thermal unit operation (cheg3113) -...
TRANSCRIPT
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