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

Lecture 7 03 CM30 /0/03 CM30 Transort I Part II: Heat Transfer Alied Heat Transfer: Heat Exchanger Modeling, Sizing, and Design Professor Faith Morrison Deartment of Chemical Engineering Michigan Technological University Before turning to radiation (last toic) we will discuss a few ractical alications How can we use Fundamental Heat Transfer to understand real devices like heat exchangers?

Lecture 7 03 CM30 /0/03 The Simlest Heat Exchanger: Double Pie Heat exchanger counter current T less T T less The heat transfer from the outside to the inside is just heat flux in an annular shell T Examle 4: Heat flux in a cylindrical shell Assumtions: long ie 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

Lecture 7 03 CM30 /0/03 BUT: The temerature difference between the fluid and the wall varies along the length of the heat exchanger. The Simlest Heat Exchanger: Double Pie Heat exchanger counter current T T T, outer bulk temerature, inner bulk temerature T T less less How can we adat h so that we can use the concet to characterize heat exchangers? L x T Let s look at the solution for radial conduction in an annulus Examle 4: Heat flux in a cylindrical shell Solution: q r c A r c T ln r c k Not constant Boundary conditions? 3

Lecture 7 03 CM30 /0/03 Examle 4: Heat flux in a cylindrical shell, Newton s law of cooling boundary Conditions Cooler Examle 4: Heat flux in a cylindrical shell R Solution for Heat Flux: Hot fluid at T b r R fluid at T b Calculate Total Heat flow: 4

Lecture 7 03 CM30 /0/03 T = driving temerature difference Q U A T Area must be secified when is reorted Q U AT 5

Lecture 7 03 CM30 /0/03 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 temerature, inner bulk temerature T T The Simlest Heat Exchanger: Double Pie Heat exchanger counter current less less We will do an oen system energy balance on a differential section to determine the correct average temerature difference to use. L x T 6

Lecture 7 03 CM30 /0/03 The Simlest Heat Exchanger: Double Pie Heat exchanger counter current Another way of looking at it: T less T T less T Inside Outside The Simlest Heat Exchanger: Double Pie Heat exchanger counter current Another way of looking at it: T less T T less Can do three balances: T. Balance on the inside system Inside Outside 7

Lecture 7 03 CM30 /0/03 The Simlest Heat Exchanger: Double Pie Heat exchanger counter current Another way of looking at it: T less T T less Can do three balances: T. Balance on the inside system. Balance on the outside system Inside Outside The Simlest Heat Exchanger: Double Pie Heat exchanger counter current Another way of looking at it: T less T T less Can do three balances: T. Balance on the inside system. Balance on the outside system 3. Overall balance Inside Outside 8

Lecture 7 03 CM30 /0/03 The Simlest Heat Exchanger: Double Pie Heat exchanger counter current Another way of looking at it: T less T T less T Inside Outside We can do: a macroscoic balances over the entire heat exchanger, or a seudo microscoic balance over a slice of the heat exchanger The Simlest Heat Exchanger: Double Pie Heat exchanger counter current Another way of looking at it: T less T T less T Inside Outside We can do: a macroscoic balances over the entire heat exchanger, or a seudo microscoic balance over a slice of the heat exchanger All the details of the algebra are here: www.chem.mtu.edu/%7efmorriso/cm30/double_ie.df 9

Lecture 7 03 CM30 /0/03 Pseudo Microscoic Energy Balance on a slice of the heat exchanger Pseudo Microscoic Energy Balance on a slice of the heat exchanger 0

Lecture 7 03 CM30 /0/03 Pseudo Microscoic Energy Balance on a slice of the heat exchanger Adiabatic Heat Exchanger > Q in = 0

Lecture 7 03 CM30 /0/03 Analysis of double ie heat exchanger (cont) Question: dq How can we write in in terms of T ' T? Answer: Use overall heat transfer coefficient, U Result: d T T dq in Cˆ m Cˆ m want to solve for T,T but this is a function of T T T (x) T T T T L T T We can write dq in / in terms of T Tif we introduce the overall heat transfer coefficient, U dq in UA( T T ) U ( R)( T T )

Lecture 7 03 CM30 /0/03 Result: d T T dq in Cˆ m Cˆ m We can write dq in / in terms of T Tif we introduce the overall heat transfer coefficient, U dq in UA( T T ) U ( R)( T T ) Result: d T T dq in Cˆ m Cˆ m 3

Lecture 7 03 CM30 /0/03 Result: d T T dq in Cˆ m Cˆ m Φ Φ Φ Φ Φ lnφ constant ΦΦ 4

Lecture 7 03 CM30 /0/03 Temerature rofile in a double ie heat exchanger: T TT T 0 e x 0 RU Cˆ m Cˆ m Useful result, but what we REALLY want is an easy way to relate Q in,overall to inlet and outlet temeratures. Begin by evaluating temerature rofile at exit conditions: T T ln T T U RL Cˆ m Cˆ m Note that the temerature curves are not necessarily linear. T, T' ( o C) 0 00 80 60 40 0 T 0 0 0 0. 0.4 0.6 0.8 x/l Distance down the heat exchanger T 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0. 0. x T T 0L L TT T T T T e 0 L RLU Cˆ m Cˆ m U 300W / mk RL 5.4m ' mc 6 kw / K mc 3 kw / K 5

Lecture 7 03 CM30 /0/03 T T ln T T U RL Cˆ m Cˆ m The terms aear in the overall macroscoic energy balances. We can therefore rearrange this equation to look like this: Q in, overall mcˆ T mcˆ T Q in, overall total heat transferred in exchanger UAT average average temerature driving force T less FINAL RESULT: Q U T T T T RL T T ln T T A T T less T Q UA T lm T lm =log mean temerature difference T lm is the correct average temerature to use for the overall heattransfer coefficients in a double ie heat exchanger. 6

Lecture 7 03 CM30 /0/03 Examle: Heat Transfer in a Double Pie Heat Exchanger: Geankolis 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 ie heat exchanger by 54,430 kg/h of 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 temerature and the heat transfer area needed. 7