Heat Exchangers

Heat Exchangers Outline • Basic ideas of heat exchangers • Overall heat transfer coefficient • Log-mean temperature difference method • Effectiveness –NTU method • Practical considerations Heat E xchangers 2 Heat Exchangers • Used to transfer energy from one fluid to another • Typically one fluid is cooled while the other is heated • May have phase change: temperature of one or both fluids is constant • Simplest is double pipe heat exchanger – Parallel flow and counter flow Figure 11-1 from Çengel, Heat and Mass Transfer 3 Compact Heat Exchangers 4 Compact Heat Exchangers II • Have large surface area for heat exchange per unit volume • Compact heat exchanger in home air conditioner – Achieved by use of fins – Car and truck radiations best example – Operate in cross flow – Fluids said to be mixed or unmixed – Refrigerant in tubes, air through fins • Mixed: one flow passage for the fluid • Unmixed: several flow passages 5 Figure 11-2 from Çengel, Heat and Mass Transfer 6 1 Heat Exchangers Compact Heat Exchangers III Shell-and-Tube Exchanger • Counter flow exchanger with larger surface area; baffles promote mixing Figure 11-3 from Çengel, Heat and Mass Transfer 7 Tube and Shell Passes Figure 11-4 from Çengel, Heat and Mass Transfer 8 Shell and Tube Passes II • Previous chart showed one shell pass and one tube pass – No cases where flow changed direction completely • Number of shell or tube passes is the number of times a fluid in the shell (or tubes) flows in a reverse direction Tube flow has one complete change of direction giving two tube passes – Examples next charts 9 Figure 11-5(a) from Çengel, Heat and Mass Transfer Shell and Tube Passes III Tube flow has three complete changes of direction giving four tube passes Shell flow changes direction to give two shell passes 11 Figure 11-5(b) from Çengel, Heat and Mass Transfer 10 Overall U • U is overall heat transfer coefficient • Analyzed here for double-pipe heat exchanger 1 1 + Rwall + hi Ai ho Ao 1 1 1 = = = U o Ao U i Ai UA R= Figure 11-7 from Çengel, Heat and Mass Transfer 12 2 Heat Exchangers Heat Exchange Analysis • Heat transfer from hot to cold fluid Analysis • For no external heat transfer, combine three equations for differential area, dA, and integrate from 1 to 2 Q& = UAΔT ( ( ) ) Q& = m& c c pc Tc,out − Tc,in • First law energy Q& = m& h c ph Th,in − Th,out balances • Assumes no heat loss to surroundings dQ& = U (Th − Tc )dA dQ& = m& c c pc dTc dQ& = − m& c dT – Subscripts c and h denote cold and hot fluids, respectively – Alternative analysis for phase change h ph 13 Analysis Result • Parallel flow Q& = UAΔTlm heat exchanger ΔTlm = Q& = UAΔTlm ΔTlm is “logmean delta T” ΔT2 − ΔT1 ΔT1 − ΔT2 = ⎛ ΔT ⎞ ⎛ ΔT ⎞ ln⎜⎜ 2 ⎟⎟ ln⎜⎜ 1 ⎟⎟ Δ T ⎝ 1⎠ ⎝ ΔT2 ⎠ ΔT1 and ΔT2 are both Th – Tc Figure 11-14 from Çengel, Heat and Mass Transfer ΔTlm = 15 – Difference in ΔT1 and ΔT2 definitions Lim (Th,out − Tc,in ) − (Th,in − Tc,out ) Figure 11-16 from Çengel, Heat and Mass Transfer ⎞ ⎟ ⎟ ⎠ 16 ΔT2 − ΔT1 ΔT2 − ΔT1 = ⎛ ΔT2 ⎞ ΔT −ΔT →0 ⎛ ΔT2 − ΔT1 + ΔT1 ⎞ 2 1 2 1 2 1 ⎟⎟ ⎟⎟ ln⎜⎜ ln⎜⎜ ΔT1 ⎝ ⎝ ΔT1 ⎠ ⎠ 1 ΔT2 − ΔT1 = = ΔT1 = 1 1 ΔT − ΔT1 ⎞ ΔT2 −ΔT1→0 ΔT2 − ΔT1 →0 ln⎛ ⎜⎜1 + 2 ⎟⎟ ΔT2 − ΔT1 ΔT1 Δ T 1 ⎝ 1+ ⎠ ΔT1 Lim ΔTlm = ΔTLim ΔT − ΔT →0 − ΔT →0 ΔT − ΔT1 = UA 2 ⎛ ΔT ⎞ ln⎜⎜ 2 ⎟⎟ ⎝ ΔT1 ⎠ ⎛T −T ⎞ ln⎜ h,out c ,in ⎟ ⎜T −T ⎟ ⎝ h,in c,out ⎠ ⎛T −T ln⎜ h,out c,out ⎜ T −T ⎝ h,in c,in • Apply l’Hopital’s rule to show that ΔTlm = ΔT1 = ΔT2 in this case • Same basic equations ΔTlm = (Th,out − Tc,out ) − (Th,in − Tc,in ) What if ΔT2 = ΔT1? Counter Flow Q& = UAΔTlm ΔT2 − ΔT1 ΔT1 − ΔT2 = ⎛ ΔT ⎞ ⎛ ΔT ⎞ ln⎜⎜ 2 ⎟⎟ ln⎜⎜ 1 ⎟⎟ ⎝ ΔT1 ⎠ ⎝ ΔT2 ⎠ Figure 11-14 from Çengel, Heat and Mass Transfer ΔT2 = Th,out – Tc,in ΔT1 = Th,in – Tc,out 14 Analysis Result II • Parallel flow heat exchanger with no external heat transfer ΔTlm = h Figure 11-14 from Çengel, Heat and Mass Transfer 17 Lim Lim – Last step takes derivative of numerator and denominator with respect to ΔT2 – ΔT1 for constant ΔT1 in denominator 18 3 Heat Exchangers Heat Exchanger Problems Problem • With ΔTlm method we want to find U or A when all temperatures are known • If we know three temperatures, we can find the fourth by an energy balance with known mass flow rates (and cp’s) Q& = m& c c pc (Tc,out − Tc,in ) ) Q& = m& c (T − T h ph h,in h,out & from two Can find Q temperatures for one stream and then find unknown temperature • An counterflow heat exchanger with U = 200 W/m2·oC is to be used to cool 1 kg/s of oil (cp =2000 J/kg·oC) from 100oC to 30oC using 3 kg/s of water (cp = 4184 J/kg·oC) at 20oC. What area is required? • Given: Th,in = 100oC, Th,out = 30oC, Tc,in = 20oC, U = 200 W/m2·oC, cp,c = 4184 & = 1 kg/s, J/kg·oC, cp,h = 2000 J/kg·oC, m h & and mc = 3 kg/s. Find: A 19 20 Solution Solution II (Th,out − Tc,in ) − (Th,in − Tc,out ) Q& = UAΔTlm = UA ⎛T −T ⎞ ln⎜ h,out c,in ⎟ ⎟ ⎜ T −T ⎝ h,in c,out ⎠ = m& c c pc (Tc,out − Tc,in ) = m& h c ph (Th,in − Th,out ) • Equations: ( ) 1 kg 2000 J J 100o C − 30o C = 140000 Q& = m& h c ph (Th,in − Th,out ) = s kg ⋅o C s Tc,out = Q& + Tc,in m& c c pc J 140000 s = 31.2 oC = 3 kg 4184 J s kg ⋅o C 21 Other Configurations (Th,out − Tc,in ) − (Th,in − Tc,out ) = ⎛T −T ⎞ ln⎜ h,out c,in ⎟ ⎜ T −T ⎟ , h in c ,out ⎠ ⎝ o o o o 30 C − 20 C − 100 C − 31.2 C = 30.5o C ⎛ 30o C − 20oC ⎞ ⎟ ln⎜⎜ o o ⎟ ⎝ 100 C − 31.2 C ⎠ ( )( ) J W ⋅s 140000 & Q s 1J A= = UΔTlm 200 W 30.5o C m 2 ⋅o C ( ) A = 22.9 m 2 22 Correction Factor • Use correction factor ΔTlm = FcorrΔTlm,CF • ΔTlm,CF = is ΔTlm for counter flow ΔTlm,CF ΔTlm = (T − T ) − (Th,in − Tc,out ) = h,out c ,in • Correction factor parameters, R and P – Shell and tube definitions below Ttube,out − Ttube,in t 2 − t1 Tshell ,in − Ttube,in T1 − t1 m& c p Tshell ,in − Ttube,in T1 − T2 = = R= Ttube,out − Ttube,in t 2 − t1 m& c p P= = ( )tube ( )shell ⎛T −T ⎞ ln⎜ h,out c,in ⎟ ⎟ ⎜ T −T ⎝ h,in c,out ⎠ • Fcorr depends on temperatures on tube side (tin and tout) and shell side (Tin and Tout) through two parameters, R and P – Correction factor charts show diagrams that illustrate the equations for P and R 23 24 4 Heat Exchangers Correction Factor Chart I P P T1 Figure 11-18 from Çengel, Heat and Mass Transfer Figure 11-18 from Çengel, Heat and Mass Transfer 25 T2 26 P Figure 11-18 from Çengel, Heat and Mass Transfer P Figure 11-18 from Çengel, Heat and Mass Transfer 27 28 Effectiveness, ε Effectiveness-NTU Method • Used when not all temperatures are known • Based on ratio of actual heat transfer to maximum possible heat transfer • Maximum possible temperature difference, ΔTmax is Th,in – Tc,in – Only one fluid, the one with the smaller value & cp, can have ΔTmax of m & cp)h & cp)c and Ch = (m – Define Cc = (m 29 ε= Q& Q& = & Qmax Cmin (Th.in − Tc,in ) Cmin = min (Ch , Cc ) • In effectiveness-NTU method we find ε, & & = εQ then find Q max & – Use CminΔTmax to find Q max because C1ΔT1 = C2ΔT2 or ΔT2 = C1ΔT1/C2 – If ΔT2 = ΔTmax and C1/C2 > 1, ΔT2 > ΔTmax – CminΔTmax is maximum heat transfer that can occur without impossible T < Tc,in 30 5 Heat Exchangers Find ε Max In example, ΔTmax = ΔToil and ΔTwater = (92) ΔTmax / 104.5 Example chart for finding effectiveness from NTU = UA/Cmin and Cmin/Cmax ratio If we used CmaxΔTmax to get Qmax = (104.5) (110) = 11,495 kW then ΔToil = (104.5)(110)/(92) = 124.9oC > ΔTmax For NTU = 1.5 and Cmin/Cmax .7 = 0.25, ε = ? Figure 11-26 from Çengel, Heat and Mass Transfer 31 Effectiveness Equations ε= 1− e • Shell and tube One shell pass and 2, 4, 6, … tube passes − NTU (1+ c ) 1+ c • Double pipe counter flow ε= More Effectiveness Equations UA NTU = Cmin • Double pipe parallel flow 32 ⎧ 1 + e − NTU ⎪ ε = 2⎨1 + c + 1 + c 2 ⎪⎩ 1 − e − NTU C c = min Cmax 1 − e − NTU (1−c ) 1 − ce − NTU (1−c ) • Any geometry with c = 0 c= 1+ c 2 1+ c 2 ⎫ ⎪ ⎬ ⎪⎭ −1 NTU = ε = 1 − e − NTU Cmin Cmax 33 UA Cmin 34 Figures from Figure 11-26 from Çengel, Heat and Mass Transfer Figure from Figure 11-26 from Çengel, Heat and Mass Transfer Problem Solution • An 25 m2 counterflow heat exchanger with U = 200 W/m2·oC is to be used to cool 1 kg/s of oil (cp = 2000 J/kg·oC) at 100oC using 3 kg/s of water (cp = 4184 J/kg·oC) at 20oC. What is the oil outlet temperature. • Given: Th,in = 100oC, Tc,in = 20oC, U = 200 W/m2·oC, A = 25 m2, cp,c = 4184 & h = 1 kg/s, J/kg·oC, cp,h = 2000 J/kg·oC, m & and mc = 3 kg/s. Find: Th,out 35 1 kg s 3 kg = s Ch = m& h c ph = Cc = m& c c pc 2000 J kg ⋅o C 4184 J kg ⋅ C o ( ) = = 2000 J s⋅o C 12552 J s⋅ C 200 W 25 m 2 2 o UA m ⋅ C NTU = = = 2.5 2000 J W ⋅ s Cmin s⋅o C 1 J o Cmin = 2000 J s⋅o C 2000 J o Cmin c= = s⋅ C Cmax 12552 J s⋅o C = 0.1593 36 6 Heat Exchangers Chart ε From chart for counter flow: Effectiveness Equation • For counterflow heat exchangers ( c = Cmin/Cmax) 1 − e − NTU (1−c ) ε= NTU = 2.5 Cmin/Cmax = 0.16 ε= 1 − ce − NTU (1−c ) 1 − e −2.5(1−0.1593) = 0.895 1 − 0.1593e −2.5(1−0.1593) ( ε = 0.89 37 Outlet Temperatures 2000 J 100o C − 20o C Q& = εQ& max = εCmin (Th.in − Tc ,in ) = 0.895 o s⋅ C J Q& = 1.43 x105 = 143 kW 38 s ) Problem Comparison • Use basic energy balance equations to & find outlet temperatures from Q • Comparing examples – The ΔTlm and ε-NTU examples had the same mass flow rates, heat capacities and U values – The ε-NTU example had A = 25 m2 compared to A = 22.9 m2 for ΔTlm – As expected, slightly larger area gives a slightly larger temperature change J 1.43x105 Q& o s = 31.4o C Tc,out = Tc,in + = 20 C + 3 kg 4184 J m& c c pc s kg ⋅o C J 1.43x105 Q& o s = 28.4o C Th,out = Th,in − = 100 C − 3 kg 4184 J m& c c pc s kg ⋅o C • Oil cools from 100oC to 30oC with A = 22.9 m2 and to 28.4oC with A = 25 m2 39 40 7