Heat Exchanger - Eng.sut.ac.th

Heat exchanger Heat exchangers are devices in which heat is transferred between two fluids at different temperatures without any mixing of fluids. Heat exchanger type 1. Direct heat transfer type 2. Storage type 3. Direct contact type 2130780490756/ Heat exchanger 1. Direct heat transfer type A direct transfer type of heat exchanger is one in which the cold and hot fluids flow simultaneously through the device and heat is transferred through a wall separating the fluids cold fluid hot fluid hot fluid cold fluid Concentric tube heat exchangers. (a) Parallel flow. (b) Counter flow.

Heat exchanger 2. Storage type heat exchanger A direct transfer type of heat exchanger is one in which the heat transfer from the hot fluid and the cold fluid occur though a coupling medium in the form of porous solid matrix. The hot and cold fluids alternatively through the matrix. The hot fluid storing heat in it and the cold fluid extracting heat from it. Heat exchanger 3. Direct contact type heat exchanger A direct transfer type of heat exchanger is one in which the two fluids are not separated. If heat is to be transferred between a gas and a fluid, the gas is either bubbled through the liquid or the liquid is sprayed in the form of droplets in the gas.

Heat exchanger Direct type heat exchanger 1. Tubular 2. Plate 3. Extended surface Heat exchanger Tubular heat exchanger 1. Concentric tube 2. Shell and tube Concentric tube Shell and tube The heat transfer area available per unit volume 100 -500 m2/m3

Heat exchanger Plate heat exchanger Series of large rectangular thin metal plates which are clamped together to form narrow parallel-plate channel. The heat transfer area available per unit volume 100-200 m2/m3 Heat exchanger Extended surface heat exchanger Fins attached on the primary heat transfer surface with the object of increasing the heat transfer area. The heat transfer area available per unit volume 700 m2/m3

Heat exchanger Classification by flow arrangement The three basic flow arrangements: o Parallel flow o Counter flow o Cross flow Heat exchanger Parallel flow Counter flow

Heat exchanger Cross flow Both fluids unmixed One fluid mixed and the other unmixed Heat exchanger Overall heat transfer coefficient (U) and fouling factor In a heat exchanger, the heat is transferred by both convection and conduction. Ta q UA (Ta –Tb) T1 T2 U is overall heat transfer coefficient hi Tb ho

Heat exchanger Plate heat exchanger Conv. Ta Cond. T1 T2 hi Tb q hi A q kA x h o A T1 1) 2) 2 T 2 T b T a T b 3) ho q U A Heat exchanger Adding above three equations (1, 2, 3) q q x q T a Tb hi A kA ho A From (4) a T1 T q Conv. T q T a Tb UA q q q x q UA hi A kA ho A 1 1 x 1 U hi k ho Across a plain wall 1 1 b 1 U h1 k h 2 4)

Heat exchanger Tubular heat exchanger r2 Tb To Ti To Ri Ti r1 Ro RT Rconv1 Rcond Rconv2 Ta Rconv1 Ra 1 hi A Rcond x kA Rconv 2 1 ho A Heat exchanger Based on inner area Ai 2πriL 1 (r r ) 1 1 o i Ui Ai h iAi kA lm h oAo 1 1 (r r )r r o i i i Ui h i krlm h o ro Based on outter area 1 (r r ) 1 1 o i UoAo h i Ai kA lm h oAo r (r r )r 1 1 o o i o U o h i ri krlm ho โดย rlm (ro ri ) ln(ro / ri )

Heat exchanger When the wall thickness of the tube is small and the thermal conductivity of the tube material is high, is as usually the case, the thermal resistance of the tube is negligible. 1 1 1 U hi ho because A i A o Heat exchanger Fouling The performance of heat exchangers usually deteriorate with time as a result of accumulation of deposits on heat transfer surfaces, representing additional resistance, called fouling.

Heat exchanger (Log) Mean Temperature Difference dA hot Th Tc cold Where: T T h Total heat transfer rate in heat exchanger dq U TdA q Tc U TdA Heat exchanger If U is assumed to be a constant q U TdA Define mean temperature difference Tm 1 A TdA area Thus: q UA T m This is the basic performance equation for a direct transfer type heat exchanger

Heat exchanger Parallel Flow Assumption 1. U is a constant 2. Heat exchanger is adequately insulated i.e. no heat loss to surrounding Heat exchanger Consider in elementary area dA (B.dx) dq U T B dx m h C ph dT h m c C pc dT c T Th Tc d ( T ) dT h dT c dq dq m h C ph m c C pc 1 1 U T B dx m C m c C pc h ph

Heat exchanger To Ti d ( T ) 1 1 m C T m c C pc h ph Where: UB L dx 0 Ti T h ,i T c ,i To T h ,o Tc ,o To 1 1 UA ln m c C pc Ti m h C ph 1 T h , i T h , o T c , o T c , i UA q Heat exchanger Ti To q UA Ti ln To This is the performance equation for a parallel-flow heat exchanger Comparing with: q UA T m Where: Tm Ti To Ti ln To

Heat exchanger For counter flow Assumption 1. U is a constant 2. Heat exchanger is adequately insulated i.e. no heat loss to surrounding Heat exchanger Consider an elementary area dA (B.dx) dq U ( T h T c ) B dx m h C ph dT h m c C pc dT c T Th Tc d ( T ) dT h dT c dq dq m h C ph m c C pc 1 1 U T B dx m C m c C pc h ph

Heat exchanger d ( T ) 1 1 m C m c C pc T h ph UB To 1 1 ln m C m c C pc Ti h ph UA To Ti L dx 0 Where: Ti T h ,i T c ,o T o T h ,o T c ,i To 1 T h , i T h , o T c , o T c , i UA ln q Ti Heat exchanger Ti To q UA Ti ln To Th,i Ti Tc,o Tc,i Comparing with: q UA T m Where: Th,o Tm Ti To Ti ln To To

Heat exchanger Special case of counter flow m h c ph m c c pc Then: T h ,i T h ,o T c ,o T c ,i Or T h ,i T c ,o T h ,o T c ,i Substituting into the expression for Tm , we get an indeterminate quantity Heat exchanger Define Ti p T o Then: T m lim p 1 T o ( p 1) ln p Apply L’ Hopital’s rule T m lim p 1 T o (1) To 1 p T m To Ti Ti To

Heat exchanger Cross flow Case 1: both fluids unmixed Hot fluid Th,i Tc,i Tc,o Cold fluid B y x Th,o L Heat exchanger Cross flow Case 2: one fluid mixed, the other unmixed Hot fluid Th,i Tc,i Tc,o Cold fluid B y x Th,o L Th f (x,y) Tc f (x)

Heat exchanger Cross flow Case -: Both fluid mixed Hot fluid Th,i Tc,i Tc,o Cold fluid B y x Th,o L Heat exchanger Both Th and Tc are functions of x and y Considering an elementary area dA ( dx dy) dq U ( T h T c ) dx dy q B 0 L 0 U ( T h T c ) dx dy Comparing with Tm 1 BL q UA T m B 0 L 0 ( T h T c ) dx dy More complicated than before but it has been done. Th f (y) Tc f (x)

Heat exchanger The integration of the three cases of cross flow has been done numerically. The results are presented in the form of a correction factor (F) F Tm cross flow Tm ifif thethearrangement arrengemen twas wasencounter-flow counter flow If the bulk exit temperatures on the hot side and cold side are Th,o and Tc,o, then Tm counter flow T T T ln T T / T h ,i h ,i c ,o c ,o h ,o h ,o Tc ,i Tc ,i Heat exchanger Mean temperature difference in cross flow F Tm cross flow Tm if the arrengement was counter flow For given values of Th,i; Th,o; Tc,i; Tc,o T counter flow Therefore: is the highest amongst all flow arrangements 0 F 1

Heat exchanger q cross flow UA Tm cross flow q cross flow UAF Tm counter flow F is plotted as a function of two parameters, R and S R T T T1,o T T T2,i 1,i 2 ,o S 2 ,o 1,i T2 ,i T2,i Heat exchanger Subscripts 1 and 2 correspond to the two fluids For case: 1 (both fluids unmixed) and case 3 (both fluids mixed) It is immaterial which subscript corresponds to the hot side and which to the cold side. For case: 1 and case 3 Subscripts 1 h 2 c or 1 c 2 h

Heat exchanger However for case 2, care must be taken to see that the mixed fluid has subscript 1 What is the parameter R? The ratio of change of temperature of the two fluids R 0 What is the parameter S? The ratio of change in temperature of one of the fluid to the difference of inlet 0 S 1 temperature of the two fluids Heat exchanger T1,i T2,i T2,o T1,o R T T 1,i 2 ,o T1,o T2 ,i Both fluids unmixed cross flow heat exchanger S T T 2 ,o 1,i T2 ,i T2 ,i

Heat exchanger T1,i mixed T2,i unmixed R T T 1,i 2 ,o T2,o T1,o T1,o T2,i One fluids mixed and the other unmixed Heat exchanger S T T 2 ,o 1,i T2,i T2 ,i

Heat exchanger Heat exchanger

Heat exchanger Heat exchanger

Heat exchanger Heat exchanger

Heat exchanger Heat exchanger

Heat exchanger Heat exchanger

Heat exchanger Heat exchanger

Heat exchanger Heat exchanger The effectiveness - NTU method Generally, we encounter two type of problems: Given: Type 1 Two fluids m h T h ,i , T h ,o Type 2 m c U T c ,i , T c ,o Find A? Given: Two fluids A heat exchanger A m h T h ,i m c T c ,i Find Th,o; Tc,o?

Heat exchanger Type 1 q UA T m A Type 2 ? q U Tm ? q UA T m We will need a trial and error approach to solve this type of problem i.e. assuming Th,o Trial and error can be avoided if we adopt the alternative method called the effectiveness -NUT Heat exchanger Effectiveness of a heat exchanger Rate of heat transfer in heat exchanger Maximum possible heat transfer rate T q Th ,i Tc ,o Th ,o q max Tc ,i Length of heat exchanger T h ,i T c ,i m h C ph ( T h , i T h , o ) ( m Cp ) s ( T h , i T c , i ) m c C pc ( T c , o T c , i ) ( m Cp ) s ( T h , i T c , i )

Heat exchanger Hence if Hence if T h ,i T h ,o T h ,i T c ,i , then m c C pc m C p s m c C pc m h C ph Note , then m h C ph m C p s m h C ph m c C pc T c ,o T c ,i T h ,i T c ,i 1) The definition are equivalent when 2) By definition 0 1 m c C pc m h C ph Heat exchanger Effectiveness – parallel flow Assume q T c ,o Tc ,i m c C pc m h C ph ( m C p ) s 1 1 T T h ,o h , i T h ,i Tc ,i T T h ,o h , i T h ,i T c ,i m h C ph m c C pc m h C ph m c C pc T c ,o T c ,i T T c ,i h ,i ( Tc ,o T c ,i ) 1 ( T h , i T h , o ) m h C ph 1 m C c pc m h C ph 1 m c C pc

Heat exchanger Effectiveness – parallel flow T Tc ,o 1 h , o T h ,i T c ,i m h C ph 1 m c C pc Derived earlier (slide 23) To 1 1 ln m C m c C pc Ti h ph T h ,o Tc ,o T T c ,i h ,i UA 1 exp 1 m C m c C pc h ph UA Heat exchanger Substituting m h C ph 1 exp 1 m c C pc UA m C h ph m h C ph 1 m c C pc

Heat exchanger If we had assumed initially m c C pc ( m C p ) s m c C pc 1 exp 1 m h C ph UA m C c pc , then m c C pc 1 m h C ph We combine the two expressions for m C p 1 exp 1 m C p UA m C s p s b Heat exchanger Define two new parameters Capacity ratio (C) C (m C p )s (m C p ) b or C min C max Number of transfer unit (NTU) NTU Note: UA (m C p )s 1) Both are dimensionless 2) 0 1 3) NTU 0 or UA C min m C p 1 m C p s b

Heat exchanger Effectiveness – parallel flow 1 exp 1 C NTU 1 C Effectiveness – counter flow 1 exp 1 C NTU 1 C exp 1 1 C NTU Heat exchanger Special cases Capacity rate (m.Cp) is infinite either on the hot side or the cold side C 0 For this solution, we obtain the relation 1 exp NTU

Heat exchanger Heat exchanger

Heat exchanger Heat exchanger

Heat exchanger Heat exchanger Example Situation: Light lubricating oil (Cp 2090 J/kg-K) is cooled with water in a small heat exchanger. Oil flow 0.5 kg/s, inlet T 375 K Water flow 0.2 kg/s, inlet T 280 K Part 1: If desired outlet temperature of the oil is 350 K, and you know the estimated overall heat transfer coefficient, U 250 W/m²-K, from manufacturer’s data for this type of heat exchanger Find: Required heat transfer area for a parallel flow heat exchanger and compare to the area needed for a counter flow heat exchanger.

Heat exchanger LMTD Toil,in 350 K Solution, Part 1: Toil,in 375 K C p ,c 4,181 J / Kg.K q C h Th ,i Th ,o Twater,out ? Twater,in 280 K and Tc ,o Tc ,i q / C c 0.5 2,090 (375 350) 26,125 W 280 26,125 /( 0.2 4,181) 311 K Heat exchanger For parallel flow, Tm, PF Ti 95 To 39 Ti To 95 39 63 ln Ti / To ln( 95 / 39) For counter flow, Ti 64 Ti 95 To 39 To 70 Ti 64 Tm, CF Ti To 64 70 67 ln Ti / To ln( 64 / 70) To 70

Heat exchanger For parallel flow, A PF q /( U Tm ,PF ) 1.66 m 2 For counter flow, A CF q /( U Tm ,CF ) 1.56 m 2 Heat exchanger Part 2: Use -NTU method to determine the required NTU and heat transfer area for parallel and counter flow Solution C p ,c 4,181 J / Kg.K To determine the minimum heat capacity rate, m h C ph 0.5 2090 1045 W / K m c C pc 0.2 4,181 836.2 W / K (m C p ) s