MEP 460 Heat Exchanger design
King Abdulaziz University
Mechanical Engineering Department
March 2019
Ch. 10 Compact Heat Exchangers
Ch. 10 Compact Heat Exchangers
1-Introduction
2-Tube-fin heat exchangers 3-Plate-fin heat exchangers 4-Examples
1- Introduction
Compact heat exchangers
Surface heat transfer area over volume 𝛼
Tube fin compact heat exchangers
Tube fin compact heat exchangers
Non-circular tubes
2- Tube fin heat exchangers
continuous fins on
flat tubes Continuous fins
on circular tube
Circular fins on circular tubes
Heat transfer and pressure drop for tube fin heat exchangers
1
𝑈𝑜𝐴𝑜 = 1
ℎ𝑖𝜂𝑖𝐴𝑖 + 𝑅𝑓𝑖
𝜂𝑖𝐴𝑖 + 𝑅𝑤 + 𝑅𝑓𝑜
𝜂𝑜𝐴𝑜 + 1 ℎ𝑜𝜂𝑜𝐴𝑜
For ho outside (gas) heat transfer coefficient use
Kays &
London book
in compact heat exchangers 𝜂𝑜 = 1 − 𝐴𝑓𝐴 (1 − 𝜂𝑓)
Overall surface efficiency 𝜂𝑜
𝜂𝑓 is the fin efficiency
Definitions
Frontal area A
frFree Flow area A
min= A
ffFin area/total area=A
f/A
o𝜎 = 𝐹𝑟𝑒𝑒 𝑓𝑙𝑜𝑤 𝑎𝑟𝑒𝑎
𝐹𝑟𝑜𝑛𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 = 𝐴𝑓𝑓 𝐴𝑓𝑟
𝛼 = Surface area/Volume Hydraulic diameter D
hMass velocity G= u
maxDefinitions
Colburn j
Hfactor 𝑗
𝐻= 𝑆𝑡𝑃𝑟
2 3Τ𝑆𝑡 = 𝑁𝑢
𝑅𝑒 𝑃𝑟 = ℎ
𝜌𝐶𝑝𝑉𝑚𝑎𝑥 = ℎ 𝐶𝑝𝐺
Mass velocity [kg/(m
2.s)
𝐺 = 𝜌𝑉𝑚𝑎𝑥 = 𝜌𝑉𝐴𝑓𝑟𝐴𝑓𝑓 = 𝑚ሶ
𝐴𝑓𝑓 = 𝑚ሶ 𝜎𝐴𝑓𝑟
Stanton Number
𝑅𝑒 = 𝐺𝐷
ℎ/𝜇
Δ𝑃 = 𝑓 𝐿
𝐷 𝜌 𝑉2
Friction coefficient f
2Major Pressure drop
𝑣𝑚 = 𝑣𝑖 + 𝑣𝑜 2 𝐴
𝐴𝑓𝑓 = 𝛼𝑉 𝜎𝐴𝑓𝑟 A: heat transfer area
Afr Frontal area Aff Free flow area
𝜎 = 𝐹𝑖𝑛 𝑎𝑟𝑒𝑎
𝑇𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 = 𝐴𝑓 𝐴
𝛼 = 𝐴 𝑉
Pressure drop gas side
𝑣
𝑖specifc volume at inlet 𝑣
𝑜specific volume at outlet
𝑣
𝑚mean specific volume =
𝑣𝑖+𝑣𝑜2
The above equation for pressure drop can be also written in terms of densities instead of specific volume
Surface information (CF-7.0-5/8J)
Circular fin on circular tubes
From Incropera 6th edition
Typical data for tube fin heat exchangers (8.0-3/8T)
Continuous fin on circular tubes
𝐹𝑖𝑛 𝑎𝑟𝑒𝑎
𝑇𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 = 𝐴
𝑓𝐴
𝛼 = 𝛽 = 𝐴 Surface density 𝑉
Surface information
Hydraulic diameter Dh
𝜎 = 𝐹𝑟𝑒𝑒 𝑓𝑙𝑜𝑤 𝑎𝑟𝑒𝑎
𝐹𝑟𝑜𝑛𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 = 𝐴𝑓𝑓 𝐴𝑓𝑟
Surface information
1
𝑈𝑜𝐴𝑜 = 1
ℎ𝑖𝜂𝑖𝐴𝑖 + 𝑅𝑓𝑖
𝜂𝑖𝐴𝑖 + 𝑅𝑤 + 𝑅𝑓𝑜
𝜂𝑜𝐴𝑜 + 1 ℎ𝑜𝜂𝑜𝐴𝑜
1
𝑈𝑜 = 1
ℎ𝑖( Τ𝐴𝑖 𝐴𝑜) + 𝐴𝑜𝑙𝑛( Τ𝑟𝑜 𝑟𝑖)
2𝜋𝑘𝐿 + 1 ℎ𝑜𝜂𝑜
Need to know the heat transfer area ratio Ai/Ao
Evaluating overall heat transfer coefficient
Neglecting fouling resistances
𝑅𝑤 = ln( Τ𝑟𝑜 𝑟𝑖) 2𝜋𝑘𝐿
1
𝑈𝑜 = 1
ℎ𝑖( Τ𝐴𝑖 𝐴𝑜) + 𝐷𝑖𝑙𝑛( Τ𝑟𝑜 𝑟𝑖)
2𝑘 ( Τ𝐴𝑖 𝐴𝑜) + 1 ℎ𝑜𝜂𝑜
Ratio of inside to outside heat transfer area
𝐴𝑖 = 𝜋𝐷𝑖𝐿 𝐴𝑜,𝑝 = 𝜋𝐷𝑜𝐿
𝐴𝑜 = 𝐴𝑢𝑓 + 𝐴𝑓 = 𝐴𝑜,𝑝 + 𝐴𝑓 𝐴𝑖
𝐴𝑜,𝑝 = 𝐷𝑖 𝐷𝑜
𝐴𝑜,𝑝 = 𝐴𝑜 − 𝐴𝑓 𝐴𝑖
𝐴𝑜 = 𝐷𝑖
𝐷𝑜 ∗ 1 − 𝐴𝑓 𝐴𝑜 𝐴𝑖 = 𝐷𝑖
𝐷𝑜 𝐴𝑜,𝑝
D
oD
iFins
Inside heat transfer area Outside heat transfer area without fins
Neglecting the area occupied by fins. i.e. Auf=Aop
h
iis given
From frontal area A
fr, and mass flow rate get G and Re
Dh Get j from Kays & London graphs and h
o Get fin efficiency for circular fins on circular pipe
Get U
ovalue
Knowing q and q
maxget the effectiveness
Knowing C
rand the effectiveness get NTU
From NTU=UA/C
mincalculate the heat transfer area A
o Calculate the volume of the heat exchanger using𝛼 = 𝛽 =
𝐴𝑜 Get the depth of the heat exchanger L form V=A
𝑉 frL
Calculate the number of rows of tubes
Example 11.6
Surface: 8.0-3/8 T
Continuous fins on circular tubes
Surface CF-8.72(c) Circular fins on circular tubes
Circular fin on circular tubes
CF-8.7-5/8J
Heat transfer
factor j and friction coefficient f for
some tube-fin and plate- fin surface
Ref.: Kays & London
Continuous fin with flat tube
Some of the data for plate-fin and tube fin compact heat exchangers
Δ𝑝 = 𝐺2
2𝜌𝑖 4𝑓 𝐿 𝐷ℎ
𝜌𝑖
𝜌 + 1 + 𝜎2 𝜌𝑖
𝜌𝑜 − 1
Δ𝑝 = 𝐺2
2𝜌𝑖 𝑓 𝐴 𝐴𝑚𝑖𝑛
𝜌𝑖
𝜌 + 1 + 𝜎2 𝜌𝑖
𝜌𝑜 − 1 𝐴
𝐴𝑚 = 4𝐿
𝐷ℎ = ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑎𝑟𝑒𝑎 𝑚𝑖𝑛. 𝑓𝑙𝑜𝑤 𝑎𝑟𝑒𝑎
Pressure drop (Gas side)
Amin min. flow area
A =heat transfer area
P=perimeter A=heat transfer area=P*L
Dh= 4Amin/P
𝐷ℎ = 4𝐴𝑚𝑖𝑛 𝑃
𝐿
𝐿 = 4𝐴𝑚𝑖𝑛𝐿 𝐴
𝐴
𝐴𝑚𝑖𝑛 = 4𝐿 𝐷ℎ L
3- Plat fin compact heat exchangers
Kays & London heat transfer and pressure drop data for plate-fin and tube-fin heat exchangers
Ch. 9
Ch. 9 Kays & London
Plain fins
Ch. 9 Kays & London
Strip fins
Ch. 9 Kays & London
Louvered fins
Ch. 9 Kays & London
Ch. 9 Kays & London
Circular tubes, continuous fins Flat tubes continuous fins
Ch. 9 Kays & London
Circular tubes- circular fins
Ch. 9 Kays & London
Flow inside circular and flattened tubes
Δ𝑝 = 𝐺2
2𝜌𝑖 𝑘𝑐 + 1 − 𝜎 + 2 𝜌𝑖
𝜌𝑜 − 1 + 𝑓 𝐴 𝐴𝑚𝑖𝑛
𝜌𝑖
𝜌 − 1 − 𝑘𝑒 − 𝜎2 𝜌𝑖 𝜌𝑜
Entrance Exit
Acceleration
Friction Major loss
Pressure drop for plat-fin heat exchanger
1
𝜌 = 1 2
1
𝜌𝑖 + 1 𝜌𝑜 Average density can be found using
Typical data for plate-fin compact heat exchanger
Surface tabulated data for plate-fin compact heat exchangers
Gas
Air
Surface 1 Surface 2
Hydaulic diameter Dh1 Plate spacing b1
Fin thickness 𝛿𝑓1
Area/ volume between plate, 𝛽1 Fin area/heat transfer area, 𝜔1 Length of the fin, 𝑙𝑓1
2 =A2/V
Hydaulic diameter Dh2 Plate spacing b2
Fin thickness 𝛿𝑓2
Area/ volume between plate, 𝛽2 Fin area/heat transfer area, 𝜔2 Length of the fin, 𝑙𝑓2
1=A1/V
Afr1
Afr2 Plate-fin compact heat exchanger (Gas-to-Gas HX)
Calculate
1-Number of passes Np
2-Calculate volume between plates for side 1 and side 2
3-Calculate heat transfer area A1, and A2 4-Calculate Amin1, Amin2
5-Calculate G1, and G2 6-Get j1,j2, f1, f2
7-Calculate h1 and h2
8-Calculate ηf1, ηf2, η01,η02 9-Calculate U value
10-Calculate Cr and NTU, then get
11-Calculate outlet temperatures 12-Calculate pressure drops
for both sides
L1 L2
L3
𝐿𝑐 = 𝑁𝑝𝑏1 + 𝑁𝑝 + 1 𝑏2 + 2 𝑁𝑝 + 1 𝑎
Assuming the number of passes in one side is Np and Np+1 on the other side, then the common edge length can be written in terms the pate spacing's b1, b2 and the plate thickness a as follows
Calculating the heat transfer areas for side (1) and side (2)
𝑁𝑝 = 𝐿𝑐 − 𝑏2 − 2𝑎 𝑏1 + 𝑏2 + 2𝑎
Volume between the plates for side (1) and side (2)
𝑉𝑝1 = 𝐿1 ∗ 𝐿2 ∗ 𝑁𝑝 ∗ 𝑏1 𝑉𝑝2 = 𝐿1 ∗ 𝐿2 ∗ 𝑁𝑝 ∗ 𝑏2
Utilizing the relation between and the volume between the plates
𝛽1 = 𝐴1
𝑉𝑝1 𝛽2 = 𝐴2 𝑉𝑝2
If 1 and 2 are calculated based on b1, b2 a, and , then one can easily find the heat
transfer areas A1 and A2
𝛼1 = 𝐴1
𝑉 𝛼2 = 𝐴2 𝑉
Where V is the total volume of the heat exchanger
𝜎1 = 𝐴𝑚𝑖𝑛,1
𝐴𝑓𝑟,1 = 𝐴𝑚𝑖𝑛,1𝐿𝑝1
𝐴𝑓𝑟,1𝐿𝑝,1 = 𝐴1 𝐷ℎ,1Τ4
𝑉 = 𝑉1𝛽1 𝐷ℎ,1Τ4 𝑉
𝑉 = 𝐿1 ∗ 𝐿2[𝑏1𝑁𝑝 + 𝑏2(𝑁𝑝+1) + 2𝑎 ∗ (𝑁𝑝 + 1)]
𝜎1 = 𝑚𝑖𝑛. 𝑓𝑙𝑜𝑤 𝑎𝑟𝑒𝑎 𝑓𝑟𝑜𝑛𝑡𝑎𝑙 𝑎𝑟𝑒𝑎
𝛽 = ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑎𝑟𝑒𝑎 𝑉𝑜𝑙𝑢𝑚𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑝𝑙𝑎𝑡𝑒𝑠
𝑉 = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 ℎ𝑒𝑎𝑡 𝑒𝑥𝑐ℎ𝑎𝑛𝑔𝑒𝑟
𝜎1 = 𝑉1𝛽1 𝐷ℎ,1Τ4
𝐿1 ∗ 𝐿2 [𝑏1𝑁𝑝 + 𝑏2(𝑁𝑝+1) + 2𝑎 ∗ (𝑁𝑝 + 1)] = 𝐿1 ∗ 𝐿2 ∗ 𝑏1 ∗ 𝑁𝑝 ∗ 𝛽1 𝐷ℎ,1Τ4
𝐿1 ∗ 𝐿2 [𝑏1𝑁𝑝 + 𝑏2(𝑁𝑝+1) + 2𝑎 ∗ (𝑁𝑝 + 1)]
𝜎1 = 𝑏1𝛽1 𝐷ℎ,1Τ4 𝑏1 + 𝑏2 + 2𝑎
𝛼1 = 𝐴1
𝑉 = 𝐴1
𝐴𝑓𝑟,1𝐿1 = 𝐴1Τ𝐿1
𝐴𝑓𝑟,1 = 4𝐴𝑚𝑖𝑛,1Τ𝐷ℎ,1
𝐴𝑓𝑟,1 = 4𝜎1
𝐷ℎ,1 = 𝑏1𝛽1 𝑏1 + 𝑏2 + 2𝑎
𝐿𝑝1 𝑓𝑙𝑜𝑤 𝑝𝑎𝑠𝑠𝑎𝑔𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑓𝑜𝑟 𝑠𝑖𝑑𝑒 (1)
𝑏1 𝑏2 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑝𝑙𝑎𝑡𝑒 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 𝑓𝑜𝑟 𝑠𝑖𝑑𝑒 1 𝑎𝑛𝑑 𝑠𝑖𝑑𝑒 2
Deducing the relation for 1 and 2
𝑁𝑝 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑠𝑠𝑒𝑠 𝐷ℎ = 4𝐴𝑚𝑖𝑛
𝑃 𝐿
𝐿 = 4𝐴𝑚𝑖𝑛𝐿 𝐴
𝜎2 = 𝑏2𝛽2 𝐷ℎ,2Τ4 𝑏1 + 𝑏2 + 2𝑎
𝛼2 = 𝐴2
𝑉 = 𝐴2
𝐴𝑓𝑟,2𝐿2 = 𝐴2Τ𝐿2
𝐴𝑓𝑟,2 = 4𝐴𝑚𝑖𝑛,2Τ𝐷ℎ,2
𝐴𝑓𝑟,2 = 4𝜎2
𝐷ℎ,2 = 𝑏2𝛽2 𝑏1 + 𝑏2 + 2𝑎
𝑏
1𝑏
2𝑎𝑟𝑒 𝑡ℎ𝑒 𝑝𝑙𝑎𝑡𝑒 𝑠𝑝𝑎𝑐𝑖𝑛𝑔 𝑓𝑜𝑟 𝑠𝑖𝑑𝑒 1 𝑎𝑛𝑑 𝑠𝑖𝑑𝑒 2
𝑁
𝑝𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑠𝑠𝑒𝑠
For side 2
1
𝑈1𝐴1 = 1
ℎ1𝐴1𝜂01 + 𝑅𝑤 + 1 ℎ2𝐴2𝜂02
𝜂01 = 1 − 𝐴𝑓1
𝐴1 (1 − 𝜂𝑓1) 𝜂02 = 1 − 𝐴𝑓2
𝐴2 (1 − 𝜂𝑓2)
𝜂𝑓 = tanh(𝑀𝑙𝑓) 𝑀𝑙𝑓
𝑀 = 2ℎ 𝑘𝑓𝛿𝑓
𝑙𝑓 is the length of the fin 𝛿𝑓 is the fin thickness
Overall heat transfer for plate-fin heat exchanger
𝐴𝑤 = 𝐿1𝐿2 ∗ 2(𝑁𝑝 + 1) 𝑅𝑤 = 𝑎
𝑘𝑤𝐴𝑤 Conduction resistance
a) Plain triangular fin
b) Plain rectangular fin c) Wavy fin d) Offset strip fin e) multi-louver fin f) Perforated fin
Some types of plate fin Compact HX
Fin types for plate fin compact heat exchanger
4-Examples
U=10 m/s
Find h and Dp
Air at p=1 atm T=400 K
Example 10.1
Example 10.1
U=20 m/s
Find h and Dp
Air at p=2 atm T=500 K
Example 10.2
Example 10.2
m=1500 kg/s Afr=0.25 m2
Find h and Dp
Air at p=1 atm T=30 C
To=100 C