NOTES: Chapter 7
NOTES: Chapter 7.1 to 7.3 – FALL 2003 CHEN 3210 – Chemical Engineering Heat Transfer Heat Transfer in External Flows External Flow – boundary layers develop without influence from nearby surfaces General Methodology for Solving Convection Problems (1) Identify the flow geometry. (2) Select a reference temperature and find or calculate fluid properties at that temperature. - This may require iteration if the reference temperature is a guess to start. (3) Calculate the Reynolds number. - For a flat plate, is the flow laminar, turbulent, or mixed? (4) Do you want a local or an average heat transfer coefficient? - A local coefficient describes the flux at a point on the surface. - An average coefficient will provide the heat transfer rate for the entire surface. (5) Select the appropriate correlation (based on steps (1), (3), and (4)). - You may need to use a momentum or mass transfer correlation and an analogy from Chapter 6 if the appropriate heat transfer correlation is not available. For most “normal” situations, the heat transfer correlation will be available. (6) Calculate the heat transfer coefficient and solve the problem. - Recheck assumptions and/or guesses and iterate or choose a different solution method if necessary. Chapter 7 focuses on finding convection coefficients for various geometries. For heat transfer, the convection coefficient is nondimensionalized as the Nusselt number:
(
)
Nu x = f x∗ , Rex ,Pr =
Nu x = f ( Rex ,Pr ) =
hx k
hx k
(local heat transfer coefficient)
(average heat transfer coefficient)
Experimental (Empirical) Approach Measure heat transfer coefficients over a plate of length L using fluids with known thermophysical properties (e.g., Pr) and plot data to fit an equation of the following form:
Nu L = CReLm Pr n
(average heat transfer coefficient over a plate of length L)
The surface of the plate is at one temperature and it is being heated or cooled by a fluid a different temperature. What temperature do you use to evaluate the fluid properties like Pr?
T f ≡ Tam =
Ts + T∞ 2
(“film temperature”)
Example: see HW#5, Problem 6.61. How would you estimate the constants, C and m, if you did not use a graphing program or solver?
Flat Plate with Parallel Flow – Theoretical Approach Instead of measuring the heat transfer coefficient, we can solve the boundary layer equations for heat, mass, and momentum transfer. For laminar flow, the momentum transfer equation is transformed with a similarity variable to obtain the following differential equation:
2
∂f u = ∂η u∞
∂3 f ∂2 f f + =0 ∂η 3 ∂η 2
η ≡ y u∞ /ν x
The solution to this equation is shown in Table 7.1. Exercise: What is the boundary layer thickness as a function of Rex?
Likewise, for laminar flow, transforming and solving the heat transfer equation yields: 1/3 Nu x = 0.332Re1/2 (Pr > 0.6) x Pr
Note: the exponent of n = 1/3 for the Prandtl number comes from this analytical solution. Exercise: Find an expression for Nu x .
Turbulent Flow (Empirical) −1/ 5
From experimental measurements, (for Rex < 107), C f , x = 0.0592 Re x
.
Exercise: Use the Chilton-Colburn analogy to find the local Nusselt number for turbulent flow.
Mixed Boundary Layer Example: Find the average heat transfer coefficient for air blowing across a uniform temperature surface with the following conditions. Tair,∞ = 350 K Ts = 300 K V = 20 m/s L=1m
Special Cases See expressions in Chapter 7.2.4: Unheated starting length (laminar or turbulent flow) Uniform surface flux (laminar or turbulent flow)
NOTES: Chapter 7.4 to 7.6 – FALL 2003 CHEN 3210 – Chemical Engineering Heat Transfer Cylinder in Cross Flow Hilpert correlation:
Nu D ≡
hD = CReDm Pr1/ 3 k
- all properties evaluated at the film temperature - for circular cross-section, and gas or liquid flow, C and m are in Table 7.2 - for non-circular cross-sections, and gas flow only, C and m are in Table 7.3 Zhukauskas correlation: 1/ 4
Pr Nu D = CRe Pr Prs m D
n
- 0.7 < Pr < 500; 1 < ReD < 106 - n = 0.37 for Pr ≤ 10; n = 0.36 for Pr > 10 - all properties evaluated at T∞ … Prs evaluated at Ts - C and m from Table 7.4 Churchill and Bernstein correlation:
Nu D = 0.3 +
0.62 Re1/D 2 Pr1/ 3 1/ 4
1 + ( 0.4 / Pr )2 / 3
5/8 Re D 1 + 282, 000
4/5
- ReDPr > 0.2 - all properties evaluated at the film temperature Mass transfer analogy – replace Nu with Sh and Pr with Sc, and neglect the Pr/Prs term (if applicable) to convert the heat transfer correlation into a mass transfer correlation.
Sphere Creeping Flow:
CD =
24 ReD
for ReD < 0.5
Whitaker correlation:
(
Nu D = 2 + 0.4 Re + 0.6 Re 1/ 2 D
2/3 D
)
1/ 4
µ Pr µs 0.4
- 0.71 < Pr < 380; 3.5 < ReD < 7.6x104 - all properties evaluated at T∞ … µs evaluated at Ts Ranz and Marshall correlation:
Nu D = 2 + 0.6 Re1/D 2 Pr1/ 3 - for heat transfer from freely falling liquid drops
Exercise: When ReD → 0, what is Nu D ? What physical situation does this describe? Banks of Tubes Grimison correlation (corrected for fluids besides air by the factor 1.13 Pr1/3): m Nu D = 1.13C1 ReD,max Pr1/ 3
ReD,max ≡ Vmax =
ρVmax D µ
ST V ST − D
- NL ≥ 10; 2000 ReD,max < 40,000; Pr ≥ 0.7 - all properties evaluated at the film temperature - C1 and m from Table 7.5 - for NL < 10, use Nu D
( N L
(
)
Nu x = f x∗ , Rex ,Pr =
Nu x = f ( Rex ,Pr ) =
hx k
hx k
(local heat transfer coefficient)
(average heat transfer coefficient)
Experimental (Empirical) Approach Measure heat transfer coefficients over a plate of length L using fluids with known thermophysical properties (e.g., Pr) and plot data to fit an equation of the following form:
Nu L = CReLm Pr n
(average heat transfer coefficient over a plate of length L)
The surface of the plate is at one temperature and it is being heated or cooled by a fluid a different temperature. What temperature do you use to evaluate the fluid properties like Pr?
T f ≡ Tam =
Ts + T∞ 2
(“film temperature”)
Example: see HW#5, Problem 6.61. How would you estimate the constants, C and m, if you did not use a graphing program or solver?
Flat Plate with Parallel Flow – Theoretical Approach Instead of measuring the heat transfer coefficient, we can solve the boundary layer equations for heat, mass, and momentum transfer. For laminar flow, the momentum transfer equation is transformed with a similarity variable to obtain the following differential equation:
2
∂f u = ∂η u∞
∂3 f ∂2 f f + =0 ∂η 3 ∂η 2
η ≡ y u∞ /ν x
The solution to this equation is shown in Table 7.1. Exercise: What is the boundary layer thickness as a function of Rex?
Likewise, for laminar flow, transforming and solving the heat transfer equation yields: 1/3 Nu x = 0.332Re1/2 (Pr > 0.6) x Pr
Note: the exponent of n = 1/3 for the Prandtl number comes from this analytical solution. Exercise: Find an expression for Nu x .
Turbulent Flow (Empirical) −1/ 5
From experimental measurements, (for Rex < 107), C f , x = 0.0592 Re x
.
Exercise: Use the Chilton-Colburn analogy to find the local Nusselt number for turbulent flow.
Mixed Boundary Layer Example: Find the average heat transfer coefficient for air blowing across a uniform temperature surface with the following conditions. Tair,∞ = 350 K Ts = 300 K V = 20 m/s L=1m
Special Cases See expressions in Chapter 7.2.4: Unheated starting length (laminar or turbulent flow) Uniform surface flux (laminar or turbulent flow)
NOTES: Chapter 7.4 to 7.6 – FALL 2003 CHEN 3210 – Chemical Engineering Heat Transfer Cylinder in Cross Flow Hilpert correlation:
Nu D ≡
hD = CReDm Pr1/ 3 k
- all properties evaluated at the film temperature - for circular cross-section, and gas or liquid flow, C and m are in Table 7.2 - for non-circular cross-sections, and gas flow only, C and m are in Table 7.3 Zhukauskas correlation: 1/ 4
Pr Nu D = CRe Pr Prs m D
n
- 0.7 < Pr < 500; 1 < ReD < 106 - n = 0.37 for Pr ≤ 10; n = 0.36 for Pr > 10 - all properties evaluated at T∞ … Prs evaluated at Ts - C and m from Table 7.4 Churchill and Bernstein correlation:
Nu D = 0.3 +
0.62 Re1/D 2 Pr1/ 3 1/ 4
1 + ( 0.4 / Pr )2 / 3
5/8 Re D 1 + 282, 000
4/5
- ReDPr > 0.2 - all properties evaluated at the film temperature Mass transfer analogy – replace Nu with Sh and Pr with Sc, and neglect the Pr/Prs term (if applicable) to convert the heat transfer correlation into a mass transfer correlation.
Sphere Creeping Flow:
CD =
24 ReD
for ReD < 0.5
Whitaker correlation:
(
Nu D = 2 + 0.4 Re + 0.6 Re 1/ 2 D
2/3 D
)
1/ 4
µ Pr µs 0.4
- 0.71 < Pr < 380; 3.5 < ReD < 7.6x104 - all properties evaluated at T∞ … µs evaluated at Ts Ranz and Marshall correlation:
Nu D = 2 + 0.6 Re1/D 2 Pr1/ 3 - for heat transfer from freely falling liquid drops
Exercise: When ReD → 0, what is Nu D ? What physical situation does this describe? Banks of Tubes Grimison correlation (corrected for fluids besides air by the factor 1.13 Pr1/3): m Nu D = 1.13C1 ReD,max Pr1/ 3
ReD,max ≡ Vmax =
ρVmax D µ
ST V ST − D
- NL ≥ 10; 2000 ReD,max < 40,000; Pr ≥ 0.7 - all properties evaluated at the film temperature - C1 and m from Table 7.5 - for NL < 10, use Nu D
( N L
