1. Typical Heat Exchange Equipment Energy balances Heat flux and heat transfer coefficients Temperature difference Overall heat transfer coefficient Heat Exchanger Analysis

2.  Heat transfer across a solid wall separating two liquids –latent heat (phase change) or sensible heat (∆T without phase change).  All such operations need heat transfer by conduction and convection  Simple tubular condenser 2

3.  If vapour entering is single component (not a mixture), not superheated, and the condensate not sub cooled then shell side temp. is constant T - point temperature difference (T 1 -inlet, T 2 - outlet)  Terminal point temperature differences = approaches  Temperature range or range, T cb – T ca 3

4. Counterflow or countercurrent flow  T ha – Temperature of hot fluid entering  T hb – Temperature of hot fluid leaving  T ca – Temperature of cold fluid entering   T ca – Temperature of cold fluid leaving 4

5.  Approaches : T 1 = T ha – T ca T 2 = T hb - T cb   Warm fluid range = T ha - T hb  Cold fluid range = T cb - T ca   Parallel flow – Fluids enter same end & flow out same end  Parallel flow: rarely used for single pass T hb >> T ca & T ha >> T cb 5

6. H/E mechanical, potential & kinetic energies << enthalpy, hence for each fluid stream:   m ( H b – H a ) = Q  q = Q/A  If the shell side fluid is hotter or colder than the ambient temperature then undesired heat loss/gain could result so lagging (insulation) is necessary. 6

7.   For the warm fluid, enthalpy balance:  m h (H hb – H ha ) = Q h  heat lost = heat gained : Q c = - Q h   Overall enthalpy balance   m c (H cb - H ca ) = m h ( H ha – H hb ) = Q   m h Cp h ( T ha – T hb ) =m c Cp c (T cb - T ca ) = Q 7

8.   For condensers with phase change:   m h λ = m c Cp c (T cb - T ca ) = Q  assuming no superheated vapour, no subcooled condensate   Otherwise :  m h [λ +Cp h ( T h – T hb )] =m c Cp c (T cb - T ca )  HEAT FLUX = Rate of heat transfer per unit area [Wm -2 ], [Btuhr -1 ft -2 ] 8

9. Important to specify whether internal or external surface areas being used (choice is arbitrary but order of magnitude different) Heat flux across solid layers proportional to driving force T  Q/A = ∆T/R also true for liquid layers   H/E driving force : T h -T c  hot fluid avg. temp cold fluid avg. temp 9

10.  ∆T, hence heat flux, varies along length of H/E   dQ/dA = U∆T = U(T h – T c )   Proportionality factor U, (overall heat transfer coefficient) [Wm -2 o C -1 ]   For external tube area A = A o, U = U o  For internal tube area A = A i, U = U i  U o = A i = D i  U i A o D o 10

11.  U o = dA i = D i U i dA o D o  For plate type H/E the areas for both sides are the same so there would be only one value for U.  Q = UA∆TQ = ∆T  U A  The value of U is important for designing any cooling or heating system  Inner diameter Outter diameter 11

12.  Assuming no accumulation of heat in the media:  Q = h 1 A ∆ T 1 Q = h 2 A∆T 2 Q = h 3 A ∆ T 3   ∆T 1 = Q/h 1 A ∆T 2 = Q/ h 2 A ∆ T 3 = Q/ h 3 A   ∆T 1 + ∆T 2 + ∆ T 3 = Q( 1 + 1 + 1) A ( h 1 h 2 h 3 )  ∆T 1 + ∆T 2 + ∆T 3 = ∆ T (Total temperature difference) 12

13.  ∆T = Q ( 1 + 1 + 1) but Q = ∆ T  A ( h 1 h 2 h 3 ) U A  I = 1 + 1 + 1  U h 1 h 2 h 3   Reciprocals of the heat transfer coefficients = resistances and are additive  13

14.  In some cases the areas are not the same:  A 1, A 2, A 3   ∆T 1 + ∆T 2 + ∆ T 3 = Q ( 1 + 1 + 1 )  (A 1 h 1 A 2 h 2 A 3 h 3 )   Using one of the areas as the basis then U will vary according to that area (say A 1 )   Q = U 1 A 1 ∆ T or ∆T = Q U 1 A  1 = 1 + A 1 1 + A 1 1  U h 1 A 2 h 2 A 3 h 3 14

15.  The U value depends on   Heat transfer mechanism  Fluid dynamics of both fluids  Properties of the H/E construction materials Geometry of fluid paths 15

16.  Deposits & scale cause performance deterioration after a period of operation  Deposits from the flow streams increase thermal resistance and decrease heat transfer rate  Fouling factor or fouling resistance used to measure the overall effect of deposits on the heat transfer  Most common fouling is accumulation of solid deposits from the fluid onto the heating surfaces 16

17.  Corrosion and chemical fouling also affect heat transfer  Glass coating and plastic pipes used to reduce chemical fouling  Algae growth in warm fluids lead to biological fouling  Chemical treatment is used to reduce biological fouling 17

18.  Fouling factor is zero for new, clean heat exchangers, R f =0  R f depends operating temperature,fluid velocity and duration of service FluidR f, m 2 o Cw -1 Distilled water, sea water, T>50 o C0.0002 Distilled water, sea water, T<50 o C0.0001 Fuel Oil0.0009 Steam (oil free)0.0001 Refrigerants (liquid)0.0002 Refrigerants (vapour)0.0004 Alcohol vapours0.0001 18

19.  Fouling factors must be obtained experimentally from the U values for both clean and dirty heat exchangersU values   Eg. Sea Water at 125.0 o C is used in a heat exchanger with fouling factor, R f =0.0002 m 2 o C/W. What is the percent reduction in the heat exchanger’s U value if in the clean state U=1961 W/m 2 o C? 19

20.  Application determines hardware and configuration  Double pipe, tube in tube or concentric tube H/E is the simplest form & as the name implies consisting of two concentric tubes.  Compact H/E are designed where there is strict limitations regarding weight and volume so as to give large heat transfer area per unit volume. The area density (ratio of area to volume) β> 700 m 2 m -3.  Examples are car radiators (β=600 m 2 m -3 ) and glass ceramic gas turbine H/E (β=15000 m 2 m -3 ) 20

21.  The most common type of H/E used in industrial applications.  Tube count can reach several hundreds with baffles.  Classifies by tube passes as, one shell pass and two tube passes, or two shell passes and four tube passes.  Fouling can be severe and the H/E must be taken out of service periodically to labourously clean the tubes. 21

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24.  Corrugated parallel plates facing each other and held firmly together by head frames with hot and cold fluids flowing between alternate plates.  Gap between plates 1.3-1.5 mm with large surface area to volume ratio.  Increasing demand for heat transfer can be met by simply increasing the number of plates.  Application prevalent in the diary and brewing industries with stainless steel the popular material of construction with rubber seal gaskets. 24

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28.  High degree of turbulence even at low flow rates  Very high heat transfer coefficients typical Heat transferWaterU Per plateFlow rateValue W/Kgal/hkW/m 2 K 15805503.70 21108504.94 264012506.13 Easy to dismantle and clean 28

29.  Sheets of metal are coiled to enclose a spiral annulus in the construction in this H/E with the advantage of good fouling characteristics and ease of cleaning.  Application confined to fluids with high solids concentration.  Velocities as high as 2.1 m/s and U values of 2.8 kW/m 2 K are achievable. 29

30. SHEs are usually relatively small in size. 30

31.  H/Es are usually selected to satisfy a particular temperature range or to predict outlet temperatures under conditions of known flow rates.  Two methods are normally used to analyze H/E performance:  LMTD method  NTU method 31

32. Counterflow or countercurrent flow T ha T h vs q T c vs q T hb T ca ΔT1ΔT1 Q QTQT ΔT2ΔT2 T cb ΔT1ΔT1 ΔT2ΔT2 ΔT vs q ΔTΔT Assumptions: i) Overall coefficient is constant ii) The specific heats of the hot & cold fluids are constant iii) Heat loss to the surroundings is negligible iv) The flows are steady and either counter current or parallel (not both) Temperature 32

33.  Based on the assumptions the temperatures of the hot and cold streams are expected to vary linearly with the heat rate.  Similarly ΔT will vary linearly with q resulting in a line of constant gradient  ΔT 1 & ΔT 2 are the temperature approaches then the gradient of the ΔT vs. q line is given by: 33

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35.  When ΔT 1 ~ ΔT 2 ΔLMTD ~ ΔT avg  With condensing fluids ΔLMTD is the same for all types of flow patterns  For non-condensing fluids in counter current flow pattern:  ΔT 2 = warm end approach  ΔT 1 = cold end approach 35

36.  For variable U values: Where U 1, U 2 are the U values at the ends of the H/E Other researchers proposed other methods of determining a representative temperature change across the H/E. According to Underwood: 36

37.  A heat exchanger is required to cool 20.0 kg/s of water from 360.0 K to 340.0 K using 25.0 kg/s of water at 300.0 K. If the overall heat transfer coefficient is constant at 2.0 kW/m 2 K what is the required surface area using;  (a) counter current concentric tube heat exchanger  (b) co-current concentric tube heat exchanger 37

38.  For multi pass H/E flow pattern is counter current in some tubes (passes) and parallel flow in others  Finding the actual temperature difference would be very difficult due to complex flow patterns  Underwood and Bowman et al introduced use of a correction factor based on graphical methods of modifying the ΔLMTD. 38

39.  Depending on equipment geometry and inlet and outlet temperatures a correction factor may be applied to the counter current flow ΔLMTD to compensate for the complex flow.  ΔT m =F*ΔLMTD CF  F<1 For cross flow & multi-pass shell & tube H/Es  F=1 Limiting value corresponds to counter flow H/Es  Charts of F vs P & R 39

40. 1 – inlet 2 – outlet T – Shellside t = Tubeside Determination of F requires the inlet and outlet temperatures of both the hot and cold fluids. 0 < P < 10 < R < infinity Phase change Tube side Phase change shell side (boiling/condensation) F = 1 For condensers or boilers F = 1 regardless of the configuration of the heat exchanger 40

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42. 1. Water at the rate of 68.0 kg/min is heated from 35 o C to 75.0 o C by an oil having a specific heat capacity of 1.9 kJ/kg o C. The fluids are used in a counter current flow double pipe heat exchanger and the oil enters at 110.0 o C and leaves at 75.0 o C. Using 320.0 W/m 2 o C for the overall heat transfer coefficient calculate the heat exchanger area. 2.If instead of a double pipe heat exchanger we use a shell and tube heat exchanger with water making one shell pass and the oil making two tube passes what would be the required area using the same U value? 42