Lecture 3 Heat Exchangers

Transcription

1 L3 Leture 3 Heat Exangers Heat Exangers. Heat Exangers Transfer eat from one fluid to anoter. Want to imise neessary ardware. Examples: boilers, ondensors, ar radiator, air-onditioning oils, uman body. Types: LIQUID-LIQUID eg. Sell and tube LIQUID-GAS Air-onditioning GAS-GAS Furnae supereater

2 Heat Exangers.. 2

3 Heat Exangers. Configurations FLUID B FLUID A FLUID B FLUID A COUNTERFLOW PARALLEL FLOW ONE-SHELL, FOUR TUBE PASSES FLUID A (unmixed CROSSFLOW FLUID B (mixed Heat Exangers.2 Table.,.2 Overall Heat Transfer Coeffiient U Internal and external termal resistanes in series. = + UA UA UA ( ( R f, R = + + R + + UA f, w ( ηo A ( ηo A ( ηo A ( η o A A is wall total surfae area ot or old side. R w wall fin Fouling fator R" f (m 2 K/W for seawater, fuel oil et. η o overall surfae effiieny (if finned 3

4 Heat Exangers.3 Heat Exanger Analysis Want a relationsip = UA T m were T m is some mean T between ot and old fluid. T ot fluid T T T old fluid dt dt dq T 2 T 2 T T 2 T 2 T2 T 2 T end end 2 Counterflow Note T,out an be < T,out end end 2 Parallel Flow T s annot ross Heat Exangers.3 Energy Balane (ounterflow on element sown. d = dt = dt mass flow rate of fluid speifi eat ( T T T dt dq dt T 2 T 2 Rate Equation T end end 2 ( d = UdA T T Now from ( dt d = (2 dt d = (dt in diretion 2 d & ( T = dq 4

5 Heat Exangers.3 Sub. d Q & from (2, ( T ( T d = U da Integrate 2 T 2 2 ln = UA T Total eat transfer rate & = ( T and = ( T Q 2 2 T T T2 T Heat Exangers.3 Sub for and put T = T END T = T END T2 T Q & = UA ln Q & = UA( LMTD ( T 2 T LOG MEAN TEMPERATURE DIFFERENCE Remember - and 2 are ends, not fluids Same formula for parallel flow (but T s are different Counterflow as igest LMTD, for given T s terefore smallest area for Q. T end end 2 T 2 5

6 Heat Exangers.3 T T2 T T 2 end end 2 Condenser end end 2 Evaporator Heat Exangers.3 Crossflow and Multipass = F UA ( LMTD ounterflow were F = f(t, T 2, T, T 2, onfiguration is orretion fator relative to ounterflow (ideal onfiguration. F< Counterflow is ideal LMTD metod best for sizing (area new H.E. wen all T s known..0 F 0.5 6

7 Heat Exangers.4 Effetiveness - NTU metod How will existing H.E. perform for given inlet onditions? Define effetiveness: ε = & Q atual max T, in? A? T, in T, in Were Q & max is for an infinitely long H.E. One fluid T T max = T,in -T,in T max = T = T and sine ( A A A ( B B B = C T A A = C T ten only te fluid wit lesser of C A, C B eat apaity rate an ave T max B B T, in end end 2 Heat Exangers.4 Table.3,.4 ie: max = C T max and ε = C or & =εc ( T Q, in, in ( T, in, in Want expression for ε wi does not ontain outlet T s. Sub. bak into Q & = UA( LMTD UA C exp C Cmax ε ounterflow = C UA C exp Cmax C Cmax C ε = ε NTU, C max and No. of transfer units (size of H.E. NTU UA C 7

8 Heat Exangers.4, Ex.4 CHARTS FOR EACH CONFIGURATION... 00% ε C / C max = (ondensor, evaporator To Use: NTU detere C max, C /C max alulate UA C ε from art & ( T Q =εc, in, in Heat Exangers.4,.5, Ex.4 Ex..4. A finned-tube ross-flow exanger as a gas-side overall U=00W/m 2 K and area 40m 2. Water enters at kg/s and 35ºC; air enters at.5kg/s and 250ºC. Wat is te eat transfer rate and te outlet temperatures? NTU metod C = m p, = kg/s 497J/kgK = 497W/K C = m p, =.5kg/s 000J/kgK = 500W/K= C C /C max = 500/497 = NTU = UA/C = 00W/m 2 K 40m 2 /500W/K = 2.67 From art ε % ε C /C max Q max = C ( T,i T,i = 500 ( = W Q = ε Q max = = W Q = m p, (T,o T,i = m p, (T,i T,o NTU T,o = /497 = 98C T,o = /500 = 73C 8