Thermal Unit Operation (ChEg3113)

Transcription

1 Thermal Unit Operation (ChEg3113) Lecture 6- Double Pipe Heat Exchanger Design Instructor: Mr. Tedla Yeshitila (M.Sc.)

2 Today Review Double pipe heat exchanger design procedure Example

3 Review Deign of heat exchanger Rating of heat exchanger Selection of heat exchanger Basic steps in typical design procedures Overall heat transfer coefficient Fouling factor

4 The calculation of heat double pipe heat exchanger consists simply of computing ho and hi to obtain Uc. Allowing a reasonable fouling resistance, a value of U D is calculated from which the surface can be found with the use of Fourier equation Q=U D AΔt Usually the first problem is to determine which fluid should be placed in the annulus and which in the inner pipe. This depend on the relative size of flow areas for both stream. For equal allowable pressure drops both the hot and cold streams, the decision rests in the arrangement producing the most nearly equal mass velocities and pressure drops.

5 In the outline and the next procedures, hot and cold fluid temperatures are represented by upper and lower case letters respectively. All fluid properties are indicated by lower case letters to eliminate the requirement for new nomenclature. Process conditions required: Hot fluid: T 1, T 2, W, C, s or ρ, μ,, ΔP, R do or R di Cold fluid: t 1,t 2, w, c, s or ρ, μ,, ΔP, R di or R do The diameter of the pipes must be given or assumed.

6 A convenient order of calculation follows: 1. Heat balance From T 1, T 2, t 1,t 2 chec the heat balance, Q, using c at T average and t average Q=WC(T 1 -T 2 ) = wc(t 2 -t 1 ) Where W and w are flowrates, C and c are specific heat capacity Radiation losses from the heat exchanger are usually insignificant compared with the heat load transfer in the exchanger. 2. LMTD LMTD = Δt 2 Δt log (Δt 2 /Δt 1 )

7 For counter flow: Hot fluid Cold temperatures T 1 Higher temperature t 2 Δ t 2 T 2 Lower temperatures t 1 Δ t 1 For parallel flow: Hot fluid Cold temperatures T 1 t 1 Δ t 2 T 2 t 2 Δ t 1 This well-nown logarithmic mean temperature difference is only applicable to sensible heat transfer in true co-current or counter-current flow (linear temperature enthalpy curves).

8 3. Caloric temperature (Tc and tc) If neither of the fluid is very viscous at the cold terminal, say not more than 1.0 centipoise, if the temperature ranges ((T 1 -T 2 ) or (t 2 -t 1 )) do not exceed F, and if the temperature difference (for counter current flow (T 1 - t 2 ) and (T 2 -t 1 )) is less than 50 0 F, the arithmetic means of T 1 and T 2 and t 1 and t 2 may be used in place of T c and t c for evaluating the physical properties. So you need to chec all these. And also for non-viscous fluids ɸ = (μ/μw) 0.14 may be assumed as 1.0. Chec from table which flow area is greater for the a given double pipe standard. Place the larger stream in the inner pipe or annulus by comparing their flow area which is highest.

9 Inner pipe 4. Flow area, a p = ΠD2 4, m2 5. Mass velocity, G p = w a p, g/ m 2.s For n parallel stream, multiply a p by n 6. Obtain μ from Figure 14 at T c or t c depending upon which the flows through the inner pipe. The unit is g/m.s Reynolds number, Re p = DG p μ, unit less 7. From the Figure 24 in which j H = h id DG p μ obtain j H. 1 3 μ μ w 0.14 vs.

10 Inner pipe 8. At T c or t c obtain c (from Figure 2), μ (from Figure 14 ) and (from Table 4) and compute 9. To obtain hi for inner pipe, h i = j H h i = h id D μ μ w μ 0.14 or 1 3 μ w 0.14 D , J/ m 2.s. To get area A, first it must calculated by using the above equation in which h io and h o depend on diameter and fluid flow area, but independent of extent of surface. 10. Convert or correct h i to the surface at OD (h io ) ; h io = h i A i A = h i ID OD

11 Annulus 4`. Flow area, a a = Π 4 (D 2 2 D 1 2 ), m 2 5`. Mass velocity, G a = W a a, g/ m 2.s For n parallel stream, multiply a a by n Equivalent diameter, D e = 4 flow area = D 2 wetted permiter 2 D 1 2 6`. Obtain μ from Figure 14 at T c or t c depending upon which the flows through the annulus. The unit is g/m.s Reynolds number, Re a = D eg a μ, unit less D 1, m 7`. From the Figure 24 in which j H = h 0D e D e G p μ obtain j H. 1 3 μ μ w 0.14 vs.

12 Annulus 8`. At T c or t c obtain c from Figure 2, μ (from Figure 14 ) and (from Table 4) and compute 9`. To obtain h o for annulus h o = j H h o = D e h 0D e 1 3 μ 1 3 μ w or μ 0.14 μ w D e , J/ m 2.s.

13 Overall-coefficient 11.Compute clean overall heat transfer coefficient U c = h ioh o h io +h o, J/ m 2.s.K 12. Compute design or dirty overall heat transfer coefficient (U D )from 1 U D = 1 U c + R d Due to R d, heat transfer is no longer transferred by the original surface A (efficiency reduced), so T 2 is higher and t 2 is lower than expected. But hi and ho remains substantially constant.

14 Let R di be the dirt factor for the inner pipe fluid at its inside on which equipment is ultimately built. diameter, and R do the dirt factor for the annulus fluid at the outside diameter of the inner pipe. These may be considered very thin for dirt but may be appreciably thic for scale, which has higher thermal conductivity than dirt. R d =R di +R do 13. Compute required surface area (A) from Q=U D A LMTD => A = Q U D LMTD which may be translated into length. The value of A correspond to U D rather than U c provides the basis

15 14. From Table 11, obtain external surface per foot length for IPS standard pipe. surface required Required length = external surface per foot length Then decide how many hairpins have to be connected in series.

16 Example 1: Double pipe Benzene - Toluene Exchanger It is desired to heat 9,820 lb/ hr of cold benzene from 80 to 120 O F using hot toluene which is cooled from 160 to 100 O F. The specific gravities at 68 O F are 0.88 and 0.87, respectively. The other fluid properties will be found from Appendix. A fouling factor of should be provided for each stream, and the allowable pressure drop on each stream is 10.0psi. A number of 20 ft hairpins of 2 by 1 ¼ in. IPS pipe are available. How many hairpins are required?

17 At the end of this class: You will be able to design double pipe heat exchanger Thermal design

18 End of lecture -6