = NTU(ε,CR) These relationships are shown in tables in standard text books. For example, we find that the relationship for a parallel flow exchanger is:

ε=

1− e 1 − CR ⋅ e − NTU ⋅(1− CR ) − NTU ⋅ ( 1− C R )

Note: These correlations are not general. Specific correlations will be given for different kind of HX.

The ε-NTU method offers a number of advantages to the designer over the traditional LMTD-Ft method. One type of calculation where the ε-NTU method may be used to clear advantage would be cases in which neither fluid outlet temperature is known.


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Module 7: Solved Problems 1. A thin-walled concentric tube heat exchanger of 0.19-m length is to be used to heat deionized water from 40 to 60°C at a flow rate of 5 kg/s. the deionized water flows through the inner tube of 30-mm diameter while hot process water at 95°C flows in the annulus formed with the outer tube of 60-mm diameter. The thermo physical properties of the fluids are:

Considering a parallel-flow configuration of the exchanger, determine the minimum flow rate required for the hot process water. Determine the overall heat transfer coefficient required for the conditions of part a. Considering a counter flow configuration, determine the minimum flow rate required for the hot process water. What is the effectiveness of the exchanger for this situation? Known: Thin-walled concentric tube, Parallel flow heat exchanger of prescribed diameter and length with process and deionized water. Inlet and outlet temperatures and flow rate of desired water. Inlet temperature and outlet temperature and flow rate of deionized water. Inlet temperature of process water. Find: (1) minimum flow rate required for the hot process water, (b) required overall heat transfer coefficient and whether it is possible to accomplish this heating, and (c) for CF arrangements minimum process water flow required and the effectiveness?


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Schematic:

Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes. Analysis: (a) from overall energy balances, .

.

q = ( m c ) h (Th,i − Th,o ) = ( m c ) h (Tc ,o − Tc ,i ) .

For a fixed term Th,i , (m )h will be a minimum when Th,o is a minimum. With the parallel flow configuration, this requires that Th,o=Tc,o=60°C. Hence, .

.

m h, min =

( mc ) c (Tc ,o − Tc ,i ) c h (Th,i − Th,o )

=

5kg / s × 4181J / kg . K (60 − 40) ° C = 2.85kg / s 4197J / kg . K (95 − 60)°C

(b)From the rate equation and the log mean temperature relation, q = UAΔTlm , PF

ΔTlm , PF =

ΔT1 - ΔT2 ΔT ⎞ ln⎛⎜ 1 ⎟ Δ T 2⎠ ⎝

And since ΔT2=0, ΔTlm=0 so that UA=∞. Since A=πDL is finite, U must be extremely large. Hence, the heating cannot be accomplished with this arrangement. .

(c) With the CF arrangements m h will be a minimum when Tho is a minimum. This requires that Th,o is a minimum. This requires that Th,o is


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a minimum. This requires that Th,o=Tc,i=40°C. Hence, from the overall energy balance, .

m=

5kg / s × 4181J / kg . K (60 − 40) K = 1.81kg / s 4197J / kg . K (95 − 40) K

For this condition, Cmin=Ch which is cooled from Th,i to Tc,i, hence ε=1 Comments: For the counter flow arrangement, the heat exchanger must be infinitely long.


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2. An automobile radiator may be viewed as a cross-flow heat exchanger with both fluids unmixed. Water, which has flow rate of 0.05kg/s, enters the radiator at 400K and is to leave at 330 K. The water is cooled by air which enters at 0.75kg/s and 300K. If the overall heat transfer coefficient is 200W/m2.K, what is the required heat transfer surface area? Known: flow rate and inlet temperature for automobile radiator. Overall heat transfer coefficient. Find: Area required to achieve a prescribed outlet temperature. Schematic:

Assumptions: (1) Negligible heat loss to surroundings and kinetic and potential energy changes, (2) Constant properties. Analysis: The required heat transfer rate is .

q = ( m c ) h (Th ,i − Th,o ) = 0.05kg / s(4209 J / kg . K )70 K = 14,732W

Using the ε-NTU method,


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C min = C h = 210.45W / K C max = C c = 755.25W / K , hence , C min / C max (Th,i − Tc ,i ) = 210.45W / K (100 K ) = 21,045W and

ε = q / q max = 14,732W / 21,045W = 0.700

From figure, NTU≈1.5, hence A = NTU (C min / U ) = 1.5 × 210.45W / K ( 200W / m 2 . K ) = 1.58m 2

Comments: (1) the air outlet temperature is Tc ,o = Tc ,i + q / C c = 300 K + (14,732W / 755.25W / K ) = 319.5 K

(2) Using the LMTD approach, ΔTlm=51.2 K, R=0.279 and P=0.7. Hence from fig F≈0.95 and A = q / FUΔTlm = (14,732W ) /[0.95( 200W / m 2 . K )51.2 K ] = 1.51m 2 .


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3. Saturated water vapor leaves a steam turbine at a flow rate of 1.5kg/s and a pressure of 0.51 bars. The vapor is to be completely condensed to saturated liquid in a shell-and –tube heat exchanger which uses city water as the cold fluid. The water enters the thin-walled tubes at 17°C and is to leave at 57°C. assuming an overall heat transfer coefficient of 200W/m2.K, determine the required heat exchanger surface area and the water flow rate. After extended operation, fouling causes the overall heat transfer coefficient to decrease to 100W/m2.K, and to completely condense the vapor, there must be an attendant reduction in the vapor flow rate. For the same water inlet temperature and flow rate, what is the new vapor flow rate required for complete condensation? Known: Pressure and initial flow rate of water vapor. Water inlet and outlet temperatures. Initial and final overall heat transfer coefficients. Find: (a) Surface area for initial U and water flow rate, (b) Vapour flow rate for final U. Schematic:

Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible wall conduction resistance. Properties: Table for sat.Water: −

(T c = 310 K ) : c p,c = 4178 J / kg . K ; (p = 0.51 bars) : Tsa t = 355K, h fg = 2304kJ/kg.

Analysis: (a) The required heat transfer rate is .

q = m h h fg = 1.5kg / s( 2.304 × 10 6 J / kg ) = 3.46 × 10 6 W

And the corresponding heat capacity rate for the water is


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C c = C min = q /(Tc ,o − Tc ,i ) = 3.48 × 10 6 W / 40 K = 86,400W / K hence , ε = q /(C min [Th,i − Tc ,i ]) = 3.46 × 10 6 W / 86,400W / K (65 K ) = 0.62 since C min /C max = 0, NTU = -ln(1 - ε ) = − ln(1 − 0.62) = 0.97

And A = NTU (C min / U ) = 0.97(86,400W / K / 2000W / m 2. K ) = 41.9m 2 .

m c = C c / c p ,c = 86,400W / K / 4178 J / kg . K = 20.7 kg / s

(b) using the final overall heat transfer coefficient, find Since C min /C max = 0, ε = 1 − exp(− NTU ) = 1 − exp(−0.485) = 0.384 hence, q = εC min (Th ,i − Tc ,i ) = 0.384(886,400W / K )65 K = 2.16106W .

m h = q / h fg = 2.16 × 10 6 W / 2.304 × 10 6 J / kg = 0.936kg / s

.

Comments: The significant reduction (38%) in m h represents a significant loss in turbine power. Periodic cleaning of condenser surfaces should be employed to minimize the adverse effects of fouling.


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4. Water at 225 kg/h is to be heated from 35 to 95°C by means of a concentric tube heat exchanger. Oil at 225kg/h and 210°C, with a specific heat of 2095 J/kg.K, is to be used as the hot fluid, If the overall heat transfer coefficient based on the outer diameter of the inner tube if 550W/m2.K,determine the length of the exchanger if the outer diameters is 100mm. Known: Concentric tube heat exchanger. Find: Length of the exchanger Schematic:

Assumptions: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes, (3) Constant properties. Properties: Table for Water: _

(Tc = ( 35 + 95) ° C / 2 = 338 K ) : c p ,c = 4188 J / kg . K

Analysis: From rate equation with Ao=πDoL, L=q/UoDoΔT A m The heat rate, q, can be evaluated from an energy balance on the cold fluid, .

q = m c c c (Tc , 0 − Tc ,i ) =

225kg / h × 4188 J / kg . K (95 − 35) K = 15,705W 3600 s / h

In order to evaluate ΔT A m, we need to know whether the exchanger is operating in CF or PF. From an energy balance on the hot fluid, find .

Th,o = Th,i − q / m h c h = 210 ° C − 15,705W /


225kg / h J × 2095 = 90.1°C 3600s / h kg . K

Prof. Pradip Dutta


Since Th,o
ΔT1 − ΔT2 ( 210 − 95) − (90.1 − 35) = °C = 81.5°C An( ΔT1 / ΔT2 ) An(115 / 55.1)

Substituting numerical values, the HXer length is L = 15,705W / 550W / m 2 . Kπ (0.10m ) × 81.4 K = 1.12m

Comments: The ε–NTU method could also be used. It would necessary to perform the hot fluid energy balance to determining operation existed. The capacity rate is Cmin/Cmax=0.50. From eq. effectiveness, and from with q evaluated from an energy balance on hot fluid, ε=

Th ,i − Th,o Th,i − Tc ,i

=

210 − 90.1 = 0.69 210 − 35

From fig, find NTU≈1.5 giving L = NTU .C min / U oπDo ≈ 1.5 × 130.94

W W 550 2 .π (0.10m ) ≈ 1.14m K m .K

Note the good agreement by both methods.


be CF for the

Prof. Pradip Dutta


5. Consider a very long, concentric tube heat exchanger having hot and cold water inlet temperatures of 85 and 15°C. The flow rate of the hot water is twice that of the cold water. Assuming equivalent hot and cold water specifies heats; determine the hot water outlet temperature for the following modes of operation (a) Counter flow, (b) Parallel flow. Known: A very long, concentric tube heat exchanger having hot and cold water inlet temperatures of 85 and 15°C, respectively: flow rate of the hot water is twice that of the cold water. Find: outlet temperatures for counter flow and parallel flow operations. Schematic:

Assumptions: (1) equivalent hot and cold water specific heats, (2) Negligible Kinetic and potential energy changes, (3) No eat loss to surroundings. Analysis: the heat rate for a concentric tube Heat exchanger with very large surface area

Operating in the counter flow mode is q = qmax = C min (Th,i − Tc ,i )

Combining the above relation and rearranging, find Th,o = −

C C min (Th,i − Tc ,i ) + Th,i = − c (Th,i − Tc ,i ) + Th,i Ch Ch

Substituting numerical values


Prof. Pradip Dutta


1 Th,o = − (85 − 15)°C + 85°C = 50°C 2

For parallel flow operation, the hot and cold outlet temperatures will be equal; that is Tc,o=Th,o. Hence C c (Tc ,o − Tc ,i ) = C h (Th,i − Th,o )

Setting Tc,o=Th,o and rearranging ⎡ ⎤ ⎡ C ⎤ C Th,o = ⎢Th,i + c Tc ,i ⎥ / ⎢1 + c ⎥ Ch ⎣ ⎦ ⎣ Ch ⎦ 1 ⎡ ⎤ ⎡ Th,o = ⎢85 + × 15⎥ °C / ⎢1 + 2 ⎦ ⎣ ⎣

1⎤ = 61.7 ° C ⎥ 2⎦

Comments: Note that while ε =1 for CF operation, for PF operation find ε= q/qmax=0.67.


Prof. Pradip Dutta


6. A concentric tube heat exchanger uses water, which is available at 15C, to cool ethylene glycol from 100 to 60C. The water and glycol flow rates are each 0.5 kg/s. What are the maximum possible heat transfer rate and effectiveness of the exchanger? Which is preferred, a parallel –flow or counter flow mode of operation? Known: Inlet temperatures and flow rate for a concentric tube heat exchanger. Find: (a) Maximum possible heat transfer rate and effectiveness, (b) Proffered mode of operation. Schematic:

Assumptions: (1) Steady-state operation, (2) Negligible KE and PE changes, (3) Negligible heat loss to surroundings, (4) Fixed overall heat transfer and coefficient. _

Properties: Table: Ethylene glycol ( T in = 80°C ); cp=2650J/kg.K; _

Water (Tm ≈ 30°C ) : c p = 4178 J / kg . K

Analysis: (a) Using the ε-NTU method, find .

C min = C h = m h c p , h = ( 0.5kg / s )( 2650 J / kg . K ) = 1325W / K

q ma x = C min (Th. i − Tc ,i ) = (1325W / K )(100 − 15)°C = 1.13 × 10 5 W .

q = m h c p ,h (Th. i − Tc ,i ) = 0.5kg / s( 2650 J / kg . K )(100 − 60)°C = 0.53 × 10 5 W

ε = q / qmax = 0.53 × 10 5 / 1.13 × 10 5 = 0.47


Prof. Pradip Dutta


(b) Tc ,o = Tc ,i +

q .

= 15°C +

m c c p ,c

0.53 × 10 5 = 40.4°C 0.5kg / s × 4178 J / kg . K

Since Tc,o
.

with (Cmin/Cmax) = ( m c p ,h / m c c p , c ) =0.63, h

From fig (NTU)PF≈0.95, (NTU)CF≈0.75 Hence (ACF/APF)= (NTU) CF/ (NTU) PF ≈(0.75/0.95)=0.79 Because of the reduced size requirement, hence capital investment, the counter flow mode of operation is proffered.


Prof. Pradip Dutta


Module 7: Short questions 1. What are the heat transfer mechanisms involved during heat transfer from the hot fluid to the cold fluid? 2. In heat exchange between air and water across a tube wall, it is proposed to use fins to enhance the overall heat transfer coefficient. Would you put the fins on the air side or on the water side? 3. When is a heat exchanger classified as compact? 4. How does a cross flow heat exchanger differ from a counter flow one? 5. What is the role of baffles in a shell-and-tube heat exchanger? What is the implication about pressure drop? 6. Under what conditions is the effectiveness NTU method preferred over LMTD method as a method of analysis of a heat exchanger? 7. Can temperature of the hot fluid drop below the inlet temperature of the cold fluid at any location in a heat exchanger? 8. Can temperature of the cold fluid rise above the inlet temperature of the hot fluid at any location in a heat exchanger? 9. Consider two double pipe counterflow heat exchangers that are identical except that one is twice as long as the other one. Which of the exchangers is more likely to have a higher effectiveness? 10. Can effectiveness be greater than one? 11. Under what conditions can a counter flow heat exchanger have an effectiveness of one? What would be your answer for a parallel flow heat exchanger?

Multiple choice questions: 1) In a thin walled heat exchanger with no fouling, the overall heat transfer coefficient is a) A(hi−1 + ho−1 )

−1

b) (hi−1 + ho−1 )

−1

c) A(hi−1 + ho−1 )

d) (hi−1 + ho−1 ) e) None of the above 2) In a liquid to gas heat exchanger, it is best to put extended surfaces on the gas side because a) This reduces fouling b) The gas side heat transfer coefficient is highest c) It reduces drag in high speed flows d) All of the first three e) None of the above 3) When applying the ε - NTU method for heat exchangers, when one fluid is condensing steam, the heat capacity ratio Cr is effectively a) 0 b) 1 c) π d) ∞ e) None of the above



4) On one side of a heat exchanger, air enters at 72.82ºC and leaves at 90ºC. On the other side of the heat exchanger is condensing steam at one atmosphere. The value for ΔTlmtd is a) 10 K b) 17.18 K c) 27ºC d) 100ºC e) None of the above 5) Select the FALSE statement concerning the ε -NTU method for heat exchangers a) qmax = Cmin (Th,i - Tc,i) b) ε = q/qmax c) NTU = UA/Cmin d) q = ε Cmin (Th,i - Tc,i) e) None of the above


Prof. Pradip Dutta

Module7.pdf

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