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C 50% PG= 6 gallon Btu⎝⎜minute⎠⎟ 0.88⎝⎜lb•º F ⎠⎟⎝⎜ε = q actual⎛q maxC water= 5 gallon ⎞ ⎛The capacitance rate for the water side is:One is called the capacitance ⎝⎜minuterate ⎠⎟ 1 Btu ⎞ ⎛ 8.33lb ⎞⎝⎜ratio. lb•º FIt’s ⎠⎟⎝⎜simply gallonthe⎠⎟ε = q actualminimum fluid capacitance rate of the two fluid streamsq⎛ maxdivided by theqlarger= ε ( Cfluid capacitance rate, stated asC water= 5 gallon ⎞ ⎛⎝⎜minute⎠⎟ 1 Btu ⎞ ⎛ 8.33lb ⎞ ⎛ 60minute⎞Btumin )( T hotin− T coldin )⎝⎜lb•º F ⎠⎟⎝⎜gallon⎠⎟⎝⎜hour ⎠⎟ = 2499hr⋅º F Formula 4-12:⎛C Formula 4-12:water= 5 gallon ⎞ ⎛The capacitance ⎝⎜minute⎠⎟ 1 Btu ⎞ ⎛ 8.33lb ⎞ ⎛ 60minute⎞Btu⎛⎝⎜rate lb•º Ffor ⎠⎟⎝⎜the gallon side ⎠⎟⎝⎜operating hour ⎠⎟ = 2499 Cwith propylene hr⋅º F50% PG= 6 gallon ⎞ ⎛ Btu ⎞ ⎛⎝⎜glycol is: ⎛C 50% PG= 6 gallon ⎞ ⎛ Btu ⎞ ⎛ 8.54lb ⎞ ⎛ 60minute⎞Btu⎝⎜minute⎠⎟ 0.88⎝⎜lb•º F ⎠⎟⎝⎜gallon⎠⎟⎝⎜hour ⎠⎟ = 2705 Chr⋅º Fratio= C minute⎠⎟ 0.88⎝⎜lb•º F ⎠⎟⎝⎜minC max⎛C 50% PG= 6 gallon ⎞ ⎛ Btu ⎞ ⎛ 8.54lb ⎞ ⎛ 60minute⎞Btu⎝⎜minute⎠⎟ 0.88⎝⎜lb•º F ⎠⎟⎝⎜gallon⎠⎟⎝⎜hour ⎠⎟ = 2705hr⋅º F Where:C ratio = capacitance rate ratio (unitless)q = εC min = smaller of the two capacitance rates for the two( COf these twomin )( Tcapacitancehotin− T coldin )q = ε ( Crates, the one associated with flow streams involved NTU (Btu/hr/ºF) = UA min )( T hotin− T coldin )the q water = ε ( C minside )( T hotinis −smaller T coldin ) and will be designated as C min . C max = larger of the two capacitance C min rates for the two flowThe larger capacitance rate will be designated C max . streams involved (Btu/hr/ºF)CIt’s important ratio= C minmaxto note that the physical properties of the The other quantity that’s needed is called NTU, whichtwo Cfluids ratio= C minwere ε = referenced q actualto an estimated average stands for number C of transfer 1− e ( − NTUunits, )(1−C ratioand )ratio= C minis defined asCtemperature maxin the qheat max exchanger. Also notice that the Formula 4-13. ε = C maxunits for capacitance rate are Btu/hr/ºF. One can interpret1− (Cthe value NTU = of UAratio)e ( − NTU )(1−C ratio )Ccapacitance rate as the rate of heat transfer Formula 4-13:min(in Btu/hr) NTU = that UA each fluid stream could carry into or out ofCthe heat exchanger minper⎛degree F change in temperatureof that stream. C1− e water ( = 5 gallon ⎞ ⎛− NTU )(1−C ratio )ε =The fluid 1− 1−having (Ce ( ⎝⎜minute⎠⎟ 1 Btu ⎞ ⎛ 8.33lb ⎞ ⎛ 60minute⎞NTU Btu = UA⎝⎜lb•º F ⎠⎟⎝⎜gallon⎠⎟⎝⎜hour ⎠⎟ = 2499hr⋅º F− NTUratio)ethe ()(1−C − NTU ratiolower )(1−C )ratio ε =NTU C minε =)of the two capacitance rates is Where:1− (Cone of the factors ratio)e ( 1+ NTU− NTUthat )(1−C ratio )limit the maximum possible rate of NTU = number of transfer units (unitless number)heat transfer. That maximum possible rate for a hypothetical U = overall heat transfer coefficient for the heat exchanger1− e ( − NTU )(1−C ratio )infinitely large heat exchanger would be the lower of the (Btu/hr/ft 2 /ºF)two capacitance NTU rates (e.g., C min ), multiplied by the inlet A = internal area of the heat exchanger (ft 2 )ε = 1+ NTU NTU Ctemperature 1+ NTUof 50%the PG= 6 gallonε =⎛ ⎞ ⎛ Btu ⎞ ⎛ 8.54lb ⎞ ⎛ 60minute⎞hot ⎝⎜fluid minute minus ⎠⎟ 0.88the ⎝⎜ inlet lb•º temperature F ⎠⎟⎝⎜gallon of ⎠⎟⎝⎜C min hour = smaller ⎠⎟ = 27051− Btu (C ratio)e ( − NTU )(1−C ratio )of the hr⋅º two capacitance rates of the two⎛ Fthe cold fluid. This allows the rate of heat transfer across flow streams (Btu/hr/ºF) C water= 5 gallon ⎞ ⎛the heat exchanger to be written as Formula 4-11.⎝⎜minute⎠⎟ 1 Btu ⎞ ⎛⎝⎜lb•º F ⎠⎟⎝⎜One can think of the NTU as the ratio of the physical ability⎛Formula 4-11:⎛⎞ ⎛of a heat exchanger to transfer heat (based on its internal⎝⎜⎠⎟ 1 Btu ⎞ ⎛ 8.33lb ⎞ ⎛ 60minute⎞BtuC water= 5 gallon ⎞ ⎛⎝⎜lb•º F ⎠⎟⎝⎜gallon⎠⎟⎝⎜hour ⎠⎟ = 2499⎝⎜minute⎠⎟ 1 Btu ⎞ ⎛ 8.33lb ⎞ ⎛ 60minute⎞Btu⎝⎜lb•º F ⎠⎟⎝⎜gallon⎠⎟⎝⎜hour ⎠⎟ = 2499hr⋅º Fhr⋅º Farea and the overall ε =NTUheat transfer coefficient) divided by1+ NTUq = ε ( C min )( T hotin− T coldin )the ability of the “least able” flow stream to convey heatinto or out of the heat exchanger. For example, a heatWhere:exchanger with a large internal area, and operating underq = actual rate of heat exchange across the heat exchanger favorable convection conditions, but also operating with a(Btu/hr)low capacitance rate on one side, would have high NTU.e = effectiveness of the heat exchanger (unitless value Because of its large internal area and high convectionbetween 0 and 1) C ratio= C ⎛minCcoefficients. it would water= 5 gallon ⎞ ⎛be very “effective” in transferringC ⎝⎜minute⎠⎟ 1 Btu ⎞ ⎛⎝⎜lb•º F ⎠⎟⎝⎜C min = the smaller of the two max fluid capacitance rates (Btu/ heat to (or from) the fluid with the lower capacitancehr/ºF)rate, possibly even approaching the performance of aT hotin = inlet temperature of the hot fluid (ºF)hypothetical infinite heat exchanger, which would have anT coldin = inlet temperature of the cold fluid (ºF)effectiveness of 1.0.NTU = UAAlthough Formula 4-11 appears C relatively simple, there Once the capacitance rate ratio (C ratio ) and number ofminis additional work involved to determine the value of transfer units (NTU) are calculated, the effectivenesseffectiveness (e).for a specific type of heat exchanger can be calculatedTwo additional quantities need to be defined in order to based on either formulas or graphs in heat transferdetermine the value of effectiveness (e).reference books.1− e ( − NTU )(1−C ratio )ε =( )(1−C ratio )421− (C ratio)e − NTUq max⎛⎞ ⎛⎞ ⎛ 8.54lb ⎞ ⎛gallon⎠⎟⎝⎜⎛⎝⎜8.33lb ⎞ ⎛gallon⎠⎟⎝⎜60miho60minutehour8.54lb ⎞ ⎛gallon⎠⎟⎝⎜8.33lb ⎞ ⎛gallon⎠⎟⎝⎜60mho60minhou60minhou
CC ratio= C ratio min= C minC maxC maxFor example, for any counterflow heat exchanger, theeffectiveness can be calculated using Formula 4-14a or4-14b. Formula 4-14a is used in the typical case whereε C= q ratio < 1. Formula NTU 4-14b is used in the less typical butpossible caseNTUwhere= UA = UAactualq maxratioC= 1 (e.g., both fluid streamsC minhave the same capacitance minrate ratio.Formula ⎛C4-14a: (use when Cratio ( <1)water= 5 gallon ⎞ ⎛⎝⎜minute⎠⎟ 1 Btu ⎞ ⎛ 8.33lb ⎞ ⎛ 60minute⎞Btu⎝⎜lb•º F ⎠⎟⎝⎜gallon⎠⎟⎝⎜hour)(1−C ⎠⎟ =ratio ) 2499hr⋅º F1− e ( − 1− NTU e − NTU)(1−C ratioε =)ε = 1− (C1− (C ratio)e − ratio NTU )e − NTU⎛C 50% PG= 6 gallon ⎞ ⎛ Btu ⎞ ⎛ 8.54lb ⎞ ⎛Formula ⎝⎜minute 4-14b: ⎠⎟ 0.88⎝⎜(use when lb•º FCratio ⎠⎟⎝⎜gallon = 1) ⎠⎟⎝⎜ε =NTU ε =NTU1+ NTU1+ NTU( )( T hotin− T coldin )( ( )(1−C ratio ))(1−C ratio )60minutehour⎞⎠⎟= 2705Btuhr⋅º FFigure 4-12effectivenessvs. NTUfor counterflowheat exchangerq = ε C min0.4The terms shown in red in Formula 4-14a are identical andthus only have to be calculated once.0.3⎛0.2The effectiveness of a counterflow heat exchanger asCa function of the number of transfer units (NTU) and the0.1capacity rate C water ratio = (C min5 gallon ⎛C ⎛ ⎞/C max ) can also be found usingFigure 4-12.⎝⎜minute⎠⎟ 1 Btuwater= 5 gallon ⎞⎛ ⎞ ⎛ 8.33lb ⎞ ⎛ 60minute⎞Btu⎝⎜minute⎝⎜ ⎠⎟ 1 Bturatio= C C 50% PG= 6 gallon ⎞ ⎛ Btu ⎞ ⎛ 8.54lb ⎞ ⎛ 60minute⎞⎝⎜minute⎠⎟ 0.88⎝⎜lb•º F ⎠⎟⎝⎜gallon⎠⎟⎝⎜hour ⎠⎟ =min⎛ ⎞ ⎛ 8.33lb ⎞ ⎛ 60minute⎞BtuC⎛max⎝⎜lb•ºlb•º F ⎠⎟⎝⎜ F ⎠⎟⎝⎜gallongallon⎠⎟⎝⎜ ⎠⎟⎝⎜C 50%hourPG= 6hour ⎠⎟ = 2499 gallon ⎞ ⎛ Btu ⎞ ⎛ 8.54lb ⎞ ⎛ 60minute⎞Btu⎠⎟ = 2499⎝⎜minute⎠⎟ 0.88⎝⎜hr⋅º lb•º FF⎠⎟⎝⎜gallon⎠⎟⎝⎜hour ⎠⎟ = 2705⎛hr⋅º FC 50% PG= 60gallon ⎞ ⎛ Btu ⎞ ⎛ 8.54lb ⎞ ⎛ 60minute⎞Btu⎝⎜minutehr⋅º ⎠⎟ 0.88⎝⎜ Flb•º F ⎠⎟⎝⎜gallon⎠⎟⎝⎜hour ⎠⎟ = 2705hr⋅º F0 1 2 3 4 5Here’s NTU = an UAexample: Consider a counterflow heat exchangerNTUCwith an internalminarea of 20 ft 2 C. It is operating with the tworatio= C min= 2499C max2705 = 0.924previously described flow streams of water at 5 gpm and C ratio= C min= 249950% propylene glycol at 6 gpm. The water stream enters The number max2705 = 0.924of transfer units is:at ε = 150ºF,1− eand ( C − NTUthe )(1−C ratiopropylene )ratio= C min= 2499Cglycol solution enters at 60ºF.max2705 = 0.924An 1− analysis (C ratio)e ( of − NTUthe )(1−C ratioconvection )coefficients and foulingNTU = UA (= 150 )20factors on both sides of the heat exchanger has beenC min2499 = 1.2NTU = UA (= 150 )20used to establish the overall heat transfer coefficient (U)C min2499 = 1.2as 150 Btu/hr/ft 2 NTU = UA (= 150 )20/ºF. Determine the rate of heat transfer These values can now be entered into Formula 4-14a:across ε =NTUC min2499 = 1.2the heat exchanger.1− e ( − NTU )(1−C ratio )1+ NTUε =(CThe two fluid capacitance rates have already beenratio)e ( = 1− e ( −1.2)(1−0.924)1− e ( − NTU )(1−C ratio ) 1− (0.924)e ( = 0.0872− NTU )(1−C ratio )ε =−1.2)(1−0.924)0.15651− (C ratio)e ( = 1− e ( −1.2)(1−0.924)0.0872− NTU )(1−C ratio ) (=1− (0.924)e )(1−0.924)1− e ( 0.1565 = 0.557− NTU )(1−C ratio )ε =determined as:1− (C ratio)e ( = 1− e ( −1.2)(1−0.924)0.0872− NTU )(1−C ratio ) (=1− (0.924)e −1.2 )(1−0.924)0.1565 = 0.557⎛The final step is to apply Formula 4-11:C water= 5 gallon ⎞ ⎛⎝⎜minute⎠⎟ 1 Btu ⎞ ⎛ 8.33lb ⎞ ⎛ 60minute⎞Btu⎝⎜lb•º F ⎠⎟⎝⎜gallon⎠⎟⎝⎜hour ⎠⎟ = 2499hr⋅º Fq = ε ( C min )( T hotinq −= T coldinε ( C) min= ) 0.557 ( T hotin( 2499 − T coldin)( 150 ) = −0.557 60) = ( 125,280 2499) ( 150 Btu − 60) = 125,28hrq = ε ( C⎛C 50% PG= 6 gallonmin )( T hotin− T coldin ) = 0.557( 2499) ( 150 − 60) = 125,280 Btu⎞ ⎛ Btu ⎞ ⎛ 8.54lb ⎞ ⎛ 60minute⎞Btuhr⎝⎜minute⎠⎟ 0.88⎝⎜lb•º F ⎠⎟⎝⎜gallon⎠⎟⎝⎜hour ⎠⎟ = 2705⎛hr⋅º FC 50% PG= 6 gallon ⎞ ⎛ Btu ⎞ ⎛ 8.54lb ⎞ ⎛ 60minute⎞Btu⎝⎜minute⎠⎟ 0.88⎝⎜lb•º F ⎠⎟⎝⎜gallon⎠⎟⎝⎜hour ⎠⎟ = 2705hr⋅º Fθ =Of these two capacitance rates, the one for the water side SOFTWARE-BASED ∆TLMTD θ =∆T HEAT EXCHANGER SELECTIONis smaller. Thus, the capacitance rate ratio is:The procedures introduced LMTD thus far — the LMTD method,C ratio= C min= 2499θ =∆Tand the effective (e) method —can be used for manualC max2705 = 0.924LMTDcalculations of heat exchanger performance. Both ofC ratio= C min= 2499these approaches demonstrate that the mathematicsC max2705 = 0.924⎡ ∆Tq 2= q 21necessary ∆T for sizing heat ⎡ ∆T exchangers can be complex. OneNTU = UA (= 150⎣⎢⎤⎡ ∆T 2 ⎦⎥q)20of the more complex 2= q 2⎤1 ⎢tasks⎥—that was not documented inC min2499 = 1.2q 2= q 2⎤1 ⎢ ⎥⎣ ∆T⎣ ∆T 2 ⎦2 ⎦NTU = UA (= 150 )201− e ( − NTU )(1−C ratio ) C 43ε =1− (C ratio)e ( = 1− e ( −1.2)(1−0.924)min2499 = 1.2 ⎡ ∆Tq 2= q 2⎤ ⎡140 − 65⎤Btu1 ⎢ ⎥ = 5,0000.0872∆T⎢− NTU )(1−C ratio ) (=1− (0.924)e −1.2 )(1−0.924)0.1565 = 0.557⎣ 1 ⎦ ⎣180 65⎥ = 3,261⎡ ∆T ∆Tq 2= q 2⎤ ⎦ hrq⎡140 − 65⎤Btu2= q 2⎤ ⎡140 − 65⎤Btu1 ⎢ ⎥ = 5,0001 ⎥ = 5,000 ∆T⎢1 ⎦ ⎣180 − 65⎥ = 3,261⎣ ∆T⎢1 ⎦ ⎣180 − 65⎥ = 3,261⎦ hr⎦ hreffectiveness (e)10.90.80.70.60.5Cmin/Cmax=0.2Cmin/Cmax=0.4Cmin/Cmax=0.6Cmin/Cmax=0.8Cmin/Cmax=1.0
Page 1 and 2: TMJOURNAL OF DESIGN INNOVATION FOR
Page 3 and 4: FROM THE GENERAL MANAGER & CEODear
Page 5 and 6: 1. INTRODUCTIONOne of humankind’s
Page 7 and 8: Figure 1-6 Figure 1-8Figure 1-9Figu
Page 9 and 10: Figure 2-4supplemental / auxiliary
Page 11 and 12: Figure 2-6solar thermalcollector ar
Page 13 and 14: Courtesy of Hydroflex Systems, Inc.
Page 15 and 16: Figure 2-15Figure 2-16Courtesy of B
Page 17 and 18: Courtesy of (a) Alfa Laval, and (b)
Page 19 and 20: Figure 2-24Courtesy of Alfa LavalLa
Page 21 and 22: how to account for plate thickness
Page 23 and 24: Figure 2-33Notice that the directio
Page 25 and 26: Figure 2-36Greywater heat exchanger
Page 27 and 28: 3. HEAT TRANSFER FUNDAMENTALSTwo fu
Page 29 and 30: Figure 3-3Heat output (Btu/hr/ft)60
Page 31 and 32: process. The value of (h) varies wi
Page 33 and 34: 4. THERMAL CHARACTERISTICS OF HEAT
Page 35 and 36: ( )q sAF T 4 4− T12 1 2⎡ 1 q =+
Page 37 and 38: 0.00032 ft ⋅sec( ) 1− ( ∆T )
Page 39 and 40: Figure 4-9enters the secondary side
Page 41: Figure 4-11FINNED-TUBE BASEBOARDout
Page 45 and 46: Figure 4-13cthe minimum surface are
Page 47 and 48: absorbing heat. The latter is a muc
Page 49 and 50: which to release its heat. However,
Page 51 and 52: 5. INSTALLATION DETAILS FOR HEAT EX
Page 53 and 54: Figure 5-4aFigure 5-5isolation /pur
Page 55 and 56: Figure 5-8temperature sensorsetpoin
Page 57 and 58: 6. HEAT EXCHANGER APPLICATIONSThis
Page 59 and 60: Figure 6-3a Figure 6-3bCourtesy of
Page 61 and 62: Figure 6-5to/fromspace heatingloadb
Page 63 and 64: Courtesy of Diversified Heat Transf
Page 65 and 66: Figure 6-10isolation / flush valves
Page 67 and 68: Courtesy of Diversified Heat Transf
Page 69 and 70: Figure 6-14modulating /condensingbo
Page 71 and 72: Figure 6-15multiple boilerstaging &
Page 73 and 74: Figure 6-17an existing boiler that
Page 75 and 76: Figure 6-19pipingbelowfrost lineret
Page 77 and 78: Figure 6-22air handlersbalancingzon
Page 79 and 80: APPENDIX B: CALEFFI PLUMBING COMPON

heating water

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