HEAT TRANSFER CHAPTER 11 Heat Exchangers 们au Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 1 HEAT TRANSFER CHAPTER 11 Heat Exchangers
Heat Exchangers. NTU-8 Method Where we’ ve been∴ Analysis of heat exchangers using log mean temperature difference (LMTD) q=y ATour -At i UA△TLMD Tr △T △T △D Where were going Computation of heat exchanger performance compared to the theoretical maximum possible for the flow conditions and hX type and size Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 2 Heat Exchangers, NTU- Method Where we’ve been …… • Analysis of heat exchangers using log mean temperature difference (LMTD) Where we’re going: • Computation of heat exchanger performance compared to the theoretical maximum possible for the flow conditions and HX type and size. LMTD i o out in UA T T T T T q UA = − = ln dq Ti To h dTc dT T
Heat Exchangers. NTU-8 Method KEY POINTS THIS LECTURE Concept of heat exchanger effectiveness,& based on the ratio of fluid heat capacity, C Concept of heat exchanger Numberof Transfer Units. NTU pplication of ntU-c method to predict the performance of a given heat exchanger ext book sections: $11.4-11.5 Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 3 Heat Exchangers, NTU- Method KEY POINTS THIS LECTURE • Concept of heat exchanger effectiveness, based on the ratio of fluid heat capacity, C. • Concept of heat exchanger Number of Transfer Units, NTU • Application of NTU- method to predict the performance of a given heat exchanger • Text book sections: §11.4 – 11.5
Recall earlier discussion For a condensing vapor or Ch Cond Out For an evaporating liquid (or Ch <<C Ev O What if Ch= Cc in a counterflow HX? △T1=△T Out Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 4 • For a condensing vapor • For an evaporating liquid • What if Ch = Cc in a counterflow HX? Recall earlier discussion ( ) or Ch Cc x T In Out x T In Out ( ) or Ch Cc x T In Out T1 = T2 TCond TEvap
Heat exchanger effectiveness Maximum possible heat transfer rate for any given inlet temperatures and flow rates occurs in a infinitely long counterflow HX c out out → C.7 Length of heat exchanger ·IfC Ch, then Tho=T Hot fluid would reach cold fuid inlet t and max Maximum△T max min(h. CI Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 5 Heat exchanger effectiveness • Maximum possible heat transfer rate for any given inlet temperatures and flow rates occurs in a infinitely long counterflow HX and : ( ) Cold fluid would reach hot fluid inlet T If max c h,i c,i c h c,o h,i q C T T C C , then: T T = − • = T Th,out Tc,in c out h in T T , , = Length of heat exchanger and : ( ) Hot fluid would reach cold fluid inlet T If max h h,i c,i c h h,o c,i q C T T C C , then: T T = − • = L → ( ) qmax = Cmin Th,i −Tc,i Maximum T
Heat exchanger effectiveness(Contd) Define: Heat exchangereffectiveness, 8 max Actual heat transfer, g, can be determined from simple energy balance g=qn=m,p,hIh,inIh, out g=mc c p, c (c,out SO Ch (Thi-T,ou) ccout C in Thus min OR= min If the heat exchanger effectiveness were known then the actual heat transferred could be found from max Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 6 Heat exchanger effectiveness (Cont’d) Define: Heat exchanger effectiveness, • Actual heat transfer, q, can be determined from simple energy balance • Thus: • If the heat exchanger effectiveness were known, then the actual heat transferred could be found from: ( ) ( ) ( ) ( ) h h i n h out c c out c i n h h p h h i n h out c c p c c out c i n so q C T T C T T q q m c T T q m c T T , , , , , , , , , , : = − = − = = − = = − ( ) ( ) ( ) ( ) min , , , , min , , , , max h i c i c c o c i h i c i h h i h o C T T C T T OR C T T C T T q q − − = − − = = max q q q qmax =
Number of Transfer units Define: Numberof Transfer Units, NTU NTu UA min nTU depends on both the heat exchanger design (UA) and the operating conditions(cmin) Define: Capacity Ratio, Cr r min max Effectiveness is a function of capacity ratio and the ntu E=f(NTU, Cr) Relationships between E, nTU and Cr can be computed Tables 11.3 and 11.4 and Figures 11.14-l1.9 in textbook Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 7 Number of Transfer Units Define: Number of Transfer Units, NTU • NTU depends on both the heat exchanger design (UA) and the operating conditions (Cmin). Define: Capacity Ratio, Cr • Effectiveness is a function of capacity ratio and the NTU • Relationships between , NTU and Cr can be computed. Tables 11.3 and 11.4 and Figures 11.14 – 11.19 in textbook C C /C (C 1) r = min max r ( ) NTU Cr = f , Cmin UA NTU
NTU-E Tables 1.0 0.6 Parallel flow 0.2 0 2 3 5 NTU 1.0 0.8 1.0o 0.75 Counter 0.6 0.50 025 flow 0.4 0.2 3 NTU Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 8 NTU - Tables Parallel flow Counter flow
Summary of Solution Method Typical scenario for using E-NTU method Given: Th in, Tcin, h, mc, hX geometry Find h out Icout, q Solution method: 1. Determine UA for this heat exchanger 1. Find u and A 2. Find C.Clc 2. Calculate NTU=UA/C 3. Determine s from tabulated formulas or plots 4. Compute actual heat transfer g=8 q max mIn 5. Find outlet temperatures from T.+ q p,c Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 9 Summary of Solution Method • Typical scenario for using -NTU method: Given: Find: Solution method: 1. Determine UA for this heat exchanger 1. Find 2. Find 2. Calculate 3. Determine from tabulated formulas or plots 4. Compute actual heat transfer 5. Find outlet temperatures from Th,in , Tc,in , m h , m c , HX geometry Th,out , Tc,out , q ( ) q qmax Cmin Th,i Tc,i = = − m q , and m q c , , , h , , , p c c o c i p h h o h i c T T c T T = + = − Cc Ch Cr , , U and A min NTU =UA/C
Calculation Methodology Performance calculation Given: Th in, Tcin, h, mc, hX geometry Find h out Icout, q Solution method: NTU Design problems: Iven hin2 hout or Find: Tou(or Thout), g, A Solution method: lMTD Heat Transfer Su Yongkang School of Mechanical Engineering
Heat Transfer Su Yongkang School of Mechanical Engineering # 10 Calculation Methodology • Performance calculation: Given: Find: Solution method: NTU • Design problems: Given: Find: Solution method: LMTD Th,in , Tc,in , m h , m c , HX geometry Th,out , Tc,out , q , , , , (or ) Th,i n Tc,i n mh mc Th,out Tc,out (or ), q, A Tc,out Th,out
