Heat Transfer Processes and Equipment
CC01, CC02, CC03
CH2043
English Program
Ho Chi Minh City University of Technology
Shuhaimi Mahadzir and Shafirah Samsuri 2021
CHAPTER 3: CONVECTIVE HEAT TRANSFER
Part 2: External Forced Convection
▪ Flow across Cylinders and Spheres
▪ Flow across Tube Banks
Flow across cylinders and spheres
• Practical applications
The tubes in a shell-and-tube heat exchanger involve both internal
flow inside the tubes and external flow outside (over) the tubes.
Many cryogenic storage tanks involve
flow over spherical shape geometry.
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Flow across cylinders and spheres
• Reynolds Number, Re
The critical Reynolds number for flow across a circular cylinder or
sphere is about Re = 2 × 105.
The boundary layer is:
Laminar for Re < 2 × 105,
Transitional for 2 × 105 < Re < 2 × 106.
Fully turbulent for Re > 2 × 106
4
Flow across cylinders and spheres
• Forced convection over a circular cylinder
(Churchill and Bernstein correlation ,1977)
The fluid properties are evaluated at the film temperature
Tf =
𝑇𝑠 +𝑇∞
2
5
Flow across cylinders and spheres
• Flow across various cylinder
geometry
The average Nusselt number for
flow across a cylinder (various
geometry) can be expressed
compactly as:
6
Flow across cylinders and spheres
• Forced convection over a sphere
(Whitaker correlation, 1972)
The fluid properties are evaluated at the free-stream temperature 𝑇∞ ,
except for s, which is evaluated at the surface temperature 𝑇𝑠 .
7
Flow across cylinders and spheres
Example 1: Heat Loss from a Steam Pipe in Windy Air
A long 10-cm diameter steam pipe whose external surface temperature is
110ºC passes through some open area that is not protected against the
winds. Determine the rate of heat loss from the pipe per unit of its length
when the air is at 1 atm pressure and 30ºC and the wind is blowing across
the pipe at a velocity of 8 m/s.
Solutions
Given: Air, T∞ = 10ºC, V = 8 m/s, D = 10 cm, Ts = 110ºC.
Determine: The rate of heat loss from the pipe per unit of its length.
8
9
Flow across cylinders and spheres
Example 1: Heat Loss from a Steam Pipe in Windy Air
Find the properties of air at Tf = 70ºC
ρ=1.028 kg/m3, n = 1.995 × 10-5 m2/s, Pr# = 0.7177, k = 0.02881 W/m∙K
Determine the Reynolds number:
8m/s×0.1m
VD
ReL =
=
= 40100
−5
2
ν
1.995×10 m /s
Determine the Nusselt number by Churchill and Bernstein correlation:
𝑁𝑢𝑐𝑦𝑙 = 0.3 +
0.62(40100)1/2 (0.7177)1/3
[1
+ (0.4/0.7177)2/3 ]1/4
40100
1+
282000
5/8 4/5
= 120.5
10
Flow across cylinders and spheres
Example 1: Heat Loss from a Steam Pipe in Windy Air
Heat transfer coefficient:
𝑁𝑢 𝑘 120.5 × 0.02881 W/m ⋅ K
ℎ=
=
= 34.7 W/m2 ⋅ K
𝐷
0.1 m
Heat loss per unit length:
Basis L = 1m,
𝑄 = ℎ𝐴𝑠 𝑇𝑠 − 𝑇∞ = 34.7 W/m2 ⋅ K × 𝜋𝐷𝐿 × 110 − 30 ℃
= 872 𝑊
#answer
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Flow across cylinders and spheres
Example 2: Heat Leaks into liquified propane (LPG) tank.
Propane is stored in a horizontal bullet tank with a
diameter of 200 cm and length 5 m. The temperature on
the surface of the tank is 10oC. On a particular day, wind
at a speed 20 km/h blows across the tank. If the average
temperature of the wind is 30oC, determine the heat
transfer coefficient for heat convection onto the surface of
the tank.
Solutions
Given: Air, T∞ = 10ºC, V = 20 km/h, D = 10 cm, Ts = 30ºC.
Determine: The heat transfer coefficient of the wind.
12
Flow across cylinders and spheres
Example 2: Heat Leaks into liquified propane (LPG) tank.
Wind velocity, V = 20 km/h = 5.56 m/s
The Reynolds Number
5.56 𝑚𝑠−1 × 2𝑚
Re = 1.516×10−5 𝑚2𝑠−1 = 7.34 × 105
The Nusselt number
Nu = 0.3 +
Nu =
ℎ𝐷
𝑘
1
5
0.62 ×(7.34 × 10 )2 ×(0.7309)1/3
0.4 2/3 1/4
1+(0.7309)
1+
... → then h = 12.1 W/m2.K
5/8 4/5
5
7.34 × 10
282000
= 964.62
#answer
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Flow across tube banks
• Practical applications
Cross-flow heat exchangers
Characteristic length, Lc = Do
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Flow across tube banks
• Tube Arrangements
1. In-line arrangement
2. Staggered arrangement
• Characteristic of Tube Pitch
1. Transverse pitch ST,
2. Longitudinal pitch SL,
3. Diagonal pitch SD.
‫٭‬Row is defined in the
direction of flow across
the tube banks
Diagonal pitch,
𝑆𝐷 =
𝑆𝐿2 + (𝑆𝑇 /2)2
15
Flow across tube banks
• Maximum Velocity
Vmax (in-line)
Vmax (staggered)
Reynolds number:
16
Flow across tube banks
• Nusselt Number correlations for cross flow over tube banks
▪ All properties except Prs are to be evaluated at arithmetic mean temperature,
Tm = (𝑇𝑖 +𝑇𝑒 )/2
▪ Pr is to be evaluated at T
s
s
17
Flow across tube banks
• Nusselt Number corrections
The average Nusselt number relations for tube banks with less than
16 rows (NL < 16 ) requires a correction factor.
𝑁𝑢𝐷,𝑁𝐿 <16 = 𝐹𝑁𝑢𝐷
The correction factor F as listed below:
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Flow across tube banks
• Log Mean Temperature Difference
▪ Temperature difference for flow over tube bank is the log mean
temperature difference and define as:
𝑇𝑠 − 𝑇𝑒 − 𝑇𝑠 − 𝑇𝑖
∆𝑇LM =
ln 𝑇𝑠 − 𝑇𝑒 / 𝑇𝑠 − 𝑇𝑖
∆𝑇𝑒 − ∆𝑇𝑖
=
ln ∆𝑇𝑒 /∆𝑇𝑖
▪ The exit temperature:
𝐴𝑠 ℎ
𝑇𝑒 = 𝑇𝑠 − 𝑇𝑠 − 𝑇𝑖 exp −
𝑚𝑐𝑝
Where 𝐴𝑠 = 𝑁𝜋𝐷𝐿 and
𝑚 = 𝜌𝑉(𝑁𝑇 𝑆𝑇 𝐿)
N = total number of tube, NT = number of tubes (transverse)
▪ The rate of heat transfer: 𝑄 = ℎ𝐴𝑠 ∆𝑇LM = 𝑚𝑐𝑝 𝑇𝑒 − 𝑇𝑖
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Flow across cylinders and spheres
Example 3: Preheating Air in a Tube Bank
Air is to be preheated before entering a furnace by
geothermal water at 120 ºC flowing through the
tubes of a tube bank located in a duct. Air enters
the duct at 20 ºC and 1 atm with an average
velocity of 4.5 m/s and flows over the tubes in
normal direction. The outer diameter of the tubes is
1.5 cm, and the tubes are arranged in-line with
longitudinal and transverse pitches of SL = ST = 5
cm. There are 6 rows in the flow direction with 10
tubes in each row.
Determine the rate of heat transfer per unit length
of the tubes.
Air
Ti = 20oC
1 atm
V = 4.5 m/s
Side view
V=
4.5 m/s
NL = 6
Flow across cylinders and spheres
Example 3: Preheating Air in a Tube Bank
The exit temperature Te and thus the mean temperature is not known.
By assuming the mean temperature is 60oC, the properties of air are:
ρ =1.059 kg/m3; n = 1.896 x 10-5 m2/s;
Pr = 0.7202; Prs = Pr at 120oC = 0.7073;
k = 0.02808 W/m∙K; Cp = 1.007 kJ/kg∙K;
#answer
21
22
Flow across cylinders and spheres
Example 3: Preheating Air in a Tube Bank
Calculate Vmax (in-line arrangement) and ReD:
ST
0.05m
Vmax =
V=
4.5m/s = 6.43m/s
ST − D
0.05m − 0.015m
Vmax D 6.43m/s × 0.015m
Re =
=
= 5087
ν
1.896 × 10−5 m2 /s
Nu = 0.27𝑅𝑒 0.63 𝑃𝑟 0.36 (𝑃𝑟Τ𝑃𝑟𝑠 )0.25
0.63
= 0.27(5087)
= 52.1
(0.7202)
0.36
0.7202
(
0.7073)0.25
#answer
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Flow across cylinders and spheres
Example 3: Preheating Air in a Tube Bank
Since the number of rows, NL = 6, the corresponding correction factor is
F = 0.945 and the Nusselt number becomes
N𝑢NL <16 = F𝑁𝑢𝐷 = 0.945 × 52.1 = 49.3
Solving for heat transfer coefficient, h:
#answer
24
Flow across cylinders and spheres
Example 3: Preheating Air in a Tube Bank
The total number of tubes is N =NL × NT = 6 × 10 = 60.
For a unit tube length (L= 1 m), the heat transfer surface area and the mass
flow rate of air (evaluated at inlet) are:
𝐴𝑠 = 𝑁𝜋𝐷𝐿 = 60 × π × 0.015m × 1m = 2.827m2
The mass flow rate mi of the fluid (inlet) at T = 20oC
→ ri = 1.204 kg/m3.
#answer
25
Flow across cylinders and spheres
Example 3: Preheating Air in a Tube Bank
The exit fluid temperature of the fluid:

 Ah
Te = Ts − (Ts − Ti ) exp s.
 mC
p






 2.827m 2  92.2W/m 2  K 

= 120 − (120 − 20) exp −
.
 2.709kg/s 1007J/kg  K 


= 29.11o C
Then the LMTD (log mean temperature difference):
(Ts − Te ) − (Ts − Ti )
(120 − 29.11) − (120 − 20)
Tlm =
=
ln[( Ts − Te ) /(Ts − Ti )] ln[(120 − 29.11) /(120 − 20)]
= 95.4 o C
#answer
26
Flow across cylinders and spheres
Example 3: Preheating Air in a Tube Bank
The rate of heat transfer become:
#answer
Finally note that:
The mean fluid temperature is Tf = (Ti+Te)/2 = (20+29.11)/ 2 = 24.6oC,
which is not close to the assumed value of 60ºC at which the fluid properties
were used in the calculations.
What do you think about the answer Q = 24866 W above?
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Summary
▪ Flow across Cylinders and Spheres
✓ Tf, Pr#, Re#, Nu#, h, Q
▪ Flow across Tube Banks
✓ In-line vs Staggered tube pitch
✓ Vmax
✓ Nu#, h, Q
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