Chapter 5 Compact Heat Exchangers (Part III)

Transcription

1 09 Chater 5 Comact Heat Exchangers (Part III) 5.8 Plate-Fin Heat Exchangers Plate-fin exchangers have various geometries of fins to comensate the high thermal resistance by increasing the heat transfer area articularly if one of fluids is air or gas. This tye of exchanger has corrugated fins sandwiched between arallel lates or formed tubes. The late fins are categorized as triangular fin, rectangular fin, wavy fin, offset stri fin (OSF), louvered fin, and erforated fin. One of the most widely used enhanced fin geometries is the offset stri fin tye, which is shown in Figure Plate-fin heat exchangers have been introduced since the 90s in the auto industry, since the 940s in the aerosace industry. They are now used widely in many industries for aircraft, cryogenics, gas turbine, nuclear, and fuel cell. Plate-fin heat exchangers are generally designed for moderate oerating ressures less than 700 kpa (gauge ressure) and have been built with a surface area density of u to 5900 m 2 /m 3. Common fin thickness ranges between 0.05 and 0.25 mm. Fin heights may range from 2 to 25 mm. Although tyical fin densities are 20 to 700 fins/m, alications exist for as many as 200 fins/m [7]. Gas Air λ L 3 b Gas (fluid ) b 2 Air (fluid 2) L L 2 Figure 5.40 Plate-fin heat exchanger, emloying offset stri fin.

2 0 0.5 ( f -) w Plate a k j b (b 2 for fluid 2) b e (b -) l f i w Fin Plate Fin c d f g h Figure 5.4 Schematic of offset stri fin (OSF) geometry 5.8. Geometric Characteristics A schematic of a single-ass crossflow late-fin heat exchanger, emloying offset stri fins, is shown in Figure The idealized fin geometry is shown in Figure 5.4. Defining the total heat transfer area for each fluid is imortant in the analysis. The total area consists of the rimary area and the fin area. The rimary area consists of the late area excet the fin base area, multi-assage side walls, and multi-assage front and back walls. It is ractical to assume that the numbers of assages for the hot fluid side and the cold fluid side are N and N + to minimize the heat loss to the ambient. The to and bottom assages in Figure 5.40 are designated to be cold fluid. The number of assages (don t confuse with the number of asses) can be obtained from an exression for L 3 as L = w (5.243) N b + N + b2 + 2 N + { 3 length of fluid length of fluid 2 thickness of total lates Solving for N gives the number of assages for hot fluid N L b 2 b + b w = (5.244) 2 w By definition, the number of assages counts the number based on one flow assage between two lates, not all the individual channels between the lates. The total number of fins for fluid (hot) is calculated by n L N f = (5.245) f For fluid 2 (cold), ( N ) L n (5.245a) 2 f 2 = + f 2

3 where f is the fin itch that is usually obtained by making inverse of the fin density. The total number of fins n f is based on the shaded area (a-c-d-e-f-g-h-j-a) in Figure 5.4 counting as the unit fin. Since total rimary area = total late areas fin base areas + assage side wall areas + assage front and back wall areas, the rimary area for fluid is exressed by A ( b + 2 ) L ( N ) = 2 LL 2N 2 L2n f + 2b L2N w + (5.246) total late area fin base area assage side wall area assage front and back wallarea The number of offset stri fins n off er the number of fins is obtained by L n off 2 = (5.247) λ2 L 2 n off = (5.247a) λ where λ,2 is the offset stri fin length for fluid and 2. The total fin area A f consists of the fin area and offset-stri edge areas. A ( b ) L n + 2( b ) n n + ( ) ( n ) = 2 2 f off f f off n f + f n f (5.248) f 2 fin surface areas (d-e and g-f) offset-stri edge areas 2(b-c-d-e) internal offset-stri edge area 2(i-j-k-l-i) st & last offset-stri edge area, 2(a-b-i-j-a) Note that the cross-shaded area (a-b-l-k-a) was not included in the offset-stri edge area because the area is closely blocked by the next stri fin as shown in Figure 5.4 so that no heat transfer at the area is exected. The total heat transfer area A t is the sum of the rimary area and the fin area as A = A + A (5.249) t f The free-flow (cross sectional) area A c is obtained by A ( b )( f ) n = (5.250) c f It is assumed in Equation (5.250) that there exists a small ga between the offset stri fins, whereby the next stri fin shown in the unit fin in Figure 5.4 is not considered as an obstructing structure to the free flow. The frontal area A fr for fluid where fluid is entering is defined by A fr = (5.25) LL 3

4 2 The hydraulic diameter for fluid is generally defined by 4A L c 2 D h = (5.252) At For the fin efficiency η f of the offset stri fin, it is assumed that the heat flow from both lates is uniform and the adiabatic lane occurs at the middle of the late sacing b. Hence, the fin rofile length L f is defined by b L f = (5.253) 2 The m value is obtained using Equation (5.96) as 2h m = + (5.254) k f λ The single fin efficiency η f is obtained using Equation (5.95) by ( m L ) tanh f η f = (5.255) m L f The overall surface (fin) efficiency η o is then obtained using Equations (5.99), (5.248) and (5.249) as Af ηo = ( η f ) (5.256) A t Correlations for Offset Stri Fin (OSF) Geometry This geometry has one of the highest heat transfer erformances relative to the friction factors. Extensive analytical, numerical and exerimental investigations have been conducted over the last 50 years. The most comrehensive correlations for j and f factors for the laminar, transition, and turbulent regions are rovided by Manglik and Bergles [27] in 995 as follows.

5 3 + = Re Re λ λ f f f f b b j (5.257) + = Re Re λ λ f f f f b b f (5.258) where the hydraulic diameter was defined by them as λ λ λ = f f f MB h b b b D 2 4 _ (5.258) This is an aroximation of Equation (5.252). However, Equation (5.258) is in excellent agreement with Equation (5.252), which indicates the general definition of hydraulic diameter, Equation (5.252), can be used in lace of Equation (5.258). The correlations of Manglik and Bergles [27] were comared with the exeriments (surface /8-9.86) reorted by Kays and London [8] in Figure The comarison for both j and f shows good agreement. Note that the comarison covers from laminar through turbulent regions.

6 4 f 0. j Re Figure 5.42 Colburn factor j and friction factor f for offset-stri-fin (OSF) tye late-fin heat exchangers. The correlation of Manglik and Bergles [27] were comared with the exeriments (/ surface) of Kays and London [8].

7 5 Examle 5.8. Plate-Fin Heat Exchanger A gas-to-air single-ass crossflow heat exchanger is designed for heat recovery from the exhaust gas to reheat incoming air in a solid oxide fuel cell (SOFC) co-generation system. Offset stri fins of the same geometry are emloyed on the gas and air sides; the geometrical roerties and surface characteristics are rovided in Figures E5.8. and E Both fins and lates (arting sheets) are made from Inconel 625 with k=8 W/mK. The anode gas flows in the heat exchanger at m 3 /s and 900 C. The cathode air on the other fluid side flows at.358 m 3 /s and 200 C. The inlet ressure of the gas is at 60 kpa absolute whereas that of air is at 200 kpa absolute. Both the gas and air ressure dros are limited to 0 kpa. It is desirable to have an equal length for L and L 2. Design a gas-to-air single-ass crossflow heat exchanger oerating at ε = Determine the core dimensions of this exchanger. Then, determine the heat transfer rate, outlet fluid temeratures and ressure dros on each fluid. Use the roerties of air for the gas. Use alsothe following geometric information given. Descrition value Fin thickness 0.02 mm Plate thickness w 0.5 mm Fin density N f 782 m - Sacing between lates b and b mm Offset stri length λ and λ mm Gas Air λ L 3 b Gas (fluid ) b 2 Air (fluid 2) L L 2 Figure E5.8. Plate-fin heat exchanger, emloying offset stri fin.

8 6 0.5 ( f -) w Plate a k j b (b 2 for fluid 2) b e (b -) l f i w Fin Plate Fin c d f g h Figure E5.8.2 Schematic of offset stri fin (OSF) geometry MathCAD format solution: Design concet is to develo a MathCAD model as a function of the dimensions of the sizing and then solve the sizing with the design requirements. Proerties: We will use the arithmetic average temeratures as the aroriate mean temerature on each side by assuming the outlet temeratures at this time. 900 C C 200 C C T gas := = 923.5K T 2 air := = K (E5.8.) Gas (subscrit ) Air (subscrit 2) ρ defined_later ρ 2 defined_later c := 26 J c kgk 2 := 073 J kgk (E5.8.2) k := W k m K 2 := W m K µ Ns := µ Ns := m 2 m 2 Pr := 0.73 Pr 2 := The thermal conductivities of the fins and walls are given as k f := 8 W k m K w := 8 W m K (E5.8.3)

9 7 Given information: T i := 900 C gas inlet temerature T 2i := 200 C air inlet temerature P i := 60kPa gas inlet ressure P 2i := 200kPa air inlet ressure Q := m3 volume flow rate at gas side s Q 2 :=.358 m3 volume flow rate at air side s (E5.8.4) (E5.8.5) (E5.8.6) (E5.8.7) (E5.8.8) (E5.8.9) Air and gas densities (inlet): The gas constants for air is known to be R 2 := J kgk (E5.8.0) We calculate the air and gas inlet densities using the ideal gas law with an assumtion that the gas constants for air and gas are equal. assuming R := R 2 P i ρ i := ρ R T i = kg i m 3 P 2i ρ 2i := ρ R 2 T 2i =.473 kg 2i m 3 (E5.8.) (E5.8.2) (E5.8.3) Hence, the mass flow rate for each fluid is calculated as mdot := ρ i Q mdot =.66 kg s mdot 2 := ρ 2i Q 2 mdot 2 = 2 kg s (E5.8.4) (E5.8.5)

10 8 Design requirements: In this design roblem, the effectiveness is major concern for heat recovery. The designers want to derive the relationshi between the effectiveness and the sizing. The frontal widths for both fluids are required to be the same (L =L 2 ). ε effectiveness P 0kPa P 2 0kPa L L 2 desirable Geometric information: ressure dro at gas side ressure dro at air side := 0.02mm fin thickness w := 0.5mm late thickness N f := 782m fin density f := fin itch N f f2 := f f =.2788mm b := 2.49mm late distance b 2 := 2.49mm λ := 3.75mm offset stri length λ 2 := 3.75mm (E5.8.6) (E5.8.7) (E5.8.8) (E5.8.9) (E5.8.20) (E5.8.2) (E5.8.22) (E5.8.22a) (E5.8.23) (E5.8.24) (E5.8.25) (E5.8.26) (E5.8.27) Assume the sizing of the heat exchanger Initially, assume the sizing to have the instant values and udate later the dimensions when the final sizing is determined. L := 0.303m L 2 := 0.303m L 3 := 0.948m (E5.8.28) It is assumed that the number of assages for the gas side and the air side are N and N +, resectively, to minimize the heat loss to the ambient. Using Equation (5.243), L 3 is calculated by

11 9 b 2 L 3 N b + N N + w (E5.8.29) Thus, the total number of assages N for the gas side is obtained using Equation (5.244) N ( L 3 ) L 3 b 2 2 w := N b + b L 3 w = (E5.8.30) Note that the total number of assages is a function of L 3. The total number of fins for each fluid is obtained n f ( L, L 3 ) n f2 ( L 2, L 3 ) L := N ( L 3 ) n f L, L 3 f L 2 := ( N ( L 3 ) + ) n f2 L 2, L 3 f2 = = (E5.8.3) (E5.8.32) The total rimary area A for each fluid is calculated using Equation (5.246). := 2 L A L, L 2, L 3 := 2 L A 2 L, L 2, L 3 L 2 N L 3 ( + ) L 2 N L 3 ( + ) 2 L 2 n f L, L 3 + 2b L 2 N L b w L N L 3 ( ) (E5.8.33) 2L n f2 L 2, L 3 + 2b 2 L N L b + 2 w L 2 N L 3 (E5.8.34) m 2 A L, L 2, L 3 = A 2 L, L 2, L 3 = 27.43m 2 (E5.8.35) The number of offset stri fins for each fluid (er the number of fins) is obtained n off ( L 2 ) L 2 := n λ off2 L L := λ 2 (E5.8.36) The total fin area A f for each fluid is obtained using Equation (5.248) as := 2 ( b ) A f L, L 2, L 3 ( ) L 2 n f L, L 3 + f n off L 2 n f L, L f n f L, L 3 (E5.8.37)

12 20 := 2 ( b 2 ) A f2 L, L 2, L 3 ( ) L n f2 L 2, L 3 + f2 n off2 L n f2 L 2, L f2 n f2 L 2, L 3 (E5.8.38) m 2 A f L, L 2, L 3 = A f2 L, L 2, L 3 = m 2 (E5.8.39) The total surface area A t for each fluid is the sum of the fin area A f and the rimary area A. := A f ( L, L 2, L 3 ) + A ( L, L 2, L 3 ) A t L, L 2, L 3 := A f2 ( L, L 2, L 3 ) + A 2 ( L, L 2, L 3 ) A t2 L, L 2, L m 2 A t L, L 2, L 3 = A t2 L, L 2, L 3 = m 2 (E5.8.40) (E5.8.4) (E5.8.42) The free-flow area A c for each side is obtained using Equation (5.250) as ( b ) ( f ) A c L, L 3 A c2 L 2, L 3 := n f L, L 3 A c L, L 3 ( b 2 ) ( f2 ) := n f2 L 2, L 3 A c2 L 2, L 3 = 0.052m 2 = 0.058m 2 (E5.8.43) (E5.8.44) The frontal area for each fluid is defined using Equation (5.25) by L L 3 A fr L, L 3 A fr2 L 2, L 3 := A fr L, L 3 L 2 L 3 := A fr2 L 2, L 3 = 0.287m 2 = 0.287m 2 (E5.8.45) (E5.8.46) The hydraulic diameter is defined using Equation (5.252) by D h L, L 2, L 3 D h2 L, L 2, L 3 4A c L, L 3 L 2 := D A t ( L, L 2, L 3 ) h L, L 2, L 3 4A c2 L 2, L 3 L := D A t2 ( L, L 2, L 3 ) h2 L, L 2, L 3 =.557 mm =.557 mm (E5.8.47) (E5.8.48) The hydraulic diameter of Equation (5.258) defined by Manglik and Bergles [27] is calculated in Equation (E5.8.49) to comare with the general definition of Equation (E5.8.47). We find that they are in good agreement. Therefore, we use Equations (E5.8.47) and (E5.8.48) in the rest of the calculations.

13 2 4 f b λ D h_mb := 2 ( f ) λ + b λ + b ( f D + ) h_mb =.556mm (E5.8.49) The orosity σ for each fluid is calculated as σ ( L, L 3 ) σ 2 ( L 2, L 3 ) A c L, L 3 := σ A fr ( L, L 3 ) L, L 3 A c2 L 2, L 3 := σ A fr2 ( L 2, L 3 ) 2 L 2, L 3 = = (E5.8.50) (E5.8.5) The volume of the exchanger for each fluid is calculated as b L V L, L 2, L 3 b 2 L V 2 L, L 2, L 3 := L 2 N L 3 V L, L 2, L 3 ( ) := L 2 N L 3 + V 2 L, L 2, L 3 = 0.036m 3 = m 3 (E5.8.52) (E5.8.53) The surface area density β for each side is calculated as β L, L 2, L 3 β 2 L, L 2, L 3 A t L, L 2, L 3 := β V ( L, L 2, L 3 ) L, L 2, L 3 A t2 L, L 2, L 3 := β V 2 ( L, L 2, L 3 ) 2 L, L 2, L 3 = 2267 m2 m 3 = 2267 m2 m 3 (E5.8.54) (E5.8.55) The mass velocities, velocities, Reynolds Numbers are calculated as G ( L, L 3 ) G 2 ( L 2, L 3 ) Re L, L 2, L 3 mdot := G A c ( L, L 3 ) L, L 3 mdot 2 := G A c2 ( L 2, L 3 ) 2 L 2, L 3 D h ( L L 2 ) = kg m 2 s = kg G L, L 3,, L 3 := Re µ L, L 2, L 3 m 2 s = (E5.8.56) (E5.8.57) (E5.8.58)

14 22 Re 2 L, L 2, L 3 D h2 ( L L 2 ) G 2 L 2, L 3,, L 3 := Re µ 2 L, L 2, L 3 2 = (E5.8.59) Colburn factors and friction factors The Reynolds numbers indicate laminar flows at both the gas and air sides. J and f factors are calculated using Equations (5.257) and (5.258) as := Re ( L, L 2, L 3 ) j a L, L 2, L 3 := j b L, L 2, L Re L, L 2, L 3 f b 0.54 f b The Colburn factor j at the gas side is calculated by λ λ f f (E5.8.60) (E5.8.6) j a ( L, L 2, L 3 ) j b ( L, L 2, L 3 ) = 0.07 j L, L 2, L 3 := j L, L 2, L 3 (E5.8.62) := Re ( L, L 2, L 3 ) f a L, L 2, L 3 := f b L, L 2, L Re L, L 2, L 3 The friction factor f at the gas side is f b f b 0.92 λ λ f f (E5.8.63) (E5.8.64) f a ( L, L 2, L 3 ) f b ( L, L 2, L 3 ) = f L, L 2, L 3 := f L, L 2, L 3 (E5.8.65) := Re 2 ( L, L 2, L 3 ) j 2a L, L 2, L f2 b λ f (E5.8.66)

15 23 := j 2b L, L 2, L 3.34 Re 2 L, L 2, L 3 The Colburn factor j 2 at the air side is j 2a ( L, L 2, L 3 ) j 2b ( L, L 2, L 3 ) f2 b λ j 2 L, L 2, L 3 := j 2 L, L 2, L 3 := Re 2 ( L, L 2, L 3 ) f 2a L, L 2, L 3 := f 2b L, L 2, L Re 2 L, L 2, L 3 The friction factor f 2 at the air side is f2 b f2 b f2 = 0.04 λ λ f2 f (E5.8.67) (E5.8.68) (E5.8.69) (E5.8.70) f 2a ( L, L 2, L 3 ) f 2b ( L, L 2, L 3 ) = 0.05 f 2 L, L 2, L 3 := f 2 L, L 2, L 3 (E5.8.7) Heat Transfer Coefficients The heat transfer coefficients in terms of the j and f factors are calculated using Equation (5.230) as h L, L 2, L 3 h 2 L, L 2, L 3 G ( L L 3 ) j L, L 2, L 3, c := h 2 L, L 2, L 3 3 Pr G 2 ( L 2 L 3 ) j 2 L, L 2, L 3, c 2 := h 2 2 L, L 2, L 3 3 Pr 2 Overall surface (fin) efficiency = W m 2 K = W m 2 K (E5.8.72) (E5.8.73) Since the offset stri fins are used on both the gas and air sides, we will use the multile fin analysis. m values are obtained using Equation (5.254) as

16 24 m L, L 2, L 3 := m 2 L, L 2, L 3 := 2h L, L 2, L 3 k f 2h 2 L, L 2, L 3 k f + + λ λ (E5.8.74) (E5.8.75) Using Equation (5.253), the fin length L f for each fluid will be b b 2 L f := L 2 f2 := 2 (E5.8.76) The single fin efficiency η f for each fluid is calculated using Equation (5.255) as η f L, L 2, L 3 ( L f ) tanh m L, L 2, L 3 := η m ( L, L 2, L 3 ) L f2 L, L 2, L 3 f ( L f2 ) L f2 tanh m 2 L, L 2, L 3 := m 2 L, L 2, L 3 (E5.8.77) The overall surface (fin) efficiencies η o are obtained using Equation (E5.256) ( η f ( L, L 2, L 3 )) A f( L, L 2, L 3 ) (, ) η o L, L 2, L 3 η o2 L, L 2, L 3 = := η A t L, L 2 L o L, L 2, L 3 3 ( η f2 ( L, L 2, L 3 )) A f2( L, L 2, L 3 ) (, ) = := η A t2 L, L 2 L o2 L, L 2, L 3 3 (E5.8.78) (E5.8.79) The conduction area for the wall thermal resistance is given by 2 L A w L, L 2, L 3 ( ) := L 2 N L 3 + A w L, L 2, L 3 = 29.85m 2 (E5.8.80) The thermal resistance of the wall is given by R w L, L 2, L 3 UA L, L 2, L 3 w := k w A w L, L 2, L 3 := η o ( L, L 2, L 3 ) h ( L, L 2, L 3 ) A t L, L 2, L 3 (E5.8.8) (E5.8.82)

17 25 UA 2 L, L 2, L 3 UA L, L 2, L 3 := η o2 ( L, L 2, L 3 ) h 2 ( L, L 2, L 3 ) A t2 L, L 2, L 3 := UA ( L, L 2, L 3 ) + R w ( L, L 2, L 3 ) + UA 2 L, L 2, L 3 (E5.8.83) (E5.8.84) The overall heat transfer coefficient times the heat transfer area is given by = 3334m 2 W UA L, L 2, L 3 ε -NTU Method m 2 K (E5.8.84a) The heat caacity rates are given as C := mdot c C W = K C 2 := mdot 2 c 2 C W = K C min := min( C, C 2 ) C min W = K C max := max( C, C 2 ) C max W = K C min C r := C C r = 0.87 max (E5.8.85) (E5.8.86) (E5.8.87) (E5.8.88) (E5.8.89) The number of transfer unit (NTU) is calculated using Equation (5.77) by NTU L, L 2, L 3 UA L, L 2, L 3 := NTU L C, L 2, L 3 min = 7.33 (E5.8.90) The effectiveness of the late-fin heat exchanger is calculated using Equation (5.38) by := ex NTU( L C, L 2, L 3 ) 0.22 r ε hx L, L 2, L ε hx L, L 2, L 3 = 0.78 ex C r NTU L, L 2, L 3 (E5.8.9)

18 26 The heat transfer rate is calculated by ε hx ( L, L 2, L 3 ) C min ( T i T 2i ) q L, L 2, L 3 := q L, L 2, L 3 The air outlet temeratures are calculated by = W (E5.8.92) T i ε hx ( L, L 2, L 3 ) C min ( T i T 2i ) T o L, L 2, L 3 T 2o L, L 2, L 3 := T C o L, L 2, L 3 T 2i ε hx ( L, L 2, L 3 ) C min + ( T i T 2i ) Pressure Dros := T C 2o L, L 2, L 3 2 = C = C (E5.8.93) (E5.8.94) The ressure dros for the late-fin heat exchanger is exressed using Equation (5.202) as P G 2 2ρ i σ 2 + K c + 2 ρ i ρ o 4fL ρ i + σ 2 K D h ρ e m ρ i ρ o (E5.8.95) Contraction and Exansion Coefficients, K c and K e WE develo equations for Figure 5.3. The jet contraction ratio for the square-tube geometry is given using Equation (5.20) by C c_tube ( σ) := e σ (E5.8.96) The friction factor used for this calculation is given in Equation (5.205). f d ( Re) := 0.049Re 0.2 if Re Re otherwise (E5.8.97) The velocity-distribution coefficient for circular tubes is given in Equation (5.207). K d_tube ( Re) := f d ( Re) f d ( Re) + if Re otherwise (E5.8.98)

19 27 which is converted to the square tubes as shown in Equation (5.209) K d_square ( Re) := +.7 K d_tube ( Re) if Re otherwise (E5.8.99) The contraction coefficient is given in Equation (5.203). K c_square ( σ, Re) 2C c_tube ( σ) + C c_tube ( σ) 2 2K d_square ( Re) := C c_tube ( σ) 2 (E5.8.00) The exansion coefficient is given in Equation (5.204). K e_square ( σ, Re) := 2K d_square ( Re) σ + σ 2 (E5.8.0) K c and K e for each fluid are determined with the orosity and the Reynolds number where the Reynolds number is assumed to be fully turbulent (Re=0 7 ) because of the frequent boundary layer interrutions due to the offset stri fins. The values may also be obtained grahically from Figure 5.3. K c_square σ ( L, L 3 ), 0 7 K c L, L 3 K c2 L 2, L 3 := K c L, L 3 K c_square σ 2 ( L 2, L 3 ), 0 7 K e L, L 3 = 0.33 := K c2 L 2, L 3 K e_square σ ( L, L 3 ), 0 7 K e2 L 2, L 3 := K e L, L 3 K e_square σ 2 ( L 2, L 3 ), 0 7 = = 0.39 := K e2 L 2, L 3 = (E5.8.02) (E5.8.03) (E5.8.04) (E5.8.05) Gas and air densities The gas and air outlet densities are a function of both the outlet temeratures and the outlet ressures that are unknown. The outlet densities are exressed using the ideal gas law as ρ o P o, L, L 2, L 3 ρ 2o P 2o, L, L 2, L 3 P o := R T o L, L 2, L 3 P 2o := R 2 T 2o L, L 2, L 3 (E5.8.06) (E5.8.07)

20 28 Since ρ m is the mean density and defined as ρ m 2 + ρ i ρ o (E5.8.08) Thus, for each fluid, ρ m P o, L, L 2, L 3 := ρ 2m P 2o, L, L 2, L 3 := ρ i ρ o P o, L, L 2, L 3 + ρ 2i ρ 2o P 2o, L, L 2, L 3 (E5.8.09) (E5.8.0) Pressure dros We divide Equation (E5.8.95) into several short equations to fit the resent age. P 2 ( L, L 3 ) := := 2 K 23 P o, L, L 2, L 3 2 G L, L 3 2ρ i P 23 P o, L, L 2, L 3 := 2 σ L, L 3 + K c L, L 3 ρ i ρ o P o, L, L 2, L 3 2 G L, L 3 2ρ i 2 L 2 (, ) 4f L, L 2, L 3 + D h L, L 2 L 3 K 23 P o, L, L 2, L 3 ρ i ρ m P o, L, L 2, L 3 (E5.8.) (E5.8.2) (E5.8.3) G P 34 ( P o, L, L 2, L 3 ) L, L 3 σ 2ρ ( L, L 3 ) 2 ρ K e ( L, L 3 ) i := i ρ o ( P o, L, L 2, L 3 ) (E5.8.4) The ressure dro at the gas side, which is Equation (E5.8.95), is the sum of the above three equations, Equations (E5.8.), (E5.8.3) and (E5.8.4) as := P 2 ( L, L 3 ) P 23 ( P o, L, L 2, L 3 ) P P o, L, L 2, L 3 + P 34 P o, L, L 2, L 3 Now the gas outlet ressure can be obtained using the MathCAD s 'root' function. (E5.8.5)

21 29 A guess value that would be a closest value to the solution is needed to be rovided for the root function as Solving P o := 60kPa := root ( P i P o ) P ( P o, L, L 2, L 3 ) P o L, L 2, L 3, P o (E5.8.6) (E5.8.7) The gas outlet ressure is finally obtained by = kPa P o L, L 2, L 3 (E5.8.8) The ressure dro at the gas side is calculated P i P o ( L, L 2, L 3 ) P L, L 2, L 3 := P L, L 2, L 3 = 0.004kPa (E5.8.9) The air outlet ressure can be obtained in a similar way. P2 2 ( L 2, L 3 ) := 2 G 2 L, L 3 2ρ 2i 2 σ 2 L 2, L 3 + K c2 L 2, L 3 (E5.8.20) := 2 K2 23 P 2o, L, L 2, L 3 P2 23 P 2o, L, L 2, L 3 := P2 34 P 2o, L, L 2, L 3 := P2 P 2o, L, L 2, L 3 ρ 2i ρ 2o P 2o, L, L 2, L 3 2 G 2 L, L 3 2ρ 2i 2 G 2 L, L 3 2ρ 2i L (, ) 4f 2 L, L 2, L 3 + D h2 L, L 2 L 3 K2 23 P 2o, L, L 2, L 3 2 σ 2 L 2, L 3 K e2 L 2, L 3 := P2 2 ( L 2, L 3 ) P2 23 ( P 2o, L, L 2, L 3 ) A guess value is needed for MathCAD as ρ 2i ρ 2m P 2o, L, L 2, L 3 ρ 2i ρ 2o P 2o, L, L 2, L 3 + P2 34 P 2o, L, L 2, L 3 (E5.8.2) (E5.8.22) (E5.8.23) (E5.8.24)

22 30 P 2o := 200kPa := root ( P 2i P 2o ) P2 ( P 2o, L, L 2, L 3 ) P 2o L, L 2, L 3 P 2o L, L 2, L 3 = 92.26kPa, P 2o (E5.8.25) (E5.8.26) (E5.8.27) The ressure dro at the air side is calculated by P 2i P 2o ( L, L 2, L 3 ) P 2 L, L 2, L 3 := P 2 L, L 2, L 3 = kpa (E5.8.28) Sizing of the Plate-Fin Heat Exchanger The calculations for the sizing usually require a number of iterations, which is a timeconsuming work. This can be easily erformed by MathCAD. The three unknowns of L, L 2, and L 3 are found using a combination of three requirements. The higher ressure dro for fluid (gas) is exected due to the higher volume flow rate. The following combination of the three requirements will be sufficient for the solution. Guess values can be udated and iterated with the new values to have a convergence. Guess values L := m L 2 := m L 3 := m Given ε hx L, L 2, L 3 0kPa P L, L 2, L 3 (E5.8.29) (E5.8.30) (E5.8.3) (E5.8.32) L L 2 (E5.8.33) L L 2 L 3 := Find L, L 2, L 3 (E5.8.34) The dimensions of the sizing with the requirements are finally found as L = 0.303m L 2 = 0.303m L 3 = 0.948m (E5.8.35) Now return to Equation (E5.8.28) and udate the values for all the correct numerical values. Summary of the results and information Given information:

23 3 T i = 900 C gas inlet temerature T 2i = 200 C air inlet temerature P i = 60 kpa gas inlet ressure P 2i = 200 kpa air inlet ressure Q Q 2 Requirements: = m3 volume flow rate at gas side s =.358 m3 volume flow rate at air side s (E5.8.36) (E5.8.37) (E5.8.38) (E5.8.39) (E5.8.40) (E5.8.4) e hx effectiveness P 0kPa ressure dro at gas side P 2 0kPa ressure dro at air side L L 2 desirable Dimensions of the sizing: (E5.8.42) (E5.8.43) (E5.8.44) (E5.8.45) L = 0.303m L 2 = 0.303m L 3 = 0.948m Outlet temeratures and ressure dros: C T o L, L 2, L 3 = gas outlet temerature C T 2o L, L 2, L 3 = air outlet temerature 0 kpa P L, L 2, L 3 = ressure dro at gas side 7.78 kpa P 2 L, L 2, L 3 = ressure dro at air side Densities of gas and air at inlet and outlet ρ i = kg ρ o P o ( L, L 2, L 3 ), L, L 2, L 3 m 3 = kg m 3 (E5.8.46) (E5.8.47) (E5.8.48) (E5.8.49) (E5.8.50) (E5.8.5)

24 32 ρ 2i =.473 kg ρ 2o P 2o ( L, L 2, L 3 ), L, L 2, L 3 m 3 Thermal quantities: 7.34 = kg NTU L, L 2, L 3 = number of transfer unit ε hx L, L 2, L 3 = effectiveness q L, L 2, L 3 = W heat transfer rate W h L, L 2, L 3 = heat transfer coefficient h 2 L, L 2, L 3 m 2 K = W Hydraulic quantities: m 2 K.557 mm D h L, L 2, L 3 = hydraulic diameter =.557 mm D h2 L, L 2, L kg G L, L 3 G 2 L 2, L 3 = mass velocity m 2 s = 8.89 kg m 2 s Re L, L 2, L 3 = Reynolds number = Re 2 L, L 2, L 3 Geometric quantities: m 3 (E5.8.52) (E5.8.53) (E5.8.54) (E5.8.55) (E5.8.56) (E5.8.57) (E5.8.58) (E5.8.59) (E5.8.60) (E5.8.6) (E5.8.62) (E5.8.63) b = 2.49 mm b 2 = 2.49 mm f =.279 mm f2 =.279 mm λ = 3.75 mm λ 2 = 3.75 mm = 0.02 mm fin thickness w = 0.5 mm wall thickness (E5.8.64) (E5.8.65) (E5.8.66) (E5.8.67) (E5.8.68)

25 33 N ( L 3 ) = number of assages at gas side N ( L 3 ) + = number of assages at air side σ ( L, L 3 ) = orosity σ 2 ( L 2, L 3 ) = m2 β L, L 2, L 3 = 2267 m2 β 2 L, L 2, L 3 = surface area density m 3 m m 2 A t L, L 2, L 3 = total heat transfer area = m 2 A t2 L, L 2, L 3 (E5.8.69) (E5.8.70) (E5.8.7) (E5.8.72) (E5.8.73) (E5.8.74) (E5.8.75) (E5.8.76) The results satisfies the design requirements.

26 34 References. Kakac, S. and Liu, H., Heat Exchangers, CRC Press, New York, Rohsennow, W. M., Hartnett, J. P., and Cho, Y. I., Handbook of Heat Transfer, 3 rd Ed., McGraw-Hill, New York, Smith, E. M., Thermal Design of Heat Exchangers, John Wiley & Sons, New York, Incroera, F. P., Dewitt, D. P., Bergman, T. L., and Lavine, A. S., Fundamentals of Heat and Mass Transfer, 6 th Ed., John Wiley & Sons, Janna, W. S., Design of Fluid Thermal Systems, 2 nd Ed., PWS Publishing Co., Sieder, E. N. and Tate, G. E., Heat Transfer and Pressure Dro if Liquids in Tubes, Ind. Eng. Chem., Vol. 28, , Gnielinski, V., New Equation for Heat and Mass Transfer in Turbulent Pie and Channel Flow, Int. Chem., Eng., Vol. 6, , Kays, W. M. and London, A. L., Comact Heat exchangers, 3 rd Ed., New York, McGraw-Hill, Mills, A. F., Heat Transfer, 2 nd Ed., Prentice Hall, New Jersey, Bejan, A., Heat Transfer, John Wiley & Sons. New York, Hesselgreaves, J. E., Comact Heat Exchangers, Pergamon, London, Kays, W. M. and London, A. L., Comact Heat Exchangers, 2 nd Ed., McGraw-Hill, New York, Kuan, T., Heat Exchanger Design Handbook, Marcel Dekker, Inc., Mason, J., 955, Heat Transfer in Cross Flow, Proc. 2 nd U.S. National Congress on Alied Mechanics, American Society of Mechanical Engineers, New York. 5. Filonenko, G.K.,Hydraulic Resistance in Pies (in Russia), Telonergetika, Vol. ¼, , Moody, L.F., Friction Factors for Pie Flow, Trans. ASME, Vol. 66, , Shah, R. K. and Sekulic, D. P., Fundamentals of Heat Exchanger Design, John Wiley and Sons, Inc., Shah, R.K. and Focke, W.W., Plate Heat Exchangers and their Design Theory in Heat Transfer Equiment Design, Ed. R.K. Shah, E.C. Subbarao, and R.A. Mashelkar, Hemishere, Washington, , Martin, H., A Theoretical Aroach to Predict the Performance of Chevron-tye Plate Heat Exchanger, Chemical Engineering and Processing, Vol. 35, , Robinson, K.K. and Briggs, D.E., Pressure Dro of Air Flowing across Triangular Pitch Banks of Finned Tubes, Chemical Engineering Progress Symosium Series, Vol. 64, , Briggs, D.E. and Young, E.H., Convection Heat Transfer and Pressure Dro of Air Flowing across Triangular Pitch Banks of Finned Tubes, Chem. Eng. Prog. Sym. Ser. 4, Vol.59,.-0, Achaichia, A. and Cowell, T.A., Heat Transfer and Pressure Dro Charactoristics of Flat Tube Louvered Plate Fin Surfaces, Exerimental Thermal and Fluid Science,Vol., , 988.

27 23. Kays, W. M., Loss Coefficients for Abrut Changes in Flow Cross Section With Low Reynolds Number Flow in Single and Multile-Tube Systems, Transaction of the American Society of Mechanical Engineers, Vol. 72, , Rouse, H., Elementary of Fluids, John Wiley and Sons, Inc, New York, Webb, R. L., Princiles of Enhanced Heat Transfer, John Wiley, Inc., New York, Wang, L., Sunden, B., and Manglik, R.M., Plate Heat Exchangers, WIT ress, Boston, Manglik, R. M. and Bergles, A. E., Heat Transfer and Pressure dro Correlations for the Rectangular Offset-Stri-Fin Comact Heat Exchanger, Ex. Thermal Fluid Sci., Vol. 0,. 7-80,

28 36 Problems Double Pie Heat Exchanger 5. A counterflow double ie heat exchanger is used to cool ethylene glycol for a chemical rocess with city water. Ethylene glycol at a flow rate of 0.63 kg/s is required to be cooled from 80 C to 65 C using water at a flow rate of.7 kg/s and 23 C, which shown in Figure P5.. The double-ie heat exchanger is comosed of 2-m long carbon-steel hairins. The inner and outer ies are 3/4 and /2 nominal schedule 40, resectively. The ethylene glycol flows through the inner tube. When the heat exchanger is initially in service (no fouling), calculate the outlet temeratures, the heat transfer rate and the ressure dros for the exchanger. How many hairins will be required? T 2o T o T i. m c L. m 2 c 2 T 2i d i d o D i Figure P5. and P5.2 Double-ie heat exchanger 5.2 A counterflow double ie heat exchanger is used to cool ethylene glycol for a chemical rocess with city water. Ethylene glycol at a flow rate of 0.63 kg/s is required to be cooled from 80 C to 65 C using water at a flow rate of.7 kg/s and 23 C, which is shown in Figure P5.2. The double-ie heat exchanger is comosed of 2-m long carbon-steel hairins. The inner and outer ies are 3/4 and /2 nominal schedule 40, resectively. The ethylene glycol flows through the inner tube. Fouling factors of 0.76x0-3 m 2 K/W for water and 0.325x0-3 m 2 K/W for ethylene glycol are secified. Calculate the outlet temeratures, the heat transfer rate and the ressure dros for the exchanger. How many hairins will be required?

29 37 Shell-and-Tube Heat Exchanger 5.3 A miniature shell-and-tube heat exchanger is designed to cool glycerin with cold water. The glycerin at a flow rate of 0.25 kg/s enters the exchanger at 60 C and leaves at 36 C. The water at a rate of 0.54 kg/s enters at 8 C, which is shown in Figure P5.3. The tube material is carbon steel. Fouling factors of 0.253x0-3 m 2 K/W for water and 0.335x0-3 m 2 K/W for glycerin are secified. Route the glycerin through the tubes. The ermissible maximum ressure dro on each side is 30 kpa. The volume of the exchanger is required to be minimized. Since the exchanger is custom designed, the tube size may be smaller than NPS /8 (DN 6 mm) that is the smallest size in Table C.6 in Aendix C, wherein the tube itch ratio of.25 and the diameter ratio of.3 can be alied. Design the shell-and-tube heat exchanger. Figure P5.3 Shell-and tube heat exchanger Plate Heat Exchanger 5.4 Hot water will be cooled by cold water using a miniature late heat exchanger as shown in Figure P5.4.. The hot water with a flow rate of.2 kg/s enters the late heat exchanger at 75 C and it will be cooled to 55 C. The cold water enters at a flow rate of.3 kg/s and 24 C. The maximum ermissible ressure dro for each stream is 30 kpa (4.5 si). Using a single ass chevron lates (Stainless steel AISI 304) with β =45, determine the rating (T, q, ε and P) and sizing (W, L, and H ) of the late heat exchanger. Descrition value Number of asses N Chevron angle β 40 degree Total number of lates N t 40 Plate thickness 0.2 mm Corrugation itch λ 4 mm Port diameter D 25 mm Thermal conductivity k w 4.9 W/mK

30 38 W Port diameter D Brazed or Gasket λ β L β Figure P5.4.Plates with chevron-tye corrugation attern for a late heat exchanger H λ β β Plate Heat Exchanger 2a (a) (b) Figure P5.4.2 (a) Brazed late heat exchanger and (b) the inter-late cross-corrugated flow. 5.5 (Time-consuming work) Cold water will be heated by wastewater using a late heat exchanger as shown in Figure P5.4.. The cold water with a flow rate of 40 kg/s enters the late heat exchanger at 22 C and it will be heated to 42 C. The hot wastewater enters at a flow rate of 30 kg/s and 65 C. The maximum ermissible ressure dro for each fluid is 70 kpa (0 si). Using a single ass chevron lates (Stainless steel AISI 304) with β =30, determine firstly the number of lates N t and the corrugation itch λ by develoing a MathCAD model as a function of not only the sizing (W, L, and H ) but also N t and λ, and then determine the rating (T, q, ε and P) and sizing of the late heat exchanger. Show your detailed work for the

31 39 determination of the number of lates and corrugation itch. The following data list is used for the calculations. Descrition value Number of asses N Chevron angle β 30 degree Plate thickness 0.6 mm Port diameter D 200 mm Thermal conductivity k w 4.9 W/mK Finned-Tube Heat Exchanger 5.6 A circular finned-tube heat exchanger is used to cool hot coolant (50% ethylene glycol) with an air stream in an engine. The coolant enters at 96 C with a flow rate of.8 kg/s and must leave at 90 C. The air stream enters at a flow rate of 2.5 kg/s and 20 C. The allowable air ressure loss is 250 Pa and the allowable coolant ressure loss is 500 Pa. The tubes of the exchanger have an equilateral triangular itch. The thermal conductivities of the aluminum fins and the coer tubes are 200 W/mK and 385 W/mK, resectively. The tube itch is cm and the fin itch is 3.43 fins/cm. The fin thickness is cm and the three diameters are d i =0.765 cm for the tube inside diameter, d o =0.965 cm for the tube outside diameter, and d e =2.337 cm for the fin outside diameter. Design the circular finned-tube heat exchanger by roviding the dimensions of the exchanger, outlet temeratures, effectiveness, and ressure dros. Air (Atmosheric ressure) T 2i =20 C. m 2 =2.5 kg/s Coolant T i =96 C. m =.8kg/s T o =90 C T 2o Figure P5.6.Flow diagram of a circular finned-tube heat exchanger

32 40 Plate-Fin Heat Exchanger 5.7 A gas-to-air single-ass crossflow heat exchanger is contemlated for heat recovery from the exhaust gas to reheat incoming air in a ortable solid oxide fuel cell (SOFC) system. Offset stri fins of the same geometry are emloyed on the gas and air sides; the geometrical roerties and surface characteristics are rovided in Figures P5.7. and P Both fins and lates (arting sheets) are made from Inconel 625 with k=8 W/m.K. The anode gas flows in the heat exchanger at m 3 /s and 750 C. The cathode air on the other fluid side flows at 0.08 m 3 /s and 90 C. The inlet ressure of the gas is at 30 kpa absolute whereas that of air is at 70 kpa absolute. Both the gas and air ressure dros are limited to 3 kpa. It is desirable to have an equal length for L and L 2. Design a gas-to-air single-ass crossflow heat exchanger oerating at ε =0.85. Determine the core dimensions of this exchanger. Then, determine the heat transfer rate, outlet fluid temeratures and ressure dros on each fluid. Use the roerties of air for the gas. Use the following geometric information. Descrition value Fin thickness mm Plate thickness w 0.25 mm Fin density N f 980 m - Sacing between lates b and b 2.3 mm Offset stri length λ and λ 2.7 mm Figure P5.7. Plate-fin heat exchanger, emloying offset stri fin.

33 4 0.5 ( f -) w Plate a k j (b2 b for fluid 2) b e (b -) l f i w Fin Plate Fin c d f g h Figure P5.7.2 Schematic of offset stri fin (OSF) geometry

34 42 Aendix C Table C.5 Thermohysical roerties of fluids Engine oil T ( C) ρ (kg/m 3 ) c (J/KgK) k (W/mK) µ x 0 2 (N.s/m 2 ) Pr % Ethylene glycol T ( C) ρ (kg/m 3 ) c (J/KgK) k (W/mK) µ x 0 2 (N.s/m 2 ) Pr Ethylene glycol T ( C) ρ (kg/m 3 ) c (J/KgK) k (W/mK) µ x 0 2 (N.s/m 2 ) Pr Glycerin T ( C) ρ (kg/m 3 ) c (J/KgK) k (W/mK) µ x 0 2 (N.s/m 2 ) Pr

35 Water T ( C) ρ (kg/m 3 ) c (J/KgK) k (W/mK) µ x 0 6 (N.s/m 2 ) Pr Table C.6 Pie Dimensions Nominal Pie Size NPS (in.) DN (mm) O.D. (in.) O.D. (mm) Schedule I.D. (in.) I.D. (mm) O.D/I.D / / / / / /

36 / / / S S S S Extra heavy S Extra heavy Standard Extra heavy Standard Extra heavy S Standard Extra heavy