2. Heat Transfer/Heat Exchanger How is the heat transfer? Mechanism of Convection Applications. Mean fluid Velocity and Boundary and their effect on the rate of heat transfer. Fundamental equation of heat transfer Logarithmic-mean temperature difference. Heat transfer Coefficients. Heat flux and Nusselt correlation Simulation program for Heat Exchanger

3. How is the heat transfer? Heat can transfer between the surface of a solid conductor and the surrounding medium whenever temperature gradient exists. Conduction Convection Natural convection Forced Convection

4. Natural and forced Convection  Natural convection occurs whenever heat flows between a solid and fluid, or between fluid layers.  As a result of heat exchange Change in density of effective fluid layers taken place, which causes upward flow of heated fluid. If this motion is associated with heat transfer mechanism only, then it is called Natural Convection

5. Forced Convection  If this motion is associated by mechanical means such as pumps, gravity or fans, the movement of the fluid is enforced.  And in this case, we then speak of Forced convection.

6. Heat Exchangers A device whose primary purpose is the transfer of energy between two fluids is named a Heat Exchanger.

7. Applications of Heat Exchangers Heat Exchangers prevent car engine overheating and increase efficiency Heat exchangers are used in Industry for heat transfer Heat exchangers are used in AC and furnaces

8. The closed-type exchanger is the most popular one. One example of this type is the Double pipe exchanger. In this type, the hot and cold fluid streams do not come into direct contact with each other. They are separated by a tube wall or flat plate.

10. Principle of Heat Exchanger First Law of Thermodynamic: “Energy is conserved.” 0 000 Control Volume Qh Cross Section Area HOT COLD Thermal Boundary Layer

11. Q hot Q cold ThTh T i,wall T o,wall TcTc Region I : Hot Liquid- Solid Convection NEWTON’S LAW OF CCOLING Region II : Conduction Across Copper Wall FOURIER’S LAW Region III: Solid – Cold Liquid Convection NEWTON’S LAW OF CCOLING THERMAL BOUNDARY LAYER Energy moves from hot fluid to a surface by convection, through the wall by conduction, and then by convection from the surface to the cold fluid.

12. Velocity distribution and boundary layer When fluid flow through a circular tube of uniform cross- suction and fully developed, The velocity distribution depend on the type of the flow. In laminar flow the volumetric flowrate is a function of the radius. V = volumetric flowrate

13.  In turbulent flow, there is no such distribution. The molecule of the flowing fluid which adjacent to the surface have zero velocity because of mass-attractive forces. Other fluid particles in the vicinity of this layer, when attempting to slid over it, are slow down by viscous forces. r Boundary layer

14. Accordingly the temperature gradient is larger at the wall and through the viscous sub-layer, and small in the turbulent core. The reason for this is 1) Heat must transfer through the boundary layer by conduction. 2) Most of the fluid have a low thermal conductivity (k) 3) While in the turbulent core there are a rapid moving eddies, which they are equalizing the temperature. heating cooling Tube wall h

15. Region I : Hot Liquid – Solid Convection Region II : Conduction Across Copper Wall Region III : Solid – Cold Liquid Convection + U = The Overall Heat Transfer Coefficient [W/m.K] roro riri

16. Calculating U using Log Mean Temperature Hot Stream : Cold Stream: Log Mean Temperature

17. 16 Heat Exchanger (HX) Design Methods HX designers usually use two well-known methods for calculating the heat transfer rate between fluid streams—the UA-LMTD and the effectiveness-NTU (number of heat transfer units) methods. Both methods can be equally employed for designing HXs. However, the  -NTU method is preferred for rating problems where at least one exit temperature is unknown. If all inlet and outlet temperatures are known, the UA-LMTD method does not require an iterative procedure and is the preferred method.

18. Energy Balance Overall Energy Balance Assume negligible heat transfer between the exchanger and its surroundings and negligible potential and kinetic energy changes for each fluid. Assuming no l/v phase change and constant specific heats, Application to the hot (h) and cold (c) fluids:

19. 18 LMTD (Log Mean Temperature Difference) The most commonly used type of heat exchanger is the recuperative heat exchanger. In this type the two fluids can flow in counter-flow, in parallel-flow, or in a combination of these, and cross-flow. The true mean temperature difference is the Logarithmic mean Temperature difference (LMTD), is defined as

20. CON CURRENT FLOW COUNTER CURRENT FLOW T1 T2 T4 T5 T3 T7 T8 T9 T10 T6 Counter - Current Flow T1 T2 T4 T5 T6 T3 T7 T8 T9 T10 Parallel Flow Log Mean Temperature evaluation T1T1 A 1 2 T2T2 T3T3 T6T6 T4T4 T6T6 T7T7 T8T8 T9T9 T 10 Wall ∆ T 1 ∆ T 2 ∆ A A 1 2

21. T1T1 A 1 2 T2T2 T3T3 T6T6 T4T4 T6T6 T7T7 T8T8 T9T9 T 10 Wall

22. Exchanger Fouling Electron microscope image showing fibers, dust, and other deposited material on a residential air conditioner coil and a fouled water line in a water heater.

23. Exchanger Fouling

24. Overall Coefficient Overall Heat Transfer Coefficient An essential requirement for heat exchanger design or performance calculations. Contributing factors include convection and conduction associated with the two fluids and the intermediate solid, as well as the potential use of fins on both sides and the effects of time-dependent surface fouling. With subscripts c and h used to designate the hot and cold fluids, respectively, the most general expression for the overall coefficient is:

25. Overall Coefficient  Assuming an adiabatic tip, the fin efficiency is  

26. 25 Heat Exchanger UA-LMTD Design Method Where U is the overall heat transfer coefficient (and is assumed to be constant over the whole surface area of the heat exchanger).

27. 26 Calculation of heat transfer——heat transfer coefficient Table - data of heat resistance from scales fluid Rs m 2 · ℃ /W fluid Rs m 2 · ℃ /W water (u<1m/s, t<50 ℃ ) vapors sea0.0001 organic0.0002 river0.0006 steam (no oil)0.0001 well0.00058waste (with oil)0.0002 distilled0.0001 freezing (with oil)0.0004 boiler supply0.00026gases for cooling tower (untreated) 0.00058 air0.0003 for cooling tower (treated) 0.00026compressed air0.0004 with solid particles0.0006 nature gas0.002 brine0.0004 coke oven gas0.002 For reducing this resistance, （ 1 ） cleaning the heat exchanger; （ 2 ） treat the fluid before use.

28. 27 2. Using empirical data or equation to calculate U Calculation of heat transfer——heat transfer coefficient Fluid typeU W/(m 2 ·K) water—gas12~60 water—water800~1800 water—kerosenearound 350 water—organic solvent280~850 gas—gas12~35 saturated steam—water1400~4700 saturated steam—gas30~300 saturated steam—oil60~350 saturated steam—boiling oil290~870 Table - empirical data of U for shell - pipe heat exchangers Empirical equations are given by manufacturers.

29. 28 Effect of h to U Calculation of heat transfer——heat transfer coefficient e. g.: using saturated steam with （ h 2 ） to heat air (h 1 ) : the pipe size is 25×2.5, air is in the pipe. The conductive heat resistance can be ignored. We have h 1 (W/m 2.k) h 2 (W/m 2.k) U O (W/m 2.k) comparison 40500031.8 U O 2 close to h 1 ; 401000031.9 h 2 doubles ， U O almost no change 80500063.2 h 1 doubles ， U O doubles —— the relatively importance of h to U the results show that the increase of relative smaller h is key to increase U—— the key resistance is important. If the value of the two h are not quite different ， the increase of any one h is effective to raising the K.

30. LMTD Method A Methodology for Heat Exchanger Design Calculations - The Log Mean Temperature Difference (LMTD) Method - A form of Newton’s Law of Cooling may be applied to heat exchangers by using a log-mean value of the temperature difference between the two fluids: Evaluation of depends on the heat exchanger type. Counter-Flow Heat Exchanger:

31. LMTD Method (cont.) Parallel-Flow Heat Exchanger:  Note that T c,o can not exceed T h,o for a PF HX, but can do so for a CF HX.  For equivalent values of UA and inlet temperatures, Shell-and-Tube and Cross-Flow Heat Exchangers:

32. Special Conditions Special Operating Conditions  Case (a): C h >>C c or h is a condensing vapor – Negligible or no change in  Case (b): C c >>C h or c is an evaporating liquid – Negligible or no change in  Case (c): C h =C c. –

33. 32 comparison of flow directions Calculation of heat transfer——temperature difference Given ： the inlet and outlet temperatures of two fluids Find ： logarithmic average T differences in different situations Solution ： cocurrent counter current hot fluid 240 ℃ → 160 ℃ cold fluid - ） 100 ℃ → 140 ℃  t 1 =140 20=  t 2 240 ℃ → 160 ℃ - ） 140 ℃ ← 100 ℃  t 1 =100 60=  t 2 < It is  clear that  t m,co <  t m,counter

34. 33 Comparison 1——heat transfer area needed At same flow rate and same T 1, T 2, t 1 and t 2, for same heat transfer task Q = w 1 C p1 (T 1 - T 2 ) = w 2 C p2 (t 2 - t 1 ) = UA △ t m task = capacity Calculation of heat transfer——temperature difference

35. 34 For a given heating task ， Q = W 2 C p2 (t 2 - t 1 ) consumption of heating medium ： W 1 = W 2 C p2 (t 2 - t 1 ) / C p1 (T 1 -T 2 ) t A t1t1 t2t2 T1T1 T2T2 cocurrent counter current ： T 2min is limited by t1t1 T2T2 t1t1 t A t2t2 T1T1 counter current cocurrent ： T 2min is limited by t2t2 It is obvious that t 2 >t 1, then T 2co > T 2counter, W 1co > W 1counter Therefore, for a given task, a counter current heat exchanger consumes little heating medium than a cocurent one. Calculation of heat transfer——temperature difference Comparison 2——consumption of heating medium For a given T 1

36. Problem: Overall Heat Transfer Coefficient Determination of heat transfer per unit length for heat recovery device involving hot flue gases and water.

37. Problem: Overall Heat Transfer Coefficient (cont.)

41. III. Rate equation and  T TM The rate equation for a shell-and-tube heat exchanger is the same as for a concentric pipe exchanger: However, U i and  T TM are evaluated somewhat differently for shell-and-tube exchangers. We will first discuss how to evaluate  T TM and then a little later in the notes we will discuss how to evaluate U i for shell-and-tube exchangers. In a shell-and-tube exchanger, the flow can be single or multipass. As a result, the temperature profiles for the two fluids in a shell-and-tube heat exchanger are more complex, as shown below.

42. Computation of  T TM : For the concentric pipe heat exchanger, we showed the following (parallel and countercurrent flow):  T TM = When a fluid flows perpendicular to a heated or cooled tube bank, and if both of the fluid temperatures are varying, then the temperature conditions do not correspond to either parallel or countercurrent. Instead, this is called crossflow.  T lm

43. F G *  T LM

44. The factor Z is the ratio of the fall in temperature of the hot fluid to the rise in temperature of the cold fluid. The factor  H is the heating effectiveness, or the ratio of the actual temperature rise of the cold fluid to the maximum possible temperature rise obtainable (if the warm-end approach were zero, based on countercurrent flow). From the given values of  H and Z, the factor F G can be read from the text book figures:

45. Therefore, as with the concentric pipe heat exchanger, the true mean temperature difference for the 1-1 exchanger is equal to the log mean temperature difference (  T LM ). For multiple pass shell-and-tube designs, the flow is complex and the  T LM is less than that for a pure countercurrent design. We must account for the smaller temperature driving force using a correction factor, F G, which is less than 1 and typically greater than 0.8. The rate of heat transfer in multiple pass heat exchangers is written as: where  T LM is the log mean temperature difference for pure countercurrent flow

46. Textbook Figures 15.6 a, b 1-2 exchangers 2-4 exchangers

47. Ten Minute Problem -- F G for multiple pass HE For a 2-4 heat exchanger with the cold fluid inside the tubes and the following temperatures: T ca = 85°FT ha = 200°F T cb = 125°FT hb = 100°F (a)What is the true mean temperature difference? (answer  T TM = 31.7°F) (b)What exchanger area is required to cool 50,000 lb m /hr of product (shell-side fluid) if the overall heat transfer coefficient is 100 Btu/hr-ft 2 -°F and C p for the product is 0.45 Btu/lb m -°F? (answer A = 710 ft 2 )