How can we use Fundamental Heat Transfer to understand real devices like heat exchangers?

Lectures 7+8 04 CM30 /30/05 CM30 Transport I Part II: Heat Transfer Applied Heat Transfer: Heat Exchanger Modeling, Sizing, and Design Professor Faith Morrison Department of Chemical Engineering Michigan Technological University Before turning to radiation (last topic) we will discuss a few practical applications How can we use Fundamental Heat Transfer to understand real devices like heat exchangers?

Lectures 7+8 04 CM30 /30/05 The Simplest Heat Exchanger: Double Pipe Heat exchanger counter current T less hot T T cold less cold The heat transfer from the outside to the inside is just heat flux in an annular shell T hot 3 Example 4: Heat flux in a cylindrical shell Assumptions: long pipe steady state k = thermal conductivity of wall h, h = heat transfer coefficients Maybe we can use heat transfer coefficient to understand forcedconvection heat exchangers... BUT... Cooler fluid at T b r R R Hot fluid at T b 4

Lectures 7+8 04 CM30 /30/05 BUT: The temperature difference between the fluid and the wall varies along the length of the heat exchanger. The Simplest Heat Exchanger: Double Pipe Heat exchanger counter current T T T, outer bulk temperature, inner bulk temperature T T cold less hot less cold How can we adapt h so that we can use the concept to characterize heat exchangers? L x T hot 5 Let s look at the solution for radial conduction in an annulus Example 4: Heat flux in a cylindrical shell Solution: q r c A r c T ln r c k Not constant Boundary conditions? 6 3

Lectures 7+8 04 CM30 /30/05 Example 4: Heat flux in a cylindrical shell, Newton s law of cooling boundary Conditions Results: Radial Heat Flux in an Annulus ln ln ln 7 Cooler Example 4: Heat flux in a cylindrical shell R Solution for Heat Flux: Hot fluid at T b r ln R fluid at T b Calculate Total Heat flow: ln Note that total heat flow is proportional to bulk Δ and (almost) area of heat transfer 8 4

Lectures 7+8 04 CM30 /30/05 T = driving temperature difference 9 Q U A T Area must be specified when is reported Q U AT 0 5

Lectures 7+8 04 CM30 /30/05 Heat flux in a cylindrical shell: But, in an actual heat exchanger, and vary along the length of the heat exchanger What kind of average do we use? T T T, outer bulk temperature, inner bulk temperature T T cold The Simplest Heat Exchanger: Double Pipe Heat exchanger counter current less hot less cold We will do an open system energy balance on a differential section to determine the correct average temperature difference to use. L x T hot 6

Lectures 7+8 04 CM30 /30/05 The Simplest Heat Exchanger: Double Pipe Heat exchanger counter current Another way of looking at it: T less hot T T cold less cold T hot Inside System Outside System 3 The Simplest Heat Exchanger: Double Pipe Heat exchanger counter current Another way of looking at it: T less hot T T cold less cold Can do three balances: T hot. Balance on the inside system Inside System Outside System 4 7

Lectures 7+8 04 CM30 /30/05 The Simplest Heat Exchanger: Double Pipe Heat exchanger counter current Another way of looking at it: T less hot T T cold less cold Can do three balances: T hot. Balance on the inside system. Balance on the outside system Inside System Outside System 5 The Simplest Heat Exchanger: Double Pipe Heat exchanger counter current Another way of looking at it: T less hot T T cold less cold Can do three balances: T hot. Balance on the inside system. Balance on the outside system 3. Overall balance Inside System Outside System 6 8

Lectures 7+8 04 CM30 /30/05 The Simplest Heat Exchanger: Double Pipe Heat exchanger counter current Another way of looking at it: T less hot T T cold less cold T hot Inside System Outside System We can do: a macroscopic balances over the entire heat exchanger, or a pseudo microscopic balance over a slice of the heat exchanger 7 The Simplest Heat Exchanger: Double Pipe Heat exchanger counter current Another way of looking at it: T less hot T T cold less cold T hot Inside System Outside System We can do: a macroscopic balances over the entire heat exchanger, or a pseudo microscopic balance over a slice of the heat exchanger All the details of the algebra are here: www.chem.mtu.edu/~fmorriso/cm30/double_pipe.pdf 8 9

Lectures 7+8 04 CM30 /30/05 Pseudo Microscopic Energy Balance on a slice of the heat exchanger 9 Pseudo Microscopic Energy Balance on a slice of the heat exchanger 0 0

Lectures 7+8 04 CM30 /30/05 Pseudo Microscopic Energy Balance on a slice of the heat exchanger Adiabatic Heat Exchanger > Q in = 0 This expression characterizes the rate of change of heat transferred with respect to distance down the heat exchanger

Lectures 7+8 04 CM30 /30/05 The Simplest Heat Exchanger: Double Pipe Heat exchanger counter current T less hot Result of inside balance: Result of outside balance: Result of overall balance: Solve for temperature derivatives, and subtract: This depends on T T cold less cold T hot All the details of the algebra are here: www.chem.mtu.edu/~fmorriso/cm30/double_pipe.pdf 3 Analysis of double pipe heat exchanger (continued) T (x) T T T T T L T Rate of change of heat transferred with respect to distance down the heat exchanger Driving force for heat transfer Question: dq How can we write in in terms of T ' T? dx Answer: Define an overall heat transfer coefficient, U 4

Lectures 7+8 04 CM30 /30/05 Want to integrate to solve for, T (x) T T T T L T T but this is a function of For the differential slice of the heat exchanger that we are considering (modeling our ideas on Newton s law of cooling),? 5 For the differential slice of the heat exchanger that we are considering (modeling our ideas on Newton s law of cooling),? This is the missing piece that we needed. We can write in terms of if we define an overall heat transfer coefficient, 6 3

Lectures 7+8 04 CM30 /30/05 7 Φ (we ll assume is constant) Φ Φ Φ Φ lnφ constant ΦΦ B.C: 0, 8 4

Lectures 7+8 04 CM30 /30/05 T less hot Temperature profile in a double pipe heat exchanger: T T cold less cold T hot T TT T 0 e x 0 RU Cˆ m p Cˆ pm 9 T less hot T T cold T hot less cold x T T 0L L TT e T, T' ( o C) 0 00 80 60 40 0 T T 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0. U 300 W / mk RL 5.4m ' mc p 6 kw / K mc 3 kw / K p Note that the temperature curves are not linear. 0 0 0 0. 0.4 0.6 0.8 x/l Distance down the heat exchanger 30 5

Lectures 7+8 04 CM30 /30/05 Temperature profile in a double pipe heat exchanger: T TT T 0 e x 0 RU Cˆ m p Cˆ pm Useful result, but what we REALLY want is an easy way to relate, to inlet and outlet temperatures. At the exit:, T T ln T T U RL Cˆ m p Cˆ pm 3 T T ln T T U RL Cˆ m p Cˆ pm The terms appear in the overall macroscopic energy balances. We can therefore rearrange this equation by replacing the terms with : total heat transferred in exchanger ln average temperature driving force 3 6

Lectures 7+8 04 CM30 /30/05 T less hot FINAL RESULT: Q U T T T T RL T T ln T T A T T cold less cold T hot Q UA T lm T lm =log mean temperature difference T lm is the correct average temperature to use for the overall heattransfer coefficients in a double pipe heat exchanger. 33 T less hot FINAL RESULT: T T cold less cold Δ Δ ln Δ Δ T hot Q UA T lm T lm =log mean temperature difference T lm is the correct average temperature to use for the overall heattransfer coefficients in a double pipe heat exchanger. 34 7

Lectures 7+8 04 CM30 /30/05 Example: Heat Transfer in a Double Pipe Heat Exchanger: Geankoplis 4 th ed. 4.5 4 Water flowing at a rate of 3.85 kg/s is to be heated from 54.5 to 87.8 o C in a double pipe heat exchanger by 54,430 kg/h of hot gas flowing counterflow and entering at 47 o C (.005 / ). The overall heat transfer coefficient based on the outer surface is U o =69. W/m K. Calculate the exit gas temperature and the heat transfer area needed. 35 Cp, water kj/kg K 4.5 4.4 4.3 4. 4. 4. 4.9 4.8 4.7 4.6 y = 0.0006x + 4.504 R = 0.9737 4.5 40 50 60 70 80 90 00 T ( o C) 36 8

Lectures 7+8 04 CM30 /30/05 Summary: Double Pipe Heat Exchanger the driving force for heat transfer changes along the length of the heat xchanger For example, T 300 o C o T T C T 340 xx8 T 50 o C T 90 0 C o T T C T 50 xx 60 70 80 300 35 38 T(x) T 430 o C The correct average driving force for the whole xchanger is the log mean temperature difference T lm 37 Optimizing Heat Exchangers double-pipe: T T T T Q UA T lm To increase Q appreciably, we must increase A, i.e. R i But: only small increases possible increasing R i decreases h 38 9

Lectures 7+8 04 CM30 /30/05 - Shell and Tube Heat Exchanger (Same as double pipe H.E.) shell tube - Shell and Tube Heat Exchanger Geankoplis 4 th ed., p9 shell tube 39 Cross Baffles in Shell-and-Tube Heat Exchangers 40 0

Lectures 7+8 04 CM30 /30/05 And other more complex arrangements: shell 4 tube 4 For double-pipe heat exchanger: Q UA T lm For shell-and-tube heat exchangers: Q UA T lm F T calculated correction factor (obtain from experimentally determined charts) T m correct mean temperature difference for shell-and-tube heat exchangers 4

Lectures 7+8 04 CM30 /30/05 - Shell and Tube H.E. F T Shell-and-Tube Heat Exchangers (- exchanger, F T = ) F T,min =0.75 Efficiency is low when is below, -4 Shell and Tube H.E. T hi = hot, in T ho = hot, out T ci = cold, in T co = cold, out F T,min =0.75 Geankoplis 4 th ed., p95 43 Heat Exchanger Design To calculate Q, we need both inlet and outlet temperatures: Q UAT T m UA F T T lm T T Δ Δ ln Δ Δ T What if the outlet temperatures are unknown? i.e. the design/spec problem. 44

Lectures 7+8 04 CM30 /30/05 Example Problem: How will this heat exchanger perform? Water flowing at a rate of 0.73 / enters the inside of a countercurrent, double-pipe heat exchanger at 300.K and is heated by an oil stream that enters at 385 at a rate of 3. /. The heat capacity of the oil is.89 /, and the average heat capacity of the water of the temperature range of interest is 4.9 /. The overall heattransfer coefficient of the exchanger is 300. /, and the area for heat transfer is 5.4. What is the total amount of heat transferred? 45 T Example Problem: How will this heat exchanger perform? T T T Water flowing at a rate of enters the inside of a countercurrent, double-pipe heat exchanger at 300.K and is heated by an oil stream that enters at at a rate of. The heat capacity of the oil is, and the average heat capacity of the water of the temperature range of interest is. The overall heattransfer coefficient of the exchanger is, and the area for heat transfer is. What is the total amount of heat transferred? You try. 46 3

Lectures 7+8 04 CM30 /30/05 Example Problem: How will this heat exchanger perform? To calculate unknown outlet temperatures: Procedure:. Guess Q. Calculate outlet temperatures 3. Calculate Δ 4. Calculate Q 5. Compare, adjust, repeat 47 Example Problem: How will this heat exchanger perform? U A T_ Tprime_ m_water m'_oil cp_water cp_oil 0.3 kw/m^k 5.4 m^ 300 K 385 K 0.73 kg/s 3. kg/s 4.9 kj/kgk.89 kg/kgk Guess Q 00 kj/s Guess Q 00 kj/s 3 Guess Q 50 kj/s T_ 333 K T_ 366 K T_ 349 K Tprime_ 368 K Tprime_ 35 K Tprime_ 360 K Delta left 68 K Delta left 5 K Delta left 60 K Delta Right 5 K Delta Right 9 K Delta Right 36 K DeltaTlm 60 K DeltaTlm 33 K DeltaTlm 47 K Q_new 76.5 kw Q_new 5.4 kw Q_new 6. kw 4 Guess Q 70 kj/s 5 Guess Q 80 kj/s 6 Guess Q 78.6 kj/s T_ 356 K T_ 359 K T_ 359 K Tprime_ 357 K Tprime_ 355 K Tprime_ 355 K Delta left 57 K Delta left 55 K Delta left 55 K Delta Right 9 K Delta Right 6 K Delta Right 6 K DeltaTlm 4 K DeltaTlm 39 K DeltaTlm 39 K Q_new 9.0 kw Q_new 78. kw Q_new 79.9 kw 03HeatExchEffecExample.xlsx 48 4

Lectures 7+8 04 CM30 /30/05 Example Problem: How will this heat exchanger perform? U A T_ Tprime_ m_water m'_oil cp_water cp_oil 0.3 kw/m^k 5.4 m^ 300 K 385 K 0.73 kg/s 3. kg/s 4.9 kj/kgk.89 kg/kgk This procedure can be sped up considerably by the use of the concept of Heat-Exchanger Effectiveness, Guess Q 00 kj/s Guess Q 00 kj/s 3 Guess Q 50 kj/s T_ 333 K T_ 366 K T_ 349 K Tprime_ 368 K Tprime_ 35 K Tprime_ 360 K Delta left 68 K Delta left 5 K Delta left 60 K Delta Right 5 K Delta Right 9 K Delta Right 36 K DeltaTlm 60 K DeltaTlm 33 K DeltaTlm 47 K Q_new 76.5 kw Q_new 5.4 kw Q_new 6. kw 4 Guess Q 70 kj/s 5 Guess Q 80 kj/s 6 Guess Q 78.6 kj/s T_ 356 K T_ 359 K T_ 359 K Tprime_ 357 K Tprime_ 355 K Tprime_ 355 K Delta left 57 K Delta left 55 K Delta left 55 K Delta Right 9 K Delta Right 6 K Delta Right 6 K DeltaTlm 4 K DeltaTlm 39 K DeltaTlm 39 K Q_new 9.0 kw Q_new 78. kw Q_new 79.9 kw 03HeatExchEffecExample.xlsx 49 Heat Exchanger Effectiveness Consider a counter-current double-pipe heat exchanger: T co T hi T ho T hi T co T ho T ci distance along the exchanger T ci 50 5

Lectures 7+8 04 CM30 /30/05 Energy balance cold side: Q in, cold Q mc T T p cold co ci Equate: Energy balance hot side: Q Q mc T T in, hot p hot ho hi mc mc p hot p cold T co Tci Tc T ho Thi Th 5 Case : mc mc c p hot T T h T hi p cold cold fluid = minimum fluid Δ Δ T co T h We want to compare the amount of heat transferred in this case to the amount of heat transferred in a PERFECT heat exchanger. T c distance along the exchanger T ho T ci 5 6

Lectures 7+8 04 CM30 /30/05 If the heat exchanger were perfect, (no heat left un-transferred) cold side: Thi T co A T ho T ci this temperature difference only depends on inlet temperatures distance along the exchanger 53 Heat Exchanger Effectiveness, Q Q A Q mc T T p cold hi ci cold fluid = minimum fluid if is known, we can calculate Q without iterations 54 7

Lectures 7+8 04 CM30 /30/05 Case : mc mc c p hot T T h p cold hot fluid = minimum fluid Δ Δ T hi T h T co T c T ho T ci distance along the exchanger 55 If the heat exchanger were perfect, (no heat left un-transferred) hot side: T hi T co A Tho T ci The same temperature difference as before (inlets) distance along the exchanger 56 8

Lectures 7+8 04 CM30 /30/05 Heat Exchanger Effectiveness Q mc T T p hot hi ci Q Q A hot fluid = minimum fluid in general, Q mc T T p min hi if is known, we can calculate without iterations ci 57 But where do we get? The same equations we use in the trial-and-error solution can be combined algebraically to give as a function of (mc p ) min, (mc p ) max. countercurrent flow: e mc mc UA mc p min p max This relation is plotted in Geankoplis, as is the relation for co-current flow. p min e mc mc UA mc p min p min p max mc mc p min p min 58 9

Lectures 7+8 04 CM30 /30/05 Heat Exchanger Effectiveness for Double-pipe or - Shell-and-Tube Heat Exchangers counter-current co-current note: Geankoplis C min =(mc p ) min Geankoplis 4 th ed., p99 59 T Example Problem: How will this heat exchanger perform? T T T Water flowing at a rate of enters the inside of a countercurrent, double-pipe heat exchanger at 300.K and is heated by an oil stream that enters at at a rate of. The heat capacity of the oil is, and the average heat capacity of the water of the temperature range of interest is. The overall heattransfer coefficient of the exchanger is, and the area for heat transfer is. What is the total amount of heat transferred? You try. 60 30

Lectures 7+8 04 CM30 /30/05 Heat Exchanger Fouling material deposits on hot surfaces rust, impurities strong effect when boiling occurs clean scale adds an additional resistance to heat transfer fouled scale 6 Heat transfer resistances add effect of fouling U i oro U i oro resistance due to interface Ri or o Ro ln hi Ri k Ri hor o resistance due to limited thermal conductivity Ri or o Ro ln hi Ri hdiri k Ri hdoro hor o see Perry s Handbook, or Geankoplis 4 th ed. Table 4.9-, page 300 for values of h d 6 3

Lectures 7+8 04 CM30 /30/05 Heat Exchanger Fouling fouled scale Geankoplis, 4 th edition, p300 63 Next: Heat transfer with phase change Evaporators Radiation DONE 64 3