Thermal Unit Operation

(ChEg3113)

Instructor: Mr. Tedla Yeshitila (M.Sc.)

Lecture 6- Double Pipe Heat Exchanger Design

Today…

• Review

• Double pipe heat exchanger design procedure

• Example

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

Chapter 4

Design of Double Pipe Heat Exchanger 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 UD is calculated

from which the surface can be found with the use of Fourier equation

Q=UDA𝛥t

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.

Chapter 4

Design of Double Pipe Heat Exchanger 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: T1, T2, W, C, s or 𝛒, μ, k, 𝜟P, Rdo or Rdi

– Cold fluid: t1,t2, w, c, s or 𝛒, μ, k, 𝜟P, Rdi or Rdo

– The diameter of the pipes must be given or assumed.

Chapter 4

Design of Double Pipe Heat Exchanger A convenient order of calculation follows:

1. Heat balance

From T1, T2, t1,t2 check the heat balance, Q , using c at T average

and t average

Q=WC(T1-T2) = wc(t2-t1)

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 =𝜟𝑡2−𝜟𝑡1

2.3 log (𝜟𝑡2/𝜟𝑡1)

Chapter 4

Design of Double Pipe Heat Exchanger For counter flow:

For parallel flow:

This well-known “logarithmic mean” temperature difference is

only applicable to sensible heat transfer in true co-current or

counter-current flow (linear temperature enthalpy curves).

Hot fluid Cold temperatures

T1 t1 𝛥 t2

T2 t2 𝛥 t1

Hot fluid Cold temperatures

T1 Higher temperature t2 𝛥 t2

T2 Lower temperatures t1 𝛥 t1

Chapter 4

Design of Double Pipe Heat Exchanger 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 ((T1-T2) or

(t2-t1)) do not exceed 50-1000F, and if the temperature difference

(for counter current flow (T1- t2) and (T2-t1)) is less than 500F,

the arithmetic means of T1 and T2 and t1 and t2 may be used in

place of Tc and t c for evaluating the physical properties. So you

need to check all these.

And also for non-viscous fluids ɸ = (μ/μw) 0.14 may be assumed

as 1.0.

Check 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.

Chapter 4

Design of Double Pipe Heat Exchanger Inner pipe

4. Flow area, 𝑎𝑝 =𝛱𝐷2

4 , m2

5. Mass velocity, 𝐺𝑝 =𝑤

𝑎𝑝 , kg/ m2 .s

For n parallel stream, multiply 𝑎𝑝 by n

6. Obtain μ from Figure 14 at Tc or t c depending upon which the

flows through the inner pipe. The unit is kg/m.s

Reynolds number, 𝑅𝑒𝑝 =𝐷𝐺𝑝

𝜇 , unit less

7. From the Figure 24 in which 𝑗𝐻 =ℎ𝑖𝐷

𝑘

𝑐𝜇

𝑘

−1

3 𝜇

𝜇𝑤

−0.14 vs.

𝐷𝐺𝑝

𝜇 obtain 𝑗𝐻.

Chapter 4

Design of Double Pipe Heat Exchanger Inner pipe

8. At Tc or t c obtain c (from Figure 2), μ (from Figure 14 ) and k

(from Table 4) and compute 𝑐𝜇

𝑘

1

3

9. To obtain hi for inner pipe,

ℎ𝑖 = 𝑗𝐻𝑘

𝐷

𝑐𝜇

𝑘

1

3 𝜇

𝜇𝑤

−0.14or

ℎ𝑖 =ℎ𝑖𝐷

𝑘

𝑐𝜇

𝑘

−1

3 𝜇

𝜇𝑤

−0.14 𝑘

𝐷

𝑐𝜇

𝑘

1

3∗ 1.0 , J/ m2.s.k

To get area A, first it must calculated by using the above

equation in which ℎ𝑖𝑜 and ℎ𝑜 depend on diameter and fluid

flow area, but independent of extent of surface.

10. Convert or correct ℎ𝑖 to the surface at OD (ℎ𝑖𝑜) ;

ℎ𝑖𝑜 = ℎ𝑖𝐴𝑖

𝐴= ℎ𝑖

𝐼𝐷

𝑂𝐷

Chapter 4

Design of Double Pipe Heat Exchanger Annulus

4`. Flow area, 𝑎𝑎 =𝛱

4(𝐷2

2 − 𝐷12) , m2

5`. Mass velocity, 𝐺𝑎 =𝑊

𝑎𝑎 , kg/ m2 .s

For n parallel stream, multiply 𝑎𝑎 by n

Equivalent diameter, 𝐷𝑒 =4∗𝑓𝑙𝑜𝑤 𝑎𝑟𝑒𝑎

𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑚𝑖𝑡𝑒𝑟=

𝐷22−𝐷1

2

𝐷1 , m

6`. Obtain μ from Figure 14 at Tc or t c depending upon which the

flows through the annulus. The unit is kg/m.s

Reynolds number, 𝑅𝑒𝑎 =𝐷𝑒𝐺𝑎

𝜇 , unit less

7`. From the Figure 24 in which 𝑗𝐻 =ℎ0𝐷𝑒

𝑘

𝑐𝜇

𝑘

−1

3 𝜇

𝜇𝑤

−0.14 vs.

𝐷𝑒𝐺𝑝

𝜇 obtain 𝑗𝐻.

Chapter 4

Design of Double Pipe Heat Exchanger Annulus

8`. At Tc or t c obtain c from Figure 2, μ (from Figure 14 ) and k

(from Table 4) and compute 𝑐𝜇

𝑘

1

3

9`. To obtain ℎ𝑜 for annulus

ℎ𝑜 = 𝑗𝐻𝑘

𝐷𝑒

𝑐𝜇

𝑘

1

3 𝜇

𝜇𝑤

−0.14 or

ℎ𝑜 =ℎ0𝐷𝑒

𝑘

𝑐𝜇

𝑘

−1

3 𝜇

𝜇𝑤

−0.14 𝑘

𝐷𝑒

𝑐𝜇

𝑘

1

3∗ 1.0 , J/ m2.s.k

Chapter 4

Design of Double Pipe Heat Exchanger Overall-coefficient

11.Compute clean overall heat transfer coefficient

𝑈𝑐 =ℎ𝑖𝑜ℎ𝑜

ℎ𝑖𝑜+ℎ𝑜, J/ m2.s.K

12. Compute design or dirty overall heat transfer coefficient

(𝑈𝐷)from

1

𝑈𝐷=

1

𝑈𝑐+ 𝑅𝑑

Due to 𝑅𝑑, heat transfer is no longer transferred by the original

surface A (efficiency reduced), so T2 is higher and t2 is lower than

expected. But hi and ho remains substantially constant.

Chapter 4

Design of Double Pipe Heat Exchanger Let Rdi be the dirt factor for the inner pipe fluid at its inside

diameter, and Rdo 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 thick for scale, which has

higher thermal conductivity than dirt.

𝑅𝑑=𝑅𝑑𝑖+𝑅𝑑𝑜

13. Compute required surface area (A) from Q=𝑈𝐷A LMTD =>

𝐴 =𝑄

𝑈𝐷𝐿𝑀𝑇𝐷 which may be translated into length.

The value of A correspond to UD rather than Uc provides the basis

on which equipment is ultimately built.

Chapter 4

Design of Double Pipe Heat Exchanger 14. From Table 11, obtain external surface per foot length for IPS

standard pipe.

Required length =surface required

external surface per foot length

Then decide how many hairpins have to be connected in series.

Chapter 4

Design of Double Pipe Heat Exchanger Example 1: Double pipe Benzene - Toluene Exchanger

It is desired to heat 9,820 lb/ hr of cold benzene from 80 to 120OF using hot toluene

which is cooled from 160 to 100OF. The specific gravities at 68 OF are 0.88 and 0.87,

respectively. The other fluid properties will be found from Appendix.

A fouling factor of 0.001 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?

At the end of this class:

• You will be able to design double pipe heat exchanger

– Thermal design

End of lecture -6