Plate Type Heat Exchanger Design

This design project aims to propose a plate type heat exchanger that can meet given heat duty and find the number of plates required. Plate type heat exchanger uses metal plates to transfer heat between two fluids. Starting point of this design is to define given properties

1. 1

REPORT TO DEPARTMENT OF CHEMICAL ENGINEERING
MIDDLE EAST TECHNICAL UNIVERSITY
FOR COURSE: CHE-327 HEAT AND MASS TRANSFER
OPERATIONS
PLATE TYPE HEAT EXCHANGER DESIGN
GROUP MEMBERS:

ALARA MELISA AYDIN
COŞKU MOLER
RESKY SAPUTRA




METU
(25.05.2016)
Ankara, TURKEY

2. 2

ABSTRACT
This design project aims to propose a plate type heat exchanger that can meet given heat
duty and find the number of plates required. Plate type heat exchanger uses metal plates to transfer
heat between two fluids. Starting point of this design is to define given properties. It is asked us to
cool the inlet fluid which is waste stream from 65
o
C to 40
o
C using cooling water at 15
o
C. Several
information of the inlet and outlet streams are given such as the inlet and outlet temperature of
waste stream, mass flow rate of inlet stream, physical properties of waste and other constructional
data for the similar heat exchanger; vertical, horizontal distances, plate thickness, length, effective
channel width, enlargement factor, chevron angle etc. Several calculations are done in 2 parts. The
first one is geometry analysis used in order to find the required number of plates. The second one
is heat transfer analysis in order to find the required heat duty for both streams and actual heat
duties for clean and fouled involving trial-error solution. Some correlations is needed such as heat
transfer coefficient calculation, correlation of Nusselt number and Reynold number in which the
empirical equation needed. Assumptions are regarded at the beginning of the design. Finally, the
required heat duty for cold and hot streams are found 1.47 x 10
7
W and the actual heat duties for
clean and fouled are 2.62 x 10
7
W and 2.32 x 10
7
W, respectively. The total required number of
plates are also found as 105 plates.

3. 3

TABLE OF CONTENT

TABLE OF CONTENTS
NOMENCLATURE…………………………………………………………………………1
1. INTRODUCTION …………………………………………………………………..3
1.1. Problem Statement…………………………………………………………… ...4
1.2. The Calculation Method………………………………………………………..5
1.3. Assumptions…………………………………………………………………… ..8
2. SAMPLE CALCULATIONS………………………………………………………… 10
2.1. Geometry Analysis……………………………………… ………………………10
2.2. Heat Transfer Analysis………………………………………………………….11
3. RESULT AND DISCUSSIONS……………………………………………… ……….14
4. CONCLUSIONS…………………………………………………………………… ….16
5. REFERENCES……………………………………………………………… …………17

4. 4

NOMENCLATURE
Thi : inlet hot stream temperature
0
c
Tho : outlet hot stream temperature
0
c
Tci : inlet cold stream temperature
0
c
Tco : outlet cold stream temperature
0
c
mc : cold stream mass flow rate kg/s
mh : hot stream mass flow rate kg/s
Gc : The cold channel mass velocity kg/m
2
s
Gh : The hot channel mass velocity kg/m
2
s
Ch : steam heat capacity J/kg.K
Cc : ipa-water mixture heat capacity J/kg.K
Qc : Amount of heat transfer under clean condition W
Qf : Amount of heat transfer under fouled condition W
Uf : Fouled overall heat transfer coefficient W/m
2
K
Uc : overall heat transfer coefficient W/m
2
K
Ae : Actual effective area m
2

A1 : Single plate efective area m
2

A1p : Single plate projected area m
2

Nt : total number of plates
Ne : The effective number of plates
Np : Number of passes
Ncp : the total number of channels per pass
Lv : Vertical distance m

5. 5

Lh : Horizontal distance m
t : Plate thickness m
Lc : Plate pack length m
Lw : Effective channel width m
p : The plate pitch m
b : the mean channel spacing m
Dh : The hydraulic diameter of the channel m
∅ : The enlargement factor
?????? : Chevron angle
o

µh : viscocity of hot fluid N.s/m
2
µc : viscocity of cold fluid N.s/m
2

Pr : prandalt number
Re : reynolds number
Nu : nusselt number
hc : convective heat transfer coefficient on clod fluid W.m
2
/K
hh : convective heat transfer coefficient on hot fluid W.m
2
/K
??????
�ℎ
: fouling factor for hot fluid m
2
.K/W
??????
��
: fouling factor for cold fluid m
2
.K/W
kw : thermal conductivity of the plate material W/m.K

6. 6

1. INTRODUCTION
Plate heat exchanger is a type of Heat Exchanger which consists of many corrugated
stainless-steel sheets separated by polymer gaskets and clamped into a steel frame. It transfers heat
by placing thin, corrugated metal sheets side by side and connecting them by gaskets. Flow of the
substances to be heated and cooled takes place between alternating sheets allowing heat to transfer
through the metal sheets.

Figure 1: Plate type heat exchanger
Some advantages using plate heat exchanger are high heat transfer area, high heat transfer
coefficient, having lower floor space requirements, multiple duties can be performed by a single
unit, most suitable type heat exchanger for lower flow rates and heat sensitive substances.
Moreover, area of heat transfer of plate heat exchanger can be increased by increasing the number
of the plates.

7. 7

1.1. Problem Statement
In this problem, a plate heat exchanger is needed to be designed for a specific purpose. This heat
exchanger should be able to cool a waste from 65°C to 40°C using cooling water which enters the
heat exchanger at 15°C. The mass flow rate of the waste stream is 140 kg/s and its properties may
be approximated as follows:
ρ = 985 kg/m3
μ = 510 x 10-6 kg/m.s
k = 0.650 W/m.K
Pr = 3.3
Cp = 4200 J/kg.K
Fouling resistance ≡ Fouling resistance of water= 0.0000069 m
2
.K/W (taken from Heat
Exchanger: Selection, Rating and Thermal Design, table 10.4)
Moreover, we are going to propose a plate type heat exchanger that can meet this heat duty and
find the number of plats required for the heat exchanger.
Some constructional data for a similar heat exchanger are given as follows:
Total effective area (Ae)= 110 m2
Vertical distance (Lv) = 1.55 m
Horizontal distance (Lh)= 0.43 m
Plate thickness (t)= 0.6 mm
Plate pack length (Lc)= 0.38 m
Effective channel width (Lw)= 0.63 m
Enlargement factor (∅)= 1.25
Chevron angle (?????? )= 45°

8. 8

Plates are stainless steel (kw = 16.5 W/m.K, taken from heat exchangers selection, Rating and
thermal design, table 10.1)

Figure 2: Main dimensions of a chevron plate and and developed and projected dimensions of a
chevron plate cross section normal to to the direction troughs.

1.2. The Calculation Method;
Calculation of this problem design are separated by 2 analysis.
The first one is geometry analysis. The channels increase the surface area of the plate as
compared to the original flat area. To express the increase of the developed length in relation to
the projected length, a surface enlargement factor,∅, is the defined as the ratio of the developed
length to the flat or projected length
∅=
??????�������� �����ℎ
??????�������� �����ℎ
=
??????���??????� ��������� ??????��??????
��������� ��??????�� ??????��??????
=A1/A1p
(1.1)

9. 9

Where Actual effective area can be calculated as (1.2)
Actual effective area (Ae) = Lp * Lw (1.2)
Or actual effective area can be calculated as Ae = A1 * Ne
Where the effective number of plates. Ne, can be estimated as Ne = Nt – 2
Also Lp and Lw can be estimated from the port distance Lv and Lh and port diameter Dp as
Lp ≈ Lv - Dp (1.3)
Lw ≈ Lh + Dp (1.4)
The value of enlargement factor is calculated the effective flow path.
From (1.3 and 1.4) we can make a new equation to find Lp.
Lp = Lv – Lw + Lh (1.5)
Flow channel is the conduit formed by two adjacent plates between the gaskets. The cross section
of a corrugated surface being very complex, the mean channel spacing, b, is defined as (1.6)
b = p –t (1.6)
The plate pitch (p) can be determined from the compressed plate pack length (Lc), which usually
specified.
p = Lc / Nt (1.7)
Where Nt is the total number of plates.
The hydraulic diameter of the channel (Dh) can be estimated as (1.8)
Dh ≈ 2b / ∅ (1.8)
Finding the total number of channels per pass (Ncp) is obtained from (1.9)
Ncp = (Nt – 1) / 2 *Np (1.9)
Where Nt is total number of plates and Np is the number of passes.

10. 10

From those correlations we can find total number of plates required. With plate type heat
exchangers, heat transfer is enhanced. The heat transfer enhancement will strongly depend on the
chevron inclination angle (??????) relative to flow direction. Moreover, the performance of a chevron
plate will also depend upon the surface enlargement factor (∅), the channel profile, the mean
channel spacing (b), the temperature dependent physical properties, and especially the variable
viscosity effects.
The second one is heat transfer analysis. In order to find heat transfer coefficient (h),
correlation of Nusselt number (Nu) and Reynold number (Re) is needed. The Reynolds number
based on channel mass velocity and the hydraulic diameter of the channel is defined as (1.10)
Re = Gc * Dh / ?????? (1.10)
Where the channel mass velocity is given by (1.11)
Gc = mch / Ncp * b * Lw (1.11)
Correlation empirical equation is needed. The correlation in the form of (1.12) are proposed by
Kumar and the values of constants Ch and n are given in table 1.1 (Heat exchangers:Selection,
Rating and Thermal design 2
nd
ed, p. 395)
Nu = Ch * Re
n
* Pr * (
??????
??????
??????
)
0.17
(1.12)
Table 1.1. Constants for single-phase heat transfer and pressure loss calculation in gasketed-plate
heat exchanger (Heat exchangers:Selection, Rating and Thermal design 2
nd
ed, p. 394).

11. 11



Overall heat transfer coefficient under fouling conditions is calculated as (1.13)
1
�
??????
=
1
ℎ
ℎ
+
1
ℎ
??????
+
�
�
??????
+??????
�ℎ
+??????
��
(1.13)
The required heat duty (Qr) for cold and hot streams is defined as (1.14)
Qr = (??????̇∗????????????)c *(Tc2 – Tc1) = (??????̇∗????????????)h * (Th1 – Th2) (1.14)
On the other hand, the actually obtained heat duty (Qf) for fouled conditions is defined as (1.15)
Qf = U*Ae*F*∆??????
��
(1.15)
In order to find ∆??????
��
, equation (1.16) is defined as
∆??????
��
=
(�
ℎ,??????�
−�
??????,���
)−(�
ℎ,���
−�
??????,??????�
)
ln (
(??????
ℎ,??????�
−????????????,���
)
(??????
ℎ,���
−??????
??????,??????�
)
(1.16)
For heat transfer analysis we are not given ??????
�,���
. So that physical properties of water cannot be
decided. Hereby trial-error solution is needed to find the correct ??????
�,���
. To determine the correct
one we need to check both the required heat of hot and cold fluid. From energy balance analysis,
the required heat of hot and cold fluid must be same. The calculation for trial-error solution stops
until it reaches the equality of the required heat of hot and cold fluid.
1.3. Assumptions;
 Physical properties are constant at 1 atm
 Heat loses to or from the surrounding are negligible.
 The kinetic and potential energy changes are negligible.
 The heat exchanger operates at steady-state conditions.
 No phase changes in the fluid streams.
 Wall thermal resistances are distributed uniformly.
 The velocity and temperature at the inlet of the heat exchanger on each fluid side are
uniform.
 The heat transfer area (A) is distributed uniformly on each fluid side.

12. 12

 The cold and hot stream mass flow rate are same
 Number of passes is one pass

13. 13

2. SAMPLE CALCULATIONS
2.1 Geometry Analysis






















The projected plate area


Single plate heat transfer area

The effective number of plates

Total number of plates

The plate pitch

the mean channel flow gap

The one channel flow area

The channel hydraulic
Ae110m
2
 Lv1.55m Lh0.43m t0.0006m Lc0.38m Lw0.63m kw16.5
W
mK
 Np1 LpLvLw Lh 1.35m A1pLpLw 0.851m
2
 A1A1p 1.063m
2
 Ne
Ae
A1
103.469 NtNe2 105.469 p
Lc
Nt
3.60310
3
 m bpt3.00310
3
 m AchbLw 1.89210
3
 m
2
 Dh
2b

4.80510
3
 m

14. 14




2.2 Heat Transfer Analysis


















Total number pf channel per pass

(Trial-error method)



water properties at 313/288 = 300.5 K



waste properties




The mass flow rate per channel



for hot fluid for cold fluid
Ncp
Nt1
2Np
52.234 assume 8.410
4
Pas k0.611
W
mK
 mc140
kg
s
 Pr
Cpc
k
5.748 Rfwater0.0000069
m
2
K
W
 Cpwaste4200
J
kgK
 kwaste0.650
W
mK
 Prwaste3.3 waste51010
6
Pas RfwasteRfwater6.910
6

s
3
K
kg
 mhmc140
kg
s
 mch
mc
Ncp
2.68
kg
s
 Gch
mch
Ach
1.41710
3

kg
sm
2

 GccGch1.41710
3

kg
sm
2

 Reh
GchDh
waste
1.33510
4
 Rec
GccDh

8.10310
3

Cpc 4185.847
J
kg K 

Tco 40 273  313 K  
Tci 15 273  288 K   Thi 65 273  338 K   Tho 40 273  313 K  

15. 15




















Since Qrh and Qrc is almost same, then Tco assumption is acceptable


Table 10.6

or hhot= 3.283x10
4
W/m
2
K
or hcold=2.669x10
4
W/m
2
K

The clean overall heat transfer coefficient

or W/m
2
K
The fouled overall heat transfer coefficient


for counter current flow

the actual heat duties for clean and fouled surfaces


The required heat

45
0
 Reh100 Rec100 ch0.3 n0.663 hhot
kwaste
Dh
chReh
n
 Prwaste
1
3
 3.28310
4

kg
s
3
K
 hcold
k
Dh
chRec
n
 Pr
1
3
 2.66810
4

kg
s
3
K
 Uc
1
1
hcold
1
hhot

t
kw

9.58710
3

kg
s
3
K
 9.58710
3
 Uf
1
1
Uc
Rfwaste Rfwater
8.46710
3

kg
s
3
K
 T2ThoTci 25K T1ThiTco 25K LMT D25K QcUcAeLMT D 2.63610
7
W QfUfAeLMT D 2.32810
7
W QrhmhCpwaste ThiTho( ) 1.4710
7
W Qrc mc Cpc  Tco Tci  ( )  1.465 10
7
 W  

16. 16
























The safety factor

The precent over surface design



The cleanliness factor

Cs
Qf
Qrh
1.584 OS100UcRfwasteRfwater( ) 13.23 CF
Uf
Uc
0.883

17. 17

3 RESULTS AND DISCUSSONS
Objective of this project was to design a proper plate type heat exchanger. Several assumptions
were made while making the calculations. These assumptions are constant physical properties at
1 atm, negligible heat losses through the surroundings, negligible kinetic and potential energy
changes, operating at steady state, no phase changes, same mass flow rates and, uniform
temperature and velocity at the inlet of the heat exchanger. Calculation of this problem design
are separated by 2 analysis. The first one is geometry analysis and the second one is heat transfer
analysis. Geometry is analyzed by the given datas and calculations were made according to these
given datas. First of all projected length was calculated as 1.35 m and then projected area was
determined as 0.851 ??????
2
, by using this value single plate heat transfer area was calculated by
enlargement factor times projected area, enlargement factor,∅, is the defined as the ratio of the
developed length to the flat or projected length and found as 1.25. Up to here effective area and
single plate heat transfer area was calculated number of effective plates was found as 103.469 by
dividing effective area by single plate heat transfer area. Then total number of plates were found
as 105.469. After that plate pitch was determined as 3.603∗10
−3
?????? and mean flow channel gap
was found by using that one as 3.003∗10
−3
??????. The one channel flow area was determined by
using mean flow channel gap and found as 1.892∗10
−3
??????
2
hydraulic diameter was calculated
by 2 mean flow channel gap divided by enlargement factor and found as 4.805∗10
−3
??????. Lastly
total number pf channel per pass was found as 52.234 and the geometry calculations were done.
In the heat transfer analyses in order to find heat transfer coefficient (h), correlation of Nusselt
number (Nu) and Reynold number (Re) is needed. The Reynolds number based on channel mass
velocity and the hydraulic diameter of the channel is defined as Re for hot fluid was determined
as 1.335∗10
4
and 8.303∗10
3
for the cold fluid. Where the channel mass velocity is given by
Gc 1.417∗10
3
��
�
2
∗�
for both hot and cold fluid. In calculation of Nusselt number Correlation
empirical equation is needed. The correlation in the form of (1.12) are proposed by Kumar and
the values of constants Ch and n are given in table 1.1. This correlation gave us the heat transfer
coefficients directly by using Nusselt number as 3.283∗10
4
??????
�
2
∗??????
for hot 2.668∗10
4
??????
�
2
∗??????
for
cold fluid. By using these heat transfer coefficients, the overall heat transfer coefficient for hot
clean and fouled heat exchangers were found 9.587∗10
3
??????
�
2
∗??????
, 8.467∗10
3
??????
�
2
∗??????
respectively.

18. 18

In the determination of Qr and Qf ∆??????
��
is required and for heat transfer analysis we are not
given ??????
�,���
. So that physical properties of water cannot be decided. Therefore trial-error solution
is needed to find the correct ??????
�,���
. To determine the correct one we need to check both the
required heat of hot and cold fluid. From energy balance analysis, the required heat of hot and
cold fluid must be same. The calculation for trial-error solution goes until it reaches the equality
of the required heat of hot and cold fluid. To be sure of the ??????
�,���
value was acceptable Qrh and
Qrc was determined and found as 1.47∗10
7
?????? and 1.465∗10
7
?????? which is really close to each
other so that Tcout value that choosen is acceptable. After calculation of ∆??????
��
Qr and Qf was
calculated as 2.637∗10
7
??????, 2.328∗10
7
?????? respectively. Lastly the safety factor was calculated
by dividing Qf to Qrh and found as 1.584 and cleanless factor as 0.883. According to the
literature Process heat transfer, 1950, typical design are based on safety factor of 1.6 which is
closer to 1.58. Moreover, based on Heat Exchanger: Selection, Rating, and Thermal Design, 2
nd
,
Typical designs are based upon a cleanliness factor of 0.85 which is quite closer with our value.

19. 19

4 CONCLUSION
After performing the required equations which are given in calculation part, it is seen that
each plate is corrugated to increase the surface area and maximize heat transfer. Within a plate
heat exchanger, the fluid paths alternate between plates allowing the two fluids to interact, but not
mix, several times in a small area. The data from the similar heat exchanger is used in order to
define the heat exchanger that is used in the project. The goal of the project is to understand the
characteristics and design of a plate heat exchanger. By geometry and heat transfer analysis, the
total number of plates, the actual with fouled surface and required heat duty are found as 105
plates, 2.63 x 10
7
W and 2.32 x 10
7
W, respectively. It is considered that no heat loss to
surroundings however in reality there should be heat loss, therefore, error due to this assumption
must be considered in real life applications.

20. 20

5 REFERENCES
 Incropera, F. (2012). Principles of heat and mass transfer (7th ed.). Singapore: John
Wiley & Sons Singapore Pte.
 Leib, T., & Pereira, C. (2008). Perry's chemical engineers' handbook (8th ed.). New York:
McGraw-Hill.
 Kakaç. S. (Sadik). Heal exchangers : selection, rating. and thermal design / Sadik Kakaç,
Hongtan Liu.-. 2nd ed
 Kern, “Process heat transfer”, McGraw Hill, 1950