UNIVERSITI TEKNOLOGI MARA
FAKULTI KEJURUTERAAN KIMIA
HEAT TRANSFER
(CHE463)
ASSIGNMENT 2
DECEMBER 2020
LECTURER: DR ISTIKAMAH SUBUKI
NAME: NURSHAMIMI NABILA BINTI AHMAD SUHAIMI
STUDENT ID: 2019291422
GROUP: EH2203F
QUESTION
1.Justification of types of Shell and Tube heat exchanger and assumption of design
2. Calculations for chemical design include: flowrate of heating medium, true temperature differences,
heat transfer area, number of tube, tube bank velocity, correction factor, heat duty, overall heat transfer
coefficient, operating pressure, fouling factor.
3. Dimension of the design must include assumption at the beginning and verified
4. Reference Data – extensive selection of references us used to support the calculation performed
I.
HEAT EXCHANGER
Heat exchanger are design to transfer heat between two or more fluids—i.e., liquids, vapors, or gases of
different temperature.
Figure 1
II.
DESIGN INPUT
Shell side (Table 1)
Tube side (Table 2)
Material
BWG number
Length of tube, L (m)
Outer Diameter, D0 (m)
Inner Diameter, Di (m)
Material thermal conductivity, k ( W/m°C)
Tube flowrate, ṁ (kg/s)
Table 3
Carbon Steel
16
5.0
0.020
0.018
49
6.73
III.
ASSUMPTIONS AND JUSTIFICATION OF HEAT EXCHANGER

Assumptions:
o Negligible heat loss to surroundings.
o Negligible kinetic and potential energy.
o Constant pressure.

Type of Heat Exchanger: Two shells & Four tube passes
Figure 2


Type of flow : Counter flow
Shape of heat exchanger : U Tube
I choose the U tube type of heat exchanger. The reason I choose to use U tube type is because it is
low in cost , it able to handle thermal expansion, no internal gasketed joint and the bundle is
replaceable. From the calculation below, 2AD > AT , thus it is in line.The number of baffle that I choose
is 6 baffles. Below are the calculations that will support my design .
IV.
CALCULATING FLOW RATE OF HEATING MEDIUM
(mCp�T)h2o = (mCp�T)steam
m = flowrate, kg/h
Cp = Specific heat, 𝑘𝐽𝑘𝑔°𝑪
�T = Temperature differences, °𝑪
Inlet
1ℎ
𝑘𝐽
(23340 kg/h)(3600𝑠)(4.0331 𝑘𝑔°𝑪)(100°C - 30°C) =ṁi(3.5241 𝑘𝐽𝑘𝑔°𝑪)(150°C - 79°C)
ṁi = 7.3152 kg/s
Outlet
(19740 kg/h)(
1ℎ
3600𝑠
)(4.0930
𝑘𝐽
)(100°C - 30°C) =ṁo(4.1962 𝑘𝐽𝑘𝑔°𝑪)(150°C - 79°C)
𝑘𝑔°𝑪
ṁo = 5.2731 kg/s
Mean
1ℎ
𝑘𝐽
(21540 kg/h)(3600𝑠)(4.0631𝑘𝑔°𝑪)(100°C - 30°C) =ṁm(3.8602 𝑘𝐽𝑘𝑔°𝑪)(150°C - 79°C)
ṁm = 6.2091 kg/s
V.
CALCULATING TRUE TEMPERATURE DIFFERENCE
𝛥𝑇 − 𝛥𝑇
1
2
ΔTlm = ln( 𝛥𝑇
/ 𝜟𝑻 )
1
𝟐
ΔT1 = Thi – Tco
ΔT1 = 150°C – 100°C
ΔT1 = 50°C
ΔT2 = Tho – Tci
ΔT2 = 79°C – 30°C
ΔT2 = 49°C
ΔTlm =
50− 49
ln( 50/ 49)
ΔTlm = 49.50°C x Correction factor, 0.90
�Tlm =44.55°C
VI.
CALCULATING HEAT TRANSFER AREA
Q = ṁCpΔT = ṁCp(Thi – Tho)
Q = (6.73
𝑘𝑔
𝑠
)(3.8602
𝑘𝐽
)(150°C - 79°C)
𝑘𝑔°𝑪
Q = 1844.5194 kJ/s
Assume U from Table 13-1 in Figure 3
Figure 3
U = 1300 W/m2°C
𝑊
1𝑘𝑊
𝑘𝑊
U = (1300 𝑚2 °C)(1000𝑊) = 1.3 𝑚2 °C
Q = UAsΔTlm
𝑄
As = 𝑈𝛥𝑇
𝑙𝑚
1𝑘𝑊
𝑘𝐽 )
1𝑠
(1844.5194 𝑘𝐽/𝑠)(
As =
𝑘𝑊
°𝐶)(44.50°𝐶)
𝑚2
(1.3
As = 31.88 m2
VI.
CALCULATING NUMBER OF TUBE
As = NπDoL
N=
N=
𝐴𝑠
𝜋𝐷𝑜𝐿
(31.88 𝑚2 )
(𝜋)(0.020𝑚)(5.0𝑚)
N = 101 = 100 tubes
N = 100 tubes
I use 2 shells, 4 tube passes,thus tube per passes (n) is
100/4 = 25
n = 25 tubes per passes
VII.
V=
CALCULATION OF TUBE BANK VELOCITY
ṁ
𝜌𝐴𝑐
ṁ = flowrate, kg/s
p = density, kg/m3
Ac = area, m2
Ac = (1/4π𝐷𝑖2)(n)
Di = inner diameter, m
N = number of tubes per passes
1
Ac = 4 π(0.018)2 (25)
Ac = 0.006362 m2
V=
6.73 𝑘𝑔/𝑠
𝑘𝑔
)(0.006362)
𝑚3
(944.6
V = 1.12 m/s
VIII.
CALCULATION OF CORRECTION FACTOR
𝑡 −𝑡
P = 𝑇1 − 𝑡2
1
P=
1
79−150
30−150
P = 0.59
R=
𝑇1 − 𝑇2
𝑡2 − 𝑡1
(ṁ𝐶𝑝)𝑡𝑢𝑏𝑒
= (ṁ𝐶𝑝)𝑠ℎ𝑒𝑙𝑙
30−100
R = 79−150
R = 0.99
Using the correction factor,F chart in Figure 4 , the correction factor can be obtained.
Figure 4
Thus, from the value of P & R obtained through calculation above, the correction factor, F value is :
F = 0.90
IX.
CALCULATION OF HEAT DUTY
Qc = mcCpc ΔT
21540
Qc = ( 3600𝑠 )(4.0631)(100-30)
Qc = 1701.76
X.
CALCULATION OF OPERATING PRESSURE
SL = ST =1.25Do
SL = ST =(1.25)(0.020)
SL = ST = 0.025
SD = √(0.025)2 + (
0.025 2
)
2
SD = 0.02795
A1 = STL = (0.025)(5) = 0.125
AT = (ST – D)L = (0.025 -0.020)(5) = 0.025
A2 = AD = (SD – D)L = (0.02795 -0.020)(5) = 0.03975
0.025
PL = SL/Do = 0.020 = 1.25
PT = ST/Do =
0.025
0.020
= 1.25
Now that the value of AD and AT are obtained, thus in line or staggered can be decided.
2(AD) ≈ AT
2(AD) > AT ≈ in line
2(AD) < AT ≈ staggered
Thus,
2(0.03975) ≈ 0.025
0.0795 ≈ 0.025
Hence , it is in line .
Vmax = (ST/ ST – D)(v)
Vmax =
0.025
0.025−0.020
× 1.12
Vmax = 5.6 m/s
Remax =
(5.6)(0.020)(944.6)
(0.0002736)
Remax = 386678.36
Using Figure 5 , the value of friction and correction factor can be obtained .
Figure 5
Thus ,
friction factor, f =0.35
correction factor, X = 1.0
PT = PL = 1.25
PT -1 =1- 0.25 =0.25
PL -1 =1- 0.25 =0.25
𝑃𝑇 −1
𝑃𝐿 −1
=1.0
To obtain the value of operating pressure, pressure drop formula can be used.
ΔP = NLfX
𝜌𝑉𝑚𝑎𝑥 2
2
NL = number of row
f = friction factor
X = correction factor
p = density
Vmax = velocity maximum
To decide the number of row , using the number of tubes per passes that is 100,
20x 5= 100 . Thus, I decide to use 5 rows .
(944.6)(5.62 )
2
ΔP = (5)( 0.35)(1.0)
ΔP = 25919.82 Pa
XI.
CALCULATION OF OVERALL HEAT COEFFICIENT
Tube saturated steam calculation.
Re =
𝜌𝑣𝐷𝑖
𝝁
P = density,kg/m3
V = tube bank velocity, m/s
Di = inner diameter,m
𝜇 = dynamic viscosity, Ns/m2
Re =
944.6𝑘𝑔 1.12𝑚
)(
)(0.018𝑚)
𝑠
𝑚3
(
0.0002736
Re = 69602.11> 2300 (turbulent flow)
Thus, Nusselt number can be determined.
Nu = 0.023 Re0.8 Prn
From table A-9, n can be determined. At Pr = 114.5 °C
Pr = 1.00
n = 0.4 for heating (Ts > Tm)
Hence,
Nu = 0.023 (69602.11)0.8 (1.00)0.4
Nu = 172.11
Nu =
ℎ𝑖𝐷
𝑘
𝑁𝑢𝐾
hi =
𝐷𝑖
(172.11)(0.3476)
hi =
0.018
hi = 3323.64 W/m2°C
𝑁
1
Db = do 𝐾𝑇𝑛1
1
Since that I choose 2 shells , 4 tubes passes , that is in line, therefore it is a square pitch
Using table 3.2 in Figure 6 , the value of K1 and n1 can be determined.
Figure 6
There are 4 tubes passes , thus :
K1= 0.158
n1 = 2.263
NT = Number of tubes = 100
do = diameter outer
Db = diameter bundle
100
1
2.263
Db = (0.020m)0.158
Db = 0.3459m
Figure 7 ( Shell bundle clearance )
Based on the Db value , the Shell inlet diameter – bundle diameter can be determined.
Ds – Db = 11mm = 0.011
Ds = 0.011 + 0.3459m
Ds = 0.3569m
Re =
Re =
𝜌𝑣𝑑
µ
(944.6)(1.12)(0.018)
0.0002736
Re = 69602.11
Based on the table A9 in figure 8 . Using the mean temperature in Table 1 . T = 65 °C
Figure 8
Pr = 1.00
L/ D = 5m/ 0.020m
L/D =250
Shell Side Calculations.
The equivalent diameter for the square-pitch layout ,De is :
De =
4 ×𝑃𝑡2 −
𝜋𝑑2
0
4
𝜋𝑑𝑜
Based on Figure 6, Pt = 1.25do
Therefore ,
Pt = 1.25 ( 0.020)
Pt = 0.025
De =
4(0.0252 )−
𝜋0.0202
4
𝜋(0.020)
De = 0.01979
Baffle spacing,B is given by :
B=
𝐿𝑇
𝑁𝐵 +1
LT = Tube length
NB = number of baffle
Assuming that NB = 6, therefore :
5
B = 6+1
B = 0.71m
Diameter of the shell,Ds are taken from tube side calculations.
Ds = 0.3569m
Volumetric flow rate,ṁ =
21540
ṁ = 3600𝑠 ×978.2972
ṁ = 0.006116 m3/s
𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
𝑑𝑒𝑛𝑠𝑖𝑡𝑦
ṁ
Shell side velocity = 𝐴
AS =
(𝑑𝑜−𝑑𝑖 )
𝑑𝑜
𝑠
× 𝐷𝑠 × 𝐵
do = outer diameter
di = inner diameter
As =
(0.020−0.018 )𝑚
0.020𝑚
× 0.3569𝑚 × 0.71𝑚
As = 0.02534m2
Shell side velocity =
0.006116 𝑚3 /𝑠
0.02534𝑚2
Shell side velocity = 0.2414 m/s
The cross flow area of the shell is,
AC =
𝐷𝑠 𝐶𝑇 𝐵
𝑃𝑇
C T = P T – do
CT = 0.025 – 0.020
CT = 0.005
Ac =
(0.3569𝑚)(0.005)(0.71𝑚)
0.025
Ac = 0.0507m2
ReD =
ṁ𝐷𝑒
𝐴𝑐 µ
21540
(
)(0.01979)
3600𝑠
ReD = (0.0507)(0.0005381)
ReD = 4340.28
4340.28>2300 , Therefore it is turbulent flow.
Therefore the Nusselt Number can be determined from Table 7.1 in Figure 9.
Figure 9
Re = 4340.28 and the shape of the shell is circle ,Thus the Nusselt Number is :
Nu = 0.193Re0.618Pr1/3
From Table- A9 in Figure 8 , at temperature 65°C, I can obtained the value of Pr.
Pr at 65°C = 2.75
Nu = 0.193(4340.28)0.618(2.75)1/3
Nu = 47.86
ho =
𝑁𝑢𝐾
𝐷𝑒
Nu = Nusselt number
K = Themal Conductivity (W/mK)
De= equivalent diameter (m)
ho =
(47.86)(0.6495)
0.01979
ho = 1570.75
The formula given to find the overall heat coefficient is :
(without fouling)
1
𝑈𝑂 𝐴𝑂
𝐷𝑜
)
𝐷𝑖
1
ln(
𝑖 𝑖
2𝜋𝐾𝐿
=ℎ𝐴 +
+ℎ
1
𝑜 𝐴𝑜
Ao =outer diameter
= πDoL = π (0.020m)(5.0m)
= 0.3142 m2
Ai = inner diameter
= πDiL = π (0.018m)(5.0m)
= 0.2827 m2
Do=outer diameter = 0.020m
Di = inner diameter = 0.018m
hi = 3675.29 W/m2°C
ho = 1610.46 W/ m2°C
1
𝑈𝑂
0.020
ln(
1
)
1
0.018
= (3323.64)(0.2827) + 2𝜋(49)(5.0)
+ (1570.75)(0.3142)
(0.3142)
Uo = 1007.52 W/m2°C
(With fouling )
1
𝑈𝑂 𝐴𝑂
=
1
ℎ𝑖 𝐴𝑖
+
𝑅𝑓𝑖
𝐴𝑖
+
𝐷𝑜
)
𝐷𝑖
ln(
2𝜋𝐾𝐿
+
𝑅𝑓𝑜
1
𝐴𝑜 ℎ𝑜 𝐴𝑜
The value of fouling factors, Rf obtained from Table 13-2 in Figure 10 is 0.0001 m2 · °C/W .
Figure 10
1
𝑈𝑂 (0.3142)
=
1
(3323.64)(0.2827)
+
0.0001
0.2827
+
0.020
)
0.018
ln(
2𝜋(49)(5)
+
0.0001
0.3142
+
1
(1570.75)(0.3142)
Uo = 830.77 W/m2°C
XII.
HEAT EXCHANGER DESIGN
Heat exchanger mechanical data.
Type
No of shell
No of tube
Tube outside diameter
Tube inside diameter
No of baffle
Tube Length
Diameter bundle
Diameter shell
Baffle spacing
2 Shell 4 Tubes Heat Exchanger (U tube)
2
4
0.020 m
0.018m
6
5.0m
0.3459m
0.3569m
0.71m